C:\book\booktex\start.DVI 12

Introduction to 

Tensor Calculus 

and 

Continuum Mechanics 

by J.H. Heinbockel 

Department of Mathematics and Statistics 

Old Dominion University

PREFACE 

This is an introductory text which presents fundamental concepts from the subject 

areas of tensor calculus, differential geometry and continuum mechanics. The material 

presented is suitable for a two semester course in applied mathematics and is flexible 

enough to be presented to either upper level undergraduate or beginning graduate students 

majoring in applied mathematics, engineering or physics. The presentation assumes the 

students have some knowledge from the areas of matrix theory, linear algebra and advanced 

calculus. Each section includes many illustrative worked examples. At the end of each 

section there is a large collection of exercises which range in difficulty. Many new ideas 

are presented in the exercises and so the students should be encouraged to read all the 

exercises. 

The purpose of preparing these notes is to condense into an introductory text the basic 

definitions and techniques arising in tensor calculus, differential geometry and continuum 

mechanics. In particular, the material is presented to (i) develop a physical understanding 

of the mathematical concepts associated with tensor calculus and (ii) develop the basic 

equations of tensor calculus, differential geometry and continuum mechanics which arise 

in engineering applications. From these basic equations one can go on to develop more 

sophisticated models of applied mathematics. The material is presented in an informal 

manner and uses mathematics which minimizes excessive formalism. 

The material has been divided into two parts. The first part deals with an introduction 

to tensor calculus and differential geometry which covers such things as the indicial 

notation, tensor algebra, covariant differentiation, dual tensors, bilinear and multilinear 

forms, special tensors, the Riemann Christoffel tensor, space curves, surface curves, curvature 

and fundamental quadratic forms. The second part emphasizes the application of 

tensor algebra and calculus to a wide variety of applied areas from engineering and physics. 

The selected applications are from the areas of dynamics, elasticity, fluids and electromagnetic 

theory. The continuum mechanics portion focuses on an introduction of the basic 

concepts from linear elasticity and fluids. The Appendix A contains units of measurements 

from the Système International d’Unitès along with some selected physical constants. The 

Appendix B contains a listing of Christoffel symbols of the second kind associated with 

various coordinate systems. The Appendix C is a summary of useful vector identities. 

J.H. Heinbockel, 1996

Copyright c○1996 by J.H. Heinbockel. All rights reserved. 

Reproduction and distribution of these notes is allowable provided it is for non-profit 

purposes only.

INTRODUCTION TO 

TENSOR CALCULUS 

AND 

CONTINUUM MECHANICS 

PART 1: INTRODUCTION TO TENSOR CALCULUS 

§1.1 INDEX NOTATION . . . . . . . . . . . . . . . . . . 1 

Exercise 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 28 

§1.2 TENSOR CONCEPTS AND TRANSFORMATIONS . . . . 35 

Exercise 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 

§1.3 SPECIAL TENSORS . . . . . . . . . . . . . . . . . . 65 

Exercise 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 

§1.4 DERIVATIVE OF A TENSOR . . . . . . . . . . . . . . 108 

Exercise 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 

§1.5 DIFFERENTIAL GEOMETRY AND RELATIVITY . . . . 129 

Exercise 1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 

PART 2: INTRODUCTION TO CONTINUUM MECHANICS 

§2.1 TENSOR NOTATION FOR VECTOR QUANTITIES . . . . 171 

Exercise 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 

§2.2 DYNAMICS . . . . . . . . . . . . . . . . . . . . . . 187 

Exercise 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 

§2.3 BASIC EQUATIONS OF CONTINUUM MECHANICS . . . 211 

Exercise 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 

§2.4 CONTINUUM MECHANICS (SOLIDS) . . . . . . . . . 243 

Exercise 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 

§2.5 CONTINUUM MECHANICS (FLUIDS) . . . . . . . . . 282 

Exercise 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 

§2.6 ELECTRIC AND MAGNETIC FIELDS . . . . . . . . . . 325 

Exercise 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . 352 

APPENDIX A UNITS OF MEASUREMENT . . . . . . . 353 

APPENDIX B CHRISTOFFEL SYMBOLS OF SECOND KIND 355 

APPENDIX C VECTOR IDENTITIES . . . . . . . . . . 362 

INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . 363

PART 1: INTRODUCTION TO TENSOR CALCULUS 

A scalar field describes a one-to-one correspondence between a single scalar number and a point. An ndimensional 

vector field is described by a one-to-one correspondence between n-numbers and a point. Let us 

generalize these concepts by assigning n-squared numbers to a single point or n-cubed numbers to a single 

point. When these numbers obey certain transformation laws they become examples of tensor fields. In 

general, scalar fields are referred to as tensor fields of rank or order zero whereas vector fields are called 

tensor fields of rank or order one. 

Closely associated with tensor calculus is the indicial or index notation. In section 1 the indicial 

notation is defined and illustrated. We also define and investigate scalar, vector and tensor fields when they 

are subjected to various coordinate transformations. It turns out that tensors have certain properties which 

are independent of the coordinate system used to describe the tensor. Because of these useful properties, 

we can use tensors to represent various fundamental laws occurring in physics, engineering, science and 

mathematics. These representations are extremely useful as they are independent of the coordinate systems 

considered. 

§1.1 INDEX NOTATION 

Two vectors A and B can be expressed in the component form 

A = A1 e1 + A2 e2 + A3 e3 and B = B1 e1 + B2 e2 + B3 e3, 

where e1, e2 and e3 are orthogonal unit basis vectors. Often when no confusion arises, the vectors A and 

B are expressed for brevity sake as number triples. For example, we can write 

A =(A1, A2, A3) and B =(B1, B2, B3) 

where it is understood that only the components of the vectors A and B are given. The unit vectors would 

be represented 

e1 =(1, 0, 0), e2 =(0, 1, 0), e3 =(0, 0, 1). 

A still shorter notation, depicting the vectors A and B is the index or indicial notation. In the index notation, 

the quantities 

Ai, i =1, 2, 3 and Bp, p =1, 2, 3 

represent the components of the vectors A and B. This notation focuses attention only on the components of 

the vectors and employs a dummy subscript whose range over the integers is specified. The symbol Ai refers 

to all of the components of the vector A simultaneously. The dummy subscript i can have any of the integer 

values 1, 2or3. For i = 1 we focus attention on the A1 component of the vector A. Setting i =2focuses 

attention on the second component A2 of the vector A and similarly when i = 3 we can focus attention on 

the third component of A. The subscript i is a dummy subscript and may be replaced by another letter, say 

p, so long as one specifies the integer values that this dummy subscript can have. 

1

2 

It is also convenient at this time to mention that higher dimensional vectors may be defined as ordered 

n−tuples. For example, the vector 

X =(X1,X2,...,XN ) 

with components Xi, i=1, 2,...,N is called a N−dimensional vector. Another notation used to represent 

this vector is 

X = X1 e1 + X2 e2 + ···+ XN eN 

where 

e1, e2,..., eN 

are linearly independent unit base vectors. Note that many of the operations that occur in the use of the 

index notation apply not only for three dimensional vectors, but also for N−dimensional vectors. 

In future sections it is necessary to define quantities which can be represented by a letter with subscripts 

or superscripts attached. Such quantities are referred to as systems. When these quantities obey certain 

transformation laws they are referred to as tensor systems. For example, quantities like 

A k ij 

e ijk 

δij 

δ j 

i 

The subscripts or superscripts are referred to as indices or suffixes. When such quantities arise, the indices 

must conform to the following rules: 

1. They are lower case Latin or Greek letters. 

2. The letters at the end of the alphabet (u, v, w, x, y, z) are never employed as indices. 

The number of subscripts and superscripts determines the order of the system. A system with one index 

is a first order system. A system with two indices is called a second order system. In general, a system with 

N indices is called a Nth order system. A system with no indices is called a scalar or zeroth order system. 

The type of system depends upon the number of subscripts or superscripts occurring in an expression. 

For example, A i jk and B m st , (all indices range 1 to N), are of the same type because they have the same 

number of subscripts and superscripts. In contrast, the systems A i jk and C mn 

p are not of the same type 

because one system has two superscripts and the other system has only one superscript. For certain systems 

the number of subscripts and superscripts is important. In other systems it is not of importance. The 

meaning and importance attached to sub- and superscripts will be addressed later in this section. 

In the use of superscripts one must not confuse “powers ”of a quantity with the superscripts. For 

example, if we replace the independent variables (x, y, z) bythesymbols(x 1 , x 2 , x 3 ), then we are letting 

y = x 2 where x 2 is a variable and not x raised to a power. Similarly, the substitution z = x 3 is the 

replacement of z by the variable x 3 and this should not be confused with x raised to a power. In order to 

write a superscript quantity to a power, use parentheses. For example, (x 2 ) 3 is the variable x 2 cubed. One 

of the reasons for introducing the superscript variables is that many equations of mathematics and physics 

can be made to take on a concise and compact form. 

There is a range convention associated with the indices. This convention states that whenever there 

is an expression where the indices occur unrepeated it is to be understood that each of the subscripts or 

superscripts can take on any of the integer values 1, 2,...,N where N is a specified integer. For example, 

A i 

Bj 

aij.

the Kronecker delta symbol δij, defined by δij =1ifi = j and δij =0fori= j, withi, j ranging over the 

values 1,2,3, represents the 9 quantities 

δ11 =1 

δ21 =0 

δ31 =0 

δ12 =0 

δ22 =1 

δ32 =0 

δ13 =0 

δ23 =0 

δ33 =1. 

The symbol δij refers to all of the components of the system simultaneously. As another example, consider 

the equation 

em · en = δmn m, n =1, 2, 3 (1.1.1) 

the subscripts m, n occur unrepeated on the left side of the equation and hence must also occur on the right 

hand side of the equation. These indices are called “free ”indices and can take on any of the values 1, 2or3 

as specified by the range. Since there are three choices for the value for m and three choices for a value of 

n we find that equation (1.1.1) represents nine equations simultaneously. These nine equations are 

e1 · e1 =1 

e2 · e1 =0 

e3 · e1 =0 

Symmetric and Skew-Symmetric Systems 

e1 · e2 =0 

e2 · e2 =1 

e3 · e2 =0 

e1 · e3 =0 

e2 · e3 =0 

e3 · e3 =1. 

A system defined by subscripts and superscripts ranging over a set of values is said to be symmetric 

in two of its indices if the components are unchanged when the indices are interchanged. For example, the 

third order system Tijk is symmetric in the indices i and k if 

Tijk = Tkji for all values of i, j and k. 

A system defined by subscripts and superscripts is said to be skew-symmetric in two of its indices if the 

components change sign when the indices are interchanged. For example, the fourth order system Tijkl is 

skew-symmetric in the indices i and l if 

Tijkl = −Tljki for all values of ijk and l. 

As another example, consider the third order system aprs, p,r,s =1, 2, 3 which is completely skewsymmetric 

in all of its indices. We would then have 

aprs = −apsr = aspr = −asrp = arsp = −arps. 

It is left as an exercise to show this completely skew- symmetric systems has 27 elements, 21 of which are 

zero. The 6 nonzero elements are all related to one another thru the above equations when (p, r, s) =(1, 2, 3). 

This is expressed as saying that the above system has only one independent component. 

3

4 

Summation Convention 

The summation convention states that whenever there arises an expression where there is an index which 

occurs twice on the same side of any equation, or term within an equation, it is understood to represent a 

summation on these repeated indices. The summation being over the integer values specified by the range. A 

repeated index is called a summation index, while an unrepeated index is called a free index. The summation 

convention requires that one must never allow a summation index to appear more than twice in any given 

expression. Because of this rule it is sometimes necessary to replace one dummy summation symbol by 

some other dummy symbol in order to avoid having three or more indices occurring on the same side of 

the equation. The index notation is a very powerful notation and can be used to concisely represent many 

complex equations. For the remainder of this section there is presented additional definitions and examples 

to illustrated the power of the indicial notation. This notation is then employed to define tensor components 

and associated operations with tensors. 

EXAMPLE 1.1-1 The two equations 

y1 = a11x1 + a12x2 

y2 = a21x1 + a22x2 

can be represented as one equation by introducing a dummy index, say k, and expressing the above equations 

as 

yk = ak1x1 + ak2x2, k =1, 2. 

The range convention states that k is free to have any one of the values 1 or 2, (k is a free index). This 

equation can now be written in the form 

yk = 

2 

i=1 

akixi = ak1x1 + ak2x2 

where i is the dummy summation index. When the summation sign is removed and the summation convention 

is adopted we have 

yk = akixi 

i, k =1, 2. 

Since the subscript i repeats itself, the summation convention requires that a summation be performed by 

letting the summation subscript take on the values specified by the range and then summing the results. 

The index k which appears only once on the left and only once on the right hand side of the equation is 

called a free index. It should be noted that both k and i are dummy subscripts and can be replaced by other 

letters. For example, we can write 

yn = anmxm 

n, m =1, 2 

where m is the summation index and n is the free index. Summing on m produces 

yn = an1x1 + an2x2 

and letting the free index n take on the values of 1 and 2 we produce the original two equations.

EXAMPLE 1.1-2. For yi = aijxj, i,j=1, 2, 3andxi = bijzj, i,j=1, 2, 3solvefortheyvariables in 

terms of the z variables. 

Solution: In matrix form the given equations can be expressed: 

⎛ 

⎝ y1 

⎞ ⎛ 

⎠ = ⎝ a11 

⎞ ⎛ 

a12 a13 

⎠ ⎝ x1 

⎞ ⎛ 

⎠ and ⎝ x1 

⎞ ⎛ 

⎠ = ⎝ b11 

⎞ ⎛ 

b12 b13 

⎠ ⎝ z1 

⎞ 

⎠ . 

y2 

y3 

a21 a22 a23 

a31 a32 a33 

Now solve for the y variables in terms of the z variables and obtain 

⎛ 

⎝ y1 

⎞ ⎛ 

⎠ = ⎝ a11 a12 

⎞ ⎛ 

a13 

⎠ 

y2 

y3 

x2 

x3 

a21 a22 a23 

a31 a32 a33 

x2 

x3 

⎝ b11 b12 b13 

b21 b22 b23 

b31 b32 b33 

b21 b22 b23 

b31 b32 b33 

⎞ ⎛ 

⎠ 

⎝ z1 

z2 

The index notation employs indices that are dummy indices and so we can write 

z3 

⎞ 

⎠ . 

yn = anmxm, n,m =1, 2, 3 and xm = bmjzj, m,j =1, 2, 3. 

Here we have purposely changed the indices so that when we substitute for xm, from one equation into the 

other, a summation index does not repeat itself more than twice. Substituting we find the indicial form of 

the above matrix equation as 

yn = anmbmjzj, m,n,j =1, 2, 3 

where n is the free index and m, j are the dummy summation indices. It is left as an exercise to expand 

both the matrix equation and the indicial equation and verify that they are different ways of representing 

the same thing. 

EXAMPLE 1.1-3. The dot product of two vectors Aq, q=1, 2, 3andBj, j=1, 2, 3 can be represented 

with the index notation by the product AiBi = AB cos θ i =1, 2, 3, A = | A|, B = | B|. Since the 

subscript i is repeated it is understood to represent a summation index. 

specified, there results 

A1B1 + A2B2 + A3B3 = AB cos θ. 

Summing on i over the range 

Observe that the index notation employs dummy indices. At times these indices are altered in order to 

conform to the above summation rules, without attention being brought to the change. As in this example, 

the indices q and j are dummy indices and can be changed to other letters if one desires. Also, in the future, 

if the range of the indices is not stated it is assumed that the range is over the integer values 1, 2and3. 

To systems containing subscripts and superscripts one can apply certain algebraic operations. We 

present in an informal way the operations of addition, multiplication and contraction. 

z2 

z3 

5

6 

Addition, Multiplication and Contraction 

The algebraic operation of addition or subtraction applies to systems of the same type and order. That 

is we can add or subtract like components in systems. For example, the sum of A i jk and B i jk is again a 

system of the same type and is denoted by C i jk = A i jk + B i jk, where like components are added. 

The product of two systems is obtained by multiplying each component of the first system with each 

component of the second system. Such a product is called an outer product. The order of the resulting 

product system is the sum of the orders of the two systems involved in forming the product. For example, 

if A i j is a second order system and B mnl is a third order system, with all indices having the range 1 to N, 

then the product system is fifth order and is denoted C imnl 

j = A i jB mnl . The product system represents N 5 

terms constructed from all possible products of the components from A i j with the components from B mnl . 

The operation of contraction occurs when a lower index is set equal to an upper index and the summation 

convention is invoked. For example, if we have a fifth order system C imnl 

j 

we form the system 

C mnl = C jmnl 

j 

= C 1mnl 

1 

+ C 2mnl 

2 

+ ···+ C Nmnl 

N . 

and we set i = j and sum, then 

Here the symbol C mnl is used to represent the third order system that results when the contraction is 

performed. Whenever a contraction is performed, the resulting system is always of order 2 less than the 

original system. Under certain special conditions it is permissible to perform a contraction on two lower case 

indices. These special conditions will be considered later in the section. 

The above operations will be more formally defined after we have explained what tensors are. 

The e-permutation symbol and Kronecker delta 

Two symbols that are used quite frequently with the indicial notation are the e-permutation symbol 

and the Kronecker delta. The e-permutation symbol is sometimes referred to as the alternating tensor. The 

e-permutation symbol, as the name suggests, deals with permutations. A permutation is an arrangement of 

things. When the order of the arrangement is changed, a new permutation results. A transposition is an 

interchange of two consecutive terms in an arrangement. As an example, let us change the digits 1 2 3 to 

3 2 1 by making a sequence of transpositions. Starting with the digits in the order 1 2 3 we interchange 2 and 

3 (first transposition) to obtain 1 3 2. Next, interchange the digits 1 and 3 ( second transposition) to obtain 

312. Finally, interchange the digits 1 and 2 (third transposition) to achieve 3 2 1. Here the total number 

of transpositions of 1 2 3 to 3 2 1 is three, an odd number. Other transpositions of 1 2 3 to 3 2 1 can also be 

written. However, these are also an odd number of transpositions.

EXAMPLE 1.1-4. The total number of possible ways of arranging the digits 1 2 3 is six. We have 

three choices for the first digit. Having chosen the first digit, there are only two choices left for the second 

digit. Hence the remaining number is for the last digit. The product (3)(2)(1) = 3! = 6 is the number of 

permutations of the digits 1, 2 and 3. These six permutations are 

1 2 3 even permutation 

1 3 2 odd permutation 


3 2 1 odd permutation 


2 1 3 odd permutation. 

Here a permutation of 1 2 3 is called even or odd depending upon whether there is an even or odd number 

of transpositions of the digits. A mnemonic device to remember the even and odd permutations of 123 

is illustrated in the figure 1.1-1. Note that even permutations of 123 are obtained by selecting any three 

consecutive numbers from the sequence 123123 and the odd permutations result by selecting any three 

consecutive numbers from the sequence 321321. 

Figure 1.1-1. Permutations of 123. 

In general, the number of permutations of n things taken m at a time is given by the relation 

P (n, m) =n(n − 1)(n − 2) ···(n − m +1). 

By selecting a subset of m objects from a collection of n objects, m ≤ n, without regard to the ordering is 

called a combination of n objects taken m at a time. For example, combinations of 3 numbers taken from 

the set {1, 2, 3, 4} are (123), (124), (134), (234). Note that ordering of a combination is not considered. That 

is, the permutations (123), (132), (231), (213), (312), (321) are considered equal. In general, the number of 

 

n 

 

n! 

combinations of n objects taken m at a time is given by C(n, m) = = 

m m!(n − m)! where n 

m are the 

binomial coefficients which occur in the expansion 

(a + b) n = 

n 

m=0 

 

n 

 

a 

m 

n−m b m . 

7

8 

The definition of permutations can be used to define the e-permutation symbol. 

Definition: (e-Permutation symbol or alternating tensor) 

The e-permutation symbol is defined 

e ijk...l ⎧ 

⎪⎨ 1 if ijk...l is an even permutation of the integers 123 ...n 

= eijk...l = −1 

⎪⎩ 

0 

if ijk...l is an odd permutation of the integers 123 ...n 

in all other cases 

EXAMPLE 1.1-5. Find e612453. 

Solution: To determine whether 612453 is an even or odd permutation of 123456 we write down the given 

numbers and below them we write the integers 1 through 6. Like numbers are then connected by a line and 

we obtain figure 1.1-2. 

Figure 1.1-2. Permutations of 123456. 

In figure 1.1-2, there are seven intersections of the lines connecting like numbers. The number of 

intersections is an odd number and shows that an odd number of transpositions must be performed. These 

results imply e612453 = −1. 

Another definition used quite frequently in the representation of mathematical and engineering quantities 

is the Kronecker delta which we now define in terms of both subscripts and superscripts. 

Definition: (Kronecker delta) The Kronecker delta is defined: 

δij = δ j 

i = 

1 if i equals j 

0 if i is different from j

EXAMPLE 1.1-6. Some examples of the e−permutation symbol and Kronecker delta are: 

e123 = e 123 =+1 

e213 = e 213 = −1 

e112 = e 112 =0 

δ 1 1 =1 

δ 1 2 =0 

δ 1 3 =0 

δ12 =0 

δ22 =1 

δ32 =0. 

EXAMPLE 1.1-7. When an index of the Kronecker delta δij is involved in the summation convention, 

the effect is that of replacing one index with a different index. For example, let aij denote the elements of an 

N × N matrix. Here i and j are allowed to range over the integer values 1, 2,...,N. Consider the product 

aijδik 

where the range of i, j, k is 1, 2,...,N. The index i is repeated and therefore it is understood to represent 

a summation over the range. The index i is called a summation index. The other indices j and k are free 

indices. They are free to be assigned any values from the range of the indices. They are not involved in any 

summations and their values, whatever you choose to assign them, are fixed. Let us assign a value of j and 

k to the values of j and k. The underscore is to remind you that these values for j and k are fixed and not 

to be summed. When we perform the summation over the summation index i we assign values to i from the 

range and then sum over these values. Performing the indicated summation we obtain 

aijδik = a1jδ1k + a2jδ2k + ···+ akjδkk + ···+ aNjδNk. 

In this summation the Kronecker delta is zero everywhere the subscripts are different and equals one where 

the subscripts are the same. There is only one term in this summation which is nonzero. It is that term 

where the summation index i was equal to the fixed value k This gives the result 

akjδkk = akj 

where the underscore is to remind you that the quantities have fixed values and are not to be summed. 

Dropping the underscores we write 

aijδik = akj 

Here we have substituted the index i by k and so when the Kronecker delta is used in a summation process 

it is known as a substitution operator. This substitution property of the Kronecker delta can be used to 

simplify a variety of expressions involving the index notation. Some examples are: 

Bijδjs = Bis 

δjkδkm = δjm 

eijkδimδjnδkp = emnp. 

Some texts adopt the notation that if indices are capital letters, then no summation is to be performed. 

For example, 

aKJδKK = aKJ 

9

10 

as δKK represents a single term because of the capital letters. Another notation which is used to denote no 

summation of the indices is to put parenthesis about the indices which are not to be summed. For example, 

a (k)jδ (k)(k) = akj, 

since δ (k)(k) represents a single term and the parentheses indicate that no summation is to be performed. 

At any time we may employ either the underscore notation, the capital letter notation or the parenthesis 

notation to denote that no summation of the indices is to be performed. To avoid confusion altogether, one 

can write out parenthetical expressions such as “(no summation on k)”. 

EXAMPLE 1.1-8. In the Kronecker delta symbol δi j we set j equal to i and perform a summation. This 

operation is called a contraction. There results δi i , which is to be summed over the range of the index i. 

Utilizing the range 1, 2,...,N we have 

δ i i = δ 1 1 + δ 2 2 + ···+ δ N N 

δ i i =1+1+···+1 

δ i i 

= N. 

In three dimension we have δi j ,i,j=1, 2, 3and 

δ k k = δ1 1 + δ2 2 + δ3 3 =3. 

In certain circumstances the Kronecker delta can be written with only subscripts. For example, 

δij, i,j =1, 2, 3. We shall find that these circumstances allow us to perform a contraction on the lower 

indices so that δii =3. 

EXAMPLE 1.1-9. The determinant of a matrix A =(aij) can be represented in the indicial notation. 

Employing the e-permutation symbol the determinant of an N × N matrix is expressed 

|A| = eij...ka1ia2j ···aNk 

where eij...k is an Nth order system. In the special case of a 2 × 2 matrix we write 

|A| = eija1ia2j 

where the summation is over the range 1,2 and the e-permutation symbol is of order 2. In the special case 

of a 3 × 3matrixwehave 

 

 

 

a11 a12 a13 

 

|A| = 

a21 a22 a23 

 

 

= eijkai1aj2ak3 = eijka1ia2ja3k 

a31 a32 a33 

where i, j, k are the summation indices and the summation is over the range 1,2,3. Here eijk denotes the 

e-permutation symbol of order 3. Note that by interchanging the rows of the 3 × 3 matrix we can obtain

more general results. Consider (p, q, r) as some permutation of the integers (1, 2, 3), and observe that the 

determinant can be expressed 

 

ap1 

∆= aq1 

 

ap2 

aq2 

 

ap3 

 

aq3 

= eijkapiaqjark. 

 

We can then write 

ar1 ar2 ar3 

I f (p, q, r) is an even permutation of (1, 2, 3) then ∆ = |A| 

I f (p, q, r) is an odd permutation of (1, 2, 3) then ∆ = −|A| 

I f (p, q, r) is not a permutation of (1, 2, 3) then ∆ = 0. 

eijkapiaqjark = epqr|A|. 

Each of the above results can be verified by performing the indicated summations. A more formal proof of 

the above result is given in EXAMPLE 1.1-25, later in this section. 

EXAMPLE 1.1-10. The expression eijkBijCi is meaningless since the index i repeats itself more than 

twice and the summation convention does not allow this. 

EXAMPLE 1.1-11. 

The cross product of the unit vectors e1, e2, e3 can be represented in the index notation by 

⎧ 

⎪⎨ ek if (i, j, k) is an even permutation of (1, 2, 3) 

ei × ej = − ek 

⎪⎩ 

0 

if (i, j, k) is an odd permutation of (1, 2, 3) 

in all other cases 

This result can be written in the form ei × ej = ekij ek. This later result can be verified by summing on the 

index k and writing out all 9 possible combinations for i and j. 

EXAMPLE 1.1-12. Given the vectors Ap, p=1, 2, 3andBp, p=1, 2, 3 the cross product of these two 

vectors is a vector Cp, p=1, 2, 3withcomponents 

The quantities Ci represent the components of the cross product vector 

Ci = eijkAjBk, i,j,k =1, 2, 3. (1.1.2) 

C = A × B = C1 e1 + C2 e2 + C3 e3. 

The equation (1.1.2), which defines the components of C, is to be summed over each of the indices which 

repeats itself. We have summing on the index k 

Ci = eij1AjB1 + eij2AjB2 + eij3AjB3. (1.1.3) 

11

12 

We next sum on the index j which repeats itself in each term of equation (1.1.3). This gives 

Ci = ei11A1B1 + ei21A2B1 + ei31A3B1 

+ ei12A1B2 + ei22A2B2 + ei32A3B2 

+ ei13A1B3 + ei23A2B3 + ei33A3B3. 

(1.1.4) 

Now we are left with i being a free index which can have any of the values of 1, 2or3. Letting i =1, then 

letting i =2, and finally letting i = 3 produces the cross product components 

C1 = A2B3 − A3B2 

C2 = A3B1 − A1B3 

C3 = A1B2 − A2B1. 

The cross product can also be expressed in the form A × B = eijkAjBk ei. This result can be verified by 

summing over the indices i,j and k. 

EXAMPLE 1.1-13. Show 

eijk = −eikj = ejki for i, j, k =1, 2, 3 

Solution: The array ikjrepresents an odd number of transpositions of the indices ijkand to each 

transposition there is a sign change of the e-permutation symbol. Similarly, jkiis an even transposition 

of ijkand so there is no sign change of the e-permutation symbol. The above holds regardless of the 

numerical values assigned to the indices i, j, k. 

The e-δ Identity 

An identity relating the e-permutation symbol and the Kronecker delta, which is useful in the simplification 

of tensor expressions, is the e-δ identity. This identity can be expressed in different forms. The 

subscript form for this identity is 

eijkeimn = δjmδkn − δjnδkm, i,j,k,m,n=1, 2, 3 

where i is the summation index and j, k, m, n are free indices. A device used to remember the positions of 

the subscripts is given in the figure 1.1-3. 

The subscripts on the four Kronecker delta’s on the right-hand side of the e-δ identity then are read 

(first)(second)-(outer)(inner). 

This refers to the positions following the summation index. Thus, j, m are the first indices after the summation 

index and k, n are the second indices after the summation index. The indices j, n are outer indices 

when compared to the inner indices k, m as the indices are viewed as written on the left-hand side of the 

identity.

Figure 1.1-3. Mnemonic device for position of subscripts. 

Another form of this identity employs both subscripts and superscripts and has the form 

e ijk eimn = δ j mδk n − δj nδk m . (1.1.5) 

One way of proving this identity is to observe the equation (1.1.5) has the free indices j, k, m, n. Each 

of these indices can have any of the values of 1, 2or3. There are 3 choices we can assign to each of j, k, m 

or n and this gives a total of 34 = 81 possible equations represented by the identity from equation (1.1.5). 

By writing out all 81 of these equations we can verify that the identity is true for all possible combinations 

that can be assigned to the free indices. 

An alternate proof of the e − δ identity is to consider the determinant 

 

 

δ 

 

 

 

1 1 δ1 2 δ1 3 

δ2 1 δ2 2 δ2 

 

 

 

3 

= 

 

 

1 0 0 

 

 

0 1 0 

 

0 0 1 

=1. 

δ 3 1 δ 3 2 δ 3 3 

By performing a permutation of the rows of this matrix we can use the permutation symbol and write 

 

 

δ 

 

 

 

i 1 δi 2 δi δ 

3 

j 

1 δ j 

2 δ j 

 

 

 

 

3 

= eijk . 

δ k 1 δ k 2 δ k 3 

By performing a permutation of the columns, we can write 

 

 

δ 

 

 

 

i r δi s δi δ 

t 

j r δj s δ j 

 

 

 

 

t 

= eijkerst. δ k r δ k s δ k t 

Now perform a contraction on the indices i and r to obtain 

 

 

 

δ 

 

 

 

i i δi s δi δ 

t 

j 

i δj s δ j 

 

 

 

 

t 

= eijkeist. δ k i δ k s δ k t 

Summing on i we have δ i i = δ1 1 + δ 2 2 + δ 3 3 = 3 and expand the determinant to obtain the desired result 

δ j sδ k t − δ j 

t δ k s = e ijk eist. 

13

14 

Generalized Kronecker delta 

The generalized Kronecker delta is defined by the (n × n) determinant 

δ ij...k 

mn...p = 

For example, in three dimensions we can write 

δ ijk 

mnp = 

 

 

 

 

 

 

 

 

 

 

 

 

. 

. 

 

δi m δi n ··· δi p 

δj m δj n ··· δj p 

. 

. .. 

. 

δ k m δk n ··· δ k p 

δi m δi n δi p 

δj m δj n δj p 

δk m δk n δk p 

 

 

 

 

 

. 

 

 

 

 

 

 

 

 

= eijkemnp. Performing a contraction on the indices k and p we obtain the fourth order system 

δ rs 

mn = δrsp mnp = erspemnp = e prs epmn = δ r mδs n − δr nδs m . 

As an exercise one can verify that the definition of the e-permutation symbol can also be defined in terms 

of the generalized Kronecker delta as 

1 2 3··· N 

ej1j2j3···jN = δj1j2j3···jN . 

Additional definitions and results employing the generalized Kronecker delta are found in the exercises. 

In section 1.3 we shall show that the Kronecker delta and epsilon permutation symbol are numerical tensors 

whichhavefixedcomponentsineverycoordinatesystem. 

Additional Applications of the Indicial Notation 

The indicial notation, together with the e − δ identity, can be used to prove various vector identities. 

EXAMPLE 1.1-14. Show, using the index notation, that A × B = − B × A 

Solution: Let 

C = A × B = C1 e1 + C2 e2 + C3 e3 = Ci ei 

D = 

and let 

B × A = D1 e1 + D2 e2 + D3 e3 = Di ei. 

We have shown that the components of the cross products can be represented in the index notation by 

Ci = eijkAjBk and Di = eijkBjAk. 

We desire to show that Di = −Ci for all values of i. Consider the following manipulations: Let Bj = Bsδsj 

and Ak = Amδmk and write 

Di = eijkBjAk = eijkBsδsjAmδmk 

(1.1.6) 

where all indices have the range 1, 2, 3. In the expression (1.1.6) note that no summation index appears 

more than twice because if an index appeared more than twice the summation convention would become 

meaningless. By rearranging terms in equation (1.1.6) we have 

Di = eijkδsjδmkBsAm = eismBsAm.

In this expression the indices s and m are dummy summation indices and can be replaced by any other 

letters. We replace s by k and m by j to obtain 

Di = eikjAjBk = −eijkAjBk = −Ci. 

Consequently, we find that D = − C or B × A = − A × B. That is, D = Di ei = −Ci ei = − C. 

Note 1. The expressions 

Ci = eijkAjBk and Cm = emnpAnBp 

with all indices having the range 1, 2, 3, appear to be different because different letters are used as subscripts. 

It must be remembered that certain indices are summed according to the summation convention 

and the other indices are free indices and can take on any values from the assigned range. Thus, after 

summation, when numerical values are substituted for the indices involved, none of the dummy letters 

used to represent the components appear in the answer. 

Note 2. A second important point is that when one is working with expressions involving the index notation, 

the indices can be changed directly. For example, in the above expression for Di we could have replaced 

j by k and k by j simultaneously (so that no index repeats itself more than twice) to obtain 

Di = eijkBjAk = eikjBkAj = −eijkAjBk = −Ci. 

Note 3. Be careful in switching back and forth between the vector notation and index notation. Observe that a 

vector A can be represented 

A = Ai ei 

or its components can be represented 

A · ei = Ai, i =1, 2, 3. 

Do not set a vector equal to a scalar. That is, do not make the mistake of writing A = Ai as this is a 

misuse of the equal sign. It is not possible for a vector to equal a scalar because they are two entirely 

different quantities. A vector has both magnitude and direction while a scalar has only magnitude. 

EXAMPLE 1.1-15. Verify the vector identity 

Solution: Let 

A · ( B × C)= B · ( C × A) 

B × C = D = Di ei where Di = eijkBjCk and let 

C × A = F = Fi ei where Fi = eijkCjAk 

where all indices have the range 1, 2, 3. To prove the above identity, we have 

A · ( B × C)= A · D = AiDi = AieijkBjCk 

= Bj(eijkAiCk) 

= Bj(ejkiCkAi) 

15

16 

since eijk = ejki. We also observe from the expression 

Fi = eijkCjAk 

that we may obtain, by permuting the symbols, the equivalent expression 

This allows us to write 

Fj = ejkiCkAi. 

A · ( B × C)=BjFj = B · F = B · ( C × A) 

which was to be shown. 

The quantity A · ( B × C) is called a triple scalar product. The above index representation of the triple 

scalar product implies that it can be represented as a determinant (See example 1.1-9). We can write 

A · ( B × 

 

 

C)= 

 

 

A1 A2 A3 

B1 B2 B3 

C1 C2 C3 

 

 

 

 

= eijkAiBjCk 

 

A physical interpretation that can be assigned to this triple scalar product is that its absolute value represents 

the volume of the parallelepiped formed by the three noncoplaner vectors A, B, C. The absolute value is 

needed because sometimes the triple scalar product is negative. This physical interpretation can be obtained 

from an analysis of the figure 1.1-4. 

Figure 1.1-4. Triple scalar product and volume

In figure 1.1-4 observe that: (i) | B × C| is the area of the parallelogram PQRS. (ii) the unit vector 

en = B × C 

| B × C| 

is normal to the plane containing the vectors B and C. (iii) The dot product 

 

 

A · en = 

A · B × C 

| B × 

 

 

= h 

C| 

equals the projection of A on en which represents the height of the parallelepiped. These results demonstrate 

that 

 

 

A · ( B × 

 

C) = | B × C| h = (area of base)(height) = volume. 

EXAMPLE 1.1-16. Verify the vector identity 

( A × B) × ( C × D)= C( D · A × B) − D( C · A × B) 

Solution: Let F = A × B = Fi ei and E = C × D = Ei ei. These vectors have the components 

Fi = eijkAjBk and Em = emnpCnDp 

where all indices have the range 1, 2, 3. The vector G = F × E = Gi ei has the components 

From the identity eqim = emqi this can be expressed 

Gq = eqimFiEm = eqimeijkemnpAjBkCnDp. 

Gq =(emqiemnp)eijkAjBkCnDp 

which is now in a form where we can use the e − δ identity applied to the term in parentheses to produce 

Simplifying this expression we have: 

Gq =(δqnδip − δqpδin)eijkAjBkCnDp. 

Gq = eijk [(Dpδip)(Cnδqn)AjBk − (Dpδqp)(Cnδin)AjBk] 

= eijk [DiCqAjBk − DqCiAjBk] 

which are the vector components of the vector 

= Cq [DieijkAjBk] − Dq [CieijkAjBk] 

C( D · A × B) − D( C · A × B). 

17

18 

Transformation Equations 

Consider two sets of N independent variables which are denoted by the barred and unbarred symbols 

x i and x i with i = 1,...,N. The independent variables x i ,i =1,...,N can be thought of as defining 

the coordinates of a point in a N−dimensional space. Similarly, the independent barred variables define a 

point in some other N−dimensional space. These coordinates are assumed to be real quantities and are not 

complex quantities. Further, we assume that these variables are related by a set of transformation equations. 

x i = x i (x 1 , x 2 ,...,x N ) i =1,...,N. (1.1.7) 

It is assumed that these transformation equations are independent. A necessary and sufficient condition that 

these transformation equations be independent is that the Jacobian determinant be different from zero, that 

is 

J( x 

x )= 

 

 

 

∂x 

 

i 

∂¯x j 

 

 

 

= 

 

∂x 

 

 

 

 

 

 

 

 

1 

∂x1 ∂x 1 

∂x2 ∂x ··· 1 

∂xN ∂x 2 

∂x1 ∂x 2 

∂x2 ∂x ··· 2 

∂xN 

 

 

 

 

 

. 

. . 

. 

.. . 

= 0. 

. 

 

 

∂x N 

∂x 1 

This assumption allows us to obtain a set of inverse relations 

∂x N 

∂x2 ∂x ··· N 

∂xN x i = x i (x 1 ,x 2 ,...,x N ) i =1,...,N, (1.1.8) 

where the x ′ s are determined in terms of the x ′ s. Throughout our discussions it is to be understood that the 

given transformation equations are real and continuous. Further all derivatives that appear in our discussions 

are assumed to exist and be continuous in the domain of the variables considered. 

EXAMPLE 1.1-17. The following is an example of a set of transformation equations of the form 

defined by equations (1.1.7) and (1.1.8) in the case N =3. Consider the transformation from cylindrical 

coordinates (r, α, z) to spherical coordinates (ρ, β, α). From the geometry of the figure 1.1-5 we can find the 

transformation equations 

r = ρ sin β 

with inverse transformation 

Now make the substitutions 

α = α 0

Figure 1.1-5. Cylindrical and Spherical Coordinates 

The resulting transformations then have the forms of the equations (1.1.7) and (1.1.8). 

Calculation of Derivatives 

We now consider the chain rule applied to the differentiation of a function of the bar variables. We 

represent this differentiation in the indicial notation. Let Φ = Φ(x1 , x2 ,...,xn ) be a scalar function of the 

variables xi , i =1,...,N and let these variables be related to the set of variables xi , with i =1,...,N by 

the transformation equations (1.1.7) and (1.1.8). The partial derivatives of Φ with respect to the variables 

xi can be expressed in the indicial notation as 

∂Φ ∂Φ 

= 

∂xi ∂xj ∂xj ∂Φ 

= 

∂xi ∂x1 ∂x1 ∂Φ 

+ 

∂xi ∂x2 ∂x2 ∂Φ 

+ ···+ 

∂xi ∂xN ∂xN ∂xi (1.1.9) 

for any fixed value of i satisfying 1 ≤ i ≤ N. 

The second partial derivatives of Φ can also be expressed in the index notation. Differentiation of 

equation (1.1.9) partially with respect to xm produces 

∂2Φ ∂xi ∂Φ 

= 

∂xm ∂xj ∂2xj ∂xi ∂ 

+ 

∂xm ∂xm 

∂Φ 

∂xj j ∂x 

. (1.1.10) 

∂xi This result is nothing more than an application of the general rule for differentiating a product of two 

quantities. To evaluate the derivative of the bracketed term in equation (1.1.10) it must be remembered that 

the quantity inside the brackets is a function of the bar variables. Let 

G = ∂Φ 

∂x j = G(x1 , x 2 ,...,x N ) 

to emphasize this dependence upon the bar variables, then the derivative of G is 

∂G ∂G 

= 

∂xm ∂xk ∂xj∂x k 

This is just an application of the basic rule from equation (1.1.9) with Φ replaced by G. Hence the derivative 

from equation (1.1.10) can be expressed 

∂x k 

∂x m = ∂2 Φ 

∂2Φ ∂xi ∂Φ 

= 

∂xm ∂xj ∂2xj ∂xi∂xm + ∂2Φ ∂xj∂x k 

where i, m arefreeindicesandj, k are dummy summation indices. 

∂xk . (1.1.11) 

∂xm ∂x j 

∂x i 

∂x k 

∂x m 

(1.1.12) 

19

20 

EXAMPLE 1.1-18. Let Φ = Φ(r, θ) wherer, θ are polar coordinates related to the Cartesian coordinates 

(x, y) by the transformation equations x = r cos θ y = r sin θ. Find the partial derivatives ∂Φ ∂ 

and 

∂x 

2Φ ∂x2 Solution: The partial derivative of Φ with respect to x is found from the relation (1.1.9) and can be written 

∂Φ 

∂x 

∂Φ ∂r 

= 

∂r ∂x 

+ ∂Φ 

∂θ 

∂θ 

. (1.1.13) 

∂x 

The second partial derivative is obtained by differentiating the first partial derivative. From the product 

rule for differentiation we can write 

∂2Φ ∂Φ ∂ 

= 

∂x2 ∂r 

2 

r ∂r ∂ ∂Φ 

+ + 

∂x2 ∂x ∂x ∂r 

∂Φ ∂ 

∂θ 

2 

θ ∂θ ∂ ∂Φ 

+ . (1.1.14) 

∂x2 ∂x ∂x ∂θ 

To further simplify (1.1.14) it must be remembered that the terms inside the brackets are to be treated as 

functions of the variables r and θ and that the derivative of these terms can be evaluated by reapplying the 

basic rule from equation (1.1.13) with Φ replaced by ∂Φ 

∂r 

∂Φ 

and then Φ replaced by ∂θ . This gives 

∂2Φ ∂Φ ∂ 

= 

∂x2 ∂r 

2 2 r ∂r ∂ Φ 

+ 

∂x2 ∂x ∂r2 ∂r 

∂x + ∂2 

Φ ∂θ 

∂r∂θ ∂x 

+ ∂Φ ∂ 

∂θ 

2 2 θ ∂θ ∂ Φ ∂r 

+ 

∂x2 ∂x ∂θ∂r ∂x + ∂2Φ ∂θ2 

∂θ 

. 

∂x 

(1.1.15) 

From the transformation equations we obtain the relations r 2 = x 2 +y 2 

and tan θ = y 

and from 

x 

these relations we can calculate all the necessary derivatives needed for the simplification of the equations 

(1.1.13) and (1.1.15). These derivatives are: 

2r ∂r 

=2x or 

∂x 

sec 2 θ ∂θ y 

= − 

∂x x2 ∂2r ∂θ 

= − sin θ 

∂x2 ∂x = sin2 θ 

r 

or 

∂r x 

= 

∂x r =cosθ 

∂θ y sin θ 

= − = − 

∂x r2 r 

∂2 ∂θ 

θ −r cos θ ∂x = 

∂x2 r2 +sinθ ∂r 

∂x 

= 2sinθ cos θ 

. 

r 

Therefore, the derivatives from equations (1.1.13) and (1.1.15) can be expressed in the form 

∂Φ 

∂x 

= ∂Φ 

∂r 

∂2Φ ∂Φ 

= 

∂x2 ∂r 

∂Φ sin θ 

cos θ − 

∂θ r 

sin 2 θ 

r +2∂Φ 

sin θ cos θ 

∂θ r2 + ∂2Φ ∂r2 cos2 θ − 2 ∂2Φ cos θ sin θ 

∂r∂θ r 

+ ∂2Φ ∂θ2 sin 2 θ 

r2 . 

By letting x 1 = r, x 2 = θ, x 1 = x, x 2 = y and performing the indicated summations in the equations (1.1.9) 

and (1.1.12) there is produced the same results as above. 

Vector Identities in Cartesian Coordinates 

Employing the substitutions x 1 = x, x 2 = y, x 3 = z, where superscript variables are employed and 

denoting the unit vectors in Cartesian coordinates by e1, e2, e3, we illustrated how various vector operations 

are written by using the index notation.

Gradient. In Cartesian coordinates the gradient of a scalar field is 

grad φ = ∂φ 

∂x e1 + ∂φ 

∂y e2 + ∂φ 

∂z e3. 

The index notation focuses attention only on the components of the gradient. In Cartesian coordinates these 

components are represented using a comma subscript to denote the derivative 

ej · grad φ = φ,j = ∂φ 

, j =1, 2, 3. 

∂xj The comma notation will be discussed in section 4. For now we use it to denote derivatives. For example 

φ ,j = ∂φ 

∂xj , φ,jk = ∂2φ ∂xj , etc. 

∂xk Divergence. In Cartesian coordinates the divergence of a vector field A is a scalar field and can be 

represented 

∇· A = div A = ∂A1 

∂x 

+ ∂A2 

∂y 

+ ∂A3 

∂z . 

Employing the summation convention and index notation, the divergence in Cartesian coordinates can be 

represented 

∇· A = div A = Ai,i = ∂Ai ∂A1 ∂A2 ∂A3 

= + + 

∂xi ∂x1 ∂x2 ∂x3 where i is the dummy summation index. 

Curl. To represent the vector B =curl A = ∇× A in Cartesian coordinates, we note that the index 

notation focuses attention only on the components of this vector. The components Bi, i =1, 2, 3of B can 

be represented 

Bi = ei · curl A = eijkAk,j, for i, j, k =1, 2, 3 

where eijk is the permutation symbol introduced earlier and Ak,j = ∂Ak 

∂x j . To verify this representation of the 

curl A we need only perform the summations indicated by the repeated indices. We have summing on j that 

Bi = ei1kAk,1 + ei2kAk,2 + ei3kAk,3. 

Now summing each term on the repeated index k gives us 

Bi = ei12A2,1 + ei13A3,1 + ei21A1,2 + ei23A3,2 + ei31A1,3 + ei32A2,3 

Here i is a free index which can take on any of the values 1, 2or3. Consequently, we have 

For i =1, B1 = A3,2 − A2,3 = ∂A3 ∂A2 

− 

∂x2 ∂x3 For i =2, B2 = A1,3 − A3,1 = ∂A1 ∂A3 

− 

∂x3 ∂x1 For i =3, B3 = A2,1 − A1,2 = ∂A2 ∂A1 

− 

∂x1 ∂x2 which verifies the index notation representation of curl A in Cartesian coordinates. 

21

22 

Other Operations. The following examples illustrate how the index notation can be used to represent 

additional vector operators in Cartesian coordinates. 

1. In index notation the components of the vector ( B ·∇) A are 

{( B ·∇) A}· ep = Ap,qBq 

p, q =1, 2, 3 

This can be verified by performing the indicated summations. We have by summing on the repeated 

index q 

Ap,qBq = Ap,1B1 + Ap,2B2 + Ap,3B3. 

The index p is now a free index which can have any of the values 1, 2or3. We have: 

for p =1, A1,qBq = A1,1B1 + A1,2B2 + A1,3B3 

= ∂A1 

∂x1 B1 + ∂A1 

∂x2 B2 + ∂A1 

for p =2, 

∂x 

A2,qBq = A2,1B1 + A2,2B2 + A2,3B3 

3 B3 

= ∂A2 

∂x1 B1 + ∂A2 

∂x2 B2 + ∂A2 

for p =3, 

∂x 

A3,qBq = A3,1B1 + A3,2B2 + A3,3B3 

= ∂A3 

∂x1 B1 + ∂A3 

∂x2 B2 + ∂A3 

∂x 

2. The scalar ( B ·∇)φ has the following form when expressed in the index notation: 

( B ·∇)φ = Biφ,i = B1φ,1 + B2φ,2 + B3φ,3 

∂φ ∂φ ∂φ 

= B1 + B2 + B3 . 

∂x1 ∂x2 ∂x3 3 B3 

3 B3 

3. The components of the vector ( B ×∇)φ is expressed in the index notation by 

 

ei · ( 

B ×∇)φ = eijkBjφ,k. 

This can be verified by performing the indicated summations and is left as an exercise. 

4. The scalar ( B ×∇) · A may be expressed in the index notation. It has the form 

( B ×∇) · A = eijkBjAi,k. 

This can also be verified by performing the indicated summations and is left as an exercise. 

5. The vector components of ∇2A in the index notation are represented 

The proof of this is left as an exercise. 

ep ·∇ 2 A = Ap,qq.

EXAMPLE 1.1-19. In Cartesian coordinates prove the vector identity 

curl (f A)=∇×(f A)=(∇f) × A + f(∇× A). 

Solution: Let B =curl(f A) and write the components as 

Bi = eijk(fAk),j 

= eijk [fAk,j + f,jAk] 

= feijkAk,j + eijkf,jAk. 

This index form can now be expressed in the vector form 

B =curl(f A)=f(∇× A)+(∇f) × A 

EXAMPLE 1.1-20. Prove the vector identity ∇·( A + B)=∇· A + ∇· B 

Solution: Let A + B = C and write this vector equation in the index notation as Ai + Bi = Ci. We then 

have 

∇· C = Ci,i =(Ai + Bi),i = Ai,i + Bi,i = ∇· A + ∇· B. 

EXAMPLE 1.1-21. In Cartesian coordinates prove the vector identity ( A ·∇)f = A ·∇f 

Solution: In the index notation we write 

( A ·∇)f = Aif,i = A1f,1 + A2f,2 + A3f,3 

∂f ∂f ∂f 

= A1 + A2 + A3 

∂x1 ∂x2 ∂x3 = A ·∇f. 

EXAMPLE 1.1-22. In Cartesian coordinates prove the vector identity 

∇×( A × B)= A(∇· B) − B(∇· A)+( B ·∇) A − ( A ·∇) B 

Solution: The pth component of the vector ∇×( A × B)is 

ep · [∇×( A × B)] = epqk[ekjiAjBi],q 

= epqkekjiAjBi,q + epqkekjiAj,qBi 

By applying the e − δ identity, the above expression simplifies to the desired result. That is, 

In vector form this is expressed 

ep · [∇×( A × B)] = (δpjδqi − δpiδqj)AjBi,q +(δpjδqi − δpiδqj)Aj,qBi 

= ApBi,i − AqBp,q + Ap,qBq − Aq,qBp 

∇×( A × B)= A(∇· B) − ( A ·∇) B +( B ·∇) A − B(∇· A) 

23

24 

EXAMPLE 1.1-23. In Cartesian coordinates prove the vector identity ∇×(∇× A)=∇(∇· A) −∇ 2 A 

Solution: We have for the ith component of ∇× A is given by ei · [∇× A]=eijkAk,j and consequently the 

pth component of ∇×(∇× A)is 

The e − δ identity produces 

ep · [∇×(∇× A)] = epqr[erjkAk,j],q 

= epqrerjkAk,jq. 

ep · [∇×(∇× A)] = (δpjδqk − δpkδqj)Ak,jq 

= Ak,pk − Ap,qq. 

Expressing this result in vector form we have ∇×(∇× A)=∇(∇· A) −∇ 2 A. 

Indicial Form of Integral Theorems 

The divergence theorem, in both vector and indicial notation, can be written 

 

 

 

div · F dτ= F · n dσ 

Fi,i dτ = Fini dσ i =1, 2, 3 (1.1.16) 

V 

S 

V 

where ni are the direction cosines of the unit exterior normal to the surface, dτ is a volume element and dσ 

is an element of surface area. Note that in using the indicial notation the volume and surface integrals are 

to be extended over the range specified by the indices. This suggests that the divergence theorem can be 

applied to vectors in n−dimensional spaces. 

The vector form and indicial notation for the Stokes theorem are 

 

(∇× 

S 

 

 

 

F ) · n dσ = F · dr 

eijkFk,jni dσ = Fi dx 

C 

S 

C 

i 

i, j, k =1, 2, 3 (1.1.17) 

and the Green’s theorem in the plane, which is a special case of the Stoke’s theorem, can be expressed 

 

∂F2 ∂F1 

− dxdy = F1 dx + F2 dy 

∂x ∂y 

C 

 

 

e3jkFk,j dS = Fi dx 

S 

C 

i 

i, j, k =1, 2 (1.1.18) 

Other forms of the above integral theorems are 

 

∇φdτ = 

V 

S 

S 

φ n dσ 

obtained from the divergence theorem by letting F = φ C where C is a constant vector. By replacing F by 

F × C in the divergence theorem one can derive 

 

∇× 

V 

 

F dτ = − F × ndσ. 

S 

In the divergence theorem make the substitution F = φ∇ψ to obtain 

 

 

2 

(φ∇ ψ +(∇φ) · (∇ψ) dτ = (φ∇ψ) · n dσ. 

V 

S

The Green’s identity 

V 

 

2 2 

φ∇ ψ − ψ∇ φ dτ = (φ∇ψ − ψ∇φ) · n dσ 

is obtained by first letting F = φ∇ψ in the divergence theorem and then letting F = ψ∇φ in the divergence 

theorem and then subtracting the results. 

Determinants, Cofactors 

For A =(aij), i,j=1,...,n an n × n matrix, the determinant of A can be written as 

det A = |A| = ei1i2i3...ina1i1a2i2a3i3 ...anin. 

This gives a summation of the n! permutations of products formed from the elements of the matrix A. The 

result is a single number called the determinant of A. 

EXAMPLE 1.1-24. Inthecasen =2wehave 

 

 

|A| = 

a11 a12 

a21 a22 

S 

 

 

 

= enma1na2m 

= e1ma11a2m + e2ma12a2m 

= e12a11a22 + e21a12a21 

= a11a22 − a12a21 

EXAMPLE 1.1-25. Inthecasen = 3 we can use either of the notations 

⎛ 

A = ⎝ a11 a12 

⎞ 

a13 

⎠ or 

⎛ 

A = 

a21 a22 a23 

a31 a32 a33 

and represent the determinant of A in any of the forms 

det A = eijka1ia2ja3k 

det A = eijkai1aj2ak3 

det A = eijka i 1a j 

2 ak 3 

det A = eijka 1 i a 2 ja 3 k. 

⎝ a11 a1 2 a1 3 

a2 1 a22 a23 a3 1 a32 a33 These represent row and column expansions of the determinant. 

An important identity results if we examine the quantity Brst = eijka i ra j sa k t . Itisaneasyexerciseto 

change the dummy summation indices and rearrange terms in this expression. For example, 

Brst = eijka i r aj s ak t = ekjia k r aj s ai t = ekjia i t aj s ak r = −eijka i t aj s ak r 

⎞ 

⎠ 

= −Btsr, 

and by considering other permutations of the indices, one can establish that Brst is completely skewsymmetric. 

In the exercises it is shown that any third order completely skew-symmetric system satisfies 

Brst = B123erst. But B123 =detA and so we arrive at the identity 

Brst = eijka i r aj s ak t 

= |A|erst. 

25

26 

Other forms of this identity are 

e ijk a r i as j at k 

= |A|erst 

Consider the representation of the determinant 

 

 

 

|A| = 

 

 

and eijkairajsakt = |A|erst. (1.1.19) 

a1 1 a12 a13 a2 1 a2 2 a2 3 

a3 1 a32 a33 by use of the indicial notation. By column expansions, this determinant can be represented 

|A| = ersta r 1 as 2 at 3 

and if one uses row expansions the determinant can be expressed as 

 

 

 

 

 

 

(1.1.20) 

|A| = e ijk a 1 i a2j a3k . (1.1.21) 

Define Ai m as the cofactor of the element am i in the determinant |A|. From the equation (1.1.20) the cofactor 

of ar 1 is obtained by deleting this element and we find 

The result (1.1.20) can then be expressed in the form 

A 1 r = ersta s 2a t 3. (1.1.22) 

|A| = a r 1A1r = a11 A11 + a21 A12 + a31 A13 . (1.1.23) 

That is, the determinant |A| is obtained by multiplying each element in the first column by its corresponding 

cofactor and summing the result. Observe also that from the equation (1.1.20) we find the additional 

cofactors 

A 2 s = ersta r 1at3 and A 3 t = ersta r 1as2 . (1.1.24) 

Hence, the equation (1.1.20) can also be expressed in one of the forms 

|A| = a s 2 A2 s = a1 2 A2 1 + a2 2 A2 2 + a3 2 A2 3 

|A| = a t 3A 3 t = a 1 3A 3 1 + a 2 3A 3 2 + a 3 3A 3 3 

The results from equations (1.1.22) and (1.1.24) can be written in a slightly different form with the indicial 

notation. From the notation for a generalized Kronecker delta defined by 

the above cofactors can be written in the form 

e ijk elmn = δ ijk 

lmn , 

A 1 r = e 123 ersta s 2a t 3 = 1 

2! e1jk ersta s ja t k = 1 

2! δ1jk 

rst a s ja t k 

A 2 r = e123 esrta s 1 at 3 

A 3 r = e123 etsra t 1 as 2 

1 

= 

2! e2jkersta s jat 1 

k = 

2! δ2jk rst asj atk 1 

= 

2! e3jkersta s jat 1 

k = 

2! δ3jk rst asj atk .

These cofactors are then combined into the single equation 

A i r 

= 1 

2! δijk 

rst as j at k 

(1.1.25) 

which represents the cofactor of ar i . When the elements from any row (or column) are multiplied by their 

corresponding cofactors, and the results summed, we obtain the value of the determinant. Whenever the 

elements from any row (or column) are multiplied by the cofactor elements from a different row (or column), 

and the results summed, we get zero. This can be illustrated by considering the summation 

a m r Ai m 

Here we have used the e − δ identity to obtain 

1 

= 

2! δijk mstasj atk am 1 

r = 

2! eijkemsta m r asj atk = 1 

2! eijkerjk|A| = 1 

2! δijk 

rjk |A| = δi r |A| 

δ ijk 

rjk = eijk erjk = e jik ejrk = δ i r δk k − δi k δk r =3δi r − δi r =2δi r 

which was used to simplify the above result. 

As an exercise one can show that an alternate form of the above summation of elements by its cofactors 

is 

a r m Am i = |A|δr i . 

27

28 

EXERCISE 1.1 

◮ 1. Simplify each of the following by employing the summation property of the Kronecker delta. Perform 

sums on the summation indices only if your are unsure of the result. 

(a) eijkδkn 

(b) eijkδisδjm 

(c) eijkδisδjmδkn 

(d) aijδin 

◮ 2. Simplify and perform the indicated summations over the range 1, 2, 3 

(a) δii 

(b) δijδij 

(c) eijkAiAjAk 

(d) eijkeijk 

(e) eijkδjk 

(e) δijδjn 

(f) δijδjnδni 

(f) AiBjδji − BmAnδmn 

◮ 3. Express each of the following in index notation. Be careful of the notation you use. Note that A = Ai 

is an incorrect notation because a vector can not equal a scalar. The notation A · ei = Ai should be used to 

express the ith component of a vector. 

(a) A · ( B × C) 

(b) A × ( B × C) 

(c) B( A · C) 

(d) B( A · C) − C( A · B) 

◮ 4. Show the e permutation symbol satisfies: (a) eijk = ejki = ekij (b) eijk = −ejik = −eikj = −ekji 

◮ 5. Use index notation to verify the vector identity A × ( B × C)= B( A · C) − C( A · B) 

◮ 6. Let yi = aijxj and xm = aimzi where the range of the indices is 1, 2 

(a) Solve for yi in terms of zi using the indicial notation and check your result 

to be sure that no index repeats itself more than twice. 

(b) Perform the indicated summations and write out expressions 

for y1,y2 in terms of z1,z2 

(c) Express the above equations in matrix form. Expand the matrix 

equations and check the solution obtained in part (b). 

◮ 7. Use the e − δ identity to simplify (a) eijkejik (b) eijkejki 

◮ 8. Prove the following vector identities: 

(a) A · ( B × C)= B · ( C × A)= C · ( A × B) triple scalar product 

(b) ( A × B) × C = B( A · C) − A( B · C) 

◮ 9. Prove the following vector identities: 

(a) ( A × B) · ( C × D)=( A · C)( B · D) − ( A · D)( B · C) 

(b) A × ( B × C)+ B × ( C × A)+ C × ( A × B)=0 

(c) ( A × B) × ( C × D)= B( A · C × D) − A( B · C × D)

◮ 10. For A =(1, −1, 0) and B =(4, −3, 2) find using the index notation, 

(a) Ci = eijkAjBk, i =1, 2, 3 

(b) AiBi 

◮ 11. Represent the differential equations 

using the index notation. 

(c) What do the results in (a) and (b) represent? 

dy1 

dt = a11y1 + a12y2 

and 

dy2 

dt = a21y1 + a22y2 

◮ 12. 

Let Φ = Φ(r, θ) wherer, θ are polar coordinates related to Cartesian coordinates (x, y) by the transformation 

equations x = r cos θ and y = r sin θ. 

(a) Find the partial derivatives 

∂Φ 

, 

∂y 

and 

∂2Φ ∂y2 (b) Combine the result in part (a) with the result from EXAMPLE 1.1-18 to calculate the Laplacian 

in polar coordinates. 

∇ 2 Φ= ∂2Φ ∂x2 + ∂2Φ ∂y2 ◮ 13. (Index notation) Let a11 =3, a12 =4, a21 =5, a22 =6. 

Calculate the quantity C = aijaij, i,j=1, 2. 

◮ 14. Show the moments of inertia Iij defined by 

 

I11 = (y 2 + z 2 )ρ(x, y, z) dτ 

R 

 

I22 = 

R 

 

I33 = 

R 

(x 2 + z 2 )ρ(x, y, z) dτ 

(x 2 + y 2 )ρ(x, y, z) dτ 

can be represented in the index notation as Iij = 

 

R 

x1 = x, x2 = y, x3 = z and dτ = dxdydz is an element of volume. 

 

I23 = I32 = − 

R 

R 

yzρ(x, y, z) dτ 

 

I12 = I21 = − xyρ(x, y, z) dτ 

 

I13 = I31 = − xzρ(x, y, z) dτ, 

◮ 15. Determine if the following relation is true or false. Justify your answer. 

Hint: Let em =(δ1m,δ2m,δ3m). 

ei · ( ej × ek) =(ei × ej) · ek = eijk, i,j,k =1, 2, 3. 

R 

x m x m δij − x i x j ρdτ, where ρ is the density, 

◮ 16. Without substituting values for i, l =1, 2, 3 calculate all nine terms of the given quantities 

(a) B il =(δ i jAk + δ i kAj)e jkl 

(b) Ail =(δ m i B k + δ k i B m )emlk 

◮ 17. Let Amnx m y n = 0 for arbitrary x i and y i , i =1, 2, 3, and show that Aij = 0 for all values of i, j. 

29

30 

◮ 18. 

(a) For amn,m,n=1, 2, 3 skew-symmetric, show that amnx m x n =0. 

(b) Let amnxmxn =0, m,n =1, 2, 3 for all values of xi ,i =1, 2, 3andshowthatamn must be skew- 

symmetric. 

◮ 19. Let A and B denote 3 × 3 matrices with elements aij and bij respectively. Show that if C = AB is a 

matrix product, then det(C) =det(A) · det(B). 

Hint: Use the result from example 1.1-9. 

◮ 20. 

(a) Let u 1 ,u 2 ,u 3 be functions of the variables s 1 ,s 2 ,s 3 . Further, assume that s 1 ,s 2 ,s 3 areinturneach 

functions of the variables x 1 , x 2 , x 3 . Let 

respect to the x ′ s. Show that 

(b) Note that ∂xi 

∂¯x j 

∂¯x j ∂xi 

= 

∂xm of the transformation (1.1.7). 

 

 

 

∂u 

 

m 

∂xn 

 

 

= ∂(u1 ,u2 , u3 ) 

 

 

 

∂u 

 

i 

∂xm 

 

 

= 

 

 

 

∂u 

 

i 

∂sj ∂sj ∂xm 

 

 

= 

 

 

 

∂u 

 

i 

∂sj 

 

 

· 

 

 

 

∂s 

 

j 

∂xm 

 

 

. 

∂(x 1 ,x 2 , x 3 ) denote the Jacobian of the u′ s with 

∂xm = δi m and show that J( x ¯x 

x 

¯x )·J( x )=1,whereJ( ¯x ) is the Jacobian determinant 

◮ 21. A third order system aℓmn with ℓ, m, n =1, 2, 3 is said to be symmetric in two of its subscripts if the 

components are unaltered when these subscripts are interchanged. When aℓmn is completely symmetric then 

aℓmn = amℓn = aℓnm = amnℓ = anmℓ = anℓm. Whenever this third order system is completely symmetric, 

then: (i) How many components are there? (ii) How many of these components are distinct? 

Hint: Consider the three cases (i) ℓ = m = n (ii) ℓ = m = n (iii) ℓ = m = n. 

◮ 22. A third order system bℓmn with ℓ, m, n =1, 2, 3 is said to be skew-symmetric in two of its subscripts 

if the components change sign when the subscripts are interchanged. A completely skew-symmetric third 

order system satisfies bℓmn = −bmℓn = bmnℓ = −bnmℓ = bnℓm = −bℓnm. (i) How many components does 

a completely skew-symmetric system have? (ii) How many of these components are zero? (iii) How many 

components can be different from zero? (iv) Show that there is one distinct component b123 and that 

bℓmn = eℓmnb123. 

Hint: Consider the three cases (i) ℓ = m = n (ii) ℓ = m = n (iii) ℓ = m = n. 

◮ 23. Let i, j, k =1, 2, 3 and assume that eijkσjk = 0 for all values of i. What does this equation tell you 

about the values σij, i, j =1, 2, 3? 

◮ 24. Assume that Amn and Bmn are symmetric for m, n =1, 2, 3. Let Amnx m x n = Bmnx m x n for arbitrary 

values of xi ,i=1, 2, 3, and show that Aij = Bij for all values of i and j. 

◮ 25. Assume Bmn is symmetric and Bmnx m x n = 0 for arbitrary values of x i ,i=1, 2, 3, show that Bij =0.

◮ 26. (Generalized Kronecker delta) Define the generalized Kronecker delta as the n×n determinant 

δ ij...k 

mn...p = 

 

 

 

 

 

 

 

. 

 

(a) Show eijk = δ 123 

ijk 

(b) Show e ijk = δ ijk 

123 

(c) Show δ ij mn = e ij emn 

δi m δi n ··· δi p 

δj m δj n ··· δj p 

. 

. .. 

δ k m δk n ··· δ k p 

(d) Define δ rs 

mn = δrsp mnp (summation on p) 

and show δ rs 

mn = δr mδs n − δr nδs m 

. 

 

 

 

 

 

 

 

 

 

where δ r s 

is the Kronecker delta. 

Note that by combining the above result with the result from part (c) 

we obtain the two dimensional form of the e − δ identity e rs emn = δ r mδ s n − δ r nδ s m. 

(e) Define δ r m 

(f) Show δ rst 

rst =3! 

◮ 27. Let A i r denote the cofactor of ar i 

= 1 

2 δrn 

mn (summation on n) and show δ rst 

pst =2δr p 

 

 

 

in the determinant 

 

 

a1 1 a12 a13 a2 1 a22 a23 a3 1 a3 2 a3 3 

 

 

 

 

as given by equation (1.1.25). 

 

(a) Show e rst A i r = eijk a s j at k (b) Show erstA r i = eijka j s ak t 

◮ 28. (a) Show that if Aijk = Ajik, i, j, k =1, 2, 3 there is a total of 27 elements, but only 18 are distinct. 

(b) Show that for i, j, k =1, 2,...,N there are N 3 elements, but only N 2 (N +1)/2 aredistinct. 

◮ 29. Let aij = BiBj for i, j =1, 2, 3whereB1,B2,B3 are arbitrary constants. Calculate det(aij) =|A|. 

◮ 30. 

(a) For A =(aij), i,j=1, 2, 3, show |A| = eijkai1aj2ak3. 

(b) For A =(a i j ),i,j=1, 2, 3, show |A| = eijka i 1 aj 

2 ak 3 . 

(c) For A =(a i j), i,j=1, 2, 3, show |A| = e ijk a 1 i a 2 ja 3 k. 

(d) For I =(δ i j), i,j=1, 2, 3, show |I| =1. 

◮ 31. Let |A| = eijkai1aj2ak3 and define Aim as the cofactor of aim. Show the determinant can be 

expressed in any of the forms: 

(a) |A| = Ai1ai1 where Ai1 = eijkaj2ak3 

(b) |A| = Aj2aj2 where Ai2 = ejikaj1ak3 

(c) |A| = Ak3ak3 where Ai3 = ejkiaj1ak2 

31

32 

◮ 32. Show the results in problem 31 can be written in the forms: 

Ai1 = 1 

2! e1steijkajsakt, Ai2 = 1 

2! e2steijkajsakt, Ai3 = 1 

2! e3steijkajsakt, or Aim = 1 

2! emsteijkajsakt 

◮ 33. Use the results in problems 31 and 32 to prove that apmAim = |A|δip. 

◮ 34. 

⎛ 

1 

Let (aij) = ⎝ 1 

2 

0 

⎞ 

1 

3⎠and 

calculate C = aijaij, i,j=1, 2, 3. 

2 3 2 

◮ 35. Let 

a111 = −1, a112 =3, a121 =4, a122 =2 

a211 =1, a212 =5, a221 =2, a222 = −2 

and calculate the quantity C = aijkaijk, i,j,k =1, 2. 

◮ 36. Let 

a1111 =2, a1112 =1, a1121 =3, a1122 =1 

a1211 =5, a1212 = −2, a1221 =4, a1222 = −2 

a2111 =1, a2112 =0, a2121 = −2, a2122 = −1 

a2211 = −2, a2212 =1, a2221 =2, a2222 =2 

and calculate the quantity C = aijklaijkl, i,j,k,l=1, 2. 

◮ 37. Simplify the expressions: 

(a) (Aijkl + Ajkli + Aklij + Alijk)xixjxkxl 

(b) (Pijk + Pjki + Pkij)x i x j x k 

(c) 

∂x i 

∂x j 

(d) aij 

∂2xi ∂xt∂xs ∂x j 

∂x 

∂ 

r − ami 

2xm ∂xs∂xt ∂xi ∂xr ◮ 38. Let g denote the determinant of the matrix having the components gij, i,j=1, 2, 3. Show that 

(a) 

 

 

g1r 

gerst = 

g2r 

 

g1s 

g2s 

 

g1t 

 

g2t 

 

 

(b) 

 

 

gir 

gersteijk = 

gjr 

 

gis 

gjs 

 

git 

 

gjt 

 

 

◮ 39. Show that e ijk emnp = δ ijk 

mnp = 

g3r g3s g3t 

 

 

 

 

 

 

δi m δi n δi δ 

p 

j m δj n δj p 

δk m δk n δk p 

◮ 40. Show that e ijk emnpA mnp = A ijk − A ikj + A kij − A jik + A jki − A kji 

Hint: Use the results from problem 39. 

◮ 41. Show that 

(a) e ij eij =2! 

(b) e ijk eijk =3! 

 

 

 

 

 

 

(c) e ijkl eijkl =4! 

gkr gks gkt 

(d) Guess at the result e i1i2...in ei1i2...in

◮ 42. Determine if the following statement is true or false. Justify your answer. eijkAiBjCk = eijkAjBkCi. 

◮ 43. Let aij, i,j=1, 2 denote the components of a 2 × 2matrixA, which are functions of time t. 

 

(a) Expand both |A| = eijai1aj2 and |A| = 

a11 

 

a12 

 

a21 a22 to verify that these representations are the same. 

(b) Verify the equivalence of the derivative relations 

d|A| 

dt 

dai1 

= eij 

dt aj2 

daj2 

+ eijai1 

dt 

and d|A| 

dt = 

 

 

 

 

da11 da12 

dt dt 

a21 a22 

 

 

 

+ 

 

 

 

a11 

da21 

dt 

(c) Let aij, i,j=1, 2, 3 denote the components of a 3 × 3matrixA, which are functions of time t. Develop 

appropriate relations, expand them and verify, similar to parts (a) and (b) above, the representation of 

a determinant and its derivative. 

◮ 44. For f = f(x1 ,x2 ,x3 )andφ = φ(f) differentiable scalar functions, use the indicial notation to find a 

formula to calculate grad φ. 

◮ 45. Use the indicial notation to prove (a) ∇×∇φ = 0 (b) ∇·∇× A =0 

◮ 46. If Aij is symmetric and Bij is skew-symmetric, i, j =1, 2, 3, then calculate C = AijBij. 

◮ 47. Assume Aij = Aij(x 1 , x 2 , x 3 )andAij = Aij(x 1 ,x 2 ,x 3 )fori, j =1, 2, 3 are related by the expression 

∂x 

Amn = Aij 

i 

∂xm ∂xj ∂Amn 

n . Calculate the derivative . 

∂x k ∂x 

◮ 48. Prove that if any two rows (or two columns) of a matrix are interchanged, then the value of the 

determinant of the matrix is multiplied by minus one. Construct your proof using 3 × 3 matrices. 

◮ 49. Prove that if two rows (or columns) of a matrix are proportional, then the value of the determinant 

of the matrix is zero. Construct your proof using 3 × 3 matrices. 

◮ 50. Prove that if a row (or column) of a matrix is altered by adding some constant multiple of some other 

row (or column), then the value of the determinant of the matrix remains unchanged. Construct your proof 

using 3 × 3 matrices. 

◮ 51. Simplify the expression φ = eijkeℓmnAiℓAjmAkn. 

◮ 52. Let Aijk denote a third order system where i, j, k =1, 2. (a) How many components does this system 

have? (b) Let Aijk be skew-symmetric in the last pair of indices, how many independent components does 

the system have? 

◮ 53. Let Aijk denote a third order system where i, j, k =1, 2, 3. (a) How many components does this 

system have? (b) In addition let Aijk = Ajik and Aikj = −Aijk and determine the number of distinct 

nonzero components for Aijk. 

a12 

da22 

dt 

 

 

 

 

33

34 

◮ 54. Show that every second order system Tij can be expressed as the sum of a symmetric system Aij and 

skew-symmetric system Bij. Find Aij and Bij in terms of the components of Tij. 

◮ 55. Consider the system Aijk, i,j,k =1, 2, 3, 4. 

(a) How many components does this system have? 

(b) Assume Aijk is skew-symmetric in the last pair of indices, how many independent components does this 

system have? 

(c) Assume that in addition to being skew-symmetric in the last pair of indices, Aijk + Ajki + Akij =0is 

satisfied for all values of i, j, and k, then how many independent components does the system have? 

◮ 56. (a) Write the equation of a line r = r0 + t A in indicial form. (b) Write the equation of the plane 

n · (r − r0) = 0 in indicial form. (c) Write the equation of a general line in scalar form. (d) Write the 

equation of a plane in scalar form. (e) Find the equation of the line defined by the intersection of the 

planes 2x +3y +6z =12and6x +3y + z =6. (f) Find the equation of the plane through the points 

(5, 3, 2), (3, 1, 5), (1, 3, 3). Find also the normal to this plane. 

◮ 57. The angle 0 ≤ θ ≤ π between two skew lines in space is defined as the angle between their direction 

vectors when these vectors are placed at the origin. Show that for two lines with direction numbers ai and 

bi i =1, 2, 3, the cosine of the angle between these lines satisfies 

cos θ = 

aibi 

√ √ 

aiai bibi 

◮ 58. Let aij = −aji for i, j =1, 2,...,N and prove that for N odd det(aij) =0. 

◮ 59. Let λ = Aijxixj where Aij = Aji and calculate (a) 

∂λ 

∂xm 

(b) 

∂ 2 λ 

∂xm∂xk 

◮ 60. Given an arbitrary nonzero vector Uk, k =1, 2, 3, define the matrix elements aij = eijkUk, whereeijk 

is the e-permutation symbol. Determine if aij is symmetric or skew-symmetric. Suppose Uk is defined by 

the above equation for arbitrary nonzero aij, thensolveforUkin terms of the aij. 

◮ 61. If Aij = AiBj = 0foralli, j values and Aij = Aji for i, j =1, 2,...,N, show that Aij = λBiBj 

where λ is a constant. State what λ is. 

◮ 62. Assume that Aijkm, withi, j, k, m =1, 2, 3, is completely skew-symmetric. How many independent 

components does this quantity have? 

◮ 63. Consider Rijkm, i, j, k, m =1, 2, 3, 4. (a) How many components does this quantity have? (b) If 

Rijkm = −Rijmk = −Rjikm then how many independent components does Rijkm have? (c) If in addition 

Rijkm = Rkmij determine the number of independent components. 

◮ 64. Let xi = aij ¯xj, i, j =1, 2, 3 denote a change of variables from a barred system of coordinates to an 

unbarred system of coordinates and assume that Āi = aijAj where aij are constants, Āi is a function of the 

∂ 

¯xj variables and Aj is a function of the xj variables. Calculate 

Āi 

. 

∂¯xm

§1.2 TENSOR CONCEPTS AND TRANSFORMATIONS 

For e1, e2, e3 independent orthogonal unit vectors (base vectors), we may write any vector A as 

A = A1 e1 + A2 e2 + A3 e3 

where (A1,A2,A3) are the coordinates of A relative to the base vectors chosen. These components are the 

projection of A onto the base vectors and 

A =( A · e1) e1 +( A · e2) e2 +( A · e3) e3. 

Select any three independent orthogonal vectors, ( E1, E2, E3), not necessarily of unit length, we can then 

write 

e1 = E1 

| E1| , e2 = E2 

| E2| , e3 = E3 

| E3| , 

and consequently, the vector A can be expressed as 

 

A 

A 

· E1 

= 

E1 · 

A · E2 

E1 + 

E1 E2 · 

A · E3 

E2 + 

E2 E3 · 

E3. 

E3 

Here we say that 

A · E (i) 

E (i) · , i =1, 2, 3 

E (i) 

are the components of A relative to the chosen base vectors E1, E2, E3. Recall that the parenthesis about 

the subscript i denotes that there is no summation on this subscript. It is then treated as a free subscript 

which can have any of the values 1, 2or3. 

Reciprocal Basis 

Consider a set of any three independent vectors ( E1, E2, E3) which are not necessarily orthogonal, nor of 

unit length. In order to represent the vector A in terms of these vectors we must find components (A1 ,A2 ,A3 ) 

such that 

A = A 1 E1 + A 2 E2 + A 3 E3. 

This can be done by taking appropriate projections and obtaining three equations and three unknowns from 

which the components are determined. A much easier way to find the components (A1 ,A2 ,A3 )istoconstruct 

a reciprocal basis ( E1 , E2 , E3 ). Recall that two bases ( E1, E2, E3) and( E1 , E2 , E3 ) are said to be reciprocal 

if they satisfy the condition 

Ei · E j = δ j 

i = 

 

1 if i = j 

0 if i = j . 

Note that E2 · E 1 = δ 1 2 =0 andE3 · E 1 = δ 1 3 = 0 so that the vector E1 is perpendicular to both the 

vectors E2 and E3. (i.e. A vector from one basis is orthogonal to two of the vectors from the other basis.) 

We can therefore write E1 = V −1E2 × E3 where V is a constant to be determined. By taking the dot 

product of both sides of this equation with the vector E1 we find that V = E1 · ( E2 × E3) isthevolume 

of the parallelepiped formed by the three vectors E1, E2, E3 when their origins are made to coincide. In a 

35

36 

similar manner it can be demonstrated that for ( E1, E2, E3) a given set of basis vectors, then the reciprocal 

basis vectors are determined from the relations 

E 1 = 1 

V E2 × E3, E 2 1 

= 

V E3 × E1, E 3 1 

= 

V E1 × E2, 

where V = E1 · ( E2 × E3) = 0 is a triple scalar product and represents the volume of the parallelepiped 

having the basis vectors for its sides. 

Let ( E1, E2, E3) and( E1 , E2 , E3 ) denote a system of reciprocal bases. We can represent any vector A 

with respect to either of these bases. If we select the basis ( E1, E2, E3) and represent A in the form 

A = A 1 E1 + A 2 E2 + A 3 E3, (1.2.1) 

then the components (A1 ,A2 ,A3 )of A relative to the basis vectors ( E1, E2, E3) are called the contravariant 

components of A. These components can be determined from the equations 

A · E 1 = A 1 , A · E 2 2 

= A , A · E 3 3 

= A . 

Similarly, if we choose the reciprocal basis ( E1 , E2 , E3 ) and represent A in the form 

A = A1 E 1 + A2 E 2 + A3 E 3 , (1.2.2) 

then the components (A1,A2,A3) relative to the basis ( E 1 , E 2 , E 3 ) are called the covariant components of 

A. These components can be determined from the relations 

A · E1 = A1, A · E2 = A2, A · E3 = A3. 

The contravariant and covariant components are different ways of representing the same vector with respect 

to a set of reciprocal basis vectors. There is a simple relationship between these components which we now 

develop. We introduce the notation 

Ei · Ej = gij = gji, and E i · E j = g ij = g ji 

(1.2.3) 

where gij are called the metric components of the space and gij are called the conjugate metric components 

of the space. We can then write 

A · E1 = A1( E 1 · E1)+A2( E 2 · E1)+A3( E 3 · E1) =A1 

or 

A · E1 = A 1 ( E1 · E1)+A 2 ( E2 · E1)+A 3 ( E3 · E1) =A1 

A1 = A 1 g11 + A 2 g12 + A 3 g13. (1.2.4) 

In a similar manner, by considering the dot products A · E2 and A · E3 one can establish the results 

A2 = A 1 g21 + A 2 g22 + A 3 g23 

These results can be expressed with the index notation as 

A3 = A 1 g31 + A 2 g32 + A 3 g33. 

Ai = gikA k . (1.2.6) 

Forming the dot products A · E 1 , A · E 2 , A · E 3 it can be verified that 

A i = g ik Ak. (1.2.7) 

The equations (1.2.6) and (1.2.7) are relations which exist between the contravariant and covariant components 

of the vector A. Similarly, if for some value j we have Ej = α E1 + β E2 + γ E3, then one can show 

that E j = g ij Ei. This is left as an exercise.

Coordinate Transformations 

Consider a coordinate transformation from a set of coordinates (x, y, z) to(u, v, w) defined by a set of 


x = x(u, v, w) 

y = y(u, v, w) 

z = z(u, v, w) 

(1.2.8) 

It is assumed that these transformations are single valued, continuous and possess the inverse transformation 

u = u(x, y, z) 

v = v(x, y, z) 

w = w(x, y, z). 

(1.2.9) 

These transformation equations define a set of coordinate surfaces and coordinate curves. The coordinate 

surfaces are defined by the equations 

u(x, y, z) =c1 

v(x, y, z) =c2 

w(x, y, z) =c3 

where c1,c2,c3 are constants. These surfaces intersect in the coordinate curves 

where 

(1.2.10) 

r(u, c2,c3), r(c1,v,c3), r(c1,c2,w), (1.2.11) 

r(u, v, w) =x(u, v, w) e1 + y(u, v, w) e2 + z(u, v, w) e3. 

The general situation is illustrated in the figure 1.2-1. 

Consider the vectors 

E 1 =gradu = ∇u, E 2 =gradv = ∇v, E 3 =gradw = ∇w (1.2.12) 

evaluated at the common point of intersection (c1,c2,c3) of the coordinate surfaces. The system of vectors 

( E1 , E2 , E3 ) can be selected as a system of basis vectors which are normal to the coordinate surfaces. 

Similarly, the vectors 

E1 = ∂r 

∂u , E2 = ∂r 

∂v , E3 = ∂r 

(1.2.13) 

∂w 

when evaluated at the common point of intersection (c1,c2,c3) forms a system of vectors ( E1, E2, E3) which 

we can select as a basis. This basis is a set of tangent vectors to the coordinate curves. It is now demonstrated 

that the normal basis ( E1 , E2 , E3 ) and the tangential basis ( E1, E2, E3) are a set of reciprocal bases. 

Recall that r = x e1 + y e2 + z e3 denotes the position vector of a variable point. By substitution for 

x, y, z from (1.2.8) there results 

r = r(u, v, w) =x(u, v, w) e1 + y(u, v, w) e2 + z(u, v, w) e3. (1.2.14) 

37

38 

A small change in r is denoted 

Figure 1.2-1. Coordinate curves and coordinate surfaces. 

dr = dx e1 + dy e2 + dz e3 = ∂r ∂r ∂r 

du + dv + dw (1.2.15) 

∂u ∂v ∂w 

where 

∂r ∂x 

= 

∂u ∂u e1 + ∂y 

∂u e2 + ∂z 

∂u e3 

∂r ∂x 

= 

∂v ∂v e1 + ∂y 

∂v e2 + ∂z 

∂v e3 

∂r ∂x 

= 

∂w ∂w e1 + ∂y 

∂w e2 + ∂z 

∂w e3. 

(1.2.16) 

In terms of the u, v, w coordinates, this change can be thought of as moving along the diagonal of a parallelepiped 

having the vector sides ∂r 

∂u du, 

∂r 

∂r 

dv, and 

∂v ∂w dw. 

Assume u = u(x, y, z) is defined by equation (1.2.9) and differentiate this relation to obtain 

du = ∂u ∂u ∂u 

dx + dy + dz. (1.2.17) 

∂x ∂y ∂z 

The equation (1.2.15) enables us to represent this differential in the form: 

du =gradu · dr 

 

∂r ∂r ∂r 

du =gradu · du + dv + 

∂u ∂v ∂w dw 

 

 

du = grad u · ∂r 

 

du + grad u · 

∂u 

∂r 

 

dv + grad u · 

∂v 

∂r 

 

dw. 

∂w 

By comparing like terms in this last equation we find that 

(1.2.18) 

E 1 · E1 =1, E 1 · E2 =0, E 1 · E3 =0. (1.2.19) 

Similarly, from the other equations in equation (1.2.9) which define v = v(x, y, z), 

can be demonstrated that 

and w = w(x, y, z) it 

 

dv = grad v · ∂r 

 

du + grad v · 

∂u 

∂r 

 

dv + grad v · 

∂v 

∂r 

 

dw 

∂w 

(1.2.20)

and 

 

dw = grad w · ∂r 

 

du + grad w · 

∂u 

∂r 

 

dv + grad w · 

∂v 

∂r 

 

dw. 

∂w 

(1.2.21) 

By comparing like terms in equations (1.2.20) and (1.2.21) we find 

E 2 · E1 =0, E 2 · E2 =1, E 2 · E3 =0 

E 3 · E1 =0, E 3 · E2 =0, E 3 · E3 =1. 

(1.2.22) 

The equations (1.2.22) and (1.2.19) show us that the basis vectors defined by equations (1.2.12) and (1.2.13) 

are reciprocal. 

Introducing the notation 

(x 1 ,x 2 ,x 3 )=(u, v, w) (y 1 ,y 2 ,y 3 )=(x, y, z) (1.2.23) 

where the x ′ s denote the generalized coordinates and the y ′ s denote the rectangular Cartesian coordinates, 

the above equations can be expressed in a more concise form with the index notation. For example, if 

x i = x i (x, y, z) =x i (y 1 ,y 2 ,y 3 ), and y i = y i (u, v, w) =y i (x 1 ,x 2 ,x 3 ), i =1, 2, 3 (1.2.24) 

then the reciprocal basis vectors can be represented 

E i =gradx i , i =1, 2, 3 (1.2.25) 

and 

Ei = ∂r 

, i =1, 2, 3. (1.2.26) 

∂xi We now show that these basis vectors are reciprocal. Observe that r = r(x 1 ,x2 ,x3 )with 

and consequently 

dr = ∂r 

dxm 

∂xm (1.2.27) 

dx i =gradx i · dr =gradx i · ∂r 

∂xm dxm 

= E i 

· Em dx m = δ i m dx m , i =1, 2, 3 (1.2.28) 

Comparing like terms in this last equation establishes the result that 

which demonstrates that the basis vectors are reciprocal. 

E i · Em = δ i m , i,m =1, 2, 3 (1.2.29) 

39

40 

Scalars, Vectors and Tensors 

Tensors are quantities which obey certain transformation laws. That is, scalars, vectors, matrices 

and higher order arrays can be thought of as components of a tensor quantity. We shall be interested in 

finding how these components are represented in various coordinate systems. We desire knowledge of these 

transformation laws in order that we can represent various physical laws in a form which is independent of 

the coordinate system chosen. Before defining different types of tensors let us examine what we mean by a 

coordinate transformation. 

Coordinate transformations of the type found in equations (1.2.8) and (1.2.9) can be generalized to 

higher dimensions. Let xi ,i = 1, 2,...,N denote N variables. These quantities can be thought of as 

representing a variable point (x1 ,x2 ,...,xN )inanNdimensional space VN . Another set of N quantities, 

call them barred quantities, xi ,i=1, 2,...,N, can be used to represent a variable point (x1 , x2 ,...,xN )in 

an N dimensional space V N . When the x ′ s are related to the x ′ s by equations of the form 

x i = x i (x 1 , x 2 ,...,x N ), i =1, 2,...,N (1.2.30) 

then a transformation is said to exist between the coordinates x i and x i ,i=1, 2,...,N. Whenever the 

relations (1.2.30) are functionally independent, single valued and possess partial derivatives such that the 

Jacobian of the transformation 

 

x 

1 2 N 

x ,x ,...,x 

J = J 

x x1 , x2 ,...,xN 

∂x 

 

 

= 

 

 

1 

∂x1 ∂x 1 

∂x2 ∂x ... 1 

∂xN 

 

 

 

. . ... . 

 

(1.2.31) 

 

 

∂x N 

∂x 1 

is different from zero, then there exists an inverse transformation 

∂x N 

∂x2 ∂x ... N 

∂xN x i = x i (x 1 ,x 2 ,...,x N ), i =1, 2,...,N. (1.2.32) 

For brevity the transformation equations (1.2.30) and (1.2.32) are sometimes expressed by the notation 

x i = x i (x), i=1,...,N and x i = x i (x), i=1,...,N. (1.2.33) 

Consider a sequence of transformations from x to ¯x andthenfrom¯xto ¯x coordinates. For simplicity 

let ¯x = y and ¯x = z. If we denote by T1,T2 and T3 the transformations 

T1 : y i = y i (x 1 ,...,x N ) i =1,...,N or T1x = y 

T2 : z i = z i (y 1 ,...,y N ) i =1,...,N or T2y = z 

Then the transformation T3 obtained by substituting T1 into T2 is called the product of two successive 

transformations and is written 

T3 : z i = z i (y 1 (x 1 ,...,x N ),...,y N (x 1 ,...,x N )) i =1,...,N or T3x = T2T1x = z. 

This product transformation is denoted symbolically by T3 = T2T1. 

The Jacobian of the product transformation is equal to the product of Jacobians associated with the 

product transformation and J3 = J2J1.

Transformations Form a Group 

AgroupGis a nonempty set of elements together with a law, for combining the elements. The combined 

elements are denoted by a product. Thus, if a and b are elements in G then no matter how you define the 

law for combining elements, the product combination is denoted ab. ThesetGandcombining law forms a 

group if the following properties are satisfied: 

(i) For all a, b ∈ G, thenab ∈ G. This is called the closure property. 

(ii) There exists an identity element I such that for all a ∈ G we have Ia = aI = a. 

(iii) There exists an inverse element. That is, for all a ∈ G there exists an inverse element a−1 such that 

aa −1 = a −1 a = I. 

(iv) The associative law holds under the combining law and a(bc) =(ab)c for all a, b, c ∈ G. 

For example, the set of elements G = {1, −1,i,−i}, wherei2 = −1 together with the combining law of 

ordinary multiplication, forms a group. This can be seen from the multiplication table. 

× 1 -1 i -i 

1 1 -1 i -i 

-1 -1 1 -i i 

-i -i i 1 -1 

i i -i -1 1 

The set of all coordinate transformations of the form found in equation (1.2.30), with Jacobian different 

from zero, forms a group because: 

(i) The product transformation, which consists of two successive transformations, belongs to the set of 

transformations. (closure) 

(ii) The identity transformation exists in the special case that x and x are the same coordinates. 

(iii) The inverse transformation exists because the Jacobian of each individual transformation is different 

from zero. 

(iv) The associative law is satisfied in that the transformations satisfy the property T3(T2T1) =(T3T2)T1. 

When the given transformation equations contain a parameter the combining law is often times represented 

as a product of symbolic operators. For example, we denote by Tα a transformation of coordinates 

having a parameter α. The inverse transformation can be denoted by T −1 

α and one can write Tαx = x or 

x = T −1 

α x. We let Tβ denote the same transformation, but with a parameter β, then the transitive property 

is expressed symbolically by TαTβ = Tγ where the product TαTβ represents the result of performing two 

successive transformations. The first coordinate transformation uses the given transformation equations and 

uses the parameter α in these equations. This transformation is then followed by another coordinate transformation 

using the same set of transformation equations, but this time the parameter value is β. The above 

symbolic product is used to demonstrate that the result of applying two successive transformations produces 

a result which is equivalent to performing a single transformation of coordinates having the parameter value 

γ. Usually some relationship can then be established between the parameter values α, β and γ. 

41

42 

Figure 1.2-2. Cylindrical coordinates. 

In this symbolic notation, we let Tθ denote the identity transformation. That is, using the parameter 

value of θ in the given set of transformation equations produces the identity transformation. The inverse 

transformation can then be expressed in the form of finding the parameter value β such that TαTβ = Tθ. 

Cartesian Coordinates 

At times it is convenient to introduce an orthogonal Cartesian coordinate system having coordinates 

y i , i =1, 2,...,N. This space is denoted EN and represents an N-dimensional Euclidean space. Whenever 

the generalized independent coordinates x i ,i=1,...,N are functions of the y ′ s, and these equations are 

functionally independent, then there exists independent transformation equations 

y i = y i (x 1 ,x 2 ,...,x N ), i =1, 2,...,N, (1.2.34) 

with Jacobian different from zero. Similarly, if there is some other set of generalized coordinates, say a barred 

system x i ,i=1,...,N where the x ′ s are independent functions of the y ′ s, then there will exist another set 

of independent transformation equations 

y i = y i (x 1 , x 2 ,...,x N ), i =1, 2,...,N, (1.2.35) 

with Jacobian different from zero. The transformations found in the equations (1.2.34) and (1.2.35) imply 

that there exists relations between the x ′ s and x ′ s of the form (1.2.30) with inverse transformations of the 

form (1.2.32). It should be remembered that the concepts and ideas developed in this section can be applied 

to a space VN of any finite dimension. Two dimensional surfaces (N = 2) and three dimensional spaces 

(N = 3) will occupy most of our applications. In relativity, one must consider spaces where N =4. 

EXAMPLE 1.2-1. (cylindrical coordinates (r, θ, z)) Consider the transformation 

x = x(r, θ, z) =r cos θ y = y(r, θ, z) =r sin θ z = z(r, θ, z) =z 

from rectangular coordinates (x, y, z) to cylindrical coordinates (r, θ, z), illustrated in the figure 1.2-2. By 

letting 

y 1 = x, y 2 = y, y 3 = z x 1 = r, x 2 = θ, x 3 = z 

the above set of equations are examples of the transformation equations (1.2.8) with u = r, v = θ, w = z as 

the generalized coordinates.

EXAMPLE 1.2.2. (Spherical Coordinates) (ρ, θ, φ) 

Consider the transformation 

x = x(ρ, θ, φ) =ρ sin θ cos φ y = y(ρ, θ, φ) =ρ sin θ sin φ z = z(ρ, θ, φ) =ρ cos θ 

from rectangular coordinates (x, y, z) to spherical coordinates (ρ, θ, φ). By letting 

y 1 = x, y 2 = y, y 3 = z x 1 = ρ, x 2 = θ,x 3 = φ 

the above set of equations has the form found in equation (1.2.8) with u = ρ, v = θ, w = φ the generalized 

coordinates. One could place bars over the x ′ s in this example in order to distinguish these coordinates from 

the x ′ s of the previous example. The spherical coordinates (ρ, θ, φ) are illustrated in the figure 1.2-3. 

Scalar Functions and Invariance 

Figure 1.2-3. Spherical coordinates. 

We are now at a point where we can begin to define what tensor quantities are. The first definition is 

for a scalar invariant or tensor of order zero. 

43

44 

Definition: ( Absolute scalar field) Assume there exists a coordinate 

transformation of the type (1.2.30) with Jacobian J different from zero. Let 

the scalar function 

f = f(x 1 ,x 2 ,...,x N ) (1.2.36) 

be a function of the coordinates xi ,i=1,...,N in a space VN . Whenever 

there exists a function 

f = f(x 1 , x 2 ,...,x N ) (1.2.37) 

which is a function of the coordinates x i ,i=1,...,N such that f = J W f, 

then f is called a tensor of rank or order zero of weight W in the space VN . 

Whenever W = 0, the scalar f is called the component of an absolute scalar 

field and is referred to as an absolute tensor of rank or order zero. 

That is, an absolute scalar field is an invariant object in the space VN with respect to the group of 

coordinate transformations. It has a single component in each coordinate system. For any scalar function 

of the type defined by equation (1.2.36), we can substitute the transformation equations (1.2.30) and obtain 

f = f(x 1 ,...,x N )=f(x 1 (x),...,x N (x)) = f(x 1 ,...,x N ). (1.2.38) 

Vector Transformation, Contravariant Components 

In VN consider a curve C defined by the set of parametric equations 

C : x i = x i (t), i =1,...,N 

where t is a parameter. The tangent vector to the curve C is the vector 

1 

T 

dx dx2 

= , 

dt dt ,...,dxN 

 

. 

dt 

In index notation, which focuses attention on the components, this tangent vector is denoted 

T i = dxi 

, i =1,...,N. 

dt 

For a coordinate transformation of the type defined by equation (1.2.30) with its inverse transformation 

defined by equation (1.2.32), the curve C is represented in the barred space by 

x i = x i (x 1 (t),x 2 (t),...,x N (t)) = x i (t), i =1,...,N, 

with t unchanged. The tangent to the curve in the barred system of coordinates is represented by 

dx i 

dt 

= ∂xi 

∂x j 

dxj , i =1,...,N. (1.2.39) 

dt

Letting T i ,i=1,...,N denote the components of this tangent vector in the barred system of coordinates, 

the equation (1.2.39) can then be expressed in the form 

T i = ∂xi 

∂x j T j , i,j =1,...,N. (1.2.40) 

This equation is said to define the transformation law associated with an absolute contravariant tensor of 

rank or order one. In the case N = 3 the matrix form of this transformation is represented 

⎛ 1 

T 

⎝ T 2 

T 3 

⎞ ⎛ 

∂x 

⎠ ⎜ 

= ⎝ 

1 

∂x1 ∂x 1 

∂x2 ∂x 1 

∂x3 ∂x 2 

∂x1 ∂x 2 

∂x2 ∂x 2 

∂x3 ⎞ ⎛ 

1 T 

⎟ 

⎠ ⎝ T 2 

T 3 

⎞ 

⎠ (1.2.41) 

A more general definition is 

∂x 3 

∂x 1 

∂x 3 

∂x 2 

∂x 3 

∂x 3 

Definition: (Contravariant tensor) Whenever N quantities A i in 

acoordinatesystem(x 1 ,...,x N ) are related to N quantities A i in a 

coordinate system (x1 ,...,xN ) such that the Jacobian J is different 

from zero, then if the transformation law 

A i = J 

W ∂xi 

Aj 

∂xj is satisfied, these quantities are called the components of a relative tensor 

of rank or order one with weight W . Whenever W = 0 these quantities 

are called the components of an absolute tensor of rank or order one. 

We see that the above transformation law satisfies the group properties. 

EXAMPLE 1.2-3. (Transitive Property of Contravariant Transformation) 

Show that successive contravariant transformations is also a contravariant transformation. 

Solution: Consider the transformation of a vector from an unbarred to a barred system of coordinates. A 

vector or absolute tensor of rank one A i = A i (x), i=1,...,N will transform like the equation (1.2.40) and 

A i (x) = ∂xi 

∂x j Aj (x). (1.2.42) 

Another transformation from x → x coordinates will produce the components 

A i 

(x) = ∂xi 

∂x j Aj (x) (1.2.43) 

Here we have used the notation A j (x) to emphasize the dependence of the components A j upon the x 

coordinates. Changing indices and substituting equation (1.2.42) into (1.2.43) we find 

A i 

(x) = ∂xi 

∂xj ∂xj ∂xm Am (x). (1.2.44) 

45

46 

From the fact that 

the equation (1.2.44) simplifies to 

∂x i 

∂x j 

∂x j 

∂x 

∂xi 

= , 

m ∂xm A i 

(x) = ∂xi 

∂x m Am (x) (1.2.45) 

and hence this transformation is also contravariant. We express this by saying that the above are transitive 

with respect to the group of coordinate transformations. 

Note that from the chain rule one can write 

∂x m 

∂x j 

∂xj ∂xm 

= 

∂xn ∂x1 Do not make the mistake of writing 

∂x m 

∂x 2 

∂x1 ∂xm 

+ 

∂xn ∂x2 ∂x2 ∂xm 

= 

∂xn ∂xn or 

∂x2 ∂xm 

+ 

∂xn ∂x3 ∂x m 

∂x 3 

∂x3 ∂xm 

= 

∂xn ∂xn = δm n . 

∂x3 ∂xm 

= 

∂xn ∂xn as these expressions are incorrect. Note that there are no summations in these terms, whereas there is a 

summation index in the representation of the chain rule. 

Vector Transformation, Covariant Components 

Consider a scalar invariant A(x) =A(x) which is a shorthand notation for the equation 

A(x 1 ,x 2 ,...,x n )=A(x 1 , x 2 ,...,x n ) 

involving the coordinate transformation of equation (1.2.30). By the chain rule we differentiate this invariant 

and find that the components of the gradient must satisfy 

Let 

Aj = ∂A 

∂xj and Ai = ∂A 

, i ∂x 

then equation (1.2.46) can be expressed as the transformation law 

∂A ∂A 

i = 

∂x ∂xj ∂xj i . (1.2.46) 

∂x 

Ai = Aj 

∂xj i . (1.2.47) 

∂x 

This is the transformation law for an absolute covariant tensor of rank or order one. A more general definition 

is

Definition: (Covariant tensor) Whenever N quantities Ai in a 

coordinate system (x1 ,...,xN ) are related to N quantities Ai in a coordinate 

system (x1 ,...,xN ), with Jacobian J different from zero, such 

that the transformation law 

Ai = J 

W ∂xj 

i 

Aj 

(1.2.48) 

∂x 

is satisfied, then these quantities are called the components of a relative 

covariant tensor of rank or order one having a weight of W . Whenever 

W = 0, these quantities are called the components of an absolute 

covariant tensor of rank or order one. 

Again we note that the above transformation satisfies the group properties. Absolute tensors of rank or 

order one are referred to as vectors while absolute tensors of rank or order zero are referred to as scalars. 

EXAMPLE 1.2-4. (Transitive Property of Covariant Transformation) 

Consider a sequence of transformation laws of the type defined by the equation (1.2.47) 

x → x 

x → x 

Ai(x) =Aj(x) ∂xj 

∂x i 

Ak(x) =Am(x) ∂xm 

∂x k 

We can therefore express the transformation of the components associated with the coordinate transformation 

x → x and 

 

Ak(x) = Aj(x) ∂xj 

∂xm m ∂x ∂xj 

= Aj(x) , 

k k 

∂x ∂x 

which demonstrates the transitive property of a covariant transformation. 

Higher Order Tensors 

We have shown that first order tensors are quantities which obey certain transformation laws. Higher 

order tensors are defined in a similar manner and also satisfy the group properties. We assume that we are 

given transformations of the type illustrated in equations (1.2.30) and (1.2.32) which are single valued and 

continuous with Jacobian J different from zero. Further, the quantities x i and x i ,i=1,...,n represent the 

coordinates in any two coordinate systems. The following transformation laws define second order and third 

order tensors. 

47

48 

Definition: (Second order contravariant tensor) Whenever N-squared quantities A ij 

in a coordinate system (x 1 ,...,x N ) are related to N-squared quantities A mn in a coordinate 

system (x 1 ,...,x N ) such that the transformation law 

A mn (x) =A ij W ∂xm 

(x)J 

∂xi ∂xn ∂xj (1.2.49) 

is satisfied, then these quantities are called components of a relative contravariant tensor of 

rank or order two with weight W . Whenever W = 0 these quantities are called the components 

of an absolute contravariant tensor of rank or order two. 

Definition: (Second order covariant tensor) Whenever N-squared quantities 

Aij in a coordinate system (x 1 ,...,x N ) are related to N-squared quantities Amn 

in a coordinate system (x 1 ,...,x N ) such that the transformation law 

W ∂xi 

Amn(x) =Aij(x)J 

∂xm ∂xj ∂xn (1.2.50) 

is satisfied, then these quantities are called components of a relative covariant tensor 

of rank or order two with weight W . Whenever W = 0 these quantities are called 

the components of an absolute covariant tensor of rank or order two. 

Definition: (Second order mixed tensor) Whenever N-squared quantities 

A i j in a coordinate system (x1 ,...,x N ) are related to N-squared quantities A m 

n in 

acoordinatesystem(x 1 ,...,x N ) such that the transformation law 

A m 

n (x) =A i W ∂xm 

j(x)J 

∂xi ∂xj ∂xn (1.2.51) 

is satisfied, then these quantities are called components of a relative mixed tensor of 

rank or order two with weight W . Whenever W = 0 these quantities are called the 

components of an absolute mixed tensor of rank or order two. It is contravariant 

of order one and covariant of order one. 

Higher order tensors are defined in a similar manner. For example, if we can find N-cubed quantities 

Am np such that 

A i 

jk(x) =A γ W ∂xi 

αβ (x)J 

∂xγ ∂xα ∂xj ∂xβ ∂xk (1.2.52) 

then this is a relative mixed tensor of order three with weight W . 

covariant of order two. 

It is contravariant of order one and

General Definition 

In general a mixed tensor of rank or order (m + n) 

T i1i2...im 

j1j2...jn 

is contravariant of order m and covariant of order n if it obeys the transformation law 

T i1i2...im 

j1j2...jn = 

 

J 

 

x 

W x 

T a1a2...am 

b1b2...bn 

∂xi1 ∂xa1 ∂xi2 ∂xim ∂xb1 

··· · a2 am ∂x ∂x ∂xj1 ∂xb2 ···∂xbn 

j2 ∂x ∂xjn (1.2.53) 

(1.2.54) 

where 

 

x 

 

 

J = 

∂x 

 

x ∂x 

= ∂(x1 ,x2 ,...,xN ) 

∂(x1 , x2 ,...,xN ) 

is the Jacobian of the transformation. When W = 0 the tensor is called an absolute tensor, otherwise it is 

called a relative tensor of weight W. 

Here superscripts are used to denote contravariant components and subscripts are used to denote covariant 

components. Thus, if we are given the tensor components in one coordinate system, then the components 

in any other coordinate system are determined by the transformation law of equation (1.2.54). Throughout 

the remainder of this text one should treat all tensors as absolute tensors unless specified otherwise. 

Dyads and Polyads 

Note that vectors can be represented in bold face type with the notation 

A = AiE i 

This notation can also be generalized to tensor quantities. Higher order tensors can also be denoted by bold 

face type. For example the tensor components Tij and Bijk can be represented in terms of the basis vectors 

E i ,i=1,...,N by using a notation which is similar to that for the representation of vectors. For example, 

T = TijE i E j 

B = BijkE i E j E k . 

Here T denotes a tensor with components Tij and B denotes a tensor with components Bijk. The quantities 

EiEj are called unit dyads and EiEjEk are called unit triads. There is no multiplication sign between the 

basis vectors. This notation is called a polyad notation. A further generalization of this notation is the 

representation of an arbitrary tensor using the basis and reciprocal basis vectors in bold type. For example, 

a mixed tensor would have the polyadic representation 

T = T ij...k 

lm...n EiEj ...EkE l E m ...E n . 

A dyadic is formed by the outer or direct product of two vectors. For example, the outer product of the 

vectors 

a = a1E 1 + a2E 2 + a3E 3 

and b = b1E 1 + b2E 2 + b3E 3 

49

50 

gives the dyad 

In general, a dyad can be represented 

ab =a1b1E 1 E 1 + a1b2E 1 E 2 + a1b3E 1 E 3 

a2b1E 2 E 1 + a2b2E 2 E 2 + a2b3E 2 E 3 

a3b1E 3 E 1 + a3b2E 3 E 2 + a3b3E 3 E 3 . 

A = AijE i E j 

i, j =1,...,N 

where the summation convention is in effect for the repeated indices. The coefficients Aij are called the 

coefficients of the dyad. When the coefficients are written as an N × N array it is called a matrix. Every 

second order tensor can be written as a linear combination of dyads. The dyads form a basis for the second 

order tensors. As the example above illustrates, the nine dyads {E1E1 , E1E2 ,...,E3E3 }, associated with 

the outer products of three dimensional base vectors, constitute a basis for the second order tensor A = ab 

having the components Aij = aibj with i, j =1, 2, 3. Similarly, a triad has the form 

T = TijkE i E j E k Sum on repeated indices 

where i, j, k have the range 1, 2,...,N.The set of outer or direct products { EiEjEk },withi, j, k =1,...,N 

constitutes a basis for all third order tensors. Tensor components with mixed suffixes like Ci jk are associated 

with triad basis of the form 

C = C i jk EiE j E k 

where i, j, k have the range 1, 2,...N.Dyads are associated with the outer product of two vectors, while triads, 

tetrads,... are associated with higher-order outer products. These higher-order outer or direct products are 

referred to as polyads. 

The polyad notation is a generalization of the vector notation. The subject of how polyad components 

transform between coordinate systems is the subject of tensor calculus. 

In Cartesian coordinates we have Ei = Ei = ei and a dyadic with components called dyads is written 

A = Aij ei ej or 

A =A11 e1 e1 + A12 e1 e2 + A13 e1 e3 

A21 e2 e1 + A22 e2 e2 + A23 e2 e3 

A31 e3 e1 + A32 e3 e2 + A33 e3 e3 

where the terms ei ej are called unit dyads. Note that a dyadic has nine components as compared with a 

vector which has only three components. The conjugate dyadic Ac is defined by a transposition of the unit 

vectors in A, toobtain 

Ac =A11 e1 e1 + A12 e2 e1 + A13 e3 e1 

A21 e1 e2 + A22 e2 e2 + A23 e3 e2 

A31 e1 e3 + A32 e2 e3 + A33 e3 e3

If a dyadic equals its conjugate A = Ac, thenAij = Aji and the dyadic is called symmetric. If a dyadic 

equals the negative of its conjugate A = −Ac, thenAij = −Aji and the dyadic is called skew-symmetric. A 

special dyadic called the identical dyadic or idemfactor is defined by 

J = e1 e1 + e2 e2 + e3 e3. 

This dyadic has the property that pre or post dot product multiplication of J with a vector V produces the 

same vector V.For example, 

V · J =(V1e1 + V2 e2 + V3 e3) · J 

= V1 e1 · e1 e1 + V2 e2 · e2 e2 + V3 e3 · e3 e3 = V 

and J · V = J · (V1 e1 + V2 e2 + V3 e3) 

= V1 e1 e1 · e1 + V2 e2 e2 · e2 + V3 e3 e3 · e3 = V 

A dyadic operation often used in physics and chemistry is the double dot product A : B where A and 

B are both dyadics. Here both dyadics are expanded using the distributive law of multiplication, and then 

each unit dyad pair ei ej : em en are combined according to the rule 

ei ej : em en =(ei · em)( ej · en). 

For example, if A = Aij ei ej and B = Bij ei ej, then the double dot product A : B is calculated as follows. 

A : B =(Aij ei ej) :(Bmn em en) =AijBmn( ei ej : em en) =AijBmn( ei · em)( ej · en) 

= AijBmnδimδjn = AmjBmj 

= A11B11 + A12B12 + A13B13 

+ A21B21 + A22B22 + A23B23 

+ A31B31 + A32B32 + A33B33 

When operating with dyads, triads and polyads, there is a definite order to the way vectors and polyad 

components are represented. For example, for A = Ai ei and B = Bi ei vectors with outer product 

A B = AmBn em en = φ 

there is produced the dyadic φ with components AmBn. In comparison, the outer product 

B A = BmAn em en = ψ 

produces the dyadic ψ with components BmAn. That is 

φ = A B =A1B1 e1 e1 + A1B2 e1 e2 + A1B3 e1 e3 

A2B1 e2 e1 + A2B2 e2 e2 + A2B3 e2 e3 

A3B1 e3 e1 + A3B2 e3 e2 + A3B3 e3 e3 

and ψ = B A =B1A1 e1 e1 + B1A2 e1 e2 + B1A3 e1 e3 

B2A1 e2 e1 + B2A2 e2 e2 + B2A3 e2 e3 

B3A1 e3 e1 + B3A2 e3 e2 + B3A3 e3 e3 

are different dyadics. 

The scalar dot product of a dyad with a vector C is defined for both pre and post multiplication as 

φ · C = A B · C = A( B · C) 

C · φ = C · A B =( C · A) B 

These products are, in general, not equal. 

51

52 

Operations Using Tensors 

The following are some important tensor operations which are used to derive special equations and to 

prove various identities. 

Addition and Subtraction 

Tensors of the same type and weight can be added or subtracted. For example, two third order mixed 

tensors, when added, produce another third order mixed tensor. Let A i jk and Bi jk 

mixed tensors. Their sum is denoted 

C i jk = A i jk + B i jk. 

denote two third order 

That is, like components are added. The sum is also a mixed tensor as we now verify. By hypothesis A i jk 

and B i jk 

are third order mixed tensors and hence must obey the transformation laws 

A i 

jk = Am ∂x 

np 

i 

∂xm ∂xn ∂xj ∂xp ∂xk B i 

jk = Bm ∂x 

np 

i 

∂xm ∂xn ∂xj ∂xp . k ∂x 

We let C i 

jk = Aijk 

+ Bijk 

denote the sum in the transformed coordinates. Then the addition of the above 

transformation equations produces 

C i 

jk = 

 

A i 

 

jk + Bijk 

= A m np + Bm ∂x 

np 

i 

∂xm ∂xn ∂xj ∂xp ∂xk = Cm ∂x 

np 

i 


Consequently, the sum transforms as a mixed third order tensor. 

Multiplication (Outer Product) 

The product of two tensors is also a tensor. The rank or order of the resulting tensor is the sum of 

the ranks of the tensors occurring in the multiplication. As an example, let Ai jk 

tensor and let Bl m 

order tensor 

denote a mixed third order 

denote a mixed second order tensor. The outer product of these two tensors is the fifth 

C il 

jkm = AijkBl m , i,j,k,l,m=1, 2,...,N. 

Here all indices are free indices as i, j, k, l, m take on any of the integer values 1, 2,...,N. Let A i 

jk and Blm 

denote the components of the given tensors in the barred system of coordinates. We define C il 

jkm as the 

outer product of these components. Observe that Cil jkm is a tensor for by hypothesis Ai jk and Bl m are tensors 

and hence obey the transformation laws 

A α 

βγ = Ai ∂x 

jk 

α 

∂xi ∂xj ∂xβ ∂xk ∂xγ B δ 

ɛ = B l m 

The outer product of these components produces 

which demonstrates that C il 

jkm 

analyzed in a similar way. 

∂xδ ∂xl ∂xm ɛ . 

∂x 

C αδ 

βγɛ = AαβγBδɛ 

= AijkBl ∂x 

m 

α 

∂xi ∂xj ∂xβ ∂xk ∂xγ ∂xδ ∂xl ∂xm ∂xɛ (1.2.55) 

= C il ∂x 

jkm 

α 

∂xi ∂xj ∂xβ ∂xk ∂xγ ∂xδ ∂xl ∂xm ∂xɛ (1.2.56) 

transforms as a mixed fifth order absolute tensor. Other outer products are

Contraction 

The operation of contraction on any mixed tensor of rank m is performed when an upper index is 

set equal to a lower index and the summation convention is invoked. When the summation is performed 

over the repeated indices the resulting quantity is also a tensor of rank or order (m − 2). For example, let 

Ai jk , i,j,k =1, 2,...,N denote a mixed tensor and perform a contraction by setting j equal to i. We obtain 

A i ik = A11k + A22k + ···+ ANNk where k is a free index. To show that Ak is a tensor, we let A i 

ik = Ak denote the contraction on the 

transformed components of Ai jk . By hypothesis Aijk is a mixed tensor and hence the components must 

= Ak 

(1.2.57) 

satisfy the transformation law 

A i 

jk = Am ∂x 

np 

i 


Now execute a contraction by setting j equal to i and perform a summation over the repeated index. We 

find 

A i 

ik = Ak = A m ∂x 

np 

i 

∂xm ∂xn ∂xi ∂xp ∂xk = Am ∂x 

np 

n 

∂xm ∂xp ∂xk = A m npδ n ∂x 

m 

p 

∂xk = An ∂x 

np 

p ∂x 

= Ap k ∂x p 

(1.2.58) 

. k ∂x 

Hence, the contraction produces a tensor of rank two less than the original tensor. Contractions on other 

mixed tensors can be analyzed in a similar manner. 

New tensors can be constructed from old tensors by performing a contraction on an upper and lower 

index. This process can be repeated as long as there is an upper and lower index upon which to perform the 

contraction. Each time a contraction is performed the rank of the resulting tensor is two less than the rank 

of the original tensor. 

Multiplication (Inner Product) 

The inner product of two tensors is obtained by: 

(i) first taking the outer product of the given tensors and 

(ii) performing a contraction on two of the indices. 

EXAMPLE 1.2-5. (Inner product) 

Let Ai and Bj denote the components of two first order tensors (vectors). The outer product of these 

tensors is 

The inner product of these tensors is the scalar 

C i j = A i Bj, i,j=1, 2,...,N. 

C = A i Bi = A 1 B1 + A 2 B2 + ···+ A N BN . 

Note that in some situations the inner product is performed by employing only subscript indices. For 

example, the above inner product is sometimes expressed as 

C = AiBi = A1B1 + A2B2 + ···AN BN . 

This notation is discussed later when Cartesian tensors are considered. 

53

54 

Quotient Law 

Assume Bqs r and Cs p are arbitrary absolute tensors. Further assume we have a quantity A(ijk) which 

we think might be a third order mixed tensor Ai jk . By showing that the equation 

A r qp Bqs 

r = Cs p 

is satisfied, then it follows that Ar qp must be a tensor. This is an example of the quotient law. Obviously, 

this result can be generalized to apply to tensors of any order or rank. To prove the above assertion we shall 

show from the above equation that A i jk is a tensor. Let xi and x i denote a barred and unbarred system of 

coordinates which are related by transformations of the form defined by equation (1.2.30). In the barred 

system, we assume that 

(1.2.59) 

A r 

qpBqs r = Csp 

where by hypothesis B ij 

k and Cl m are arbitrary absolute tensors and therefore must satisfy the transformation 

equations 

We substitute for B qs 

r 

B qs 

r 

C s 

∂x 

= Bij 

k 

q 

p = Cl m 

∂xi ∂xs ∂xj ∂xk ∂xr ∂xs ∂xl ∂xm p . 

∂x 

and Csp 

in the equation (1.2.59) and obtain the equation 

A r 

 

qp B ij ∂x 

k 

q 

∂xi ∂xs ∂xj ∂xk ∂xr 

= C l ∂x 

m 

s 

∂xl ∂xm ∂xp 

= A r qmBql ∂x 

r 

s 

∂xl ∂xm p . 

∂x 

Since the summation indices are dummy indices they can be replaced by other symbols. We change l to j, 

q to i and r to k and write the above equation as 

∂xs ∂xj 

A r ∂x 

qp 

q 

∂xi ∂xk ∂xr − Ak ∂x 

im 

m 

∂xp 

B ij 

k =0. 

Use inner multiplication by ∂xn 

s ∂x and simplify this equation to the form 

Because B in 

k 

δ n j 

 

A r 

qp 

 

A r 

qp 

∂xq ∂xi ∂xq ∂xi ∂xk ∂xr − Akim ∂xk ∂xr − Akim ∂xm ∂xp ∂x m 

∂x p 

 

B ij 

k =0 or 

 

B in 

k =0. 

is an arbitrary tensor, the quantity inside the brackets is zero and therefore 

A r 

qp 

∂xk ∂xr − Ak ∂x 

im 

m 

p =0. 

∂x 

This equation is simplified by inner multiplication by ∂xi 

∂xj ∂x l 

∂xk to obtain 

∂xq ∂xi δ q 

j δl rArqp − Ak ∂x 

im 

m 

∂xp ∂xi ∂xj A l 

jp = A k im 

∂x m 

∂x p 

∂x i 

∂x j 

∂x l 

=0 or 

∂xk ∂x l 

∂x k 

which is the transformation law for a third order mixed tensor.

EXERCISE 1.2 

◮ 1. Consider the transformation equations representing a rotation of axes through an angle α. 

Tα : 

1 x 1 2 = x cos α − x sin α 

x2 = x1 sin α + x2 cos α 

Treat α as a parameter and show this set of transformations constitutes a group by finding the value of α 

which: 

(i) gives the identity transformation. 

(ii) gives the inverse transformation. 

(iii) show the transformation is transitive in that a transformation with α = θ1 followed by a transformation 

with α = θ2 is equivalent to the transformation using α = θ1 + θ2. 

◮ 2. Show the transformation 

1 1 x = αx 

Tα : 

x2 = 1 

αx2 forms a group with α as a parameter. Find the value of α such that: 

(i) the identity transformation exists. 

(ii) the inverse transformation exists. 

(iii) the transitive property is satisfied. 

◮ 3. Show the given transformation forms a group with parameter α. 

Tα : 

x 1 = x 1 

1−αx 1 

x 2 = x2 

1−αx 1 

◮ 4. Consider the Lorentz transformation from relativity theory having the velocity parameter V, c is the 

speed of light and x4 = t is time. 

⎧ 

TV : 

⎪⎨ x2 = x2 ⎪⎩ 

x1 = x1−Vx 4 

 

V 1− 2 

c2 x 3 = x 3 

x4 = x4− Vx1 

c2 

V 1− 2 

c2 Show this set of transformations constitutes a group, by establishing: 

(i) V = 0 gives the identity transformation T0. 

(ii) TV2 · TV1 = T0 requires that V2 = −V1. 

(iii) TV2 · TV1 = TV3 requires that 

V3 = V1 + V2 

1+ V1V2 

c2 . 

◮ 5. For ( E1, E2, E3) an arbitrary independent basis, (a) Verify that 

E 1 = 1 

V E2 × E3, 

E 2 = 1 

V E3 × E1, 

E 3 = 1 

V E1 × E2 

is a reciprocal basis, where V = E1 · ( E2 × E3) (b) Show that E j = g ij Ei. 

55

56 

Figure 1.2-4. Cylindrical coordinates (r, β, z). 

◮ 6. For the cylindrical coordinates (r, β, z) illustrated in the figure 1.2-4. 

(a) Write out the transformation equations from rectangular (x, y, z) coordinates to cylindrical (r, β, z) 

coordinates. Also write out the inverse transformation. 

(b) Determine the following basis vectors in cylindrical coordinates and represent your results in terms of 

cylindrical coordinates. 

(i) The tangential basis E1, E2, E3. (ii)The normal basis E1 , E2 , E3 . (iii) êr, êβ, êz 

where êr, êβ, êz are normalized vectors in the directions of the tangential basis. 

(c) A vector A = Ax e1 + Ay e2 + Az e3 can be represented in any of the forms: 

A = A 1 E1 + A 2 E2 + A 3 E3 

A = A1 E 1 + A2 E 2 + A3 E 3 

A = Arêr + Aβêβ + Azêz 

depending upon the basis vectors selected . In terms of the components Ax,Ay,Az 

(i) Solve for the contravariant components A1 ,A2 ,A3 . 

(ii) Solve for the covariant components A1,A2,A3. 

(iii) Solve for the components Ar,Aβ,Az. Express all results in cylindrical coordinates. (Note the 

components Ar,Aβ,Az are referred to as physical components. Physical components are considered in 

more detail in a later section.)

Figure 1.2-5. Spherical coordinates (ρ, α, β). 

◮ 7. For the spherical coordinates (ρ, α, β) illustrated in the figure 1.2-5. 

(a) Write out the transformation equations from rectangular (x, y, z) coordinates to spherical (ρ, α, β) coordinates. 

Also write out the equations which describe the inverse transformation. 

(b) Determine the following basis vectors in spherical coordinates 

(i) The tangential basis E1, E2, E3. 

(ii) The normal basis E1 , E2 , E3 . 

(iii) êρ, êα, êβ which are normalized vectors in the directions of the tangential basis. Express all results 

in terms of spherical coordinates. 

(c) A vector A = Ax e1 + Ay e2 + Az e3 can be represented in any of the forms: 

A = A 1 E1 + A 2 E2 + A 3 E3 

A = A1 E 1 + A2 E 2 + A3 E 3 

A = Aρêρ + Aαêα + Aβêβ 

depending upon the basis vectors selected . Calculate, in terms of the coordinates (ρ, α, β) andthe 

components Ax,Ay,Az 

(i) The contravariant components A1 ,A2 ,A3 . 

(ii) The covariant components A1,A2,A3. 

(iii) The components Aρ,Aα,Aβ which are called physical components. 

◮ 8. Work the problems 6,7 and then let (x1 ,x2 ,x3 )=(r, β, z) denote the coordinates in the cylindrical 

system and let (x1 , x2 , x3 )=(ρ, α, β) denote the coordinates in the spherical system. 

(a) Write the transformation equations x → x from cylindrical to spherical coordinates. Also find the 

inverse transformations. ( Hint: See the figures 1.2-4 and 1.2-5.) 

(b) Use the results from part (a) and the results from problems 6,7 to verify that 

∂x 

Ai = Aj 

j 

∂xi for i =1, 2, 3. 

(i.e. Substitute Aj from problem 6 to get Āi given in problem 7.) 

57

58 

(c) Use the results from part (a) and the results from problems 6,7 to verify that 

A i j ∂xi 

= A 

∂xj for i =1, 2, 3. 

(i.e. Substitute Aj from problem 6 to get Āi given by problem 7.) 

◮ 9. Pick two arbitrary noncolinear vectors in the x, y plane, say 

V1 =5e1 + e2 and V2 = e1 +5e2 

and let V3 = e3 be a unit vector perpendicular to both V1 and V2. The vectors V1 and V2 can be thought of 

as defining an oblique coordinate system, as illustrated in the figure 1.2-6. 

(a) Find the reciprocal basis ( V 1 , V 2 , V 3 ). 

(b) Let 

r = x e1 + y e2 + z e3 = α V1 + β V2 + γ V3 

and show that 

(c) Show 

α = 5x y 

− 

24 24 

β = − x 5y 

+ 

24 24 

γ = z 

x =5α + β 

y = α +5β 

z = γ 

(d) For γ = γ0 constant, show the coordinate lines are described by α = constant 

and sketch some of these coordinate lines. (See figure 1.2-6.) 

and β = constant, 

(e) Find the metrics gij and conjugate metrices gij associated with the (α, β, γ) space. 

Figure 1.2-6. Oblique coordinates.

◮ 10. Consider the transformation equations 

substituted into the position vector 

Define the basis vectors 

with the reciprocal basis 

where 

E 1 = 1 

V E2 × E3, 

x = x(u, v, w) 

y = y(u, v, w) 

z = z(u, v, w) 

r = x e1 + y e2 + z e3. 

( E1, E2, E3) = 

Let v = E1 · ( E2 × E3 )andshowthatv · V =1. 

◮ 11. Given the coordinate transformation 

 

∂r ∂r ∂r 

, , 

∂u ∂v ∂w 

E 2 = 1 

V E3 × E1, 

V = E1 · ( E2 × E3). 

x = −u − 2v y = −u − v z = z 

(a) Find and illustrate graphically some of the coordinate curves. 

(b) For r = r(u, v, z) a position vector, define the basis vectors 

E1 = ∂r 

∂u , 

E2 = ∂r 

∂v , 

E 3 = 1 

V E1 × E2. 

E3 = ∂r 

∂z . 

Calculate these vectors and then calculate the reciprocal basis E1 , E2 , E3 . 

(c) With respect to the basis vectors in (b) find the contravariant components Ai associated with the vector 

A = α1 e1 + α2 e2 + α3 e3 

where (α1,α2,α3) areconstants. 

(d) Find the covariant components Ai associated with the vector A given in part (c). 

(e) Calculate the metric tensor gij and conjugate metric tensor gij . 

(f) From the results (e), verify that gijg jk = δ k i 

(g) Use the results from (c)(d) and (e) to verify that Ai = gikA k 

(h) Use the results from (c)(d) and (e) to verify that A i = g ik Ak 

(i) Find the projection of the vector A on unit vectors in the directions E1, E2, E3. 

(j) Find the projection of the vector A on unit vectors the directions E1 , E2 , E3 . 

59

60 

◮ 12. For r = y i ei where y i = y i (x 1 ,x 2 ,x 3 ), i =1, 2, 3 we have by definition 


Ej = ∂r ∂yi 

= 

∂xj gij = Ei · Ej = ∂ym 

∂x i 

∂x j ei. From this relation show that E m = ∂x m 

∂y m 

∂x j , and gij = E i · E j = ∂xi 

∂y m 

◮ 13. Consider the set of all coordinate transformations of the form 

y i = a i j xj + b i 

where a i j and bi are constants and the determinant of a i j 

tions forms a group. 

ej 

∂yj ∂xj , i,j,m=1,...,3 

∂ym is different from zero. Show this set of transforma- 

◮ 14. For αi , βi constants and t a parameter, xi = αi + tβi,i =1, 2, 3 is the parametric representation of 

a straight line. Find the parametric equation of the line which passes through the two points (1, 2, 3) and 

(14, 7, −3). What does the vector dr 

dt represent? 

◮ 15. A surface can be represented using two parameters u, v by introducing the parametric equations 

x i = x i (u, v), i =1, 2, 3, a 

x i = αi u + βi v + γi 

i =1, 2, 3, 

where αi βi and γi are constants. Find the equation of the plane which passes through the points (1, 2, 3), 

(14, 7, −3) and (5, 5, 5). What does this problem have to do with the position vector r(u, v), the vectors 

∂r ∂r 

∂u , ∂v 

and r(0, 0)? Hint: See problem 15. 

◮ 18. Determine the points of intersection of the curve x 1 = t, x 2 =(t) 2 , x 3 =(t) 3 with the plane 

8 x 1 − 5 x 2 + x 3 − 4=0. 

◮ 19. Verify the relations Veijk E k = Ei × Ej and v −1 e ijk Ek = E i × E j where v = E1 · ( E2 × E3 )and 

V = E1 · ( E2 × E3).. 

◮ 20. Let ¯x i and x i , i =1, 2, 3 be related by the linear transformation ¯x i = c i j xj ,wherec i j 

such that the determinant c = det(c i j ) is different from zero. Let γn m denote the cofactor of cm n 

the determinant c. 

(a) Show that c i j γj 

k = γi j cj 

k = δi k . 

(b) Show the inverse transformation can be expressed x i = γ i j ¯x j . 

(c) Show that if A i is a contravariant vector, then its transformed components are Āp = c p qA q . 

(d) Show that if Ai is a covariant vector, then its transformed components are Āi = γ p 

i Ap. 

are constants 

divided by 

◮ 21. Show that the outer product of two contravariant vectors Ai and Bi , i =1, 2, 3 results in a second 

order contravariant tensor. 

◮ 22. Show that for the position vector r = y i (x 1 ,x 2 ,x 3 ) ei the element of arc length squared is 

ds 2 = dr · dr = gijdx i dx j where gij = Ei · Ej = ∂ym 

∂x i 

∂ym . 

∂xj ◮ 23. For Ai jk ,Bm n and C p 

tq absolute tensors, show that if Ai jkBk n = Ci jn then AijkB 

k 

n = C i 

jn. 

◮ 24. Let Aij denote an absolute covariant tensor of order 2. Show that the determinant A = det(Aij) is 

an invariant of weight 2 and (A) is an invariant of weight 1. 

◮ 25. Let Bij denote an absolute contravariant tensor of order 2. Show that the determinant B = det(Bij ) 

is an invariant of weight −2 and √ B is an invariant of weight −1. 

◮ 26. 

(a) Write out the contravariant components of the following vectors 

(i) E1 (ii) E2 (iii) E3 where Ei = ∂r 

∂x i for i =1, 2, 3. 

(b) Write out the covariant components of the following vectors 

(i) E 1 

(ii) E 2 

(ii) E 3 where E i =gradx i , for i =1, 2, 3. 

61

62 

◮ 27. Let Aij and A ij denote absolute second order tensors. Show that λ = AijA ij is a scalar invariant. 

◮ 28. Assume that aij, i, j =1, 2, 3, 4 is a skew-symmetric second order absolute tensor. (a) Show that 

bijk = ∂ajk 

∂x 

∂x 

∂x k 

∂aki ∂aij 

+ + i j 

is a third order tensor. (b) Show bijk is skew-symmetric in all pairs of indices and (c) determine the number 

of independent components this tensor has. 

◮ 29. Show the linear forms A1x + B1y + C1 and A2x + B2y + C2, with respect to the group of rotations 

and translations x = x cos θ − y sin θ + h and y = x sin θ + y cos θ + k, have the forms A1x + B1y + C1 and 

A2x + B2y + C2. Also show that the quantities A1B2 − A2B1 and A1A2 + B1B2 are invariants. 

◮ 30. Show that the curvature of a curve y = f(x) isκ = ± y ′′ (1 + y ′2 ) −3/2 and that this curvature remains 

invariant under the group of rotations given in the problem 1. Hint: Calculate dy dy dx 

dx = dx 

◮ 31. Show that when the equation of a curve is given in the parametric form x = x(t), y= y(t), then 

˙x¨y − ˙y¨x 

the curvature is κ = ± 

(˙x 2 +˙y2 and remains invariant under the change of parameter t = t(t), where 

) 3/2 

˙x = dx 

dt , etc. 

◮ 32. Let A ij 

k 

denote a third order mixed tensor. (a) Show that the contraction Aij 

i is a first order 

which is not a tensor. This shows 

that in general, the process of contraction does not always apply to indices at the same level. 

contravariant tensor. (b) Show that contraction of i and j produces A ii 

k 

◮ 33. Let φ = φ(x 1 ,x 2 ,...,x N ) denote an absolute scalar invariant. (a) Is the quantity ∂φ 

∂x i atensor?(b) 

Is the quantity ∂2 φ 

∂x i ∂x j atensor? 

◮ 34. Consider the second order absolute tensor aij, i,j=1, 2wherea11 =1,a12 =2,a21 =3anda22 =4. 

Find the components of aij under the transformation of coordinates x1 = x1 + x2 and x2 = x1 − x2 . 

◮ 35. Let Ai, Bi denote the components of two covariant absolute tensors of order one. Show that 

Cij = AiBj is an absolute second order covariant tensor. 

◮ 36. Let A i denote the components of an absolute contravariant tensor of order one and let Bi denote the 

components of an absolute covariant tensor of order one, show that C i j = Ai Bj transforms as an absolute 

mixed tensor of order two. 

◮ 37. (a) Show the sum and difference of two tensors of the same kind is also a tensor of this kind. (b) Show 

that the outer product of two tensors is a tensor. Do parts (a) (b) in the special case where one tensor Ai is a relative tensor of weight 4 and the other tensor B j 

k is a relative tensor of weight 3. What is the weight 

of the outer product tensor T ij 

k = AiB j 

k in this special case? 

◮ 38. Let A ij 

km denote the components of a mixed tensor of weight M. Form the contraction Bj m = Aij im 

and determine how Bj m transforms. What is its weight? 

◮ 39. Let A i j 

contraction S = A i i 

denote the components of an absolute mixed tensor of order two. Show that the scalar 

is an invariant. 

dx .

◮ 40. Let A i = A i (x 1 ,x 2 ,...,x N ) denote the components of an absolute contravariant tensor. Form the 

quantity B i j 

∂Ai = ∂xj and determine if Bi j transforms like a tensor. 

◮ 41. Let Ai denote the components of a covariant vector. (a) Show that aij = ∂Ai 

components of a second order tensor. (b) Show that 

∂xj ∂Aj 

− 

∂xi are the 

∂aij ∂ajk ∂aki 

+ + =0. 

∂xk ∂xi ∂xj ◮ 42. Show that xi = KeijkAjBk, withK= 0 and arbitrary, is a general solution of the system of equations 

Aix i =0,Bix i =0,i=1, 2, 3. Give a geometric interpretation of this result in terms of vectors. 

◮ 43. Given the vector A = y e1 + z e2 + x e3 where e1, e2, e3 denote a set of unit basis vectors which 

define a set of orthogonal x, y, z axes. Let E1 =3e1 +4e2, E2 =4e1 +7e2 and E3 = e3 denote a set of 

basis vectors which define a set of u, v, w axes. (a) Find the coordinate transformation between these two 

sets of axes. (b) Find a set of reciprocal vectors E1 , E3 , E3 . (c) Calculate the covariant components of A. 

(d) Calculate the contravariant components of A. 

◮ 44. Let A = Aij ei ej denote a dyadic. Show that 

A : Ac = A11A11 + A12A21 + A13A31 + A21A12 + A22A22 + A23A32 + A31A13 + A32A23 + A23A33 

◮ 45. Let A = Ai ei, B = Bi ei, C = Ci ei, D = Di ei denote vectors and let φ = A B, ψ = C D denote 

dyadics which are the outer products involving the above vectors. Show that the double dot product satisfies 

φ : ψ = A B : C D =( A · C)( B · D) 

◮ 46. Show that if aij is a symmetric tensor in one coordinate system, then it is symmetric in all coordinate 

systems. 

◮ 47. Write the transformation laws for the given tensors. (a) A k ij (b) A ij 

k (c) A ijk 

m 

∂x 

◮ 48. Show that if Ai = Aj 

j 

andunbarredsystems. 

∂x i ,thenAi = Aj ∂xj 

◮ 49. 

(a) Show that under the linear homogeneous transformation 

the quadratic form 

∂x i . Note that this is equivalent to interchanging the bar 

x1 =a 1 1 x1 + a 2 1 x2 

x2 =a 1 2x1 + a 2 2x2 

Q(x1,x2) =g11(x1) 2 +2g12x1x2 + g22(x2) 2 becomes Q(x1, x2) =g 11 (x1) 2 +2g 12 x1x2 + g 22 (x2) 2 

where gij = g11a j 

1ai1 + g12(a i 1aj2 + aj1 

ai2 )+g22a i 2aj2 . 

(b) Show F = g11g22 − (g12) 2 is a relative invariant of weight 2 of the quadratic form Q(x1,x2) with respect 

to the group of linear homogeneous transformations. i.e. Show that F =∆ 2 F where F = g 11g 22 −(g 12) 2 

and ∆ = (a 1 1a 2 2 − a 2 1a 1 2). 

63

64 

◮ 50. Let ai and bi for i =1,...,n denote arbitrary vectors and form the dyadic 

By definition the first scalar invariant of Φ is 

Φ=a1b1 + a2b2 + ···+ anbn. 

φ1 = a1 · b1 + a2 · b2 + ···+ an · bn 

where a dot product operator has been placed between the vectors. The first vector invariant of Φ is defined 

φ = a1 × b1 + a2 × b2 + ···+ an × bn 

where a vector cross product operator has been placed between the vectors. 

(a) Show that the first scalar and vector invariant of 

are respectively 1 and e1 + e3. 

Φ= e1 e2 + e2 e3 + e3 e3 

(b) From the vector f = f1 e1 + f2 e2 + f3 e3 one can form the dyadic ∇f having the matrix components 

⎛ ∂f1 ∂f2 ∂f3 ⎞ 

∂x ∂x ∂x 

∂f1 ∂f2 ∂f3 

∇f = ⎝ 

⎠ 

∂y ∂y ∂y . 

∂f1 

∂z 

∂f2 

∂z 

∂f3 

∂z 

Show that this dyadic has the first scalar and vector invariants given by 

∇·f = ∂f1 ∂f2 ∂f3 

+ + 

∂x ∂y ∂z 

 

∂f3 ∂f2 ∂f1 

∇×f = − e1 + 

∂y ∂z ∂z 

 

∂f3 ∂f2 ∂f1 

− e2 + − e3 

∂x ∂x ∂y 

◮ 51. Let Φ denote the dyadic given in problem 50. The dyadic Φ2 defined by 

Φ2 = 1 

ai × ajbi × bj 

2 

i,j 

is called the Gibbs second dyadic of Φ, where the summation is taken over all permutations of i and j. When 

i = j the dyad vanishes. Note that the permutations i, j and j, i give the same dyad and so occurs twice 

in the final sum. The factor 1/2 removes this doubling. Associated with the Gibbs dyad Φ2 are the scalar 

invariants 

φ2 = 1 

(ai × aj) · (bi × bj) 

2 

Show that the dyad 

has 

φ3 = 1 

6 

i,j 

 

(ai × aj · ak)(bi × bj · bk) 

i,j,k 

Φ=as+ tq+ cu 

the first scalar invariant φ1 = a · s + b · t + c · u 

the first vector invariant φ = a × s + b × t + c × u 

Gibbs second dyad Φ2 = b × ct × u + c × au × s + a × bs × t 

second scalar of Φ φ2 =(b × c) · (t · u)+(c × a) · (u × s)+(a × b) · (s × t) 

third scalar of Φ φ3 =(a × b · c)(s × t · u)

◮ 52. (Spherical Trigonometry) Construct a spherical triangle ABC on the surface of a unit sphere with 

sides and angles less than 180 degrees. Denote by a,b cthe unit vectors from the origin of the sphere to the 

vertices A,B and C. Make the construction such that a·(b×c) is positive with a, b, c forming a right-handed 

system. Let α, β, γ denote the angles between these unit vectors such that 

a · b =cosγ c · a =cosβ b · c =cosα. (1) 

The great circles through the vertices A,B,C then make up the sides of the spherical triangle where side α 

is opposite vertex A, side β is opposite vertex B and side γ is opposite the vertex C. The angles A,B and C 

between the various planes formed by the vectors a, b and c are called the interior dihedral angles of the 

spherical triangle. Note that the cross products 

a × b =sinγ c b× c =sinα a c× a =sinβ b (2) 

define unit vectors a, b and c perpendicular to the planes determined by the unit vectors a, b and c. The 

dot products 

a · b =cosγ b · c =cosα c · a =cosβ (3) 

define the angles α,β and γ which are called the exterior dihedral angles at the vertices A,B and C and are 

such that 

α = π − A β = π − B γ = π − C. (4) 

(a) Using appropriate scaling, show that the vectors a, b, c and a, b, c form a reciprocal set. 

(b) Show that a · (b × c) =sinαa · a =sinβb · b =sinγc · c 

(c) Show that a · (b × c) =sinαa · a =sinβb · b =sinγc · c 

(d) Using parts (b) and (c) show that 

sin α sin β sin γ 

= = 


(e) Use the results from equation (4) to derive the law of sines for spherical triangles 


= = 

sin A sin B sin C 

(f) Using the equations (2) show that 

and hence show that 

In a similar manner show also that 

sin β sin γb · c =(c × a) · (a × b) =(c · a)(a · b) − b · c 

cos α =cosβ cos γ − sin β sin γ cos α. 

cos α =cosβ cos γ − sin β sin γ cos α. 

(g) Using part (f) derive the law of cosines for spherical triangles 

cos α =cosβ cos γ +sinβ sin γ cos A 

cos A = − cos B cos C +sinB sin C cos α 

A cyclic permutation of the symbols produces similar results involving the other angles and sides of the 

spherical triangle. 

65

§1.3 SPECIAL TENSORS 

Knowing how tensors are defined and recognizing a tensor when it pops up in front of you are two 

different things. Some quantities, which are tensors, frequently arise in applied problems and you should 

learn to recognize these special tensors when they occur. In this section some important tensor quantities 

are defined. We also consider how these special tensors can in turn be used to define other tensors. 

Metric Tensor 

Define yi ,i=1,...,N as independent coordinates in an N dimensional orthogonal Cartesian coordinate 

system. The distance squared between two points y i 

and y i + dy i expression 

, i =1,...,N is defined by the 

ds 2 = dy m dy m =(dy 1 ) 2 +(dy 2 ) 2 + ···+(dy N ) 2 . (1.3.1) 

Assume that the coordinates y i are related to a set of independent generalized coordinates x i ,i=1,...,N 

by a set of transformation equations 

y i = y i (x 1 ,x 2 ,...,x N ), i =1,...,N. (1.3.2) 

To emphasize that each y i depends upon the x coordinates we sometimes use the notation y i = y i (x), for 

i =1,...,N. The differential of each coordinate can be written as 

dy m = ∂ym 

∂x j dxj , m =1,...,N, (1.3.3) 

and consequently in the x-generalized coordinates the distance squared, found from the equation (1.3.1), 

becomes a quadratic form. Substituting equation (1.3.3) into equation (1.3.1) we find 

where 

ds 2 = ∂ym 

∂x i 

gij = ∂ym 

∂x i 

∂y m 

∂x j dxi dx j = gij dx i dx j 

(1.3.4) 

∂ym , i,j =1,...,N (1.3.5) 

∂xj are called the metrices of the space defined by the coordinates xi ,i=1,...,N. Here the gij are functions of 

the x coordinates and is sometimes written as gij = gij(x). Further, the metrices gij are symmetric in the 

indices i and j so that gij = gji for all values of i and j over the range of the indices. If we transform to 

another coordinate system, say xi ,i=1,...,N, then the element of arc length squared is expressed in terms 

of the barred coordinates and ds2 = gij dxidxj , where gij = gij(x) is a function of the barred coordinates. 

The following example demonstrates that these metrices are second order covariant tensors. 

65

66 

EXAMPLE 1.3-1. Show the metric components gij are covariant tensors of the second order. 

Solution: In a coordinate system xi ,i=1,...,N the element of arc length squared is 

ds 2 = gijdx i dx j 

while in a coordinate system x i ,i=1,...,N the element of arc length squared is represented in the form 

(1.3.6) 

ds 2 = g mn dx m dx n . (1.3.7) 

The element of arc length squared is to be an invariant and so we require that 

g mndx m dx n = gijdx i dx j 

(1.3.8) 

Here it is assumed that there exists a coordinate transformation of the form defined by equation (1.2.30) 

together with an inverse transformation, as in equation (1.2.32), which relates the barred and unbarred 

coordinates. In general, if x i = x i (x), then for i =1,...,N we have 

dx i = ∂xi 

∂xm dxm and dx j = ∂xj 

∂x 

Substituting these differentials in equation (1.3.8) gives us the result 

gmndx m dx n ∂x 

= gij 

i 

∂xm ∂x j 

∂x n dxm dx n 

For arbitrary changes in dx m this equation implies that g mn = gij 

as a second order absolute covariant tensor. 

or 

n dxn 

 

∂x 

gmn − gij 

i 

∂xm ∂xj ∂xn 

dx m dx n =0 

∂xi ∂xm (1.3.9) 

∂x j 

∂x n and consequently gij transforms 

EXAMPLE 1.3-2. (Curvilinear coordinates) Consider a set of general transformation equations from 

rectangular coordinates (x, y, z) to curvilinear coordinates (u, v, w). These transformation equations and the 

corresponding inverse transformations are represented 

x = x(u, v, w) 

y = y(u, v, w) 

z = z(u, v, w). 

u = u(x, y, z) 

v = v(x, y, z) 

w = w(x, y, z) 

(1.3.10) 

Here y 1 = x, y 2 = y, y 3 = z and x 1 = u, x 2 = v, x 3 = w are the Cartesian and generalized coordinates 

and N =3. The intersection of the coordinate surfaces u = c1,v = c2 and w = c3 define coordinate curves 

of the curvilinear coordinate system. The substitution of the given transformation equations (1.3.10) into 

the position vector r = x e1 + y e2 + z e3 produces the position vector which is a function of the generalized 

coordinates and 

r = r(u, v, w) =x(u, v, w) e1 + y(u, v, w) e2 + z(u, v, w) e3

and consequently dr = ∂r ∂r ∂r 

du + dv + dw, where 

∂u ∂v ∂w 

E1 = ∂r ∂x 

= 

∂u ∂u e1 + ∂y 

∂u e2 + ∂z 

∂u e3 

E2 = ∂r ∂x 

= 

∂v ∂v e1 + ∂y 

∂v e2 + ∂z 

∂v e3 

E3 = ∂r ∂x 

= 

∂w ∂w e1 + ∂y 

∂w e2 + ∂z 

∂w e3. 

are tangent vectors to the coordinate curves. The element of arc length in the curvilinear coordinates is 

ds 2 = dr · dr = ∂r ∂r ∂r ∂r ∂r ∂r 

· dudu + · dudv + · 

∂u ∂u ∂u ∂v ∂u ∂w dudw 

+ ∂r ∂r ∂r ∂r ∂r ∂r 

· dvdu + · dvdv + · 

∂v ∂u ∂v ∂v ∂v ∂w dvdw 

+ ∂r ∂r ∂r ∂r ∂r ∂r 

· dwdu + · dwdv + · 

∂w ∂u ∂w ∂v ∂w ∂w dwdw. 

(1.3.11) 


Utilizing the summation convention, the above can be expressed in the index notation. 

quantities 

Define the 

g11 = ∂r ∂r 

· 

∂u ∂u 

g21 = ∂r ∂r 

· 

∂v ∂u 

g31 = ∂r 

g12 = 

∂r 

· 

∂w ∂u 

∂r ∂r 

· 

∂u ∂v 

g22 = ∂r ∂r 

· 

∂v ∂v 

g32 = ∂r 

g13 = 

∂r 

· 

∂w ∂v 

∂r ∂r 

· 

∂u ∂w 

g23 = ∂r ∂r 

· 

∂v ∂w 

g33 = ∂r ∂r 

· 

∂w ∂w 

and let x 1 = u, x 2 = v, x 3 = w. Then the above element of arc length can be expressed as 

where 

ds 2 = Ei · Ej dx i dx j = gijdx i dx j , i,j =1, 2, 3 

gij = Ei · Ej = ∂r ∂r 

· 

∂xi ∂x 

∂ym , i,j free indices (1.3.13) 

∂xj ∂ym 

= j ∂xi are called the metric components of the curvilinear coordinate system. The metric components may be 

thought of as the elements of a symmetric matrix, since gij = gji. In the rectangular coordinate system 

x, y, z, the element of arc length squared is ds2 = dx2 + dy2 + dz2 . In this space the metric components are 

⎛ 

1 

gij = ⎝ 0 

0 

1 

⎞ 

0 

0⎠ 

. 

0 0 1 

67

68 

EXAMPLE 1.3-3. (Cylindrical coordinates (r, θ, z)) 

The transformation equations from rectangular coordinates to cylindrical coordinates can be expressed 

as x = r cos θ, y = r sin θ, z = z. Here y 1 = x, y 2 = y, y 3 = z and x 1 = r, x 2 = θ, x 3 = z, and the 

position vector can be expressed r = r(r, θ, z) =r cos θ e1 + r sin θ e2 + z e3. The derivatives of this position 

vector are calculated and we find 

E1 = ∂r 

∂r =cosθe1 +sinθe2, E2 = ∂r 

∂θ = −r sin θ e1 + r cos θ e2, E3 = ∂r 

= e3. 

∂z 

From the results in equation (1.3.13), the metric components of this space are 

⎛ 

1 

gij = ⎝ 0 

0 

r 

0 

2 ⎞ 

0 ⎠ . 

0 0 1 

Wenotethatsincegij =0wheni = j, the coordinate system is orthogonal. 

Given a set of transformations of the form found in equation (1.3.10), one can readily determine the 

metric components associated with the generalized coordinates. For future reference we list several different 

coordinate systems together with their metric components. Each of the listed coordinate systems are 

orthogonal and so gij =0fori= j. The metric components of these orthogonal systems have the form 

and the element of arc length squared is 

1. Cartesian coordinates (x, y, z) 

⎛ 

gij = ⎝ h21 0 0 

0 h2 2 0 

0 0 h2 ⎞ 

⎠ 

3 

ds 2 = h 2 1(dx 1 ) 2 + h 2 2(dx 2 ) 2 + h 2 3(dx 3 ) 2 . 

x = x 

y = y 

z = z 

h1 =1 

h2 =1 

h3 =1 

The coordinate curves are formed by the intersection of the coordinate surfaces 

x =Constant, y =Constant and z =Constant.

2. Cylindrical coordinates (r, θ, z) 

Figure 1.3-1. Cylindrical coordinates. 

x = r cos θ 

y = r sin θ 

z = z 

r ≥ 0 

0 ≤ θ ≤ 2π 

−∞

70 

Figure 1.3-2. Spherical coordinates. 

The coordinate curves, illustrated in the figure 1.3-3, are formed by the intersection of the coordinate 

surfaces 

5. Parabolic coordinates (ξ,η,φ) 

x 2 = −2ξ 2 (y − ξ2 

) Parabolic cylinders 

2 

x 2 =2η 2 (y + η2 

) 

2 

Parabolic cylinders 

z = Constant Planes. 

Figure 1.3-3. Parabolic cylindrical coordinates in plane z =0. 

x = ξη cos φ 

y = ξη sin φ 

z = 1 

2 (ξ2 − η 2 ) 

ξ ≥ 0 

η ≥ 0 

0


surfaces 

6. Elliptic cylindrical coordinates (ξ,η,z) 

x 2 + y 2 = −2ξ 2 (z − ξ2 

) Paraboloids 

2 

x 2 + y 2 =2η 2 (z + η2 

) 

2 

Paraboloids 

y = x tan φ Planes. 

Figure 1.3-4. Parabolic coordinates, φ = π/4. 

x =coshξ cos η 

y =sinhξ sin η 

z = z 

ξ ≥ 0 

0 ≤ η ≤ 2π 

−∞

72 

7. Elliptic coordinates (ξ,η,φ) 

Figure 1.3-5. Elliptic cylindrical coordinates in the plane z =0. 

x = (1 − η 2 )(ξ 2 − 1) cos φ 

y = (1 − η 2 )(ξ 2 − 1) sin φ 

z = ξη 

1 ≤ ξ

Figure 1.3-6. Elliptic coordinates φ = π/4. 

Figure 1.3-7. Bipolar coordinates. 


surfaces 

(x − a coth v) 2 + y 2 = a2 

sinh 2 v 

x 2 +(y − a cot u) 2 = a2 

Cylinders 

sin 2 u 

Cylinders 

z = Constant Planes. 

73

74 

9. Conical coordinates (u, v, w) 

x = uvw 

y = u 

 

(v2 − a2 )(w2 − a2 ) 

a a2 − b2 z = u 

 

(v2 − b2 )(w2 − b2 ) 

b b2 − a2 ab , b2 >v 2 >a 2 >w 2 , u ≥ 0 

h 2 1 =1 

h 2 2 = 

h 2 3 = 

u 2 (v 2 − w 2 ) 

(v 2 − a 2 )(b 2 − v 2 ) 

u 2 (v 2 − w 2 ) 

(w 2 − a 2 )(w 2 − b 2 ) 


surfaces 

x 2 + y 2 + z 2 = u 2 

Spheres 

x2 y2 

+ 

v2 v2 z2 

+ 

− a2 v2 =0, 

− b2 Cones 

x2 + 

w2 y2 

10. Prolate spheroidal coordinates (u, v, φ) 

w2 + 

− a2 z2 

w2 =0, Cones. 

− b2 Figure 1.3-8. Conical coordinates. 

x = a sinh u sin v cosφ , u ≥ 0 

y = a sinh u sin v sin φ, 0 ≤ v ≤ π 

z = a cosh u cos v, 0 ≤ φ

11. Oblate spheroidal coordinates (ξ,η,φ) 

Figure 1.3-9. Prolate spheroidal coordinates 

x = a cosh ξ cos η cosφ , ξ ≥ 0 

y = a cosh ξ cos η sin φ, − π π 

≤ η ≤ 

2 2 

z = a sinh ξ sin η, 0 ≤ φ ≤ 2π 

h 2 1 = h2 2 

h 2 2 = a2 (sinh 2 ξ +sin 2 η) 

h 2 3 = a 2 cosh 2 ξ cos 2 η 


surfaces 

x2 + 

(a cosh ξ) 2 

x2 + 

(a cos η) 2 

12. Toroidal coordinates (u, v, φ) 

y2 + 

(a cosh ξ) 2 

y2 − 

(a cos η) 2 

a sinh v cos φ 

x = , 

cosh v − cos u 

0 ≤ u

76 

Figure 1.3-10. Oblate spheroidal coordinates 

Figure 1.3-11. Toroidal coordinates 

is a mixed second order tensor. 

EXAMPLE 1.3-4. Show the Kronecker delta δi j 

Solution: Assume we have a coordinate transformation xi = xi (x),i =1,...,N of the form (1.2.30) and 

possessing an inverse transformation of the form (1.2.32). Let δ i 

j and δi j denote the Kronecker delta in the 

barred and unbarred system of coordinates. By definition the Kronecker delta is defined 

δ i 

j = δi j = 

 

0, 

1, 

if 

if 

i = j 

i = j .

Employing the chain rule we write 

∂xm ∂xm 

n = 

∂x ∂xi ∂xi ∂xm 

n = 

∂x ∂xi ∂xk ∂xn δi k 

(1.3.14) 

By hypothesis, the xi ,i=1,...,N are independent coordinates and therefore we have ∂xm 

n ∂x = δmn 

and (1.3.14) 

simplifies to 

δ m 

n = δi ∂x 

k 

m 

∂xi ∂xk n . 

∂x 

Therefore, the Kronecker delta transforms as a mixed second order tensor. 

Conjugate Metric Tensor 

Let g denote the determinant of the matrix having the metric tensor gij,i,j =1,...,N as its elements. 

In our study of cofactor elements of a matrix we have shown that 

cof(g1j)g1k + cof(g2j)g2k + ...+ cof(gNj)gNk = gδ j 

k . (1.3.15) 

We can use this fact to find the elements in the inverse matrix associated with the matrix having the 

components gij. Theelementsofthisinversematrixare 

g ij = 1 

g cof(gij) (1.3.16) 

and are called the conjugate metric components. We examine the summation g ij gik and find: 

The equation 

g ij gik = g 1j g1k + g 2j g2k + ...+ g Nj gNk 

= 1 

g [cof(g1j)g1k + cof(g2j)g2k + ...+ cof(gNj)gNk] 

= 1 

 

gδ 

g 

j 

 

k = δ j 

k 

g ij gik = δ j 

k 

(1.3.17) 

is an example where we can use the quotient law to show gij is a second order contravariant tensor. Because 

of the symmetry of gij and gij the equation (1.3.17) can be represented in other forms. 

EXAMPLE 1.3-5. Let Ai and Ai denote respectively the covariant and contravariant components of a 

vector A. Show these components are related by the equations 

Ai = gijA j 

A k = g jk Aj 

where gij and g ij are the metric and conjugate metric components of the space. 

(1.3.18) 

(1.3.19) 

77

78 

Solution: We multiply the equation (1.3.18) by g im (inner product) and use equation (1.3.17) to simplify 

the results. This produces the equation g im Ai = g im gijA j = δ m j A j = A m . Changing indices produces the 

result given in equation (1.3.19). Conversely, if we start with equation (1.3.19) and multiply by gkm (inner 

product) we obtain gkmA k = gkmg jk Aj = δ j mAj = Am which is another form of the equation (1.3.18) with 

the indices changed. 

Notice the consequences of what the equations (1.3.18) and (1.3.19) imply when we are in an orthogonal 

Cartesian coordinate system where 

In this special case, we have 

⎛ 

1 

gij = ⎝ 0 

0 

1 

⎞ 

0 

0⎠ 

and g 

0 0 1 

ij ⎛ 

1 

= ⎝ 0 

0 

1 

⎞ 

0 

0⎠ 

. 

0 0 1 

A1 = g11A 1 + g12A 2 + g13A 3 = A 1 

A2 = g21A 1 + g22A 2 + g23A 3 = A 2 

A3 = g31A 1 + g32A 2 + g33A 3 = A 3 . 

These equations tell us that in a Cartesian coordinate system the contravariant and covariant components 

are identically the same. 

EXAMPLE 1.3-6. We have previously shown that if Ai is a covariant tensor of rank 1 its components in 

a barred system of coordinates are 

Ai = Aj 

Solve for the Aj in terms of the Aj. (i.e. find the inverse transformation). 

Solution: Multiply equation (1.3.20) by ∂xi 

∂xm (inner product) and obtain 

Ai 

∂x i 

= Aj 

∂xm In the above product we have ∂xj 

∂xi ∂xi ∂xj 

= 

∂xm coordinates. This reduces equation (1.3.21) to the form 

∂xj i . (1.3.20) 

∂x 

∂x j 

∂x i 

∂xi . (1.3.21) 

∂xm ∂x m = δj m since xj and x m are assumed to be independent 

∂x i 

Ai 

∂xm = Ajδ j m = Am 

(1.3.22) 

which is the desired inverse transformation. 

This result can be obtained in another way. Examine the transformation equation (1.3.20) and ask the 

question, “When we have two coordinate systems, say a barred and an unbarred system, does it matter which 

system we call the barred system?” With some thought it should be obvious that it doesn’t matter which 

system you label as the barred system. Therefore, we can interchange the barred and unbarred symbols in 

equation (1.3.20) and obtain the result Ai = Aj 

a different set of indices. 

∂xj which is the same form as equation (1.3.22), but with 

∂xi

Associated Tensors 

Associated tensors can be constructed by taking the inner product of known tensors with either the 

metric or conjugate metric tensor. 

Definition: (Associated tensor) Any tensor constructed by multiplying (inner 

product) a given tensor with the metric or conjugate metric tensor is called an 

associated tensor. 

Associated tensors are different ways of representing a tensor. The multiplication of a tensor by the 

metric or conjugate metric tensor has the effect of lowering or raising indices. For example the covariant 

and contravariant components of a vector are different representations of the same vector in different forms. 

These forms are associated with one another by way of the metric and conjugate metric tensor and 

g ij Ai = A j 

gijA j = Ai. 

EXAMPLE 1.3-7. The following are some examples of associated tensors. 

A j = g ij Ai 

A m .jk = g mi Aijk 

A .nm 

i.. = g mk g nj Aijk 

Aj = gijA i 

A i.k 

m = gmjA ijk 

Amjk = gimA i .jk 

Sometimes ‘dots’are used as indices in order to represent the location of the index that was raised or lowered. 

If a tensor is symmetric, the position of the index is immaterial and so a dot is not needed. For example, if 

Amn is a symmetric tensor, then it is easy to show that An .m and A.n m 

as An m without confusion. 

are equal and therefore can be written 

Higher order tensors are similarly related. For example, if we find a fourth order covariant tensor Tijkm 

we can then construct the fourth order contravariant tensor T pqrs from the relation 

T pqrs = g pi g qj g rk g sm Tijkm. 

This fourth order tensor can also be expressed as a mixed tensor. Some mixed tensors associated with 

the given fourth order covariant tensor are: 

T p 

.jkm = gpi Tijkm, T pq 

..km = gqj T p 

.jkm . 

79

80 

Riemann Space VN 

A Riemannian space VN is said to exist if the element of arc length squared has the form 

ds 2 = gijdx i dx j 

(1.3.23) 

where the metrices gij = gij(x 1 ,x 2 ,...,x N ) are continuous functions of the coordinates and are different 

from constants. In the special case gij = δij the Riemannian space VN reduces to a Euclidean space EN . 

The element of arc length squared defined by equation (1.3.23) is called the Riemannian metric and any 

geometry which results by using this metric is called a Riemannian geometry. A space VN is called flat if 

it is possible to find a coordinate transformation where the element of arclength squared is ds 2 = ɛi(dx i ) 2 

where each ɛi is either +1 or −1. A space which is not flat is called curved. 

Geometry in VN 

Given two vectors A = A i Ei and B = B j Ej, then their dot product can be represented 

A · B = A i B j Ei · Ej = gijA i B j = AjB j = A i Bi = g ij AjBi = | A|| B| cos θ. (1.3.24) 

Consequently, in an N dimensional Riemannian space VN the dot or inner product of two vectors A and B 

is defined: 

gijA i B j = AjB j = A i Bi = g ij AjBi = AB cos θ. (1.3.25) 

In this definition A is the magnitude of the vector Ai , the quantity B is the magnitude of the vector Bi and 

θ is the angle between the vectors when their origins are made to coincide. In the special case that θ =90◦ we have gijAiB j = 0 as the condition that must be satisfied in order that the given vectors Ai and Bi are 

orthogonal to one another. Consider also the special case of equation (1.3.25) when Ai = Bi and θ =0. In 

this case the equations (1.3.25) inform us that 

g in AnAi = A i Ai = ginA i A n =(A) 2 . (1.3.26) 

From this equation one can determine the magnitude of the vector A i . The magnitudes A and B can be 

written A =(ginA i A n ) 1 

2 and B =(gpqB p B q ) 1 

2 and so we can express equation (1.3.24) in the form 

cos θ = 

gijA i B j 

(gmnA m A n ) 1 

2 (gpqB p B q ) 1 

2 

. (1.3.27) 

An import application of the above concepts arises in the dynamics of rigid body motion. Note that if a 

vector A i has constant magnitude and the magnitude of dAi 

dt is different from zero, then the vectors A i and 

dA i 

dt 

i dAj 

must be orthogonal to one another due to the fact that gijA 

dt =0. As an example, consider the unit 

vectors e1, e2 and e3 on a rotating system of Cartesian axes. We have for constants ci, i =1, 6that 

d e1 

dt = c1 e2 + c2 e3 

d e2 

dt = c3 e3 + c4 e1 

d e3 

dt = c5 e1 + c6 e2 

because the derivative of any ei (i fixed) constant vector must lie in a plane containing the vectors ej and 

ek, (j = i , k = i and j = k), since any vector in this plane must be perpendicular to ei.

The above definition of a dot product in VN can be used to define unit vectors in VN . 

EXAMPLE 1.3-8. (Unit vectors) 

Definition: (Unit vector) Whenever the magnitude of a vector 

Ai is unity, the vector is called a unit vector. In this case we 

have 

gijA i A j =1. (1.3.28) 

In VN the element of arc length squared is expressed ds 2 = gij dx i dx j which can be expressed in the 

dx 

form 1 = gij 

i dx 

ds 

j 

dxi 

. This equation states that the vector ,i=1,...,N is a unit vector. One application 

ds ds 

of this equation is to consider a particle moving along a curve in VN which is described by the parametric 

equations xi = xi (t), for i =1,...,N. The vector V i = dxi 

dt ,i=1,...,N represents a velocity vector of the 

particle. By chain rule differentiation we have 

where V = ds 

dt 

V i = dxi 

dt 

= dxi 

ds 

ds 

dt 

dxi 

= V , (1.3.29) 

ds 

dxi 

is the scalar speed of the particle and ds is a unit tangent vector to the curve. The equation 

(1.3.29) shows that the velocity is directed along the tangent to the curve and has a magnitude V. That is 

2 ds 

=(V ) 

dt 

2 = gijV i V j . 

EXAMPLE 1.3-9. (Curvilinear coordinates) 

Find an expression for the cosine of the angles between the coordinate curves associated with the 


x = x(u, v, w), y = y(u, v, w), z = z(u, v, w). 

81

82 

Figure 1.3-12. Angles between curvilinear coordinates. 

Solution: Let y 1 = x, y 2 = y, y 3 = z and x 1 = u, x 2 = v, x 3 = w denote the Cartesian and curvilinear 

coordinates respectively. With reference to the figure 1.3-12 we can interpret the intersection of the surfaces 

v = c2 and w = c3 as the curve r = r(u, c2,c3) which is a function of the parameter u. By moving only along 

thiscurvewehavedr = ∂r 

du and consequently 

∂u 

or 

This equation shows that the vector dx1 

ds 

be represented by tr 1 

(1) = √ δ g11 r 1 . 

ds 2 = dr · dr = ∂r ∂r 

· 

∂u ∂u dudu = g11(dx 1 ) 2 , 

1= dr dr 

· = g11 

ds ds 

dx 1 

ds 

2 

. 

= 1 

√ g11 is a unit vector along this curve. This tangent vector can 

The curve which is defined by the intersection of the surfaces u = c1 and w = c3 has the unit tangent 

= 1 

√ g22 δ r 2. Similarly, the curve which is defined as the intersection of the surfaces u = c1 and 

vector tr (2) 

v = c2 has the unit tangent vector tr (3) 

unit vectors tr (1) and tr (2) , is obtained from the result of equation (1.3.25). We find 

cos θ12 = gpqt p 

(1) tq 

(2) 

For θ13 the angle between the directions t i (1) and ti (3) 

= 1 

√ g33 δ r 3. The cosine of the angle θ12, which is the angle between the 

1 

= gpq √ δ 

g11 

p 1 

1 √ 

g22 

we find 

g13 

cos θ13 = √ √ . 

g11 g33 

Finally, for θ23 the angle between the directions t i (2) and ti (3) 

δ q 

2 = 

we find 

g12 

√ √ . 

g11 g22 

cos θ23 = √ √ . 

g22 g33 

When θ13 = θ12 = θ23 =90◦ , we have g12 = g13 = g23 = 0 and the coordinate curves which make up the 

curvilinear coordinate system are orthogonal to one another. 

In an orthogonal coordinate system we adopt the notation 

g23 

g11 =(h1) 2 , g22 =(h2) 2 , g33 =(h3) 2 

and gij =0,i= j.

Epsilon Permutation Symbol 

Associated with the e−permutation symbols there are the epsilon permutation symbols defined by the 

relations 

ɛijk = √ geijk and ɛ ijk = 1 

√ g e ijk 

(1.3.30) 

where g is the determinant of the metrices gij. 

It can be demonstrated that the eijk permutation symbol is a relative tensor of weight −1 whereasthe 

ɛijk permutation symbol is an absolute tensor. Similarly, the eijk permutation symbol is a relative tensor of 

weight +1 and the corresponding ɛijk permutation symbol is an absolute tensor. 

EXAMPLE 1.3-10. (ɛ permutation symbol) 

Show that eijk is a relative tensor of weight −1 and the corresponding ɛijk permutation symbol is an 

absolute tensor. 

Solution: Examine the Jacobian 

and make the substitution 

J 

 

x 

 

= 

x 

From the definition of a determinant we may write 

 

 

 

 

 

 

 

∂x 1 

∂x1 ∂x 2 

∂x1 ∂x 3 

∂x1 ∂x 1 

∂x2 ∂x 2 

∂x2 ∂x 3 

∂x2 ∂x 1 

∂x3 ∂x 2 

∂x3 ∂x 3 

∂x3 a i j = ∂xi 

j ,i,j=1, 2, 3. 

∂x 

 

 

 

 

 

 

 

eijka i ma j na k p = J( x 

x )emnp. (1.3.31) 

By definition, emnp = emnp in all coordinate systems and hence equation (1.3.31) can be expressed in the 

form 

 

J( x 

x ) 

−1 ∂x 

eijk 

i 

∂xm ∂xj ∂xn ∂xk p = emnp 

∂x 

(1.3.32) 

which demonstrates that eijk transforms as a relative tensor of weight −1. 

Wehavepreviouslyshownthemetrictensorgij 

according to the rule gij = gmn 

is a second order covariant tensor and transforms 

∂x m 

∂x i 

∂xn j . Taking the determinant of this result we find 

∂x 

 

 

g = |gij| = |gmn| 

 

∂x m 

∂x i 

 

 

 

 

2 

 

= g J( x 

x ) 

2 (1.3.33) 

where g is the determinant of (gij) andg is the determinant of (g ij ). This result demonstrates that g is a 

scalar invariant of weight +2. Taking the square root of this result we find that 

√ x 

g = gJ( ). (1.3.34) 

x 

Consequently, we call √ g a scalar invariant of weight +1. Now multiply both sides of equation (1.3.32) by 

√ 

g and use (1.3.34) to verify the relation 

√ ∂x 

geijk 

i 

∂xm ∂xj ∂xn ∂xk ∂xp = g emnp. (1.3.35) 

This equation demonstrates that the quantity ɛijk = √ geijk transforms like an absolute tensor. 

83

84 

Figure 1.3-14. Translation followed by rotation of axes 

In a similar manner one can show e ijk is a relative tensor of weight +1 and ɛ ijk = 1 

√ g e ijk is an absolute 

tensor. This is left as an exercise. 

Another exercise found at the end of this section is to show that a generalization of the e − δ identity 

is the epsilon identity 

g ij ɛiptɛjrs = gprgts − gpsgtr. (1.3.36) 

Cartesian Tensors 

Consider the motion of a rigid rod in two dimensions. No matter how complicated the movement of 

the rod is we can describe the motion as a translation followed by a rotation. Consider the rigid rod AB 

illustrated in the figure 1.3-13. 

Figure 1.3-13. Motion of rigid rod 

In this figure there is a before and after picture of the rod’s position. By moving the point B to B ′ we 

have a translation. This is then followed by a rotation holding B fixed.

Figure 1.3-15. Rotation of axes 

A similar situation exists in three dimensions. Consider two sets of Cartesian axes, say a barred and 

unbarred system as illustrated in the figure 1.3-14. Let us translate the origin 0 to 0 and then rotate the 

(x, y, z) axes until they coincide with the (x, y, z) axes. We consider first the rotation of axes when the 

origins 0 and 0 coincide as the translational distance can be represented by a vector bk ,k=1, 2, 3. When 

the origin 0 is translated to 0 we have the situation illustrated in the figure 1.3-15, where the barred axes 

can be thought of as a transformation due to rotation. 

Let 

r = x e1 + y e2 + z e3 

(1.3.37) 

denote the position vector of a variable point P with coordinates (x, y, z) with respect to the origin 0 and the 

unit vectors e1, e2, e3. This same point, when referenced with respect to the origin 0 and the unit vectors 

ê1, ê2, ê3, has the representation 

r = x ê1 + y ê2 + z ê3. (1.3.38) 

By considering the projections of r upon the barred and unbarred axes we can construct the transformation 

equations relating the barred and unbarred axes. We calculate the projections of r onto the x, y and z axes 

and find: 

r · e1 = x = x( ê1 · e1)+y( ê2 · e1)+z( ê3 · e1) 

r · e2 = y = x( ê1 · e2)+y( ê2 · e2)+z( ê3 · e2) 

r · e3 = z = x( ê1 · e3)+y( ê2 · e3)+z( ê3 · e3). 

We also calculate the projection of r onto the x, y, z axes and find: 

r · ê1 = x = x( e1 · ê1)+y( e2 · ê1)+z( e3 · ê1) 

r · ê2 = y = x( e1 · ê2)+y( e2 · ê2)+z( e3 · ê2) 

r · ê3 = z = x( e1 · ê3)+y( e2 · ê3)+z( e3 · ê3). 

(1.3.39) 

(1.3.40) 

By introducing the notation (y1,y2,y3) =(x, y, z) (y1, y2, y3)=(x, y, z) and defining θij as the angle 

between the unit vectors ei and êj, we can represent the above transformation equations in a more concise 

85

86 

form. We observe that the direction cosines can be written as 

ℓ11 = e1 · ê1 =cosθ11 

ℓ21 = e2 · ê1 =cosθ21 

ℓ31 = e3 · ê1 =cosθ31 

ℓ12 = e1 · ê2 =cosθ12 

ℓ22 = e2 · ê2 =cosθ22 

ℓ32 = e3 · ê2 =cosθ32 

which enables us to write the equations (1.3.39) and (1.3.40) in the form 

Using the index notation we represent the unit vectors as: 

ℓ13 = e1 · ê3 =cosθ13 

ℓ23 = e2 · ê3 =cosθ23 

ℓ33 = e3 · ê3 =cosθ33 

(1.3.41) 

yi = ℓijy j and y i = ℓjiyj. (1.3.42) 

êr = ℓpr ep or ep = ℓpr êr (1.3.43) 

where ℓpr are the direction cosines. In both the barred and unbarred system the unit vectors are orthogonal 

and consequently we must have the dot products 

êr · êp = δrp and em · en = δmn (1.3.44) 

where δij is the Kronecker delta. Substituting equation (1.3.43) into equation (1.3.44) we find the direction 

cosines ℓij must satisfy the relations: 

The relations 

êr · ês = ℓpr ep · ℓms em = ℓprℓms ep · em = ℓprℓmsδpm = ℓmrℓms = δrs 

and er · es = ℓrm êm · ℓsn ên = ℓrmℓsn êm · ên = ℓrmℓsnδmn = ℓrmℓsm = δrs. 

ℓmrℓms = δrs and ℓrmℓsm = δrs, (1.3.45) 

with summation index m, are important relations which are satisfied by the direction cosines associated with 

a rotation of axes. 

Combining the rotation and translation equations we find 

yi = ℓijy j 

 

+ bi 

 

. (1.3.46) 

rotation translation 

We multiply this equation by ℓik and make use of the relations (1.3.45) to find the inverse transformation 

These transformations are called linear or affine transformations. 

y k = ℓik(yi − bi). (1.3.47) 

Consider the xi axes as fixed, while the xi axes are rotating with respect to the xi axes where both sets 

of axes have a common origin. Let A = Ai ei denote a vector fixed in and rotating with the xi axes. We 

denote by d 

A 

and 

dt 

d 

A 

the derivatives of 

dt 

A with respect to the fixed (f) and rotating (r) axes. We can 

f 

r

write, with respect to the fixed axes, that d 

A 

 

dt = 

f 

dAi 

dt ei 

i d ei 

d ei 

+ A . Note that is the derivative of a 

dt dt 

vector with constant magnitude. Therefore there exists constants ωi, i =1,...,6 such that 

d e1 

dt = ω3 e2 − ω2 e3 

d e2 

dt = ω1 e3 − ω4 e1 

d e3 

dt = ω5 e1 − ω6 e2 

i.e. see page 80. From the dot product e1 · e2 = 0 we obtain by differentiation e1 · de2 

dt 

de1 

+ dt · e2 =0 

which implies ω4 = ω3. Similarly, from the dot products e1 · e3 and e2 · e3 we obtain by differentiation the 

additional relations ω5 = ω2 and ω6 = ω1. The derivative of A with respect to the fixed axes can now be 

represented 

d 

A 

 

dt 

= dAi 

dt ei +(ω2A3 − ω3A2) e1 +(ω3A1 − ω1A3) e2 +(ω1A2 − ω2A1) e3 = d 

A 

 

dt 

+ ω × A 

f 

where ω = ωi ei is called an angular velocity vector of the rotating system. The term ω × A represents the 

velocity of the rotating system relative to the fixed system and d 

A 

 

dt = 

r 

dAi 

dt ei represents the derivative with 

respect to the rotating system. 

Employing the special transformation equations (1.3.46) let us examine how tensor quantities transform 

when subjected to a translation and rotation of axes. These are our special transformation laws for Cartesian 

tensors. We examine only the transformation laws for first and second order Cartesian tensor as higher order 

transformation laws are easily discerned. We have previously shown that in general the first and second order 

tensor quantities satisfy the transformation laws: 

∂yj 

Ai = Aj 

∂yi A i = A j ∂y i 

∂yj 

A mn = A ij ∂y m 

∂yi 

Amn = Aij 

∂y n 

∂yj 

∂yi ∂yj 

∂ym ∂yn A m 

n = Ai ∂ym ∂yj 

j 

∂yi ∂yn r 

(1.3.48) 

(1.3.49) 

(1.3.50) 

(1.3.51) 

(1.3.52) 

For the special case of Cartesian tensors we assume that yi and yi,i=1, 2, 3 are linearly independent. We 

differentiate the equations (1.3.46) and (1.3.47) and find 

∂yj = ℓij 

∂yk ∂yi 

∂yk ∂yi 

= ℓijδjk = ℓik, and = ℓik = ℓikδim = ℓmk. 

∂yk ∂ym ∂ym 

Substituting these derivatives into the transformation equations (1.3.48) through (1.3.52) we produce the 


Ai = Ajℓji 

A i = A j ℓji 

A mn = A ij ℓimℓjn 

Amn = Aijℓimℓjn 

A m 

n = Aij ℓimℓjn. 

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88 

Figure 1.3-16. Transformation to curvilinear coordinates 

These are the transformation laws when moving from one orthogonal system to another. In this case the 

direction cosines ℓim are constants and satisfy the relations given in equation (1.3.45). The transformation 

laws for higher ordered tensors are similar in nature to those given above. 

In the unbarred system (y1,y2,y3) the metric tensor and conjugate metric tensor are: 

gij = δij and g ij = δij 

where δij is the Kronecker delta. In the barred system of coordinates, which is also orthogonal, we have 

From the orthogonality relations (1.3.45) we find 

Weexaminetheassociatedtensors 

g ij = ∂ym 

∂y i 

∂ym 

. 

∂yj g ij = ℓmiℓmj = δij and g ij = δij. 

A i = g ij Aj 

A ij = g im g jn Amn 

A i n = gim Amn 

Ai = gijA j 

Amn = gmignjA ij 

A i ij 

n = gnjA 

and find that the contravariant and covariant components are identical to one another. This holds also in 

the barred system of coordinates. Also note that these special circumstances allow the representation of 

contractions using subscript quantities only. This type of a contraction is not allowed for general tensors. It 

is left as an exercise to try a contraction on a general tensor using only subscripts to see what happens. Note 

that such a contraction does not produce a tensor. These special situations are considered in the exercises. 

Physical Components 

We have previously shown an arbitrary vector A can be represented in many forms depending upon 

the coordinate system and basis vectors selected. For example, consider the figure 1.3-16 which illustrates a 

Cartesian coordinate system and a curvilinear coordinate system.

Figure 1.3-17. Physical components 

In the Cartesian coordinate system we can represent a vector A as 

A = Ax e1 + Ay e2 + Az e3 

where ( e1, e2, e3) are the basis vectors. Consider a coordinate transformation to a more general coordinate 

system, say (x 1 ,x 2 ,x 3 ). The vector A can be represented with contravariant components as 

A = A 1 E1 + A 2 E2 + A 3 E3 

(1.3.53) 

with respect to the tangential basis vectors ( E1, E2, E3). Alternatively, the same vector A can be represented 

in the form 

A = A1 E 1 + A2 E 2 + A3 E 3 

(1.3.54) 

having covariant components with respect to the gradient basis vectors ( E 1 , E 2 , E 3 ). These equations are 

just different ways of representing the same vector. In the above representations the basis vectors need not 

be orthogonal and they need not be unit vectors. In general, the physical dimensions of the components Ai and Aj are not the same. 

The physical components of the vector A in a direction is defined as the projection of A upon a unit 

vector in the desired direction. For example, the physical component of A in the direction E1 is 

Similarly, the physical component of A in the direction E 1 is 

A · E1 

| A1 

= 

E1| | E1| = projection of A on E1. (1.3.58) 

A · E 1 

| E 1 | 

= A1 

| E 1 | = projection of A on E 1 . (1.3.59) 

EXAMPLE 1.3-11. (Physical components) Let α, β, γ denote nonzero positive constants such that the 

product relation αγ = 1 is satisfied. Consider the nonorthogonal basis vectors 


E1 = α e1, E2 = β e1 + γ e2, E3 = e3 

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90 

It is readily verified that the reciprocal basis is 

E 1 = γ e1 − β e2, E 2 = α e2, E 3 = e3. 

Consider the problem of representing the vector A = Ax e1 + Ay e2 in the contravariant vector form 

This vector has the contravariant components 

A = A 1 E1 + A 2 E2 or tensor form A i ,i=1, 2. 

A 1 = A · E 1 = γAx − βAy and A 2 = A · E 2 = αAy. 

Alternatively, this same vector can be represented as the covariant vector 

A = A1 E 1 + A2 E 2 which has the tensor form Ai, i=1, 2. 

The covariant components are found from the relations 

A1 = A · E1 = αAx 

A2 = A · E2 = βAx + γAy. 

The physical components of A in the directions E 1 and E 2 are found to be: 

A · E1 | E1 | 

E 2 

A · 

| E2 | 

A1 

= 

| E1 | = γAx − βAy 

 

γ2 + β2 = A(1) 

A2 

= 

| E2 αAy 

= 

| α = Ay = A(2). 

Note that these same results are obtained from the dot product relations using either form of the vector A. 

For example, we can write 

A · E 1 

and A · E 2 

| E1 | = A1( E1 · E1 )+A2( E2 · E1 ) 

| E1 | 

| E2 | = A1( E1 · E2 )+A2( E2 · E2 ) 

| E2 | 

= A(1) 

= A(2). 

In general, the physical components of a vector A in a direction of a unit vector λi is the generalized 

dot product in VN . This dot product is an invariant and can be expressed 

gijA i λ j = A i λi = Aiλ i = projection of A in direction of λ i

Physical Components For Orthogonal Coordinates 

where 

In orthogonal coordinates observe the element of arc length squared in V3 is 

ds 2 = gijdx i dx j =(h1) 2 (dx 1 ) 2 +(h2) 2 (dx 2 ) 2 +(h3) 2 (dx 3 ) 2 

In this case the curvilinear coordinates are orthogonal and 

⎛ 

gij = ⎝ (h1) 2 0 0 

0 (h2) 2 0 

0 0 (h3) 2 

⎞ 

⎠ . (1.3.60) 

h 2 (i) = g (i)(i) i not summed and gij =0,i= j. 

At an arbitrary point in this coordinate system we take λi ,i =1, 2, 3 as a unit vector in the direction 

of the coordinate x1 . We then obtain 

This is a unit vector since 

λ 1 = dx1 

ds , λ2 =0, λ 3 =0. 

1=gijλ i λ j = g11λ 1 λ 1 = h 2 1 (λ1 ) 2 

or λ1 = 1 . Here the curvilinear coordinate system is orthogonal and in this case the physical component 

h1 

of a vector Ai , in the direction xi , is the projection of Ai on λi in V3. The projection in the x1 direction is 

determined from 

A(1) = gijA i λ j = g11A 1 λ 1 = h 2 1 1 

1A = h1A 

h1 

1 . 

Similarly, we choose unit vectors µ i and ν i ,i=1, 2, 3inthex 2 and x 3 directions. These unit vectors 

can be represented 

µ 1 =0, 

ν 1 =0, 

µ 2 = dx2 

ds 

ν 2 =0, 

1 

= , 

h2 

µ 3 =0 

ν 3 = dx3 

ds 

= 1 

h3 

and the physical components of the vector A i in these directions are calculated as 

A(2) = h2A 2 

and A(3) = h3A 3 . 

In summary, we can say that in an orthogonal coordinate system the physical components of a contravariant 

tensor of order one can be determined from the equations 

A(i) =h (i)A (i) = √ g (i)(i)A (i) , i =1, 2 or 3 no summation on i, 

which is a short hand notation for the physical components (h1A 1 ,h2A 2 ,h3A 3 ). In an orthogonal coordinate 

system the nonzero conjugate metric components are 

g (i)(i) = 1 

, i =1, 2, or3 no summation on i. 

g (i)(i) 

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92 

These components are needed to calculate the physical components associated with a covariant tensor of 

order one. For example, in the x1−direction, we have the covariant components 

1 

λ1 = g11λ 1 = h 2 1 

h1 

and consequently the projection in V3 can be represented 

= h1, λ2 =0, λ3 =0 

gijA i λ j = gijA i g jm λm = Ajg jm λm = A1λ1g 11 = A1h1 

h2 1 

In a similar manner we calculate the relations 

A(2) = A2 

h2 

and A(3) = A3 

h3 

1 

= A1 

h1 

= A(1). 

for the other physical components in the directions x 2 and x 3 . These physical components can be represented 

in the short hand notation 

A(i) = A (i) 

h (i) 

= A (i) 

√ , i =1, 2 or 3 no summation on i. 

g(i)(i) 

In an orthogonal coordinate system the physical components associated with both the contravariant and 

covariant components are the same. To show this we note that when Aigij = Aj is summed on i we obtain 

Since gij =0fori = j this equation reduces to 

Another form for this equation is 

A 1 g1j + A 2 g2j + A 3 g3j = Aj. 

A (i) g (i)(i) = A (i), i not summed. 

A(i) =A (i)√ g(i)(i) = A (i) 

√ g(i)(i) 

i not summed, 

which demonstrates that the physical components associated with the contravariant and covariant components 

are identical. 

NOTATION The physical components are sometimes expressed by symbols with subscripts which represent 

the coordinate curve along which the projection is taken. For example, let Hi denote the contravariant 

components of a first order tensor. The following are some examples of the representation of the physical 

components of Hi in various coordinate systems: 

orthogonal coordinate tensor physical 

coordinates system components components 

general (x 1 ,x 2 ,x 3 ) H i H(1),H(2),H(3) 

rectangular (x, y, z) H i Hx,Hy,Hz 

cylindrical (r, θ, z) H i Hr,Hθ,Hz 

spherical (ρ, θ, φ) H i Hρ,Hθ,Hφ 

general (u, v, w) H i Hu,Hv,Hw

Higher Order Tensors 

The physical components associated with higher ordered tensors are defined by projections in VN just 

like the case with first order tensors. For an nth ordered tensor Tij...k we can select n unit vectors λ i ,µ i ,...,ν i 

and form the inner product (projection) 

Tij...kλ i µ j ...ν k . 

When projecting the tensor components onto the coordinate curves, there are N choices for each of the unit 

vectors. This produces N n physical components. 

The above inner product represents the physical component of the tensor Tij...k along the directions of 

the unit vectors λ i ,µ i ,...,ν i . The selected unit vectors may or may not be orthogonal. In the cases where 

the selected unit vectors are all orthogonal to one another, the calculation of the physical components is 

greatly simplified. By relabeling the unit vectors λi (m) ,λi (n) ,...,λi (p) where (m), (n), ..., (p) representoneof 

the N directions, the physical components of a general nth order tensor is represented 

T (mn...p)=Tij...kλ i (m) λj 

(n) ...λk (p) 

EXAMPLE 1.3-12. (Physical components) 

In an orthogonal curvilinear coordinate system V3 with metric gij, i,j=1, 2, 3, find the physical components 

of 

(i) the second order tensor Aij. (ii) the second order tensor A ij . (iii) the second order tensor A i j . 

Solution: The physical components of Amn,m,n =1, 2, 3 along the directions of two unit vectors λi and 

µ i is defined as the inner product in V3. These physical components can be expressed 

A(ij) =Amnλ m (i) µn (j) 

i, j =1, 2, 3, 

where the subscripts (i) and(j) represent one of the coordinate directions. Dropping the subscripts (i) and 

(j), we make the observation that in an orthogonal curvilinear coordinate system there are three choices for 

the direction of the unit vector λi and also three choices for the direction of the unit vector µ i . These three 

choices represent the directions along the x1 ,x2 or x3 coordinate curves which emanate from a point of the 

curvilinear coordinate system. This produces a total of nine possible physical components associated with 

the tensor Amn. 

For example, we can obtain the components of the unit vector λ i ,i=1, 2, 3inthex 1 direction directly 

from an examination of the element of arc length squared 

By setting dx 2 = dx 3 = 0, we find 

ds 2 =(h1) 2 (dx 1 ) 2 +(h2) 2 (dx 2 ) 2 +(h3) 2 (dx 3 ) 2 . 

dx 1 

ds 

1 

= = λ 

h1 

1 , λ 2 =0, λ 3 =0. 

This is the vector λ i (1) ,i =1, 2, 3. Similarly, if we choose to select the unit vector λi ,i =1, 2, 3inthex 2 

direction, we set dx 1 = dx 3 = 0 in the element of arc length squared and find the components 

λ 1 =0, λ 2 = dx2 

ds 

1 

= , λ 

h2 

3 =0. 

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94 

This is the vector λ i (2) ,i=1, 2, 3. Finally, if we select λi ,i=1, 2, 3inthex 3 direction, we set dx 1 = dx 2 =0 

in the element of arc length squared and determine the unit vector 

λ 1 =0, λ 2 =0, λ 3 = dx3 

ds 

1 

= . 

h3 

This is the vector λ i (3) ,i =1, 2, 3. Similarly, the unit vector µi can be selected as one of the above three 

directions. Examining all nine possible combinations for selecting the unit vectors, we calculate the physical 

components in an orthogonal coordinate system as: 

A(11) = A11 

h1h1 

A(21) = A21 

h1h2 

A(31) = A31 

h3h1 

A(12) = A12 

h1h2 

A(22) = A22 

h2h2 

A(32) = A32 

h3h2 

These results can be written in the more compact form 

For mixed tensors we have 

A(ij) = A (i)(j) 

h (i)h (j) 

A(13) = A13 

h1h3 

A(23) = A23 

h2h3 

A(33) = A33 

h3h3 

no summation on i or j . (1.3.61) 

A i j = gim Amj = g i1 A1j + g i2 A2j + g i3 A3j. (1.3.62) 

From the fact gij =0fori= j, together with the physical components from equation (1.3.61), the equation 

(1.3.62) reduces to 

A (i) 

(j) = g(i)(i) A (i)(j) = 1 

h2 · h (i)h (j)A(ij) no summation on i and i, j =1, 2or3. 

(i) 

This can also be written in the form 

A(ij) =A (i) h (i) 

(j) h (j) 

Hence, the physical components associated with the mixed tensor A i j 

canbeexpressedas 

A(11) = A 1 1 

A(21) = A 2 h2 

1 

h1 

A(31) = A 3 h3 

1 

h1 

no summation on i or j. (1.3.63) 

A(12) = A 1 h1 

2 

h2 

A(22) = A 2 2 

A(32) = A 3 h3 

2 

h2 

For second order contravariant tensors we may write 

in an orthogonal coordinate system 

A(13) = A 1 h1 

3 

h3 

A(23) = A 2 h2 

3 

h3 

A(33) = A 3 3. 

A ij gjm = A i m = Ai1 g1m + A i2 g2m + A i3 g3m.

Weusethefactgij =0fori = j together with the physical components from equation (1.3.63) to reduce the 

above equation to the form A (i) 

(m) = A(i)(m) g (m)(m) 

we have 

no summation on m . In terms of physical components 

h (m) 

A(im) =A 

h (i) 

(i)(m) h 2 (m) or A(im) =A (i)(m) h (i)h (m). no summation i, m =1, 2, 3 (1.3.64) 

Examining the results from equation (1.3.64) we find that the physical components associated with the 

contravariant tensor Aij , in an orthogonal coordinate system, can be written as: 

A(11) = A 11 h1h1 

A(21) = A 21 h2h1 

A(31) = A 31 h3h1 

Physical Components in General 

A(12) = A 12 h1h2 

A(22) = A 22 h2h2 

A(32) = A 32 h3h2 

A(13) = A 13 h1h3 

A(23) = A 23 h2h3 

A(33) = A 33 h3h3. 

In an orthogonal curvilinear coordinate system, the physical components associated with the nth order 

tensor Tij...kl along the curvilinear coordinate directions can be represented: 

T (ij...kl)= 

T (i)(j)...(k)(l) 

h (i)h (j) ...h (k)h (l) 

no summations. 

These physical components can be related to the various tensors associated with Tij...kl. For example, in 

an orthogonal coordinate system, the physical components associated with the mixed tensor T ij...m 

n...kl can be 

expressed as: 

T (ij...mn...kl)=T (i)(j)...(m) h (i)h (j) ...h (m) 

(n)...(k)(l) 

no summations. (1.3.65) 

h (n) ...h (k)h (l) 

EXAMPLE 1.3-13. (Physical components) Let xi = xi (t),i =1, 2, 3 denote the position vector of a 

particle which moves as a function of time t. Assume there exists a coordinate transformation xi = xi (x), for 

i =1, 2, 3, of the form given by equations (1.2.33). The position of the particle when referenced with respect 

to the barred system of coordinates can be found by substitution. The generalized velocity of the particle 

in the unbarred system is a vector with components 

v i = dxi 

,i=1, 2, 3. 

dt 

The generalized velocity components of the same particle in the barred system is obtained from the chain 

rule. We find this velocity is represented by 

v i = dxi 

dt 

∂xi 

= 

∂xj dxj dt 

This equation implies that the contravariant quantities 

(v 1 ,v 2 ,v 3 )=( dx1 

dt 

= ∂xi 

∂x j vj . 

dx2 dx3 

, , 

dt dt ) 

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96 

are tensor quantities. These quantities are called the components of the generalized velocity. The coordinates 

x 1 ,x 2 ,x 3 are generalized coordinates. This means we can select any set of three independent variables for 

the representation of the motion. The variables selected might not have the same dimensions. For example, 

in cylindrical coordinates we let (x 1 = r, x 2 = θ, x 3 = z). Here x 1 and x 3 have dimensions of distance but x 2 

has dimensions of angular displacement. The generalized velocities are 

v 1 = dx1 

dt 

dr 

= 

dt , v2 = dx2 

dt 

dθ 

= 

dt , v3 = dx3 

dt 

= dz 

dt . 

Here v 1 and v 3 have units of length divided by time while v 2 has the units of angular velocity or angular 

change divided by time. Clearly, these dimensions are not all the same. Let us examine the physical 

components of the generalized velocities. We find in cylindrical coordinates h1 =1,h2 = r, h3 =1andthe 

physical components of the velocity have the forms: 

vr = v(1) = v 1 h1 = dr 

dt , vθ = v(2) = v 2 h2 = r dθ 

dt , vz = v(3) = v 3 h3 = dz 

dt . 

Now the physical components of the velocity all have the same units of length divided by time. 

Additional examples of the use of physical components are considered later. For the time being, just 

remember that when tensor equations are derived, the equations are valid in any generalized coordinate 

system. In particular, we are interested in the representation of physical laws which are to be invariant and 

independent of the coordinate system used to represent these laws. Once a tensor equation is derived, we 

can chose any type of generalized coordinates and expand the tensor equations. Before using any expanded 

tensor equations we must replace all the tensor components by their corresponding physical components in 

order that the equations are dimensionally homogeneous. It is these expanded equations, expressed in terms 

of the physical components, which are used to solve applied problems. 

Tensors and Multilinear Forms 

Tensors can be thought of as being created by multilinear forms defined on some vector space V. Let 

us define on a vector space V a linear form, a bilinear form and a general multilinear form. We can then 

illustrate how tensors are created from these forms. 

Definition: (Linear form) Let V denote a vector space which 

contains vectors x, x1,x2,....A linear form in x is a scalar function 

ϕ(x) having a single vector argument x which satisfies the linearity 

properties: 

(i) ϕ(x1 + x2) =ϕ(x1)+ϕ(x2) 

(ii) ϕ(µx1) =µϕ(x1) 

for all arbitrary vectors x1,x2 in V and all real numbers µ. 

(1.3.66)

An example of a linear form is the dot product relation 

ϕ(x) = A · x (1.3.67) 

where A is a constant vector and x is an arbitrary vector belonging to the vector space V. 

Note that a linear form in x can be expressed in terms of the components of the vector x and the base 

vectors ( e1, e2, e3) used to represent x. To show this, we write the vector x in the component form 

x = x i ei = x 1 e1 + x 2 e2 + x 3 e3, 

where xi ,i =1, 2, 3arethecomponentsofxwith respect to the basis vectors ( e1, e2, e3). By the linearity 

property of ϕ we can write 

ϕ(x) =ϕ(x i ei) =ϕ(x 1 e1 + x 2 e2 + x 3 e3) 

= ϕ(x 1 e1)+ϕ(x 2 e2)+ϕ(x 3 e3) 

= x 1 ϕ( e1)+x 2 ϕ( e2)+x 3 ϕ( e3) =x i ϕ( ei) 

Thus we can write ϕ(x) =xiϕ( ei) and by defining the quantity ϕ( ei) =ai as a tensor we obtain ϕ(x) =xiai. Note that if we change basis from ( e1, e2, e3) to( E1, E2, E3) then the components of x also must change. 

Letting xi denote the components of x with respect to the new basis, we would have 

x = x i Ei and ϕ(x) =ϕ(x i Ei) =x i ϕ( Ei). 

The linear form ϕ defines a new tensor ai = ϕ( Ei) sothatϕ(x) =xiai. Whenever there is a definite relation 

between the basis vectors ( e1, e2, e3) and( E1, E2, E3), say, 

Ei = ∂xj 

ej, i ∂x 

then there exists a definite relation between the tensors ai and ai. This relation is 

ai = ϕ( Ei) =ϕ( ∂xj 

∂xi ej) = ∂xj 

∂xi ϕ( ej) = ∂xj 

i aj. 

∂x 

This is the transformation law for an absolute covariant tensor of rank or order one. 

The above idea is now extended to higher order tensors. 

Definition: ( Bilinear form) A bilinear form in x and y is a 

scalar function ϕ(x, y) with two vector arguments, which satisfies 

the linearity properties: 

(i) ϕ(x1 + x2,y1) =ϕ(x1,y1)+ϕ(x2,y1) 

(ii) ϕ(x1,y1 + y2) =ϕ(x1,y1)+ϕ(x1,y2) 

(iii) ϕ(µx1,y1) =µϕ(x1,y1) 

(iv) ϕ(x1,µy1) =µϕ(x1,y1) 

(1.3.68) 

for arbitrary vectors x1,x2,y1,y2 in the vector space V and for all 

real numbers µ. 

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98 

Note in the definition of a bilinear form that the scalar function ϕ is linear in both the arguments x and 

y. An example of a bilinear form is the dot product relation 

ϕ(x, y) =x · y (1.3.69) 

where both x and y belong to the same vector space V. 

The definition of a bilinear form suggests how multilinear forms can be defined. 

Definition: (Multilinear forms) A multilinear form of degree M or a M degree 

linear form in the vector arguments 

is a scalar function 

x1,x2,...,xM 

ϕ(x1,x2,...,xM ) 

of M vector arguments which satisfies the property that it is a linear form in each of its 

arguments. That is, ϕ must satisfy for each j =1, 2,...,M the properties: 

(i) ϕ(x1,...,xj1 + xj2,...xM )=ϕ(x1,...,xj1,...,xM )+ϕ(x1,...,xj2,...,xM ) 

(ii) ϕ(x1,...,µxj,...,xM )=µϕ(x1,...,xj,...,xM ) 

(1.3.70) 

for all arbitrary vectors x1,...,xM in the vector space V and all real numbers µ. 

An example of a third degree multilinear form or trilinear form is the triple scalar product 

ϕ(x, y,z) =x · (y × z). (1.3.71) 

Note that multilinear forms are independent of the coordinate system selected and depend only upon the 

vector arguments. In a three dimensional vector space we select the basis vectors ( e1, e2, e3) and represent 

all vectors with respect to this basis set. For example, if x, y,z are three vectors we can represent these 

vectors in the component forms 

x = x i ei, y = y j ej, z = z k ek (1.3.72) 

where we have employed the summation convention on the repeated indices i, j and k. Substituting equations 

(1.3.72) into equation (1.3.71) we obtain 

ϕ(x i ei,y j ej,z k ek) =x i y j z k ϕ( ei, ej, ek), (1.3.73) 

since ϕ is linear in all its arguments. By defining the tensor quantity 

ϕ( ei, ej, ek) =eijk 

(1.3.74)

(See exercise 1.1, problem 15) the trilinear form, given by equation (1.3.71), with vectors from equations 

(1.3.72), can be expressed as 

ϕ(x, y,z) =eijkx i y j z k , i,j,k =1, 2, 3. (1.3.75) 

The coefficients eijk of the trilinear form is called a third order tensor. It is the familiar permutation symbol 

considered earlier. 

In a multilinear form of degree M, ϕ(x, y,...,z), the M arguments can be represented in a component 

form with respect to a set of basis vectors ( e1, e2, e3). Let these vectors have components xi ,yi ,zi ,i=1, 2, 3 

with respect to the selected basis vectors. We then can write 

x = x i ei, y = y j ej, z = z k ek. 

Substituting these vectors into the M degree multilinear form produces 

ϕ(x i ei,y j ej,...,z k ek) =x i y j ···z k ϕ( ei, ej,..., ek). (1.3.76) 

Consequently, the multilinear form defines a set of coefficients 

aij...k = ϕ( ei, ej,..., ek) (1.3.77) 

which are referred to as the components of a tensor of order M. The tensor is thus created by the multilinear 

form and has M indices if ϕ is of degree M. 

Note that if we change to a different set of basis vectors, say, ( E1, E2, E3) the multilinear form defines 

anewtensor 

aij...k = ϕ( Ei, Ej,..., Ek). (1.3.78) 

This new tensor has a bar over it to distinguish it from the previous tensor. A definite relation exists between 

the new and old basis vectors and consequently there exists a definite relation between the components of 

the barred and unbarred tensors components. Recall that if we are given a set of transformation equations 

y i = y i (x 1 ,x 2 ,x 3 ),i=1, 2, 3, (1.3.79) 

from rectangular to generalized curvilinear coordinates, we can express the basis vectors in the new system 

by the equations 

Ei = ∂yj 

∂xi ej, i =1, 2, 3. (1.3.80) 

For example, see equations (1.3.11) with y1 = x, y2 = y, y3 = z,x1 = u, x2 = v, x3 = w. Substituting 

equations (1.3.80) into equations (1.3.78) we obtain 

aij...k = ϕ( ∂yα ∂yβ ∂yγ 

eα, eβ,..., 

∂xi ∂xj ∂x 

By the linearity property of ϕ, this equation is expressible in the form 

∂yβ ∂yγ 

... 

∂xj k eγ). 

aij...k = ∂yα 

∂xi ∂xk ϕ( eα, eβ,..., eγ) 

aij...k = ∂yα 

∂xi ∂yβ ∂yγ 

... aαβ...γ 

∂xj ∂xk This is the familiar transformation law for a covariant tensor of degree M. By selecting reciprocal basis 

vectors the corresponding transformation laws for contravariant vectors can be determined. 

The above examples illustrate that tensors can be considered as quantities derivable from multilinear 

forms defined on some vector space. 

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100 

Dual Tensors 

The e-permutation symbol is often used to generate new tensors from given tensors. For Ti1i2...im 

skew-symmetric tensor, we define the tensor 

a 

ˆT j1j2...jn−m = 1 

m! ej1j2...jn−mi1i2...imTi1i2...im m ≤ n (1.3.81) 

as the dual tensor associated with Ti1i2...im. Note that the e-permutation symbol or alternating tensor has 

a weight of +1 and consequently the dual tensor will have a higher weight than the original tensor. 

The e-permutation symbol has the following properties 

e i1i2...iN ei1i2...iN = N! 

e i1i2...iN ej1j2...jN = δ i1i2...iN 

j1j2...jN 

ek1k2...kmi1i2...iN −me j1j2...jmi1i2...iN −m =(N − m)!δ j1j2...jm 

k1k2...km 

δ j1j2...jm 

k1k2...km Tj1j2...jm = m!Tk1k2...km. 

(1.3.82) 

Using the above properties we can solve for the skew-symmetric tensor in terms of the dual tensor. We find 

1 

Ti1i2...im = 

(n − m)! ei1i2...imj1j2...jn−m ˆ T j1j2...jn−m . (1.3.83) 

For example, if Aij i, j =1, 2, 3 is a skew-symmetric tensor, we may associate with it the dual tensor 

V i = 1 

2! eijkAjk, which is a first order tensor or vector. Note that Aij has the components 

⎛ 

⎞ 

⎝ 0 A12 A13 

⎠ (1.3.84) 

−A12 0 A23 

−A13 −A23 0 

and consequently, the components of the vector V are 

(V 1 ,V 2 ,V 3 )=(A23,A31,A12). (1.3.85) 

Note that the vector components have a cyclic order to the indices which comes from the cyclic properties 

of the e-permutation symbol. 

As another example, consider the fourth order skew-symmetric tensor Aijkl, i,j,k,l=1,...,n.Wecan 

associate with this tensor any of the dual tensor quantities 

V = 1 

4! eijkl Aijkl 

V i = 1 

4! eijklm Ajklm 

V ij = 1 

4! eijklmn Aklmn 

V ijk = 1 

4! eijklmnp Almnp 

V ijkl = 1 

4! eijklmnpr Amnpr 

Applications of dual tensors can be found in section 2.2. 

(1.3.86)

EXERCISE 1.3 

◮ 1. 

∂x 

(a) From the transformation law for the second order tensor gij = gab 

a 

∂xi ∂xb ∂xj solve for the gab in terms of gij. (b) Show that if gij is symmetric in one coordinate system it is symmetric in all coordinate systems. 

(c) Let g = det(gij )andg = det(gij) andshowthatg = gJ2 ( x 

x ) and consequently g = √ gJ( x 

). This 

x 

shows that g is a scalar invariant of weight 2 and √ g is a scalar invariant of weight 1. 

◮ 2. For 

gij = ∂ym 

∂x i 

∂ym ∂xj show that g ij = ∂xi 

∂ym ∂xj ∂ym ◮ 3. Show that in a curvilinear coordinate system which is orthogonal we have: 

 

 

◮ 4. Show that g = det(gij) = 

∂y 

 

i 

∂xj 

 

 

 

(a) g = det(gij) =g11g22g33 

(b) gmn = g mn =0 form= n 

(c) g NN = 1 

for N =1, 2, 3 (no summation on N) 

2 

gNN 

= J 2 , where J is the Jacobian. 

◮ 5. Define the quantities h1 = hu = | ∂r 

∂u |, h2 = hv = | ∂r 

∂v |, h3 = hw = | ∂r 

| and construct the unit 

∂w 

vectors 

eu = 1 ∂r 

h1 ∂u , ev = 1 ∂r 

h2 ∂v , ew = 1 ∂r 

h3 ∂w . 

(a) Assume the coordinate system is orthogonal and show that 

g11 = h 2 1 = 

g22 = h 2 2 = 

g33 = h 2 3 = 

 

∂x 

∂u 

 

∂x 

∂v 

 

∂x 

∂w 

2 

2 

2 

 

∂y 

+ 

∂u 

 

∂y 

+ 

∂v 

 

∂y 

+ 

∂w 

2 

2 

2 ∂z 

+ , 

∂u 

2 ∂z 

+ , 

∂v 

2 ∂z 

+ . 

∂w 

(b) Show that dr can be expressed in the form dr = h1 eu du + h2 ev dv + h3 ew dw. 

(c) Show that the volume of the elemental parallelepiped having dr as diagonal can be represented 

Hint: 

dτ = √ gdudvdw= Jdudvdw= 

| A · ( B × 

 

 

C)| = 

 

 

2 

∂(x, y, z) 

∂(u, v, w) dudvdw. 

A1 A2 A3 

B1 B2 B3 

C1 C2 C3 

 

 

 

 

 

 

101

102 

Figure 1.3-18 Oblique cylindrical coordinates. 

◮ 6. For the change dr given in problem 5, show the elemental parallelepiped with diagonal dr has: 

(a) the element of area dS1 = 

dvdw in the u =constant surface. 

 

(b) The element of area dS2 = 

 

(c) the element of area dS3 = 

g22g33 − g 2 23 

g33g11 − g 2 13 

g11g22 − g 2 12 

dudw in the v =constant surface. 

dudv in the w =constant surface. 

(d) What do the above elements of area reduce to in the special case the curvilinear coordinates are orthogonal? 

Hint: 

| A × 

B| = ( A × B) · ( A × B) 

 

= ( A · A)( B · B) − ( A · B)( A · . 

B) 

◮ 7. In Cartesian coordinates you are given the affine transformation. xi = ℓijxj where 

x1 = 1 

15 (5x1 − 14x2 +2x3), x2 = − 1 

3 (2x1 + x2 +2x3), x3 = 1 

15 (10x1 +2x2 − 11x3) 

(a) Show the transformation is orthogonal. 

(b) A vector A(x1,x2,x3) in the unbarred system has the components 

A1 =(x1) 2 , A2 =(x2) 2 

A3 =(x3) 2 . 

Find the components of this vector in the barred system of coordinates. 

◮ 8. Calculate the metric and conjugate metric tensors in cylindrical coordinates (r, θ, z). 

◮ 9. Calculate the metric and conjugate metric tensors in spherical coordinates (ρ, θ, φ). 

◮ 10. Calculate the metric and conjugate metric tensors in parabolic cylindrical coordinates (ξ,η,z). 

◮ 11. Calculate the metric and conjugate metric components in elliptic cylindrical coordinates (ξ,η,z). 

◮ 12. Calculate the metric and conjugate metric components for the oblique cylindrical coordinates (r, φ, η), 

illustrated in figure 1.3-18, where x = r cosφ , y = r sin φ + η cos α, z = η sin α and α is a parameter 

0

◮ 13. Calculate the metric and conjugate metric tensor associated with the toroidal surface coordinates 

(ξ,η) illustrated in the figure 1.3-19, where 

x =(a + b cos ξ)cosη 

y =(a + b cos ξ)sinη 

z = b sin ξ 

a>b>0 

0

104 

Figure 1.3-20. Spherical surface coordinates 

◮ 18. Given the fourth order tensor Cikmp = λδikδmp + µ(δimδkp + δipδkm)+ν(δimδkp − δipδkm) whereλ, µ 

and ν are scalars and δij is the Kronecker delta. Show that under an orthogonal transformation of rotation of 

axes with xi = ℓijxj where ℓrsℓis = ℓmrℓmi = δri the components of the above tensor are unaltered. Any 

tensor whose components are unaltered under an orthogonal transformation is called an ‘isotropic’ tensor. 

Another way of stating this problem is to say “Show Cikmp is an isotropic tensor.” 

◮ 19. Assume Aijl is a third order covariant tensor and B pqmn is a fourth order contravariant tensor. Prove 

that AiklB klmn is a mixed tensor of order three, with one covariant and two contravariant indices. 

◮ 20. Assume that Tmnrs is an absolute tensor. Show that if Tijkl + Tijlk = 0 in the coordinate system x r 

then T ijkl + T ijlk = 0 in any other coordinate system x r . 


Hint: See problem 38, Exercise 1.1 

 

 

 

ɛijkɛrst = 

 

 

gir gis git 

gjr gjs gjt 

gkr gks gkt 

◮ 22. Determine if the tensor equation ɛmnpɛmij +ɛmnjɛmpi = ɛmniɛmpj is true or false. Justify your answer. 

◮ 23. Prove the epsilon identity g ij ɛiptɛjrs = gprgts − gpsgtr. Hint: See problem 38, Exercise 1.1 

◮ 24. Let A rs denote a skew-symmetric contravariant tensor and let cr = 1 

2 ɛrmnA mn where 

ɛrmn = √ germn. Show that cr are the components of a covariant tensor. Write out all the components. 

◮ 25. Let Ars denote a skew-symmetric covariant tensor and let c r = 1 

2 ɛrmnAmn where ɛ rmn = 1 

√ e 

g rmn . 

Show that cr are the components of a contravariant tensor. Write out all the components.

◮ 26. Let ApqB qs 

r = Cs pr where Bqs r is a relative tensor of weight ω1 and Cs pr is a relative tensor of weight 

ω2. Prove that Apq is a relative tensor of weight (ω2 − ω1). 

◮ 27. When Ai j is an absolute tensor prove that √ gAi j is a relative tensor of weight +1. 

◮ 28. When A i j 

1 

is an absolute tensor prove that √ 

g Ai j is a relative tensor of weight −1. 

◮ 29. 

(a) Show eijk is a relative tensor of weight +1. 

(b) Show ɛ ijk = 1 

√ g e ijk is an absolute tensor. Hint: See example 1.1-25. 

◮ 30. The equation of a surface can be represented by an equation of the form Φ(x 1 ,x 2 ,x 3 )=constant. 

Show that a unit normal vector to the surface can be represented by the vector 

n i = 

ij ∂Φ g ∂xj (g mn ∂Φ 

∂x m 

∂Φ 

∂x n ) 1 

2 

◮ 31. Assume that g ij = λgij with λ a nonzero constant. Find and calculate g ij in terms of g ij . 

◮ 32. Determine if the following tensor equation is true. Justify your answer. 

Hint: See problem 21, Exercise 1.1. 

ɛrjkA r i + ɛirkA r j + ɛijrA r k = ɛijkA r r . 

◮ 33. Show that for Ci and C i associated tensors, and C i = ɛ ijk AjBk, then Ci = ɛijkA j B k 

◮ 34. Prove that ɛ ijk and ɛijk are associated tensors. Hint: Consider the determinant of gij. 

◮ 35. Show ɛ ijk AiBjCk = ɛijkA i B j C k . 

◮ 36. Let T i j ,i,j=1, 2, 3 denote a second order mixed tensor. Show that the given quantities are scalar 

invariants. 

◮ 37. 

(i) I1 = T i 

i 

(ii) I2 = 1 

2 

(iii) I3 = det|T i j | 

. 

(T i 

i ) 2 − T i m 

(a) Assume Aij and Bij ,i,j=1, 2, 3 are absolute contravariant tensors, and determine if the inner product 

C ik = A ij B jk is an absolute tensor? 

(b) Assume that the condition ∂xj 

∂xn ∂xj ∂xm = δnm is satisfied, and determine whether the inner product in 

part (a) is a tensor? 

(c) Consider only transformations which are a rotation and translation of axes yi = ℓijyj + bi, where ℓij are 

direction cosines for the rotation of axes. Show that ∂yj ∂yj = δnm 

∂yn ∂ym 

m 

T 

i 

105

106 

◮ 38. For Aijk a Cartesian tensor, determine if a contraction on the indices i and j is allowed. That 

is, determine if the quantity Ak = Aiik, (summation on i) is a tensor. Hint: See part(c) of the previous 

problem. 

◮ 39. Prove the e-δ identity e ijk eimn = δ j mδ k n − δ j nδ k m. 

◮ 40. Consider the vector Vk, k=1, 2, 3 and define the matrix (aij) having the elements aij = eijkVk, 

where eijk is the e−permutation symbol. 

(a) Solve for Vi in terms of amn by multiplying both sides of the given equation by eijl and note the e − δ 

identity allows us to simplify the result. 

(b) Sum the given expression on k and then assign values to the free indices (i,j=1,2,3) and compare your 

results with part (a). 

(c) Is aij symmetric, skew-symmetric, or neither? 

◮ 41. It can be shown that the continuity equation of fluid dynamics can be expressed in the tensor form 

1 

√ g 

∂ 

∂x r (√ gϱV r )+ ∂ϱ 

∂t =0, 

where ϱ is the density of the fluid, t is time, V r ,withr =1, 2, 3 are the velocity components and g = |gij| 

is the determinant of the metric tensor. Employing the summation convention and replacing the tensor 

components of velocity by their physical components, express the continuity equation in 

(a) Cartesian coordinates (x, y, z) with physical components Vx,Vy,Vz. 

(b) Cylindrical coordinates (r, θ, z) with physical components Vr,Vθ,Vz. 

(c) Spherical coordinates (ρ, θ, φ) with physical components Vρ,Vθ,Vφ. 

◮ 42. Let x 1 ,x 2 ,x 3 denote a set of skewed coordinates with respect to the Cartesian coordinates y 1 ,y 2 ,y 3 . 

Assume that E1, E2, E3 are unit vectors in the directions of the x 1 ,x 2 and x 3 axes respectively. If the unit 

vectors satisfy the relations 

E1 · E1 =1 

E2 · E2 =1 

E3 · E3 =1 

then calculate the metrices gij and conjugate metrices g ij . 

E1 · E2 =cosθ12 

E1 · E3 =cosθ13 

E2 · E3 =cosθ23, 

◮ 43. Let Aij, i,j=1, 2, 3, 4 denote the skew-symmetric second rank tensor 

⎛ 

0 a b 

⎞ 

c 

⎜ −a 

Aij = ⎝ 

−b 

0 

−d 

d 

0 

e⎟ 

⎠ , 

f 

−c −e −f 0 

where a, b, c, d, e, f are complex constants. Calculate the components of the dual tensor 

V ij = 1 

2 eijkl Akl.

◮ 44. In Cartesian coordinates the vorticity tensor at a point in a fluid medium is defined 

ωij = 1 

 

∂Vj ∂Vi 

− 

2 ∂xi ∂xj 

where Vi are the velocity components of the fluid at the point. The vorticity vector at a point in a fluid 

medium in Cartesian coordinates is defined by ω i = 1 

2 eijkωjk. Show that these tensors are dual tensors. 

◮ 45. Write out the relation between each of the components of the dual tensors 

ˆT ij = 1 

2 eijklTkl i, j, k, l =1, 2, 3, 4 

and show that if ijkl is an even permutation of 1234, then ˆ T ij = Tkl. 

◮ 46. Consider the general affine transformation ¯xi = aijxj where (x1 ,x2 ,x3 )=(x, y, z) withinverse 

transformation xi = bij ¯xj. Determine (a) the image of the plane Ax + By + Cz + D = 0 under this 

transformation and (b) the image of a second degree conic section 

Ax 2 +2Bxy + Cy 2 + Dx + Ey + F =0. 

◮ 47. Using a multilinear form of degree M, derive the transformation law for a contravariant vector of 

degree M. 

◮ 48. Let g denote the determinant of gij and show that ∂g ∂gij 

= ggij . 

∂xk ∂xk ◮ 49. We have shown that for a rotation of xyz axes with respect to a set of fixed ¯x¯y¯z axes, the derivative 

of a vector A with respect to an observer on the barred axes is given by 

d 

A 

 

dt = 

f 

d 

A 

 

dt + ω × 

r 

A. 

Introduce the operators 

Df A = d 

A 

 

dt = derivative in fixed system 

f 

Dr A = d 

A 

 

dt = derivative in rotating system 

r 

(a) Show that Df A =(Dr + ω×) A. 

(b) Consider the special case that the vector A is the position vector r. Show that Dfr =(Dr + ω×)r 

produces 

 

V 

= 

f 

 

 

V 

+ ω × r where 

r 

 

 

V 

represents the velocity of a particle relative to the fixed system 

 

f 

 

and V 

represents the velocity of a particle with respect to the rotating system of coordinates. 

r 

 

 

 

(c) Show that a 

= a 

+ ω × (ω × r) wherea 

represents the acceleration of a particle relative to the 

f r 

f 

 

fixed system and a represents the acceleration of a particle with respect to the rotating system. 

r 

(d) Show in the special case ω is a constant that 

 

 

a =2ω × V + ω × (ω × r) 

f 

where V is the velocity of the particle relative to the rotating system. The term 2ω × V is referred to 

as the Coriolis acceleration and the term ω × (ω × r) is referred to as the centripetal acceleration. 

107

108 

§1.4 DERIVATIVE OF A TENSOR 

In this section we develop some additional operations associated with tensors. Historically, one of the 

basic problems of the tensor calculus was to try and find a tensor quantity which is a function of the metric 

tensor gij and some of its derivatives ∂gij 

∂ 2 gij 

, 

∂xm ∂xm , .... A solution of this problem is the fourth order 

∂xn Riemann Christoffel tensor Rijkl to be developed shortly. In order to understand how this tensor was arrived 

at, we must first develop some preliminary relationships involving Christoffel symbols. 

Christoffel Symbols 

Let us consider the metric tensor gij which we know satisfies the transformation law 

Define the quantity 

(α, β, γ) = ∂gαβ ∂gab 

γ = 

∂x ∂xc ∂xc ∂xγ ∂xa ∂xα ∂x 

gαβ = gab 

a 

∂xα ∂xb . β ∂x 

∂x b 

∂x 

∂ 

+ gab β 2xa ∂xα∂xγ ∂x b 

∂x 

β + gab 

∂xa ∂xα ∂ 2 x b 

∂x β ∂x γ 

and form the combination of terms 1 

[(α, β, γ)+(β,γ,α) − (γ,α,β)] to obtain the result 

2 

 

1 ∂gαβ 

2 ∂xγ + ∂gβγ ∂xα − ∂gγα ∂xβ 

= 1 

 

∂gab 

2 ∂x 

∂x 

∂x b 

∂gbc ∂gca 

+ − c a 

∂x a 

∂x α 

∂x b 

∂x β 

∂x c 

∂x 

γ + gab 

∂xb ∂xβ ∂2xa ∂xα γ . (1.4.1) 

∂x 

In this equation the combination of derivatives occurring inside the brackets is called a Christoffel symbol 

of the first kind and is defined by the notation 

[ac, b] =[ca, b] = 1 

 

∂gab ∂gbc ∂gac 

+ − 

2 ∂xc ∂xa ∂xb 

. (1.4.2) 

The equation (1.4.1) defines the transformation for a Christoffel symbol of the first kind and can be expressed 

as 

[α γ,β]=[ac, b] ∂xa 

∂xα ∂xb ∂xβ ∂xc ∂ 

γ + gab 

∂x 2xa ∂xα∂xγ ∂xb . (1.4.3) 

β ∂x 

Observe that the Christoffel symbol of the first kind [ac, b] does not transform like a tensor. However, it is 

symmetric in the indices a and c. 

At this time it is convenient to use the equation (1.4.3) to develop an expression for the second derivative 

term which occurs in that equation as this second derivative term arises in some of our future considerations. 

To solve for this second derivative we can multiply equation (1.4.3) by ∂xβ 

∂xd gde and simplify the result to the 

form 

∂2xe ∂xα∂x γ = −gde [ac, d] ∂xa 

∂xα ∂xc γ + [αγ,β]∂xβ 

∂x ∂xd gde . (1.4.4) 

The transformation g de λµ ∂xd 

= g 

∂xλ ∂xe µ allows us to express the equation (1.4.4) in the form 

∂x 

∂2xe ∂xα∂xγ = −gde [ac, d] ∂xa 

∂xα ∂xc ∂xγ + gβµ [αγ,β] ∂xe 

µ . (1.4.5) 

∂x

Define the Christoffel symbol of the second kind as 

 

i i 

= = g 

jk kj 

iα [jk,α]= 1 

2 giα 

 

∂gkα 

∂x 

∂x 

∂x α 

∂gjα ∂gjk 

+ − j k 

 

. (1.4.6) 

This Christoffel symbol of the second kind is symmetric in the indices j and k and from equation (1.4.5) we 

see that it satisfies the transformation law 

e 

a 

µ ∂x e ∂x 

µ = 

αγ ∂x ac ∂xα ∂xc ∂xγ + ∂2xe ∂xα γ . (1.4.7) 

∂x 

Observe that the Christoffel symbol of the second kind does not transform like a tensor quantity. We can use 

the relation defined by equation (1.4.7) to express the second derivative of the transformation equations in 

terms of the Christoffel symbols of the second kind. At times it will be convenient to represent the Christoffel 

symbols with a subscript to indicate the metric from which they are calculated. Thus, an alternative notation 

 

 

i 

i 

for is the notation . 

jk 

jk 

g 

EXAMPLE 1.4-1. (Christoffel symbols) Solve for the Christoffel symbol of the first kind in terms of 

the Christoffel symbol of the second kind. 

Solution: By the definition from equation (1.4.6) we have 

 

i 

= g 

jk 

iα [jk,α]. 

We multiply this equation by gβi and find 

and so 

[jk,α]=gαi 

gβi 

 

i 

= δ 

jk 

α β [jk,α]=[jk,β] 

 

 

i 

1 

N 

= gα1 + ···+ gαN . 

jk jk 

jk 

EXAMPLE 1.4-2. (Christoffel symbols of first kind) 

Derive formulas to find the Christoffel symbols of the first kind in a generalized orthogonal coordinate 

system with metric coefficients 

gij =0 for i = j and g (i)(i) = h 2 (i) , i =1, 2, 3 

where i is not summed. 

Solution: In an orthogonal coordinate system where gij =0fori= j we observe that 

[ab, c] = 1 

 

∂gac 

2 ∂x 

∂x 

∂gbc ∂gab 

+ − b a 

Here there are 3 3 = 27 quantities to calculate. We consider the following cases: 

∂x c 

 

. (1.4.8) 

109

110 

CASE I Let a = b = c = i, then the equation (1.4.8) simplifies to 

[ab, c] =[ii, i] = 1 ∂gii 

2 ∂xi (no summation on i). (1.4.9) 

From this equation we can calculate any of the Christoffel symbols 

[11, 1], [22, 2], or [33, 3]. 

CASE II Let a = b = i = c, then the equation (1.4.8) simplifies to the form 

[ab, c] =[ii, c] =− 1 ∂gii 

2 ∂xc (no summation on i and i = c). (1.4.10) 

since, gic =0fori= c. This equation shows how we may calculate any of the six Christoffel symbols 

[11, 2], [11, 3], [22, 1], [22, 3], [33, 1], [33, 2]. 

CASE III Let a = c = i = b, and noting that gib =0fori= b, it can be verified that the equation (1.4.8) 

simplifies to the form 

[ab, c] =[ib, i] =[bi, i] = 1 ∂gii 

2 ∂xb (no summation on i and i = b). (1.4.11) 

From this equation we can calculate any of the twelve Christoffel symbols 

[12, 1] = [21, 1] 

[32, 3] = [23, 3] 

[13, 1] = [31, 1] 

[31, 3] = [13, 3] 

[21, 2] = [12, 2] 

[23, 2] = [32, 2] 

CASE IV Let a = b = c and show that the equation (1.4.8) reduces to 

This represents the six Christoffel symbols 

[ab, c] =0, (a = b = c.) 

[12, 3] = [21, 3] = [23, 1] = [32, 1] = [31, 2] = [13, 2] = 0. 

From the Cases I,II,III,IV all twenty seven Christoffel symbols of the first kind can be determined. In 

practice, only the nonzero Christoffel symbols are listed. 

EXAMPLE 1.4-3. (Christoffel symbols of the first kind)Find the nonzero Christoffel symbols of the 

first kind in cylindrical coordinates. 

Solution: From the results of example 1.4-2 we find that for x 1 = r, x 2 = θ, x 3 = z and 

g11 =1, g22 =(x 1 ) 2 = r 2 , g33 =1 

the nonzero Christoffel symbols of the first kind in cylindrical coordinates are: 

[22, 1] = − 1 ∂g22 

2 ∂x1 = −x1 = −r 

[21, 2] = [12, 2] = 1 ∂g22 

2 ∂x1 = x1 = r.

EXAMPLE 1.4-4. (Christoffel symbols of the second kind) 

Find formulas for the calculation of the Christoffel symbols of the second kind in a generalized orthogonal 

coordinate system with metric coefficients 

gij =0 for i = j and g (i)(i) = h 2 (i) , i =1, 2, 3 

where i is not summed. 

Solution: By definition we have 

 

i 

= g 

jk 

im [jk,m]=g i1 [jk,1] + g i2 [jk,2] + g i3 [jk,3] (1.4.12) 

By hypothesis the coordinate system is orthogonal and so 

g ij =0 for i = j and g ii = 1 

gii 

i not summed. 

The only nonzero term in the equation (1.4.12) occurs when m = i and consequently 

 

i 

= g 

jk 

ii [jk,i]= [jk,i] 

gii 

no summation on i. (1.4.13) 

We can now consider the four cases considered in the example 1.4-2. 

CASE I Let j = k = i and show 

 

i 

= 

ii 

[ii, i] 

= 

gii 

1 ∂gii 1 ∂ 

= 

2gii ∂xi 2 ∂xi ln gii no summation on i. (1.4.14) 

CASE II Let k = j = i and show 

 

i 

= 

jj 

[jj,i] 

= 

gii 

−1 ∂gjj 

2gii ∂xi no summation on i or j. (1.4.15) 

CASE III Let i = j = k and verify that 

 

j j 

= = 

jk kj 

[jk,j] 

= 

gjj 

1 ∂gjj 1 ∂ 

= 

2gjj ∂xk 2 ∂xk ln gjj no summation on i or j. (1.4.16) 

CASE IV For the case i = j = k we find 

 

i 

= 

jk 

[jk,i] 

=0, 

gii 

i = j = k no summation on i. 

The above cases represent all 27 terms. 

111

112 

EXAMPLE 1.4-5. (Notation) In the case of cylindrical coordinates we can use the above relations and 

find the nonzero Christoffel symbols of the second kind: 

 

1 

= − 

22 

1 ∂g22 

2g11 ∂x1 = −x1 = −r 

 

2 2 

1 1 

= 

= = 

12 21 

1 x1 r 

= 1 ∂g22 

2g22 ∂x 

Note 1: The notation for the above Christoffel symbols are based upon the assumption that x 1 = r, x 2 = θ 

and x 3 = z. However, in tensor calculus the choice of the coordinates can be arbitrary. We could just as well 

have defined x 1 = z,x 2 = r and x 3 = θ. In this latter case, the numbering system of the Christoffel symbols 

changes. To avoid confusion, an alternate method of writing the Christoffel symbols is to use coordinates in 

place of the integers 1,2 and 3. For example, in cylindrical coordinates we can write 

 

θ θ 

= = 

rθ θr 

1 

 

r 

and = −r. 

r θθ 

If we define x 1 = r, x 2 = θ, x 3 = z, then the nonzero Christoffel symbols are written as 

 

2 2 

= = 


1 

r 

and 

 

1 

= −r. 

22 

In contrast, if we define x 1 = z,x 2 = r, x 3 = θ, then the nonzero Christoffel symbols are written 

 

3 3 

= = 

23 32 

1 

r 

and 

 

2 

= −r. 

33 

Note 2: Some textbooks use the notation Γa,bc for Christoffel symbols of the first kind and Γ d bc = gda Γa,bc for 

Christoffel symbols of the second kind. This notation is not used in these notes since the notation suggests 

that the Christoffel symbols are third order tensors, which is not true. The Christoffel symbols of the first 

and second kind are not tensors. This fact is clearly illustrated by the transformation equations (1.4.3) and 

(1.4.7). 

Covariant Differentiation 

Let Ai denote a covariant tensor of rank 1 which obeys the transformation law 

∂x 

Aα = Ai 

i 

α . (1.4.17) 

∂x 

Differentiate this relation with respect to xβ and show 

∂Aα 

∂x 

β = Ai 

∂ 2 x i 

∂xα ∂Ai 

+ β ∂x ∂xj ∂x j 

∂x β 

∂xi α . (1.4.18) 

∂x 

Now use the relation from equation (1.4.7) to eliminate the second derivative term from (1.4.18) and express 

it in the form 

i 

j 

∂Aα σ ∂x i ∂x 

= Ai β σ − 

∂x αβ ∂x jk ∂xα ∂xk ∂xβ 

+ ∂Ai 

∂xj ∂xj ∂xβ ∂xi α . 

∂x 

(1.4.19)

Employing the equation (1.4.17), with α replaced by σ, the equation (1.4.19) is expressible in the form 

 

∂Aα σ 

− Aσ = β ∂x αβ 

∂Aj 

∂xk ∂xj ∂xα ∂xk j i ∂x 

− Ai β ∂x jk ∂xα ∂xk ∂xβ (1.4.20) 

or alternatively 

∂Aα 

− Aσ β ∂x 

 

σ 

αβ 

j 

∂Aj i ∂x 

= − Ai 

∂xk jk ∂xα ∂xk . (1.4.21) 

β ∂x 


Aj,k = ∂Aj 

 

i 

− Ai 

(1.4.22) 

∂xk jk 

as the covariant derivative of Aj with respect to xk . The equation (1.4.21) demonstrates that the covariant 

derivative of a covariant tensor produces a second order tensor which satisfies the transformation law 

Aα,β = Aj,k 

∂xj ∂xα Other notations frequently used to denote the covariant derivative are: 

∂xk . (1.4.23) 

β ∂x 

Aj,k = Aj;k = A j/k = ∇kAj = Aj|k. (1.4.24) 

In the special case where gij are constants the Christoffel symbols of the second kind are zero, and consequently 

the covariant derivative reduces to Aj,k = ∂Aj 

. That is, under the special circumstances where the 

∂xk Christoffel symbols of the second kind are zero, the covariant derivative reduces to an ordinary derivative. 

Covariant Derivative of Contravariant Tensor 

form 

A contravariant tensor A i obeys the transformation law A i = A 

A i α ∂xi 

= A 

∂xα α ∂xi 

which can be expressed in the 

∂xα (1.4.24) 

by interchanging the barred and unbarred quantities. We write the transformation law in the form of equation 

(1.4.24) in order to make use of the second derivative relation from the previously derived equation (1.4.7). 

Differentiate equation (1.4.24) with respect to xj to obtain the relation 

∂Ai ∂xj = Aα ∂2xi ∂xα∂x β 

∂xβ ∂Aα 

+ 

∂xj ∂xβ ∂xβ ∂xj ∂xi α . (1.4.25) 

∂x 

Changing the indices in equation (1.4.25) and substituting for the second derivative term, using the relation 

from equation (1.4.7), produces the equation 

∂Ai i 

m 

σ ∂x i ∂x 

= Aα 

∂xj σ − 

αβ ∂x mk ∂xα ∂xk ∂xβ 

∂xβ ∂Aα 

+ 

∂xj ∂xβ ∂xβ ∂xj ∂xi α . (1.4.26) 

∂x 

Applying the relation found in equation (1.4.24), with i replaced by m, together with the relation 

∂x β 

∂x j 

∂x k 

∂x β = δk j , 

113

114 

we simplify equation (1.4.26) to the form 

i ∂A 

+ 

∂xj 

i 

A 

mj 

m 

 

= 

∂A σ 

+ β ∂x 

 

σ 

A 

αβ 

α 

 

∂xβ ∂xj ∂xi σ . (1.4.27) 

∂x 


A i ,j = ∂Ai 

 

i 

+ A 

∂xj mj 

m 

(1.4.28) 

as the covariant derivative of the contravariant tensor Ai . The equation (1.4.27) demonstrates that a covariant 

derivative of a contravariant tensor will transform like a mixed second order tensor and 

A i ,j = A σ ∂x 

,β 

β 

∂xj ∂xi σ . (1.4.29) 

∂x 

Again it should be observed that for the condition where gij are constants we have A i ∂Ai 

,j = and the 

∂xj covariant derivative of a contravariant tensor reduces to an ordinary derivative in this special case. 

In a similar manner the covariant derivative of second rank tensors can be derived. We find these 

derivatives have the forms: 

Aij,k = ∂Aij 

 

σ 

σ 

− Aσj − Aiσ 

∂xk ik jk 

A i j,k = ∂Aij ∂xk + Aσ 

i 

j − A 

σk 

i 

σ 

σ 

jk 

A ij 

 

∂Aij i 

,k = + Aσj + A 

∂xk σk 

iσ 

 

j 

. 

σk 

(1.4.30) 

In general, the covariant derivative of a mixed tensor 

of rank n has the form 

A ij...k 

lm...p,q 

= ∂Aij...k 

lm...p 

∂x q 

+ Aσj...k 

lm...p 

− A ij...k 

σm...p 

A ij...k 

lm...p 

 

i 

+ A 

σq 

iσ...k 

 

j 

lm...p + ···+ A 

σq 

ij...σ 

 

k 

lm...p σq 

 

σ 

− A 

lq 

ij...k 

 

σ 

lσ...p −···−A 

mq 

ij...k 

 

σ 

lm...σ pq 

(1.4.31) 

and this derivative is a tensor of rank n +1. Note the pattern of the + signs for the contravariant indices 

and the − signs for the covariant indices. 

Observe that the covariant derivative of an nth order tensor produces an n+1st order tensor, the indices 

of these higher order tensors can also be raised and lowered by multiplication by the metric or conjugate 

metric tensor. For example we can write 

g im Ajk|m = Ajk| i 

and g im A jk |m = A jk | i

Rules for Covariant Differentiation 

The rules for covariant differentiation are the same as for ordinary differentiation. That is: 

(i) The covariant derivative of a sum is the sum of the covariant derivatives. 

(ii) The covariant derivative of a product of tensors is the first times the covariant derivative of the second 

plus the second times the covariant derivative of the first. 

(iii) Higher derivatives are defined as derivatives of derivatives. Be careful in calculating higher order derivativesasingeneral 

Ai,jk = Ai,kj. 

EXAMPLE 1.4-6. (Covariant differentiation) Calculate the second covariant derivative Ai,jk. 

Solution: The covariant derivative of Ai is 

Ai,j = ∂Ai 

 

σ 

− Aσ . 

∂xj ij 

By definition, the second covariant derivative is the covariant derivative of a covariant derivative and hence 

Ai,jk =(Ai,j) ,k = ∂ 

∂xk 

∂Ai σ 

m 

m 

− Aσ − Am,j − Ai,m . 

∂xj ij 

ik 

jk 

Simplifying this expression one obtains 

Ai,jk = ∂2 Ai 

− 

∂xj ∂Aσ 

− 

∂xk ∂xk 

σ ∂ 

− Aσ 

ij ∂xk 

σ 

ij 

 

∂Am σ m ∂Ai σ m 

− Aσ 

− − Aσ 

. 

∂xj mj ik ∂xm im jk 

Rearranging terms, the second covariant derivative can be expressed in the form 

∂xj ∂Aσ 

− 

∂xk ∂xk 

σ 

− 

ij 

∂Am 

∂xj 

m 

− 

ik 

∂Ai 

∂xm 

m 

jk 

 

∂ 

− Aσ 

∂xk 

σ σ m m σ 

− 

− 

. 

ij im jk ik mj 

Ai,jk = ∂2 Ai 

(1.4.32) 

115

116 

Riemann Christoffel Tensor 

where 

Utilizing the equation (1.4.32), it is left as an exercise to show that 

R σ ijk 

Ai,jk − Ai,kj = AσR σ ijk 

∂ 

= 

∂xj 

σ 

− 

ik 

∂ 

∂xk 

σ 

+ 

ij 

 

m σ 

− 

ik mj 

is called the Riemann Christoffel tensor. The covariant form of this tensor is 

 

m σ 

 

ij mk 

(1.4.33) 

Rhjkl = gihR i jkl. (1.4.34) 

It is an easy exercise to show that this covariant form can be expressed in either of the forms 

Rinjk = ∂ 

 

∂ 

s 

s 

[nk, i] − [nj, i]+[ik, s] − [ij, s] 

∂xj ∂xk nj 

nk 

or Rijkl = 1 

2 ∂ gil 

2 ∂xj∂xk − ∂2gjl ∂xi∂xk − ∂2gik ∂xj∂xl + ∂2gjk ∂xi∂xl 

+ g αβ ([jk,β][il, α] − [jl,β][ik, α]) . 

From these forms we find that the Riemann Christoffel tensor is skew symmetric in the first two indices 

and the last two indices as well as being symmetric in the interchange of the first pair and last pairs of 

indices and consequently 

Rjikl = −Rijkl Rijlk = −Rijkl Rklij = Rijkl. 

In a two dimensional space there are only four components of the Riemann Christoffel tensor to consider. 

These four components are either +R1212 or −R1212 since they are all related by 

R1212 = −R2112 = R2121 = −R1221. 

In a Cartesian coordinate system Rhijk = 0. The Riemann Christoffel tensor is important because it occurs 

in differential geometry and relativity which are two areas of interest to be considered later. Additional 

properties of this tensor are found in the exercises of section 1.5.

Physical Interpretation of Covariant Differentiation 

In a system of generalized coordinates (x 1 ,x 2 ,x 3 ) we can construct the basis vectors ( E1, E2, E3). These 

basis vectors change with position. That is, each basis vector is a function of the coordinates at which they 

are evaluated. We can emphasize this dependence by writing 

Ei = Ei(x 1 ,x 2 ,x 3 )= ∂r 

∂x i 

i =1, 2, 3. 

Associated with these basis vectors we have the reciprocal basis vectors 

E i = E i (x 1 ,x 2 ,x 3 ), i =1, 2, 3 

which are also functions of position. A vector A can be represented in terms of contravariant components as 

A = A 1 E1 + A 2 E2 + A 3 E3 = A j Ej 

or it can be represented in terms of covariant components as 

A change in the vector A is represented as 

where from equation (1.4.35) we find 

(1.4.35) 

A = A1 E 1 + A2 E 2 + A3 E 3 = Aj E j . (1.4.36) 

d A = ∂ A 

dxk 

∂xk ∂ A 

∂xk = Aj ∂ Ej ∂Aj 

+ 

∂xk or alternatively from equation (1.4.36) we may write 

∂ A 

= Aj 

∂xk ∂x k Ej 

We define the covariant derivative of the covariant components as 

Ai,k = ∂ A 

∂xk · Ei = ∂Ai 

∂x 

(1.4.37) 

∂ Ej ∂Aj 

+ 

∂xk ∂xk E j . (1.4.38) 

k + Aj 

The covariant derivative of the contravariant components are defined by the relation 

Introduce the notation 

∂ 

Ej m 

= Em 

∂xk jk 

We then have 

∂ E j 

∂x k · Ei. (1.4.39) 

A i ,k = ∂ A 

∂x k · E i = ∂Ai 

∂x k + Aj ∂ Ej 

∂x k · E i . (1.4.40) 

and 

E i · ∂ 

Ej m 

= Em · 

∂xk jk 

E i 

m 

= δ 

jk 

i m = 

 

i 

jk 

∂ Ej 

j 

= − E 

∂xk mk 

m . (1.4.41) 

(1.4.42) 

117

118 

and 

Ei · ∂ Ej 

j 

= − E 

∂xk mk 

m · 

j 

Ei = − δ 

mk 

m 

j 

i = − . 

ik 

(1.4.43) 

Then equations (1.4.39) and (1.4.40) become 

Ai,k = ∂Ai 

 

j 

− 

∂xk ik 

A i ,k 

∂Ai 

= + 

∂xk i 

jk 

 

Aj 

 

A j , 

which is consistent with our earlier definitions from equations (1.4.22) and (1.4.28). Here the first term of 

the covariant derivative represents the rate of change of the tensor field as we move along a coordinate curve. 

The second term in the covariant derivative represents the change in the local basis vectors as we move 

along the coordinate curves. This is the physical interpretation associated with the Christoffel symbols of 

the second kind. 

We make the observation that the derivatives of the basis vectors in equations (1.4.39) and (1.4.40) are 

related since 


Hence we can express equation (1.4.39) in the form 

Ei · E j = δ j 

i 

∂ 

∂xk ( Ei · E j )= Ei · ∂ Ej ∂xk + ∂ Ei 

∂xk · E j =0 

or 

Ai,k = ∂Ai 

Ei · ∂ Ej ∂xk = − E j · ∂ Ei 

∂xk ∂xk − Aj E j · ∂ Ei 

∂x 

We write the first equation in (1.4.41) in the form 

∂ 

Ej m 

= gim 

∂xk jk 

E i =[jk,i] E i 


and 

∂ Ej 

∂xk · E m 

i 

= Ei · 

jk 

E m 

i 

= δ 

jk 

m 

m 

i = 

jk 

∂ Ej 

∂x k · Em =[jk,i] E i · Em =[jk,i]δ i m =[jk,m]. 

k . (1.4.44) 

(1.4.45) 

(1.4.46) 

These results also reduce the equations (1.4.40) and (1.4.44) to our previous forms for the covariant derivatives. 

The equations (1.4.41) are representations of the vectors ∂ Ei 

∂x k and ∂ E j 

∂x k in terms of the basis vectors and 

reciprocal basis vectors of the space. The covariant derivative relations then take into account how these 

vectors change with position and affect changes in the tensor field. 

The Christoffel symbols in equations (1.4.46) are symmetric in the indices j and k since 

∂ Ej ∂ 

= 

∂xk ∂xk 

∂r 

∂xj 

= ∂ 

∂xj 

∂r 

∂xk 

= ∂ Ek 

. (1.4.47) 

∂xj

The equations (1.4.46) and (1.4.47) enable us to write 

[jk,m]= Em · ∂ 

Ej 1 

= Em · 

∂xk 2 

∂ Ej 

∂xk + Em · ∂ Ek 

∂xj 

= 1 

 

∂ 

2 ∂xk 

Em · 

Ej + ∂ 

∂xj 

Em · 

Ek − Ej · ∂ Em 

∂xk − Ek · ∂ Em 

∂xj 

= 1 

 

∂ 

2 ∂xk 

Em · 

Ej + ∂ 

∂xj 

Em · 

Ek − Ej · ∂ Ek 

∂xm − Ek · ∂ Ej 

∂xm 

= 1 

 

∂ 

2 ∂xk 

Em · 

Ej + ∂ 

∂xj 

Em · 

Ek − ∂ 

∂xm 

Ej · 

Ek 

 

= 1 

 

∂gmj ∂gmk ∂gjk 

+ − 

2 ∂xk ∂xj ∂xm 

=[kj, m] 

which again agrees with our previous result. 

For future reference we make the observation that if the vector A is represented in the form A = AjEj, 

involving contravariant components, then we may write 

d A = ∂ A 

∂xk dxk 

∂A 

= 

j 

∂xk Ej + A j ∂ Ej 

∂xk 

dx k 

j ∂A 

= 

∂xk Ej + A j 

 

i 

Ei dx 

jk 

k 

 

j ∂A j 

= + A 

∂xk mk 

m 

 

Ej dx k = A j 

,k dxk (1.4.48) 

Ej. 

Similarly, if the vector A is represented in the form A = Aj Ej involving covariant components it is left as 

an exercise to show that 

Ricci’s Theorem 

d A = Aj,k dx k E j 

Ricci’s theorem states that the covariant derivative of the metric tensor vanishes and gik,l =0. 

Proof: We have 

gik,l = ∂gik 

 

m m 

− gim − gmk 

∂xl kl il 

(1.4.49) 

gik,l = ∂gik 

− [kl, i] − [il, k] 

∂xl gik,l = ∂gik 

 

1 ∂gik ∂gil ∂gkl 

− + − 

∂xl 2 ∂xl ∂xk ∂xi 

− 1 

 

∂gik ∂gkl ∂gil 

+ − 

2 ∂xl ∂xi ∂xk 

=0. 

Because of Ricci’s theorem the components of the metric tensor can be regarded as constants during covariant 

differentiation. 

EXAMPLE 1.4-7. (Covariant differentiation) Show that δ i j,k =0. 

Solution 

δ i j,k = ∂δi j 

∂xk + δσ 

i 

j − δ 

σk 

i 

σ i i 

σ = − =0. 

jk jk jk 

119


EXAMPLE 1.4-8. (Covariant differentiation) Show that g ij 

,k =0. 

Solution: Since gijg jk = δk i we take the covariant derivative of this expression and find 

(gijg jk ),l = δ k i,l =0 

gijg jk 

,l + gij,lg jk =0. 

But gij,l = 0 by Ricci’s theorem and hence gijg jk 

,l =0. We multiply this expression by gim and obtain 

g im gijg jk 

,l = δm j g jk 

,l = gmk ,l =0 

which demonstrates that the covariant derivative of the conjugate metric tensor is also zero. 

EXAMPLE 1.4-9. (Covariant differentiation) Some additional examples of covariant differentiation 

are: 

Intrinsic or Absolute Differentiation 

(i) (gilA l ),k = gilA l ,k = Ai,k 

(ii) (gimgjnA ij ) ,k = gimgjnA ij 

,k = Amn,k 

The intrinsic or absolute derivative of a covariant vector Ai taken along a curve x i = x i (t),i =1,...,N 

is defined as the inner product of the covariant derivative with the tangent vector to the curve. The intrinsic 

derivative is represented 

δAi dx 

= Ai,j 

δt j 

dt 

δAi 

δt = 

j 

∂Ai α dx 

− Aα 

∂xj ij dt 

j 

δAi dAi α dx 

= − Aα 

δt dt ij dt . 

(1.4.50) 

Similarly, the absolute or intrinsic derivative of a contravariant tensor A i is represented 

δAi δt = Ai dx 

,j 

j 

dt 

dAi 

= 

dt + 

 

i k dxj 

A 

jk dt . 

The intrinsic or absolute derivative is used to differentiate sums and products in the same manner as used 

in ordinary differentiation. Also if the coordinate system is Cartesian the intrinsic derivative becomes an 

ordinary derivative. 

The intrinsic derivative of higher order tensors is similarly defined as an inner product of the covariant 

derivative with the tangent vector to the given curve. For example, 

δA ij 

klm 

δt 

dx 

= Aij 

klm,p 

p 

dt 

is the intrinsic derivative of the fifth order mixed tensor A ij 

klm .

EXAMPLE 1.4-10. (Generalized velocity and acceleration) Let t denote time and let xi = xi (t) 

for i =1,...,N, denote the position vector of a particle in the generalized coordinates (x1 ,...,xN ). From 

the transformation equations (1.2.30), the position vector of the same particle in the barred system of 

coordinates, (x 1 , x 2 ,...,x N ), is 

x i = x i (x 1 (t),x 2 (t),...,x N (t)) = x i (t), i =1,...,N. 

The generalized velocity is v i = dxi 

dt ,i=1,...,N. The quantity vi transforms as a tensor since by definition 

v i = dxi 

dt 

∂xi 

= 

∂xj dxj dt 

= ∂xi 

∂x j vj . (1.4.51) 

Let us now find an expression for the generalized acceleration. Write equation (1.4.51) in the form 

and differentiate with respect to time to obtain 

The equation (1.4.53) demonstrates that dvi 

dt 

v j i ∂xj 

= v 

∂xi dvj dt = vi ∂2xj ∂xi∂x k 

dxk dvi ∂x 

+ 

dt dt 

j 

∂xi (1.4.52) 

(1.4.53) 

does not transform like a tensor. From the equation (1.4.7) 

previously derived, we change indices and write equation (1.4.53) in the form 

Rearranging terms we find 

dv j 

dt 

= vi dxk 

dt 

j 

a 

σ ∂x j ∂x 

σ − 

ik ∂x ac ∂xi ∂xc ∂xk 

+ ∂xj 

∂xi dvi dt . 

∂vj ∂xk dxk dt + 

a 

c 

j ∂x ∂x 

vi 

ac i ∂x ∂xk dxk 

= 

dt 

∂xj 

∂xi ∂vi ∂xk dxk dt + 

 

σ i ∂xj 

v 

ik ∂xσ dxk dt 

 

j ∂v j 

+ v 

∂xk ak 

a 

k dx 

dt = 

 

∂vσ 

σ 

+ v k ∂x ik 

i 

 

dxk ∂x 

dt 

j 

∂xσ δv j 

δt 

= δvσ 

δt 

∂xj σ . 

∂x 

The above equation illustrates that the intrinsic derivative of the velocity is a tensor quantity. This derivative 

is called the generalized acceleration and is denoted 

f i = δvi 

δt = vi dx 

,j 

j 

dt 

dvi 

= 

dt + 

 

i 

mn 

To summarize, we have shown that if 

v m v n = d2xi + 

dt2 

i 

m dx 

mn dt 

x i = x i (t), i =1,...,N is the generalized position vector, then 

v i = dxi 

, i =1,...,N is the generalized velocity, and 

dt 

f i = δvi 

δt = vi ,j 

dxj , i =1,...,N is the generalized acceleration. 

dt 

or 

dxn , i =1,...,N (1.4.54) 

dt 

121


Parallel Vector Fields 

Let y i = y i (t), i =1, 2, 3 denote a space curve Cin a Cartesian coordinate system and let Y i define a 

constant vector in this system. Construct at each point of the curve C the vector Y i . This produces a field 

of parallel vectors along the curve C. What happens to the curve and the field of parallel vectors when we 

transform to an arbitrary coordinate system using the transformation equations 

with inverse transformation 

y i = y i (x 1 ,x 2 ,x 3 ), i =1, 2, 3 

x i = x i (y 1 ,y 2 ,y 3 ), i =1, 2, 3? 

The space curve Cin the new coordinates is obtained directly from the transformation equations and can 

be written 

x i = x i (y 1 (t),y 2 (t),y 3 (t)) = x i (t), i =1, 2, 3. 

The field of parallel vectors Y i become X i in the new coordinates where 

Y i = X 

j ∂yi 

. (1.4.55) 

∂xj Since the components of Y i are constants, their derivatives will be zero and consequently we obtain by 

differentiating the equation (1.4.55), with respect to the parameter t, that the field of parallel vectors Xi must satisfy the differential equation 

dX j 

dt 

∂yi ∂xj + Xj ∂2y i 

∂xj∂xm dxm dt 

= dY i 

dt 

=0. (1.4.56) 

Changing symbols in the equation (1.4.7) and setting the Christoffel symbol to zero in the Cartesian system 

of coordinates, we represent equation (1.4.7) in the form 

∂2y i 

∂xj i α ∂y 

= 

∂xm jm ∂xα and consequently, the equation (1.4.56) can be reduced to the form 

δX j 

δt 

j dX 

= 

dt + 

 

j k dxm 

X 

km dt 

=0. (1.4.57) 

The equation (1.4.57) is the differential equation which must be satisfied by a parallel field of vectors X i 

along an arbitrary curve x i (t).

EXERCISE 1.4 

◮ 1. Find the nonzero Christoffel symbols of the first and second kind in cylindrical coordinates 

(x 1 ,x 2 ,x 3 )=(r, θ, z), where x = r cos θ, y = r sin θ, z = z. 

◮ 2. Find the nonzero Christoffel symbols of the first and second kind in spherical coordinates 

(x 1 ,x 2 ,x 3 )=(ρ, θ, φ), where x = ρ sin θ cos φ, y = ρ sin θ sin φ, z = ρ cos θ. 

◮ 3. Find the nonzero Christoffel symbols of the first and second kind in parabolic cylindrical coordinates 

(x 1 ,x 2 ,x 3 )=(ξ,η,z), where x = ξη, y = 1 

2 (ξ2 − η 2 ), z = z. 

◮ 4. Find the nonzero Christoffel symbols of the first and second kind in parabolic coordinates 

(x 1 ,x 2 ,x 3 )=(ξ,η,φ), where x = ξη cos φ, y = ξη sin φ, z = 1 

2 (ξ2 − η 2 ). 

◮ 5. Find the nonzero Christoffel symbols of the first and second kind in elliptic cylindrical coordinates 

(x 1 ,x 2 ,x 3 )=(ξ,η,z), where x =coshξ cos η, y =sinhξ sin η, z = z. 

◮ 6. Find the nonzero Christoffel symbols of the first and second kind for the oblique cylindrical coordinates 

(x 1 ,x 2 ,x 3 )=(r, φ, η), where x = r cos φ, y = r sin φ+η cos α, z = η sin α with 0


◮ 13. 

(a) Show √ g is a relative scalar of weight +1. 

(b) Use the results from problem 9(c) and problem 43 of the exercises to show that 

( √ g),k = ∂√ 

g m √g 

− 

=0. 

∂xk 

km 

m 

(c) Show that = 

km 

∂ 

∂xk ln(√g)= 1 ∂g 

. 

2g ∂xk 

m 

◮ 14. Use the result from problem 9(b) to show = 

km 

∂ 

∂xk ln(√g)= 1 ∂g 

. 

2g ∂xk Hint: Expand the covariant derivative ɛrst,p and then substitute ɛrst = √ gerst. Simplify by inner 

multiplication with erst 

√ 

g and note the Exercise 1.1, problem 26. 

◮ 15. Calculate the covariant derivative A i ,m and then contract on m and i to show that 

A i ,i = 1 

√ g 

∂ 

∂xi √ i 

gA . 

◮ 16. Show 1 ∂ 

√ 

g ∂xj √ ij 

gg 

i 

+ g 

pq 

pq =0. Hint: See problem 14. 

◮ 17. Prove that the covariant derivative of a sum equals the sum of the covariant derivatives. 

Hint: Assume Ci = Ai + Bi and write out the covariant derivative for Ci,j. 

◮ 18. Let Ci j = AiBj and prove that the covariant derivative of a product equals the first term times the 

covariant derivative of the second term plus the second term times the covariant derivative of the first term. 

◮ 19. Start with the transformation law Āij 

∂x 

= Aαβ 

α 

∂¯x i 

∂xβ and take an ordinary derivative of both sides 

∂¯x j 

with respect to ¯x k and hence derive the relation for Aij,k given in (1.4.30). 

◮ 20. Start with the transformation law A ij = 

∂xi 

Āαβ 

∂¯x α 

with respect to xk and hence derive the relation for A ij 

,k 

◮ 21. Find the covariant derivatives of 

(a) A ijk 

∂xj and take an ordinary derivative of both sides 

∂¯x β 

given in (1.4.30). 

(b) A ij 

k (c) A i jk (d) Aijk 

◮ 22. Find the intrinsic derivative along the curve x i = x i (t), i =1,...,N for 

◮ 23. 

(a) A ijk 

(a) Assume A = AiEi and show that d A = Ai ,k dxk Ei. 

(b) Assume A = Ai Ei and show that d A = Ai,k dxk E i . 

(b) A ij 

k (c) A i jk (d) Aijk

◮ 24. (parallel vector field) Imagine a vector field A i = A i (x 1 ,x 2 ,x 3 ) which is a function of position. 

Assume that at all points along a curve x i = x i (t),i =1, 2, 3 the vector field points in the same direction, 

we would then have a parallel vector field or homogeneous vector field. Assume A is a constant, then 

d A = ∂ A 

∂xk dxk =0. Show that for a parallel vector field the condition Ai,k = 0 must be satisfied. 


∂[ik, n] 

∂x j 

∂ 

= gnσ 

∂xj ◮ 26. Show Ar,s − As,r = ∂Ar ∂As 

− . 

∂xs ∂xr 

 

σ 

σ 

+([nj, σ]+[σj, n]) . 

ik 

ik 

◮ 27. In cylindrical coordinates you are given the contravariant vector components 

A 1 = r A 2 =cosθ A 3 = z sin θ 

(a) Find the physical components Ar, Aθ, and Az. 

(b) Denote the physical components of A i ,j 

Find these physical components. 

, i, j =1, 2, 3, by 

◮ 28. Find the covariant form of the contravariant tensor C i = ɛ ijk Ak,j. 

Express your answer in terms of A k ,j . 

Arr Arθ Arz 

Aθr Aθθ Aθz 

Azr Azθ Azz. 

◮ 29. In Cartesian coordinates let x denote the magnitude of the position vector xi. Show that (a) x ,j = 1 

x xj 

(b) x ,ij = 1 

x δij − 1 

x3 xixj (c) x ,ii = 2 

1 

. (d) LetU = 

x x , x = 0,andshowthatU ,ij = −δij 3xixj 

+ 

x3 x5 and 

U ,ii =0. 

◮ 30. Consider a two dimensional space with element of arc length squared 

ds 2 = g11(du 1 ) 2 + g22(du 2 ) 2 

and metric 

 

g11 

gij = 

0 

 

0 

g22 

where u 1 ,u 2 are surface coordinates. 

(a) Find formulas to calculate the Christoffel symbols of the first kind. 

(b) Find formulas to calculate the Christoffel symbols of the second kind. 

◮ 31. Find the metric tensor and Christoffel symbols of the first and second kind associated with the 

two dimensional space describing points on a cylinder of radius a. Let u 1 = θ and u 2 = z denote surface 

coordinates where 

x = a cos θ = a cos u 1 

y = a sin θ = a sin u 1 

z = z = u 2 




two dimensional space describing points on a sphere of radius a. Let u 1 = θ and u 2 = φ denote surface 

coordinates where 

x = a sin θ cos φ = a sin u 1 cos u 2 

y = a sin θ sin φ = a sin u 1 sin u 2 

z = a cos θ = a cos u 1 


two dimensional space describing points on a torus having the parameters a and b and surface coordinates 

u 1 = ξ, u 2 = η. illustrated in the figure 1.3-19. The points on the surface of the torus are given in terms 

of the surface coordinates by the equations 

x =(a + b cos ξ)cosη 

y =(a + b cos ξ)sinη 

z = b sin ξ 

◮ 34. Prove that eijka m b j c k u i ,m + eijka i b m c k u j ,m + eijka i b j c m u k ,m = ur ,reijka i b j c k . Hint: See Exercise 1.3, 

problem 32 and Exercise 1.1, problem 21. 

◮ 35. Calculate the second covariant derivative A i ,jk. 

◮ 36. Show that σ ij 1 ∂ 

,j = √ 

g ∂xj √ ij 

gσ + σ mn 

 

i 

mn 

◮ 37. Find the contravariant, covariant and physical components of velocity and acceleration in (a) Cartesian 

coordinates and (b) cylindrical coordinates. 

◮ 38. Find the contravariant, covariant and physical components of velocity and acceleration in spherical 

coordinates. 

◮ 39. In spherical coordinates (ρ, θ, φ) show that the acceleration components can be represented in terms 

of the velocity components as 

Hint: Calculate ˙vρ, ˙vθ, ˙vφ. 

fρ =˙vρ − v2 θ + v2 φ 

, fθ =˙vθ + 

ρ 

vρvθ 

ρ − v2 φ 

ρ tan θ , fφ =˙vφ + vρvφ vθvφ 

+ 

ρ ρ tan θ 

◮ 40. The divergence of a vector Ai is Ai ,i . That is, perform a contraction on the covariant derivative 

Ai ,j to obtain Ai ,i . Calculate the divergence in (a) Cartesian coordinates (b) cylindrical coordinates and (c) 

spherical coordinates. 

◮ 41. If S is a scalar invariant of weight one and Ai jk is a third order relative tensor of weight W ,show 

is an absolute tensor. 

that S −W A i jk

◮ 42. Let ¯ Y i ,i =1, 2, 3 denote the components of a field of parallel vectors along the curve C defined by 

the equations y i =¯y i (t), i =1, 2, 3inaspacewithmetrictensor¯gij, i, j =1, 2, 3. Assume that ¯ Y i and d¯yi 

dt 

are unit vectors such that at each point of the curve ¯ C we have 

¯gij ¯ i d¯yj 

Y 

dt 

=cosθ = Constant. 

(i.e. The field of parallel vectors makes a constant angle θ with the tangent to each point of the curve ¯ C.) 

Show that if ¯ Y i and ¯y i (t) undergo a transformation x i = x i (¯y 1 , ¯y 2 , ¯y 3 ), i =1, 2, 3 then the transformed 

vector Xm = ¯ i ∂xm Y ∂ ¯y j makes a constant angle with the tangent vector to the transformed curve C given by 

xi = xi (¯y 1 (t), ¯y 2 (t), ¯y 3 (t)). 

◮ 43. Let J denote the Jacobian determinant | ∂xi 

∂x j |. Differentiate J with respect to xm and show that 

Hint: See Exercise 1.1, problem 27 and (1.4.7). 

p 

∂J α ∂x r 

= J 

− J . 

∂xm αp ∂xm rm 

◮ 44. Assume that φ is a relative scalar of weight W so that φ = J W φ. Differentiate this relation with 

respect to xk . Use the result from problem 42 to obtain the transformation law: 

 

∂φ α 

− W φ = J k ∂x αk 

W 

m 

∂φ r ∂x 

− W φ . 

∂xm mr 

k ∂x 

The quantity inside the brackets is called the covariant derivative of a relative scalar of weight W. The 

covariant derivative of a relative scalar of weight W is defined as 

φ ,k = ∂φ 

 

r 

− W φ 

∂xk kr 

and this definition has an extra term involving the weight. 

It can be shown that similar results hold for relative tensors of weight W. For example, the covariant 

derivative of first and second order relative tensors of weight W have the forms 

T i ,k 

i ∂T i 

= + 

∂xk T i j,k = ∂T i j 

+ 

∂xk km 

i 

kσ 

 

T m 

r 

− W 

 

T σ j − 

σ 

jk 

kr 

 

T i 

 

T i σ − W 

 

r 

T 

kr 

i j 

When the weight term is zero these covariant derivatives reduce to the results given in our previous definitions. 

◮ 45. Let dxi 

dt = vi denote a generalized velocity and define the scalar function of kinetic energy T of a 

particle with mass m as 

T = 1 

2 mgij v i v j = 1 

2 mgij ˙x i ˙x j . 

Show that the intrinsic derivative of T is the same as an ordinary derivative of T. (i.e. Show that δT 

δT 

= dT 

dt .) 



◮ 46. Verify the relations 

∂gij 

∂xk = −gmj 

∂g 

gni 

nm 

∂xk ∂gin ∂xk = −gmn ij ∂gjm 

g 

∂xk ◮ 47. Assume that Bijk is an absolute tensor. Is the quantity T jk = 1 ∂ 

√ 

g ∂xi √ ijk 

gB a tensor? Justify 

your answer. If your answer is “no”, explain your answer and determine if there any conditions you can 

impose upon Bijk such that the above quantity will be a tensor? 

◮ 48. The e-permutation symbol can be used to define various vector products. Let Ai,Bi,Ci,Di 

i =1,...,N denote vectors, then expand and verify the following products: 

(a) In two dimensions 

R =eijAiBj a scalar determinant. 

(b) In three dimensions 

(c) In four dimensions 

with similar products in higher dimensions. 

◮ 49. Expand the curl operator for: 

(a) Two dimensions B = eijAj,i 

(b) Three dimensions Bi = eijkAk,j 

(c) Four dimensions Bij = eijkmAm,k 

Ri =eijAj a vector (rotation). 

S =eijkAiBjCk a scalar determinant. 

Si =eijkBjCk a vector cross product. 

Sij =eijkCk a skew-symmetric matrix 

T =eijkmAiBjCkDm a scalar determinant. 

Ti =eijkmBjCkDm 4-dimensional cross product. 

Tij =eijkmCkDm skew-symmetric matrix. 

Tijk =eikmDm skew-symmetric tensor.

§1.5 DIFFERENTIAL GEOMETRY AND RELATIVITY 

In this section we will examine some fundamental properties of curves and surfaces. In particular, at 

each point of a space curve we can construct a moving coordinate system consisting of a tangent vector, a 

normal vector and a binormal vector which is perpendicular to both the tangent and normal vectors. How 

these vectors change as we move along the space curve brings up the subjects of curvature and torsion 

associated with a space curve. The curvature is a measure of how the tangent vector to the curve is changing 

and the torsion is a measure of the twisting of the curve out of a plane. We will find that straight lines have 

zero curvature and plane curves have zero torsion. 

In a similar fashion, associated with every smooth surface there are two coordinate surface curves and 

a normal surface vector through each point on the surface. The coordinate surface curves have tangent 

vectors which together with the normal surface vectors create a set of basis vectors. These vectors can be 

used to define such things as a two dimensional surface metric and a second order curvature tensor. The 

coordinate curves have tangent vectors which together with the surface normal form a coordinate system at 

each point of the surface. How these surface vectors change brings into consideration two different curvatures. 

A normal curvature and a tangential curvature (geodesic curvature). How these curvatures are related to 

the curvature tensor and to the Riemann Christoffel tensor, introduced in the last section, as well as other 

interesting relationships between the various surface vectors and curvatures, is the subject area of differential 

geometry. 

Also presented in this section is a brief introduction to relativity where again the Riemann Christoffel 

tensor will occur. Properties of this important tensor are developed in the exercises of this section. 

Space Curves and Curvature 

For xi = xi (s),i =1, 2, 3, a 3-dimensional space curve in a Riemannian space Vn with metric tensor gij, 

and arc length parameter s, the vector T i = dxi 

ds represents a tangent vector to the curve at a point P on 

the curve. The vector T i is a unit vector because 

gijT i T j dx 

= gij 

i dx 

ds 

j 

ds 

=1. (1.5.1) 

Differentiate intrinsically, with respect to arc length, the relation (1.5.1) and verify that 

which implies that 

j 

i δT δT 

gijT + gij 

δs i 

δs T j =0, (1.5.2) 

i 

j δT 

gijT =0. 

δs 

(1.5.3) 

i 

δT 

δs is perpendicular to the tangent vector T i . Define the unit normal vector N i to the 

i 

δT 

δs and write 

N i = 1 δT 

κ 

i 

δs 

(1.5.4) 

Hence, the vector 

space curve to be in the same direction as the vector 

where κ is a scale factor, called the curvature, and is selected such that 

gijN i N j δT 

= 1 which implies gij 

i δT 

δs 

j 

δs = κ2 . (1.5.5) 


130 

The reciprocal of curvature is called the radius of curvature. The curvature measures the rate of change of 

the tangent vector to the curve as the arc length varies. By differentiating intrinsically, with respect to arc 

length s, the relation gijT i N j = 0 we find that 

Consequently, the curvature κ can be determined from the relation 

i δNj 

gijT 

δs 

i δNj δT 

gijT + gij 

δs i 

δs N j =0. (1.5.6) 

δT 

= −gij 

i 

δs N j = −gijκN i N j = −κ (1.5.7) 

which defines the sign of the curvature. In a similar fashion we differentiate the relation (1.5.5) and find that 

i δNj 

gijN =0. 

δs 

(1.5.8) 

This later equation indicates that the vector δNj 

δs is perpendicular to the unit normal N i . The equation 

(1.5.3) indicates that T i is also perpendicular to N i and hence any linear combination of these vectors will 

also be perpendicular to N i . The unit binormal vector is defined by selecting the linear combination 

δNj δs 

and then scaling it into a unit vector by defining 

B j = 1 

j δN 

τ δs 

+ κT j 

 

j 

+ κT 

(1.5.9) 

(1.5.10) 

where τ is a scalar called the torsion. The sign of τ is selected such that the vectors T i ,Ni and Bi form a 

right handed system with ɛijkT iN jB k = 1 and the magnitude of τ is selected such that Bi is a unit vector 

satisfying 

gijB i B j =1. (1.5.11) 

The triad of vectors T i ,Ni ,Bi at a point on the curve form three planes. The plane containing T i and Bi is 

called the rectifying plane. The plane containing N i and Bi is called the normal plane. The plane containing 

T i and N i is called the osculating plane. The reciprocal of the torsion is called the radius of torsion. The 

torsion measures the rate of change of the osculating plane. The vectors T i ,Ni and Bi form a right-handed 

orthogonal system at a point on the space curve and satisfy the relation 

B i = ɛ ijk TjNk. (1.5.12) 

By using the equation (1.5.10) it can be shown that B i is perpendicular to both the vectors T i and N i since 

gijB i T j =0 and gijB i N j =0. 

It is left as an exercise to show that the binormal vector B i satisfies the relation δBi 

δs = −τNi . The three 

relations 

δT i 

i 

= κN 

δs 

δN i 

δs = τBi − κT i 

δB i 

δs 

= −τNi 

(1.5.13)

are known as the Frenet-Serret formulas of differential geometry. 

Surfaces and Curvature 

Let us examine surfaces in a Cartesian frame of reference and then later we can generalize our results 

to other coordinate systems. A surface in Euclidean 3-dimensional space can be defined in several different 

ways. Explicitly, z = f(x, y), implicitly, F (x, y, z) = 0 or parametrically by defining a set of parametric 

equations of the form 

x = x(u, v), y = y(u, v), z = z(u, v) 

which contain two independent parameters u, v called surface coordinates. For example, the equations 

x = a sin θ cos φ, y = a sin θ sin φ, z = a cos θ 

are the parametric equations which define a spherical surface of radius a with parameters u = θ and v = φ. 

See for example figure 1.3-20 in section 1.3. By eliminating the parameters u, v one can derive the implicit 

form of the surface and by solving for z one obtains the explicit form of the surface. Using the parametric 

form of a surface we can define the position vector to a point on the surface which is then represented in 

terms of the parameters u, v as 

r = r(u, v) =x(u, v) e1 + y(u, v) e2 + z(u, v) e3. (1.5.14) 

The coordinates (u, v) are called the curvilinear coordinates of a point on the surface. The functions 

x(u, v),y(u, v),z(u, v) are assumed to be real and differentiable such that ∂r ∂r 

∂u × ∂v =0. The curves 

r(u, c2) and r(c1,v) (1.5.15) 

with c1,c2 constants, then define two surface curves called coordinate curves, which intersect at the surface 

coordinates (c1,c2). The family of curves defined by equations (1.5.15) with equally spaced constant values 

evaluated at the 

surface coordinates (c1,c2) on the surface, are tangent vectors to the coordinate curves through the point 

ci,ci +∆ci,ci +2∆ci,... define a surface coordinate grid system. The vectors ∂r ∂r 

∂u and ∂v 

and are basis vectors for any vector lying in the surface. Letting (x, y, z) =(y 1 ,y 2 ,y 3 )and(u, v) =(u 1 ,u 2 ) 

and utilizing the summation convention, we can write the position vector in the form 

r = r(u 1 ,u 2 )=y i (u 1 ,u 2 ) ei. (1.5.16) 

The tangent vectors to the coordinate curves at a point P can then be represented as the basis vectors 

Eα = ∂r ∂yi 

= 

∂uα ∂uα ei, α =1, 2 (1.5.17) 

where the partial derivatives are to be evaluated at the point P where the coordinate curves on the surface 

intersect. From these basis vectors we construct a unit normal vector to the surface at the point P by 

calculating the cross product of the tangent vector ru = ∂r 

. A unit normal is then 

∂u and rv = ∂r 

∂v 

n = n(u, v) = E1 × E2 

| E1 × E2| = ru × rv 

|ru × rv| 

(1.5.18) 

131

132 

and is such that the vectors E1, E2 and n form a right-handed system of coordinates. 

If we transform from one set of curvilinear coordinates (u, v) toanotherset(ū, ¯v), which are determined 

by a set of transformation laws 

u = u(ū, ¯v), v = v(ū, ¯v), 

the equation of the surface becomes 

r = r(ū, ¯v) =x(u(ū, ¯v),v(ū, ¯v)) e1 + y(u(ū, ¯v),v(ū, ¯v)) e2 + z(u(ū, ¯v),v(ū, ¯v)) e3 

and the tangent vectors to the new coordinate curves are 

∂r ∂r ∂u ∂r ∂v 

= + 

∂ū ∂u ∂ū ∂v ∂ū and 

∂r ∂r ∂u ∂r ∂v 

= + 

∂¯v ∂u ∂¯v ∂v ∂¯v . 

Using the indicial notation this result can be represented as 

∂yi ∂yi 

= 

∂ūα ∂uβ ∂uβ . 

∂ūα This is the transformation law connecting the two systems of basis vectors on the surface. 

A curve on the surface is defined by a relation f(u, v) = 0 between the curvilinear coordinates. Another 

way to represent a curve on the surface is to represent it in a parametric form where u = u(t) andv = v(t), 

where t is a parameter. The vector 

dr ∂r du ∂r dv 

= + 

dt ∂u dt ∂v dt 

is tangent to the curve on the surface. 

An element of arc length with respect to the surface coordinates is represented by 

ds 2 = dr · dr = ∂r ∂r 

· 

∂uα ∂uβ duα du β = aαβdu α du β 

(1.5.19) 

where aαβ = ∂r 

∂uα · ∂r 

∂uβ with α, β =1, 2 defines a surface metric. This element of arc length on the surface is 

often written as the quadratic form 

A = ds 2 = E(du) 2 +2Fdudv+ G(dv) 2 = 1 

E (Edu+ Fdv)2 2 EG − F 

+ dv 

E 

2 

(1.5.20) 

and called the first fundamental form of the surface. Observe that for ds2 to be positive definite the quantities 

E and EG − F 2 must be positive. 

The surface metric associated with the two dimensional surface is defined by 

aαβ = Eα · Eβ = ∂r ∂r ∂yi 

· = 

∂uα ∂uβ ∂uα ∂yi , α,β =1, 2 (1.5.21) 

∂uβ with conjugate metric tensor a αβ defined such that a αβ aβγ = δ α γ . Here the surface is embedded in a three 

dimensional space with metric gij and aαβ is the two dimensional surface metric. In the equation (1.5.20) 

the quantities E,F,G are functions of the surface coordinates u, v and are determined from the relations 

E =a11 = ∂r ∂r ∂yi 

· = 

∂u ∂u ∂u1 ∂yi ∂u1 F =a12 = ∂r ∂r ∂yi 

· = 

∂u ∂v ∂u1 G =a22 = ∂r ∂r ∂yi 

· = 

∂v ∂v ∂u2 ∂yi ∂u2 ∂y i 

∂u 2 

(1.5.22)

Here and throughout the remainder of this section, we adopt the convention that Greek letters have the 

range 1,2, while Latin letters have the range 1,2,3. 

Construct at a general point P on the surface the unit normal vector n at this point. Also construct a 

plane which contains this unit surface normal vector n. Observe that there are an infinite number of planes 

which contain this unit surface normal. For now, select one of these planes, then later on we will consider 

all such planes. Let r = r(s) denote the position vector defining a curve C which is the intersection of the 

selected plane with the surface, where s is the arc length along the curve, which is measured from some fixed 

point on the curve. Let us find the curvature of this curve of intersection. The vector T = dr 

ds , evaluated 

at the point P, is a unit tangent vector to the curve C and lies in the tangent plane to the surface at the 

point P. Here we are using ordinary differentiation rather than intrinsic differentiation because we are in 

a Cartesian system of coordinates. Differentiating the relation T · T =1, with respect to arc length s we 

find that T · dT 

dT 

ds = 0 which implies that the vector ds is perpendicular to the tangent vector T. Since the 

coordinate system is Cartesian we can treat the curve of intersection C as a space curve, then the vector 

K = dT 

ds , evaluated at point P, is defined as the curvature vector with curvature | K| = κ and radius of 

curvature R =1/κ. A unit normal N to the space curve is taken in the same direction as dT 

ds so that the 

curvature will always be positive. We can then write K = κ N = d T 

. Consider the geometry of figure 1.5-1 

ds 

and define on the surface a unit vector u = n × T which is perpendicular to both the surface tangent vector 

T and the surface normal vector n, such that the vectors T i ,ui and ni forms a right-handed system. 

Figure 1.5-1 Surface curve with tangent plane and a normal plane. 

133

134 

The direction of u in relation to T is in the same sense as the surface tangents E1 and E2. Note that 

the vector dT 

ds is perpendicular to the tangent vector T and lies in the plane which contains the vectors n 

and u. We can therefore write the curvature vector K in the component form 

K = d T 

ds = κ (n) n + κ (g) u = Kn + Kg 

(1.5.23) 

where κ (n) is called the normal curvature and κ (g) is called the geodesic curvature. The subscripts are not 

indices. These curvatures can be calculated as follows. From the orthogonality condition n · T =0weobtain 

by differentiation with respect to arc length s the result n · d T 

ds + T · dn 

=0. Consequently, the normal 

ds 

curvature is determined from the dot product relation 

n · K = κ (n) = − T · dn 

ds 

= − dr 

ds 

dn 

· . (1.5.24) 

ds 

By taking the dot product of u with equation (1.5.23) we find that the geodesic curvature is determined 

from the triple scalar product relation 

Normal Curvature 

κ (g) = u · d T 

ds =(n × T ) · d T 

. (1.5.25) 

ds 

The equation (1.5.24) can be expressed in terms of a quadratic form by writing 

κ (n) ds 2 = −dr · dn. (1.5.26) 

The unit normal to the surface n and position vector r are functions of the surface coordinates u, v with 

dr = ∂r 

∂u 

du + ∂r 

∂v 

dv and dn = ∂n 

∂u 

We define the quadratic form 

 

∂r ∂r 

B = −dr · dn = − du + 

∂u ∂v dv 

 

∂n ∂n 

· du + 

∂u ∂v dv 

 

where 

e = − ∂r 

∂u 

B = e(du) 2 +2fdudv + g(dv) 2 = bαβ du α du β 

 

∂n 

∂r ∂n ∂n ∂r 

· , 2f = − · + · , g = − 

∂u ∂u ∂v ∂u ∂v 

∂r ∂n 

· 

∂v ∂v 

and bαβ α, β =1, 2 is called the curvature tensor and a αγ bαβ = b γ 

β 

∂n 

du + dv. (1.5.27) 

∂v 

(1.5.28) 

(1.5.29) 

is an associated curvature tensor. 

The quadratic form of equation (1.5.28) is called the second fundamental form of the surface. Alternative 

methods for calculating the coefficients of this quadratic form result from the following considerations. The 

unit surface normal is perpendicular to the tangent vectors to the coordinate curves at the point P and 

therefore we have the orthogonality relationships 

∂r 

· n =0 and 

∂u 

∂r 

· n =0. (1.5.30) 

∂v

Observe that by differentiating the relations in equation (1.5.30), with respect to both u and v, one can 

derive the results 

e = ∂2r ∂r ∂n 

· n = − · = b11 

∂u2 ∂u ∂u 

f = ∂2r ∂r ∂n ∂n ∂r 

· n = − · = − · 

∂u∂v ∂u ∂v ∂u ∂v = b21 = b12 

g = ∂2 (1.5.31) 

r ∂r ∂n 

· n = − · = b22 

∂v2 ∂v ∂v 

and consequently the curvature tensor can be expressed as 

bαβ = − ∂r ∂n 

· . (1.5.32) 

∂uα ∂uβ The quadratic forms from equations (1.5.20) and (1.5.28) enable us to represent the normal curvature 

in the form of a ratio of quadratic forms. We find from equation (1.5.26) that the normal curvature in the 

direction du 

dv is 

κ (n) = B 

A = e(du)2 +2fdudv+ g(dv) 2 

E(du) 2 . 

+2Fdudv+ G(dv) 2 (1.5.33) 

If we write the unit tangent vector to the curve in the form T = dr ∂r 

ds = ∂uα du α 

ds and express the derivative 

of the unit surface normal with respect to arc length as dn ∂n 

ds = ∂uβ du β 

ds , then the normal curvature can be 

expressed in the form 

κ (n) = − T · dn 

 

∂r ∂n 

= − · 

ds ∂uα ∂uβ α du du 

ds 

β 

ds 

(1.5.34) 

= bαβdu α du β 

ds 2 

= bαβduαduβ aαβduα . 

duβ Observe that the curvature tensor is a second order symmetric tensor. 

In the previous discussions, the plane containing the unit normal vector was arbitrary. Let us now 

consider all such planes that pass through this unit surface normal. As we vary the plane containing the unit 

surface normal n at P we get different curves of intersection with the surface. Each curve has a curvature 

associated with it. By examining all such planes we can find the maximum and minimum normal curvatures 

associated with the surface. We write equation (1.5.33) in the form 

κ (n) = 

e +2fλ+ gλ2 

E +2Fλ+ Gλ 2 

where λ = dv 

du . From the theory of proportions we can also write this equation in the form 

κ (n) = 

(1.5.35) 

(e + fλ)+λ(f + gλ) f + gλ e + fλ 

= = . (1.5.36) 

(E + Fλ)+λ(F + Gλ) F + Gλ E + Fλ 

Consequently, the curvature κ will satisfy the differential equations 

(e − κE)du +(f − κF )dv =0 and (f − κF )du +(g − κG)dv =0. (1.5.37) 

The maximum and minimum curvatures occur in those directions λ where dκ (n) 

dλ =0. Calculating the derivative 

of κ (n) with respect to λ and setting the derivative to zero we obtain a quadratic equation in λ 

(Fg− Gf)λ 2 +(Eg − Ge)λ +(Ef − Fe)=0, (Fg − Gf) = 0. 

135

136 

This equation has two roots λ1 and λ2 which satisfy 

Eg − Ge 

λ1 + λ2 = − 

Fg− Gf 

and λ1λ2 = 

Ef − Fe 

, (1.5.38) 

Fg− Gf 

where Fg − Gf = 0. The curvatures κ (1),κ (2) corresponding to the roots λ1 and λ2 are called the principal 

curvatures at the point P. Several quantities of interest that are related to κ (1) and κ (2) are: (1) the principal 

radii of curvature Ri =1/κi,i =1, 2; (2) H = 1 

2 (κ (1) + κ (2)) called the mean curvature and K = κ (1)κ (2) 

called the total curvature or Gaussian curvature of the surface. Observe that the roots λ1 and λ2 determine 

two directions on the surface 

dr1 

du 

∂r ∂r 

= + 

∂u ∂v λ1 

If these directions are orthogonal we will have 

This requires that 

dr1 

du 

· dr2 

du =(∂r 

∂u 

and 

dr2 

du 

∂r ∂r 

= + 

∂u ∂v λ2. 

∂r ∂r ∂r 

+ λ1)( + 

∂v ∂u ∂v λ2) =0. 

Gλ1λ2 + F (λ1 + λ2)+E =0. (1.5.39) 

It is left as an exercise to verify that this is indeed the case and so the directions determined by the principal 

curvatures must be orthogonal. In the case where Fg− Gf =0wehavethatF =0andf = 0 because the 

coordinate curves are orthogonal and G must be positive. In this special case there are still two directions 

determined by the differential equations (1.5.37) with dv =0,du arbitrary, and du =0,dv arbitrary. From 

the differential equations (1.5.37) we find these directions correspond to 

κ (1) = e 

E 

and κ (2) = g 

G . 

Weletλ α = duα 

ds denote a unit vector on the surface satisfying aαβλ α λ β =1. Then the equation (1.5.34) 

can be written as κ (n) = bαβλ α λ β or we can write (bαβ − κ (n) aαβ)λ α λ β =0. The maximum and minimum 

normal curvature occurs in those directions λ α where 

(bαβ − κ (n) aαβ)λ α =0 

and so κ (n) must be a root of the determinant equation |bαβ − κ (n) aαβ| =0or 

|a αγ bαβ − κ (n)δ γ 

β | = 

 

 

 

b11 − κ (n) 

b 2 1 

b1 2 

b2 2 − κ (n) 

 

 

 

= κ2 (n) − bαβa αβ κ (n) + b 

=0. (1.5.40) 

a 

This is a quadratic equation in κ (n) of the form κ2 (n) − (κ (1) + κ (2))κ (n) + κ (1)κ (2) =0. In other words the 

principal curvatures κ (1) and κ (2) are the eigenvalues of the matrix with elements b γ 

β = aαγbαβ. Observe that 

from the determinant equation in κ (n) we can directly find the total curvature or Gaussian curvature which 

is an invariant given by K = κ (1)κ (2) = |bα β | = |aαγbγβ| = b/a. The mean curvature is also an invariant 

obtained from H = 1 

2 (κ (1) + κ (2)) = 1 

2aαβbαβ, where a = a11a22 − a12a21 and b = b11b22 − b12b21 are the 

determinants formed from the surface metric tensor and curvature tensor components.

The equations of Gauss, Weingarten and Codazzi 

At each point on a space curve we can construct a unit tangent T , a unit normal N and unit binormal 

B. The derivatives of these vectors, with respect to arc length, can also be represented as linear combinations 

of the base vectors T, N, B. See for example the Frenet-Serret formulas from equations (1.5.13). In a similar 

fashion the surface vectors ru,rv, n form a basis and the derivatives of these basis vectors with respect to 

the surface coordinates u, v can also be expressed as linear combinations of the basis vectors ru,rv, n. For 

example, the derivatives ruu,ruv,rvv can be expressed as linear combinations of ru,rv, n. Wecanwrite 

ruu = c1ru + c2rv + c3n 

ruv = c4ru + c5rv + c6n 

rvv = c7ru + c8rv + c9n 

(1.5.41) 

where c1,...,c9 are constants to be determined. It is an easy exercise (see exercise 1.5, problem 8) to show 

that these equations can be written in the indicial notation as 

∂2r ∂uα 

γ ∂r 

= 

∂uβ αβ ∂uγ + bαβn. 

These equations are known as the Gauss equations. 

(1.5.42) 

In a similar fashion the derivatives of the normal vector can be represented as linear combinations of 

the surface basis vectors. If we write 

∂n 

∂u = c1ru + c2rv 

∂n 

∂v = c3ru + c4rv 

or 

∂r 

∂u = c∗ ∂n 

1 

∂u + c∗ ∂n 

2 

∂v 

∂r 

∂v = c∗ ∂n 

3 

∂u + c∗ ∂n 

4 

∂v 

(1.5.43) 

where c1,...,c4 and c∗ 1 ,...,c∗4 are constants. These equations are known as the Weingarten equations. It 

is easily demonstrated (see exercise 1.5, problem 9) that the Weingarten equations can be written in the 

indicial form 

∂n 

∂uα = −bβ ∂r 

α 

∂uβ (1.5.44) 

where b β α = a βγ bγα is the mixed second order form of the curvature tensor. 

The equations of Gauss produce a system of partial differential equations defining the surface coordinates 

xi as a function of the curvilinear coordinates u and v. The equations are not independent as certain 

compatibility conditions must be satisfied. In particular, it is required that the mixed partial derivatives 

must satisfy 

Wecalculate 

∂3r ∂uα∂uβ = 

∂uδ ∂3r ∂uα∂uβ 2 γ ∂ r 

= 

∂uδ αβ ∂uγ ∂ 

+ 

∂uδ ∂3r ∂uα∂uδ . 

∂uβ 

γ 

αβ 

∂uδ ∂r ∂n ∂bαβ 

+ bαβ + n 

∂uγ ∂uδ ∂uδ and use the equations of Gauss and Weingarten to express this derivative in the form 

∂3r ∂uα∂uβ ⎡ 

ω 

⎢ 

∂ 

αβ 

= ⎢ 

∂uδ ⎣ ∂uδ + 

 

γ ω 

− bαβb 

αβ γδ 

ω ⎤ 

 

⎥ 

∂r γ 

δ ⎦ + bγδ + 

∂uω αβ 

∂bαβ 

∂uδ 

n. 

137

138 

Forming the difference 

∂3r ∂uα∂uβ ∂ 

− 

∂uδ 3r ∂uα∂uδ =0 

∂uβ we find that the coefficients of the independent vectors n and ∂r 

∂uω equal to zero produces the Codazzi equations 

must be zero. Setting the coefficient of n 

 

γ 

γ 

bγδ − bγβ + 

αβ αδ 

∂bαβ 

∂uδ ∂bαδ 

− =0. 

∂uβ (1.5.45) 

These equations are sometimes referred to as the Mainardi-Codazzi equations. Equating to zero the coefficient 

of ∂r 

∂uω we find that R δ αγβ = bαβb δ γ − bαγb δ β or changing indices we have the covariant form 

aωδR δ αβγ = Rωαβγ = bωβbαγ − bωγbαβ, (1.5.46) 

where 

R δ ∂ 

αγβ = 

∂uγ 

δ 

− 

αβ 

∂ 

∂uβ 

δ ω δ ω δ 

+ 

− 

αγ αβ ωγ αγ ωβ 

is the mixed Riemann curvature tensor. 

EXAMPLE 1.5-1 

(1.5.47) 

Show that the Gaussian or total curvature K = κ (1)κ (2) depends only upon the metric aαβ and is 

K = R1212 

where a = det(aαβ). 

a 

Solution: 

Utilizing the two-dimensional alternating tensor eαβ and the property of determinants we can write 

e γδ K = e αβ b γ α bδ β 

ζ and δ to obtain 

where from page 137, K = |bγ 

β | = |aαγ bαβ|. Now multiply by eγζ and then contract on 

eγδe γδ K = eγδe αβ b γ α bδ β =2K 

2K = eγδe αβ (a γµ bαu) a δν 

bβν 

But eγδa γµ a δν = ae µν so that 2K = e αβ ae µν bαµbβν. Using √ ae µν = ɛ µν we have 2K = ɛ µν ɛ αβ bαµbβν. 

Interchanging indices we can write 

2K = ɛ βγ ɛ ωα bωβbαγ and 2K = ɛ γβ ɛ ωα bωγbαβ. 

Adding these last two results we find that 4K = ɛ βγ ɛ ωγ (bωβbαγ − bωγbαβ) =ɛ βγ ɛ ωγ Rωαβγ. Now multiply 

both sides by ɛστɛλν to obtain 4Kɛστɛλν = δ βγ 

στ δ ωα 

λν Rωαβγ. From exercise 1.5, problem 16, the Riemann 

curvature tensor Rijkl is skew symmetric in the (i, j), (k, l) as well as being symmetric in the (ij), (kl) pair 

of indices. Consequently, δ βγ 

στ δ ωα 

λν Rωαβγ =4Rλνστ and hence Rλνστ = Kɛστɛλν and we have the special case 

where K √ √ 

ae12 ae12 = R1212 or K = R1212 

b 

. A much simpler way to obtain this result is to observe K = 

a a 

(bottom of page 137) and note from equation (1.5.46) that R1212 = b11b22 − b12b21 = b. 

Note that on a surface ds2 = aαβduαduβ where aαβ are the metrices for the surface. This metric is a 

∂u 

tensor and satisfies āγδ = aαβ 

α 

∂ūγ ∂uβ and by taking determinants we find 

∂ūδ ā = 

 

 

āγδ 

 

 

 

∂uα 

∂ūγ 

 

 

∂uβ 

∂ūδ 

 

= aJ 2

where J is the Jacobian of the surface coordinate transformation. Here the curvature tensor for the surface 

Rαβγδ has only one independent component since R1212 = R2121 = −R1221 = −R2112 (See exercises 20,21). 

From the transformation law 

∂u 

¯Rɛηλµ = Rαβγδ 

α 

∂ūɛ ∂uβ ∂ūη ∂uγ ∂ūλ ∂uδ ∂ū µ 

one can sum over the repeated indices and show that ¯ R1212 = R1212J 2 and consequently 

¯R1212 

ā 

= R1212 

a 

= K 

which shows that the Gaussian curvature is a scalar invariant in V2. 

Geodesic Curvature 

For C an arbitrary curve on a given surface the curvature vector K, associated with this curve, is 

the vector sum of the normal curvature κ (n) n and geodesic curvature κ (g) u and lies in a plane which 

is perpendicular to the tangent vector to the given curve on the surface. The geodesic curvature κ (g) is 

obtained from the equation (1.5.25) and can be represented 

Substituting into this expression the vectors 

κ (g) = u · K = u · d T 

ds =(n × T ) · d T 

ds = 

 

T × d 

T 

· n. 

ds 

T = dr 

ds 

du 

= ru 

ds 

dv 

+ rv 

ds 

d T 

ds = K = ruu(u ′ ) 2 +2ruvu ′ v ′ + rvv(v ′ ) 2 + ruu ′′ + rvv ′′ , 

where ′ = d 

ds , and by utilizing the results from problem 10 of the exercises following this section, we find 

that the geodesic curvature can be represented as 

 

2 

κ (g) = (u 

11 

′ ) 3 

2 1 

+ 2 − 


 

 

2 

1 

− 2 

22 

22 

 

1 

u 


′ (v ′ ) 2 − 

(u ′ ) 2 v ′ + 

 

(v ′ ) 3 +(u ′ v ′′ − u ′′ v ′ 

EG 

) − F 2 . 

(1.5.48) 

This equation indicates that the geodesic curvature is only a function of the surface metrices E,F,G and 

the derivatives u ′ ,v ′ ,u ′′ ,v ′′ . When the geodesic curvature is zero the curve is called a geodesic curve. Such 

curves are often times, but not always, the lines of shortest distance between two points on a surface. For 

example, the great circle on a sphere which passes through two given points on the sphere is a geodesic curve. 

If you erase that part of the circle which represents the shortest distance between two points on the circle 

you are left with a geodesic curve connecting the two points, however, the path is not the shortest distance 

between the two points. 

For plane curves we let u = x and v = y so that the geodesic curvature reduces to 

kg = u ′ v ′′ − u ′′ v ′ = dφ 

ds 

139

140 

where φ is the angle between the tangent T to the curve and the unit vector e1. 

Geodesics are curves on the surface where the geodesic curvature is zero. Since kg = 0 along a geodesic 

surface curve, then at every point on this surface curve the normal N to the curve will be in the same 

direction as the normal n to the surface. In this case, we have ru · n =0andrv · n = 0 which reduces to 

since the vectors n and d T 

ds 

d T 

ds · ru =0 and d T 

ds · rv =0, (1.5.49) 

have the same direction. In particular, we may write 

∂r du ∂r dv 

= + 

∂u ds ∂v ds = ru u ′ + rv v ′ 

d T 

ds = ruu (u ′ ) 2 +2ruv u ′ v ′ + rvv (v ′ ) 2 + ru u ′′ + rv v ′′ 

T = dr 

ds 

Consequently, the equations (1.5.49) become 

d T 

ds · ru =(ruu · ru)(u ′ ) 2 +2(ruv · ru) u ′ v ′ +(rvv · ru)(v ′ ) 2 + Eu ′′ + Fv ′′ =0 

d T 

ds · rv =(ruu · rv)(u ′ ) 2 +2(ruv · rv) u ′ v ′ +(rvv · rv)(v ′ ) 2 + Fu ′′ + Gv ′′ . (1.5.50) 

=0. 

Utilizing the results from exercise 1.5,(See problems 4,5 and 6), we can eliminate v ′′ from the equations 

(1.5.50) to obtain 

d2 2 

u 1 du 1 du dv 

+ 

+2 

ds2 11 ds 12 ds ds + 

2 1 dv 

=0 

22 ds 

and eliminating u ′′ from the equations (1.5.50) produces the equation 

d2 2 

v 2 du 2 du dv 

+ 

+2 

ds2 11 ds 12 ds ds + 

2 2 dv 

=0. 

22 ds 

In tensor form, these last two equations are written 

d2uα 

α du 

+ 

ds2 βγ a 

β 

ds 

du γ 

ds 

=0, α,β,γ =1, 2 (1.5.51) 

where u = u 1 and v = u 2 . The equations (1.5.51) are the differential equations defining a geodesic curve on 

a surface. We will find that these same type of equations arise in considering the shortest distance between 

two points in a generalized coordinate system. See for example problem 18 in exercise 2.2.

Tensor Derivatives 

Let uα = uα (t) denote the parametric equations of a curve on the surface defined by the parametric 

equations xi = xi (u1 ,u2 ). We can then represent the surface curve in the spatial geometry since the surface 

curve can be represented in the spatial coordinates through the representation x i = x i (u 1 (t),u 2 (t)) = x i (t). 

Recall that for xi = xi (t) a given curve C , the intrinsic derivative of a vector field Ai along C is defined as 

the inner product of the covariant derivative of the vector field with the tangent vector to the curve. This 

intrinsic derivative is written 

δAi δt = Ai dx 

,j 

j 

dt = 

 

∂Ai 

i 

+ A 

∂xj jk g 

k 

 

dxj dt 

or 

δA i 

δt 

dAi 

= 

dt + 

 

i k dxj 

A 

jk g dt 

where the subscript g indicates that the Christoffel symbol is formed from the spatial metric gij. If Aα is a 

surface vector defined along the curve C, the intrinsic derivative is represented 

δAα δt = Aα du 

,β 

β 

dt = 

 

α ∂A α 

+ A 

∂uβ βγ a 

γ 

β du 

dt 

or 

δAα dAα 

= 

δt dt + 

 

α γ duβ 

A 

βγ a dt 

where the subscript a denotes that the Christoffel is formed from the surface metric aαβ. 

Similarly, the formulas for the intrinsic derivative of a covariant spatial vector Ai or covariant surface 

vector Aα are given by 

δAi dAi 

= 

δt dt − 

 

k dx 

Ak 

ij g 

j 

dt 

and 

δAα dAα 

= 

δt dt − 

 

γ du 

Aα 

αβ a 

β 

dt . 

Consider a mixed tensor T i α which is contravariant with respect to a transformation of space coordinates 

xi and covariant with respect to a transformation of surface coordinates uα . For T i α defined over the surface 

curve C, which can also be viewed as a space curve C, define the scalar invariant Ψ = Ψ(t) =T i αAiB α where 

Ai is a parallel vector field along the curve C when it is viewed as a space curve and Bα is also a parallel 

vector field along the curve C when it is viewed as a surface curve. Recall that these parallel vector fields 

must satisfy the differential equations 

δAi 

δt 

dAi 

= 

dt − 

 

k dx 

Ak 

ij g 

j 

dt 

=0 and 

δB α 

δt 

dBα 

= 

dt + 

 

α γ duβ 

B 

βγ a dt 

=0. (1.5.52) 

The scalar invariant Ψ is a function of the parameter t of the space curve since both the tensor and the 

parallel vector fields are to be evaluated along the curve C. By differentiating the function Ψ with respect 

to the parameter t there results 

dΨ 

dt = dT i α 

dt AiB α + T i dAi 

α 

dt Bα + T i αAi dBα . (1.5.53) 

dt 

141

142 

But the vectors Ai and Bα are parallel vector fields and must satisfy the relations given by equations (1.5.52). 

This implies that equation (1.5.53) can be written in the form 

dΨ 

dt = 

 

dT i α 

dt + 

 

i 

T 

kj g 

k dx 

α 

j 

dt − 

 

γ 

T 

βα a 

i du 

γ 

β 

 

AiB 

dt 

α . (1.5.54) 

The quantity inside the brackets of equation (1.5.54) is defined as the intrinsic tensor derivative with respect 

to the parameter t along the curve C. This intrinsic tensor derivative is written 

δT i α 

dt = dT i α 

dt + 

 

i 

T 

kj g 

k dx 

α 

j 

dt − 

 

γ 

T 

βα a 

i du 

γ 

β 

. (1.5.55) 

dt 

The spatial representation of the curve C is related to the surface representation of the curve C through the 

defining equations. Therefore, we can express the equation (1.5.55) in the form 

δT i α 

dt = 

 

∂T i 

α i 

+ T 

∂uβ kj g 

k ∂x 

α 

j 

γ 

− T 

∂uβ βα a 

i 

du 

γ 

β 

(1.5.56) 

dt 

The quantity inside the brackets is a mixed tensor which is defined as the tensor derivative of T i α with 

respect to the surface coordinates uβ . The tensor derivative of the mixed tensor T i α with respect to the 

surface coordinates u β is written 

T i α,β = ∂T i α 

+ 

∂uβ In general, given a mixed tensor T i...j 

α...β 

 

i 

 

kj g 

T k α 

∂x j 

− 

∂uβ 

γ 

T 

βα a 

i γ. 

which is contravariant with respect to transformations of the 

space coordinates and covariant with respect to transformations of the surface coordinates, then we can 

define the scalar field along the surface curve C as 

Ψ(t) =T i...j 

α...β Ai ···AjB α ···B β 

(1.5.57) 

where Ai,...,Aj and B α ,...,B β are parallel vector fields along the curve C. The intrinsic tensor derivative 

is then derived by differentiating the equation (1.5.57) with respect to the parameter t. 

Tensor derivatives of the metric tensors gij,aαβ and the alternating tensors ɛijk,ɛαβ and their associated 

tensors are all zero. Hence, they can be treated as constants during the tensor differentiation process. 

Generalizations 

In a Riemannian space Vn with metric gij and curvilinear coordinates x i ,i =1, 2, 3, the equations of a 

surface can be written in the parametric form x i = x i (u 1 ,u 2 )whereu α ,α =1, 2 are called the curvilinear 

coordinates of the surface. Since 

dx i = ∂xi 

duα 

(1.5.58) 

∂uα then a small change duα on the surface results in change dxi in the space coordinates. Hence an element of 

arc length on the surface can be represented in terms of the curvilinear coordinates of the surface. This same 

element of arc length can also be represented in terms of the curvilinear coordinates of the space. Thus, an 

element of arc length squared in terms of the surface coordinates is represented 

ds 2 = aαβdu α du β 

(1.5.59)

where aαβ is the metric of the surface. This same element when viewed as a spatial element is represented 

By equating the equations (1.5.59) and (1.5.60) we find that 

gijdx i dx j = gij 

ds 2 = gijdx i dx j . (1.5.60) 

∂xi ∂uα ∂x j 

∂u β duα du β = aαβdu α du β . (1.5.61) 

The equation (1.5.61) shows that the surface metric is related to the spatial metric and can be calculated 

∂x 

from the relation aαβ = gij 

i 

∂uα ∂xj . This equation reduces to the equation (1.5.21) in the special case of 

∂uβ Cartesian coordinates. In the surface coordinates we define the quadratic form A = aαβdu α du β as the first 

fundamental form of the surface. The tangent vector to the coordinate curves defining the surface are given 

by ∂xi 

∂u α and can be viewed as either a covariant surface vector or a contravariant spatial vector. We define 

this vector as 

x i α = ∂xi 

, i =1, 2, 3, α =1, 2. (1.5.62) 

∂uα Any vector which is a linear combination of the tangent vectors to the coordinate curves is called a surface 

vector. A surface vector A α can also be viewed as a spatial vector A i . The relation between the spatial 

representation and surface representation is Ai = Aαxi α. The surface representation Aα ,α =1, 2andthe 

spatial representation Ai ,i=1, 2, 3 define the same direction and magnitude since 

gijA i A j = gijA α x i αA β x j 

β = gijx i αx j 

β Aα A β = aαβA α A β . 

Consider any two surface vectors A α and B α and their spatial representations A i and B i where 

A i = A α x i α and B i = B α x i α. (1.5.63) 

These vectors are tangent to the surface and so a unit normal vector to the surface can be defined from the 

cross product relation 

niAB sin θ = ɛijkA j B k 

(1.5.64) 

where A, B are the magnitudes of A i ,B i and θ is the angle between the vectors when their origins are made 

to coincide. Substituting equations (1.5.63) into the equation (1.5.64) we find 

niAB sin θ = ɛijkA α x j αBβ x k β . (1.5.65) 

In terms of the surface metric we have AB sin θ = ɛαβA α B β so that equation (1.5.65) can be written in the 

form 

which for arbitrary surface vectors implies 

(niɛαβ − ɛijkx j αx k β)A α B β =0 (1.5.66) 

niɛαβ = ɛijkx j αxkβ or ni = 1 

2 ɛαβɛijkx j αxkβ . (1.5.67) 

The equation (1.5.67) defines a unit normal vector to the surface in terms of the tangent vectors to the 

coordinate curves. This unit normal vector is related to the covariant derivative of the surface tangents as 

143

144 

is now demonstrated. By using the results from equation (1.5.50), the tensor derivative of equation (1.5.59), 

with respect to the surface coordinates, produces 

x i α,β = ∂2xi ∂uα 

i 

+ x 

∂uβ pq g 

p αx q 

β − 

 

σ 

x 

αβ a 

i σ 

(1.5.68) 

where the subscripts on the Christoffel symbols refer to the metric from which they are calculated. Also the 

tensor derivative of the equation (1.5.57) produces the result 

gijx i α,γx j 

β + gijx i αx j 

β,γ = aαβ,γ =0. (1.5.69) 

Interchanging the indices α, β, γ cyclically in the equation (1.5.69) one can verify that 

gijx i α,βx j γ =0. (1.5.70) 

The equation (1.5.70) indicates that in terms of the space coordinates, the vector xi α,β is perpendicular to 

the surface tangent vector xi γ and so must have the same direction as the unit surface normal ni . Therefore, 

there must exist a second order tensor bαβ such that 

bαβn i = x i α,β . (1.5.71) 

By using the relation gijn i n j = 1 we can transform equation (1.5.71) to the form 

bαβ = gijn j x i α,β 

1 

= 

2 ɛγδɛijkx i α,βxjγ xkδ . (1.5.72) 

The second order symmetric tensor bαβ is called the curvature tensor and the quadratic form 

B = bαβdu α du β 

(1.5.73) 

is called the second fundamental form of the surface. 

Consider also the tensor derivative with respect to the surface coordinates of the unit normal vector to 

the surface. This derivative is 

n i 

∂ni i 

,α = + n 

∂uα jk g 

j x k α . (1.5.74) 

Taking the tensor derivative of gijninj = 1 with respect to the surface coordinates produces the result 

gijninj ,α = 0 which shows that the vector nj ,α is perpendicular to ni and must lie in the tangent plane to the 

surface. It can therefore be expressed as a linear combination of the surface tangent vectors xi α and written 

in the form 

n i ,α = η β αx i β 

(1.5.75) 

where the coefficients η β α can be written in terms of the surface metric components aαβ and the curvature 

components bαβ as follows. The unit vector n i is normal to the surface so that 

gijn i x j α 

=0. (1.5.76)

The tensor derivative of this equation with respect to the surface coordinates gives 

gijn i βxjα + gijn i x j 

α,β =0. (1.5.77) 

Substitute into equation (1.5.77) the relations from equations (1.5.57), (1.5.71) and (1.5.75) and show that 

Solving the equation (1.5.78) for the coefficients η γ 

β we find 

bαβ = −aαγη γ 

β . (1.5.78) 

η γ 

β = −aαγ bαβ. (1.5.79) 

Now substituting equation (1.5.79) into the equation (1.5.75) produces the Weingarten formula 

n i ,α = −aγβbγαx i β . (1.5.80) 

This is a relation for the derivative of the unit normal in terms of the surface metric, curvature tensor and 

surface tangents. 

A third fundamental form of the surface is given by the quadratic form 

where cαβ is defined as the symmetric surface tensor 

C = cαβdu α du β 

(1.5.81) 

cαβ = gijn i ,αn j 

,β . (1.5.82) 

By using the Weingarten formula in the equation (1.5.81) one can verify that 

Geodesic Coordinates 

cαβ = a γδ bαγbβδ. (1.5.83) 

In a Cartesian coordinate system the metric tensor gij is a constant and consequently the Christoffel 

symbols are zero at all points of the space. This is because the Christoffel symbols are dependent upon 

the derivatives of the metric tensor which is constant. If the space VN is not Cartesian then the Christoffel 

symbols do not vanish at all points of the space. However, it is possible to find a coordinate system where 

the Christoffel symbols will all vanish at a given point P of the space. Such coordinates are called geodesic 

coordinates of the point P. 

Consider a two dimensional surface with surface coordinates uα and surface metric aαβ. If we transform 

to some other two dimensional coordinate system, say ūα with metric āαβ, where the two coordinates are 

related by transformation equations of the form 

u α = u α (ū 1 , ū 2 ), α =1, 2, (1.5.84) 

145

146 

then from the transformation equation (1.4.7) we can write, after changing symbols, 

 

δ 

βγ 

ā 

∂u α 

= 

∂ū δ 

 

α 

δɛ 

a 

∂uδ ∂ū β 

∂uɛ ∂ū γ + ∂2uα ∂ū β . (1.5.85) 

∂ū γ 

This is a relationship between the Christoffel symbols in the two coordinate systems. If 

apointP , then for that particular point the equation (1.5.85) reduces to 

∂2uα ∂ū β 

α 

= − 

∂ū γ δɛ 

a 

∂uδ ∂ū β 

∂u ɛ 

∂ū γ 

 

δ 

vanishes at 

βγ 

ā 

(1.5.86) 

where all terms are evaluated at the 

point P. Conversely, if the equation (1.5.86) is satisfied at the point P, 

δ 

then the Christoffel symbol must be zero at this point. Consider the special coordinate transforma- 

βγ 

ā 

tion 

u α = u α 0 +ū α − 1 

 

α 

ū 

2 βγ a 

β ū α 

(1.5.87) 

where uα 0 are the surface coordinates of the point P. The point P in the new coordinates is given by 

ū α =0. We now differentiate the relation (1.5.87) to see if it satisfies the equation (1.5.86). We calculate 

the derivatives 

∂uα ∂ū τ = δα τ − 1 

 

α 

ū 

2 βτ a 

β − 1 

 

α 

ū 

2 τγ a 

γ 

 

 

(1.5.88) 

uα =0 

and 

∂2uα ∂ū τ 

α 

 

 

= − 

∂ū σ (1.5.89) 

τσ a 

uα =0 

where these derivative are evaluated at ū α =0. We find the derivative equations (1.5.88) and (1.5.89) do 

satisfy the equation (1.5.86) locally at the point P. Hence, the Christoffel symbols will all be zero at this 

particular point. The new coordinates can then be called geodesic coordinates. 

Riemann Christoffel Tensor 

Consider the Riemann Christoffel tensor defined by the equation (1.4.33). Various properties of this 

tensor are derived in the exercises at the end of this section. We will be particularly interested in the 

Riemann Christoffel tensor in a two dimensional space with metric aαβ and coordinates uα . We find the 

Riemann Christoffel tensor has the form 

R δ .αβγ = ∂ 

∂uβ 

δ 

− 

αγ 

∂ 

∂uγ 

δ τ δ τ δ 

+ 

− 

(1.5.90) 

αβ αγ βτ αβ γτ 

where the Christoffel symbols are evaluated with respect to the surface metric. The above tensor has the 

associated tensor 

Rσαβγ = aσδR δ .αβγ 

which is skew-symmetric in the indices (σ, α) and(β,γ) such that 

(1.5.91) 

Rσαβγ = −Rασβγ and Rσαβγ = −Rσαγβ. (1.5.92) 

The two dimensional alternating tensor is used to define the constant 

K = 1 

4 ɛαβɛ γδ Rαβγδ 

(1.5.93)

(see example 1.5-1) which is an invariant of the surface and called the Gaussian curvature or total curvature. 

In the exercises following this section it is shown that the Riemann Christoffel tensor of the surface can be 

expressed in terms of the total curvature and the alternating tensors as 

Consider the second tensor derivative of x r α 

which can be shown to satisfy the relation 

Rαβγδ = Kɛαβɛγδ. (1.5.94) 

which is given by 

x r α,βγ = ∂xr 

α,β r 

+ x 

∂uγ mn g 

r α,βxnγ − 

 

δ 

x 

αγ a 

r δ,β − 

 

δ 

x 

βγ a 

r α,γ 

(1.5.95) 

x r α,βγ − x r α,γβ = R δ .αβγx r δ. (1.5.96) 

Using the relation (1.5.96) we can now derive some interesting properties relating to the tensors aαβ,bαβ, 

cαβ, Rαβγδ, the mean curvature H and the total curvature K. 

Consider the tensor derivative of the equation (1.5.71) which can be written 

x i α,βγ = bαβ,γn i + bαβn i ,γ 

(1.5.97) 

where 

bαβ,γ = ∂bαβ 

 

σ 

σ 

− bσβ − bασ. (1.5.98) 

∂uα αγ a βγ a 

By using the Weingarten formula, given in equation (1.5.80), the equation (1.5.97) can be expressed in the 

form 

x i α,βγ = bαβ,γn i − bαβa τσ bτγx i σ 

and by using the equations (1.5.98) and (1.5.99) it can be established that 

(1.5.99) 

x r α,βγ − xr α,γβ =(bαβ,γ − bαγ,β)n r − a τδ (bαβbτγ − bαγbτβ)x r δ . (1.5.100) 

Now by equating the results from the equations (1.5.96) and (1.5.100) we arrive at the relation 

R δ .αβγx r δ =(bαβ,γ − bαγ,β)n r − a τδ (bαβbτγ − bαγbτβ)x r δ. (1.5.101) 

Multiplying the equation (1.5.101) by nr and using the results from the equation (1.5.76) there results the 

Codazzi equations 

bαβ,γ − bαγ,β =0. (1.5.102) 

Multiplying the equation (1.5.101) by grmx m σ and simplifying one can derive the Gauss equations of the 

surface 

Rσαβγ = bαγbσβ − bαβbσγ. (1.5.103) 

By using the Gauss equations (1.5.103) the equation (1.5.94) can be written as 

Kɛσαɛβγ = bαγbσβ − bαβbσγ. (1.5.104) 

147

148 

Another form of equation (1.5.104) is obtained by using the equation (1.5.83) together with the relation 

aαβ = −a σγ ɛσαɛβγ. It is left as an exercise to verify the resulting form 

−Kaαβ = cαβ − a σγ bσγbαβ. (1.5.106) 


H = 1 

2 aσγbσγ as the mean curvature of the surface, then the equation (1.5.106) can be written in the form 

By multiplying the equation (1.5.108) by du α du β and summing, we find 

is a relation connecting the first, second and third fundamental forms. 

(1.5.107) 

cαβ − 2Hbαβ + Kaαβ =0. (1.5.108) 

C − 2HB+ KA=0 (1.5.109) 

EXAMPLE 1.5-2 

In a two dimensional space the Riemann Christoffel tensor has only one nonzero independent component 

R1212. ( See Exercise 1.5, problem number 21.) Consequently, the equation (1.5.104) can be written in the 

form K √ √ 

ae12 ae12 = b22b11 − b21b12 and solving for the Gaussian curvature K we find 

Surface Curvature 

K = b22b11 − b12b21 

= 

a11a22 − a12a21 

b 

a 

R1212 

= . (1.5.110) 

a 

For a surface curve u α = u α (s),α =1, 2 lying upon a surface x i = x i (u 1 ,u 2 ),i =1, 2, 3, we have a two 

dimensional space embedded in a three dimensional space. Thus, if t α = duα 

is a unit tangent vector to 

ds 

du 

the surface curve then aαβ 

α du 

ds 

β 

vector to the space curve x i = x i (u 1 (s),u 2 (s)) with T i = dxi 

ds 

The surface vector tα and the space vector T i are related by 

ds = aαβt α t β =1. This same vector can be represented as the unit tangent 

. That is we will have gij 

dx i 

ds 

dx j 

ds = gijT i T j =1. 

T i = ∂xi 

∂uα duα ds = xiα tα . (1.5.111) 

The surface vector t α is a unit vector so that aαβt α t β =1. If we differentiate this equation intrinsically with 

α δtβ 

respect to the parameter s, we find that aαβt 

δs 

=0. This shows that the surface vector δtα 

δs 

is perpendicular 

to the surface vector tα . Let uα denote a unit normal vector in the surface plane which is orthogonal to the 

tangent vector tα . The direction of uα is selected such that ɛαβtαuβ =1. Therefore, there exists a scalar κ (g) 

such that 

δt α 

δs = κ (g)u α 

(1.5.112)

where κ (g) is called the geodesic curvature of the curve. In a similar manner it can be shown that δuα 

δs 

is a surface vector orthogonal to t α . Let δuα 

δs = αtα where α is a scalar constant to be determined. By 

differentiating the relation aαβt α u β = 0 intrinsically and simplifying we find that α = −κ (g) and therefore 

δu α 

δs = −κ (g)t α . (1.5.113) 

The equations (1.5.112) and (1.5.113) are sometimes referred to as the Frenet-Serret formula for a curve 

relative to a surface. 

Taking the intrinsic derivative of equation (1.5.111), with respect to the parameter s, we find that 

δT i 

δs = xi δt 

α 

α 

δs + xi du 

α,β 

β 

ds tα . (1.5.114) 

Treating the curve as a space curve we use the Frenet formulas (1.5.13). If we treat the curve as a surface 

curve, then we use the Frenet formulas (1.5.112) and (1.5.113). In this way the equation (1.5.114) can be 

writtenintheform 

κN i = x i α κ (g)u α + x i α,β tβ t α . (1.5.115) 

By using the results from equation (1.5.71) in equation (1.5.115) we obtain 

κN i = κ (g)u i + bαβn i t α t β 

(1.5.116) 

where ui is the space vector counterpart of the surface vector uα . Let θ denote the angle between the surface 

normal ni and the principal normal N i ,thenwehavethatcosθ = niN i . Hence, by multiplying the equation 

(1.5.116) by ni we obtain 

κ cos θ = bαβt α t β . (1.5.117) 

Consequently, for all curves on the surface with the same tangent vector tα ,thequantityκcos θ will remain 

constant. This result is known as Meusnier’s theorem. Note also that κ cos θ = κ (n) is the normal component 

of the curvature and κ sin θ = κ (g) is the geodesic component of the curvature. Therefore, we write the 

equation (1.5.117) as 

κ (n) = bαβt α t β 

(1.5.118) 

which represents the normal curvature of the surface in the direction tα . The equation (1.5.118) can also be 

writtenintheform 

du 

κ (n) = bαβ 

α du 

ds 

β 

which is a ratio of quadratic forms. 

B 

= 

ds A 

(1.5.119) 

The surface directions for which κ (n) has a maximum or minimum value is determined from the equation 

(1.5.119) which is written as 

(bαβ − κ (n)aαβ)λ α λ β =0. (1.5.120) 

The direction giving a maximum or minimum value to κ (n) must then satisfy 

(bαβ − κ (n)aαβ)λ β =0 (1.5.121) 

149

150 

so that κ (n) must be a root of the determinant equation 

The expanded form of equation (1.5.122) can be written as 

det(bαβ − κ (n)aαβ) =0. (1.5.122) 

κ 2 (n) − aαβbαβκ (n) + b 

=0 (1.5.123) 

a 

where a = a11a22 − a12a21 and b = b11b22 − b12b21. Using the definition given in equation (1.5.107) and using 

the result from equation (1.5.110), the equation (1.5.123) can be expressed in the form 

The roots κ (1) and κ (2) of the equation (1.5.124) then satisfy the relations 

and 

κ 2 (n) − 2Hκ (n) + K =0. (1.5.124) 

H = 1 

2 (κ (1) + κ (2)) (1.5.125) 

K = κ (1)κ (2). (1.5.126) 

Here H is the mean value of the principal curvatures and K is the Gaussian or total curvature which is the 

product of the principal curvatures. It is readily verified that 

H = 

Eg − 2fF + eG 

2(EG − F 2 ) 

and K = 

are invariants obtained from the surface metric and curvature tensor. 

eg − f 2 

EG − F 2 

Relativity 

Sir Isaac Newton and Albert Einstein viewed the world differently when it came to describing gravity and 

the motion of the planets. In this brief introduction to relativity we will compare the Newtonian equations 

with the relativistic equations in describing planetary motion. We begin with an examination of Newtonian 

systems. 

Newton’s viewpoint of planetary motion is a multiple bodied problem, but for simplicity we consider 

only a two body problem, say the sun and some planet where the motion takes place in a plane. Newton’s 

law of gravitation states that two masses m and M are attracted toward each other with a force of magnitude 

GmM 

ρ2 ,whereGisaconstant, ρ is the distance between the masses, m is the mass of the planet and M is the 

mass of the sun. One can construct an x, y plane containing the two masses with the origin located at the 

center of mass of the sun. Let eρ =cosφe1 +sinφe2 denote a unit vector at the origin of this coordinate 

systemandpointinginthedirectionofthemassm. The vector force of attraction of mass M on mass m is 

given by the relation 

F = −GmM 

ρ 2 

eρ. (1.5.127)

Figure 1.5-2. Parabolic and elliptic conic sections 

The equation of motion of mass m with respect to mass M is obtained from Newton’s second law. Let 

ρ = ρ eρ denote the position vector of mass m with respect to the origin. Newton’s second law can then be 

written in any of the forms 

F = −GmM 

ρ 2 

eρ = m d2 ρ 

dt 2 = md V 

dt 

= −GmM 

ρ 3 ρ (1.5.128) 

and from this equation we can show that the motion of the mass m can be described as a conic section. 

Recall that a conic section is defined as a locus of points p(x, y) such that the distance of p from a fixed 

point (or points), called a focus (foci), is proportional to the distance of the point p from a fixed line, called 

a directrix, that does not contain the fixed point. The constant of proportionality is called the eccentricity 

and is denoted by the symbol ɛ. For ɛ = 1 a parabola results; for 0 ≤ ɛ ≤ 1 an ellipse results; for ɛ>1a 

hyperbola results; and if ɛ = 0 the conic section is a circle. 

With reference to figure 1.5-2, a conic section is defined in terms of the ratio FP = ɛ where FP = ρ and 

PD 

PD =2q− ρ cos φ. Fromtheɛratio we solve for ρ and obtain the polar representation for the conic section 

ρ = 

p 

1+ɛ cos φ 

(1.5.129) 

151

152 

where p =2qɛ and the angle φ is known as the true anomaly associated with the orbit. The quantity p is 

called the semi-parameter of the conic section. (Note that when φ = π 

2 ,thenρ = p.) A more general form 

of the above equation is 

ρ = 

p 

1+ɛ cos(φ − φ0) 

or u = 1 

ρ = A[1 + ɛ cos(φ − φ0)], (1.5.130) 

where φ0 is an arbitrary starting anomaly. An additional symbol a, knownasthesemi-majoraxesofan 

elliptical orbit can be introduced where q, p, ɛ, a are related by 

p 

1+ɛ = q = a(1 − ɛ) or p = a(1 − ɛ2 ). (1.5.131) 

To show that the equation (1.5.128) produces a conic section for the motion of mass m with respect to 

mass M we will show that one form of the solution of equation (1.5.128) is given by the equation (1.5.129). 

To verify this we use the following vector identities: 

From the equation (1.5.128) we find that 

 

d 

ρ × 

dt 

dρ 

 

dt 

so that an integration of equation (1.5.133) produces 

ρ × eρ =0 

 

d 

ρ × 

dt 

dρ 

 

=ρ × 

dt 

d2ρ dt2 d eρ 

eρ · 

dt =0 

 

d eρ d eρ 

eρ × eρ × = − 

dt dt . 

(1.5.132) 

= ρ × d2ρ GM 

= − 

dt2 ρ2 ρ × eρ = 0 (1.5.133) 

ρ × dρ 

dt = h = constant. (1.5.134) 

The quantity H = ρ × m V = ρ × m dρ 

dt is the angular momentum of the mass m so that the quantity h 

represents the angular momentum per unit mass. The equation (1.5.134) tells us that h is a constant for our 

two body system. Note that because h is constant we have 

d 

 

V × h = 

dt 

d V 

dt × h = − GM 

ρ2 eρ 

 

× ρ × dρ 

 

dt 

= − GM 

ρ2 eρ 

d eρ dρ 

× [ρ eρ × (ρ + 

dt dt eρ)] 

and consequently an integration produces 

= − GM 

ρ 2 eρ × ( eρ × 

V × h = GM eρ + C 

d eρ 

dt )ρ2 = GM 

d eρ 

dt

where C is a vector constant of integration. The triple scalar product formula gives us 

ρ · ( V × h)= h · (ρ × dρ 

dt )=h2 = GM ρ · eρ + ρ · C 

or 

h 2 = GMρ + Cρcos φ (1.5.135) 

where φ is the angle between the vectors C and ρ. From the equation (1.5.135) we find that 

ρ = 

p 

1+ɛ cos φ 

(1.5.136) 

where p = h2 /GM and ɛ = C/GM. This result is known as Kepler’s first law and implies that when ɛ

154 

The substitution ρ = 1 

u 

Figure 1.5-3. Relative motion of two inertial systems. 

can be used to represent the equation (1.5.142) in the form 

2 du 

dφ 

+ u 2 − 2GM 

h 

E 

u + =0 (1.5.143) 

2 h2 which is a form we will return to later in this section. Note that we can separate the variables in equations 

(1.5.142) or (1.5.143). The results can then be integrate to produce the equation (1.5.130). 

Newton also considered the relative motion of two inertial systems, say S and S. Consider two such 

systems as depicted in the figure 1.5-3 where the S system is moving in the x−direction with speed v relative 

to the system S. 

For a Newtonian system, if at time t = 0 we have clocks in both systems which coincide, than at time t 

apointP (x, y, z) intheSsystem can be described by the transformation equations 

x =x + vt 

y =y 

z =z 

t =t 

or 

x =x − vt 

y =y 

z =z 

t =t. 

(1.5.144) 

These are the transformation equation of Newton’s relativity sometimes referred to as a Galilean transformation. 

Before Einstein the principle of relativity required that velocities be additive and obey Galileo’s velocity 

addition rule 

V P/R = V P/Q + V Q/R. (1.5.145)

That is, the velocity of P with respect to R equals the velocity of P with respect to Q plus the velocity of Q 

with respect to R. For example, a person (P ) running north at 3 km/hr on a train (Q) moving north at 60 

km/hr with respect to the ground (R) has a velocity of 63 km/hr with respect to the ground. What happens 

when (P ) is a light wave moving on a train (Q) which is moving with velocity V relative to the ground? Are 

the velocities still additive? This type of question led to the famous Michelson-Morley experiment which 

has been labeled as the starting point for relativity. Einstein’s answer to the above question was ”NO” and 

required that V P/R = V P/Q = c =speed of light be a universal constant. 

In contrast to the Newtonian equations, Einstein considered the motion of light from the origins 0 and 

0 of the systems S and S. IftheSsystem moves with velocity v relative to the S system and at time t =0 

a light signal is sent from the S system to the S system, then this light signal will move out in a spherical 

wave front and lie on the sphere 

x 2 + y 2 + z 2 = c 2 t 2 

(1.5.146) 

where c is the speed of light. Conversely, if a light signal is sent out from the S system at time t = 0, it will 

lie on the spherical wave front 

x 2 + y 2 + z 2 = c 2 t 2 . (1.5.147) 

Observe that the Newtonian equations (1.5.144) do not satisfy the equations (1.5.146) and (1.5.147) identically. 

If y = y and z = z then the space variables (x, x) andtimevariables(t, t) must somehow be related. 

Einstein suggested the following transformation equations between these variables 

x = γ(x − vt) and x = γ(x + vt) (1.5.148) 

where γ is a constant to be determined. The differentials of equations (1.5.148) produce 

from which we obtain the ratios 

dx γ(dx − vdt) 

= 

γ(dx + vdt) dx 

When dx 

dt 

= dx 

dt 

dx = γ(dx − vdt) and dx = γ(dx + vdt) (1.5.149) 

or 

1 

γ(1 + v dx 

dt 

= c, the speed of light, the equation (1.5.150) requires that 

γ 2 =(1− v2 

)−1 

c2 From the equations (1.5.148) we eliminate x and find 

) = γ(1 − v 

dx 

dt 

). (1.5.150) 

or γ =(1− v2 

c 2 )−1/2 . (1.5.151) 

t = γ(t − v 

x). (1.5.152) 

c2 We can now replace the Newtonian equations (1.5.144) by the relativistic transformation equations 

x =γ(x + vt) 

y =y 

z =z 

t =γ(t + v 

x) 

c2 or 

x =γ(x − vt) 

y =y 

z =z 

t =γ(t − v 

x) 

c2 (1.5.153) 

155

156 

where γ is given by equation (1.5.151). These equations are also known as the Lorentz transformation. 

Note that for v

where g11 = −1, g22 = −ρ2 , g33 = −ρ2 sin 2 θ, g44 = c2 and gij =0fori= j. The negative signs are 

2 2 2 = c − v is positive when v

158 

Subtracting the first equation from the third equation gives 

The second equation in (1.5.164) then becomes 

du dv 

+ 

dρ dρ =0 or u + v = c1 = constant. (1.5.165) 

ρ du 

=1− eu 

dρ 

Separate the variables in equation (1.5.166) and integrate to obtain the result 

e u = 

where c2 is a constant of integration and consequently 

e v = e c1−u = e c1 

1 

1 − c2 

ρ 

(1.5.166) 

(1.5.167) 

 

1 − c2 

 

. (1.5.168) 

ρ 

The constant c1 is selected such that g44 approaches c2 as ρ increases without bound. This produces the 

metrices 

g11 = −1 

1 − c2 , g22 = −ρ 

ρ 

2 , g33 = −ρ 2 sin 2 θ, g44 = c 2 (1 − c2 

) (1.5.169) 

ρ 

where c2 is a constant still to be determined. The metrices given by equation (1.5.169) are now used to 

expand the equations (1.5.157) representing the geodesics in this four dimensional space. The differential 

equations representing the geodesics are found to be 

d2 2 ρ 1 du dρ 

+ − ρe 

ds2 2 dρ ds 

−u 

2 dθ 

− ρe 

ds 

−u sin 2 2 dφ 

θ + 

ds 

1 

2 dv dt 

ev−u =0 

2 dρ ds 

d 

(1.5.170) 

2 2 θ 2 dθ dρ 

dφ 

+ − sin θ cos θ =0 

ds2 ρ ds ds ds 

(1.5.171) 

d2φ 2 dφ dρ θ dφ dθ 

+ +2cos =0 

ds2 ρ ds ds sin θ ds ds 

(1.5.172) 

d2t dv dt dρ 

+ =0. 

ds2 dρ ds ds 

(1.5.173) 

is a constant. This 

value of θ also simplifies the equations (1.5.170) and (1.5.172). The equation (1.5.172) becomes an exact 

differential equation 

 

d 2 dφ 

ρ =0 

ds ds 

or 

2 dφ 

ρ 

ds = c4, (1.5.174) 

and the equation (1.5.173) also becomes an exact differential 

 

d dt 

ds ds ev 

 

=0 or 

dt 

ds ev = c5, (1.5.175) 

The equation (1.5.171) is identically satisfied if we examine planar orbits where θ = π 

2 

where c4 and c5 are constants of integration. This leaves the equation (1.5.170) which determines ρ. Substituting 

the results from equations (1.5.174) and (1.5.175), together with the relation (1.5.161), the equation 

(1.5.170) reduces to 

d2ρ c2 

+ 

ds2 2ρ2 + c2c2 4 c2 

− (1 − 

2ρ4 ρ ) c24 =0. (1.5.176) 

ρ3

By the chain rule we have 

d2ρ ds2 = d2ρ dφ2 2 dφ 

+ 

ds 

dρ d 

dφ 

2φ ds2 = d2ρ dφ2 and so equation (1.5.176) can be written in the form 


u 

d2ρ 2 

− 

dφ2 ρ 

c2 4 

+ 

ρ4 dρ 

dφ 

2 2 −2c4 ρ5 

2 dρ 

+ 

dφ 

c2 ρ 

2 

2 

c2 + 

4 

c2 

2 − 

 

1 − c2 

 

ρ =0. (1.5.177) 

ρ 

reduces the equation (1.5.177) to the form 

d2u c2 

+ u − 

dφ2 2c2 = 

4 

3 

2 c2u 2 . (1.5.178) 

Multiply the equation (1.5.178) by 2 du 

dφ and integrate with respect to φ to obtain 

2 du 

dφ 

+ u 2 − c2 

c2 u = c2u 

4 

3 + c6. (1.5.179) 

where c6 is a constant of integration. To determine the constant c6 we write the equation (1.5.161) in the 

special case θ = π 

2 and use the substitutions from the equations (1.5.174) and (1.5.175) to obtain 

e u 

2 dρ 

= e 

ds 

u 

 

dρ 

dφ 

2 dφ 

=1− ρ 

ds 

2 

2 dφ 

+ e 

ds 

v 

2 dt 

ds 

or 2 

dρ 

+ 1 − 

dφ 

c2 

 

ρ 

ρ 

2 

+ 1 − c2 

ρ − c25 c2 4 ρ 

c2 =0. (1.5.180) 

4 

reduces the equation (1.5.180) to the form 

2 du 

+ u 

dφ 

2 − c2u 3 + 1 

c2 4 

Now comparing the equations (1.5.181) and (1.5.179) we select 

2 c5 1 

c6 = − 1 

c2 c2 4 

so that the equation (1.5.179) takes on the form 

2 du 

+ u 

dφ 

2 − c2 

c2 

u + 1 − 

4 

c25 c2 

1 

c2 4 

= c2u 3 

(1.5.182) 

Now we can compare our relativistic equation (1.5.182) with our Newtonian equation (1.5.143). In order 

that the two equations almost agree we select the constants c2,c4,c5 so that 


u 

c2 

c2 4 

= 2GM 

h 2 

and 

− c2 

c2 u − 

4 

c25 c2c2 =0. (1.5.181) 

4 

1 − c2 5 

c 2 

c 2 4 

= E 

. (1.5.183) 

h2 The equations (1.5.183) are only two equations in three unknowns and so we use the additional equation 

dφ 

lim ρ2 = lim 

ρ→∞ dt ρ→∞ 

dφ ds 

ρ2 = h (1.5.184) 

ds dt 

159

160 

which is obtained from equation (1.5.141). Substituting equations (1.5.174) and (1.5.175) into equation 

(1.5.184), rearranging terms and taking the limit we find that 

From equations (1.5.183) and (1.5.185) we obtain the results that 

c 2 5 

c4c2 = h. (1.5.185) 

c5 

c2 

= 

1+ E 

c2 , c2 = 2GM 

c2 

1 

1+E/c 2 

 

, c4 = 

h 

c 1+E/c 2 

These values substituted into equation (1.5.181) produce the differential equation 

Let α = c2 

c2 = 

4 

2GM 

h2 2 du 

dφ 

+ u 2 − 2GM 

h 

(1.5.186) 

E 2GM 

u + = 2 h2 c2 

1 

1+E/c2 

u 3 . (1.5.187) 

and β = c2 = 2GM 

c 2 ( 1 

1+E/c 2 ) then the differential equation (1.5.178) can be written as 

We know the solution to equation (1.5.143) is given by 

d2u α 3 

+ u − = 

dφ2 2 2 βu2 . (1.5.188) 

u = 1 

ρ = A(1 + ɛ cos(φ − φ0)) (1.5.189) 

and so we assume a solution to equation (1.5.188) of this same general form. We know that A is small and so 

we make the assumption that the solution of equation (1.5.188) given by equation (1.5.189) is such that φ0 is 

approximately constant and varies slowly as a function of Aφ. Observethatifφ0 = φ0(Aφ), then dφ0 

dφ = φ′ 0A 

and d2φ0 dφ2 = φ ′′ 

0A2 , where primes denote differentiation with respect to the argument of the function. (i.e. 

Aφ for this problem.) The derivatives of equation (1.5.189) produce 

du 

dφ = − ɛA sin(φ − φ0)(1 − φ ′ 0A) 

d2u dφ2 =ɛA3 sin(φ − φ0)φ ′′ 

0 − ɛA cos(φ − φ0)(1 − 2Aφ ′ 0 + A 2 (φ ′ 0) 2 ) 

= − ɛA cos(φ − φ0)+2ɛA 2 φ ′ 0 cos(φ − φ0)+O(A 3 ). 

Substituting these derivatives into the differential equation (1.5.188) produces the equations 

2ɛA 2 φ ′ 0 cos(φ − φ0)+A − α 

2 

= 3β 

2 

A 2 +2ɛA 2 cos(φ − φ0)+ɛ 2 A 2 cos 2 (φ − φ0) + O(A 3 ). 

Now A is small so that terms O(A3 ) can be neglected. Equating the constant terms and the coefficient of 

the cos(φ − φ0) terms we obtain the equations 

A − α 

2 

= 3β 

2 A2 

2ɛA 2 φ ′ 0 =3βɛA 2 + 3β 

2 ɛ2 A 2 cos(φ − φ0). 

Treating φ0 as essentially constant, the above system has the approximate solutions 

A ≈ α 

2 

φ0 ≈ 3β 3β 

Aφ + 

2 4 Aɛ sin(φ − φ0) (1.5.190)

The solutions given by equations (1.5.190) tells us that φ0 varies slowly with time. For ɛ less than 1, the 

elliptical motion is affected by this change in φ0. It causes the semi-major axis of the ellipse to slowly rotate 

at a rate given by dφ0 

dt . Using the following values for the planet Mercury 

G =6.67(10 −8 )dynecm 2 /g 2 

M =1.99(10 33 )g 

a =5.78(10 12 )cm 

ɛ =0.206 

c =3(10 10 )cm/sec 

β ≈ 2GM 

c2 =2.95(105 )cm 

h ≈ GMa(1 − ɛ2 )=2.71(10 19 )cm 2 /sec 

dφ 

dt ≈ 

 

GM 

a3 1/2 sec −1 Kepler’s third law 

we calculate the slow rate of rotation of the semi-major axis to be approximately 

dφ0 

dt 

= dφ0 

dφ 

dφ 

dt 

3 

≈ 

2 βAdφ 

 

GM 

≈ 3 

dt ch 

2 GM 

a 3 

1/2 

=6.628(10 −14 )rad/sec 

=43.01 seconds of arc per century. 

(1.5.191) 

(1.5.192) 

This slow variation in Mercury’s semi-major axis has been observed and measured and is in agreement with 

the above value. Newtonian mechanics could not account for the changes in Mercury’s semi-major axis, but 

Einstein’s theory of relativity does give this prediction. The resulting solution of equation (1.5.188) can be 

viewed as being caused by the curvature of the space-time continuum. 

The contracted curvature tensor Gij set equal to zero is just one of many conditions that can be assumed 

in order to arrive at a metric for the space-time continuum. Any assumption on the value of Gij relates to 

imposing some kind of curvature on the space. Within the large expanse of our universe only our imaginations 

limit us as to how space, time and matter interact. You can also imagine the existence of other tensor metrics 

in higher dimensional spaces where the geodesics within the space-time continuum give rise to the motion 

of other physical quantities. 

This short introduction to relativity is concluded with a quote from the NASA News@hg.nasa.gov news 

release, spring 1998, Release:98-51. “An international team of NASA and university researchers has found 

the first direct evidence of a phenomenon predicted 80 years ago using Einstein’s theory of general relativity– 

that the Earth is dragging space and time around itself as it rotates.”The news release explains that the 

effect is known as frame dragging and goes on to say “Frame dragging is like what happens if a bowling 

ball spins in a thick fluid such as molasses. As the ball spins, it pulls the molasses around itself. Anything 

stuck in the molasses will also move around the ball. Similarly, as the Earth rotates it pulls space-time in 

its vicinity around itself. This will shift the orbits of satellites near the Earth.”This research is reported in 

the journal Science. 

161

162 

EXERCISE 1.5 

◮ 1. Let κ = δ T 

δs · N and τ = δ N 

δs · B. Assume in turn that each of the intrinsic derivatives of T, N, B are 

some linear combination of T, N, B and hence derive the Frenet-Serret formulas of differential geometry. 

◮ 2. Determine the given surfaces. Describe and sketch the curvilinear coordinates upon each surface. 

(a) r(u, v) =u e1 + v e2 (b) r(u, v) =u cos v e1 + u sin v e2 (c) r(u, v) = 2uv2 

u2 + v2 e1 + 2u2v u2 e2. 

+ v2 ◮ 3. Determine the given surfaces and describe the curvilinear coordinates upon the surface. Use some 

graphics package to plot the surface and illustrate the coordinate curves on the surface. Find element of 

area dS in terms of u and v. 

(a) r(u, v) =a sin u cos v e1 + b sin u sin v e2 + c cos u e3 a, b, c constants 0 ≤ u, v ≤ 2π 

(b) r(u, v) =(4+vsin u 

2 )cosue1 +(4+v sin u 

2 )sinue2 + v cos u 

2 e3 − 1 ≤ v ≤ 1, 0 ≤ u ≤ 2π 

(c) r(u, v) =au cos v e1 + bu sin v e2 + cu e3 

(d) r(u, v) =u cos v e1 + u sin v e2 + αv e3 α constant 

(e) r(u, v) =a cos v e1 + b sin v e2 + u e3 a, b constant 

(f) r(u, v) =u cos v e1 + u sin v e2 + u 2 e3 

◮ 4. 

 

E 

Consider a two dimensional space with metric tensor (aαβ) = 

F 

 

F 

. Assume that the surface is 

G 

described by equations of the form y i = y i (u, v) and that any point on the surface is given by the position 

vector r = r(u, v) =yi ei. Show that the metrices E,F,G are functions of the parameters u, v and are given 

by 

E = ru · ru, F = ru · rv, G = rv · rv where ru = ∂r 

∂u and rv = ∂r 

∂v . 

◮ 5. For the metric given in problem 4 show that the Christoffel symbols of the first kind are given by 

[1 1, 1] = ru · ruu [1 2, 1] = [2 1, 1] = ru · ruv [2 2, 1] = ru · rvv 

[1 1, 2] = rv · ruu 

[1 2, 2] = [2 1, 2] = rv · ruv 

which can be represented [αβ,γ]= ∂2r ∂uα ∂r 

· , 

∂uβ ∂uγ α,β,γ =1, 2. 

◮ 6. Show that the results in problem 5 can also be written in the form 

[1 1, 1] = 1 

2 Eu [1 2, 1] = [2 1, 1] = 1 

2 Ev 

[1 1, 2] = Fu − 1 

2 Ev 

where the subscripts indicate partial differentiation. 

[1 2, 2] = [2 1, 2] = 1 

2 Gu 

[2 2, 2] = rv · rvv 

[2 2, 1] = Fv − 1 

2 Gu 

[2 2, 2] = 1 

2 Gv 

◮ 7. For the metric 

 

given 

 

in problem 4, show that the Christoffel symbols of the second kind can be 

γ 

expressed in the form = a 

αβ 

γδ [αβ,δ], α,β,γ =1, 2 and produce the results 

 

1 

= 

11 

GEu − 2FFu + FEv 

2(EG − F 2 ) 

 

1 

= 

22 

2GFv − GGu − FGv 

2(EG − F 2 

1 1 

= = 


) 

GEv − FGu 

2(EG − F 2 ) 

 

2 2 

= = 


EGu − FEv 

2(EG − F 2 

2 

= 

11 

) 

2EFu − EEv − FEu 

2(EG − F 2 ) 

 

2 

= 

22 

EGv − 2FFv + FGu 

2(EG − F 2 ) 

where the subscripts indicate partial differentiation.

◮ 8. Derive the Gauss equations by assuming that 

ruu = c1ru + c2rv + c3 n , ruv = c4ru + c5rv + c6 n , rvv = c7ru + c8rv + c9n 

where c1,...,c9 are constants 

 

determined 

 

by 

 

taking dot products 

 

of 

 

the above 

 

vectors 

 

with the vectors ru,rv, 

1 

2 

1 

2 

and n. Show that c1 = , c2 = , c3 = e, c4 = , c5 = , c6 = f, 

 

11 11 

12 12 

1 

2 

∂ 

c7 = , c8 = , c9 = g Show the Gauss equations can be written 

22 22 

2r ∂uα 

γ ∂r 

= 

+bαβn. 

∂uβ αβ ∂uγ ◮ 9. Derive the Weingarten equations 

and show 

fF − eG 

c1 = 

EG − F 2 

eF − fE 

c2 = 

EG − F 2 

nu = c1ru + c2rv 

nv = c3ru + c4rv 

gF − fG 

c3 = 

EG − F 2 

fF − gE 

c4 = 

EG − F 2 

and 

c ∗ 1 = 

c ∗ 2 

ru = c ∗ 1 nu + c ∗ 2 nv 

rv = c ∗ 3 nu + c ∗ 4 nv 

fF − gE 

eg − f 2 

= fE − eF 

eg − f 2 

c ∗ 3 = 

c ∗ 4 

fG− gF 

eg − f 2 

= fF − eG 

eg − f 2 

The constants in the above equations are determined in a manner similar to that suggested in problem 8. 

Show that the Weingarten equations can be written in the form 

∂n 

∂u α = −bβ α 

∂r 

. 

∂uβ ◮ 10. Using n = ru × rv 

√ , the results from exercise 1.1, problem 9(a), and the results from problem 5, 

EG − F 2 

verify that 

 

2 EG 

(ru × ruu) · n = 

− F 2 

11 

 

2 EG 

(ru × ruv) · n = 

− F 2 


 

1 EG 

(rv × ruu) · n = − 

− F 2 

11 

 

2 EG 

(ru × rvv) · n = 

− F 2 

22 

 

1 EG 

(rv × ruv) · n = − 

− F 2 

21 

 

1 EG 

(rv × rvv) · n = − 

− F 2 

22 

(ru × rv) · n = EG − F 2 

and then derive the formula for the geodesic curvature given by equation (1.5.48). 

Hint:(n × T ) · d T 

ds =( T × d T 

ds ) · n and aαδ ]βγ,δ]= 

α 

βγ 

 

. 

163

164 

◮ 11. Verify the equation (1.5.39) which shows that the normal curvature directions are orthogonal. i.e. 

verify that Gλ1λ2 + F (λ1 + λ2)+E =0. 

◮ 12. Verify that δ βγ 

στ δ ωα 

λν Rωαβγ =4Rλνστ . 

◮ 13. Find the first fundamental form and unit normal to the surface defined by z = f(x, y). 

◮ 14. Verify 

Ai,jk − Ai,kj = AσR σ .ijk 

where 

R σ .ijk = ∂ 

∂xj 

σ 

− 

ik 

∂ 

∂xk 

σ n σ n σ 

+ 

− 

. 

ij ik nj ij nk 

which is sometimes written 

 

 

∂ 

Rinjk = ∂x 

 

 

j 

∂ 

∂xk [nj, k] 

 

 

 

 

 

[nk, i] + 

 

 

s 

nj 

 

[ij, s] 

 

s 

 

nk 

 

[ik, s] 

◮ 15. For Rijkl = giσR σ .jkl show 

which is sometimes written 

◮ 16. Show 

Rijkl = 1 

2 

Rinjk = ∂ 

 

∂ 

s 

s 

[nk, i] − [nj, i]+[ik, s] − [ij, s] 

∂xj ∂xk nj 

nk 

R σ .ijk = 

 

 

 

 

 

 

 

 

∂ 

∂xj 

σ 

ij 

∂ 

∂xk σ 

ik 

 

 

 

 

 

+ 

 

 

 

 

n 

ik 

 

σ 

nk 

 

n 

 

ij 

 

σ 

nj 

2 ∂ gil 

∂xj∂xk − ∂2gjl ∂xi∂xk − ∂2gik ∂xj∂xl + ∂2gjk ∂xi∂xl 

+ g αβ ([jk,β][il, α] − [jl,β][ik, α]) . 

◮ 17. Use the results from problem 15 to show 

(i) Rjikl = −Rijkl, (ii) Rijlk = −Rijkl, (iii) Rklij = Rijkl 

Hence, the tensor Rijkl is skew-symmetric in the indices i, j and k, l. Also the tensor Rijkl is symmetric with 

respect to the (ij) and(kl) pair of indices. 

◮ 18. Verify the following cyclic properties of the Riemann Christoffel symbol: 

(i) Rnijk + Rnjki + Rnkij = 0 first index fixed 

(ii) Rinjk + Rjnki + Rknij = 0 second index fixed 

(iii) Rijnk + Rjkni + Rkinj = 0 third index fixed 

(iv) Rikjn + Rkjin + Rjikn = 0 fourth index fixed 

◮ 19. By employing the results from the previous problems, show all components of the form: 

Riijk, Rinjj, Riijj, Riiii, (no summation on i or j) must be zero.

◮ 20. Find the number of independent components associated with the Riemann Christoffel tensor 

Rijkm, i,j,k,m=1, 2,...,N.There are N 4 components to examine in an N−dimensional space. Many of 

these components are zero and many of the nonzero components are related to one another by symmetries 

or the cyclic properties. Verify the following cases: 

CASE I We examine components of the form Rinin, i = n with no summation of i or n. The first index 

can be chosen in N ways and therefore with i = n the second index can be chosen in N − 1 ways. Observe 

that Rinin = Rnini, (no summation on i or n) and so one half of the total combinations are repeated. This 

leaves M1 = 1 

2 N(N − 1) components of the form Rinin. The quantity M1 can also be thought of as the 

number of distinct pairs of indices (i, n). 

CASE II We next examine components of the form Rinji, i = n = j where there is no summation on 

the index i. We have previously shown that the first pair of indices can be chosen in M1 ways. Therefore, 

the third index can be selected in N − 2 ways and consequently there are M2 = 1 

2N(N − 1)(N − 2) distinct 

components of the form Rinji with i = n = j. 

CASE III Next examine components of the form Rinjk where i = n = j = k. From CASE I the first pairs 

of indices (i, n) can be chosen in M1 ways. Taking into account symmetries, it can be shown that the second 

pair of indices can be chosen in 1 

1 

2 (N − 2)(N − 3) ways. This implies that there are 4N(N − 1)(N − 2)(N − 3) 

ways of choosing the indices i, n, j and k with i = n = j = k. By symmetry the pairs (i, n) and(j, k) canbe 

interchanged and therefore only one half of these combinations are distinct. This leaves 

1 

N(N − 1)(N − 2)(N − 3) 

8 

distinct pairs of indices. Also from the cyclic relations we find that only two thirds of the above components 

are distinct. This produces 

N(N − 1)(N − 2)(N − 3) 

M3 = 


distinct components of the form Rinjk with i = n = j = k. 

Adding the above components from each case we find there are 

distinct and independent components. 

Verify the entries in the following table: 

M4 = M1 + M2 + M3 = N 2 (N 2 − 1) 


Dimension of space N 1 2 3 4 5 

Number of components N 4 1 16 81 256 625 

M4 = Independent components of Rijkm 0 1 6 20 50 

Note 1: A one dimensional space can not be curved and all one dimensional spaces are Euclidean. (i.e. if we have 

an element of arc length squared given by ds2 = f(x)(dx) 2 , we can make the coordinate transformation 

 

2 2 f(x)dx = du and reduce the arc length squared to the form ds = du .) 

Note 2: In a two dimensional space, the indices can only take on the values 1 and 2. In this special case there 

are 16 possible components. It can be shown that the only nonvanishing components are: 

R1212 = −R1221 = −R2112 = R2121. 

165

166 

For these nonvanishing components only one independent component exists. By convention, the component 

R1212 is selected as the single independent component and all other nonzero components are 

expressed in terms of this component. 

Find the nonvanishing independent components Rijkl for i, j, k, l =1, 2, 3, 4andshowthat 

R1212 

R1313 

R2323 

R1414 

R2424 

R3434 

R1231 

R1421 

R1341 

R2132 

can be selected as the twenty independent components. 

R2142 

R2342 

R3213 

R3243 

R3143 

R4124 

R4314 

R4234 

R1324 

R1432 

◮ 21. 

(a) For N =2showR1212 is the only nonzero independent component and 

R1212 = R2121 = −R1221 = −R2112. 

(b) Show that on the surface of a sphere of radius r0 we have R1212 = r2 0 sin 2 θ. 

◮ 22. Show for N =2that 

R1212 = R1212J 2 = R1212 

 

 

 

∂x 

 

∂x 

 

◮ 23. Define Rij = Rs .ijs as the Ricci tensor and Gij = Ri 1 

j − 2δi jR as the Einstein tensor, where Ri j = gikRkj and R = Ri i . Show that 

(a) Rjk = g ab Rjabk 

(b) Rij = ∂2 log √ g 

∂xi − 

∂xj (c) R i ijk =0 

√ 

b ∂ log g ∂ 

− 

ij ∂xb ∂xa 

a b a 

+ 

ij ia jb 

◮ 24. By employing the results from the previous problem show that in the case N =2wehave 

R11 

g11 

= R22 

g22 

= R12 

g12 

= − R1212 

g 

where g is the determinant of gij. 

◮ 25. Consider the case N =2wherewehaveg12 = g21 = 0 and show that 

(a) R12 = R21 =0 

(b) R11g22 = R22g11 = R1221 

2 

(c) R = 2R1221 

g11g22 

(d) Rij = 1 

2 Rgij, where R = g ij Rij 

The scalar invariant R is known as the Einstein curvature of the surface and the tensor G i j = Ri j 

1 − 2δi jR is 

known as the Einstein tensor. 

◮ 26. For N =3showthatR1212,R1313,R2323,R1213,R2123,R3132 are independent components of the 

Riemann Christoffel tensor.

a11 0 

◮ 27. For N =2andaαβ = 

show that 

0 a22 

◮ 28. For N =2andaαβ = 

K = 1 

2 √ 

∂ 

a ∂u1 K = R1212 

a 

a12 

a11 

a11 a12 

√ a 

1 

= − 

2 √ 

∂ 

a ∂u1 

1 

√a 

a21 a22 

∂a11 1 

− √ 

∂u2 a 

 

show that 

∂a22 

∂u1 

+ ∂ 

∂u2 

1 

√a 

∂a22 

∂u1 

+ ∂ 

∂u2 

2 ∂a12 1 

√a − √ 

∂u1 a 

∂a11 

∂u2 

. 

∂a11 a12 

− 

∂u2 a11 

√ a 

∂a11 

∂u1 

. 

Check your results by setting a12 = a21 = 0 and comparing this answer with that given in the problem 27. 

◮ 29. Write out the Frenet-Serret formulas (1.5.112)(1.5.113) for surface curves in terms of Christoffel 

symbols of the second kind. 

◮ 30. 

(a) Use the fact that for n =2wehaveR1212 = R2121 = −R2112 = −R1221 together with eαβ, eαβ the two 

dimensional alternating tensors to show that the equation (1.5.110) can be written as 

Rαβγδ = Kɛαβɛγδ where ɛαβ = √ aeαβ and ɛ αβ = 1 

√ a e αβ 

are the corresponding epsilon tensors. 

(b) Show that from the result in part (a) we obtain 1 

4 Rαβγδɛ αβ ɛ γδ = K. 

Hint: See equations (1.3.82),(1.5.93) and (1.5.94). 

◮ 31. Verify the result given by the equation (1.5.100). 

◮ 32. Show that a αβ cαβ =4H 2 − 2K. 

◮ 33. Find equations for the principal curvatures associated with the surface 

x = u, y = v, z = f(u, v). 

◮ 34. Geodesics on a sphere Let (θ, φ) denote the surface coordinates of the sphere of radius ρ defined 

by the parametric equations 

x = ρ sin θ cos φ, y = ρ sin θ sin φ, z = ρ cos θ. (1) 

Consider also a plane which passes through the origin with normal having the direction numbers (n1,n2,n3). 

This plane is represented by n1x+n2y +n3z = 0 and intersects the sphere in a great circle which is described 

by the relation 

n1 sin θ cos φ + n2 sin θ sin φ + n3 cos θ =0. (2) 

This is an implicit relation between the surface coordinates θ, φ which describes the great circle lying on the 

sphere. We can write this later equation in the form 

n1 cos φ + n2 sin φ = −n3 

tan θ 

(3) 

167

168 

and in the special case where n1 =cosβ, n2 =sinβ,n3 = − tan α is expressible in the form 

cos(φ − β) = 

tan α 

tan θ 

or φ − β =cos −1 

 

tan α 

. (4) 

tan θ 

The above equation defines an explicit relationship between the surface coordinates which defines a great 

circle on the sphere. The arc length squared relation satisfied by the surface coordinates together with the 

equation obtained by differentiating equation (4) with respect to arc length s gives the relations 

sin 2 θ dφ 

ds = 

tan α 

 

1 − tan2 α 

tan 2 θ 

dθ 

ds 

ds 2 = ρ 2 dθ 2 + ρ 2 sin 2 θdφ 2 

The above equations (1)-(6) are needed to consider the following problem. 

(a) Show that the differential equations defining the geodesics on the surface of a sphere (equations (1.5.51)) 

are 

d2θ − sin θ cos θ 

ds2 (b) Multiply equation (8) by sin 2 θ and integrate to obtain 

where c1 is a constant of integration. 

(c) Multiply equation (7) by dθ 

ds 

where c 2 2 

is a constant of integration. 

(5) 

(6) 

2 dφ 

=0 (7) 

ds 

d2φ dθ dφ 

+2cotθ =0 (8) 

ds2 ds ds 

sin 2 θ dφ 

ds 

(d) Use the equations (5)(6) to show that c2 =1/ρ and c1 = 

(e) Show that equations (9) and (10) imply that 

= c1 

(9) 

and use the result of equation (9) to show that an integration produces 

2 dθ 

= 

ds 

−c21 sin 2 θ + c22 (10) 

tan α 

tan θ 

dφ 

dθ 

= tan α 

tan 2 θ 

sin α 

ρ . 

sec 2 θ 

 

1 − tan2 α 

tan 2 θ 

and making the substitution u = this equation can be integrated to obtain the equation (4). We 

can now expand the equation (4) and express the results in terms of x, y, z to obtain the equation (3). 

This produces a plane which intersects the sphere in a great circle. Consequently, the geodesics on a 

sphere are great circles.

◮ 35. Find the differential equations defining the geodesics on the surface of a cylinder. 

◮ 36. Find the differential equations defining the geodesics on the surface of a torus. (See problem 13, 

Exercise 1.3) 

◮ 37. Find the differential equations defining the geodesics on the surface of revolution 

x = r cos φ, y = r sin φ, z = f(r). 

Note the curve z = f(x) gives a profile of the surface. The curves r = Constant are the parallels, while the 

curves φ = Constant are the meridians of the surface and 

ds 2 =(1+f ′2 ) dr 2 + r 2 dφ 2 . 

◮ 38. Find the unit normal and tangent plane to an arbitrary point on the right circular cone 

x = u sin α cos φ, y = u sin α sin φ, z = u cos α. 

This is a surface of revolution with r = u sin α and f(r) =r cot α with α constant. 

◮ 39. Let s denote arc length and assume the position vector r(s) is analytic about a point s0. Show that 

the Taylor series r(s) =r(s0)+hr ′ (s0)+ h2 

2! r ′′ (s0)+ h3 

3! r ′′′ (s0)+··· about the point s0, withh = s − s0 is 

given by r(s) =r(s0)+h T + 1 

2 κh2 N + 1 

6 h3 (−κ 2 T + κ ′ N + κτ B)+··· which is obtained by differentiating 

the Frenet formulas. 

◮ 40. 

(a) Show that the circular helix defined by x = a cos t, y = a sin t, z = bt with a, b constants, has the 

property that any tangent to the curve makes a constant angle with the line defining the z-axis. 

(i.e. T · e3 =cosα = constant.) 

(b) Show also that N · e3 = 0 and consequently e3 is parallel to the rectifying plane, which implies that 

e3 = T cos α + B sin α. 

(c) Differentiate the result in part (b) and show that κ/τ =tanαis a constant. 

◮ 41. Consider a space curve xi = xi(s) in Cartesian coordinates. 

 

 

(a) Show that κ = d T 

 

 

= 

ds 

x ′ ix′ i 

(b) Show that τ = 1 

κ 2 eijkx ′ i x′′ 

j x′′′ 

k . Hint: Consider r ′ · r ′′ × r ′′′ 

◮ 42. 

(a) Find the direction cosines of a normal to a surface z = f(x, y). 

(b) Find the direction cosines of a normal to a surface F (x, y, z) =0. 

(c) Find the direction cosines of a normal to a surface x = x(u, v),y = y(u, v),z = z(u, v). 

◮ 43. Show that for a smooth surface z = f(x, y) the Gaussian curvature at a point on the surface is given 

by 

K = fxxfyy − f 2 xy 

(f 2 x + f 2 . 

y +1)2 

169

170 

◮ 44. Show that for a smooth surface z = f(x, y) themeancurvatureatapointonthesurfaceisgivenby 

H = (1 + f 2 y )fxx − 2fxfyfxy +(1+f2 x )fyy 

2(f 2 x + f 2 y +1) 3/2 

. 

◮ 45. Express the Frenet-Serret formulas (1.5.13) in terms of Christoffel symbols of the second kind. 

◮ 46. Verify the relation (1.5.106). 

◮ 47. In Vn assume that Rij = ρgij and show that ρ = R 

n where R = gijRij. This result is known as 

Einstein’s gravitational equation at points where matter is present. It is analogous to the Poisson equation 

∇2V = ρ from the Newtonian theory of gravitation. 

◮ 48. In Vn assume that Rijkl = K(gikgjl − gilgjk) andshowthatR = Kn(1 − n). (Hint: See problem 23.) 

◮ 49. Assume gij =0fori= j and verify the following. 

(a) Rhijk =0forh= i = j = k 

(b) Rhiik = √ 2 ∂ 

gii 

√ gii 

∂xh∂xk − ∂√gii ∂xh ∂ log √ ghh 

∂xk − ∂√gii ∂xk ∂ log √ gkk 

∂xh 

for h, i, k unequal. 

(c) Rhiih = √ ⎡ 

√ ⎢ 

gii ghh ⎣ ∂ 

∂xh 

1 ∂ √ gii 

∂xh 

+ ∂ 

∂xi 

1 ∂ √ ghh 

∂xi n ∂ 

+ 

√ gii 

∂xm ∂ √ ghh 

∂xm ⎤ 

⎥ 

⎦ where h = i. 

√ghh 

√gii 

m=1 

m=h m=i 

◮ 50. Consider a surface of revolution where x = r cos θ, y = r sin θ and z = f(r) is a given function of r. 

(a) Show in this V2 we have ds 2 =(1+(f ′ ) 2 )dr 2 + r 2 dθ 2 where ′ = d 

ds . 

(b) Show the geodesic equations in this V2 are 

d2r ds2 + f ′ f ′′ 

1+(f ′ ) 2 

d2θ 2 dθ dr 

+ 

ds2 r ds ds =0 

2 dr 

− 

ds 

(c) Solve the second equation in part (b) to obtain dθ 

ds 

r 

1+(f ′ ) 2 

dθ = ± a1+(f ′ ) 2 

r √ r2 dr which theoretically can be integrated. 

− a2 2 dθ 

=0 

ds 

a 

= . Substitute this result for ds in part (a) to show 

r2

PART 2: INTRODUCTION TO CONTINUUM MECHANICS 

In the following sections we develop some applications of tensor calculus in the areas of dynamics, 

elasticity, fluids and electricity and magnetism. We begin by first developing generalized expressions for the 

vector operations of gradient, divergence, and curl. Also generalized expressions for other vector operators 

are considered in order that tensor equations can be converted to vector equations. We construct a table to 

aid in the translating of generalized tensor equations to vector form and vice versa. 

The basic equations of continuum mechanics are developed in the later sections. These equations are 

developed in both Cartesian and generalized tensor form and then converted to vector form. 

§2.1 TENSOR NOTATION FOR SCALAR AND VECTOR QUANTITIES 

We consider the tensor representation of some vector expressions. Our goal is to develop the ability to 

convert vector equations to tensor form as well as being able to represent tensor equations in vector form. 

In this section the basic equations of continuum mechanics are represented using both a vector notation and 

the indicial notation which focuses attention on the tensor components. In order to move back and forth 

between these notations, the representation of vector quantities in tensor form is now considered. 

Gradient 

For Φ = Φ(x 1 ,x 2 ,...,x N ) a scalar function of the coordinates x i ,i =1,...,N , the gradient of Φ is 

defined as the covariant vector 

The contravariant form of the gradient is 

Φ,i = ∂Φ 

, i =1,...,N. (2.1.1) 

∂xi g im Φ,m. (2.1.2) 

Note, if C i = g im Φ,m, i =1, 2, 3 are the tensor components of the gradient then in an orthogonal coordinate 

system we will have 

C 1 = g 11 Φ,1, C 2 = g 22 Φ,2, C 3 = g 33 Φ,3. 

We note that in an orthogonal coordinate system that gii =1/h2 i , (no sum on i), i =1, 2, 3 and hence 

replacing the tensor components by their equivalent physical components there results the equations 

C(1) 

h1 

= 1 

h 2 1 

∂Φ 

, 

∂x1 C(2) 

h2 

= 1 

h 2 2 

∂Φ 

, 

∂x2 Simplifying, we find the physical components of the gradient are 

C(1) = 1 

h1 

∂Φ 

1 

, C(2) = 

∂x1 h2 

C(3) 

h3 

= 1 

h 2 3 

∂Φ 

1 

, C(3) = 

∂x2 h3 

∂Φ 

. 

∂x3 ∂Φ 

. 

∂x3 These results are only valid when the coordinate system is orthogonal and gij =0fori = j and gii = h 2 i , 

with i =1, 2, 3, and where i is not summed. 

171

172 

Divergence 

The divergence of a contravariant tensor Ar is obtained by taking the covariant derivative with respect 

to xk and then performing a contraction. This produces 

div A r = A r ,r. (2.1.3) 

Still another form for the divergence is obtained by simplifying the expression (2.1.3). The covariant derivative 


A r 

∂Ar r 

,k = + A 

∂xk mk 

m . 

Upon contracting the indices r and k and using the result from Exercise 1.4, problem 13, we obtain 

A r ,r 

∂Ar 1 ∂( 

= + 

∂xr √ 

g 

√ g) 

Am 

∂xm 

√g ∂Ar ∂xr + Ar ∂√g ∂xr 

A r ,r = 1 

√ 

g 

A r ,r = 1 ∂ 

√ 

g ∂xr (√gA r ) . 

(2.1.4) 

EXAMPLE 2.1-1. (Divergence) Find the representation of the divergence of a vector Ar in spherical 

coordinates (ρ, θ, φ). Solution: In spherical coordinates we have 

x 1 = ρ, x 2 = θ, x 3 = φ with gij =0 for i = j and 

g11 = h 2 1 =1, g22 = h 2 2 = ρ2 , g33 = h 2 3 = ρ2 sin 2 θ. 

The determinant of gij is g = |gij| = ρ 4 sin 2 θ and √ g = ρ 2 sin θ. Employing the relation (2.1.4) we find 

div A r = 1 

 

∂ 

√ 

g ∂x1 (√gA 1 )+ ∂ 

∂x2 (√gA 2 )+ ∂ 

∂x3 (√gA 3 

) . 

In terms of the physical components this equation becomes 

div A r = 1 

 

∂ 

√ 

g ∂ρ (√g A(1) 

)+ 

h1 

∂ 

∂θ (√g A(2) 

)+ 

h2 

∂ 

∂φ (√g A(3) 

 

) . 

h3 

By using the notation 

A(1) = Aρ, A(2) = Aθ, A(3) = Aφ 

for the physical components, the divergence can be expressed in either of the forms: 

∂ 

∂ρ (ρ2 sin θAρ)+ ∂ 

div A r 1 

= 

ρ2 sin θ 

∂θ (ρ2 sin θ Aθ 

ρ 

div A r = 1 

ρ2 ∂ 

∂ρ (ρ2Aρ)+ 1 ∂ 

1 ∂Aφ 

(sin θAθ)+ 

ρ sin θ ∂θ ρ sin θ ∂φ . 

∂ 

)+ 

∂φ (ρ2 sin θ Aφ 

ρ sin θ ) 

 

or

Curl 

The contravariant components of the vector C =curl A are represented 

In expanded form this representation becomes: 

C 1 = 1 

 

∂A3 ∂A2 

√ − 

g ∂x2 ∂x3 

C 2 = 1 

√ g 

C 3 = 1 

√ g 

C i = ɛ ijk Ak,j. (2.1.5) 

 

∂A1 ∂A3 

− 

∂x3 ∂x1 

∂A2 ∂A1 

− 

∂x1 ∂x2 

 

. 

(2.1.6) 

EXAMPLE 2.1-2. (Curl) Find the representation for the components of curl A in spherical coordinates 

(ρ, θ, φ). 

Solution: In spherical coordinates we have :x 1 = ρ, x 2 = θ, x 3 = φ with gij =0fori = j and 

g11 = h 2 1 =1, g22 = h 2 2 = ρ2 , g33 = h 2 3 = ρ2 sin 2 θ. 

The determinant of gij is g = |gij| = ρ 4 sin 2 θ with √ g = ρ 2 sin θ. The relations (2.1.6) are tensor equations 

representing the components of the vector curl A. To find the components of curl A in spherical components 

we write the equations (2.1.6) in terms of their physical components. These equations take on the form: 

We employ the notations 

C(1) 

h1 

C(2) 

h2 

C(3) 

h3 

= 1 

 

∂ 

√ 

g ∂θ (h3A(3)) − ∂ 

∂φ (h2A(2)) 

 

= 1 

 

∂ 

√ 

g ∂φ (h1A(1)) − ∂ 

∂ρ (h3A(3)) 

 

= 1 

 

∂ 

√ 

g ∂ρ (h2A(2)) − ∂ 

∂θ (h1A(1)) 

 

. 

C(1) = Cρ, C(2) = Cθ, C(3) = Cφ, A(1) = Aρ, A(2) = Aθ, A(3) = Aφ 

(2.1.7) 

to denote the physical components, and find the components of the vector curl A, in spherical coordinates, 

are expressible in the form: 

1 

Cρ = 

ρ2 

∂ 

sin θ ∂θ (ρ sin θAφ) − ∂ 

∂φ (ρAθ) 

 

Cθ = 1 

 

∂ 

ρ sin θ ∂φ (Aρ) − ∂ 

 

(ρ sin θAφ) 

∂ρ 

Cφ = 1 

 

∂ 

ρ ∂ρ (ρAθ) − ∂ 

∂θ (Aρ) 

 

. 

(2.1.8) 

173

174 

Laplacian 

The Laplacian ∇ 2 U has the contravariant form 

∇ 2 U = g ij U,ij =(g ij U,i),j = 

Expanding this expression produces the equations: 

∇ 2 U = ∂ 

∂xj 

ij ∂U 

g 

∂xi 

im ∂U 

+ g 

∂xi 

j 

mj 

∇ 2 U = ∂ 

∂xj 

ij ∂U 

g 

∂xi 

+ 1 ∂ 

√ 

g 

√ g ∂U 

gij 

∂xj ∂xi ∇ 2 U = 1 

 

√g ∂ 

√ 

g ∂xj 

ij ∂U 

g 

∂xi 

ij ∂U 

+ g 

∂xi ∂ √ g 

∂xj 

∇ 2 U = 1 ∂ 

√ 

g ∂xj 

√ggij ∂U 

∂xi 

. 

In orthogonal coordinates we have g ij =0fori = j and 

and so (2.1.10) when expanded reduces to the form 

∇ 2 U = 

1 

h1h2h3 

g11 = h 2 1 , g22 = h 2 2 , g33 = h 2 3 

 

ij ∂U 

g 

∂xi 

. (2.1.9) 

,j 

(2.1.10) 

 

∂ 

∂x1 

h2h3 ∂U 

h1 ∂x1 

+ ∂ 

∂x2 

h1h3 ∂U 

h2 ∂x2 

+ ∂ 

∂x3 

h1h2 ∂U 

h3 ∂x3 

. (2.1.11) 

This representation is only valid in an orthogonal system of coordinates. 

EXAMPLE 2.1-3. (Laplacian) Find the Laplacian in spherical coordinates. 

Solution: Utilizing the results given in the previous example we find the Laplacian in spherical coordinates 

has the form 

∇ 2 1 

U = 

ρ2 

∂ 

ρ 

sin θ ∂ρ 

2 sin θ ∂U 

 

+ 

∂ρ 

∂ 

 

sin θ 

∂θ 

∂U 

 

+ 

∂θ 

∂ 

This simplifies to 

 

1 ∂U 

. 

∂φ sin θ ∂φ 


∇ 2 U = ∂2U 2 ∂U 1 

+ + 

∂ρ2 ρ ∂ρ ρ2 ∂2U cot θ 

+ 

∂θ2 ρ2 ∂U 

∂θ + 

1 

ρ 2 sin 2 θ 

The table 1 gives the vector and tensor representation for various quantities of interest. 

∂2U . (2.1.13) 

∂φ2

VECTOR GENERAL TENSOR CARTESIAN TENSOR 

A A i or Ai Ai 

A · B A i Bi = gijA i B j = AiB i 

A i Bi = g ij AiBj 

C = A × B C i = 1 

√ g e ijk AjBk 

AiBi 

Ci = eijkAjBk 

∇ Φ = grad Φ g im Φ,m Φ,i = ∂Φ 

∂x i 

∇· A =div A g mn Am,n = A r ,r = 1 ∂ 

√ 

g ∂xr (√gA r ) Ai,i = ∂Ai 

∂xi ∇× A = C =curl A C i = ɛ ijk ∂Ak 

Ak,j Ci = eijk 

∂xj ∇ 2 U g mn U ,mn = 1 

√ g 

∂ 

∂xj 

√ggij ∂U 

∂xi 

∂ 

∂xi 

∂U 

∂xi 

C =(A ·∇) B C i = A m B i ,m Ci = Am 

C = A(∇· B) C i = A i B j 

,j 

∂Bi 

∂xm ∂Bm 

Ci = Ai 

∂xm C = ∇ 2 A i jm i 

C = g A ,mj or Ci = g jm Ai,mj Ci = ∂ 

∂xm 

∂Ai 

∂xm 

 

A ·∇ φ g im A i φ ,m Aiφ,i 

 

∇ ∇· 

A 

 

∇× ∇× 

A 

g im A r 

,r ,m 

ɛijkg jm ɛ kst 

At,s ,m 

Table 1 Vector and tensor representations. 

∂ 2 Aj 

∂xj∂xi 

∂ 2 Ar 

∂xi∂xr 

− ∂2 Ai 

∂xj∂xj 

175

176 

EXAMPLE 2.1-4. (Maxwell’s equations) In the study of electrodynamics there arises the following 

vectors and scalars: 

E =Electric force vector, [ E]=Newton/coulomb 

B =Magnetic force vector, [ B]=Weber/m 2 

D =Displacement vector, [ D]=coulomb/m 2 

H =Auxilary magnetic force vector, [ H]=ampere/m 

J =Free current density, [ J]=ampere/m 2 

ϱ =free charge density, [ϱ] =coulomb/m 3 

The above quantities arise in the representation of the following laws: 

Faraday’s Law This law states the line integral of the electromagnetic force around a loop is proportional 

to the rate of flux of magnetic induction through the loop. This gives rise to the first electromagnetic field 

equation: 

∇× E = − ∂ B 

or ɛ 

∂t 

ijk Ek,j = − ∂Bi 

. (2.1.15) 

∂t 

Ampere’s Law This law states the line integral of the magnetic force vector around a closed loop is 

proportional to the sum of the current through the loop and the rate of flux of the displacement vector 

through the loop. This produces the second electromagnetic field equation: 

∇× H = J + ∂ D 

∂t 

or ɛ ijk Hk,j = J i + ∂Di 

. (2.1.16) 

∂t 

Gauss’s Law for Electricity This law states that the flux of the electric force vector through a closed 

surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic 

field equation: 

∇· D = ϱ or 

1 ∂ 

√ 

g ∂xi √ i 

gD = ϱ. (2.1.17) 

Gauss’s Law for Magnetism This law states the magnetic flux through any closed volume is zero. This 

produces the fourth electromagnetic field equation: 

∇· B =0 or 

1 ∂ 

√ 

g ∂xi √ i 

gB =0. (2.1.18) 

The four electromagnetic field equations are referred to as Maxwell’s equations. These equations arise 

in the study of electrodynamics and can be represented in other forms. These other forms will depend upon 

such things as the material assumptions and units of measurements used. Note that the tensor equations 

(2.1.15) through (2.1.18) are representations of Maxwell’s equations in a form which is independent of the 

coordinate system chosen. 

In applications, the tensor quantities must be expressed in terms of their physical components. In a 

general orthogonal curvilinear coordinate system we will have 

g11 = h 2 1 , g22 = h 2 2 , g33 = h 2 3 , and gij =0 for i = j. 

This produces the result √ g = h1h2h3. Further, if we represent the physical components of 

Di,Bi,Ei,Hi by D(i),B(i),E(i), and H(i)

the Maxwell equations can be represented by the equations in table 2. The tables 3, 4 and 5 are the 

representation of Maxwell’s equations in rectangular, cylindrical, and spherical coordinates. These latter 

tables are special cases associated with the more general table 2. 

1 

h1h2h3 

1 

h1h2h3 

1 

h1h2h3 

1 

h1h2h3 

1 

h1h2h3 

1 

h1h2h3 

1 

h1h2h3 

1 

h1h2h3 

 

∂ 

∂x2 (h3E(3)) − ∂ 

 

(h2E(2)) = − 

∂x3 1 

 

∂ 

∂x3 (h1E(1)) − ∂ 

 

(h3E(3)) = − 

∂x1 1 

 

∂ 

= − 1 

∂x1 (h2E(2)) − ∂ 

(h1E(1)) 

∂x2 h1 

h2 

h3 

∂B(1) 

∂t 

∂B(2) 

∂t 

∂B(3) 

∂t 

 

∂ 

∂x2 (h3H(3)) − ∂ 

 

(h2H(2)) = 

∂x3 J(1) 

+ 

h1 

1 ∂D(1) 

h1 ∂t 

 

∂ 

∂x3 (h1H(1)) − ∂ 

 

(h3H(3)) = 

∂x1 J(2) 

+ 

h2 

1 ∂D(2) 

h2 ∂t 

 

∂ 

∂x1 (h2H(2)) − ∂ 

 

(h1H(1)) = 

∂x2 J(3) 

+ 

h3 

1 ∂D(3) 

h3 ∂t 

 

∂ 

∂x1 

 

D(1) 

h1h2h3 + ∂ 

∂x2 

 

D(2) 

h1h2h3 + ∂ 

∂x3 

 

D(3) 

h1h2h3 = ϱ 

h1 

 

∂ 

∂x1 

 

B(1) 

h1h2h3 + ∂ 

∂x2 

 

B(2) 

h1h2h3 + ∂ 

∂x3 

 

B(3) 

h1h2h3 =0 

h1 

Table 2 Maxwell’s equations in generalized orthogonal coordinates. 

Note that all the tensor components have been replaced by their physical components. 

h2 

h2 

h3 

h3 

177

178 

∂Ez 

∂y 

∂Ex 

∂z 

∂Ey 

∂x 

− ∂Ey 

∂z 

− ∂Ez 

∂x 

− ∂Ex 

∂y 

= − ∂Bx 

∂t 

= − ∂By 

∂t 

= − ∂Bz 

∂t 

∂Hz 

∂y 

∂Hx 

∂z 

∂Hy 

∂x 

Here we have introduced the notations: 

Dx = D(1) 

Dy = D(2) 

Dz = D(3) 

Bx = B(1) 

By = B(2) 

Bz = B(3) 

∂Hy 

− 

∂z = Jx + ∂Dx 

∂t 

∂Hz 

− 

∂x = Jy + ∂Dy 

∂t 

∂Hx 

− 

∂y = Jz + ∂Dz 

∂t 

Hx = H(1) 

Hy = H(2) 

Hz = H(3) 

with x 1 = x, x 2 = y, x 3 = z, h1 = h2 = h3 =1 

Jx = J(1) 

Jy = J(2) 

Jz = J(3) 

∂Dx 

∂x 

∂Bx 

∂x 

Table 3 Maxwell’s equations Cartesian coordinates 

1 ∂Ez 

r ∂θ 

− ∂Eθ 

∂z 

= − ∂Br 

∂t 

∂Er ∂Ez ∂Bθ 

− = − 

∂z ∂r ∂t 

1 ∂ 

r ∂r (rEθ) − 1 ∂Er ∂Bz 

= − 

r ∂θ ∂t 

1 ∂ 1 ∂Dθ ∂Dz 

(rDr)+ + = ϱ 

r ∂r r ∂θ ∂z 


Dr = D(1) 

Dθ = D(2) 

Dz = D(3) 

Br = B(1) 

Bθ = B(2) 

Bz = B(3) 

1 

r 

Hr = H(1) 

Hθ = H(2) 

Hz = H(3) 

1 ∂Hz 

r ∂θ 

∂Hr 

∂z 

∂ 

∂r (rHθ) − 1 

with x 1 = r, x 2 = θ, x 3 = z, h1 =1, h2 = r, h3 =1. 

+ ∂Dy 

∂y 

+ ∂By 

∂y 

Ex = E(1) 

Ey = E(2) 

Ez = E(3) 

+ ∂Dz 

∂z 

∂Hθ 

− 

∂z = Jr + ∂Dr 

∂t 

∂Hz 

− 

∂r = Jθ + ∂Dθ 

∂t 

∂Hr 

r ∂θ = Jz + ∂Dz 

∂t 

1 ∂ 1 ∂Bθ ∂Bz 

(rBr)+ + 

r ∂r r ∂θ ∂z =0 

Jr = J(1) 

Jθ = J(2) 

Jz = J(3) 

Er = E(1) 

Eθ = E(2) 

Ez = E(3) 

Table 4 Maxwell’s equations in cylindrical coordinates. 

= ϱ 

+ ∂Bz 

∂z =0

1 ∂ 

ρ sin θ ∂θ (sin θEφ) − ∂Eθ 

 

= − 

∂φ 

∂Bρ 

∂t 

1 ∂Eρ 1 ∂ 

− 

ρ sin θ ∂φ ρ ∂ρ (ρEφ) =− ∂Bθ 

∂t 

1 ∂ 

ρ ∂ρ (ρEθ) − 1 ∂Eρ ∂Bφ 

= − 

ρ ∂θ ∂t 

 

1 ∂ 

ρ sin θ ∂θ (sin θHφ) − ∂Hθ 

 

= Jρ + 

∂φ 

∂Dρ 

∂t 

1 ∂Hρ 1 ∂ 

− 

ρ sin θ ∂φ ρ ∂ρ (ρHφ) =Jθ + ∂Dθ 

∂t 

1 ∂ 

ρ ∂ρ (ρHθ) − 1 ∂Hρ 

ρ ∂θ = Jφ + ∂Dφ 

∂t 

1 

ρ2 ∂ 

∂ρ (ρ2Dρ)+ 1 ∂ 

1 ∂Dφ 

(sin θDθ)+ 

ρ sin θ ∂θ ρ sin θ ∂φ =ϱ 

1 

ρ2 ∂ 

∂ρ (ρ2Bρ)+ 1 ∂ 

1 ∂Bφ 

(sin θBθ)+ 

ρ sin θ ∂θ ρ sin θ ∂φ =0 


Dρ = D(1) 

Dθ = D(2) 

Dφ = D(3) 

Bρ = B(1) 

Bθ = B(2) 

Bφ = B(3) 

Hρ = H(1) 

Hθ = H(2) 

Hφ = H(3) 

Jρ = J(1) 

Jθ = J(2) 

Jφ = J(3) 

with x 1 = ρ, x 2 = θ, x 3 = φ, h1 =1, h2 = ρ, h3 = ρ sin θ 

Table 5 Maxwell’s equations spherical coordinates. 

Eigenvalues and Eigenvectors of Symmetric Tensors 

Consider the equation 

Eρ = E(1) 

Eθ = E(2) 

Eφ = E(3) 

TijAj = λAi, i,j =1, 2, 3, (2.1.19) 

where Tij = Tji is symmetric, Ai are the components of a vector and λ is a scalar. Any nonzero solution 

Ai of equation (2.1.19) is called an eigenvector of the tensor Tij and the associated scalar λ is called an 

eigenvalue. When expanded these equations have the form 

(T11 − λ)A1 + T12A2 + T13A3 =0 

T21A1 +(T22 − λ)A2 + T23A3 =0 

T31A1 + T32A2 +(T33 − λ)A3 =0. 

The condition for equation (2.1.19) to have a nonzero solution Ai is that the characteristic equation 

should be zero. This equation is found from the determinant equation 

 

 

 

T11 − λ T12 T13 

 

f(λ) = 

T21 T22 − λ T23 

=0, (2.1.20) 

T31 T32 T33 − λ 

179

180 

which when expanded is a cubic equation of the form 

where I1,I2 and I3 are invariants defined by the relations 

f(λ) =−λ 3 + I1λ 2 − I2λ + I3 =0, (2.1.21) 

I1 = Tii 

I2 = 1 

2 TiiTjj − 1 

2 TijTij 

I3 = eijkTi1Tj2Tk3. 

When Tij is subjected to an orthogonal transformation, where ¯ Tmn = Tijℓimℓjn, then 

ℓimℓjn (Tmn − λδmn) = ¯ Tij − λδij and det (Tmn − λδmn) =det 

Tij ¯ − λδij . 

Hence, the eigenvalues of a second order tensor remain invariant under an orthogonal transformation. 

If Tij is real and symmetric then 

• the eigenvalues of Tij will be real, and 

• the eigenvectors corresponding to distinct eigenvalues will be orthogonal. 

(2.1.22) 

Proof: To show a quantity is real we show that the conjugate of the quantity equals the given quantity. If 

(2.1.19) is satisfied, we multiply by the conjugate Ai and obtain 

AiTijAj = λAiAi. (2.1.25) 

The right hand side of this equation has the inner product AiAi which is real. It remains to show the left 

hand side of equation (2.1.25) is also real. Consider the conjugate of this left hand side and write 

AiTijAj = AiT ijAj = AiTjiAj = AiTijAj. 

Consequently, the left hand side of equation (2.1.25) is real and the eigenvalue λ can be represented as the 

ratio of two real quantities. 

Assume that λ (1) and λ (2) are two distinct eigenvalues which produce the unit eigenvectors ˆ L1 and ˆ L2 

with components ℓi1 and ℓi2,i=1, 2, 3 respectively. We then have 

Consider the products 

Tijℓj1 = λ (1)ℓi1 and Tijℓj2 = λ (2)ℓi2. (2.1.26) 

λ (1)ℓi1ℓi2 = Tijℓj1ℓi2, 

λ (2)ℓi1ℓi2 = ℓi1Tijℓj2 = ℓj1Tjiℓi2. 

and subtract these equations. We find that 

(2.1.27) 

[λ (1) − λ (2)]ℓi1ℓi2 =0. (2.1.28) 

By hypothesis, λ (1) is different from λ (2) and consequently the inner product ℓi1ℓi2 must be zero. Therefore, 

the eigenvectors corresponding to distinct eigenvalues are orthogonal.

Therefore, associated with distinct eigenvalues λ (i),i=1, 2, 3 there are unit eigenvectors 

ˆL (i) = ℓi1 ê1 + ℓi2 ê2 + ℓi3 ê3 

with components ℓim,m=1, 2, 3 which are direction cosines and satisfy 

The unit eigenvectors satisfy the relations 

and can be written as the single equation 

Consider the transformation 

ℓinℓim = δmn and ℓijℓjm = δim. (2.1.23) 

Tijℓj1 = λ (1)ℓi1 Tijℓj2 = λ (2)ℓi2 Tijℓj3 = λ (3)ℓi3 

Tijℓjm = λ (m)ℓim, m =1, 2, or3 m not summed. 

xi = ℓijxj or xm = ℓmjxj 

which represents a rotation of axes, where ℓij are the direction cosines from the eigenvectors of Tij. This is a 

linear transformation where the ℓij satisfy equation (2.1.23). Such a transformation is called an orthogonal 

transformation. In the new x coordinate system, called principal axes, we have 

∂x 

T mn = Tij 

i 

∂xm ∂x j 

∂x n = Tijℓimℓjn = λ (n)ℓinℓim = λ (n)δmn (no sum on n). (2.1.24) 

This equation shows that in the barred coordinate system there are the components 

 

T mn = 

⎡ 

⎣ λ (1) 0 0 

0 λ (2) 0 

0 0 λ (3) 

That is, along the principal axes the tensor components Tij are transformed to the components T ij where 

T ij =0fori = j. The elements T (i)(i) , i not summed, represent the eigenvalues of the transformation 

(2.1.19). 

⎤ 

⎦ . 

181

182 

EXERCISE 2.1 

◮ 1. In cylindrical coordinates (r, θ, z) withf = f(r, θ, z) find the gradient of f. 

◮ 2. In cylindrical coordinates (r, θ, z) with A = A(r, θ, z) find div A. 

◮ 3. In cylindrical coordinates (r, θ, z) for A = A(r, θ, z) find curl A. 

◮ 4. In cylindrical coordinates (r, θ, z) forf = f(r, θ, z) find ∇ 2 f. 

◮ 5. In spherical coordinates (ρ, θ, φ) withf = f(ρ, θ, φ) find the gradient of f. 

◮ 6. In spherical coordinates (ρ, θ, φ) with A = A(ρ, θ, φ) find div A. 

◮ 7. In spherical coordinates (ρ, θ, φ) for A = A(ρ, θ, φ) find curl A. 

◮ 8. In spherical coordinates (ρ, θ, φ) forf = f(ρ, θ, φ) find ∇ 2 f. 

◮ 9. Let r = x ê1 +y ê2 +z ê3 denote the position vector of a variable point (x, y, z) in Cartesian coordinates. 

Let r = |r| denote the distance of this point from the origin. Find in terms of r and r: 

(a) grad(r) (b) grad(r m ) (c) grad( 1 

) (d) grad(lnr) (e) grad(φ) 

r 

where φ = φ(r) is an arbitrary function of r. 

◮ 10. Let r = x ê1+y ê2+z ê3 denote the position vector of a variable point (x, y, z) in Cartesian coordinates. 

Let r = |r| denote the distance of this point from the origin. Find: 

where φ = φ(r) is an arbitrary function or r. 

(a) div (r) (b) div (r m r) (c) div (r −3 r) (d) div (φr) 

◮ 11. Let r = x ê1 + y ê2 + z ê3 denote the position vector of a variable point (x, y, z) in Cartesian 

coordinates. Let r = |r| denote the distance of this point from the origin. Find: (a) 

where φ = φ(r) is an arbitrary function of r. 

curl r (b) curl (φr) 

◮ 12. Expand and simplify the representation for curl (curl A). 

◮ 13. Show that the curl of the gradient is zero in generalized coordinates. 

◮ 14. Write out the physical components associated with the gradient of φ = φ(x 1 ,x 2 ,x 3 ). 


g im Ai,m = 1 

√ g 

∂ 

∂xi √ im i 

gg Am = A ,i = 1 

√ 

g 

∂ 

∂xi √ i 

gA .

◮ 16. Let r =(r · r) 1/2 = x 2 + y 2 + z 2 ) and calculate (a) ∇ 2 (r) (b) ∇ 2 (1/r) (c) ∇ 2 (r 2 ) (d) ∇ 2 (1/r 2 ) 

◮ 17. Given the tensor equations Dij = 1 

2 (vi,j + vj,i), i,j =1, 2, 3. Let v(1),v(2),v(3) denote the 

physical components of v1,v2,v3 and let D(ij) denote the physical components associated with Dij. Assume 

the coordinate system (x1 ,x2 ,x3 ) is orthogonal with metric coefficients g (i)(i) = h2 i ,i=1, 2, 3andgij =0 

for i = j. 

(a) Find expressions for the physical components D(11),D(22) and D(33) in terms of the physical components 

v(i),i=1, 2, 3. Answer: D(ii) = 1 ∂V (i) V (j) ∂hi 

+ no sum on i. 

hi ∂xi hihj ∂xj j=i 

(b) Find expressions for the physical components D(12),D(13) and D(23) in terms of the physical components 

v(i),i=1, 2, 3. Answer: D(ij) = 1 

 

hi ∂ 

2 hj ∂xj 

V (i) 

+ 

hi 

hj ∂ 

hi ∂xi 

V (j) 

hj 

◮ 18. Write out the tensor equations in problem 17 in Cartesian coordinates. 

◮ 19. Write out the tensor equations in problem 17 in cylindrical coordinates. 

◮ 20. Write out the tensor equations in problem 17 in spherical coordinates. 

◮ 21. Express the vector equation (λ +2µ)∇Φ − 2µ∇×ω + F = 0 intensorform. 

◮ 22. Write out the equations in problem 21 for a generalized orthogonal coordinate system in terms of 

physical components. 

◮ 23. Write out the equations in problem 22 for cylindrical coordinates. 

◮ 24. Write out the equations in problem 22 for spherical coordinates. 

◮ 25. Use equation (2.1.4) to represent the divergence in parabolic cylindrical coordinates (ξ,η,z). 

◮ 26. Use equation (2.1.4) to represent the divergence in parabolic coordinates (ξ,η,φ). 

◮ 27. Use equation (2.1.4) to represent the divergence in elliptic cylindrical coordinates (ξ,η,z). 

Change the given equations from a vector notation to a tensor notation. 

◮ 28. B = v ∇· A +(∇·v) A 

◮ 29. 

◮ 30. 

◮ 31. 

◮ 32. 

d 

dt [ A · ( B × C)] = d A 

dt · ( B × C)+ A · ( d B 

dt × C)+ A · ( B × d C 

dt ) 

dv ∂v 

= +(v ·∇)v 

dt ∂t 

1 ∂ 

c 

H 

∂t = −curl E 

d B 

dt − ( B ·∇)v + B(∇·v) =0 

183

184 

Change the given equations from a tensor notation to a vector notation. 

◮ 33. ɛ ijk Bk,j + F i =0 

◮ 34. gijɛ jkl Bl,k + Fi =0 

◮ 35. 

∂ϱ 

∂t +(ϱvi),i=0 

◮ 36. ϱ( ∂vi ∂vi ∂P 

+ vm )=− 

∂t ∂xm ∂xi + µ ∂2vi ∂xm∂x 

1 

12bh3 − 1 

24b2h 2 

− 1 

24b2 2 1 h 12b3h m + Fi 

 

◮ 37. The moment of inertia of an area or second moment of area is defined by Iij = (ymymδij −yiyj) dA 

A 

where dA is an element of area. Calculate the moment of inertia 

 

Iij, i,j=1, 2 for the triangle illustrated in 

the figure 2.1-1 and show that Iij = 

. 

Figure 2.1-1 Moments of inertia for a triangle 

◮ 38. Use the results from problem 37 and rotate the axes in figure 2.1-1 through an angle θ to a barred 

system of coordinates. 

(a) Show that in the barred system of coordinates 

 

I11 + I22 I11 − I22 

I11 = 

+ 

cos 2θ + I12 sin 2θ 

2 

2 

 

I11 − I22 

I12 = I21 = − 

sin 2θ + I12 cos 2θ 

2 

 

I11 + I22 I11 − I22 

I22 = 

− 

cos 2θ − I12 sin 2θ 

2 

2 

(b) For what value of θ will I11 have a maximum value? 

(c) Show that when I11 is a maximum, we will have I22 a minimum and I12 = I21 =0.

Figure 2.1-2 Mohr’s circle 

◮ 39. Otto Mohr1 gave the following physical interpretation to the results obtained in problem 38: 

• Plot the points A(I11,I12) andB(I22, −I12) as illustrated in the figure 2.1-2 

• Draw the line AB and calculate the point C where this line intersects the I axes. Show the point C 

has the coordinates 

( I11 + I22 

, 0) 

2 

• Calculate the radius of the circle with center at the point C and with diagonal AB and show this 

radius is 

 

I11 

2 − I22 

r = 

+ I 

2 

2 12 

• Show the maximum and minimum values of I occur where the constructed circle intersects the I axes. 

Show that Imax = I11 = I11 + I22 

2 

+ r Imin = I22 = I11 + I22 

2 

◮ 40. Show directly that the eigenvalues of the symmetric matrix Iij = 

λ2 = Imin where Imax and Imin are given in problem 39. 

− r. 

I11 I12 

I21 I22 

 

are λ1 = Imax and 

◮ 41. Find the principal axes and moments of inertia for the triangle given in problem 37 and summarize 

your results from problems 37,38,39, and 40. 

◮ 42. Verify for orthogonal coordinates the relations 

or 

 

∇× 

A · ê (i) = 

∇× A = 

1 

h1h2h3 

◮ 43. Verify for orthogonal coordinates the relation 

 

∇×(∇× 

A) · ê (i) = 

3 

e (i)jk 

h1h2h3 

k=1 

∂(h (k)A(k)) 

h (i) 

∂xj 

 

 

 

h1 ê1 h2 ê2 h3 ê3 

 

∂ ∂ 

∂ 

∂x1 ∂x2 ∂x3 

h1A(1) h2A(2) h3A(3) . 

3 

e (i)jrersm 

m=1 

h (i) 

h1h2h3 

1 Christian Otto Mohr (1835-1918) German civil engineer. 

∂ 

∂xj 

h 2 (r) 

h1h2h3 

 

∂(h (m)A(m)) 

∂xs 

185

186 

◮ 44. Verify for orthogonal coordinates the relation 

 

∇ ∇· 

A · ê (i) = 1 

 

∂ 1 ∂(h2h3A(1)) 

h (i) ∂x (i) h1h2h3 ∂x1 

◮ 45. Verify the relation 

 

( A ·∇) 

B · ê (i) = 

3 

k=1 

A(k) ∂B(i) 

+ 

h (k) ∂xk 

 

k=i 

◮ 46. The Gauss divergence theorem is written 

 

1 

 

∂F ∂F2 ∂F3 

+ + dτ = 

∂x ∂y ∂z 

B(k) 

hkh (i) 

+ ∂(h1h3A(2)) 

∂x2 

+ ∂(h1h2A(3)) 

∂x3 

 

A(i) ∂h (i) 

− A(k) 

∂xk 

∂hk 

 

∂x (i) 

n1F 1 + n2F 2 + n3F 3 dσ 

V 

S 

where V is the volume within a simple closed surface S. Here it is assumed that F i = F i (x, y, z) are 

continuous functions with continuous first order derivatives throughout V and ni are the direction cosines 

of the outward normal to S, dτ is an element of volume and dσ is an element of surface area. 

(a) Show that in a Cartesian coordinate system 

F i ∂F1 

,i = 

∂x 

 

and that the tensor form of this theorem is 

(b) Write the vector form of this theorem. 

(c) Show that if we define 

V 

+ ∂F2 

∂y 

F i 

,i dτ = 

∂F3 

+ 

∂z 

 

F 

S 

i ni dσ. 

ur = ∂u 

∂xr , vr = ∂v 

∂xr and Fr = grmF m = uvr 

then F i 

,i = gimFi,m = g im (uvi,m + umvi) 

(d) Show that another form of the Gauss divergence theorem is 

 

g im 

umvi dτ = uvmn m 

dσ − 

V 

S 

ug 

V 

im vi,m dτ 

Write out the above equation in Cartesian coordinates. 

⎛ 

1 

◮ 47. Find the eigenvalues and eigenvectors associated with the matrix A = ⎝ 1 

1 

2 

⎞ 

2 

1⎠ 

. 

2 1 1 

Show that the eigenvectors are orthogonal. 

⎛ 

1 


2 

1 

⎞ 

1 

0⎠ 

. 

1 0 1 


⎛ 

1 


1 

1 

⎞ 

0 

1⎠ 

. 

0 


1 1 

◮ 50. The harmonic and biharmonic functions or potential functions occur in the mathematical modeling 

of many physical problems. Any solution of Laplace’s equation ∇2Φ = 0 is called a harmonic function and 

any solution of the biharmonic equation ∇4Φ = 0 is called a biharmonic function. 

(a) Expand the Laplace equation in Cartesian, cylindrical and spherical coordinates. 

(b) Expand the biharmonic equation in two dimensional Cartesian and polar coordinates. 

Hint: Consider ∇ 4 Φ=∇ 2 (∇ 2 Φ). In Cartesian coordinates ∇ 2 Φ=Φ,ii and ∇ 4 Φ=Φ,iijj.

§2.2 DYNAMICS 

Dynamics is concerned with studying the motion of particles and rigid bodies. By studying the motion 

of a single hypothetical particle, one can discern the motion of a system of particles. This in turn leads to 

the study of the motion of individual points in a continuous deformable medium. 

Particle Movement 

The trajectory of a particle in a generalized coordinate system is described by the parametric equations 

x i = x i (t), i =1,...,N (2.2.1) 

where t is a time parameter. If the coordinates are changed to a barred system by introducing a coordinate 

transformation 

x i = x i (x 1 ,x 2 ,...,x N ), i =1,...,N 

then the trajectory of the particle in the barred system of coordinates is 

x i = x i (x 1 (t),x 2 (t),...,x N (t)), i =1,...,N. (2.2.2) 

The generalized velocity of the particle in the unbarred system is defined by 

v i = dxi 

, i =1,...,N. (2.2.3) 

dt 

By the chain rule differentiation of the transformation equations (2.2.2) one can verify that the velocity in 

the barred system is 

v r = dxr ∂xr 

= 

dt ∂xj dxj ∂xr 

= 

dt ∂xj vj , r =1,...,N. (2.2.4) 

Consequently, the generalized velocity vi is a first order contravariant tensor. The speed of the particle is 

obtained from the magnitude of the velocity and is 

v 2 = gijv i v j . 

The generalized acceleration f i of the particle is defined as the intrinsic derivative of the generalized velocity. 

The generalized acceleration has the form 

f i = δvi 

δt = vi dx 

,n 

n 

dt 

and the magnitude of the acceleration is 

dvi 

= 

dt + 

 

i 

mn 

f 2 = gijf i f j . 

v m v n = d2xi + 

dt2 

i 

m dx dx 

mn dt 

n 

dt 

(2.2.5) 

187

188 

Frenet-Serret Formulas 

Figure 2.2-1 Tangent, normal and binormal to point P on curve. 

The parametric equations (2.2.1) describe a curve in our generalized space. With reference to the figure 

2.2-1 we wish to define at each point P of the curve the following orthogonal unit vectors: 

T i = unit tangent vector at each point P. 

N i = unit normal vector at each point P. 

B i = unit binormal vector at each point P. 

These vectors define the osculating, normal and rectifying planes illustrated in the figure 2.2-1. 

In the generalized coordinates the arc length squared is 

Define T i = dxi 

ds 

ds 2 = gijdx i dx j . 

as the tangent vector to the parametric curve defined by equation (2.2.1). This vector is a 

unit tangent vector because if we write the element of arc length squared in the form 

dx 

1=gij 

i dx 

ds 

j 

ds = gijT i T j , (2.2.6) 

we obtain the generalized dot product for T i . This generalized dot product implies that the tangent vector 

is a unit vector. Differentiating the equation (2.2.6) intrinsically with respect to arc length s along the curve 

produces 

which simplifies to 

δT 

gmn 

m 

δs T n n 

m δT 

+ gmnT 

δs =0, 

m 

n δT 

gmnT =0. (2.2.7) 

δs

The equation (2.2.7) is a statement that the vector 

vector is defined as 

N i = 1 δT 

κ 

i 

δs 

δT m 

δs is orthogonal to the vector T m . The unit normal 

or Ni = 1 

κ 

δTi 

, (2.2.8) 

δs 

where κ is a scalar called the curvature and is chosen such that the magnitude of N i is unity. The reciprocal 

of the curvature is R = 1 

κ , which is called the radius of curvature. The curvature of a straight line is zero 

while the curvature of a circle is a constant. The curvature measures the rate of change of the tangent vector 

as the arc length varies. 

The equation (2.2.7) can be expressed in the form 

gijT i N j =0. (2.2.9) 

Taking the intrinsic derivative of equation (2.2.9) with respect to the arc length s produces 

or 

i δNj 

gijT 

δs 

The generalized dot product can be written 

i δNj δT 

gijT + gij 

δs i 

δs N j =0 

δT 

= −gij 

i 

δs N j = −κgijN i N j = −κ. (2.2.10) 

gijT i T j =1, 

and consequently we can express equation (2.2.10) in the form 

Consequently, the vector 

i δNj 

gijT 

δs = −κgijT i T j 

δN j 

δs 

or gijT i 

 

j δN j 

+ κT =0. (2.2.11) 

δs 

+ κT j 


is orthogonal to T i . In a similar manner, we can use the relation gijN i N j = 1 and differentiate intrinsically 

with respect to the arc length s to show that 

i δNj 

gijN 

δs =0. 

This in turn can be expressed in the form 

gijN i 

 

j δN j 

+ κT =0. 

δs 

This form of the equation implies that the vector represented in equation (2.2.12) is also orthogonal to the 

unit normal N i . We define the unit binormal vector as 

B i = 1 

 

i δN i 

+ κT or Bi = 

τ δs 1 

 

δNi 

+ κTi 

(2.2.13) 

τ δs 

where τ is a scalar called the torsion. The torsion is chosen such that the binormal vector is a unit vector. 

The torsion measures the rate of change of the osculating plane and consequently, the torsion τ is a measure 

189

190 

of the twisting of the curve out of a plane. The value τ = 0 corresponds to a plane curve. The vectors 

T i ,N i ,B i , i =1, 2, 3 satisfy the cross product relation 

B i = ɛ ijk TjNk. 

If we differentiate this relation intrinsically with respect to arc length s we find 

δB i 

δs 

= ɛijk 

 

Tj 

δNk 

δs 

δTj 

+ 

δs Nk 

 

= ɛ ijk [Tj(τBk − κTk)+κNjNk] 

= τɛ ijk TjBk = −τɛ ikj BkTj = −τN i . 

The relations (2.2.8),(2.2.13) and (2.2.14) are now summarized and written 

δT i 

i 

= κN 

δs 

δN i 

δs = τBi − κT i 

δB i 

δs = −τNi . 

These equations are known as the Frenet-Serret formulas of differential geometry. 

Velocity and Acceleration 

Chain rule differentiation of the generalized velocity is expressible in the form 

v i = dxi 

dt 

= dxi 

ds 

(2.2.14) 

(2.2.15) 

ds 

dt = T i v, (2.2.16) 

where v = ds 

dt is the speed of the particle and is the magnitude of vi . The vector T i is the unit tangent vector 

to the trajectory curve at the time t. The equation (2.2.16) is a statement of the fact that the velocity of a 

particle is always in the direction of the tangent vector to the curve and has the speed v. 

By chain rule differentiation, the generalized acceleration is expressible in the form 

f r = δvr 

δt 

dv 

= 

dt T r r δT 

+ v 

δt 

= dv 

dt T r + v 

δT r 

δs 

ds 

dt 

= dv 

dt T r + κv 2 N r . 

(2.2.17) 

The equation (2.2.17) states that the acceleration lies in the osculating plane. Further, the equation (2.2.17) 

indicates that the tangential component of the acceleration is dv 

dt , while the normal component of the acceleration 

is κv 2 .

Work and Potential Energy 

Define M as the constant mass of the particle as it moves along the curve defined by equation (2.2.1). 

Also let Qr denote the components of a force vector (in appropriate units of measurements) which acts upon 

the particle. Newton’s second law of motion can then be expressed in the form 

Q r = Mf r 

or Qr = Mfr. (2.2.18) 

The work done W in moving a particle from a point P0 to a point P1 along a curve x r = x r (t),r =1, 2, 3, 

with parameter t, is represented by a summation of the tangential components of the forces acting along the 

path and is defined as the line integral 

P1 

dx 

W = Qr 

P0 

r P1 

ds = Qr dx 

ds P0 

r t1 

dx 

= Qr 

t0 

r t1 

dt = Qrv 

dt t0 

r dt (2.2.19) 

where Qr = grsQs is the covariant form of the force vector, t is the time parameter and s is arc length along 

the curve. 

Conservative Systems 

If the force vector is conservative it means that the force is derivable from a scalar potential function 

V = V (x 1 ,x 2 ,...,x N ) such that Qr = −V ,r = − ∂V 

, r =1,...,N. (2.2.20) 

∂xr In this case the equation (2.2.19) can be integrated and we find that to within an additive constant we will 

have V = −W. The potential function V is called the potential energy of the particle and the work done 

becomes the change in potential energy between the starting and end points and is independent of the path 

connecting the points. 

Lagrange’s Equations of Motion 

The kinetic energy T of the particle is defined as one half the mass times the velocity squared and can 

be expressed in any of the forms 

T = 1 

2 M 

2 ds 

= 

dt 

1 

2 Mv2 = 1 

2 Mgmnv m v n = 1 

2 Mgmn ˙x m ˙x n , (2.2.21) 

where the dot notation denotes differentiation with respect to time. It is an easy exercise to calculate the 

derivatives 

∂T 

and thereby verify the relation 

d 

dt 

∂ ˙x r = Mgrm ˙x m 

∂T 

∂ ˙x r 

∂T 

∂x 

 

= M 

1 

= r 2 

 

grm¨x m + ∂grm 

∂xn ˙xn ˙x m 

 

M ∂gmn 

∂x r ˙xm ˙x n , 

(2.2.22) 

 

d ∂T 

dt ∂ ˙x r 

 

− ∂T 

∂xr = Mfr = Qr, r =1,...,N. (2.2.23) 

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192 

This equation is called the Lagrange’s form of the equations of motion. 

EXAMPLE 2.2-1. (Equations of motion in spherical coordinates) Find the Lagrange’s form of 

the equations of motion in spherical coordinates. 

Solution: Let x 1 = ρ, x 2 = θ, x 3 = φ then the element of arc length squared in spherical coordinates has 

the form 

ds 2 =(dρ) 2 + ρ 2 (dθ) 2 + ρ 2 sin 2 θ(dφ) 2 . 

The element of arc length squared can be used to construct the kinetic energy. For example, 

T = 1 

2 M 

2 ds 

= 

dt 

1 

2 M 

 

(˙ρ) 2 + ρ 2 ( ˙ θ) 2 + ρ 2 sin 2 θ( ˙ φ) 2 

. 

The Lagrange form of the equations of motion of a particle are found from the relations (2.2.23) and are 

calculated to be: 

Mf1 = Q1 = d 

 

∂T 

− 

dt ∂ ˙ρ 

∂T 

 

= M ¨ρ − ρ( 

∂ρ ˙ θ) 2 − ρ sin 2 θ( ˙ φ) 2 

Mf2 = Q2 = d 

 

∂T 

dt ∂ ˙ 

− 

θ 

∂T 

 

d 

 

= M ρ 

∂θ dt 

2 

θ˙ 

− ρ 2 sin θ cos θ( ˙ φ) 2 

 

Mf3 = Q3 = d 

 

∂T 

dt ∂ ˙ 

− 

φ 

∂T 

 

d 

 

= M ρ 

∂φ dt 

2 sin 2 θ ˙ 

φ 

 

. 

In terms of physical components we have 

 

Qρ = M ¨ρ − ρ( ˙ θ) 2 − ρ sin 2 θ( ˙ φ) 2 

Qθ = M 

 

d 

 

ρ 

ρ dt 

2 

θ˙ 

− ρ 2 sin θ cos θ( ˙ φ) 2 

 

Qφ = M 

 

d 

 

ρ 

ρ sin θ dt 

2 sin 2 θ ˙ 

φ 

 

. 

Euler-Lagrange Equations of Motion 

Starting with the Lagrange’s form of the equations of motion from equation (2.2.23), we assume that 

the external force Qr is derivable from a potential function V as specified by the equation (2.2.20). That is, 

we assume the system is conservative and express the equations of motion in the form 

 

d ∂T 

dt ∂ ˙x r 

 

− ∂T ∂V 

= − 

∂xr ∂xr = Qr, r =1,...,N (2.2.24) 

The Lagrangian is defined by the equation 

L = T − V = T (x 1 ,...,x N , ˙x 1 ,..., ˙x N ) − V (x 1 ,...,x N )=L(x i , ˙x i ). (2.2.25) 

Employing the defining equation (2.2.25), it is readily verified that the equations of motion are expressible 

in the form 

 

d ∂L 

dt ∂ ˙x r 

 

− ∂L 

=0, r =1,...,N, (2.2.26) 

∂xr which are called the Euler-Lagrange form for the equations of motion.

Figure 2.2-2 Simply pulley system 

EXAMPLE 2.2-2. (Simple pulley system) 


Find the equation of motion for the simply pulley system 

Solution: The given system has only one degree of freedom, say y1. It is assumed that 

The kinetic energy of the system is 

y1 + y2 = ℓ =aconstant. 

T = 1 

2 (m1 + m2)˙y 2 1 . 

Let y1 increase by an amount dy1 and show the work done by gravity can be expressed as 

dW = m1gdy1 + m2gdy2 

dW = m1gdy1 − m2gdy1 

dW =(m1 − m2)gdy1 = Q1 dy1. 

Here Q1 =(m1− m2)g is the external force acting on the system where g is the acceleration of gravity. The 

Lagrange equation of motion is 

 

d ∂T 

− 

dt ∂ ˙y1 

∂T 

∂y1 

= Q1 

or 

(m1 + m2)¨y1 =(m1 − m2)g. 

Initial conditions must be applied to y1 and ˙y1 before this equation can be solved. 

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194 

EXAMPLE 2.2-3. (Simple pendulum) Find the equation of motion for the pendulum system illustrated 

in the figure 2.2-3. 

Solution: Choose the angle θ illustrated in the figure 2.2-3 as the generalized coordinate. If the pendulum 

is moved from a vertical position through an angle θ, we observe that the mass m moves up a distance 

h = ℓ − ℓ cos θ. The work done in moving this mass a vertical distance h is 

W = −mgh = −mgℓ(1 − cos θ), 

since the force is −mg in this coordinate system. In moving the pendulum through an angle θ, the arc length 

s swept out by the mass m is s = ℓθ. This implies that the kinetic energy can be expressed 

T = 1 

2 m 

2 ds 

= 

dt 

1 

2 m 

 

ℓ ˙ 2 θ = 1 

2 mℓ2 ( ˙ θ) 2 . 

The Lagrangian of the system is 

Figure 2.2-3 Simple pendulum system 

L = T − V = 1 

2 mℓ2 ( ˙ θ) 2 − mgℓ(1 − cos θ) 

and from this we find the equation of motion 

 

d ∂L 

dt ∂ ˙ 

− 

θ 

∂L 

d 

 

=0 or mℓ 

∂θ dt 

2 

θ˙ 

− mgℓ(− sin θ) =0. 

This in turn simplifies to the equation 

¨θ + g 

sin θ =0. 

ℓ 

This equation together with a set of initial conditions for θ and ˙ θ represents the nonlinear differential equation 

which describes the motion of a pendulum without damping.

EXAMPLE 2.2-4. (Compound pendulum) Find the equations of motion for the compound pendulum 


Solution: Choose for the generalized coordinates the angles x 1 = θ1 and x 2 = θ2 illustrated in the figure 

2.2-4. To find the potential function V for this system we consider the work done as the masses m1 and 

m2 are moved. Consider independent motions of the angles θ1 and θ2. Imagine the compound pendulum 

initially in the vertical position as illustrated in the figure 2.2-4(a). Now let m1 be displaced due to a change 

in θ1 and obtain the figure 2.2-4(b). The work done to achieve this position is 

W1 = −(m1 + m2)gh1 = −(m1 + m2)gL1(1 − cos θ1). 

Starting from the position in figure 2.2-4(b) we now let θ2 undergo a displacement and achieve the configuration 

in the figure 2.2-4(c). 

Figure 2.2-4 Compound pendulum 

The work done due to the displacement θ2 can be represented 

W2 = −m2gh2 = −m2gL2(1 − cos θ2). 

Since the potential energy V satisfies V = −W to within an additive constant, we can write 

V = −W = −W1 − W2 = −(m1 + m2)gL1 cos θ1 − m2gL2 cos θ2 + constant, 

where the constant term in the potential energy has been neglected since it does not contribute anything to 

the equations of motion. (i.e. the derivative of a constant is zero.) 

The kinetic energy term for this system can be represented 

T = 1 

2 m1 

2 ds1 

+ 

dt 

1 

2 m2 

2 ds2 

dt 

T = 1 

2 m1(˙x 2 1 +˙y2 1 )+1 

2 m2(˙x 2 2 +˙y2 2 ), 

(2.2.27) 

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196 

where 

(x1,y1) =(L1 sin θ1 , −L1 cos θ1) 

(x2,y2) =(L1 sin θ1 + L2 sin θ2, −L1 cos θ1 − L2 cos θ2) 

(2.2.28) 

are the coordinates of the masses m1 and m2 respectively. Substituting the equations (2.2.28) into equation 

(2.2.27) and simplifying produces the kinetic energy expression 

T = 1 

2 (m1 + m2)L 2 1 ˙ θ 2 1 + m2L1L2 ˙ θ1 ˙ θ2 cos(θ1 − θ2)+ 1 

2 m2L 2 2 ˙ θ 2 2. (2.2.29) 

Writing the Lagrangian as L = T − V , the equations describing the motion of the compound pendulum 

are obtained from the Lagrangian equations 

 

d ∂L 

dt ∂ ˙ 

− 

θ1 

∂L 

 

d ∂L 

=0 and 

∂θ1 

dt ∂ ˙ 

− 

θ2 

∂L 

=0. 

∂θ2 

Calculating the necessary derivatives, substituting them into the Lagrangian equations of motion and then 

simplifying we derive the equations of motion 

L1 ¨ θ1 + 

m2 

m1 + m2 

L2 ¨ θ2 cos(θ1 − θ2)+ 

m2 

m1 + m2 

L2( ˙ θ2) 2 sin(θ1 − θ2)+g sin θ1 =0 

L1 ¨ θ1 cos(θ1 − θ2)+L2 ¨ θ2 − L1( ˙ θ1) 2 sin(θ1 − θ2)+g sin θ2 =0. 

These equations are a set of coupled, second order nonlinear ordinary differential equations. These equations 

are subject to initial conditions being imposed upon the angular displacements (θ1,θ2) and the angular 

velocities ( ˙ θ1, ˙ θ2). 

Alternative Derivation of Lagrange’s Equations of Motion 

Let c denote a given curve represented in the parametric form 

x i = x i (t), i =1,...,N, t0 ≤ t ≤ t1 

and let P0,P1 denote two points on this curve corresponding to the parameter values t0 and t1 respectively. 

Let c denote another curve which also passes through the two points P0 and P1 as illustrated in the figure 

2.2-5. 

The curve c is represented in the parametric form 

x i = x i (t) =x i (t)+ɛη i (t), i =1,...,N, t0 ≤ t ≤ t1 

in terms of a parameter ɛ. In this representation the function η i (t) must satisfy the end conditions 

η i (t0) =0 and η i (t1) =0 i =1,...,N 

since the curve c is assumed to pass through the end points P0 and P1. 

Consider the line integral 

I(ɛ) = 

t1 

L(t, x i + ɛη i , ˙x i + ɛ ˙η i ) dt, (2.2.30) 

t0

where 

Figure 2.2-5. Motion along curves c and c 

L = T − V = L(t, x i , ˙x i ) 

is the Lagrangian evaluated along the curve c. We ask the question, “What conditions must be satisfied by 

the curve c in order that the integral I(ɛ) haveanextremumvaluewhenɛiszero?”If the integral I(ɛ) has 

a minimum value when ɛ is zero it follows that its derivative with respect to ɛ will be zero at this value and 

we will have 

 

dI(ɛ) 

 

dɛ 

ɛ=0 

=0. 

Employing the definition 

 

dI 

 

I(ɛ) − I(0) 

dɛ = lim 

= I 

ɛ→0 

ɛ=0 

ɛ 

′ (0) = 0 

we expand the Lagrangian in equation (2.2.30) in a series about the point ɛ =0. Substituting the expansion 

L(t, x i + ɛη i , ˙x i + ɛ ˙η i )=L(t, x i , ˙x i 

∂L 

)+ɛ 

 

+ ɛ 2 [ ]+··· 

into equation (2.2.30) we calculate the derivative 

I ′ I(ɛ) − I(0) 

(0) = lim 

ɛ→0 ɛ 

∂xi ηi + ∂L 

˙ηi 

∂ ˙x i 

t1 

∂L 

= lim 

ɛ→0 

t0 ∂xi ηi (t)+ ∂L 

∂ ˙x i ˙ηi 

(t) dt + ɛ [ ]+···=0, 

where we have neglected higher order powers of ɛ since ɛ is approaching zero. Analysis of this equation 

informs us that the integral I has a minimum value at ɛ = 0 provided that the integral 

t1 

∂L 

δI = 

∂xi ηi (t)+ ∂L 

∂ ˙x i ˙ηi 

(t) dt =0 (2.2.31) 

t0 

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198 

is satisfied. Integrating the second term of this integral by parts we find 

δI = 

t1 

t0 

∂L 

∂xi ηi 

∂L 

dt + 

∂ ˙x i ηi t1 (t) 

t0 

t1 

− 

t0 

d 

dt 

 

∂L 

∂ ˙x i 

 

η i (t) dt =0. (2.2.32) 

The end condition on ηi (t) makes the middle term in equation (2.2.32) vanish and we are left with the 

integral 

t1 

δI = η i 

∂L d ∂L 

(t) − 

∂xi dt ∂ ˙x i 

 

dt =0, (2.2.33) 

t0 

which must equal zero for all ηi (t). Since ηi (t) is arbitrary, the only way the integral in equation (2.2.33) can 

be zero for all ηi (t) is for the term inside the brackets to vanish. This produces the result that the integral 

of the Lagrangian is an extremum when the Euler-Lagrange equations 

d 

dt 

 

∂L 

∂ ˙x i 

 

− ∂L 

=0, i =1,...,N (2.2.34) 

∂xi are satisfied. This is a necessary condition for the integral I(ɛ) to have a minimum value. 

In general, any line integral of the form 

I = 

t1 

φ(t, x i , ˙x i ) dt (2.2.35) 

t0 

has an extremum value if the curve c defined by x i = x i (t), i = 1,...,N satisfies the Euler-Lagrange 

equations 

 

d ∂φ 

dt ∂ ˙x i 

 

− ∂φ 

=0, i =1,...,N. (2.2.36) 

∂xi The above derivation is a special case of (2.2.36) when φ = L. Note that the equations of motion equations 

(2.2.34) are just another form of the equations (2.2.24). Note also that 

δT 

δt 

 

δ 1 

= 

δt 2 mgijv i v j 

 

= mgijv i f j = mfiv i = mfi ˙x i 

and if we assume that the force Qi is derivable from a potential function V ,thenmfi = Qi = − ∂V 

,so 

∂xi that δT 

δt = mfi ˙x i = Qi ˙x i = − ∂V 

∂xi ˙x i = − δV δ 

or (T + V )=0orT + V = h = constant called the energy 

δt δt 

constant of the system. 

Action Integral 

The equations of motion (2.2.34) or (2.2.24) are interpreted as describing geodesics in a space whose 

line-element is 

ds 2 =2m(h − V )gjkdx j dx k 

where V is the potential function for the force system and T + V = h is the energy constant of the motion. 

The integral of ds along a curve C between two points P1 and P2 is called an action integral and is 

A = √ P2 

2m 

P1 

 

dx 

(h − V )gjk 

j 

dτ 

dxk 1/2 

dτ 

dτ

where τ is a parameter used to describe the curve C. The principle of stationary action states that of all 

curves through the points P1 and P2 the one which makes the action an extremum is the curve specified by 

Newton’s second law. The extremum is usually a minimum. To show this let 

φ = √ 

dx 

2m (h − V )gjk 

j 

dτ 

in equation (2.2.36). Using the notation ˙x k = dxk 

dτ 

∂φ 

=2m 

∂ ˙x i 

∂φ 

=2m (h − V )∂gjk 

∂xi 2φ 

φ (h − V )gik ˙x k 

we find that 

dxk 1/2 

dτ 

∂V 

∂xi ˙x j ˙x k − 2m 

2φ ∂xi gjk ˙x j ˙x k . 

The equation (2.2.36) which describe the extremum trajectories are found to be 

 

d 2m 

dt φ (h − V )gik ˙x k 

 

− 2m 

(h − V )∂gjk 

2φ 

∂V 

∂xi ˙x j ˙x k + 2m 

φ ∂xi gjk ˙x j ˙x k =0. 

By changing variables from τ to t where dt 

dτ = √ mφ 

√ 2(h−V ) we find that the trajectory for an extremum must 

satisfy the equation 

m d 

 

gik 

dt 

dxk 

dt 

− m ∂gjk 

2 ∂xi dxj dx 

dt 

k 

dt 

∂V 

+ =0 

∂xi which are the same equations as (2.2.24). (i.e. See also the equations (2.2.22).) 

Dynamics of Rigid Body Motion 

Let us derive the equations of motion of a rigid body which is rotating due to external forces acting 

upon it. We neglect any translational motion of the body since this type of motion can be discerned using 

our knowledge of particle dynamics. The derivation of the equations of motion is restricted to Cartesian 

tensors and rotational motion. 

Consider a system of N particles rotating with angular velocity ωi, i=1, 2, 3, about a line L through 

the center of mass of the system. Let V (α) denote the velocity of the αth particle which has mass m (α) and 

position x (α) 

i , i =1, 2, 3 with respect to an origin on the line L. Without loss of generality we can assume 

that the origin of the coordinate system is also at the center of mass of the system of particles, as this choice 

of an origin simplifies the derivation. The velocity components for each particle is obtained by taking cross 

products and we can write 

V (α) = ω × r (α) 

or V (α) 

i 

(α) 

= eijkωjx . (2.2.37) 

The kinetic energy of the system of particles is written as the sum of the kinetic energies of each 

individual particle and is 

T = 1 

2 

N 

m (α)V 

α=1 

(α) 

i i 

(α) 1 

V = 

2 

N 

α=1 

m (α)eijkωjx (α) 

k 

k 

(α) 

eimnωmx n 

. (2.2.38) 

199

200 

Employing the e − δ identity the equation (2.2.38) can be simplified to the form 

T = 1 

2 

N 

m (α) 

α=1 

 

ωmωmx (α) 

k x(α) 

(α) 

k − ωnωkx k x(α) 

 

n . 

Define the second moments and products of inertia by the equation 

Iij = 

N 

m (α) 

α=1 

 

x (α) 

k x(α) 

k δij − x (α) 

i x(α) 

 

j 

(2.2.39) 

and write the kinetic energy in the form 

T = 1 

2 Iijωiωj. (2.2.40) 

Similarly, the angular momentum of the system of particles can also be represented in terms of the 

second moments and products of inertia. The angular momentum of a system of particles is defined as a 

summation of the moments of the linear momentum of each individual particle and is 

Hi = 

N 

α=1 

m (α)eijkx (α) 

j v(α) 

k = 

N 

α=1 

The e − δ identity simplifies the equation (2.2.41) to the form 

Hi = ωj 

α=1 

N 

m (α) 

m (α)eijkx (α) 

j 

(α) 

ekmnωmx n . (2.2.41) 

 

x (α) 

n x(α) n δij − x (α) 

j x(α) 

 

i = ωjIji. (2.2.42) 

The equations of motion of a rigid body is obtained by applying Newton’s second law of motion to the 

system of N particles. The equation of motion of the αth particle is written 

m (α)¨x (α) 

i 

Summing equation (2.2.43) over all particles gives the result 

N 

α=1 

m (α)¨x (α) 

i 

(α) 

= F i . (2.2.43) 

= 

N 

α=1 

F (α) 

i . (2.2.44) 

This represents the translational equations of motion of the rigid body. The equation (2.2.44) represents the 

rate of change of linear momentum being equal to the total external force acting upon the system. Taking 

produces 

the cross product of equation (2.2.43) with the position vector x (α) 

j 

and summing over all particles we find the equation 

N 

α=1 

m (α)¨x (α) 

t erstx (α) 

s = erstx (α) 

s F (α) 

t 

m (α)erstx (α) 

s ¨x(α) t = 

N 

α=1 

erstx (α) (α) 

s F t . (2.2.45)

The equations (2.2.44) and (2.2.45) represent the conservation of linear and angular momentum and can be 

writtenintheforms 

 

N 

d 

m (α) ˙x 

dt 

α=1 

(α) 

 

N 

r = F 

α=1 

(α) 

r 

(2.2.46) 

and 

 

N 

d 

m (α)erstx 

dt 

α=1 

(α) 

s ˙x (α) 

 

N 

t = erstx 

α=1 

(α) (α) 

s F t . (2.2.47) 

By definition we have Gr = m (α) ˙x (α) 

r 

representing the linear momentum, Fr = F (α) 

r 

the total force 

acting on the system of particles, Hr = m (α)erstx (α) 

s ˙x (α) 

t is the angular momentum of the system relative 

to the origin, and Mr = erstx (α) 

s F (α) 

t is the total moment of the system relative to the origin. We can 

therefore express the equations (2.2.46) and (2.2.47) in the form 

dGr 

dt = Fr (2.2.48) 

and 

dHr 

dt = Mr. (2.2.49) 

The equation (2.2.49) expresses the fact that the rate of change of angular momentum is equal to the 

moment of the external forces about the origin. These equations show that the motion of a system of 

particles can be studied by considering the motion of the center of mass of the system (translational motion) 

and simultaneously considering the motion of points about the center of mass (rotational motion). 

We now develop some relations in order to express the equations (2.2.49) in an alternate form. Toward 

this purpose we consider first the concepts of relative motion and angular velocity. 

Relative Motion and Angular Velocity 

Consider two different reference frames denoted by S and S. Both reference frames are Cartesian 

coordinates with axes xi and xi , i =1, 2, 3, respectively. The reference frame S is fixed in space and is 

called an inertial reference frame or space-fixed reference system of axes. The reference frame S is fixed 

to and rotates with the rigid body and is called a body-fixed system of axes. Again, for convenience, it 

is assumed that the origins of both reference systems are fixed at the center of mass of the rigid body. 

Further, we let the system S have the basis vectors ei,i=1, 2, 3, while the reference system S has the basis 

vectors êi ,i =1, 2, 3. The transformation equations between the two sets of reference axes are the affine 

transformations 

xi = ℓjixj and xi = ℓijxj (2.2.50) 

where ℓij = ℓij(t) are direction cosines which are functions of time t (i.e. the ℓij are the cosines of the 

angles between the barred and unbarred axes where the barred axes are rotating relative to the space-fixed 

unbarred axes.) The direction cosines satisfy the relations 

ℓijℓik = δjk and ℓijℓkj = δik. (2.2.51) 

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202 

EXAMPLE 2.2-5. (Euler angles φ, θ, ψ) Consider the following sequence of transformations which 

are used in celestial mechanics. First a rotation about the x3 axis taking the xi axes to the yi axes 

⎛ 

⎝ y1 

⎞ ⎛ 

cos φ 

y2 ⎠ = ⎝ − sin φ 

sin φ 

cos φ 

⎞ ⎛ 

0 

0 ⎠ ⎝ 

y3 0 0 1 

x1 

⎞ 

x2 ⎠ 

x3 

where the rotation angle φ is called the longitude of the ascending node. Second, a rotation about the y1 

axis taking the yi axes to the y ′ i axes 

⎛ 

⎝ y′ 1 

y ′ 2 

y ′ ⎞ ⎛ 

1 

⎠ = ⎝ 0 

3 0 

0 

cosθ 

− sin θ 

⎞ ⎛ 

0 

sin θ ⎠ ⎝ 

cos θ 

y1 

⎞ 

y2 ⎠ 

y3 

where the rotation angle θ is called the angle of inclination of the orbital plane. Finally, a rotation about 

the y ′ 3 axis taking the y′ i axes to the ¯xi axes 

⎛ 

⎝ ¯x1 

⎞ ⎛ 

⎞ ⎛ 

cos ψ sin ψ 0 

¯x2 ⎠ = ⎝ − sin ψ cos ψ 0 ⎠ ⎝ 

¯x3 0 0 1 

y′ 1 

y ′ 2 

y ′ ⎞ 

⎠ 

3 

where the rotation angle ψ is called the argument of perigee. The Euler angle θ is the angle ¯x30x3, the angle 

φ is the angle x10y1 and ψ is the angle y10¯x1. These angles are illustrated in the figure 2.2-6. Note also that 

the rotation vectors associated with these transformations are vectors of magnitude ˙ φ, ˙ θ, ˙ ψ in the directions 

indicated in the figure 2.2-6. 

Figure 2.2-6. Euler angles. 

By combining the above transformations there results the transformation equations (2.2.50) 

⎛ 

⎝ ¯x1 

⎞ ⎛ 

⎞ ⎛ 

cos ψ cos φ − cos θ sin φ sin ψ cos ψ sin φ +cosθcos φ sin ψ sin ψ sin θ 

¯x2 ⎠ = ⎝ − sin ψ cos φ − cos θ sin φ cos ψ − sin ψ sin φ +cosθcos φ cos ψ cos ψ sin θ ⎠ ⎝ 

¯x3 

sin θ sin φ − sin θ cos φ cos θ 

x1 

⎞ 

x2 ⎠ . 

x3 

It is left as an exercise to verify that the transformation matrix is orthogonal and the components ℓji 

satisfy the relations (2.2.51).

Consider the velocity of a point which is rotating with the rigid body. Denote by vi = vi(S), for 

i =1, 2, 3, the velocity components relative to the S reference frame and by vi = vi(S), i =1, 2, 3the 

velocity components of the same point relative to the body-fixed axes. In terms of the basis vectors we can 

write 

V = v1(S) ê1 + v2(S) ê2 + v3(S) ê3 = dxi 

dt êi 

as the velocity in the S reference frame. Similarly, we write 

(2.2.52) 

 

V = v1(S)e1 + v2(S)e2 + v3(S)e3 = dxi 

dt ei 

(2.2.53) 

as the velocity components relative to the body-fixed reference frame. There are occasions when it is desirable 

to represent V in the S frame of reference and V in the S frame of reference. In these instances we can write 

and 

V = v1(S)e1 + v2(S)e2 + v3(S)e3 

(2.2.54) 

 

V = v1(S) ê1 + v2(S) ê2 + v3(S) ê3. (2.2.55) 

Here we have adopted the notation that vi(S) are the velocity components relative to the S reference frame 

and vi(S) are the same velocity components relative to the S reference frame. Similarly, vi(S) denotes the 

velocity components relative to the S reference frame, while vi(S) denotes the same velocity components 

relative to the S reference frame. 

Here both V and V are vectors and so their components are first order tensors and satisfy the transformation 

laws 

vi(S) =ℓjivj(S) =ℓji ˙xj and vi(S) =ℓijvj(S) =ℓij ˙xj. (2.2.56) 

The equations (2.2.56) define the relative velocity components as functions of time t. By differentiating the 

equations (2.2.50) we obtain 

dxi 

dt = vi(S) =ℓji ˙xj + ˙ ℓjixj 

(2.2.57) 

and 

dxi 

dt = vi(S) =ℓij ˙xj + ˙ ℓijxj. (2.2.58) 

Multiply the equation (2.2.57) by ℓmi and multiply the equation (2.2.58) by ℓim and derive the relations 

and 

vm(S) =vm(S)+ℓmi ˙ ℓjixj 

(2.2.59) 

vm(S) =vm(S)+ℓim ˙ ℓijxj. (2.2.60) 

The equations (2.2.59) and (2.2.60) describe the transformation laws of the velocity components upon changing 

from the S to the S reference frame. These equations can be expressed in terms of the angular velocity 

by making certain substitutions which are now defined. 

The first order angular velocity vector ωi is related to the second order skew-symmetric angular velocity 

tensor ωij by the defining equation 

ωmn = eimnωi. (2.2.61) 

203

204 

The equation (2.2.61) implies that ωi and ωij are dual tensors and 

ωi = 1 

2 eijkωjk. 

Also the velocity of a point which is rotating about the origin relative to the S frame of reference is vi(S) = 

eijkωjxk which can also be written in the form vm(S) =−ωmkxk. Since the barred axes rotate with the rigid 

body, then a particle in the barred reference frame will have vm(S) = 0, since the coordinates of a point 

in the rigid body will be constants with respect to this reference frame. Consequently, we write equation 

(2.2.59) in the form 0 = vm(S)+ℓmi ˙ ℓjixj which implies that 

vm(S) =−ℓmi ˙ ℓjixj = −ωmkxk or ωmj = ωmj(S,S)=ℓmi ˙ ℓji. 

This equation is interpreted as describing the angular velocity tensor of S relative to S. Sinceωij is a tensor, 

it can be represented in the barred system by 

ωmn(S,S)=ℓimℓjnωij(S,S) 

= ℓimℓjnℓis ˙ ℓjs 

= δmsℓjn ˙ ℓjs 

= ℓjn ˙ ℓjm 

(2.2.62) 

By differentiating the equations (2.2.51) it is an easy exercise to show that ωij is skew-symmetric. The 

second order angular velocity tensor can be used to write the equations (2.2.59) and (2.2.60) in the forms 

vm(S) =vm(S)+ωmj(S,S)xj 

vm(S) =vm(S)+ωjm(S,S)xj 

(2.2.63) 

The above relations are now employed to derive the celebrated Euler’s equations of motion of a rigid body. 

Euler’s Equations of Motion 

We desire to find the equations of motion of a rigid body which is subjected to external forces. These 

equations are the formulas (2.2.49), and we now proceed to write these equations in a slightly different form. 

Similar to the introduction of the angular velocity tensor, given in equation (2.2.61), we now introduce the 

following tensors 

1. The fourth order moment of inertia tensor Imnst which is related to the second order moment of 

inertia tensor Iij by the equations 

Imnst = 1 

2 ejmneistIij or Iij = 1 

2 Ipqrseipqejrs 

(2.2.64) 

2. The second order angular momentum tensor Hjk which is related to the angular momentum vector 

Hi by the equation 

Hi = 1 

2 eijkHjk or Hjk = eijkHi (2.2.65) 

3. The second order moment tensor Mjk which is related to the moment Mi by the relation 

Mi = 1 

2 eijkMjk or Mjk = eijkMi. (2.2.66)

Now if we multiply equation (2.2.49) by erjk, thenitcanbewrittenintheform 

dHij 

dt = Mij. (2.2.67) 

Similarly, if we multiply the equation (2.2.42) by eimn, then it can be expressed in the alternate form 

Hmn = eimnωjIji = Imnstωst 

and because of this relation the equation (2.2.67) can be expressed as 

d 

dt (Iijstωst) =Mij. (2.2.68) 

We write this equation in the barred system of coordinates where Ipqrs will be a constant and consequently 

its derivative will be zero. We employ the transformation equations 

Iijst = ℓipℓjqℓsrℓtkIpqrk 

ωij = ℓsiℓtjωst 

M pq = ℓipℓjqMij 

and then multiply the equation (2.2.68) by ℓipℓjq and simplify to obtain 

d 

ℓipℓjq ℓiαℓjβIαβrkωrk = M pq. 

dt 

Expand all terms in this equation and take note that the derivative of the Iαβrk is zero. The expanded 

equation then simplifies to 

dωrk 

Ipqrk 

dt +(δαuδpvδβq + δpαδβuδqv) Iαβrkωrkωuv = M pq. (2.2.69) 

Substitute into equation (2.2.69) the relations from equations (2.2.61),(2.2.64) and (2.2.66), and then multiply 

by empq and simplify to obtain the Euler’s equations of motion 

dωi 

Iim 

dt − etmjIijωiωt = M m. (2.2.70) 

Dropping the bar notation and performing the indicated summations over the range 1,2,3 we find the 

Euler equations have the form 

dω1 

I11 

dt 

dω1 

I12 

dt 

dω2 

+ I21 

dt 

dω2 

+ I22 

dt 

dω3 

+ I31 

dω3 

+ I32 

dt +(I13ω1 + I23ω2 + I33ω3) ω2 − (I12ω1 + I22ω2 + I32ω3) ω3 = M1 

dt +(I11ω1 + I21ω2 + I31ω3) ω3 − (I13ω1 + I23ω2 + I33ω3) ω1 = M2 

(2.2.71) 

dω1 dω2 dω3 

I13 + I23 + I33 

dt dt dt +(I12ω1 + I22ω2 + I32ω3) ω1 − (I11ω1 + I21ω2 + I31ω3) ω2 = M3. 

In the special case where the barred axes are principal axes, then Iij =0fori= j and the Euler’s 

equations reduces to the system of nonlinear differential equations 

dω1 

I11 

dt +(I33 − I22)ω2ω3 = M1 

dω2 

I22 

dt +(I11 − I33)ω3ω1 = M2 

dω3 

I33 

(2.2.72) 

dt +(I22 − I11)ω1ω2 = M3. 

In the case of constant coefficients and constant moments the solutions of the above differential equations 

can be expressed in terms of Jacobi elliptic functions. 

205

206 

EXERCISE 2.2 

◮ 1. Find a set of parametric equations for the straight line which passes through the points P1(1, 1, 1) and 

P2(2, 3, 4). Find the unit tangent vector to any point on this line. 

◮ 2. Consider the space curve x = 1 

2 sin2 t, y = 1 1 

2t − 4 sin 2t, z =sintwhere t is a parameter. Find the unit 

vectors T i ,Bi ,Ni ,i=1, 2, 3 at the point where t = π. 

◮ 3. A claim has been made that the space curve x = t, y = t2 ,z= t3 intersects the plane 11x-6y+z=6 in 

three distinct points. Determine if this claim is true or false. Justify your answer and find the three points 

of intersection if they exist. 

◮ 4. Find a set of parametric equations xi = xi(s1,s2),i =1, 2, 3 for the plane which passes through the 

points P1(3, 0, 0), P2(0, 4, 0) and P3(0, 0, 5). Find a unit normal to this plane. 

◮ 5. For the helix x =sint y =cost z = 2 

t find the equation of the tangent plane to the curve at the 

π 

point where t = π/4. Find the equation of the tangent line to the curve at the point where t = π/4. 

◮ 6. Express the generalized velocity and acceleration in cylindrical coordinates. Find the physical components 

of velocity and acceleration in cylindrical coordinates. 

◮ 7. Express the generalized velocity and acceleration in spherical coordinates. Find the physical components 

of velocity and acceleration in spherical coordinates. 

◮ 8. Verify the derivative ∂T 

∂ ˙x r = Mgrm ˙x m . 

◮ 9. Verify the derivative d 

 

∂T 

dt ∂ ˙x r 

 

◮ 10. Verify the derivative ∂T 

∂x 

1 

= r 2 

= M 

 

grm¨x m + ∂grm 

∂xn ˙xn ˙x m 

 

. 

M ∂gmn 

∂x r ˙xm ˙x n . 

◮ 11. Use the results from problems 8,9 and 10 to derive the Lagrange’s form for the equations of motion 

defined by equation (2.2.23). 

◮ 12. Expand equation (2.2.39) and write out all the components of the moment of inertia tensor Iij. 

◮ 13. For ρ the density of a continuous material and dτ an element of volume inside a region R where the 

material is situated, we write ρdτ as an element of mass inside R. Find an equation which describes the 

center of mass of the region R. 

◮ 14. Use the equation (2.2.68) to derive the equation (2.2.69). 

◮ 15. Drop the bar notation and expand the equation (2.2.70) and derive the equations (2.2.71). 

◮ 16. Verify the Euler transformation, given in example 2.2-5, is orthogonal.

Figure 2.2-7. Pulley and mass system 

◮ 17. For the pulley and mass system illustrated in the figure 2.2-7 let 

a = the radius of each pulley. 

ℓ1 = the length of the upper chord. 

ℓ2 = the length of the lower chord. 

Neglect the weight of the pulley and find the equations of motion for the pulley mass system. 

◮ 18. Let φ = ds 

dt , where s is the arc length between two points on a curve in generalized coordinates. 

(a) Write the arc length in general coordinates as ds = gmn ˙x m ˙x ndt and show the integral I, defined by 

equation (2.2.35), represents the distance between two points on a curve. 

(b) Using the Euler-Lagrange equations (2.2.36) show that the shortest distance between two points in a 

i 

jk 

 

˙x j ˙x k =˙x i 

generalized space is the curve defined by the equations: ¨x i + 

dt 

(c) Show in the special case t = s the equations in part (b) reduce to d2xi j i dx dx 

+ 

ds2 jk ds 

k 

ds =0,for 

i =1,...,N. An examination of equation (1.5.51) shows that the above curves are geodesic curves. 

(d) Show that the shortest distance between two points in a plane is a straight line. 

(e) Consider two points on the surface of a cylinder of radius a. Let u 1 = θ and u 2 = z denote surface 

coordinates in the two dimensional space defined by the surface of the cylinder. Show that the shortest 

distance between the points where θ =0,z=0andθ = π, z = H is L = a2π2 + H2 . 

◮ 19. For T = 1 

2mgijviv j the kinetic energy of a particle and V the potential energy of the particle show 

that T + V = constant. 

Hint: mfi = Qi = − ∂V 

∂x i , i =1, 2, 3 and dx i 

dt =˙xi = v i ,i=1, 2, 3. 

d 2 s 

dt 2 

ds 

207

208 

◮ 20. Define H = T + V as the sum of the kinetic energy and potential energy of a particle. The quantity 

H = H(xr ,pr) is called the Hamiltonian of the particle and it is expressed in terms of: 

• the particle position xi and 

• the particle momentum pi = mvi = mgij ˙x j . Here x r and pr are treated as independent variables. 

(a) Show that the particle momentum is a covariant tensor of rank 1. 

(b) Express the kinetic energy T in terms of the particle momentum. 

(c) Show that pi = ∂T 

. 

∂ ˙x i 

(d) Show that dxi ∂H 

dpi ∂H 

= and = − . These are a set of differential equations describing the 

dt ∂pi 

dt ∂xi position change and momentum change of the particle and are known as Hamilton’s equations of motion 

for a particle. 

◮ 21. Let 

δT i 

δs = κN i and δNi 

δs = τBi − κT i and calculate the intrinsic derivative of the cross product 

B i = ɛ ijk TjNk and find δBi 

δs in terms of the unit normal vector. 

◮ 22. For T the kinetic energy of a particle and V the potential energy of a particle, define the Lagrangian 

L = L(x i , ˙x i )=T − V = 1 

2 Mgij ˙x i ˙x j − V as a function of the independent variables x i , ˙x i . Define the 

Hamiltonian H = H(x i ,pi) =T + V = 1 

2M gij pipj + V, as a function of the independent variables x i ,pi, 

where pi is the momentum vector of the particle and M is the mass of the particle. 

(a) Show that pi = ∂T 

. 

∂ ˙x i 

(b) Show that ∂H ∂L 

= − 

∂xi ∂xi ◮ 23. When the Euler angles, figure 2.2-6, are applied to the motion of rotating objects, θ is the angle 

of nutation, φ is the angle of precession and ψ is the angle of spin. Take projections and show that the 

time derivative of the Euler angles are related to the angular velocity vector components ωx,ωy,ωz by the 

relations 

ωx = ˙ θ cos ψ + ˙ φ sin θ sin ψ 

ωy = − ˙ θ sin ψ + ˙ φ sin θ cos ψ 

ωz = ˙ ψ + ˙ φ cos θ 

where ωx,ωy,ωz are the angular velocity components along the x1, x2, x3 axes. 

◮ 24. Find the equations of motion for the compound pendulum illustrated in the figure 2.2-8. 

◮ 25. Let F = − GMm 

r3 r denote the inverse square law force of attraction between the earth and sun, with 

G a universal constant, M the mass of the sun, m the mass of the earth and r 

r a unit vector from origin 

at the center of the sun pointing toward the earth. (a) Write down Newton’s second law, in both vector 

and tensor form, which describes the motion of the earth about the sun. (b) Show that d 

dt (r × v) =0 and 

consequently r × v = r × dr 

dt = h =aconstant.

Figure 2.2-8. Compound pendulum 

◮ 26. Construct a set of axes fixed and attached to an airplane. Let the x axis be a longitudinal axis running 

from the rear to the front of the plane along its center line. Let the y axis run between the wing tips and 

let the z axis form a right-handed system of coordinates. The y axis is called a lateral axis and the z axis is 

called a normal axis. Define pitch as any angular motion about the lateral axis. Define roll as any angular 

motion about the longitudinal axis. Define yaw as any angular motion about the normal axis. Consider two 

sets of axes. One set is the x, y, z axes attached to and moving with the aircraft. The other set of axes is 

denoted X, Y, Z and is fixed in space ( an inertial set of axes). Describe the pitch, roll and yaw of an aircraft 

with respect to the inertial set of axes. Show the transformation is orthogonal. Hint: Consider pitch with 

respect to the fixed axes, then consider roll with respect to the pitch axes and finally consider yaw with 

respect to the roll axes. This produces three separate transformation matrices which can then be combined 

to describe the motions of pitch, roll and yaw of an aircraft. 

◮ 27. In Cartesian coordinates let Fi = Fi(x1 ,x2 ,x3 ) denote a force field and let xi = xi (t) denote a curve 

C. (a) Show Newton’s second law implies that along the curve C d 

 

1 

dt 2 m 

2 i dx 

dt 

 

= Fi(x 1 ,x 2 ,x 3 ) dxi 

dt 

(no summation on i) and hence 

 

d 1 

dt 2 m 

dx 2 2 2 1 

2 

3 

dx dx 

+ + 

dt dt dt 

 

= d 

 

1 

dt 2 mv2 

 

dx 

= F1 

1 dx 

+ F2 

dt 2 dx 

+ F3 

dt 3 

dt 

(b) Consider two points on the curve C, saypointA, x i (tA) andpointB, x i (tB) and show that the work 

done in moving from A to B in the force field Fi is 

1 

2 mv2 

tB B 

= Fidx 

tA A 

1 + F2dx 2 + F3dx 3 

209

210 

where the right hand side is a line integral along the path C from A to B. (c) Show that if the force field is 

derivable from a potential function U(x 1 ,x 2 ,x 3 ) by taking the gradient, then the work done is independent 

of the path C and depends only upon the end points A and B. 

◮ 28. Find the Lagrangian equations of motion of a spherical pendulum which consists of a bob of mass m 

suspended at the end of a wire of length ℓ, which is free to swing in any direction subject to the constraint 

that the wire length is constant. Neglect the weight of the wire and show that for the wire attached to the 

origin of a right handed x, y, z coordinate system, with the z axis downward, φ the angle between the wire 

and the z axis and θ the angle of rotation of the bob from the y axis, that there results the equations of 

 

d 

motion sin 

dt 

2 φ dθ 

 

d 

=0 and 

dt 

2 2 φ dθ 

− sin φ cos φ + 

dt2 dt 

g 

sin φ =0 

ℓ 

◮ 29. In Cartesian coordinates show the Frenet formulas can be written 

d T 

ds = δ × T, 

d N 

ds = δ × N, 

where δ is the Darboux vector and is defined δ = τ T + κ B. 

d B 

ds = δ × B 

◮ 30. Consider the following two cases for rigid body rotation. 

Case 1: Rigid body rotation about a fixed line which is called the fixed axis of rotation. Select a point 0 

on this fixed axis and denote by e a unit vector from 0 in the direction of the fixed line and denote by êR 

a unit vector which is perpendicular to the fixed axis of rotation. The position vector of a general point 

in the rigid body can then be represented by a position vector from the point 0 given by r = h e + r0 êR 

where h, r0 and e are all constants and the vector êR is fixed in and rotating with the rigid body. 

Denote by ω = dθ 

dt the scalar angular change with respect to time of the vector êR as it rotates about 

the fixed line and define the vector angular velocity as ω = d 

(θ e) =dθ e where θ e is defined as the 

dt dt 

vector angle of rotation. 

d êR 

(a) Show that = e × êR. 

dθ 

(b) Show that V = dr d êR d êR dθ 

= r0 = r0 

dt dt dθ dt = ω × (r0 êR) =ω × (h e + r0 êR) =ω × r. 

Case 2: Rigid body rotation about a fixed point 0. Construct at point 0 the unit vector ê1 which is 

d ê1 

fixed in and rotating with the rigid body. From pages 80,87 we know that must be perpendicular 

dt 

d ê1 

to ê1 and so we can define the vector ê2 as a unit vector which is in the direction of such that 

dt 

d ê1 

dt = α ê2 for some constant α. We can then define the unit vector ê3 from ê3 = ê1 × ê2. 

d ê3 

(a) Show that 

dt , which must be perpendicular to ê3, is also perpendicular to ê1. 

d ê3 

d ê3 

(b) Show that can be written as 

dt dt = β ê2 for some constant β. 

d ê2 

(c) From ê2 = ê3 × ê1 show that 

dt =(αê3 − β ê1) × ê2 

d ê1 

(d) Define ω = α ê3 − β ê1 and show that = ω × ê1, 

dt 

d ê2 

dt 

= ω × ê2, 

d ê3 

dt 

= ω × ê3

(e) Let r = x ê1 + y ê2 + z ê3 denote an arbitrary point within the rigid body with respect to the point 0. 

Show that dr 

= ω × r. 

dt 

Note that in Case 2 the direction of ω is not fixed as the unit vectors ê3 and ê1 are constantly changing. 

In this case the direction ω is called an instantaneous axis of rotation and ω, which also can change in 

magnitude and direction, is called the instantaneous angular velocity. 

211

§2.3 BASIC EQUATIONS OF CONTINUUM MECHANICS 

Continuum mechanics is the study of how materials behave when subjected to external influences. 

External influences which affect the properties of a substance are such things as forces, temperature, chemical 

reactions, and electric phenomena. Examples of forces are gravitational forces, electromagnetic forces, and 

mechanical forces. Solids deform under external forces and so deformations are studied. Fluids move under 

external forces and so the velocity of the fluid is studied. 

A material is considered to be a continuous media which is a collection of material points interconnected 

by internal forces (forces between the atoms making up the material). We concentrate upon the macroscopic 

properties rather than the microscopic properties of the material. We treat the material as a body which is 

homogeneous and continuous in its makeup. 

In this introduction we will only consider solid media and liquid media. In general, most of the ideas 

and concepts developed in this section can be applied to any type of material which is assumed to be a 

collection of material points held together by some kind of internal forces. 

An elastic material is one which deforms under applied forces in such a way that it will return to its 

original unloaded state when the applied forces are removed. When a linear relation exists between the 

applied forces and material displacements, then the material is called a linear elastic material. In contrast, a 

plastic material is one which deforms under applied forces in such a way that it does not return to its original 

state after removal of the applied forces. Plastic materials will always exhibit some permanent deformation 

after removal of the applied forces. An elastic material is called homogeneous if it has the same properties 

throughout. An isotropic material has the same properties, at a point, in all directions about the point. 

In this introduction we develop the basic mathematical equations which describe how a continuum 

behaves when subjected to external forces. We shall discover that there exists a set of basic equations 

associated with all continuous material media. These basic equations are developed for linear elastic materials 

and applied to solids and fluids in later sections. 

Introduction to Elasticity 

Take a rubber band, which has a rectangular cross section, and mark on it a parallelepiped having a 

length ℓ, awidthwand a height h, as illustrated in the figure 2.3-1. 

Now apply a force F to both ends of the parallelepiped cross section on the rubber band and examine 

what happens to the parallelepiped. You will see that: 

1. ℓ increases by an amount ∆ℓ. 

2. w decreases by an amount ∆w. 

3. h decreases by an amount ∆h. 

There are many materials which behave in a manner very similar to the rubber band. Most materials, 

when subjected to tension forces will break if the change ∆ℓ is only one or two percent of the original length. 

The above example introduces us to several concepts which arise in the study of materials when they are 

subjected to external forces. The first concept is that of strain which is defined as 

strain = 

change in length 

, (dimensionless). 

original length 

211


Figure 2.3-1. Section of a rubber band 

When the force F is applied to our rubber band example there arises the strains 

∆ℓ 

ℓ , 

∆w 

w , 

The second concept introduced by our simple example is stress. Stress is defined as a force per unit area. In 

particular, 

Force 

force 

stress = 

, with dimension of 

Area over which force acts unit area . 

We will be interested in studying stress and strain in homogeneous, isotropic materials which are in equilibrium 

with respect to the force system acting on the material. 

Hooke’s Law 

For linear elastic materials, where the forces are all one dimensional, the stress and strains are related 

by Hooke’s law which has two parts. The Hooke’s law, part one, states that stress is proportional to strain 

in the stretch direction, where the Young’s modulus E is the proportionality constant. This is written 

 

F ∆ℓ 

Hooke’s law part 1 

= E . (2.3.1) 

A ℓ 

A graph of stress vs strain is a straight line with slope E in the linear elastic range of the material. 

The Hooke’s law, part two, involves the fact that there is a strain contraction perpendicular to the 

stretch direction. The strain contraction is the same for both the width and height and is proportional to 

the strain in the stretch direction. The proportionality constant being the Poisson’s ratio ν. 

Hooke’s law part 2 

∆w 

w 

= ∆h 

h 

∆h 

h . 

= −ν ∆ℓ 

ℓ 

, 0

Figure 2.3-2. Typical Stress-strain curve. 

Consider a typical stress-strain curve, such as the one illustrated in the figure 2.3-2, which is obtained 

by placing a material in the shape of a rod or wire in a machine capable of performing tensile straining at a 

low rate. The engineering stress is the tensile force F divided by the original cross sectional area A0. Note 

that during a tensile straining the cross sectional area A of the sample is continually changing and getting 

smaller so that the actual stress will be larger than the engineering stress. Observe in the figure 2.3-2 that 

the stress-strain relation remains linear up to a point labeled the proportional limit. For stress-strain points 

in this linear region the Hooke’s law holds and the material will return to its original shape when the loading 

is removed. For points beyond the proportional limit, but less than the yield point, the material no longer 

obeys Hooke’s law. In this nonlinear region the material still returns to its original shape when the loading 

is removed. The region beyond the yield point is called the plastic region. At the yield point and beyond, 

there is a great deal of material deformation while the loading undergoes only small changes. For points 

in this plastic region, the material undergoes a permanent deformation and does not return to its original 

shape when the loading is removed. In the plastic region there usually occurs deformation due to slipping of 

atomic planes within the material. In this introductory section we will restrict our discussions of material 

stress-strain properties to the linear region. 

EXAMPLE 2.3-1. (One dimensional elasticity) Consider a circular rod with cross sectional area A 

which is subjected to an external force F applied to both ends. The figure 2.3-3 illustrates what happens to 

the rod after the tension force F is applied. Consider two neighboring points P and Q on the rod, where P 

is at the point x and Q is at the point x +∆x. When the force F is applied to the rod it is stretched and 

P moves to P ′ and Q moves to Q ′ . We assume that when F is applied to the rod there is a displacement 

function u = u(x, t) which describes how each point in the rod moves as a function of time t. If we know the 

displacement function u = u(x, t) we would then be able to calculate the following distances in terms of the 

displacement function 

PP ′ = u(x, t), 0P ′ = x + u(x, t), QQ ′ = u(x +∆x, t) 0Q ′ = x +∆x + u(x +∆x, t). 

213

214 

Figure 2.3-3. One dimensional rod subjected to tension force 

The strain associated with the distance ℓ =∆x = PQ is 

e = ∆ℓ 

ℓ = P ′ Q ′ − PQ 

= 

PQ 

(0Q′ − 0P ′ ) − (0Q − 0P ) 

PQ 

[x +∆x + u(x +∆x, t) − (x + u(x, t))] − [(x +∆x) − x] 

e = 

∆x 

u(x +∆x, t) − u(x, t) 

e = . 

∆x 

Use the Hooke’s law part(i) and write 

F 

A 

Taking the limit as ∆x approaches zero we find that 

u(x +∆x, t) − u(x, t) 

= E . 

∆x 

F 

A 

∂u(x, t) 

= E . 

∂x 

Hence, the stress is proportional to the spatial derivative of the displacement function. 

Normal and Shearing Stresses 

Let us consider a more general situation in which we have some material which can be described as 

having a surface area S which encloses a volume V. Assume that the density of the material is ϱ and the 

material is homogeneous and isotropic. Further assume that the material is subjected to the forces b and t (n) 

where b is a body force per unit mass [force/mass], and t (n) is a surface traction per unit area [force/area]. 

The superscript (n) on the vector is to remind you that we will only be interested in the normal component 

of the surface forces. We will neglect body couples, surface couples, and concentrated forces or couples that 

act at a single point. If the forces described above are everywhere continuous we can calculate the resultant 

force F and resultant moment M acting on the material by constructing various surface and volume integrals 

which sum the forces acting upon the material. In particular, the resultant force F acting on our material 

can be described by the surface and volume integrals: 

 

F = t (n) 

dS + ϱbdτ (2.3.3) 

S 

V

Figure 2.3-4. Stress vectors acting upon an element of volume 

which is a summation of all the body forces and surface tractions acting upon our material. Here ϱ is the 

density of the material, dS is an element of surface area, and dτ is an element of volume. 

The resultant moment M about the origin is similarly expressed as 

 

M = 

S 

r × t (n) 

dS + ϱ(r × 

V 

b) dτ. (2.3.4) 

The global motion of the material is governed by the Euler equations of motion. 

• The time rate of change of linear momentum equals the resultant force or 

 

d 

ϱvdτ = 

dt V 

 

F = t 

S 

(n) 

dS + ϱ 

V 

bdτ. (2.3.5) 

This is a statement concerning the conservation of linear momentum. 

• The time rate of change of angular momentum equals the resultant moment or 

 

d 

ϱr × vdτ = 

dt V 

 

M = r × t 

S 

(n) 

dS + ϱ(r × 

V 

b) dτ. (2.3.6) 

This is a statement concerning conservation of angular momentum. 

The Stress Tensor 

Define the stress vectors 

t 1 = σ 11 ê1 + σ 12 ê2 + σ 13 ê3 

t 2 = σ 21 ê1 + σ 22 ê2 + σ 23 ê3 

t 3 = σ 31 ê1 + σ 32 ê2 + σ 33 ê3, 

(2.3.7) 

where σ ij ,i,j=1, 2, 3 is the stress tensor acting at each point of the material. The index i indicates the 

coordinate surface x i = a constant, upon which t i acts. The second index j denotes the direction associated 

with the components of t i . 

215

216 

Figure 2.3-5. Stress distribution at a point 

For i =1, 2, 3 we adopt the convention of sketching the components of t i in the positive directions if 

the exterior normal to the surface xi = constant also points in the positive direction. This gives rise to the 

figure 2.3-4 which illustrates the stress vectors acting upon an element of volume in rectangular Cartesian 

coordinates. The components σ 11 ,σ 22 ,σ 33 are called normal stresses while the components σ ij ,i= j are 

called shearing stresses. The equations (2.3.7) can be written in the more compact form using the indicial 

notation as 

t i = σ ij êj, i,j =1, 2, 3. (2.3.8) 

If we know the stress distribution at three orthogonal interfaces at a point P in a solid body, we can then 

determine the stress at the point P with respect to any plane passing through the point P. With reference to 

the figure 2.3-5, consider an arbitrary plane passing through the point P which lies within the material body 

being considered. Construct the elemental tetrahedron with orthogonal axes parallel to the x 1 = x, x 2 = y 

and x 3 = z axes. In this figure we have the following surface tractions: 

−t 1 

−t 2 

−t 3 

t (n) 

on the surface 0BC 

on the surface 0AC 

on the surface 0AB 

on the surface ABC 

The superscript parenthesis n is to remind you that this surface traction depends upon the orientation of 

the plane ABC which is determined by a unit normal vector having the direction cosines n1,n2 and n3.

Let 

These surface areas are related by the relations 

∆S1 = the surface area 0BC 

∆S2 = the surface area 0AC 

∆S3 = the surface area 0AB 

∆S = the surface area ABC . 

∆S1 = n1∆S, ∆S2 = n2∆S, ∆S3 = n3∆S (2.3.9) 

which can be thought of as projections of ∆S upon the planes xi =constant for i =1, 2, 3. 

Cauchy Stress Law 

Let t j (n) denote the components of the surface traction on the surface ABC. Thatis,welet 

t (n) = t 1(n) ê1 + t 2(n) ê2 + t 3(n) ê3 = t j (n) êj. (2.3.10) 

It will be demonstrated that the components t j (n) of the surface traction forces t (n) associated with a plane 

through P and having the unit normal with direction cosines n1,n2 and n3, must satisfy the relations 

t j (n) = ni σ ij , i,j =1, 2, 3. (2.3.11) 

This relation is known as the Cauchy stress law. 

Proof: Sum the forces acting on the elemental tetrahedron in the figure 2.3-5. If the body is in equilibrium, 

then the sum of these forces must equal zero or 

(−t 1 ∆S1)+(−t 2 ∆S2)+(−t 3 ∆S3)+t (n) ∆S =0. (2.3.12) 

The relations in the equations (2.3.9) are used to simplify the sum of forces in the equation (2.3.12). It is 

readily verified that the sum of forces simplifies to 

Substituting in the relations from equation (2.3.8) we find 

or in component form 

which is the Cauchy stress law. 

t (n) = n1 t 1 + n2 t 2 + n3 t 3 = ni t i . (2.3.13) 

t (n) = t j (n) êj = niσ ij êj, i,j =1, 2, 3 (2.3.14) 

t j (n) = niσ ij 

(2.3.15) 

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218 

Conservation of Linear Momentum 

Let R denote a region in space where there exists a material volume with density ϱ having surface 

tractions and body forces acting upon it. Let vi denote the velocity of the material volume and use Newton’s 

second law to set the time rate of change of linear momentum equal to the forces acting upon the volume as 

in (2.3.5). We find 

 

δ 

ϱv 

δt R 

j 

dτ = σ 

S 

ij 

ni dS + ϱb 

R 

j dτ. 

Here dτ is an element of volume, dS is an element of surface area, bj are body forces per unit mass, and σij are the stresses. Employing the Gauss divergence theorem, the surface integral term is replaced by a volume 

integral and Newton’s second law is expressed in the form 

 

R 

j j ij 

ϱf − ϱb − σ ,i dτ =0, (2.3.16) 

where f j is the acceleration from equation (1.4.54). Since R is an arbitrary region, the equation (2.3.16) 

implies that 

σ ij ,i + ϱb j = ϱf j . (2.3.17) 

This equation arises from a balance of linear momentum and represents the equations of motion for material 

in a continuum. If there is no velocity term, then equation (2.3.17) reduces to an equilibrium equation which 

can be written 

This equation can also be written in the covariant form 

σ ij ,i + ϱb j =0. (2.3.18) 

g si σms,i + ϱbm =0, 

which reduces to σij,j + ϱbi = 0 in Cartesian coordinates. The equation (2.3.18) is an equilibrium equation 

and is one of our fundamental equations describing a continuum. 

Conservation of Angular Momentum 

The conservation of angular momentum equation (2.3.6) has the Cartesian tensors representation 

 

 

 

d 

ϱeijkxjvk dτ = eijkxjσpknp dS + ϱeijkxjbk dτ. (2.3.19) 

dt R 

S 

R 

Employing the Gauss divergence theorem, the surface integral term is replaced by a volume integral to obtain 

 

eijkϱ d 

dt (xjvk) 

 

− eijk ϱxjbk + ∂ 

 

(xjσpk) dτ =0. 

∂xp (2.3.20) 

R 

Since equation (2.3.20) must hold for all arbitrary volumes R we conclude that 

eijkϱ d 

dt (xjvk) 

 

 

∂σpk 

=eijk ϱxjbk + xj + σjk 

∂xp

Figure 2.3-6. Shearing parallel to the y axis 

which can be rewritten in the form 

 

eijk σjk + xj( ∂σpk 

∂xp + ϱbk − ϱ dvk 

 

) − ϱvjvk =0. 

dt 

(2.3.21) 

In the equation (2.3.21) the middle term is zero because of the equation (2.3.17). Also the last term in 

(2.3.21) is zero because eijkvjvk represents the cross product of a vector with itself. The equation (2.3.21) 

therefore reduces to 

eijkσjk =0, (2.3.22) 

which implies (see exercise 1.1, problem 22) that σij = σji for all i and j. Thus, the conservation of angular 

momentum requires that the stress tensor be symmetric. Consequently, there are only 6 independent stress 

components to be determined. This is another fundamental law for a continuum. 

Strain in Two Dimensions 

Consider the matrix equation 

x 

= 

y 

1 0 

β 1 

 

x 

y 

(2.3.23) 

which can be used to transform points (x, y) topoints(x, y). When this transformation is applied to the 

unit square illustrated in the figure 2.3-6(a) we obtain the geometry illustrated in the figure 2.3-6(b) which 

represents a shearing parallel to the y axis. If β is very small, we can use the approximation tan β ≈ β and 

then this transformation can be thought of as a rotation of the element P1P2 through an angle β to the 

position P ′ 1 P ′ 2 

when the barred axes are placed atop the unbarred axes. 

Similarly, the matrix equation 

x 

= 

y 

1 α 

0 1 

 

x 

y 

(2.3.24) 

can be used to represent a shearing of the unit square parallel to the x axis as illustrated in the figure 

2.3-7(b). 

219

220 

Figure 2.3-7. Shearing parallel to the x axis 

Figure 2.3-8. Shearing parallel to x and y axes 

Again, if α is very small, we may use the approximation tan α ≈ α and interpret α as an angular rotation 

of the element P1P4 to the position P ′ 1P ′ 4 . Now let us multiply the matrices given in equations (2.3.23) and 

(2.3.24). Note that the order of multiplication is important as can be seen by an examination of the products 

 

x 1 0 1 α x 1 α x 

= 

= 

y β 1 0 1 y β 1+αβ y 

 

x 1 α 1 0 x 1+αβ α x 

= 

= 

. 

y 0 1 β 1 y β 1 y 

(2.3.25) 

In equation (2.3.25) we will assume that the product αβ is very, very small and can be neglected. Then the 

order of matrix multiplication will be immaterial and the transformation equation (2.3.25) will reduce to 

 

x 1 α x 

= 

. (2.3.26) 

y β 1 y 

Applying this transformation to our unit square we obtain the simultaneous shearing parallel to both the x 

and y axes as illustrated in the figure 2.3-8. 

This transformation can then be interpreted as the superposition of the two shearing elements depicted 


For comparison, we consider also the transformation equation 

 

x 1 0 x 

= 

(2.3.27) 

y −α 1 y

Figure 2.3-9. Superposition of shearing elements 

Figure 2.3-10. Rotation of element P1P2 

where α is very small. Applying this transformation to the unit square previously considered we obtain the 

results illustrated in the figure 2.3-10. 

Note the difference in the direction of shearing associated with the transformation equations (2.3.27) 

and (2.3.23) illustrated in the figures 2.3-6 and 2.3-10. If the matrices appearing in the equations (2.3.24) 

and (2.3.27) are multiplied and we neglect product terms because α is assumed to be very small, we obtain 

the matrix equation 

 

x 1 α x 1 0 x 0 α x 

= 

= 

+ 

. (2.3.28) 

y −α 1 y 0 1 y −α 0 y 

 

identity 

rotation 

This can be interpreted as a superposition of the transformation equations (2.3.24) and (2.3.27) which 

represents a rotation of the unit square as illustrated in the figure 2.3-11. 

The matrix on the right-hand side of equation (2.3.28) is referred to as a rotation matrix. The ideas 

illustrated by the above simple transformations will appear again when we consider the transformation of an 

arbitrary small element in a continuum when it under goes a strain. In particular, we will be interested in 

extracting the rigid body rotation from a deformed element and treating this rotation separately from the 

strain displacement. 

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222 

Transformation of an Arbitrary Element 

Figure 2.3-11. Rotation of unit square 

In two dimensions, we consider a rectangular element ABCD as illustrated in the figure 2.3-12. 

Let the points ABCD have the coordinates 

and denote by 

A(x, y), B(x +∆x, y), C(x, y +∆y), D(x +∆x, y +∆y) (2.3.29) 

u = u(x, y), v = v(x, y) 

the displacement field associated with each of the points in the material continuum when it undergoes a 

deformation. Assume that the deformation of the element ABCD in figure 2.3-12 can be represented by the 

matrix equation 

x b11 b12 x 

= 

(2.3.30) 

y b21 b22 y 

where the coefficients bij,i,j =1, 2, 3 are to be determined. Let us define u = u(x, y) as the horizontal 

displacement of the point (x, y) andv = v(x, y) as the vertical displacement of the same point. We can now 

express the displacement of each of the points A, B, C and D in terms of the displacement field u = u(x, y) 

and v = v(x, y). Consider first the displacement of the point A to A ′ . Here the coordinates (x, y) deformto 

the new coordinates 

x = x + u, y = y + v. 

That is, the coefficients bij must be chosen such that the equation 

 

x + u 

= 

y + v 

b11 b12 

b21 b22 

 

x 

y 

(2.3.31) 

is satisfied. We next examine the displacement of the point B to B ′ . This displacement is described by the 

coordinates (x +∆x, y) transforming to (x, y), where 

x = x +∆x + u(x +∆x, y), y = y + v(x +∆x, y). (2.3.32)

Figure 2.3-12. Displacement of element ABCD to A ′ B ′ C ′ D ′ 

Expanding u and v in (2.3.32) in Taylor series about the point (x, y) we find 

x = x +∆x + u + ∂u 

∆x + h.o.t. 

∂x 

y = y + v + ∂v 

∆x + h.o.t., 

∂x 

∂u 

x + u +∆x + ∂x∆x y + v + ∂v 

∂x∆x 

= 

(2.3.33) 

where h.o.t. denotes higher order terms which have been neglected. The equations (2.3.33) require that the 

coefficients bij satisfy the matrix equation 

 

b11 b12 x +∆x 

. (2.3.34) 

y 

b21 b22 

223

224 

The displacement of the point C to C ′ is described by the coordinates (x, y +∆y) transforming to (x, y) 

where 

x = x + u(x, y +∆y), y = y +∆y + v(x, y +∆y). (2.3.35) 

Again we expand the displacement field components u and v in a Taylor series about the point (x, y) and 

find 

x = x + u + ∂u 

∆y + h.o.t. 

∂y 

y = y +∆y + v + ∂v 

∆y + h.o.t. 

∂y 

(2.3.36) 

This equation implies that the coefficients bij must be chosen such that 

∂u x + u + ∂y ∆y 

y + v +∆y + ∂v 

∂y∆y 

= 


b21 b22 

 

x 

. (2.3.37) 

y +∆y 

Finally, it can be verified that the point D with coordinates (x +∆x, y +∆y) moves to the point D ′ with 

coordinates 

x = x +∆x + u(x +∆x, y +∆y), y = y +∆y + v(x +∆x, y +∆y). (2.3.38) 

Expanding u and v in a Taylor series about the point (x, y) we find the coefficients bij must be chosen to 

satisfy the matrix equation 

∂u ∂u 

x +∆x + u + ∂x∆x + 

y +∆y + v + ∂v 

∂x 

∂y∆y 

= 

∆x + ∂v 

∂y ∆y 


b21 b22 

 

x +∆x 

. (2.3.39) 

y +∆y 

The equations (2.3.31),(2.3.34),(2.3.37) and (2.3.39) give rise to the simultaneous equations 

b11x + b12y = x + u 

b21x + b22y = y + v 

b11(x +∆x)+b12y = x + u +∆x + ∂u 

∂x ∆x 

b21(x +∆x)+b22y = y + v + ∂v 

∂x ∆x 

b11x + b12(y +∆y) =x + u + ∂u 

∂y ∆y 

b21x + b22(y +∆y) =y + v +∆y + ∂v 

∂y ∆y 

b11(x +∆x)+b12(y +∆y) =x +∆x + u + ∂u ∂u 

∆x + 

∂x ∂y ∆y 

b21(x +∆x)+b22(y +∆y) =y +∆y + v + ∂v ∂v 

∆x + 

∂x ∂y ∆y. 

It is readily verified that the system of equations (2.3.40) has the solution 

b11 =1+ ∂u 

∂x 

b21 = ∂v 

∂x 

b12 = ∂u 

∂y 

b22 =1+ ∂v 

∂y . 

(2.3.40) 

(2.3.41)

Figure 2.3-13. Change in 45 ◦ line 

Hence the transformation equation (2.3.30) can be written as 

∂u ∂u 

x 1+ ∂x ∂y 

= 

y 

∂v 

∂x 

1+ ∂v 

∂y 

 

x 

. (2.3.42) 

y 

A physical interpretation associated with this transformation is obtained by writing it in the form: 

 

x 1 0 x e11 e12 x ω11 ω12 x 

= 

+ 

+ 

, (2.3.43) 

y 0 1 y e21 e22 y ω21 ω22 y 

 

identity 

strain matrix 

rotation matrix 

where 

e11 = ∂u 

∂x 

e12 = 1 

 

∂v ∂u 

+ 

2 ∂x ∂y 

e21 = 1 

 

∂u ∂v 

+ 

2 ∂y ∂x 

e22 = ∂v 

∂y 

are the elements of a symmetric matrix called the strain matrix and 

ω11 =0 

ω21 = 1 

ω12 = 

∂v ∂u 

− 

2 ∂x ∂y 

1 

 

∂u ∂v 

− 

2 ∂y ∂x 

ω22 =0 

are the elements of a skew symmetric matrix called the rotation matrix. 

The strain per unit length in the x-direction associated with the point A in the figure 2.3-12 is 

e11 = 

∂u ∆x + ∂x∆x − ∆x 

= 

∆x 

∂u 

∂x 

and the strain per unit length of the point A in the y direction is 

e22 = 

∂v ∆y + ∂y∆y − ∆y 

∆y 

(2.3.44) 

(2.3.45) 

(2.3.46) 

= ∂v 

. (2.3.47) 

∂y 

These are the terms along the main diagonal in the strain matrix. The geometry of the figure 2.3-12 implies 

that 

∂v 

∂x tan β = 

∆x 

∆x + ∂u 

∂u 

∂y 

, and tan α = 

∂x∆x ∆y 

∆y + ∂v . (2.3.48) 

∂y∆y For small derivatives associated with the displacements u and v it is assumed that the angles α and β are 

small and the equations (2.3.48) therefore reduce to the approximate equations 

tan β ≈ β = ∂v 

tan α ≈ α = 

∂x 

∂u 

. (2.3.49) 

∂y 

For a physical interpretation of these terms we consider the deformation of a small rectangular element which 

undergoes a shearing as illustrated in the figure 2.3-13. 

225

226 

The quantity 

Figure 2.3-14. Displacement field due to state of strain 

α + β = 

 

∂u ∂v 

+ =2e12 =2e21 

∂y ∂x 

(2.3.50) 

is the change from a ninety degree angle due to the deformation and hence we can write 1 

2 (α+β) =e12 = e21 

as representing a change from a 45 ◦ angle due to the deformation. The quantities e21,e12 are called the 

shear strains and the quantity 

is called the shear angle. 

γ12 =2e12 

(2.3.51) 

In the equation (2.3.45), the quantities ω21 = −ω12 are the elements of the rigid body rotation matrix 

and are interpreted as angles associated with a rotation. The situation is analogous to the transformations 

and figures for the deformation of the unit square which was considered earlier. 

Strain in Three Dimensions 

The development of strain in three dimensions is approached from two different viewpoints. The first 

approach considers the derivation using Cartesian tensors and the second approach considers the derivation 

of strain using generalized tensors. 

Cartesian Tensor Derivation of Strain. 

Consider a material which is subjected to external forces such that all the points in the material undergo 

a deformation. Let (y1,y2,y3) denote a set of orthogonal Cartesian coordinates, fixed in space, which is 

used to describe the deformations within the material. Further, let ui = ui(y1,y2,y3),i =1, 2, 3denotea 

displacement field which describes the displacement of each point within the material. With reference to the 

figure 2.3-14 let P and Q denote two neighboring points within the material while it is in an unstrained state. 

These points move to the points P ′ and Q ′ when the material is in a state of strain. We let yi,i =1, 2, 3 

represent the position vector to the general point P in the material, which is in an unstrained state, and 

denote by yi + ui,i=1, 2, 3 the position vector of the point P ′ when the material is in a state of strain.

For Q a neighboring point of P which moves to Q ′ when the material is in a state of strain, we have 

from the figure 2.3-14 the following vectors: 

position of P : yi, i =1, 2, 3 

position of P ′ : yi + ui(y1,y2,y3), i =1, 2, 3 

position of Q : yi +∆yi, i =1, 2, 3 

position of Q ′ : yi +∆yi + ui(y1 +∆y1,y2 +∆y2,y3 +∆y3), i =1, 2, 3 

(2.3.52) 

Employing our earlier one dimensional definition of strain, we define the strain associated with the point P 

L − L0 

in the direction PQ as e = , where L0 = PQ and L = P ′ Q ′ . To calculate the strain we need to first 

L0 

calculate the distances L0 and L. The quantities L2 0 and L2 are easily calculated by considering dot products 

of vectors. For example, we have L 2 0 =∆yi∆yi, and the distance L = P ′ Q ′ is the magnitude of the vector 

yi +∆yi + ui(y1 +∆y1,y2 +∆y2,y3 +∆y3) − (yi + ui(y1,y2,y3)), i =1, 2, 3. 

Expanding the quantity ui(y1 +∆y1,y2 +∆y2,y3 +∆y3) in a Taylor series about the point P and neglecting 

higher order terms of the expansion we find that 

L 2 =(∆yi + ∂ui 

∆ym)(∆yi + 

∂ym 

∂ui 

∆yn). 

∂yn 

Expanding the terms in this expression produces the equation 

L 2 =∆yi∆yi + ∂ui 

∆yi∆yn + 

∂yn 

∂ui 

∆ym∆yi + 

∂ym 

∂ui 

∂ym 

∂ui 

∆ym∆yn. 

∂yn 

Note that L and L0 are very small and so we express the difference L 2 − L 2 0 in terms of the strain e. We can 

write 

L 2 − L 2 0 =(L + L0)(L − L0) =(L − L0 +2L0)(L − L0) =(e +2)eL 2 0 . 

Now for e very small, and e2 negligible, the above equation produces the approximation 

eL 2 0 ≈ L2 − L2 0 

= 

2 

1 

 

∂um 

+ 

2 ∂yn 

∂un 

+ 

∂ym 

∂ur 

 

∂ur 

∆ym∆yn. 

∂ym ∂yn 

The quantities 

emn = 1 

 

∂um 

+ 

2 ∂yn 

∂un 

+ 

∂ym 

∂ur 

 

∂ur 

(2.3.53) 

∂ym ∂yn 

is called the Green strain tensor or Lagrangian strain tensor. To show that eij is indeed a tensor, we consider 

the transformation yi = ℓijyj +bi, where ℓjiℓki = δjk = ℓijℓik. Note that from the derivative relation ∂yi 

∂y = ℓij 

j 

and the transformation equations ui = ℓijuj,i =1, 2, 3 we can express the strain in the barred system of 

coordinates. Performing the necessary calculations produces 

eij = 1 

 

∂ui 

+ 

2 ∂yj ∂uj 

+ 

∂yi ∂ur 

 

∂ur 

∂yi ∂yj = 1 

 

∂ 

(ℓikuk) 

2 ∂yn 

∂yn 

+ 

∂yj ∂ 

(ℓjkuk) 

∂ym 

∂ym 

+ 

∂yi ∂ 

(ℓrsus) 

∂yk 

∂yk ∂ 

(ℓrmum) 

∂yi ∂yt 

∂yt 

 

∂yj = 1 

 

 

∂um ∂uk 

∂us ∂up 

ℓimℓnj + ℓjkℓmi + ℓrsℓrpℓkiℓtj 

2 ∂yn ∂ym 

∂yk ∂yt 

= 1 

 

∂um 

+ 

2 ∂yn 

∂un 

+ 

∂ym 

∂us 

 

∂us 

ℓimℓnj 

∂ym ∂yn 

or eij = emnℓimℓnj.Consequently, the strain eij transforms like a second order Cartesian tensor. 

227

228 

Lagrangian and Eulerian Systems 

Let xi denote the initial position of a material particle in a continuum. Assume that at a later time the 

particle has moved to another point whose coordinates are xi . Both sets of coordinates are referred to the 

same coordinate system. When the final position can be expressed as a function of the initial position and 

time we can write x i = x i (x 1 , x 2 , x 3 ,t). Whenever the changes of any physical quantity is represented in terms 

of its initial position and time, the representation is referred to as a Lagrangian or material representation of 

the quantity. This can be thought of as a transformation of the coordinates. When the Jacobian J( x 

x )ofthis 

transformation is different from zero, the above set of equations have a unique inverse x i = x i (x 1 ,x 2 ,x 3 ,t), 

where the position of the particle is now expressed in terms of its instantaneous position and time. Such a 

representation is referred to as an Eulerian or spatial description of the motion. 

Let (x1, x2, x3) denote the initial position of a particle whose motion is described by xi = xi(x1, x2, x3,t), 

then ui = xi − xi denotes the displacement vector which can by represented in a Lagrangian or Eulerian 

form. For example, if 

x1 =2(x1 − x2)(e t − 1) + (x2 − x1)(e −t − 1) + x1 

x2 =(x1 − x2)(e t − 1) + (x2 − x1)(e −t − 1) + x2 

x3 = x3 

then the displacement vector can be represented in the Lagrangian form 

or the Eulerian form 

u1 =2(x1 − x2)(e t − 1) + (x2 − x1)(e −t − 1) 

u2 =(x1 − x2)(e t − 1) + (x2 − x1)(e −t − 1) 

u3 =0 

u1 = x1 − (2x2 − x1)(1 − e −t ) − (x1 − x2)(e −2t − e −t ) − x1e −t 

u2 = x2 − (2x2 − x1)(1 − e −t ) − (x2 − x1)(e −2t − e −t ) − x2e −t 

u3 =0. 

Note that in the Lagrangian system the displacements are expressed in terms of the initial position and 

time, while in the Eulerian system the independent variables are the position coordinates and time. Euler 

equations describe, as a function of time, how such things as density, pressure, and fluid velocity change at 

a fixed point in the medium. In contrast, the Lagrangian viewpoint follows the time history of a moving 

individual fluid particle as it moves through the medium.

General Tensor Derivation of Strain. 

With reference to the figure 2.3-15 consider the deformation of a point P within a continuum. Let 

(y1 ,y2 ,y3 ) denote a Cartesian coordinate system which is fixed in space. We can introduce a coordinate 

transformation yi = yi (x1 ,x2 ,x3 ), i =1, 2, 3 and represent all points within the continuum with respect 

to a set of generalized coordinates (x1 ,x2 ,x3 ). Let P denote a general point in the continuum while it is 

in an unstrained state and assume that this point gets transformed to a point P ′ when the continuum 

experiences external forces. If P moves to P ′ , then all points Q which are near P will move to points Q ′ 

near P ′ . We can imagine that in the unstrained state all the points of the continuum are referenced with 

respect to the set of generalized coordinates (x 1 ,x 2 ,x 3 ). After the strain occurs, we can imagine that it will 

be convenient to represent all points of the continuum with respect to a new barred system of coordinates 

(x 1 , x 2 , x 3 ). We call the original set of coordinates the Lagrangian system of coordinates and the new set 

of barred coordinates the Eulerian coordinates. The Eulerian coordinates are assumed to be described by 

a set of coordinate transformation equations xi = xi (x1 ,x2 ,x3 ), i =1, 2, 3 with inverse transformations 

xi = xi (x1 , x2 , x3 ), i =1, 2, 3, which are assumed to exist. The barred and unbarred coordinates can 

be related to a fixed set of Cartesian coordinates yi ,i = 1, 2, 3, and we may assume that there exists 


y i = y i (x 1 ,x 2 ,x 3 ), i =1, 2, 3 and y i = y i (x 1 , x 2 , x 3 ), i =1, 2, 3 

which relate the barred and unbarred coordinates to the Cartesian axes. In the discussion that follows 

be sure to note whether there is a bar over a symbol, as we will be jumping back and forth between the 

Lagrangian and Eulerian reference frames. 

Figure 2.3-15. Strain in generalized coordinates 

In the Lagrangian system of unbarred coordinates we have the basis vectors Ei = ∂r 

which produce 

∂xi the metrices gij = Ei · Ej. Similarly, in the Eulerian system of barred coordinates we have the basis vectors 

 

Ei = ∂r 

∂xi which produces the metrices Gij = Ei · Ej. These basis vectors are illustrated in the figure 2.3-15. 

229

230 

We assume that an element of arc length squared ds2 in the unstrained state is deformed to the element 

of arc length squared ds2 in the strained state. An element of arc length squared can be expressed in terms 

of the barred or unbarred coordinates. For example, in the Lagrangian system, let dr = PQ so that 

L 2 0 = dr · dr = ds2 = gijdx i dx j , (2.3.54) 

where gij are the metrices in the Lagrangian coordinate system. This same element of arc length squared 

canbeexpressedinthebarredsystemby 

L 2 0 = ds 2 = gijdx i dx j ∂x 

, where gij = gmn 

m 

∂xi Similarly, in the Eulerian system of coordinates the deformed arc length squared is 

∂xn j . (2.3.55) 

∂x 

L 2 = dr · dr = ds 2 = Gijdx i dx j , (2.3.56) 

where Gij are the metrices in the Eulerian system of coordinates. This same element of arc length squared 

can be expressed in the Lagrangian system by the relation 

where 

In the Lagrangian system we have 

L 2 = ds 2 = Gijdx i dx j ∂x 

, where Gij = Gmn 

m 

∂xi ∂xn . (2.3.57) 

∂xj ds 2 − ds 2 =(Gij − gij)dx i dx j =2eijdx i dx j 

eij = 1 

2 (Gij − gij) (2.3.58) 

is called the Green strain tensor or Lagrangian strain tensor. Alternatively, in the Eulerian system of 

coordinates we may write 

ds 2 − ds 2 = i j 

Gij − gij dx dx =2eijdx i dx j 

where 

eij = 1 

Gij − gij 2 

is called the Almansi strain tensor or Eulerian strain tensor. 

(2.3.59)

Note also in the figure 2.3-15 there is the displacement vector u. This vector can be represented in any 

of the following forms: 

u = u i Ei contravariant, Lagrangian basis 

u = ui E i 

covariant, Lagrangian reciprocal basis 

u = u i Ei contravariant, Eulerian basis 

 

u = uiE 

i 

covariant, Eulerian reciprocal basis. 

By vector addition we have r + u = r and consequently dr + du = dr. In the Lagrangian frame of reference 

at the point P we represent u in the contravariant form u = uiEi and write dr in the form dr = dxiEi. By 

use of the equation (1.4.48) we can express du in the form du = ui ,kdxkEi. These substitutions produce the 

representation dr =(dx i + u i ,kdx k ) Ei in the Lagrangian coordinate system. We can then express ds2 in the 

Lagrangian system. We find 

dr · dr = ds 2 =(dx i + u i ,k dxk ) Ei · (dx j + u j ,m dxm ) Ej 

=(dx i dx j + u j ,mdx m dx i + u i ,kdx k dx j + u i ,ku j ,mdx k dx m )gij 

and consequently from the relation (2.3.58) we derive the representation 

eij = 1 

ui,j + uj,i + um,iu 

2 

m 

,j . (2.3.60) 

This is the representation of the Lagrangian strain tensor in any system of coordinates. The strain tensor 

eij is symmetric. We will restrict our study to small deformations and neglect the product terms in equation 

(2.3.60). Under these conditions the equation (2.3.60) reduces to eij = 1 

2 (ui,j + uj,i). 

If instead, we chose to represent the displacement u with respect to the Eulerian basis, then we can 

write 

These relations imply that 

u = u i Ei with du = u i ,k dxk Ei. 

dr = dr − du =(dx i − u i ,k dxk ) Ei. 

This representation of dr in the Eulerian frame of reference can be used to calculate the strain eij from the 

relation ds 2 − ds 2 . It is left as an exercise to show that there results 

eij = 1 

ui,j + uj,i − um,iu 

2 

m 

,j . (2.3.61) 

The equation (2.3.61) is the representation of the Eulerian strain tensor in any system of coordinates. Under 

conditions of small deformations both the equations (2.3.60) and (2.3.61) reduce to the linearized Lagrangian 

and Eulerian strain tensor eij = 1 

2 (ui,j + uj,i). In the case of large deformations the equations (2.3.60) and 

(2.3.61) describe the strains. In the case of linear elasticity, where the deformations are very small, the 

product terms in equations (2.3.60) and (2.3.61) are neglected and the Lagrangian and Eulerian strains 

reduce to their linearized forms 

eij = 1 

2 [ui,j + uj,i] eij = 1 

2 [ui,j + uj,i] . (2.3.62) 

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232 

Figure 2.3-16. Displacement due to strain 

Compressible and Incompressible Material With reference to figure 2.3-16, let xi , i =1, 2, 3denote 

the position vector of an arbitrary point P in a continuum before there is a state of strain. Let Q be 

a neighboring point of P with position vector xi + dxi , i =1, 2, 3. Also in the figure 2.3-16 there is the 

displacement vector u. Here it is assumed that u = u(x 1 ,x2 ,x3 ) denotes the displacement field when the 

continuum is in a state of strain. The figure 2.3-16 illustrates that in a state of strain P moves to P ′ and Q 

moves to Q ′ . Let us find a relationship between the distance PQbefore the strain and the distance P ′ Q ′ when 

the continuum is in a state of strain. For E1, E2, E3 basis functions constructed at P we have previously 

shown that if 

u(x 1 ,x 2 ,x 3 )=u i Ei then du = u i ,jdx j Ei. 

Now for u + du the displacement of the point Q we may use vector addition and write 

PQ+ u + du = u + P ′ Q ′ . (2.3.63) 

Let PQ = dxiEi = aiEi denote an arbitrary small change in the continuum. This arbitrary displacement 

gets deformed to P ′ Q ′ = A i Ei due to the state of strain in the continuum. Employing the equation (2.3.63) 

we write 

which can be written in the form 

dx i + u i ,jdx j = a i + u i ,ja j = A i 

δa i = A i − a i = u i ,j aj where dx i = a i ,i=1, 2, 3 (2.3.64) 

denotes an arbitrary small change. The tensor ui ,j and the associated tensor ui,j = gitut ,j are in general 

not symmetric tensors. However, we know we can express ui,j as the sum of a symmetric (eij) andskew- 

symmetric(ωij) tensor. We therefore write 

where 

ui,j = eij + ωij or u i ,j = ei j + ωi j , 

eij = 1 

2 (ui,j + uj,i) = 1 

2 (gimu m ,j + gjmu m ,i) and ωij = 1 

2 (ui,j − uj,i) = 1 

2 (gimu m ,j − gjmu m ,i) . 

The deformation of a small quantity ai can therefore be represented by a pure strain Ai − ai = ei sas followed 

by a rotation Ai − ai = ωi sas .

Consider now a small element of volume inside a material medium. With reference to the figure 2.3- 

17(a) we let a, b,c denote three small arbitrary independent vectors constructed at a general point P within 

the material before any external forces are applied. We imagine a, b,c as representing the sides of a small 

parallelepiped before any deformation has occurred. When the material is placed in a state of strain the 

point P will move to P ′ and the vectors a, b,c will become deformed to the vectors A, B, C as illustrated in 

the figure 2.3-17(b). The vectors A, B, C represent the sides of the parallelepiped after the deformation. 

Figure 2.3-17. Deformation of a parallelepiped 

Let ∆V denote the volume of the parallelepiped with sides a, b,c at P before the strain and let ∆V ′ 

denote the volume of the deformed parallelepiped after the strain, when it then has sides A, B, C at the 

point P ′ . We define the ratio of the change in volume due to the strain divided by the original volume as 

the dilatation at the point P. The dilatation is thus expressed as 

Θ= ∆V ′ − ∆V 

∆V 

= dilatation. (2.3.65) 

Since ui ,i =1, 2, 3 represents the displacement field due to the strain, we use the result from equation 

(2.3.64) and represent the displaced vectors A, B, C in the form 

A i = a i + u i ,j aj 

B i = b i + u i ,j bj 

C i = c i + u i ,jc j 

(2.3.66) 

where a, b,c are arbitrary small vectors emanating from the point P in the unstrained state. The element of 

volume ∆V, before the strain, is calculated from the triple scalar product relation 

∆V = a · ( b × c) =eijka i b j c k . 

The element of volume ∆V ′ , which occurs due to the strain, is calculated from the triple scalar product 

∆V ′ = A · ( B × C)=eijkA i B j C k . 

233

234 

Substituting the relations from the equations (2.3.66) into the triple scalar product gives 

∆V ′ = eijk(a i + u i ,m am )(b j + u j ,n bn )(c k + u k ,p cp ). 

Expanding the triple scalar product and employing the result from Exercise 1.4, problem 34, we find the 

simplified result gives us the dilatation 

Θ= ∆V ′ − ∆V 

∆V 

= u r ,r 

=div(u). (2.3.67) 

That is, the dilatation is the divergence of the displacement field. If the divergence of the displacement field 

is zero, there is no volume change and the material is said to be incompressible. If the divergence of the 

displacement field is different from zero, the material is said to be compressible. 

Note that the strain eij is expressible in terms of the displacement field by the relation 

eij = 1 

2 (ui,j + uj,i), and consequently g mn emn = u r ,r . (2.3.68) 

Hence, for an orthogonal system of coordinates the dilatation can be expressed in terms of the strain elements 

along the main diagonal. 

Conservation of Mass 

Consider the material in an arbitrary region R of a continuum. Let ϱ = ϱ(x, y, z, t) denote the density 

of the material within the region. Assume that the dimension of the density ϱ is gm/cm3 in the cgs system 

of units. We shall assume that the region R is bounded by a closed surface S with exterior unit normal n 

defined everywhere on the surface. Further, we let v = v(x, y, z, t) denote a velocity field associated with all 

points within the continuum. The velocity field has units of cm/sec in the cgs system of units. Neglecting 

sources and sinks, the law of conservation of mass examines all the material entering and leaving a region R. 

Enclosed within R is the material mass m where m = ϱdτ with dimensions of gm in the cgs system of 

R 

units. Here dτ denotes an element of volume inside the region R. The change of mass with time is obtained 

by differentiating the above relation. Differentiating the mass produces the equation 

∂m 

∂t = 

 

∂ϱ 

dτ (2.3.69) 

∂t 

and has the dimensions of gm/sec. 

Consider also the surface integral 

 

I = 

S 

R 

ϱv · ˆndσ (2.3.70) 

where dσ is an element of surface area on the surface S which encloses R and ˆn is the exterior unit normal 

vector to the surface S. The dimensions of the integral I is determined by examining the dimensions of each 

term in the integrand of I. We find that 

[I] = gm cm 

· 

cm3 sec · cm2 = gm 

sec 

and so the dimension of I is the same as the dimensions for the change of mass within the region R. The 

surface integral I is the flux rate of material crossing the surface of R and represents the change of mass

entering the region if v · ˆn is negative and the change of mass leaving the region if v · ˆn is positive, as ˆn is 

always an exterior unit normal vector. Equating the relations from equations (2.3.69) and (2.3.70) we obtain 

a mathematical statement for mass conservation 

∂m 

∂t = 

 

 

∂ϱ 

dτ = − ϱv · ndσ. (2.3.71) 

R ∂t S 

The equation (2.3.71) implies that the rate at which the mass contained in R increases must equal the rate 

at which the mass flows into R through the surface S. The negative sign changes the direction of the exterior 

normal so that we consider flow of material into the region. Employing the Gauss divergence theorem, the 

surface integral in equation (2.3.71) can be replaced by a volume integral and the law of conservation of 

mass is then expressible in the form 

 

∂ϱ 

∂t +div(ϱv) 

 

dτ =0. (2.3.72) 

R 

Since the region R is an arbitrary volume we conclude that the term inside the brackets must equal zero. 

This gives us the continuity equation 

∂ϱ 

+div(ϱv) =0 (2.3.73) 

∂t 

which represents the mass conservation law in terms of velocity components. This is the Eulerian representation 

of continuity of mass flow. 

Equivalent forms of the continuity equation are: 

∂ϱ 

+ v · grad ϱ + ϱ div v =0 

∂t 

∂ϱ 

∂t 

∂ϱ ∂vi 

+ vi + ϱ =0 

∂xi ∂xi Dϱ ∂vi 

+ ϱ =0 

Dt ∂xi where Dϱ ∂ϱ ∂ϱ 

= + 

Dt ∂t ∂xi dxi ∂ϱ ∂ϱ 

= + 

dt ∂t ∂xi vi is called the material derivative of the density ϱ. Note that the 

material derivative contains the expression ∂ϱ 

∂xi vi which is known as the convective or advection term. If the 

density ϱ = ϱ(x, y, z, t) is a constant we have 

Dϱ 

Dt 

∂ϱ ∂ϱ dx 

= + 

∂t ∂x dt 

∂ϱ dy 

+ 

∂y dt 

∂ϱ dz ∂ϱ ∂ϱ 

+ = + 

∂z dt ∂t ∂xi dxi dt 

=0 (2.3.74) 

and hence the continuity equation reduces to div (v) =0. Thus, if div (v) is zero, then the material is 

incompressible. 

EXAMPLE 2.3-2. (Continuity Equation) Find the Lagrangian representation of mass conservation. 

Solution: Let (X, Y, Z) denote the initial position of a fluid particle and denote the density of the fluid by 

ϱ(X, Y, Z, t) sothatϱ(X, Y, Z, 0) denotes the density at the time t =0. Consider a simple closed region in 

our continuum and denote this region by R(0) at time t =0andbyR(t) atsomelatertimet. That is, all 

the points in R(0) move in a one-to-one fashion to points in R(t). Initially the mass of material in R(0) is 

m(0) = ϱ(X, Y, Z, 0) dτ(0) where dτ(0) = dXdY dZ is an element of volume in R(0). We have after a 

R(0) 

235

236 

 

time t has elapsed the mass of material in the region R(t) givenbym(t) = ϱ(X, Y, Z, t) dτ(t) where 

R(t) 

x,y,z 

dτ(t) =dxdydz is a deformed element of volume related to the dτ(0) by dτ(t) =J X,Y,Z dτ(0) where J is 

the Jacobian of the Eulerian (x, y, z) variables with respect to the Lagrangian (X, Y, Z) representation.For 

mass conservation we require that m(t) =m(0) for all t. This implies that 

ϱ(X, Y, Z, t)J = ϱ(X, Y, Z, 0) (2.3.75) 

for all time, since the initial region R(0) is arbitrary. The right hand side of equation (2.3.75) is independent 

of time and so 

d 

(ϱ(X, Y, Z, t)J) =0. (2.3.76) 

dt 

This is the Lagrangian form of the continuity equation which expresses mass conservation. Using the result 

that dJ 

dt = Jdiv V,(see problem 28, Exercise 2.3), the equation (2.3.76) can be expanded and written in the 

form 

Dϱ 

Dt + ϱ div V =0 (2.3.77) 

where Dϱ 

Dt is from equation (2.3.74). The form of the continuity equation (2.3.77) is one of the Eulerian forms 

previously developed. 

In the Eulerian coordinates the continuity equation is written ∂ϱ 

∂t +div(ϱv) = 0, while in the Lagrangian 

system the continuity equation is written d(ϱJ) 

dt =0. Note that the velocity carries the Lagrangian axes and 

the density change grad ϱ. This is reflective of the advection term v · grad ϱ. Thus, in order for mass to 

be conserved it need not remain stationary. The mass can flow and the density can change. The material 

derivative is a transport rule depicting the relation between the Eulerian and Lagrangian viewpoints. 

In general, from a Lagrangian viewpoint, any quantity Q(x, y, z, t) which is a function of both position 

and time is seen as being transported by the fluid velocity (v1,v2,v3) toQ(x + v1dt, y + v2dt, z + v3dt, t + dt). 

Then the time derivative of Q contains both ∂Q 

∂t and the advection term v ·∇Q. In terms of mass flow, the 

Eulerian viewpoint sees flow into and out of a fixed volume in space, as depicted by the equation (2.3.71), 

In contrast, the Lagrangian viewpoint sees the same volume moving with the fluid and consequently 

 

D 

ρdτ =0, 

Dt R(t) 

where R(t) represents the volume moving with the fluid. Both viewpoints produce the same continuity 

equation reflecting the conservation of mass. 

Summary of Basic Equations 

Let us summarize the basic equations which are valid for all types of a continuum. We have derived: 

• Conservation of mass (continuity equation) 

∂ϱ 

∂t +(ϱvi ),i =0

• Conservation of linear momentum sometimes called the Cauchy equation of motion. 

• Conservation of angular momentum 

• Strain tensor for linear elasticity 

σ ij ,i + ϱb j = ϱf j , j =1, 2, 3. 

σij = σji 

eij = 1 

2 (ui,j + uj,i). 

If we assume that the continuum is in equilibrium, and there is no motion, then the velocity and 

acceleration terms above will be zero. The continuity equation then implies that the density is a constant. 

The conservation of angular momentum equation requires that the stress tensor be symmetric and we need 

find only six stresses. The remaining equations reduce to a set of nine equations in the fifteen unknowns: 

3 displacements u1,u2,u3 

6strains e11,e12,e13,e22,e23,e33 

6 stresses σ11,σ12,σ13,σ22,σ23,σ33 

Consequently, we still need additional information if we desire to determine these unknowns. 

Note that the above equations do not involve any equations describing the material properties of the 

continuum. We would expect solid materials to act differently from liquid material when subjected to external 

forces. An equation or equations which describe the material properties are called constitutive equations. 

In the following sections we will investigate constitutive equations for solids and liquids. We will restrict 

our study to linear elastic materials over a range where there is a linear relationship between the stress and 

strain. We will not consider plastic or viscoelastic materials. Viscoelastic materials have the property that 

the stress is not only a function of strain but also a function of the rates of change of the stresses and strains 

and consequently properties of these materials are time dependent. 

237

238 

EXERCISE 2.3 

◮ 1. Assume an orthogonal coordinate system with metric tensor gij =0fori= j and g (i)(i) = h2 i (no 

summation on i). Use the definition of strain 

and show that in terms of the physical components 

there results the equations: 

 

t ∂u t 

eii = git + u 

∂xi mi 

m 

 

∂u 

2eij = git 

t ∂u 

+ gjt 

∂xj t 

, i = j 

∂xi e(ii) = ∂ 

∂x i 

2e(ij) = hi 

hj 

u(i) 

∂ 

∂x j 

hi 

ers = 1 

2 (ur,s + us,r) = 1 

grtu 

2 

t ,s + gstu t 

,r 

e(ij) = eij 

hihj 

no summation on i or j 

u(i) =hiu i no summation on i 

u(i) 

 

+ 1 

hi 

no summation on i 

2h2 3 u(m) ∂ 

i 

hm ∂x 

m=1 

m 

2 

hi no summation on i 

 

+ hj ∂ 

hi ∂xi 

u(j) 

, no summation on i or j, i = j. 

hj 

◮ 2. Use the results from problem 1 to write out all components of the strain tensor in Cartesian coordinates. 

Use the notation u(1) = u,u(2) = v,u(3) = w and 

to verify the relations: 

e(11) = exx, e(22) = eyy, e(33) = ezz, e(12) = exy, e(13) = exz, e(23) = eyz 

exx = ∂u 

∂x 

eyy = ∂v 

∂y 

ezz = ∂w 

∂z 

exy = 1 

 

∂v ∂u 

+ 

2 ∂x ∂y 

 

∂u ∂w 

+ 

∂z ∂x 

ezy = 1 

 

∂w ∂v 

+ 

2 ∂y ∂z 

exz = 1 

2 

◮ 3. Use the results from problem 1 to write out all components of the strain tensor in cylindrical coordinates. 

Use the notation u(1) = ur , u(2) = uθ, u(3) = uz and 

to verify the relations: 

e(11) = err, e(22) = eθθ, e(33) = ezz, e(12) = erθ, e(13) = erz, e(23) = eθz 

err = ∂ur 

∂r 

eθθ = 1 ∂uθ 

r ∂θ 

ezz = ∂uz 

∂z 

+ ur 

r 

erθ = 1 

 

1 ∂ur ∂uθ uθ 

+ − 

2 r ∂θ ∂r r 

erz = 1 

 

∂uz ∂ur 

+ 

2 ∂r ∂z 

eθz = 1 

 

∂uθ 1 ∂uz 

+ 

2 ∂z r ∂θ

◮ 4. Use the results from problem 1 to write out all components of the strain tensor in spherical coordinates. 

Use the notation u(1) = uρ,u(2) = uθ,u(3) = uφ and 

to verify the relations 

e(11) = eρρ, e(22) = eθθ, e(33) = eφφ, e(12) = eρθ, e(13) = eρφ, e(23) = eθφ 

eρρ = ∂uρ 

∂ρ 

eθθ = 1 ∂uθ uρ 

+ 

ρ ∂θ ρ 

eφφ = 1 ∂uφ uρ uθ 

+ + cot θ 

ρ sin θ ∂φ ρ ρ 

eρθ = 1 

 

1 ∂uρ uθ ∂uθ 

− + 

2 ρ ∂θ ρ ∂ρ 

eρφ = 1 

 

1 ∂uρ uφ ∂uφ 

− + 

2 ρ sin θ ∂φ ρ ∂ρ 

eθφ = 1 

 

1 ∂uφ uφ 1 ∂uθ 

− cot θ + 

2 ρ ∂θ ρ ρ sin θ ∂φ 

◮ 5. Expand equation (2.3.67) and find the dilatation in terms of the physical components of an orthogonal 

system and verify that 

Θ= 

1 

 

∂(h2h3u(1)) 

∂x1 + ∂(h1h3u(2)) 

∂x2 + ∂(h1h2u(3)) 

∂x3 

h1h2h3 

◮ 6. Verify that the dilatation in Cartesian coordinates is 

Θ=exx + eyy + ezz = ∂u ∂v ∂w 

+ + 

∂x ∂y ∂z . 

◮ 7. Verify that the dilatation in cylindrical coordinates is 

Θ=err + eθθ + ezz = ∂ur 

∂r 

◮ 8. Verify that the dilatation in spherical coordinates is 

Θ=eρρ + eθθ + eφφ = ∂uρ 

∂ρ 

1 ∂uθ 1 

+ + 

r ∂θ r ur + ∂uz 

∂z . 

1 ∂uθ 2 

+ + 

ρ ∂θ ρ uρ + 1 ∂uφ 

ρ sin θ ∂φ + uθ cot θ 

ρ 

◮ 9. Show that in an orthogonal set of coordinates the rotation tensor ωij canbewrittenintermsofphysical 

components in the form 

Hint: See problem 1. 

ω(ij) = 1 

 

∂(hiu(i)) 

2hihj ∂xj − ∂(hju(j)) 

∂xi 

, no summations 

◮ 10. Use the result from problem 9 to verify that in Cartesian coordinates 

ωyx = 1 

 

∂v ∂u 

− 

2 ∂x ∂y 

ωxz = 1 

 

∂u ∂w 

− 

2 ∂z ∂x 

ωzy = 1 

 

∂w ∂v 

− 

2 ∂y ∂z 

. 

239

240 

◮ 11. Use the results from problem 9 to verify that in cylindrical coordinates 

ωθr = 1 

 

∂(ruθ) ∂ur 

− 

2r ∂r ∂θ 

ωrz = 1 

 

∂ur ∂uz 

− 

2 ∂z ∂r 

ωzθ = 1 

 

1 ∂uz ∂uθ 

− 

2 r ∂θ ∂z 

◮ 12. Use the results from problem 9 to verify that in spherical coordinates 

ωθρ = 1 

 

∂(ρuθ) ∂uρ 

− 

2ρ ∂ρ ∂θ 

ωρφ = 1 

 

1 ∂uρ ∂(ρuφ) 

− 

2ρ sin θ ∂φ ∂ρ 

 

1 ∂(uφ sin θ) 

ωφθ = 

− 

2ρ sin θ ∂θ 

∂uθ 

 

∂φ 

◮ 13. The conditions for static equilibrium in a linear elastic material are determined from the conservation 

law 

σ j 

i,j + ϱbi =0, i,j =1, 2, 3, 

where σ i j are the stress tensor components, bi are the external body forces per unit mass and ϱ is the density 

of the material. Assume an orthogonal coordinate system and verify the following results. 

(a) Show that 

∂ 

) − [ij, m]σmj 

(b) Use the substitutions 

σ j 1 

i,j = √ 

g 

σ(ij) =σ j hj 

i 

hi 

b(i) = bi 

hi 

∂xj (√gσ j 

i 

no summation on i or j 

no summation on i 

σ(ij) =σ ij hihj no summation on i or j 

and express the equilibrium equations in terms of physical components and verify the relations 

3 

j=1 

1 ∂ 

√ 

g ∂xj where there is no summation on i. 

√ ghiσ(ij) 

hj 

 

− 1 

2 

3 

j=1 

σ(jj) 

h2 ∂(h 

j 

2 j ) 

∂xi + hiϱb(i) =0, 

◮ 14. Use the results from problem 13 and verify that the equilibrium equations in Cartesian coordinates 

can be expressed 

∂σxx 

∂x 

∂σyx 

∂x 

∂σzx 

∂x 

+ ∂σxy 

∂y 

+ ∂σyy 

∂y 

+ ∂σzy 

∂y 

+ ∂σxz 

∂z + ϱbx =0 

+ ∂σyz 

∂z + ϱby =0 

+ ∂σzz 

∂z + ϱbz =0

◮ 15. Use the results from problem 13 and verify that the equilibrium equations in cylindrical coordinates 


∂σrr 1 ∂σrθ ∂σrz 1 

+ + + 

∂r r ∂θ ∂z r (σrr − σθθ)+ϱbr =0 

∂σθr 1 ∂σθθ ∂σθz 2 

+ + + 

∂r r ∂θ ∂z r σθr + ϱbθ =0 

∂σzr 1 ∂σzθ ∂σzz 1 

+ + + 

∂r r ∂θ ∂z r σzr + ϱbz =0 

◮ 16. Use the results from problem 13 and verify that the equilibrium equations in spherical coordinates 


∂σρρ 

∂ρ 

1 ∂σρθ 1 ∂σρφ 1 

+ + + 

ρ ∂θ ρ sin θ ∂φ ρ (2σρρ − σθθ − σφφ + σρθ cot θ)+ϱbρ =0 

∂σθρ 

∂ρ 

1 ∂σθθ 1 ∂σθφ 1 

+ + + 

ρ ∂θ ρ sin θ ∂φ ρ (3σρθ +[σθθ − σφφ]cotθ)+ϱbθ =0 

∂σφρ 

∂ρ 

1 ∂σφθ 1 ∂σφφ 1 

+ + + 

ρ ∂θ ρ sin θ ∂φ ρ (3σρφ +2σθφ cot θ)+ϱbφ =0 

◮ 17. Derive the result for the Lagrangian strain defined by the equation (2.3.60). 

◮ 18. Derive the result for the Eulerian strain defined by equation (2.3.61). 

◮ 19. The equation δa i = u i ,j aj , describes the deformation in an elastic solid subjected to forces. The 

quantity δai denotes the difference vector Ai − ai between the undeformed and deformed states. 

(a) Let |a| denote the magnitude of the vector ai and show that the strain e in the direction ai can be 

represented 

e = δ|a| 

|a| 

= eij 

 

i j a a 

|a| 

|a| 

= eijλ i λ j , 

where λi is a unit vector in the direction ai . 

(b) Show that for λ1 =1,λ2 =0,λ3 = 0 there results e = e11, with similar results applying to vectors λi in 

the y and z directions. 

Hint: Consider the magnitude squared |a| 2 = gija i a j . 

◮ 20. At the point (1, 2, 3) of an elastic solid construct the small vector a = ɛ( 2 

3 ê1 + 2 

3 ê2 + 1 

3 ê3), where 

ɛ>0 is a small positive quantity. The solid is subjected to forces such that the following displacement field 

results. 

u =(xy ê1 + yz ê2 + xz ê3) × 10 −2 

Calculate the deformed vector A after the displacement field has been imposed. 

◮ 21. For the displacement field 

u =(x 2 + yz) ê1 +(xy + z 2 ) ê2 + xyz ê3 

(a) Calculate the strain matrix at the point (1, 2, 3). 

(b) Calculate the rotation matrix at the point (1, 2, 3). 

241

242 

◮ 22. Show that for an orthogonal coordinate system the ith component of the convective operator can be 

written 

[( V ·∇) 3 

3 

 

V (m) ∂A(i) A(m) 

A]i = 

+ 

V (i) 

∂xm ∂hi 

− V (m)∂hm 

∂xm ∂xi 

m=1 

hm 

m=1 

m=i 

◮ 23. Consider a parallelepiped with dimensions ℓ, w, h which has a uniform pressure P applied to each 

face. Show that the volume strain can be expressed as 

∆V 

V 

= ∆ℓ 

ℓ 

+ ∆w 

w 

+ ∆h 

h 

hmhi 

−3P (1 − 2ν) 

= . 

E 

The quantity k = E/3(1 − 2ν) is called the bulk modulus of elasticity. 

◮ 24. Show in Cartesian coordinates the continuity equation is 

where (u, v, w) arethevelocitycomponents. 

∂ϱ ∂(ϱu) ∂(ϱv) ∂(ϱw) 

+ + + 

∂t ∂x ∂y ∂z =0, 

◮ 25. Show in cylindrical coordinates the continuity equation is 

∂ϱ 

∂t 

1 ∂(rϱVr) 

+ 

r ∂r 

where Vr,Vθ,Vz are the velocity components. 

1 

+ 

r 

∂(ϱVθ) 

∂θ 

◮ 26. Show in spherical coordinates the continuity equation is 

where Vρ,Vθ,Vφ are the velocity components. 

+ ∂(ϱVz) 

∂z =0 

∂ϱ 1 

+ 

∂t ρ2 ∂(ρ2ϱVρ) + 

∂ρ 

1 ∂(ϱVθ sin θ) 

+ 

ρ sin θ ∂θ 

1 ∂(ϱVφ) 

ρ sin θ ∂φ =0 

◮ 27. (a) Apply a stress σyy to both ends of a square element in a x, y continuum. Illustrate and label 

all changes that occur due to this stress. (b) Apply a stress σxx to both ends of a square element in a 

x, y continuum. Illustrate and label all changes that occur due to this stress. (c) Use superposition of your 

results in parts (a) and (b) and explain each term in the relations 

exx = σxx 

E 

− ν σyy 

E 

◮ 28. Show that the time derivative of the Jacobian J = J 

div V = ∂V1 

∂x 

+ ∂V2 

∂y 

+ ∂V3 

∂z 

and eyy = σyy 

E 

− ν σxx 

E . 

 

x, y, z 

satisfies 

X, Y, Z 

dJ 

dt = J div V where 

and V1 = dx 

dt , V2 = dy 

dt , V3 = dz 

dt . 

Hint: Let (x, y, z) =(x1,x2,x3) and(X, Y, Z) =(X1,X2,X3), then note that 

eijk 

∂V1 ∂x2 ∂x3 ∂V1 ∂xm ∂x2 ∂x3 ∂x1 ∂x2 ∂x3 ∂V1 

= eijk 

= eijk 

, etc. 

∂Xi ∂Xj ∂Xk ∂xm ∂Xi ∂Xj ∂Xk ∂Xi ∂Xj ∂Xk ∂x1

§2.4 CONTINUUM MECHANICS (SOLIDS) 

In this introduction to continuum mechanics we consider the basic equations describing the physical 

effects created by external forces acting upon solids and fluids. In addition to the basic equations that 

are applicable to all continua, there are equations which are constructed to take into account material 

characteristics. These equations are called constitutive equations. For example, in the study of solids the 

constitutive equations for a linear elastic material is a set of relations between stress and strain. In the study 

of fluids, the constitutive equations consists of a set of relations between stress and rate of strain. Constitutive 

equations are usually constructed from some basic axioms. The resulting equations have unknown material 

parameters which can be determined from experimental investigations. 

One of the basic axioms, used in the study of elastic solids, is that of material invariance. This axiom 

requires that certain symmetry conditions of solids are to remain invariant under a set of orthogonal 

transformations and translations. This axiom is employed in the next section to simplify the constitutive 

equations for elasticity. We begin our study of continuum mechanics by investigating the development of 

constitutive equations for linear elastic solids. 

Generalized Hooke’s Law 

If the continuum material is a linear elastic material, we introduce the generalized Hooke’s law in 

Cartesian coordinates 

σij = cijklekl, i,j,k,l=1, 2, 3. (2.4.1) 

The Hooke’s law is a statement that the stress is proportional to the gradient of the deformation occurring 

in the material. These equations assume a linear relationship exists between the components of the stress 

tensor and strain tensor and we say stress is a linear function of strain. Such relations are referred to as a 

set of constitutive equations. Constitutive equations serve to describe the material properties of the medium 

when it is subjected to external forces. 

Constitutive Equations 

The equations (2.4.1) are constitutive equations which are applicable for materials exhibiting small 

deformations when subjected to external forces. The 81 constants cijkl are called the elastic stiffness of the 

material. The above relations can also be expressed in the form 

eij = sijklσkl, i,j,k,l=1, 2, 3 (2.4.2) 

where sijkl are constants called the elastic compliance of the material. Since the stress σij and strain eij 

have been shown to be tensors we can conclude that both the elastic stiffness cijkl and elastic compliance 

sijkl are fourth order tensors. Due to the symmetry of the stress and strain tensors we find that the elastic 

stiffness and elastic compliance tensor must satisfy the relations 

cijkl = cjikl = cijlk = cjilk 

sijkl = sjikl = sijlk = sjilk 

(2.4.3) 

and consequently only 36 of the 81 constants are actually independent. If all 36 of the material (crystal) 

constants are independent the material is called triclinic and there are no material symmetries. 

243

244 

Restrictions on Elastic Constants due to Symmetry 

The equations (2.4.1) and (2.4.2) can be replaced by an equivalent set of equations which are easier to 

analyze. This is accomplished by defining the quantities 

where ⎛ 

and ⎛ 

e1, e2, e3, e4, e5, e6 

σ1, σ2, σ3, σ4, σ5, σ6 

⎝ e1 e4 e5 

e4 e2 e6 

e5 e6 e3 

⎝ σ1 σ4 σ5 

σ4 σ2 σ6 

σ5 σ6 σ3 

⎞ ⎛ 

⎠ = 

⎞ ⎛ 

⎠ = 

⎝ e11 e12 e13 

e21 e22 e23 

e31 e32 e33 

⎝ σ11 σ12 σ13 

σ21 σ22 σ23 

σ31 σ32 σ33 

Then the generalized Hooke’s law from the equations (2.4.1) and (2.4.2) can be represented in either of 

the forms 

σi = cijej or ei = sijσj where i, j =1,...,6 (2.4.4) 

where cij are constants related to the elastic stiffness and sij are constants related to the elastic compliance. 

These constants satisfy the relation 

Here 

and similarly 

relations 

where 

⎞ 

⎠ 

⎞ 

⎠ . 

smicij = δmj where i, m, j =1,...,6 (2.4.5) 

 

ei, i = j =1, 2, 3 

eij = 

e1+i+j, i = j, and i =1, or, 2 

 

σi, i = j =1, 2, 3 

σij = 

σ1+i+j, i = j, and i =1, or, 2. 

These relations show that the constants cij are related to the elastic stiffness coefficients cpqrs by the 

cm1 = cij11 

cm2 = cij22 

cm3 = cij33 

cm4 =2cij12 

cm5 =2cij13 

cm6 =2cij23 

 

i, if i = j =1, 2, or 3 

m = 

1+i + j, if i = j and i =1or2. 

A similar type relation holds for the constants sij and spqrs. The above relations can be verified by expanding 

the equations (2.4.1) and (2.4.2) and comparing like terms with the expanded form of the equation (2.4.4).

The generalized Hooke’s law can now be expressed in a form where the 36 independent constants can 

be examined in more detail under special material symmetries. We will examine the form 

⎛ ⎞ 

e1 

⎛ 

s11 s12 s13 s14 s15 

⎞ ⎛ ⎞ 

s16 σ1 

⎜ e2 ⎟ ⎜ s21 

⎜ ⎟ ⎜ 

⎜ e3 ⎟ ⎜ s31 

⎜ ⎟ = ⎜ 

⎜ e4 ⎟ ⎜ s41 

⎝ ⎠ ⎝ 

s22 

s32 

s42 

s23 

s33 

s43 

s24 

s34 

s44 

s25 

s35 

s45 

s26 ⎟ ⎜ σ2 ⎟ 

⎟ ⎜ ⎟ 

s36 ⎟ ⎜ σ3 ⎟ 

⎟ ⎜ ⎟ . 

s46 ⎟ ⎜ σ4 ⎟ 

⎠ ⎝ ⎠ 

(2.4.6) 

e5 

e6 

s51 s52 s53 s54 s55 s56 

s61 s62 s63 s64 s65 s66 

Alternatively, in the arguments that follow, one can examine the equivalent form 

⎛ ⎞ 

σ1 

⎛ 

c11 c12 c13 c14 c15 

⎞ ⎛ ⎞ 

c16 e1 

⎜ σ2 ⎟ ⎜ c21 

⎜ ⎟ ⎜ 

⎜ σ3 ⎟ ⎜ c31 

⎜ ⎟ = ⎜ 

⎜ σ4 ⎟ ⎜ c41 

⎝ ⎠ ⎝ 

c22 

c32 

c42 

c23 

c33 

c43 

c24 

c34 

c44 

c25 

c35 

c45 

c26 ⎟ ⎜ e2 ⎟ 

⎟ ⎜ ⎟ 

c36 ⎟ ⎜ e3 ⎟ 

⎟ ⎜ ⎟ . 

c46 ⎟ ⎜ e4 ⎟ 

⎠ ⎝ ⎠ 

Material Symmetries 

σ5 

σ6 

c51 c52 c53 c54 c55 c56 

c61 c62 c63 c64 c65 c66 

A material (crystal) with one plane of symmetry is called an aelotropic material. If we let the x1x2 

plane be a plane of symmetry then the equations (2.4.6) must remain invariant under the coordinate 

transformation ⎛ 

⎝ x1 

⎞ ⎛ 

1 

x2 ⎠ = ⎝ 0 

0 

1 

0 

0 

⎞ ⎛ 

⎠ ⎝ 

x3 0 0 −1 

x1 

⎞ 

x2 ⎠ (2.4.7) 

x3 

which represents an inversion of the x3 axis. That is, if the x1-x2 plane is a plane of symmetry we should be 

able to replace x3 by −x3 and the equations (2.4.6) should remain unchanged. This is equivalent to saying 

that a transformation of the type from equation (2.4.7) changes the Hooke’s law to the form ei = sijσj where 

the sij remain unaltered because it is the same material. Employing the transformation equations 

we examine the stress and strain transformation equations 

x1 = x1, x2 = x2, x3 = −x3 (2.4.8) 

∂xp ∂xq 

σij = σpq 

∂xi ∂xj 

and 

∂xp ∂xq 

eij = epq . 

∂xi ∂xj 

(2.4.9) 

If we expand both of the equations (2.4.9) and substitute in the nonzero derivatives 

we obtain the relations 

∂x1 

=1, 

∂x1 

σ11 = σ11 

σ22 = σ22 

σ33 = σ33 

σ21 = σ21 

σ31 = −σ31 

σ23 = −σ23 

∂x2 

=1, 

∂x2 

e11 = e11 

e22 = e22 

e33 = e33 

e21 = e21 

e31 = −e31 

e23 = −e23. 

σ5 

σ6 

e5 

e6 

∂x3 

= −1, (2.4.10) 

∂x3 

(2.4.11) 

245

246 

We conclude that if the material undergoes a strain, with the x1-x2 plane as a plane of symmetry then 

e5 and e6 change sign upon reversal of the x3 axis and e1,e2,e3,e4 remain unchanged. Similarly, we find σ5 

and σ6 change sign while σ1,σ2,σ3,σ4 remain unchanged. The equation (2.4.6) then becomes 

⎛ 

e1 

⎞ ⎛ 

s11 s12 s13 s14 s15 

⎞ ⎛ 

s16 σ1 

⎞ 

⎜ e2 ⎟ ⎜ s21 

⎜ ⎟ ⎜ 

⎜ e3 ⎟ ⎜ s31 

⎜ ⎟ = ⎜ 

⎜ e4 ⎟ ⎜ s41 

⎝ ⎠ ⎝ 

−e5 s51 

s22 

s32 

s42 

s52 

s23 

s33 

s43 

s53 

s24 

s34 

s44 

s54 

s25 

s35 

s45 

s55 

s26 ⎟ ⎜ σ2 ⎟ 

⎟ ⎜ ⎟ 

s36 ⎟ ⎜ σ3 ⎟ 

⎟ ⎜ ⎟ . 

s46 ⎟ ⎜ σ4 ⎟ 

⎠ ⎝ ⎠ 

s56 −σ5 


−e6 s61 s62 s63 s64 s65 s66 −σ6 

If the stress strain relation for the new orientation of the x3 axis is to have the same form as the 

old orientation, then the equations (2.4.6) and (2.4.12) must give the same results. Comparison of these 

equations we find that 

s15 = s16 =0 

s25 = s26 =0 

s35 = s36 =0 

s45 = s46 =0 

s51 = s52 = s53 = s54 =0 

s61 = s62 = s63 = s64 =0. 

(2.4.13) 

In summary, from an examination of the equations (2.4.6) and (2.4.12) we find that for an aelotropic 

material (crystal), with one plane of symmetry, the 36 constants sij reduce to 20 constants and the generalized 

Hooke’s law (constitutive equation) has the form 

⎛ ⎞ 

e1 

⎛ 

s11 s12 s13 s14 0 0 

⎞ ⎛ ⎞ 

σ1 

⎜ e2 ⎟ ⎜ s21 

⎜ ⎟ ⎜ 

⎜ e3 ⎟ ⎜ s31 

⎜ ⎟ = ⎜ 

⎜ e4 ⎟ ⎜ s41 

⎝ ⎠ ⎝ 

s22 

s32 

s42 

s23 

s33 

s43 

s24 

s34 

s44 

0 

0 

0 

0 

0 

0 

⎟ ⎜ σ2 ⎟ 

⎟ ⎜ ⎟ 

⎟ ⎜ σ3 ⎟ 

⎟ ⎜ ⎟ . 

⎟ ⎜ σ4 ⎟ 

⎠ ⎝ ⎠ 

(2.4.14) 

e5 

e6 

0 0 0 0 s55 s56 

0 0 0 0 s65 s66 

Alternatively, the Hooke’s law can be represented in the form 

⎛ ⎞ ⎛ 

σ1 

⎜ σ2 ⎟ ⎜ 

⎜ ⎟ ⎜ 

⎜ σ3 ⎟ ⎜ 

⎜ ⎟ = ⎜ 

⎜ σ4 ⎟ ⎜ 

⎝ ⎠ ⎝ 

σ5 

σ6 

c11 c12 c13 c14 0 0 

c21 c22 c23 c24 0 0 

c31 c32 c33 c34 0 0 

c41 c42 c43 c44 0 0 

0 0 0 0 c55 c56 

0 0 0 0 c65 c66 

⎞ ⎛ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎠ ⎝ 

σ5 

σ6 

e1 

e2 

e3 

e4 

e5 

e6 

⎞ 

⎟ . 

⎟ 

⎠

Additional Symmetries 

If the material (crystal) is such that there is an additional plane of symmetry, say the x2-x3 plane, then 

reversal of the x1 axis should leave the equations (2.4.14) unaltered. If there are two planes of symmetry 

then there will automatically be a third plane of symmetry. Such a material (crystal) is called orthotropic. 

Introducing the additional transformation 

x1 = −x1, x2 = x2, x3 = x3 

which represents the reversal of the x1 axes, the expanded form of equations (2.4.9) are used to calculate the 

effect of such a transformation upon the stress and strain tensor. We find σ1,σ2,σ3,σ6,e1,e2,e3,e6 remain 

unchanged while σ4,σ5,e4,e5 change sign. The equation (2.4.14) then becomes 

⎛ 

e1 

⎞ ⎛ 

⎜ e2 ⎟ ⎜ 

⎜ ⎟ ⎜ 

⎜ e3 ⎟ ⎜ 

⎜ ⎟ = ⎜ 

⎜ −e4 ⎟ ⎜ 

⎝ ⎠ ⎝ 

−e5 

e6 

s11 s12 s13 s14 0 0 

s21 s22 s23 s24 0 0 

s31 s32 s33 s34 0 0 

s41 s42 s43 s44 0 0 

0 0 0 0 s55 s56 

0 0 0 0 s65 s66 

⎞ ⎛ 

σ1 

⎞ 

⎟ ⎜ σ2 ⎟ 

⎟ ⎜ ⎟ 

⎟ ⎜ σ3 ⎟ 

⎟ ⎜ ⎟ . 

⎟ ⎜ −σ4 ⎟ 

⎠ ⎝ ⎠ 

−σ5 

(2.4.15) 

Note that if the constitutive equations (2.4.14) and (2.4.15) are to produce the same results upon reversal 

of the x1 axes, then we require that the following coefficients be equated to zero: 

This then produces the constitutive equation 

or its equivalent form 

⎛ 

⎜ 

⎝ 

⎛ 

⎜ 

⎝ 

e1 

e2 

e3 

e4 

e5 

e6 

σ1 

σ2 

σ3 

σ4 

σ5 

σ6 

⎞ ⎛ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ = ⎜ 

⎟ ⎜ 

⎠ ⎝ 

⎞ ⎛ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ = ⎜ 

⎟ ⎜ 

⎠ ⎝ 

s14 = s24 = s34 =0 

s41 = s42 = s43 =0 

s56 = s65 =0. 

s11 s12 s13 0 0 0 

s21 s22 s23 0 0 0 

s31 s32 s33 0 0 0 

0 0 0 s44 0 0 

0 0 0 0 s55 0 

0 0 0 0 0 s66 

c11 c12 c13 0 0 0 

c21 c22 c23 0 0 0 

c31 c32 c33 0 0 0 

0 0 0 c44 0 0 

0 0 0 0 c55 0 

0 0 0 0 0 c66 

⎞ ⎛ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎠ ⎝ 

⎞ ⎛ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎠ ⎝ 

σ6 

σ1 

σ2 

σ3 

σ4 

σ5 

σ6 

e1 

e2 

e3 

e4 

e5 

e6 

⎞ 

⎟ 

⎠ 

⎞ 

⎟ 

⎠ 

(2.4.16) 

and the original 36 constants have been reduced to 12 constants. This is the constitutive equation for 

orthotropic material (crystals). 

247

248 

Axis of Symmetry 

If in addition to three planes of symmetry there is an axis of symmetry then the material (crystal) is 

termed hexagonal. Assume that the x1 axis is an axis of symmetry and consider the effect of the transformation 

x 1 = x 1 , x 2 = x 3 

x 3 = −x 2 

upon the constitutive equations. It is left as an exercise to verify that the constitutive equations reduce to 

the form where there are 7 independent constants having either of the forms 

⎛ ⎞ 

e1 

⎛ 

s11 s12 s12 0 0 0 

⎞ ⎛ ⎞ 

σ1 

⎜ e2 ⎟ ⎜ s21 

⎜ ⎟ ⎜ 

⎜ e3 ⎟ ⎜ 

⎜ ⎟ = ⎜ 

⎜ e4 ⎟ ⎜ 

⎝ ⎠ ⎝ 

s22 s23 0 0 0 ⎟ ⎜ σ2 ⎟ 

⎟ ⎜ ⎟ 

⎟ ⎜ σ3 ⎟ 

⎟ ⎜ ⎟ 

⎟ ⎜ σ4 ⎟ 

⎠ ⎝ ⎠ 

or ⎛ 

⎜ 

⎝ 

e5 

e6 

σ1 

σ2 

σ3 

σ4 

σ5 

σ6 

⎞ ⎛ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ = ⎜ 

⎟ ⎜ 

⎠ ⎝ 

s21 s23 s22 0 0 0 

0 0 0 s44 0 0 

0 0 0 0 s44 0 

0 0 0 0 0 s66 

c11 c12 c12 0 0 0 

c21 c22 c23 0 0 0 

c21 c23 c22 0 0 0 

0 0 0 c44 0 0 

0 0 0 0 c44 0 

0 0 0 0 0 c66 

Finally, if the material is completely symmetric, the x2 axis is also an axis of symmetry and we can 

consider the effect of the transformation 

x 1 = −x 3 , x 2 = x 2 , x 3 = x 1 

upon the constitutive equations. 

constitutive equation to the form 

It can be verified that these transformations reduce the Hooke’s law 

⎛ ⎞ 

e1 

⎛ 

s11 s12 s12 0 0 0 

⎞ ⎛ ⎞ 

σ1 

⎜ e2 ⎟ ⎜ s12 

⎜ ⎟ ⎜ 

⎜ e3 ⎟ ⎜ 

⎜ ⎟ = ⎜ 

⎜ e4 ⎟ ⎜ 

⎝ ⎠ ⎝ 

s11 s12 0 0 0 ⎟ ⎜ σ2 ⎟ 

⎟ ⎜ ⎟ 

⎟ ⎜ σ3 ⎟ 

⎟ ⎜ ⎟ . 

⎟ ⎜ σ4 ⎟ 

⎠ ⎝ ⎠ 

(2.4.17) 

e5 

e6 

s12 s12 s11 0 0 0 

0 0 0 s44 0 0 

0 0 0 0 s44 0 

0 0 0 0 0 s44 

Materials (crystals) with atomic arrangements that exhibit the above symmetries are called isotropic 

materials. An equivalent form of (2.4.17) is the relation 

⎛ ⎞ 

σ1 

⎛ 

c11 c12 c12 0 0 0 

⎞ ⎛ ⎞ 

e1 

⎜ σ2 ⎟ ⎜ c12 

⎜ ⎟ ⎜ 

⎜ σ3 ⎟ ⎜ 

⎜ ⎟ = ⎜ 

⎜ σ4 ⎟ ⎜ 

⎝ ⎠ ⎝ 

c11 c12 0 0 0 ⎟ ⎜ e2 ⎟ 

⎟ ⎜ ⎟ 

⎟ ⎜ e3 ⎟ 

⎟ ⎜ ⎟ . 

⎟ ⎜ e4 ⎟ 

⎠ ⎝ ⎠ 

σ5 

σ6 

c12 c12 c11 0 0 0 

0 0 0 c44 0 0 

0 0 0 0 c44 0 

0 0 0 0 0 c44 

⎞ ⎛ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎟ ⎜ 

⎠ ⎝ 

The figure 2.4-1 lists values for the elastic stiffness associated with some metals which are isotropic 1 

1Additional constants are given in “International Tables of Selected Constants”, Metals: Thermal and 

Mechanical Data, Vol. 16, Edited by S. Allard, Pergamon Press, 1969. 

σ5 

σ6 

e1 

e2 

e3 

e4 

e5 

e6 

σ5 

σ6 

e5 

e6 

⎞ 

⎟ . 

⎟ 

⎠

Metal c11 c12 c44 

Na 0.074 0.062 0.042 

Pb 0.495 0.423 0.149 

Cu 1.684 1.214 0.754 

Ni 2.508 1.500 1.235 

Cr 3.500 0.678 1.008 

Mo 4.630 1.610 1.090 

W 5.233 2.045 1.607 

Figure 2.4-1. Elastic stiffness coefficients for some metals which are cubic. 

Constants are given in units of 10 12 dynes/cm 2 

Under these conditions the stress strain constitutive relations can be written as 

Isotropic Material 

σ1 = σ11 =(c11 − c12)e11 + c12(e11 + e22 + e33) 



σ4 = σ12 = c44e12 

σ5 = σ13 = c44e13 

σ6 = σ23 = c44e23. 

(2.4.18) 

Materials (crystals) which are elastically the same in all directions are called isotropic. We have shown 

that for a cubic material which exhibits symmetry with respect to all axes and planes, the constitutive 

stress-strain relation reduces to the form found in equation (2.4.17). Define the quantities 

s11 = 1 

E , s12 = − ν 

E , s44 = 1 

2µ 

where E is the Young’s Modulus of elasticity, ν is the Poisson’s ratio, and µ is the shear or rigidity modulus. 

For isotropic materials the three constants E,ν,µ are not independent as the following example demonstrates. 

EXAMPLE 2.4-1. (Elastic constants) For an isotropic material, consider a cross section of material in 

the x 1 -x 2 plane which is subjected to pure shearing so that σ4 = σ12 is the only nonzero stress as illustrated 


For the above conditions, the equation (2.4.17) reduces to the single equation 

e4 = e12 = s44σ4 = s44σ12 or µ = σ12 

and so the shear modulus is the ratio of the shear stress to the shear angle. Now rotate the axes through a 

45 degree angle to a barred system of coordinates where 

γ12 

x 1 = x 1 cos α − x 2 sin α x 2 = x 1 sin α + x 2 cos α 

249

250 

Figure 2.4-2. Element subjected to pure shearing 

where α = π 

4 . Expanding the transformation equations (2.4.9) we find that 

and similarly 

In the barred system, the Hooke’s law becomes 

σ1 = σ11 =cosα sin ασ12 +sinα cos ασ21 = σ12 = σ4 

σ2 = σ22 = − sin α cos ασ12 − sin α cos ασ21 = −σ12 = −σ4, 

e1 = e11 = e4, e2 = e22 = −e4. 

e1 = s11σ1 + s12σ2 

or 

e4 = s11σ4 − s12σ4 = s44σ4. 

Hence, the constants s11,s12,s44 are related by the relation 

s11 − s12 = s44 

or 

1 ν 

+ 

E E 

1 

= . (2.4.19) 

2µ 

This is an important relation connecting the elastic constants associated with isotropic materials. The 

above transformation can also be applied to triclinic, aelotropic, orthotropic, and hexagonal materials to 

find relationships between the elastic constants. 

Observe also that some texts postulate the existence of a strain energy function U ∗ which has the 

property that σij = ∂U∗ . In this case the strain energy function, in the single index notation, is written 

∂eij 

U ∗ = cijeiej where cij and consequently sij are symmetric. In this case the previous discussed symmetries 

give the following results for the nonzero elastic compliances sij : 13 nonzero constants instead of 20 for 

aelotropic material, 9 nonzero constants instead of 12 for orthotropic material, and 6 nonzero constants 

instead of 7 for hexagonal material. This is because of the additional property that sij = sji be symmetric.

The previous discussion has shown that for an isotropic material the generalized Hooke’s law (constitutive 

equations) have the form 

e11 = 1 

E [σ11 − ν(σ22 + σ33)] 

e22 = 1 

E [σ22 − ν(σ33 + σ11)] 

e33 = 1 

E [σ33 − ν(σ11 + σ22)] 

e21 = e12 = 1+ν 

E σ12 

e32 = e23 = 1+ν 

E σ23 

e31 = e13 = 1+ν 

E σ13 

, (2.4.20) 

where equation (2.4.19) holds. These equations can be expressed in the indicial notation and have the form 

eij = 1+ν 

E σij − ν 

E σkkδij, (2.4.21) 

where σkk = σ11 + σ22 + σ33 is a stress invariant and δij is the Kronecker delta. We can solve for the stress 

in terms of the strain by performing a contraction on i and j in equation (2.4.21). This gives the dilatation 

eii = 1+ν 

E σii − 3ν 

E σkk 

1 − 2ν 

= 

E σkk. 

Note that from the result in equation (2.4.21) we are now able to solve for the stress in terms of the strain. 

We find 

eij = 1+ν 

E σij − ν 

1 − 2ν ekkδij 

E 

1+ν eij 

νE 

= σij − 

(1 + ν)(1 − 2ν) ekkδij 

or σij = E 

1+ν eij 

νE 

+ 

(1 + ν)(1 − 2ν) ekkδij. 

The tensor equation (2.4.22) represents the six scalar equations 

σ11 = 

σ22 = 

σ33 = 

E 

(1 + ν)(1 − 2ν) [(1 − ν)e11 + ν(e22 + e33)] 

E 

(1 + ν)(1 − 2ν) [(1 − ν)e22 + ν(e33 + e11)] 

E 

(1 + ν)(1 − 2ν) [(1 − ν)e33 + ν(e22 + e11)] 

σ12 = E 

1+ν e12 

σ13 = E 

1+ν e13 

σ23 = E 

1+ν e23. 

(2.4.22) 

251

252 

Alternative Approach to Constitutive Equations 

The constitutive equation defined by Hooke’s generalized law for isotropic materials can be approached 

from another point of view. Consider the generalized Hooke’s law 

σij = cijklekl, i,j,k,l=1, 2, 3. 

If we transform to a barred system of coordinates, we will have the new Hooke’s law 

For an isotropic material we require that 

σij = cijklekl, i,j,k,l=1, 2, 3. 

cijkl = cijkl. 

Tensors whose components are the same in all coordinate systems are called isotropic tensors. We have 

previously shown in Exercise 1.3, problem 18, that 

cpqrs = λδpqδrs + µ(δprδqs + δpsδqr)+κ(δprδqs − δpsδqr) 

is an isotropic tensor when we consider affine type transformations. If we further require the symmetry 

conditions found in equations (2.4.3) be satisfied, we find that κ = 0 and consequently the generalized 

Hooke’s law must have the form 

σpq = cpqrsers =[λδpqδrs + µ(δprδqs + δpsδqr)] ers 

σpq = λδpqerr + µ(epq + eqp) 

or σpq =2µepq + λerrδpq, 

(2.4.23) 

where err = e11 + e22 + e33 = Θ is the dilatation. The constants λ and µ are called Lame’s constants. 

Comparing the equation (2.4.22) with equation (2.4.23) we find that the constants λ and µ satisfy the 

relations 

E 

νE 

µ = 

λ = 

. (2.4.24) 

2(1 + ν) 

(1 + ν)(1 − 2ν) 

In addition to the constants E,ν,µ,λ, it is sometimes convenient to introduce the constant k, called the bulk 

modulus of elasticity, (Exercise 2.3, problem 23), defined by 

k = 

E 

. (2.4.25) 

3(1 − 2ν) 

The stress-strain constitutive equation (2.4.23) was derived using Cartesian tensors. To generalize the 

equation (2.4.23) we consider a transformation from a Cartesian coordinate system yi ,i=1, 2, 3toageneral 

coordinate system xi ,i=1, 2, 3. We employ the relations 

and 

σmn = σij 

g ij = ∂ym 

∂x i 

∂ym ∂xj , gij = ∂xi 

∂ym ∂xj ∂ym ∂yi ∂xm ∂yj ∂xn , emn 

∂y 

= eij 

i 

∂xm ∂yj ∂xn , or erq 

∂x 

= eij 

i 

∂yr ∂x j 

∂y q

and convert equation (2.4.23) to a more generalized form. Multiply equation (2.4.23) by ∂yp 

∂xm the result 

which can be simplified to the form 

Dropping the bar notation, we have 

The contravariant form of this equation is 

σmn = λ ∂yq 

∂xm ∂yq ∂xn err + µ (emn + enm) , 

σmn = λg mneijg ij + µ (emn + enm) . 

σmn = λgmng ij eij + µ (emn + enm) . 

σ sr = λg sr g ij eij + µ (g ms g nr + g ns g mr ) emn. 

Employing the equations (2.4.24) the above result can also be expressed in the form 

σ rs = 

∂yq n and verify 

∂x 

 

E 

g 

2(1 + ν) 

ms g nr + g ns g mr + 2ν 

1 − 2ν gsrg mn 

 

emn. (2.4.26) 

This is a more general form for the stress-strain constitutive equations which is valid in all coordinate systems. 

Multiplying by gsk and employing the use of associative tensors, one can verify 

σ i j 

 

E 

= e 

1+ν 

i ν 

j + 

1 − 2ν emm δi 

j 

or σ i j =2µe i j + λe m mδ i j, 

are alternate forms for the equation (2.4.26). As an exercise, solve for the strains in terms of the stresses 

and show that 

Ee i j =(1+ν)σ i j − νσ m mδ i j. 

EXAMPLE 2.4-2. (Hooke’s law) Let us construct a simple example to test the results we have 

developed so far. Consider the tension in a cylindrical bar illustrated in the figure 2.4-3. 

Figure 2.4-3. Stress in a cylindrical bar 

253

254 

Assume that 

⎛ 

F 

A 

σij = ⎝ 0 

0 

0 

⎞ 

0 

0⎠ 

0 0 0 

where F is the constant applied force and A is the cross sectional area of the cylinder. Consequently, the 

generalized Hooke’s law (2.4.21) produces the nonzero strains 

From these equations we obtain: 

The first part of Hooke’s law 

The second part of Hooke’s law 

e11 = 1+ν 

E σ11 − ν 

E (σ11 + σ22 + σ33) = σ11 

E 

e22 = −ν 

E σ11 

e33 = −ν 

E σ11 

σ11 = Ee11 or F 

= Ee11. 

A 

lateral contraction −e22 

= = 

longitudinal extension e11 

−e33 

= ν = Poisson’s ratio. 

e11 

This example demonstrates that the generalized Hooke’s law for homogeneous and isotropic materials 

reduces to our previous one dimensional result given in (2.3.1) and (2.3.2). 

Basic Equations of Elasticity 

Assuming the density ϱ is constant, the basic equations of elasticity reduce to the equations representing 

conservation of linear momentum and angular momentum together with the strain-displacement relations 

and constitutive equations. In these equations the body forces are assumed known. These basic equations 

produce 15 equations in 15 unknowns and are a formidable set of equations to solve. Methods for solving 

these simultaneous equations are: 1) Express the linear momentum equations in terms of the displacements 

ui and obtain a system of partial differential equations. Solve the system of partial differential equations 

for the displacements ui and then calculate the corresponding strains. The strains can be used to calculate 

the stresses from the constitutive equations. 2) Solve for the stresses and from the stresses calculate the 

strains and from the strains calculate the displacements. This converse problem requires some additional 

considerations which will be addressed shortly.

Navier’s Equations 

Basic Equations of Linear Elasticity 

• Conservation of linear momentum. 

σ ij 

,i + ϱbj = ϱf j 

j =1, 2, 3. (2.4.27(a)) 

where σ ij is the stress tensor, b j is the body force per unit mass and f j is 

the acceleration. If there is no motion, then f j = 0 and these equations 

reduce to the equilibrium equations 

• Conservation of angular momentum. σij = σji 

• Strain tensor. 

σ ij 

,i + ϱbj =0 j =1, 2, 3. (2.4.27(b)) 

eij = 1 

2 (ui,j + uj,i) (2.4.28) 

where ui denotes the displacement field. 

• Constitutive equation. For a linear elastic isotropic material we have 

σ i j 

= E 

1+ν ei j + 

or its equivalent form 

E 

(1 + ν)(1 − 2ν) ek k δi j i, j =1, 2, 3 (2.4.29(a)) 

σ i j =2µei j + λer r δi j i, j =1, 2, 3, (2.4.29(b)) 

where e r r is the dilatation. This produces 15 equations for the 15 unknowns 

u1,u2,u3,σ11,σ12,σ13,σ22,σ23,σ33,e11,e12,e13,e22,e23,e33, 

which represents 3 displacements, 6 strains and 6 stresses. In the above 

equations it is assumed that the body forces are known. 

The equations (2.4.27) through (2.4.29) can be combined and written as one set of equations. The 

resulting equations are known as Navier’s equations for the displacements ui over the range i =1, 2, 3. To 

derive the Navier’s equations in Cartesian coordinates, we write the equations (2.4.27),(2.4.28) and (2.4.29) 

in Cartesian coordinates. We then calculate σij,j in terms of the displacements ui and substitute the results 

into the momentum equation (2.4.27(a)). Differentiation of the constitutive equations (2.4.29(b)) produces 

σij,j =2µeij,j + λekk,jδij. (2.4.30) 

255

256 

A contraction of the strain produces the dilatation 

err = 1 

2 (ur,r + ur,r) =ur,r 

From the dilatation we calculate the covariant derivative 

(2.4.31) 

ekk,j = uk,kj. (2.4.32) 

Employing the strain relation from equation (2.4.28), we calculate the covariant derivative 

eij,j = 1 

2 (ui,jj + uj,ij). (2.4.33) 

These results allow us to express the covariant derivative of the stress in terms of the displacement field. We 

find 

σij,j = µ [ui,jj + uj,ij]+λδijuk,kj 

or σij,j =(λ + µ)uk,ki + µui,jj. 

Substituting equation (2.4.34) into the linear momentum equation produces the Navier equations: 

In vector form these equations can be expressed 

(2.4.34) 

(λ + µ)uk,ki + µui,jj + ϱbi = ϱfi, i =1, 2, 3. (2.4.35) 

(λ + µ)∇ (∇·u)+µ∇ 2 u + ϱ b = ϱ f, (2.4.36) 

where u is the displacement vector, b is the body force per unit mass and f is the acceleration. In Cartesian 

coordinates these equations have the form: 

2 ∂ u1 

(λ + µ) + 

∂x1∂xi 

∂2u2 + 

∂x2∂xi 

∂2 

u3 

+ µ∇ 

∂x3∂xi 

2 ui + ϱbi = ϱ ∂2ui , 

∂t2 for i =1, 2, 3, where 

∇ 2 ui = ∂2ui ∂x1 2 + ∂2ui ∂x2 2 + ∂2ui . 

∂x3 

2 

The Navier equations must be satisfied by a set of functions ui = ui(x1,x2,x3) which represent the 

displacement at each point inside some prescribed region R. Knowing the displacement field we can calculate 

the strain field directly using the equation (2.4.28). Knowledge of the strain field enables us to construct the 

corresponding stress field from the constitutive equations. 

In the absence of body forces, such as gravity, the solution to equation (2.4.36) can be represented 

in the form u = u (1) + u (2) , where u (1) satisfies div u (1) = ∇·u (1) = 0 and the vector u (2) satisfies 

curl u (2) = ∇×u (2) =0. The vector field u (1) is called a solenoidal field, while the vector field u (2) is 

called an irrotational field. Substituting u into the equation (2.4.36) and setting b =0, we find in Cartesian 

coordinates that 

2 (1) ∂ u 

ϱ 

∂t2 + ∂2u (2) 

∂t2 

 

=(λ + µ)∇ ∇·u (2) 

+ µ∇ 2 u (1) + µ∇ 2 u (2) . (2.4.37)

The vector field u (1) can be eliminated from equation (2.4.37) by taking the divergence of both sides of the 

equation. This produces 

ϱ ∂2 ∇·u (2) 

∂t 2 

=(λ + µ)∇ 2 (∇·u (2) )+µ∇·∇ 2 u (2) . 

The displacement field is assumed to be continuous and so we can interchange the order of the operators ∇ 2 

and ∇ and write 

This last equation implies that 

 

∇· ϱ ∂2u (2) 

∂t2 − (λ +2µ)∇2u (2) 

 

=0. 

ϱ ∂2 u (2) 

∂t 2 

=(λ +2µ)∇2 u (2) 

and consequently, u (2) is a vector wave which moves with the speed (λ +2µ)/ϱ. Similarly, when the vector 

field u (2) is eliminated from the equation (2.4.37), by taking the curl of both sides, we find the vector u (1) 

also satisfies a wave equation having the form 

ϱ ∂2 u (1) 

∂t 2 

= µ∇2 u (1) . 

This later wave moves with the speed µ/ϱ. The vector u (2) is a compressive wave, while the wave u (1) is 

ashearingwave. 

The exercises 30 through 38 enable us to write the Navier’s equations in Cartesian, cylindrical or 

spherical coordinates. In particular, we have for cartesian coordinates 

(λ + µ)( ∂2u ∂x2 + ∂2v ∂x∂y + ∂2w ∂x∂z )+µ(∂2 u 

∂x2 + ∂2u ∂y2 + ∂2u ∂z2 )+ϱbx =ϱ ∂2u ∂t2 (λ + µ)( ∂2u ∂x∂y + ∂2v ∂y2 + ∂2w ∂y∂z )+µ( ∂2v ∂x2 + ∂2v ∂y2 + ∂2v ∂z2 )+ϱby =ϱ ∂2v ∂t2 (λ + µ)( ∂2u ∂x∂z + ∂2v ∂y∂z + ∂2w ∂z2 )+µ(∂2 w 

∂x2 + ∂2w ∂y2 + ∂2w ∂z2 )+ϱbz =ϱ ∂2w ∂t2 and in cylindrical coordinates 

(λ + µ) ∂ 

 

1 

∂r r 

µ( ∂2ur 1 ∂ur 1 

+ + 

∂r2 r ∂r r2 ∂2ur µ( ∂2uθ 1 ∂uθ 1 

+ + 

∂r2 r ∂r r2 ∂2uθ ∂ 

∂r 

 

1 ∂uθ ∂uz 

(rur)+ + + 

r ∂θ ∂z 

∂θ2 + ∂2ur ur 2 

− − 

∂z2 r2 r2 ∂uθ 

∂θ )+ϱbr =ϱ ∂2ur ∂t2 (λ + µ) 1 

 

∂ 1 ∂ 1 ∂uθ ∂uz 

(rur)+ + + 

r ∂θ r ∂r r ∂θ ∂z 

∂θ2 + ∂2uθ 2 

+ 

∂z2 r2 ∂ur uθ 

− 

∂θ r2 )+ϱbθ =ϱ ∂2uθ ∂t2 (λ + µ) ∂ 

 

1 ∂ 1 ∂uθ ∂uz 

(rur)+ + + 

∂z r ∂r r ∂θ ∂z 

µ( ∂2uz 1 ∂uz 1 

+ + 

∂r2 r ∂r r2 ∂2uz ∂θ2 + ∂2uz ∂z2 )+ϱbz =ϱ ∂2uz ∂t2 257

258 

and in spherical coordinates 

(λ + µ) ∂ 

 

1 

∂ρ ρ2 ∂ 

∂ρ (ρ2uρ)+ 1 ∂ 

ρ sin θ ∂θ (uθ sin θ)+ 1 

 

∂uφ 

+ 

ρ sin θ ∂φ 

µ(∇ 2 uρ − 2 

ρ2 uρ − 2 

ρ2 ∂uθ 

∂θ − 2uθ cot θ 

ρ2 2 

− 

ρ2 ∂uφ 

sin θ ∂φ )+ϱbρ =ϱ ∂2uρ ∂t2 (λ + µ) 1 

 

∂ 1 

ρ ∂θ ρ2 ∂ 

∂ρ (ρ2uρ)+ 1 ∂ 


 

∂uφ 

+ 


µ(∇ 2 uθ + 2 

ρ2 ∂uρ 

∂θ − 

uθ 

ρ2 sin 2 2 

− 

θ ρ2 cos θ 

sin 2 ∂uφ 

θ ∂φ )+ϱbθ =ϱ ∂2uθ ∂t2 

1 ∂ 1 

(λ + µ) 

ρ sin θ ∂φ ρ2 ∂ 

∂ρ (ρ2uρ)+ 1 ∂ 


 

∂uφ 

+ 


µ(∇ 2 uφ − 

1 

ρ2 sin 2 θ uφ 

2 

+ 

ρ2 ∂uρ 2cosθ 

+ 

sin θ ∂φ ρ2 sin 2 ∂uθ 

θ ∂φ )+ϱbφ =ϱ ∂2uφ ∂t2 where ∇ 2 is determined from either equation (2.1.12) or (2.1.13). 

Boundary Conditions 

In elasticity the body forces per unit mass (bi,i =1, 2, 3) are assumed known. In addition one of the 

following type of boundary conditions is usually prescribed: 

• The displacements ui, i =1, 2, 3 are prescribed on the boundary of the region R over which a solution 

is desired. 

• The stresses (surface tractions) are prescribed on the boundary of the region R over which a solution is 

desired. 

• The displacements ui,i = 1, 2, 3 are given over one portion of the boundary and stresses (surface 

tractions) are specified over the remaining portion of the boundary. This type of boundary condition is 

known as a mixed boundary condition. 

General Solution of Navier’s Equations 

There has been derived a general solution to the Navier’s equations. It is known as the Papkovich-Neuber 

solution. In the case of a solid in equilibrium one must solve the equilibrium equations 

(λ + µ)∇ (∇·u)+µ∇ 2 u + ϱb =0 or 

∇ 2 u + 1 

ϱ 

∇(∇·u)+ 

1 − 2ν µ b =0 (ν = 1 

2 ) 

(2.4.38)

THEOREM A general elastostatic solution of the equation (2.4.38) in terms of harmonic potentials φ, is 

ψ 

u =grad(φ + r · ψ) − 4(1 − ν) ψ (2.4.39) 

where φ and ψ are continuous solutions of the equations 

∇ 2 φ = −ϱr · b 

4µ(1 − ν) 

and ∇ 2 ψ = 

ϱ b 

4µ(1 − ν) 

with r = x ê1 + y ê2 + z ê3 a position vector to a general point (x, y, z) within the continuum. 

Proof: First we write equation (2.4.38) in the tensor form 

Now our problem is to show that equation (2.4.39), in tensor form, 

(2.4.40) 

ui,kk + 1 

1 − 2ν (uj,j) ,i + ϱ 

µ bi =0 (2.4.41) 

ui = φ,i +(xjψj),i − 4(1 − ν)ψi 

is a solution of equation (2.4.41). Toward this purpose, we differentiate equation (2.4.42) 

and then contract on i and k giving 

ui,k = φ,ik +(xjψj),ik − 4(1 − ν)ψi,k 

(2.4.42) 

(2.4.43) 

ui,i = φ,ii +(xjψj),ii − 4(1 − ν)ψi,i. (2.4.44) 

Employing the identity (xjψj),ii =2ψi,i + xiψi,kk the equation (2.4.44) becomes 

By differentiating equation (2.4.43) we establish that 

We use the hypothesis 

φ,kk = −ϱxjFj 

4µ(1 − ν) 

and simplify the equation (2.4.46) to the form 

ui,i = φ,ii +2ψi,i + xiψi,kk − 4(1 − ν)ψi,i. (2.4.45) 

ui,kk = φ,ikk +(xjψj),ikk − 4(1 − ν)ψi,kk 

Also by differentiating (2.4.45) one can establish that 

=(φ,kk),i +((xjψj),kk) ,i − 4(1 − ν)ψi,kk 

=[φ,kk +2ψj,j + xjψj,kk] ,i − 4(1 − ν)ψi,kk. 

and ψj,kk = 

ϱFj 

4µ(1 − ν) , 

(2.4.46) 

ui,kk =2ψj,ji − 4(1 − ν)ψi,kk. (2.4.47) 

uj,ji =(φ,jj),i +2ψj,ji +(xjψj,kk),i − 4(1 − ν)ψj,ji 

= 

 

−ϱxjFj 

4µ(1 − ν) 

= −2(1 − 2ν)ψj,ji. 

+2ψj,ji + 

,i 

 

ϱxjFj 

4µ(1 − ν) 

 

,i 

− 4(1 − ν)ψj,ji 

(2.4.48) 

259

260 

Finally, from the equations (2.4.47) and (2.4.48) we obtain the desired result that 

ui,kk + 1 

1 − 2ν uj,ji + ϱFi 

µ =0. 

Consequently, the equation (2.4.39) is a solution of equation (2.4.38). 

As a special case of the above theorem, note that when the body forces are zero, the equations (2.4.40) 

become 

∇ 2 φ =0 and ∇ 2 ψ = 0. 

In this case, we find that equation (2.4.39) is a solution of equation (2.4.38) provided φ and each component of 

ψ are harmonic functions. The Papkovich-Neuber potentials are used together with complex variable theory 

to solve various two-dimensional elastostatic problems of elasticity. Note also that the Papkovich-Neuber 

potentials are not unique as different combinations of φ and ψ can produce the same value for u. 

Compatibility Equations 

If we know or can derive the displacement field ui,i =1, 2, 3 we can then calculate the components of 

the strain tensor 

eij = 1 

2 (ui,j + uj,i). (2.4.49) 

Knowing the strain components, the stress is found using the constitutive relations. 

Consider the converse problem where the strain tensor is given or implied due to the assigned stress 

field and we are asked to determine the displacement field ui,i=1, 2, 3. Is this a realistic request? Is it even 

possible to solve for three displacements given six strain components? It turns out that certain mathematical 

restrictions must be placed upon the strain components in order that the inverse problem have a solution. 

These mathematical restrictions are known as compatibility equations. That is, we cannot arbitrarily assign 

six strain components eij and expect to find a displacement field ui,i=1, 2, 3 with three components which 

satisfies the strain relation as given in equation (2.4.49). 

EXAMPLE 2.4-3. Suppose we are given the two partial differential equations, 

∂u 

∂u 

= x + y and 

∂x ∂y = x3 . 

Can we solve for u = u(x, y)? The answer to this question is “no”, because the given equations are inconsistent. 

The inconsistency is illustrated if we calculate the mixed second derivatives from each equation. We 

find from the first equation that ∂2u ∂x∂y = 1 and from the second equation we calculate ∂2u ∂y∂x =3x2 . These 

mixed second partial derivatives are unequal for all x different from √ 3/3. In general, if we have two first 

order partial differential equations ∂u 

∂u 

= f(x, y) and = g(x, y), then for consistency (integrability of 

∂x ∂y 

the equations) we require that the mixed partial derivatives 

∂2u ∂f 

= 

∂x∂y ∂y = ∂2u ∂g 

= 

∂y∂x ∂x 

be equal to one another for all x and y values over the domain for which the solution is desired. This is an 

example of a compatibility equation.

A similar situation occurs in two dimensions for a material in a state of strain where ezz = ezx = ezy =0, 

called plane strain. In this case, are we allowed to arbitrarily assign values to the strains exx,eyy and exy and 

from these strains determine the displacement field u = u(x, y) andv = v(x, y) inthex− and y−directions? 

Let us try to answer this question. Assume a state of plane strain where ezz = ezx = ezy =0. Further, let 

us assign 3 arbitrary functional values f,g,h such that 

exx = ∂u 

∂x = f(x, y), exy = 1 

 

∂u ∂v 

+ = g(x, y), 

2 ∂y ∂x 

eyy = ∂v 

= h(x, y). 

∂y 

We must now decide whether these equations are consistent. That is, will we be able to solve for the 

displacement field u = u(x, y)andv = v(x, y)? To answer this question, let us derive a compatibility equation 

(integrability condition). From the given equations we can calculate the following partial derivatives 

∂2exx ∂y2 = ∂3u ∂x∂y2 = ∂2f ∂y2 ∂2eyy ∂x2 = ∂3v ∂y∂x2 = ∂2h ∂x2 2 ∂2 exy 

∂x∂y = ∂3 u 

∂x∂y 2 + ∂3 v 

∂y∂x 2 =2 ∂2 g 

∂x∂y . 

This last equation gives us the compatibility equation 

or the functions g, f, h must satisfy the relation 

2 ∂2exy ∂x∂y = ∂2exx ∂y2 + ∂2eyy ∂x2 2 ∂2g ∂x∂y = ∂2f ∂y2 + ∂2h . 

∂x2 Cartesian Derivation of Compatibility Equations 

If the displacement field ui,i=1, 2, 3 is known we can derive the strain and rotation tensors 

eij = 1 

2 (ui,j + uj,i) and ωij = 1 

2 (ui,j − uj,i). (2.4.50) 

Now work backwards. Assume the strain and rotation tensors are given and ask the question, “Is it possible 

to solve for the displacement field ui,i=1, 2, 3?” If we view the equation (2.4.50) as a system of equations 

with unknowns eij,ωij and ui and if by some means we can eliminate the unknowns ωij and ui then we 

will be left with equations which must be satisfied by the strains eij. These equations are known as the 

compatibility equations and they represent conditions which the strain components must satisfy in order 

that a displacement function exist and the equations (2.4.37) are satisfied. Let us see if we can operate upon 

the equations (2.4.50) to eliminate the quantities ui and ωij and hence derive the compatibility equations. 

Addition of the equations (2.4.50) produces 

ui,j = ∂ui 

∂xj 

= eij + ωij. (2.4.51) 

261

262 

Differentiate this expression with respect to xk and verify the result 

∂ 2 ui 

∂xj∂xk 

= ∂eij 

∂xk 

+ ∂ωij 

. 

∂xk 

(2.4.52) 

We further assume that the displacement field is continuous so that the mixed partial derivatives are equal 

and 

∂2ui ∂xj∂xk 

= ∂2ui . 

∂xk∂xj 

(2.4.53) 

Interchanging j and k in equation (2.4.52) gives us 

∂ 2 ui 

∂xk∂xj 

= ∂eik 

∂xj 

+ ∂ωik 

. 

∂xj 

(2.4.54) 

Equating the second derivatives from equations (2.4.54) and (2.4.52) and rearranging terms produces the 

result 

∂eij 

∂xk 

− ∂eik 

∂xj 

= ∂ωik 

∂xj 

− ∂ωij 

∂xk 

(2.4.55) 

form 

Making the observation that ωij satisfies ∂ωik 

∂eij 

∂xj 

− ∂ωij 

∂xk 

= ∂ωjk 

∂xi 

, the equation (2.4.55) simplifies to the 

∂xk 

− ∂eik 

∂xj 

= ∂ωjk 

. 

∂xi 

(2.4.56) 

The term involving ωjk can be eliminated by using the mixed partial derivative relation 

∂ 2 ωjk 

= 

∂xi∂xm 

∂2ωjk . (2.4.57) 

∂xm∂xi 

To derive the compatibility equations we differentiate equation (2.4.56) with respect to xm and then 

interchanging the indices i and m and substitute the results into equation (2.4.57). This will produce the 

compatibility equations 

∂2eij + 

∂xm∂xk 

∂2emk − 

∂xi∂xj 

∂2eik − 

∂xm∂xj 

∂2emj =0. (2.4.58) 

∂xi∂xk 

This is a set of 81 partial differential equations which must be satisfied by the strain components. Fortunately, 

due to symmetry considerations only 6 of these 81 equations are distinct. These 6 distinct equations are 

known as the St. Venant’s compatibility equations and can be written as 

∂ 2 e11 

∂x2∂x3 

∂ 2 e22 

∂x1∂x3 

∂ 2 e33 

∂x1∂x2 

= ∂2e12 − 

∂x1∂x3 

∂2e23 ∂x1 2 + ∂2e31 ∂x1∂x2 

= ∂2e23 − 

∂x2∂x1 

∂2e31 ∂x2 2 + ∂2e12 ∂x2∂x3 

= ∂2e31 − 

∂x3∂x2 

∂2e12 ∂x3 2 + ∂2e23 ∂x3∂x1 

2 ∂2e12 = 

∂x1∂x2 

∂2e11 ∂x2 2 + ∂2e22 ∂x1 2 

2 ∂2e23 = 

∂x2∂x3 

∂2e22 ∂x3 2 + ∂2e33 ∂x2 2 

2 ∂2e31 = 

∂x3∂x1 

∂2e33 ∂x1 2 + ∂2e11 . 

∂x3 

2 

Observe that the fourth compatibility equation is the same as that derived in the example 2.4-3. 

These compatibility equations can also be expressed in the indicial form 

(2.4.59) 

eij,km + emk,ji − eik,jm − emj,ki =0. (2.4.60)

Compatibility Equations in Terms of Stress 

In the generalized Hooke’s law, equation (2.4.29), we can solve for the strain in terms of stress. This 

in turn will give rise to a representation of the compatibility equations in terms of stress. The resulting 

equations are known as the Beltrami-Michell equations. Utilizing the strain-stress relation 

eij = 1+ν 

E σij − ν 

E σkkδij 

we substitute for the strain in the equations (2.4.60) and rearrange terms to produce the result 

σij,km + σmk,ji − σik,jm − σmj,ki = 

ν 

1+ν [δijσnn,km + δmkσnn,ji − δikσnn,jm − δmjσnn,ki] . 

(2.4.61) 

Now only 6 of these 81 equations are linearly independent. It can be shown that the 6 linearly independent 

equations are equivalent to the equations obtained by setting k = m and summing over the repeated indices. 

We then obtain the equations 

σij,mm + σmm,ij − (σim,m) ,j − (σmj,m) ,i = ν 

1+ν [δijσnn,mm + σnn,ij] . 

Employing the equilibrium equation σij,i + ϱbj = 0 the above result can be written in the form 

σij,mm + 1 

1+ν σkk,ij − ν 

1+ν δijσnn,mm = −(ϱbi),j − (ϱbj),i 

or 

∇ 2 σij + 1 

1+ν σkk,ij − ν 

1+ν δijσnn,mm = −(ϱbi),j − (ϱbj),i. 

This result can be further simplified by observing that a contraction on the indices k and i in equation 

(2.4.61) followed by a contraction on the indices m and j produces the result 

σij,ij = 

1 − ν 

1+ν σnn,jj. 

Consequently, the Beltrami-Michell equations can be written in the form 

∇ 2 σij + 1 

1+ν σpp,ij = − ν 

1 − ν δij(ϱbk) ,k − (ϱbi) ,j − (ϱbj) ,i. (2.4.62) 

Their derivation is left as an exercise. The Beltrami-Michell equations together with the linear momentum 

(equilibrium) equations σij,i + ϱbj = 0 represent 9 equations in six unknown stresses. This combinations 

of equations is difficult to handle. An easier combination of equations in terms of stress functions will be 

developed shortly. 

The Navier equations with boundary conditions are difficult to solve in general. Let us take the momentum 

equations (2.4.27(a)), the strain relations (2.4.28) and constitutive equations (Hooke’s law) (2.4.29) 

and make simplifying assumptions so that a more tractable systems results. 

263

264 

Plane Strain 

The plane strain assumption usually is applied in situations where there is a cylindrical shaped body 

whose axis is parallel to the z axis and loads are applied along the z−direction. In any x-y plane we assume 

that the surface tractions and body forces are independent of z. We set all strains with a subscript z equal 

to zero. Further, all solutions for the stresses, strains and displacements are assumed to be only functions 

of x and y and independent of z. Note that in plane strain the stress σzz is different from zero. 

In Cartesian coordinates the strain tensor is expressible in terms of its physical components which can 

be represented in the matrix form 

⎛ 

⎝ e11 

⎞ ⎛ 

e12 e13 

⎠ = ⎝ exx 

⎞ 

exy exz 

⎠ . 

e21 e22 e23 

e31 e32 e33 

eyx eyy eyz 

ezx ezy ezz 

If we assume that all strains which contain a subscript z are zero and the remaining strain components are 

functions of only x and y, we obtain a state of plane strain. For a state of plane strain, the stress components 

are obtained from the constitutive equations. The condition of plane strain reduces the constitutive equations 

to the form: 

exx = 1 

E [σxx − ν(σyy + σzz)] 

eyy = 1 

E [σyy − ν(σzz + σxx)] 

0= 1 

E [σzz − ν(σxx + σyy)] 

exy = eyx = 1+ν 

E σxy 

ezy = eyz = 1+ν 

E σyz =0 

ezx = exz = 1+ν 

E σxz 

E 

σxx = 

(1 + ν)(1 − 2ν) 

=0 

[(1 − ν)exx + νeyy] 

E 

σyy = 

(1 + ν)(1 − 2ν) [(1 − ν)eyy + νexx] 

E 

σzz = 

(1 + ν)(1 − 2ν) [ν(eyy + exx)] 

σxy = E 

1+ν exy 

(2.4.63) 

σxz =0 

σyz =0 

where σxx, σyy, σzz, σxy, σxz, σyz are the physical components of the stress. The above constitutive 

equations imply that for a state of plane strain we will have 

σzz = ν(σxx + σyy) 

exx = 1+ν 

E [(1 − ν)σxx − νσyy] 

eyy = 1+ν 

E [(1 − ν)σyy − νσxx] 

exy = 1+ν 

E σxy. 

Also under these conditions the compatibility equations reduce to 

∂ 2 exx 

∂y 2 + ∂2 eyy 

∂x 2 =2∂2 exy 

∂x∂y .

Plane Stress 

An assumption of plane stress is usually applied to thin flat plates. The plate thinness is assumed to be 

in the z−direction and loads are applied perpendicular to z. Under these conditions all stress components 

with a subscript z are assumed to be zero. The remaining stress components are then treated as functions 

of x and y. 

In Cartesian coordinates the stress tensor is expressible in terms of its physical components and can be 

represented by the matrix ⎛ 

⎝ σ11 

⎞ ⎛ 

⎞ 

σ12 σ13 

⎠ = 

⎠ . 

σ21 σ22 σ23 

σ31 σ32 σ33 

⎝ σxx σxy σxz 

σyx σyy σyz 

σzx σzy σzz 

If we assume that all the stresses with a subscript z are zero and the remaining stresses are only functions of 

x and y we obtain a state of plane stress. The constitutive equations simplify if we assume a state of plane 

stress. These simplified equations are 

exx = 1 

E σxx − ν 

E σyy 

eyy = 1 

E σyy − ν 

E σxx 

ezz = − ν 

E (σxx + σyy) 

exy = 1+ν 

E σxy 

σxx = 

exz =0 

E 

1 − ν2 [exx + νeyy] 

σyy = E 

1 − ν2 [eyy + νexx] 

σzz =0=(1−ν)ezz + ν(exx + eyy) 

σxy = E 

1+ν exy 

(2.4.64) 

σyz =0 

eyz =0. 

σxz =0 

For a state of plane stress the compatibility equations reduce to 

and the three additional equations 

∂2ezz =0, 

∂x2 ∂2exx ∂y2 + ∂2eyy ∂x2 =2∂2 exy 

∂x∂y 

∂2ezz =0, 

∂y2 ∂ 2 ezz 

∂x∂y =0. 

These three additional equations complicate the plane stress problem. 

Airy Stress Function 

(2.4.65) 

In Cartesian coordinates we examine the equilibrium equations (2.4.25(b)) under the conditions of plane 

strain. In terms of physical components we find that these equations reduce to 

∂σxx 

∂x 

∂σxy 

+ 

∂y + ϱbx =0, 

∂σyx ∂σyy 

+ 

∂x ∂y + ϱby =0, 

∂σzz 

∂z =0. 

The last equation is satisfied since σzz is a function of x and y. If we further assume that the body forces 

are conservative and derivable from a potential function V by the operation ϱ b = −grad V or ϱbi = −V ,i 

we can express the above equilibrium equations in the form: 

∂σxx 

∂x 

∂σyx 

∂x 

+ ∂σxy 

∂y 

+ ∂σyy 

∂y 

− ∂V 

∂x =0 

− ∂V 

∂y =0 

(2.4.66) 

265

266 

We will consider these equations together with the compatibility equations (2.4.65). The equations 

(2.4.66) will be automatically satisfied if we introduce a scalar function φ = φ(x, y) and assume that the 

stresses are derivable from this function and the potential function V according to the rules: 

σxx = ∂2φ ∂y2 + V σxy = − ∂2φ σyy = 

∂x∂y 

∂2φ + V. (2.4.67) 

∂x2 The function φ = φ(x, y) is called the Airy stress function after the English astronomer and mathematician 

Sir George Airy (1801–1892). Since the equations (2.4.67) satisfy the equilibrium equations we need only 

consider the compatibility equation(s). 

For a state of plane strain we substitute the relations (2.4.63) into the compatibility equation (2.4.65) 

and write the compatibility equation in terms of stresses. We then substitute the relations (2.4.67) and 

express the compatibility equation in terms of the Airy stress function φ. These substitutions are left as 

exercises. After all these substitutions the compatibility equation, for a state of plane strain, reduces to the 

form 

∂4φ ∂x4 +2 ∂4φ ∂x2∂y2 + ∂4 2 φ 1 − 2ν ∂ V 

+ 

∂y4 1 − ν ∂x2 + ∂2V ∂y2 

=0. (2.4.68) 

In the special case where there are no body forces we have V = 0 and equation (2.4.68) is further simplified 

to the biharmonic equation. 

∇ 4 φ = ∂4φ ∂x4 +2 ∂4φ ∂x2∂y2 + ∂4φ =0. (2.4.69) 

∂y4 In polar coordinates the biharmonic equation is written 

∇ 4 φ = ∇ 2 (∇ 2 2 ∂ 1 ∂ 1 

φ)= + + 

∂r2 r ∂r r2 ∂2 ∂θ2 2 ∂ φ 1 ∂φ 1 

+ + 

∂r2 r ∂r r2 ∂2φ ∂θ2 

=0. 

For conditions of plane stress, we can again introduce an Airy stress function using the equations (2.4.67). 

However, an exact solution of the plane stress problem which satisfies all the compatibility equations is 

difficult to obtain. By removing the assumptions that σxx,σyy,σxy are independent of z, and neglecting 

body forces, it can be shown that for symmetrically distributed external loads the stress function φ can be 

represented in the form 

φ = ψ − νz2 

2(1 + ν) ∇2ψ (2.4.70) 

where ψ is a solution of the biharmonic equation ∇4ψ =0. Observe that if z is very small, (the condition 

of a thin plate), then equation (2.4.70) gives the approximation φ ≈ ψ. Under these conditions, we obtain 

the approximate solution by using only the compatibility equation (2.4.65) together with the stress function 

defined by equations (2.4.67) with V =0. Note that the solution we obtain from equation (2.4.69) does not 

satisfy all the compatibility equations, however, it does give an excellent first approximation to the solution 

in the case where the plate is very thin. 

In general, for plane strain or plane stress problems, the equation (2.4.68) or (2.4.69) must be solved for 

the Airy stress function φ which is defined over some region R. In addition to specifying a region of the x, y 

plane, there are certain boundary conditions which must be satisfied. The boundary conditions specified for 

the stress will translate through the equations (2.4.67) to boundary conditions being specified for φ. In the 

special case where there are no body forces, both the problems for plane stress and plane strain are governed 

by the biharmonic differential equation with appropriate boundary conditions.

EXAMPLE 2.4-4 Assume there exist a state of plane strain with zero body forces. For F11,F12,F22 

constants, consider the function defined by 

φ = φ(x, y) = 1 

F22 x 

2 

2 − 2F12 xy + F11 y 2 . 

This function is an Airy stress function because it satisfies the biharmonic equation ∇4φ = 0. The resulting 

stress field is 

σxx = ∂2φ = F11 σyy = 

∂y2 ∂2φ = F22 σxy = − 

∂x2 ∂2φ = F12. 

∂x∂y 

This example, corresponds to stresses on an infinite flat plate and illustrates a situation where all the stress 

components are constants for all values of x and y. In this case, we have σzz = ν(F11+F22). The corresponding 

strain field is obtained from the constitutive equations. We find these strains are 

exx = 1+ν 

E [(1 − ν)F11 − νF22] eyy = 1+ν 

E [(1 − ν)F22 − νF11] exy = 1+ν 

E F12. 

The displacement field is found to be 

u = u(x, y) = 1+ν 

E [(1 − ν)F11 

 

1+ν 

− νF22] x + F12y + c1y + c2 

E 

v = v(x, y) = 1+ν 

E [(1 − ν)F22 

 

1+ν 

− νF11] y + F12x − c1x + c3, 

E 

with c1,c2,c3 constants, and is obtained by integrating the strain displacement equations given in Exercise 

2.3, problem 2. 

EXAMPLE 2.4-5. A special case from the previous example is obtained by setting F22 = F12 =0. 

This is the situation of an infinite plate with only tension in the x−direction. In this special case we have 

φ = 1 

2 F11y 2 . Changing to polar coordinates we write 

φ = φ(r, θ) = F11 

2 r2 sin 2 θ = F11 

4 r2 (1 − cos 2θ). 

The Exercise 2.4, problem 20, suggests we utilize the Airy equations in polar coordinates and calculate the 

stresses 

σrr = 1 ∂φ 1 

+ 

r ∂r r2 ∂2φ ∂θ2 = F11 cos 2 θ = F11 

2 

σθθ = ∂2 φ 

σrθ = 1 

r 2 

∂r2 = F11 sin 2 θ = F11 

2 

∂φ 

∂θ 

− 1 

r 

∂2φ ∂r∂θ 

= − F11 

2 

(1 − cos 2θ) 

sin 2θ. 

(1 + cos 2θ) 

267

268 

EXAMPLE 2.4-6. We now consider an infinite plate with a circular hole x2 + y2 = a2 which is traction 

free. Assume the plate has boundary conditions at infinity defined by σxx = F11, σyy =0, σxy =0. Find 

the stress field. 

Solution: 

The traction boundary condition at r = a is ti = σminm or 

t1 = σ11n1 + σ12n2 and t2 = σ12n1 + σ22n2. 

For polar coordinates we have n1 = nr =1,n2 = nθ = 0 and so the traction free boundary conditions at 

the surface of the hole are written σrr|r=a =0 and σrθ|r=a =0. The results from the previous example 

are used as the boundary conditions at infinity. 

Our problem is now to solve for the Airy stress function φ = φ(r, θ) which is a solution of the biharmonic 

equation. The previous example 2.4-5 and the form of the boundary conditions at infinity suggests that we 

assume a solution to the biharmonic equation which has the form φ = φ(r, θ) =f1(r)+f2(r)cos2θ, where 

f1,f2 are unknown functions to be determined. Substituting the assumed solution into the biharmonic 

equation produces the equation 

 

2 d 1 d 

+ f 

dr2 r dr 

′′ 

1 + 1 

r f ′ 2 d 1 d 4 

1 + + − 

dr2 r dr r2 

f ′′ 

2 + 1 

r f ′ 2 − 4 f2 

r2 

cos 2θ =0. 

We therefore require that f1,f2 be chosen to satisfy the equations 

 

2 d 1 d 

+ f 

dr2 r dr 

′′ 1 

1 + 

r f ′ 2 d 1 d 

1 =0 + 

dr2 r dr 

or r 4 f (iv) 

1 

+2r3 f ′′′ 

1 − r2 f ′′ 

1 + rf ′ 1 =0 

r 4 f (iv) 

2 

4 

− 

r2 

f ′′ 1 

2 + 

r f ′ f2 

2 − 4 

r2 

=0 

+2r3 f ′′′ 

2 − 9r2 f ′′ 

2 +9rf ′ 2 =0 

These equations are Cauchy type equations. Their solutions are obtained by assuming a solution of the form 

f1 = r λ and f2 = r m and then solving for the constants λ and m. We find the general solutions of the above 

equations are 

f1 = c1r 2 ln r + c2r 2 + c3 ln r + c4 and f2 = c5r 2 + c6r 4 + c7 

+ c8. 

r2 The constants ci,i =1,...,8 are now determined from the boundary conditions. The constant c4 can be 

arbitrary since the derivative of a constant is zero. The remaining constants are determined from the stress 

conditions. Using the results from Exercise 2.4, problem 20, we calculate the stresses 

σrr = c1(1 + 2 ln r)+2c2 + c3 

r 

σθθ = c1(3 + 2 ln r)+2c2 − c3 

r 

 

σrθ = 

2c5 +6c6r 2 − 6 c7 c8 

− 2 

r4 r2 

− 2c5 +6 2 c7 

+4c8 

r4 r2 

cos 2θ 

 

+ 2c5 +12c6r 2 2 +6 c7 

r4 

cos 2θ 

 

sin 2θ.

The stresses are to remain bounded for all values of r and consequently we require c1 and c6 to be zero 

to avoid infinite stresses for large values of r. The stress σrr|r=a =0requiresthat 

2c2 + c3 

a2 =0 and 2c5 +6 c7 

+4c8 =0. 

a4 a2 The stress σrθ|r=a =0requiresthat 

2c5 − 6 c7 c8 

− 2 =0. 

a4 a2 In the limit as r →∞we require that the stresses must satisfy the boundary conditions from the previous 

example 2.4-5. This leads to the equations 2c2 = F11 

2 and 2c5 = − F11 

. Solving the above system of equations 

2 

produces the Airy stress function 

φ = φ(r, θ) = F11 

4 

and the corresponding stress field is 

F11 

+ 

4 r2 − a2 

2 F11 

 

F11a 

ln r + c4 + 

2 

F11 

− 

2 4 r2 − 

σrr = F11 

2 

σrθ = − F11 

σθθ = F11 

2 

2 

 

1 − a2 

 

+ F11 

 

2 

r2 1+3 a4 

r r2 

1 − 3 a4 

+2a2 

r4 r2 

sin 2θ 

 

1+ a2 

r2 

− F11 

 

1+3 

2 

a4 

r4 

cos 2θ. 

4 − 4 a2 

 

cos 2θ 

F11a 4 

4r 2 

 

cos 2θ 

There is a maximum stress σθθ =3F11 at θ = π/2, 3π/2 and a minimum stress σθθ = −F11 at θ =0,π. 

The effect of the circular hole has been to magnify the applied stress. The factor of 3 is known as a stress 

concentration factor. In general, sharp corners and unusually shaped boundaries produce much higher stress 

concentration factors than rounded boundaries. 

EXAMPLE 2.4-7. Consider an infinite cylindrical tube, with inner radius R1 and the outer radius R0, 

which is subjected to an internal pressure P1 andanexternalpressureP0asillustrated in the figure 2.4-7. 

Find the stress and displacement fields. 

Solution: Let ur,uθ,uz denote the displacement field. We assume that uθ =0anduz = 0 since the 

cylindrical surface r equal to a constant does not move in the θ or z directions. The displacement ur = ur(r) 

is assumed to depend only upon the radial distance r. Under these conditions the Navier equations become 

(λ +2µ) d 

 

1 d 

dr r dr (rur) 

 

=0. 

r c2 

This equation has the solution ur = c1 + 

2 r 

and the strain components are found from the relations 

err = dur 

dr , eθθ = ur 

r , ezz = erθ = erz = ezθ =0. 

The stresses are determined from Hooke’s law (the constitutive equations) and we write 

σij = λδijΘ+2µeij, 

269

270 

where 

Θ= ∂ur 

∂r 

is the dilatation. These stresses are found to be 

+ ur 

r 

= 1 

r 

∂ 

∂r (rur) 

σrr =(λ + µ)c1 − 2µ 

r 2 c2 σθθ =(λ + µ)c1 + 2µ 

r 2 c2 σzz = λc1 σrθ = σrz = σzθ =0. 

We now apply the boundary conditions 

 

σrr|r=R1nr = − 

(λ + µ)c1 − 2µ 

R2 c2 

1 

Solving for the constants c1 and c2 we find 

This produces the displacement field 

ur = 

and stress fields 

 

 

=+P1 and σrr|r=R0nr = 

c1 = R2 1P1 − R2 0P0 (λ + µ)(R2 0 − R2 1 ) , c2 = R2 1R2 0 (P1 − P0) 

2µ(R2 0 − R2 1 

(λ + µ)c1 − 2µ 

R2 c2 

0 

) . 

 

= −P0. 

R2 1P1 

2(R2 0 − R2 1 ) 

 

r 

λ + µ + R2 

0 R 

− 

µr 

2 0P0 

2(R2 0 − R2 1 ) 

 

r 

λ + µ + R2 

1 

, uθ =0, uz =0, 

µr 

σrr = R2 1P1 R2 0 − R2 

1 − 

1 

R2 0 

r2 σθθ = R2 1P1 R2 0 − R2 1 

σzz = 

 

− R2 0 P0 

 

1 − R2 

1 

R2 0 − R2 1 r2 

1+ R2 0 

r2 

− R2 0P0 R2 0 − R2 

1+ 

1 

R2 1 

r2 

λ 

λ + µ 

 

2 R1P1 − R2 0P0 σrz = σzθ = σrθ =0 

R 2 0 − R2 1 

EXAMPLE 2.4-8. By making simplifying assumptions the Navier equations can be reduced to a more 

tractable form. For example, we can reduce the Navier equations to a one dimensional problem by making 

the following assumptions 

1. Cartesian coordinates x1 = x, x2 = y, x3 = z 

2. u1 = u1(x, t), u2 = u3 =0. 

3. There are no body forces. 

4. Initial conditions of u1(x, 0) = 0 and 

∂u1(x, 0) 

=0 

∂t 

5. Boundary conditions of the displacement type u1(0,t)=f(t), 

where f(t) is a specified function. These assumptions reduce the Navier equations to the single one dimensional 

wave equation 

∂2u1 ∂t2 = α2 ∂2u1 ∂x2 , α2 λ +2µ 

= . 

ρ 

The solution of this equation is 

 

f(t − x/α), 

u1(x, t) = 

0, 

x ≤ αt 

x > αt .

The solution represents a longitudinal elastic wave propagating in the x−direction with speed α. The stress 

wave associated with this displacement is determined from the constitutive equations. We find 

This produces the stress wave 

σxx = 

σxx =(λ + µ)exx =(λ + µ) ∂u1 

∂x . 

− (λ+µ) 

α f ′ (t − x/α), x ≤ αt 

0, x > αt . 

Here there is a discontinuity in the stress wave front at x = αt. 

Summary of Basic Equations of Elasticity 

The equilibrium equations for a continuum have been shown to have the form σ ij 

,j + ϱbi =0, where 

bi are the body forces per unit mass and σij is the stress tensor. In addition to the above equations we 

have the constitutive equations σij = λekkδij +2µeij which is a generalized Hooke’s law relating stress to 

strain for a linear elastic isotropic material. The strain tensor is related to the displacement field ui by 

the strain equations eij = 1 

2 (ui,j + uj,i) . These equations can be combined to obtain the Navier equations 

µui,jj +(λ + µ)uj,ji + ϱbi =0. 

The above equations must be satisfied at all interior points of the material body. A boundary value 

problem results when conditions on the displacement of the boundary are specified. That is, the Navier 

equations must be solved subject to the prescribed displacement boundary conditions. If conditions on 

the stress at the boundary are specified, then these prescribed stresses are called surface tractions and 

must satisfy the relations t i (n) = σ ij nj, where ni is a unit outward normal vector to the boundary. For 

surface tractions, we need to use the compatibility equations combined with the constitutive equations and 

equilibrium equations. This gives rise to the Beltrami-Michell equations of compatibility 

σij,kk + 1 

1+ν σkk,ij + ϱ(bi,j + bj,i)+ ν 

1 − ν ϱbk,k =0. 

Here we must solve for the stress components throughout the continuum where the above equations hold 

subject to the surface traction boundary conditions. Note that if an elasticity problem is formed in terms of 

the displacement functions, then the compatibility equations can be ignored. 

For mixed boundary value problems we must solve a system of equations consisting of the equilibrium 

equations, constitutive equations, and strain displacement equations. We must solve these equations subject 

to conditions where the displacements ui are prescribed on some portion(s) of the boundary and stresses are 

prescribed on the remaining portion(s) of the boundary. Mixed boundary value problems are more difficult 

to solve. 

For elastodynamic problems, the equilibrium equations are replaced by equations of motion. In this 

case we need a set of initial conditions as well as boundary conditions before attempting to solve our basic 

system of equations. 

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272 

EXERCISE 2.4 

◮ 1. Verify the generalized Hooke’s law constitutive equations for hexagonal materials. 

In the following problems the Young’s modulus E, Poisson’s ratio ν, the shear modulus or modulus 

of rigidity µ (sometimes denoted by G in Engineering texts), Lame’s constant λ and the bulk modulus of 

elasticity k are assumed to satisfy the equations (2.4.19), (2.4.24) and (2.4.25). Show that these relations 

imply the additional relations given in the problems 2 through 6. 

◮ 2. 

◮ 3. 

◮ 4. 

◮ 5. 

◮ 6. 

µ(3λ +2µ) 

E = 

µ + λ 

λ(1 + ν)(1 − 2ν) 

E = 

ν 

3k − E 

ν = 

6k 

λ 

ν = 

2(µ + λ) 

ν = 

ν = 

E = 

9k(k − λ) 

3k − λ 

E =2µ(1 + ν) 

(E + λ) 2 +8λ 2 − (E + λ) 

3k − 2µ 

2(µ +3k) 

 

(E + λ) 2 +8λ2 +(E +3λ) 

k = 

6 

2µ +3λ 

k = 

3 

3(k − λ) 

µ = 

2 

λ(1 − 2ν) 

µ = 

2ν 

λ = 3kν 

1+ν 

µ(2µ − E) 

λ = 

E − 3µ 

3k(1 − 2ν) 

µ = 

2(1 + ν) 

µ = 3Ek 

9k − E 

4λ 

E 

k = 

3(1 − 2ν) 

µE 

k = 

3(3µ − E) 

µ = 

µ = 

3k − 2µ 

λ = 

3 

3k(3k − E) 

λ = 

9k − E 

E = 9kµ 

µ +3k 

E =3(1− 2ν)k 

E − 2µ 

ν = 

2µ 

ν = 

λ 

3k − λ 

2µ(1 + ν) 

k = 

3(1 − 2ν) 

λ(1 + ν) 

k = 

3ν 

(E + λ) 2 +8λ 2 +(E − 3λ) 

E 

2(1 + ν) 

4 

νE 

λ = 

(1 + ν)(1 − 2ν) 

λ = 2µν 

1 − 2ν 

◮ 7. The previous exercises 2 through 6 imply that the generalized Hooke’s law 

σij =2µeij + λδijekk 

is expressible in a variety of forms. From the set of constants (µ,λ,ν,E,k) we can select any two constants 

and then express Hooke’s law in terms of these constants. 

(a) Express the above Hooke’s law in terms of the constants E and ν. 

(b) Express the above Hooke’s law in terms of the constants k and E. 

(c) Express the above Hooke’s law in terms of physical components. Hint: The quantity ekk is an invariant 

hence all you need to know is how second order tensors are represented in terms of physical components. 

See also problems 10,11,12.

◮ 8. Verify the equations defining the stress for plane strain in Cartesian coordinates are 

E 

σxx = 

(1 + ν)(1 − 2ν) [(1 − ν)exx + νeyy] 

E 

σyy = 

(1 + ν)(1 − 2ν) [(1 − ν)eyy + νexx] 

Eν 

σzz = 

(1 + ν)(1 − 2ν) [exx + eyy] 

σxy = E 

1+ν exy 

σyz = σxz =0 

◮ 9. Verify the equations defining the stress for plane strain in polar coordinates are 

E 

σrr = 

(1 + ν)(1 − 2ν) [(1 − ν)err + νeθθ] 

E 

σθθ = 

(1 + ν)(1 − 2ν) [(1 − ν)eθθ + νerr] 

νE 

σzz = 

(1 + ν)(1 − 2ν) [err + eθθ] 

σrθ = E 

1+ν erθ 

σrz = σθz =0 

◮ 10. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and 

stress in terms of strain, in Cartesian coordinates. Express your results using the parameters ν and E. 

(Assume a linear elastic, homogeneous, isotropic material.) 

◮ 11. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and 

stress in terms of strain, in cylindrical coordinates. Express your results using the parameters ν and E. 

(Assume a linear elastic, homogeneous, isotropic material.) 

◮ 12. Write out the independent components of Hooke’s generalized law for strain in terms of stress, and 

stress in terms of strain in spherical coordinates. Express your results using the parameters ν and E. (Assume 

a linear elastic, homogeneous, isotropic material.) 

◮ 13. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in 

Cartesian coordinates. Verify the equilibrium equations are 


∂σxx 

∂x 

∂σyx 

∂x 

+ ∂σxy 

∂y + ϱbx =0 

+ ∂σyy 

∂y + ϱby =0 

∂σzz 

∂z + ϱbz =0 

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274 

◮ 14 . For a linear elastic, homogeneous, isotropic material assume there exists a state of plane strain in 

polar coordinates. Verify the equilibrium equations are 


∂σrr 

∂r 

1 ∂σrθ 1 

+ + 

r ∂θ r (σrr − σθθ)+ϱbr =0 

∂σrθ 

∂r 

1 ∂σθθ 2 

+ + 

r ∂θ r σrθ + ϱbθ =0 

∂σzz 

∂z + ϱbz =0 

◮ 15. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in 

Cartesian coordinates. Verify the equilibrium equations are 

∂σxx 

∂x 

∂σyx 

∂x 

+ ∂σxy 

∂y + ϱbx =0 

+ ∂σyy 

∂y + ϱby =0 

◮ 16. Determine the compatibility equations in terms of the Airy stress function φ when there exists a state 

of plane stress. Assume the body forces are derivable from a potential function V. 

◮ 17. For a linear elastic, homogeneous, isotropic material assume there exists a state of plane stress in 

polar coordinates. Verify the equilibrium equations are 

∂σrr 

∂r 

1 ∂σrθ 1 

+ + 

r ∂θ r (σrr − σθθ)+ϱbr =0 

∂σrθ 

∂r 

1 ∂σθθ 2 

+ + 

r ∂θ r σrθ + ϱbθ =0

◮ 18. Figure 2.4-4 illustrates the state of equilibrium on an element in polar coordinates assumed to be of 

unit length in the z-direction. Verify the stresses given in the figure and then sum the forces in the r and θ 

directions to derive the same equilibrium laws developed in the previous exercise. 

Figure 2.4-4. Polar element in equilibrium. 

Hint: Resolve the stresses into components in the r and θ directions. Use the results that sin dθ 

2 

cos dθ 

2 

≈ dθ 

2 and 

≈ 1 for small values of dθ. Sum forces and then divide by rdr dθ and take the limit as dr → 0and 

dθ → 0. 

◮ 19. Express each of the physical components of plane stress in polar coordinates, σrr, σθθ, andσrθ 

in terms of the physical components of stress in Cartesian coordinates σxx, σyy, σxy. Hint: Consider the 

∂x 

transformation law σij = σab 

a 

∂xi ∂xb j . 

∂x 

◮ 20. Use the results from problem 19 and assume the stresses are derivable from the relations 

σxx = V + ∂2φ ∂y2 , σxy = − ∂2φ ∂x∂y , σyy = V + ∂2φ ∂x2 where V is a potential function and φ is the Airy stress function. Show that upon changing to polar 

coordinates the Airy equations for stress become 

σrr = V + 1 ∂φ 1 

+ 

r ∂r r2 ∂2φ ∂θ2 , σrθ = 1 

r2 ∂φ 

∂θ 

− 1 

r 

∂2φ ∂r∂θ , σθθ = V + ∂2φ . 

∂r2 ◮ 21. Verify that the Airy stress equations in polar coordinates, given in problem 20, satisfy the equilibrium 

equations in polar coordinates derived in problem 17. 

275

276 

◮ 22. In Cartesian coordinates show that the traction boundary conditions, equations (2.3.11), can be 

written in terms of the constants λ and µ as 

 

 

∂u1 ∂u1 ∂u2 

T1 = λn1ekk + µ 2n1 + n2 + 

∂x1 ∂x2 ∂x1 

∂u1 ∂u3 

+ n3 + 

∂x3 ∂x1 

 

∂u2 ∂u1 

T2 = λn2ekk + µ n1 + 

∂x1 ∂x2 

 

∂u2 ∂u2 ∂u3 

+2n2 + n3 + 

∂x2 ∂x3 ∂x2 

 

∂u3 ∂u1 

T3 = λn3ekk + µ n1 + 

∂x1 ∂x3 

∂u3 ∂u2 

+ n2 + 

∂x2 ∂x3 

∂u3 

+2n3 

∂x3 

where (n1,n2,n3) are the direction cosines of the unit normal to the surface, u1,u2,u3 are the components 

of the displacements and T1,T2,T3 are the surface tractions. 

◮ 23. Consider an infinite plane subject to tension in the x−direction only. Assume a state of plane strain 

and let σxx = T with σxy = σyy =0. Find the strain components exx, eyy and exy. Also find the displacement 

field u = u(x, y) andv = v(x, y). 

◮ 24. Consider an infinite plane subject to tension in the y-direction only. Assume a state of plane strain 

and let σyy = T with σxx = σxy =0. Find the strain components exx, eyy and exy. Also find the displacement 

field u = u(x, y) andv = v(x, y). 

◮ 25. Consider an infinite plane subject to tension in both the x and y directions. Assume a state of plane 

strain and let σxx = T , σyy = T and σxy =0. Find the strain components exx,eyy and exy. Also find the 

displacement field u = u(x, y) andv = v(x, y). 

◮ 26. An infinite cylindrical rod of radius R0 has an external pressure P0 as illustrated in figure 2.5-5. Find 

the stress and displacement fields. 

Figure 2.4-5. External pressure on a rod.

Figure 2.4-6. Internal pressure on circular hole. 

Figure 2.4-7. Tube with internal and external pressure. 

◮ 27. An infinite plane has a circular hole of radius R1 with an internal pressure P1 as illustrated in the 

figure 2.4-6. Find the stress and displacement fields. 

◮ 28. A tube of inner radius R1 and outer radius R0 has an internal pressure of P1 andanexternalpressure 

of P0 as illustrated in the figure 2.4-7. Verify the stress and displacement fields derived in example 2.4-7. 

◮ 29. Use Cartesian tensors and combine the equations of equilibrium σij,j + ϱbi =0, Hooke’s law σij = 

λekkδij +2µeij and the strain tensor eij = 1 

2 (ui,j + uj,i) and derive the Navier equations of equilibrium 

where Θ = e11 + e22 + e33 is the dilatation. 

σij,j + ϱbi =(λ + µ) ∂Θ 

∂x i + µ ∂2 ui 

∂x k ∂x k + ϱbi =0, 

◮ 30. Show the Navier equations in problem 29 can be written in the tensor form 

or the vector form 

µui,jj +(λ + µ)uj,ji + ϱbi =0 

µ∇ 2 u +(λ + µ)∇ (∇·u)+ϱ b = 0. 

277

278 

◮ 31. Show that in an orthogonal coordinate system the components of ∇(∇·u) can be expressed in terms 

of physical components by the relation 

[∇ (∇·u)] i = 1 ∂ 

∂xi 

1 

 

∂(h2h3u(1)) 

∂x1 + ∂(h1h3u(2)) 

∂x2 + ∂(h1h2u(3)) 

∂x3 

hi 

h1h2h3 

◮ 32. Show that in orthogonal coordinates the components of ∇ 2 u can be written 

∇ 2 u 

i = gjk ui,jk = Ai 

and in terms of physical components one can write 

3 1 

hiA(i) = 

h 

j=1 

2 ⎡ 

⎣ 

j 

∂2 (hiu(i)) 

∂xj∂xj 3 

 

m ∂(hmu(m)) 

− 2 

ij ∂x 

m=1 

j 

3 

 

m ∂(hiu(i)) 

− 

jj ∂x 

m=1 

m 

3 

 

∂ 

− hmu(m) 

∂xj 3 

3 

⎤ 

m m p 

m p 

− 

− 

⎦ 

ij ip jj jp ij 

m=1 

◮ 33. Use the results in problem 32 to show in Cartesian coordinates the physical components of [∇ 2 u]i = Ai 


p=1 

2 

∇ u · ê1 = A(1) = ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 2 

∇ u · ê2 = A(2) = ∂2v ∂x2 + ∂2v ∂y2 + ∂2v ∂z2 2 

∇ u · ê3 = A(3) = ∂2w ∂x2 + ∂2w ∂y2 + ∂2w ∂z2 where (u, v, w) are the components of the displacement vector u. 

◮ 34. Use the results in problem 32 to show in cylindrical coordinates the physical components of [∇ 2 u]i = Ai 


p=1 

2 

∇ u · êr = A(1) = ∇ 2 ur − 1 

r2 ur − 2 

r2 ∂uθ 

∂θ 

2 

∇ u · êθ = A(2) = ∇ 2 uθ + 2 

r2 ∂ur 1 

− uθ 

∂θ r2 2 

∇ u · êz = A(3) = ∇ 2 uz 

∂α 1 

+ 

∂r r2 ∂2α ∂θ2 + ∂2α ∂z2 where ur,uθ,uz are the physical components of u and ∇ 2 α = ∂2α 1 

+ 

∂r2 r 

◮ 35. Use the results in problem 32 to show in spherical coordinates the physical components of [∇2u]i = Ai 


2 

∇ u · êρ = A(1) = ∇ 2 uρ − 2 

ρ2 uρ − 2 

ρ2 ∂uθ 

∂θ 

2 

∇ u · êθ = A(2) = ∇ 2 uθ + 2 

ρ2 ∂uρ 

∂θ − 

1 

ρ2 sin θ uθ − 2cosθ 

ρ2 sin 2 θ 

2 

∇ u · êφ = A(3) = ∇ 2 uφ − 

1 

ρ 2 sin 2 θ uφ + 

where uρ,uθ,uφ are the physical components of u and where 

∇ 2 α = ∂2α 2 ∂α 1 

+ + 

∂ρ2 ρ ∂ρ ρ2 ∂2α cot θ 

+ 

∂θ2 ρ2 ∂α 

∂θ + 

2cotθ 

− 

ρ2 uθ 

2 

− 

ρ2 ∂uφ 

sin θ ∂φ 

∂uθ 

∂φ 

2 

ρ2 ∂uρ 2cosθ 

+ 

sin θ ∂φ ρ2 sin 2 ∂uθ 

θ ∂φ 

1 

ρ2 sin 2 ∂ 

θ 

2α ∂φ2

◮ 36. Combine the results from problems 30,31,32 and 33 and write the Navier equations of equilibrium 

in Cartesian coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical 

components and then use these results, together with the results from Exercise 2.3, problems 2 and 14, to 

derive the Navier equations. 


in cylindrical coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical 




in spherical coordinates. Alternatively, write the stress-strain relations (2.4.29(b)) in terms of physical 



◮ 39. Assume ϱ b = −grad V and let φ denote the Airy stress function defined by 

σxx =V + ∂2 φ 

∂y 2 

σyy =V + ∂2 φ 

∂x 2 

σxy = − ∂2φ ∂x∂y 

(a) Show that for conditions of plane strain the equilibrium equations in two dimensions are satisfied by the 

above definitions. (b) Express the compatibility equation 

in terms of φ and V and show that 

∂2exx ∂y2 + ∂2eyy ∂x2 =2∂2 exy 

∂x∂y 

∇ 4 1 − 2ν 

φ + 

1 − ν ∇2V =0. 

◮ 40. Consider the case where the body forces are conservative and derivable from a scalar potential function 

such that ϱbi = −V,i. Show that under conditions of plane strain in rectangular Cartesian coordinates the 

compatibility equation e11,22 + e22,11 =2e12,12 can be reduced to the form ∇ 2 σii = 1 

1 − ν ∇2 V ,i =1, 2 

involving the stresses and the potential. Hint: Differentiate the equilibrium equations. 

◮ 41. Use the relation σ i j =2µe i j + λe m mδ i j and solve for the strain in terms of the stress. 

◮ 42. Derive the equation (2.4.26) from the equation (2.4.23). 

◮ 43. In two dimensions assume that the body forces are derivable from a potential function V and 

ϱbi = −gijV ,j. Also assume that the stress is derivable from the Airy stress function and the potential 

function by employing the relations σ ij = ɛ im ɛ jn um,n + g ij V i,j,m,n =1, 2whereum = φ ,m and 

ɛ pq is the two dimensional epsilon permutation symbol and all indices have the range 1,2. 

(a) Show that ɛimɛjn (φm) ,nj =0. 

(b) Show that σ ij 

,j = −ϱbi . 

(c) Verify the stress laws for cylindrical and Cartesian coordinates given in problem 20 by using the above 

expression for σ ij . Hint: Expand the contravariant derivative and convert all terms to physical compo- 

nents. Also recall that ɛ ij = 1 

√ g e ij . 

279

280 

◮ 44. Consider a material with body forces per unit volume ρF i ,i=1, 2, 3 and surface tractions denoted by 

σ r = σ rj nj, where nj is a unit surface normal. Further, let δui denote a small displacement vector associated 

with a small variation in the strain δeij. 

 

(a) Show the work done during a small variation in strain is δW = δWB + δWS where δWB = ρF 

V 

i 

δui dτ 

is a volume integral representing the work done by the body forces and δWS = σ 

S 

r δur dS is a surface 

integral representing the work done by the surface forces. 

(b) Using the Gauss divergence theorem show that the work done can be represented as 

δW = 1 

 

2 

c ijmn δ[emneij] dτ or W = 1 

 

2 

σ ij eij dτ. 

V 

The scalar quantity 1 

2 σij eij is called the strain energy density or strain energy per unit volume. 

Hint: Interchange subscripts, add terms and calculate 2W = 

V σij [δui,j + δuj,i] dτ. 

◮ 45. Consider a spherical shell subjected to an internal pressure pi and external pressure po. Letadenote the inner radius and b the outer radius of the spherical shell. Find the displacement and stress fields in 

spherical coordinates (ρ, θ, φ). 

Hint: Assume symmetry in the θ and φ directions and let the physical components of displacements satisfy 

the relations uρ = uρ(ρ), uθ = uφ =0. 

◮ 46. (a) Verify the average normal stress is proportional to the dilatation, where the proportionality 

constant is the bulk modulus of elasticity. i.e. Show that 1 

3σi E 

i = 1−2ν 

V 

1 

3eii = keii where k is the bulk modulus 

of elasticity. 

(b) Define the quantities of strain deviation and stress deviation in terms of the average normal stress 

s = 1 

3σi 1 

i and average cubic dilatation e = 3eii as follows 

strain deviator ε i j = eij − eδi j 

stress deviator s i j = σ i j − sδ i j 

Show that zero results when a contraction is performed on the stress and strain deviators. (The above 

definitions are used to split the strain tensor into two parts. One part represents pure dilatation and 

the other part represents pure distortion.) 

(c) Show that (1 − 2ν)s = Ee or s =(3λ +2µ)e 

(d) Express Hooke’s law in terms of the strain and stress deviator and show 

which simplifies to s i j =2µε i j. 

E(ε i j + eδ i j)=(1+ν)s i j +(1− 2ν)sδ i j 

◮ 47. Show the strain energy density (problem 44) can be written in terms of the stress and strain deviators 

(problem 46) and 

W = 1 

 

σ 

2 V 

ij eij dτ = 1 

 

(3se + s 

2 V 

ij εij) dτ 

and from Hooke’s law 

W = 3 

 

((3λ +2µ)e 

2 V 

2 + 2µ 

3 εijεij) dτ.

◮ 48. Find the stress σrr,σrθ and σθθ in an infinite plate with a small circular hole, which is traction free, 

when the plate is subjected to a pure shearing force F12. Determine the maximum stress. 

◮ 49. Show that in terms of E and ν 

C1111 = 

E(1 − ν) 

(1 + ν)(1 − 2ν) 

C1122 = 

◮ 50. Show that in Cartesian coordinates the quantity 

Eν 

(1 + ν)(1 − 2ν) 

C1212 = 

S = σxxσyy + σyyσzz + σzzσxx − (σxy) 2 − (σyz) 2 − (σxz) 2 

E 

2(1 + ν) 

is a stress invariant. Hint: First verify that in tensor form S = 1 

2 (σiiσjj − σijσij). 

◮ 51. Show that in Cartesian coordinates for a state of plane strain where the displacements are given by 

u = u(x, y),v = v(x, y) andw = 0, the stress components must satisfy the equations 

∂σxx 

∂x 

∂σyx 

∂x 

+ ∂σxy 

∂y + ϱbx =0 

+ ∂σyy 

∂y + ϱby =0 

∇ 2 (σxx + σyy) = −ϱ 

1 − ν 

∂bx 

∂x 

 

∂by 

+ 

∂y 

◮ 52. Show that in Cartesian coordinates for a state of plane stress where σxx = σxx(x, y), σyy = σyy(x, y), 

σxy = σxy(x, y) andσxz = σyz = σzz = 0 the stress components must satisfy 

∂σxx 

∂x 

∂σyx 

∂x 

+ ∂σxy 

∂y + ϱbx =0 

+ ∂σyy 

∂y + ϱby =0 

∇ 2 (σxx + σyy) =− ϱ(ν +1) 

∂bx 

∂x 

 

∂by 

+ 

∂y 

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282 

§2.5 CONTINUUM MECHANICS (FLUIDS) 

Let us consider a fluid medium and use Cartesian tensors to derive the mathematical equations that 

describe how a fluid behaves. A fluid continuum, like a solid continuum, is characterized by equations 

describing: 

1. Conservation of linear momentum 

2. Conservation of angular momentum σij = σji. 

3. Conservation of mass (continuity equation) 

∂ϱ ∂ϱ 

+ vi + ϱ 

∂t ∂xi 

∂vi 

∂xi 

σij,j + ϱbi = ϱ ˙vi 

(2.5.1) 

=0 or Dϱ 

Dt + ϱ∇· V =0. (2.5.2) 

In the above equations vi,i=1, 2, 3 is a velocity field, ϱ is the density of the fluid, σij is the stress tensor 

and bj is an external force per unit mass. In the cgs system of units of measurement, the above quantities 

have dimensions 

[˙vj] =cm/sec 2 , [bj] =dynes/g, [σij] =dyne/cm 2 , [ϱ] =g/cm 3 . (2.5.3) 

The displacement field ui,i =1, 2, 3 can be represented in terms of the velocity field vi,i =1, 2, 3, by 

the relation 

t 

ui = vi dt. (2.5.4) 

The strain tensor components of the medium can then be represented in terms of the velocity field as 

where 

eij = 1 

2 (ui,j + uj,i) = 

t 

0 

0 

1 

2 (vi,j + vj,i) dt = 

t 

0 

Dij dt, (2.5.5) 

Dij = 1 

2 (vi,j + vj,i) (2.5.6) 

is called the rate of deformation tensor , velocity strain tensor, orrate of strain tensor. 

Note the difference in the equations describing a solid continuum compared with those for a fluid 

continuum. In describing a solid continuum we were primarily interested in calculating the displacement 

field ui,i =1, 2, 3 when the continuum was subjected to external forces. In describing a fluid medium, we 

calculate the velocity field vi,i =1, 2, 3 when the continuum is subjected to external forces. We therefore 

replace the strain tensor relations by the velocity strain tensor relations in all future considerations concerning 

the study of fluid motion. 

Constitutive Equations for Fluids 

In addition to the above basic equations, we will need a set of constitutive equations which describe the 

material properties of the fluid. Toward this purpose consider an arbitrary point within the fluid medium 

and pass an imaginary plane through the point. The orientation of the plane is determined by a unit normal 

ni , i =1, 2, 3 to the planar surface. For a fluid at rest we wish to determine the stress vector t (n) 

i acting 

on the plane element passing through the selected point P. We desire to express t (n) 

i in terms of the stress 

tensor σij. The superscript (n) on the stress vector is to remind you that the stress acting on the planar 

element depends upon the orientation of the plane through the point.

is colinear with the normal vector to the surface passing through 

the selected point. It is also assumed that for fluid elements at rest, there are no shear forces acting on the 

planar element through an arbitrary point and therefore the stress tensor σij should be independent of the 

orientation of the plane. That is, we desire for the stress vector σij to be an isotropic tensor. This requires 

We make the assumption that t (n) 

i 

σij to have a specific form. To find this specific form we let σij denote the stress components in a general 

coordinate system x i , i =1, 2, 3andletσij denote the components of stress in a barred coordinate system 

x i ,i=1, 2, 3. Since σij is a tensor, it must satisfy the transformation law 

σmn = σij 

∂xi ∂xm ∂xj n , i,j,m,n=1, 2, 3. (2.5.7) 

∂x 

We desire for the stress tensor σij to be an invariant under an arbitrary rotation of axes. Consider 

therefore the special coordinate transformations illustrated in the figures 2.5-1(a) and (b). 

Figure 2.5-1. Coordinate transformations due to rotations 

For the transformation equations given in figure 2.5-1(a), the stress tensor in the barred system of 

coordinates is 

σ11 = σ22 σ21 = σ32 σ31 = σ12 

σ12 = σ23 σ22 = σ33 σ32 = σ13 

σ13 = σ21 σ23 = σ31 σ33 = σ11. 

(2.5.8) 

If σij is to be isotropic, we desire that σ11 = σ11, σ22 = σ22 and σ33 = σ33. If the equations (2.5.8) are 

to produce these results, we require that σ11, σ22 and σ33 must be equal. We denote these common values 

by (−p). In particular, the equations (2.5.8) show that if σ11 = σ11, σ22 = σ22 and σ33 = σ33, then we must 

require that σ11 = σ22 = σ33 = −p. If σ12 = σ12 and σ23 = σ23, then we also require that σ12 = σ23 = σ31. 

We note that if σ13 = σ13 and σ32 = σ32, then we require that σ21 = σ32 = σ13. If the equations (2.5.7) are 

expanded using the transformation given in figure 2.5-1(b), we obtain the additional requirements that 

σ11 = σ22 σ21 = −σ12 σ31 = σ32 

σ12 = −σ21 σ22 = σ11 σ32 = −σ31 

σ13 = σ23 σ23 = −σ13 σ33 = σ33. 

(2.5.9) 

283

284 

Analysis of these equations implies that if σij is to be isotropic, then σ21 = σ21 = −σ12 = −σ21 

or σ21 = 0 which implies σ12 = σ23 = σ31 = σ21 = σ32 = σ13 =0. (2.5.10) 

The above analysis demonstrates that if the stress tensor σij is to be isotropic, it must have the form 

Use the traction condition (2.3.11), and express the stress vector as 

σij = −pδij. (2.5.11) 

t (n) 

j = σijni = −pnj. (2.5.12) 

This equation is interpreted as representing the stress vector at a point on a surface with outward unit 

normal ni, wherepis the pressure (hydrostatic pressure) stress magnitude assumed to be positive. The 

negative sign in equation (2.5.12) denotes a compressive stress. 

Imagine a submerged object in a fluid medium. We further imagine the object to be covered with unit 

normal vectors emanating from each point on its surface. The equation (2.5.12) shows that the hydrostatic 

pressure always acts on the object in a compressive manner. A force results from the stress vector acting on 

the object. The direction of the force is opposite to the direction of the unit outward normal vectors. It is 

a compressive force at each point on the surface of the object. 

The above considerations were for a fluid at rest (hydrostatics). For a fluid in motion (hydrodynamics) 

a different set of assumptions must be made. Hydrodynamical experiments show that the shear stress 

components are not zero and so we assume a stress tensor having the form 

σij = −pδij + τij, i,j =1, 2, 3, (2.5.13) 

where τij is called the viscous stress tensor. Note that all real fluids are both viscous and compressible. 

Definition: (Viscous/inviscid fluid) If the viscous stress tensor 

τij is zero for all i, j, then the fluid is called an inviscid, nonviscous, 

ideal or perfect fluid. The fluid is called viscous when τij 

is different from zero. 

In these notes it is assumed that the equation (2.5.13) represents the basic form for constitutive equations 

describing fluid motion.

Viscosity 

Figure 2.5-2. Viscosity experiment. 

Most fluids are characterized by the fact that they cannot resist shearing stresses. That is, if you put a 

shearing stress on the fluid, the fluid gives way and flows. Consider the experiment illustrated in the figure 

2.5-2 which illustrates a fluid moving between two parallel plane surfaces. Let S denote the distance between 

the two planes. Now keep the lower surface fixed or stationary and move the upper surface parallel to the 

lower surface with a constant velocity V0. If you measure the force F required to maintain the constant 

velocity of the upper surface, you discover that the force F varies directly as the area A of the surface and 

the ratio V0/S. This is expressed in the form 

F 

A 

V0 

= µ∗ . (2.5.14) 

S 

The constant µ ∗ is a proportionality constant called the coefficient of viscosity. The viscosity usually depends 

upon temperature, but throughout our discussions we will assume the temperature is constant. A dimensional 

analysis of the equation (2.5.14) implies that the basic dimension of the viscosity is [µ ∗ ]=ML −1 T −1 . For 

example, [µ ∗ ]=gm/(cmsec) in the cgs system of units. The viscosity is usually measured in units of 

centipoise where one centipoise represents one-hundredth of a poise, where the unit of 1poise= 1gram 

per centimeter per second. The result of the above experiment shows that the stress is proportional to the 

change in velocity with change in distance or gradient of the velocity. 

Linear Viscous Fluids 

The above experiment with viscosity suggest that the viscous stress tensor τij is dependent upon both 

the gradient of the fluid velocity and the density of the fluid. 

In Cartesian coordinates, the simplest model suggested by the above experiment is that the viscous 

stress tensor τij is proportional to the velocity gradient vi,j and so we write 

τik = cikmpvm,p, (2.5.15) 

where cikmp is a proportionality constant which is dependent upon the fluid density. 

The viscous stress tensor must be independent of any reference frame, and hence we assume that the 

proportionality constants cikmp can be represented by an isotropic tensor. Recall that an isotropic tensor 

has the basic form 

cikmp = λ ∗ δikδmp + µ ∗ (δimδkp + δipδkm)+ν ∗ (δimδkp − δipδkm) (2.5.16) 

285

286 

where λ ∗ ,µ ∗ and ν ∗ are constants. Examining the results from equations (2.5.11) and (2.5.13) we find that if 

the viscous stress is symmetric, then τij = τji. This requires ν∗ be chosen as zero. Consequently, the viscous 

stress tensor reduces to the form 

τik = λ ∗ δikvp,p + µ ∗ (vk,i + vi,k). (2.5.17) 

The coefficient µ ∗ is called the first coefficient of viscosity and the coefficient λ ∗ is called the second coefficient 

of viscosity. Sometimes it is convenient to define 

ζ = λ ∗ + 2 

3 µ∗ 

(2.5.18) 

as “another second coefficient of viscosity,” or “bulk coefficient of viscosity.” The condition of zero bulk 

viscosity is known as Stokes hypothesis. Many fluids problems assume the Stoke’s hypothesis. This requires 

that the bulk coefficient be zero or very small. Under these circumstances the second coefficient of viscosity 

is related to the first coefficient of viscosity by the relation λ ∗ = − 2 

3 µ∗ . In the study of shock waves and 

acoustic waves the Stoke’s hypothesis is not applicable. 

There are many tables and empirical formulas where the viscosity of different types of fluids or gases 

can be obtained. For example, in the study of the kinetic theory of gases the viscosity can be calculated 

from the Sutherland formula µ ∗ 3/2 C1gT 

= where C1,C2 are constants for a specific gas. These constants 

T + C2 

can be found in certain tables. The quantity g is the gravitational constant and T is the temperature in 

degrees Rankine ( o R = 460 + o F ). Many other empirical formulas like the above exist. Also many graphs 

and tabular values of viscosity can be found. The table 5.1lists the approximate values of the viscosity of 

some selected fluids and gases. 

Table 5.1 

Viscosity of selected fluids and gases 

in units of gram 

cm−sec =Poise 

at Atmospheric Pressure. 

Substance 0 ◦ C 20 ◦ C 60 ◦ C 100 ◦ C 

Water 0.01798 0.01002 0.00469 0.00284 

Alcohol 0.01773 

Ethyl Alcohol 0.012 0.00592 

Glycol 0.199 0.0495 0.0199 

Mercury 0.017 0.0157 0.013 0.0100 

Air 1.708(10 −4 ) 2.175(10 −4 ) 

Helium 1.86(10 −4 ) 1.94(10 −4 ) 2.28(10 −4 ) 

Nitrogen 1.658(10 −4 ) 1.74(10 −4 ) 1.92(10 −4 ) 2.09(10 −4 ) 

The viscous stress tensor given in equation (2.5.17) may also be expressed in terms of the rate of 

deformation tensor defined by equation (2.5.6). This representation is 

τij = λ ∗ δijDkk +2µ ∗ Dij, (2.5.19) 

where 2Dij = vi,j + vj,i and Dkk = D11 + D22 + D33 = v1,1 + v2,2 + v3,3 = vi,i = Θ is the rate of change 

of the dilatation considered earlier. In Cartesian form, with velocity components u, v, w, the viscous stress

tensor components are 

τxx =(λ ∗ +2µ ∗ ) ∂u 

 

∂v ∂w 

+ λ∗ + 

∂x ∂y ∂z 

τyy =(λ ∗ +2µ ∗ ) ∂v 

+ λ∗ 

∂y 

τzz =(λ ∗ +2µ ∗ ) ∂w 

∂z 

+ λ∗ 

 

∂u ∂w 

+ 

∂x ∂z 

 

∂u ∂v 

+ 

∂x ∂y 

τyx = τxy =µ ∗ 

 

∂u ∂v 

+ 

∂y ∂x 

τzx = τxz =µ ∗ 

 

∂w ∂u 

+ 

∂x ∂z 

τzy = τyz =µ ∗ 

 

∂v ∂w 

+ 

∂z ∂y 

In cylindrical form, with velocity components vr,vθ,vz, the viscous stess tensor components are 

∗ ∂vr 

τrr =2µ 

where ∇· V = 1 

r 

∂r + λ∗∇· V 

τθθ =2µ ∗ 

 

1 ∂vθ vr 

+ + λ 

r ∂θ r 

∗ ∇·V 

∗ ∂vz 

τzz =2µ 

∂z + λ∗∇· V 

∂ 

∂r 

1 ∂vθ ∂vz 

(rvr)+ + 

r ∂θ ∂z 

τθr = τrθ =µ ∗ 

 

1 ∂vr ∂vθ vθ 

+ − 

r ∂θ ∂r r 

τrz = τzr =µ ∗ 

 

∂vr ∂vz 

+ 

∂z ∂r 

τzθ = τθz =µ ∗ 

 

1 ∂vz ∂vθ 

+ 

r ∂θ ∂z 

In spherical coordinates, with velocity components vρ,vθ,vφ, the viscous stress tensor components have the 

form 

∗ ∂vρ 

τρρ =2µ 

∂ρ + λ∗∇· V 

τθθ =2µ ∗ 

 

1 ∂vθ vρ 

+ + λ 

ρ ∂θ ρ 

∗ ∇· V 

τφφ =2µ ∗ 

 

1 ∂vφ vρ 

+ 

ρ sin θ ∂φ ρ + vθ 

 

cot θ 

+ λ 

ρ 

∗ ∇· V 

where ∇· V = 1 

ρ2 ∂ 

2 1 ∂ 

1 ∂vφ 

ρ vρ + (sin θvθ)+ 

∂ρ 

ρ sin θ ∂θ ρ sin θ ∂φ 

τρθ = τθρ =µ ∗ 

 

ρ ∂ 

 

vθ 

+ 

∂ρ ρ 

1 

 

∂vρ 

ρ ∂θ 

τφρ = τρφ =µ ∗ 

 

1 ∂vr ∂ vθ 

+ ρ 

ρ sin θ ∂φ ∂ρ ρ 

τθφ = τφθ =µ ∗ 

 

sin θ ∂ vφ 

ρ ∂θ sin θ 

 

 

+ 1 

ρ sin θ 

Note that the viscous stress tensor is a linear function of the rate of deformation tensor Dij. Such a 

fluidiscalledaNewtonian fluid. In cases where the viscous stress tensor is a nonlinear function of Dij the 

fluid is called non-Newtonian. 

Definition: (Newtonian Fluid) If the viscous stress tensor τij 

is expressible as a linear function of the rate of deformation tensor 

Dij, the fluid is called a Newtonian fluid. Otherwise, the fluid is 

called a non-Newtonian fluid. 

Important note: Do not assume an arbitrary form for the constitutive equations unless there is experimental 

evidence to support your assumption. A constitutive equation is a very important step in the 

modeling processes as it describes the material you are working with. One cannot arbitrarily assign a form 

to the viscous stress and expect the mathematical equations to describe the correct fluid behavior. The form 

of the viscous stress is an important part of the modeling process and by assigning different forms to the 

viscous stress tensor then various types of materials can be modeled. We restrict our study in these notes 

to Newtonian fluids. 

In Cartesian coordinates the rate of deformation-stress constitutive equations for a Newtonian fluid can 

be written as 

σij = −pδij + λ ∗ δijDkk +2µ ∗ Dij 

 

∂vθ 

∂φ 

(2.5.20) 

287

288 

which can also be written in the alternative form 

σij = −pδij + λ ∗ δijvk,k + µ ∗ (vi,j + vj,i) (2.5.21) 

involving the gradient of the velocity. 

Upon transforming from a Cartesian coordinate system yi ,i = 1, 2, 3 to a more general system of 

coordinates xi ,i=1, 2, 3, we write 

∂y 

σmn = σij 

i 

∂xm ∂yj n . (2.5.22) 

∂x 

Now using the divergence from equation (2.1.3) and substituting equation (2.5.21) into equation (2.5.22) we 

obtain a more general expression for the constitutive equation. Performing the indicated substitutions there 

results 

σmn = −pδij + λ ∗ δijv k ,k + µ∗ (vi,j + vj,i) ∂y i 

∂x m 

σmn = −pg mn + λ ∗ g mn v k ,k + µ∗ (vm,n + vn,m). 

Dropping the bar notation, the stress-velocity strain relationships in the general coordinates x i ,i=1, 2, 3, is 

Summary 

∂y j 

∂x n 

σmn = −pgmn + λ ∗ gmng ik vi,k + µ ∗ (vm,n + vn,m). (2.5.23) 

The basic equations which describe the motion of a Newtonian fluid are : 

Continuity equation (Conservation of mass) 

∂ϱ 

∂t + ϱv i 

Dϱ 

=0, or ,i Dt + ϱ∇· V =0 1equation. (2.5.24) 

Conservation of linear momentum σ ij 

,j + ϱbi = ϱ ˙v i , 3equations 

or in vector form ϱ D V 

Dt = ϱ b + ∇·σ = ϱ b −∇p + ∇·τ (2.5.25) 

where σ = 3 3 i=1 j=1 (−pδij + τij)êi êj and τ = 3 3 i=1 j=1 τij êi êj are second order tensors. Conservation 

of angular momentum σ ij = σ ji , (Reduces the set of equations (2.5.23) to 6 equations.) Rate of 

deformation tensor (Velocity strain tensor) 

Constitutive equations 

Dij = 1 

2 (vi,j + vj,i) , 6equations. (2.5.26) 

σmn = −pgmn + λ ∗ gmng ik vi,k + µ ∗ (vm,n + vn,m), 6equations. (2.5.27)

In the cgs system of units the above quantities have the following units of measurements in Cartesian 

coordinates 

vi is the velocity field ,i=1, 2, 3, [vi] =cm/sec 

b i 

σij is the stress tensor, i,j=1, 2, 3, [σij] = dyne/cm 2 

ϱ is the fluid density [ϱ] =gm/cm 3 

is the external body forces per unit mass [b i ] = dyne/gm 

Dij is the rate of deformation tensor [Dij] =sec −1 

λ ∗ ,µ ∗ 

p is the pressure [p] = dyne/cm 2 

are coefficients of viscosity [λ ∗ ]=[µ ∗ ]=Poise 

where 1Poise = 1gm/cm sec 

If we assume the external body forces per unit mass are known, then the equations (2.5.24), (2.5.25), 

(2.5.26), and (2.5.27) represent 16 equations in the 16 unknowns 

ϱ, v1,v2,v3,σ11,σ12,σ13,σ22,σ23,σ33,D11,D12,D13,D22,D23,D33. 

Navier-Stokes-Duhem Equations of Fluid Motion 

Substituting the stress tensor from equation (2.5.27) into the linear momentum equation (2.5.25), and 

assuming that the viscosity coefficients are constants, we obtain the Navier-Stokes-Duhem equations for fluid 

motion. In Cartesian coordinates these equations can be represented in any of the equivalent forms 

ϱ ˙vi = ϱbi − p,jδij +(λ ∗ + µ ∗ )vk,ki + µ ∗ vi,jj 

ϱ ∂vi 

∂t + ϱvjvi,j = ϱbi +(−pδij + τij) ,j 

∂ϱvi 

∂t +(ϱvivj + pδij − τij) ,j = ϱbi 

ϱ Dv 

Dt = ϱ b −∇p +(λ ∗ + µ ∗ )∇ (∇·v)+µ ∗ ∇ 2 v 

(2.5.28) 

where Dv ∂v 

= +(v ·∇) v is the material derivative, substantial derivative or convective derivative. This 

Dt ∂t 

derivative is represented as 

˙vi = ∂vi 

∂t 

∂vi 

+ 

∂xj dxj dt 

= ∂vi 

∂t 

+ ∂vi 

∂x j vj = ∂vi 

∂t + vi,jv j . (2.5.29) 

In the vector form of equations (2.5.28), the terms on the right-hand side of the equation represent force 

terms. The term ϱb represents external body forces per unit volume. If these forces are derivable from a 

potential function φ, then the external forces are conservative and can be represented in the form −ϱ∇ φ. 

The term −∇ p is the gradient of the pressure and represents a force per unit volume due to hydrostatic 

pressure. The above statement is verified in the exercises that follow this section. The remaining terms can 

be written 

fviscous =(λ ∗ + µ ∗ )∇ (∇·v)+µ ∗ ∇ 2 v (2.5.30) 

289

290 

and are given the physical interpretation of an internal force per unit volume. These internal forces arise 

from the shearing stresses in the moving fluid. If fviscous is zero the vector equation in (2.5.28) is called 

Euler’s equation. 

If the viscosity coefficients are nonconstant, then the Navier-Stokes equations can be written in the 

Cartesian form 

ϱ[ ∂vi ∂vi 

+ vj ]=ϱbi + 

∂t ∂xj 

∂ 

 

−pδij + λ 

∂xj 

∗ ∂vk 

δij + µ 

∂xk 

∗ 

 

∂vi 

+ 

∂xj 

∂vj 

 

∂xi 

=ϱbi − ∂p 

+ 

∂xi 

∂ 

 

∗ ∂vk 

λ + 

∂xi ∂xk 

∂ 

∂xj 

µ ∗ 

 

∂vi 

+ 

∂xj 

∂vj 

 

∂xi 

which can also be written in terms of the bulk coefficient of viscosity ζ = λ ∗ + 2 

3 µ∗ as 

ϱ[ ∂vi 

∂t 

∂vi 

+ vj ]=ϱbi − 

∂xj 

∂p 

∂xi 

=ϱbi − ∂p 

+ 

∂xi 

∂ 

∂xi 

+ ∂ 

 

(ζ − 

∂xi 

2 

These equations form the basics of viscous flow theory. 

3 µ∗ ) ∂vk 

 

+ 

∂xk 

∂ 

∂xj 

ζ ∂vk 

 

+ 

∂xk 

∂ 

∂xj 

µ ∗ 

 

∂vi 

∂xj 

 

µ ∗ 

 

∂vi 

+ 

∂xj 

∂vj 

 

∂xi 

+ ∂vj 

− 

∂xi 

2 

3 δij 

 

∂vk 

∂xk 

In the case of orthogonal coordinates, where g (i)(i) = h 2 i (no summation) and gij =0fori = j, general 

expressions for the Navier-Stokes equations in terms of the physical components v(1),v(2),v(3) are: 

Navier-Stokes-Duhem equations for compressible fluid in terms of physical components: (i = j = k) 

 

∂v(i) v(1) ∂v(i) 

ϱ + + 

∂t h1 ∂x1 

v(2) ∂v(i) 

+ 

h2 ∂x2 

v(3) ∂v(i) 

h3 ∂x3 

− v(j) 

 

v(j) 

hihj 

∂hj 

− v(i) 

∂xi 

∂hi 

 

+ 

∂xj 

v(k) 

 

v(i) 

hihk 

∂hi 

− v(k) 

∂xk 

∂hk 

 

∂xi 

 

= 

ϱ b(i) 

− 

hi 

1 ∂p 

+ 

hi ∂xi 

1 ∂ 

∗ 

λ ∇· 

µ 

V + 

hi ∂xi 

∗ 

 

hj ∂ v(j) 

+ 

hihj hi ∂xi hj 

hi 

 

∂ v(i) 

hj ∂xj hi 

∂hi 

∂hj 

+ µ∗ 

 

hi ∂ v(i) 

+ 

hihk hk ∂xk hi 

hk 

 

∂ v(k) ∂hi 

− 

hi ∂xi hk ∂xk 

2µ∗ 

 

1 ∂v(j) 

+ 

hihj hj ∂xj 

v(k) ∂hj 

+ 

hjhk ∂xk 

v(i) 

 

∂hj 

hihj ∂xi 

− 2µ∗ 

 

1 ∂v(k) 

+ 

hihk hk ∂xk 

v(i) ∂hk 

+ 

hihk ∂xi 

v(k) 

 

∂hk ∂hk 1 ∂ 

+ 

2µ 

hkhj ∂xi ∂xi hihjhk ∂xi 

∗ 

1 ∂v(i) 

hjhk 

+ 

hi ∂xi 

v(j) ∂hi 

+ 

hihj ∂hj 

v(k) 

 

∂hi 

hihk ∂xk 

+ ∂ 

 

µ 

∂xj 

∗ 

hj ∂ v(j) 

hihk 

+ 

hi ∂xi hj 

hi 

 

∂ v(i) 

hj ∂xj hi 

 

+ ∂ 

 

µ 

∂xk 

∗ 

hi ∂ v(i) 

hihj 

+ 

hk ∂xk hi 

hk 

 

∂ v(k) 

hi ∂xi hk 

 

(2.5.31) 

where ∇·v is found in equation (2.1.4). 

In the above equation, cyclic values are assigned to i, j and k. That is, for the x1 components assign 

the values i =1,j =2,k =3;forthex2components assign the values i =2,j =3,k = 1; and for the x3 

components assign the values i =3,j =1,k =2. 

The tables 5.2, 5.3 and 5.4 show the expanded form of the Navier-Stokes equations in Cartesian, cylindrical 

and spherical coordinates respectively.

ϱ 

ϱ DVx 

Dt 

ϱ DVy 

Dt 

ϱ DVz 

Dt 

 

∂p ∂ ∗ ∂Vx 

=ϱbx − + 2µ 

∂x ∂x ∂x + λ∗∇· 

V + ∂ 

 

µ 

∂y 

∗ 

 

∂Vx 

∂y 

 

∂p ∂ 

=ϱby − + µ 

∂y ∂x 

∗ 

 

∂Vy 

∂x 

 

∂p ∂ 

=ϱbz − + µ 

∂z ∂x 

∗ 

 

∂Vz 

∂x 

where D 

Dt ()=∂() 

∂() ∂() ∂() 

+ Vx + Vy + Vz 

∂t ∂x ∂y ∂z 

DVr 

Dt 

 

DVθ 

ϱ 

Dt 

and ∇· V = ∂Vx 

∂x 

+ ∂Vy 

∂y 

+ ∂Vz 

∂z 

 

∂Vy 

+ + 

∂x 

∂ 

 

µ 

∂z 

∗ 

 

∂Vx 

∂z 

 

∂Vx 

+ + 

∂y 

∂ 

 

∗ ∂Vy 

2µ 

∂y ∂y + λ∗∇· 

V + ∂ 

 

µ 

∂z 

∗ 

 

∂Vy 

∂z 

 

∂Vx 

+ + 

∂z 

∂ 

 

µ 

∂y 

∗ 

 

∂Vz 

∂y 

 

∂Vz 

+ 

∂x 

 

∂Vz 

+ 

∂y 

 

∂Vy 

+ + 

∂z 

∂ 

 

∗ ∂Vz 

2µ 

∂z ∂z + λ∗∇· 

V 

Table 5.2 Navier-Stokes equations for compressible fluids in Cartesian coordinates. 

 

V 2 

θ 

− =ϱbr − 

r 

∂p 

∂r 

 

VrVθ 

+ 

r 

ϱ DVz 

Dt 

+ ∂ 

∂z 

 

∂ ∗ ∂Vr 

+ 2µ 

∂r ∂r + λ∗∇· 

V + 1 

 

∂ 

µ 

r ∂θ 

∗ 

 

1 ∂Vr 

r ∂θ 

 

µ ∗ 

 

∂Vr ∂Vz 

+ + 

∂z ∂r 

2µ∗ 

 

∂Vr 1 ∂Vθ Vr 

− − 

r ∂r r ∂θ r 

=ϱbθ − 1 

 

∂p ∂ 

+ µ 

r ∂θ ∂r 

∗ 

 

1 ∂Vr ∂Vθ Vθ 

+ − + 

r ∂θ ∂r r 

1 

 

∂ 

2µ 

r ∂θ 

∗ 

+ ∂ 

 

µ 

∂z 

∗ 

 

1 ∂Vz ∂Vθ 

+ + 

r ∂θ ∂z 

2µ∗ 

 

1 ∂Vr ∂Vθ Vθ 

+ − 

r r ∂θ ∂r r 

=ϱbz − ∂p 

∂z 

+ 1 

r 

∂ 

∂r 

 

µ ∗ 

∂Vr 

r 

∂z 

where D 

Dt ()=∂() 

∂() Vθ ∂() ∂() 

+ Vr + + Vz 

∂t ∂r r ∂θ ∂z 

and ∇·V = 1 ∂(rVr) 

r ∂r 

1 ∂Vθ ∂Vz 

+ + 

r ∂θ ∂z 

 

∂Vz 

+ 

∂r 

+ 1 

r 

∂ 

∂θ 

 

µ ∗ 

 

1 ∂Vz 

r ∂θ 

+ ∂Vθ 

∂r 

 

1 ∂Vθ 

r ∂θ 

 

Vθ 

− 

r 

 

Vr 

+ + λ 

r 

∗ ∇· 

V 

(2.5.31a) 

 

∂Vθ 

+ + 

∂z 

∂ 

 

∗ ∂Vz 

2µ 

∂z ∂z + λ∗∇· 

V 

Table 5.3 Navier-Stokes equations for compressible fluids in cylindrical coordinates. 

(2.5.31b) 

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292 

Observe that for incompressible flow Dϱ 

Dt = 0 which implies ∇· V =0. Therefore, the assumptions 

of constant viscosity and incompressibility of the flow will simplify the above equations. If on the other 

hand the viscosity is temperature dependent and the flow is compressible, then one should add to the above 

equations the continuity equation, an energy equation and an equation of state. The energy equation comes 

from the first law of thermodynamics applied to a control volume within the fluid and will be considered 

in the sections ahead. The equation of state is a relation between thermodynamic variables which is added 

so that the number of equations equals the number of unknowns. Such a system of equations is known as 

a closed system. An example of an equation of state is the ideal gas law where pressure p is related to gas 

density ϱ and temperature T by the relation p = ϱRT where R is the universal gas constant. 

 

DVρ V 2 

θ 

ϱ − 

Dt + V 2 

φ 

= ϱbρ − 

ρ 

∂p 

 

∂ ∗ ∂Vρ 

+ 2µ 

∂ρ ∂ρ ∂ρ + λ∗∇· 

V + 1 

 

∂ 

µ 

ρ ∂θ 

∗ ρ ∂ 

 

Vθ 

+ 

∂ρ ρ 

µ∗ 

 

∂Vρ 

ρ ∂θ 

+ 1 

 

∂ µ ∗ ∂Vρ 

ρ sin θ ∂φ ρ sin θ ∂φ + µ∗ρ ∂ 

 

Vφ 

∂ρ ρ 

+ µ∗ 

 

4 

ρ 

∂Vρ 2 ∂Vθ 4Vρ 2 ∂Vφ 

− − − 

∂ρ ρ ∂θ ρ ρ sin θ ∂φ − 2Vθ cot θ 

+ ρ cot θ 

ρ 

∂ 

 

Vθ 

+ 

∂ρ ρ 

cot θ 

 

∂Vρ 

ρ ∂θ 

 

DVθ VρVθ 

ϱ + 

Dt ρ − V 2 

φ cot θ 

= ϱbθ − 

ρ 

1 

 

∂p ∂ 

+ µ 

ρ ∂θ ∂ρ 

∗ ρ ∂ 

 

Vθ 

+ 

∂ρ ρ 

µ∗ 

 

∂Vρ 

ρ ∂θ 

+ 1 

 

∂ 2µ ∗ 

∂Vθ 

+ Vρ + λ 

ρ ∂θ ρ ∂θ ∗ 

∇·V 

+ 1 

 

∂ µ ∗ 

sin θ ∂ Vφ 

+ 

ρ sin θ ∂φ ρ ∂θ sin θ 

µ∗ 

 

∂Vθ 


+ µ∗ 

 

1 ∂Vθ 1 ∂Vφ 

2cotθ − 

ρ 

ρ ∂θ ρ sin θ ∂φ − Vθ 

 

cot θ 

+3 ρ 

ρ 

∂ 

 

Vθ 

+ 

∂ρ ρ 

1 

 

∂Vρ 

ρ ∂θ 

 

DVφ VθVφ 

ϱ + 

Dt ρ + VθVφ 

 

cot θ 

= ϱbφ − 

ρ 

1 

 

∂p ∂ µ ∗ ∂Vρ 

+ 

ρ sin θ ∂φ ∂ρ ρ sin θ ∂φ + µ∗ρ ∂ 

 

Vφ 

∂ρ ρ 

+ 1 

 

∂ µ ∗ 


+ 

ρ ∂θ ρ ∂θ sin θ 

µ∗ 

 

∂Vθ 


+ 1 

 

∂ 2µ ∗ 

1 ∂Vφ 

ρ sin θ ∂φ ρ sin θ ∂φ + Vρ + Vθ 

 

cot θ + λ ∗ ∇· 

V 

+ µ∗ 

 

 

3 ∂Vρ ∂ Vφ 


+3ρ +2cotθ 

+ 

ρ ρ sin θ ∂φ ∂ρ ρ 

ρ ∂θ sin θ 

1 

 

∂Vθ 


where D 

Dt ()=∂() 

∂() Vθ ∂() Vφ ∂() 

+ Vρ + + 

∂t ∂ρ ρ ∂θ ρ sin θ ∂φ 

and ∇· V = 1 

ρ2 ∂(ρ2Vρ) 1 ∂Vθ sin θ 

+ + 

∂ρ ρ sin θ ∂θ 

1 ∂Vφ 


Table 5.4 Navier-Stokes equations for compressible fluids in spherical coordinates. 

(2.5.31c)

We now consider various special cases of the Navier-Stokes-Duhem equations. 

Special Case 1: Assume that b is a conservative force such that b = −∇ φ. Also assume that the viscous 

force terms are zero. Consider steady flow ( ∂v 

∂t 

= 0) and show that equation (2.5.28) reduces to the equation 

(v ·∇) v = −1 

∇ p −∇φ ϱ is constant. (2.5.32) 

ϱ 

Employing the vector identity 

(v ·∇) v =(∇×v) × v + 1 

∇(v · v), (2.5.33) 

2 

we take the dot product of equation (2.5.32) with the vector v. Noting that v · [(∇×v) × v] =0 weobtain 

 

p 1 

v ·∇ + φ + 

ϱ 2 v2 

 

=0. (2.5.34) 

This equation shows that for steady flow we will have 

p 1 

+ φ + 

ϱ 2 v2 = constant (2.5.35) 

along a streamline. This result is known as Bernoulli’s theorem. In the special case where φ = gh is a 

force due to gravity, the equation (2.5.35) reduces to p v2 

+ + gh = constant. This equation is known as 

ϱ 2 

Bernoulli’s equation. It is a conservation of energy statement which has many applications in fluids. 

Special Case 2: Assume that b = −∇ φ is conservative and define the quantity Ω by 

Ω =∇×v =curlv ω = 1 

Ω (2.5.36) 

2 

as the vorticity vector associated with the fluid flow and observe that its magnitude is equivalent to twice 

the angular velocity of a fluid particle. Then using the identity from equation (2.5.33) we can write the 

Navier-Stokes-Duhem equations in terms of the vorticity vector. We obtain the hydrodynamic equations 

∂v 

∂t + Ω × v + 1 

2 ∇ v2 = − 1 

1 

∇ p −∇φ + 

ϱ ϱ fviscous, (2.5.37) 

where fviscous is defined by equation (2.5.30). In the special case of nonviscous flow this further reduces to 

the Euler equation 

∂v 

∂t + Ω × v + 1 

2 ∇ v2 = − 1 

∇ p −∇φ. 

ϱ 

If the density ϱ is a function of the pressure only it is customary to introduce the function 

then the Euler equation becomes 

P = 

p 

c 

dp 

ϱ 

so that ∇P = dP 1 

∇p = 

dp ϱ ∇p 

∂v 

∂t + Ω × v = −∇(P + φ + 1 

2 v2 ). 

Some examples of vorticies are smoke rings, hurricanes, tornadoes, and some sun spots. You can create 

a vortex by letting water stand in a sink and then remove the plug. Watch the water and you will see that 

a rotation or vortex begins to occur. Vortices are associated with circulating motion. 

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294 

Pick an arbitrary simple closed curve C and place it in the fluid flow and define the line integral 

K = v · êt ds, where ds is an element of arc length along the curve C, v is the vector field defining the 

C 

velocity, and êt is a unit tangent vector to the curve C. The integral K is called the circulation of the fluid 

around the closed curve C. The circulation is the summation of the tangential components of the velocity 

field along the curve C. The local vorticity at a point is defined as the limit 

Circulation around C 

lim 

Area→0 Area inside C 

= circulation per unit area. 

By Stokes theorem, if curlv = 0, then the fluid is called irrotational and the circulation is zero. Otherwise 

the fluid is rotational and possesses vorticity. 

If we are only interested in the velocity field we can eliminate the pressure by taking the curl of both 

sides of the equation (2.5.37). If we further assume that the fluid is incompressible we obtain the special 

equations 

∇·v = 0 Incompressible fluid, ϱ is constant. 

Ω =curlv Definition of vorticity vector. 

∂ Ω 

∂t + ∇×( Ω × v) = µ∗ 

ϱ ∇2 (2.5.38) 

Ω Results because curl of gradient is zero. 

Note that when Ω is identically zero, we have irrotational motion and the above equations reduce to the 

Cauchy-Riemann equations. Note also that if the term ∇×( Ω × v) is neglected, then the last equation in 

equation (2.5.38) reduces to a diffusion equation. This suggests that the vorticity diffuses through the fluid 

once it is created. 

Vorticity can be caused by a rigid rotation or by shear flow. For example, in cylindrical coordinates let 

V = rω êθ, with r, ω constants, denote a rotational motion, then curl V = ∇× V =2ωêz, which shows the 

vorticity is twice the rotation vector. Shear can also produce vorticity. For example, consider the velocity 

field V = y ê1 with y ≥ 0. Observe that this type of flow produces shear because | V | increases as y increases. 

For this flow field we have curl V = ∇× V = − ê3. The right-hand rule tells us that if an imaginary paddle 

wheel is placed in the flow it would rotate clockwise because of the shear effects. 

Scaled Variables 

In the Navier-Stokes-Duhem equations for fluid flow we make the assumption that the external body 

forces are derivable from a potential function φ and write b = −∇ φ [dyne/gm] Wealsowanttowritethe 

Navier-Stokes equations in terms of scaled variables 

v = v 

v0 

p = p 

p0 

ϱ = ϱ 

ϱ0 

t = t 

τ 

φ = φ 

gL , 

x = x 

L 

y = y 

L 

z = z 

L 

which can be referred to as the barred system of dimensionless variables. Dimensionless variables are introduced 

by scaling each variable associated with a set of equations by an appropriate constant term called a 

characteristic constant associated with that variable. Usually the characteristic constants are chosen from 

various parameters used in the formulation of the set of equations. The characteristic constants assigned to 

each variable are not unique and so problems can be scaled in a variety of ways. The characteristic constants

assigned to each variable are scales, of the appropriate dimension, which act as reference quantities which 

reflect the order of magnitude changes expected of that variable over a certain range or area of interest 

associated with the problem. An inappropriate magnitude selected for a characteristic constant can result 

in a scaling where significant information concerning the problem can be lost. This is analogous to selecting 

an inappropriate mesh size in a numerical method. The numerical method might give you an answer but 

details of the answer might be lost. 

In the above scaling of the variables occurring in the Navier-Stokes equations we let v0 denote some 

characteristic speed, p0 a characteristic pressure, ϱ0 a characteristic density, L a characteristic length, g the 

acceleration of gravity and τ a characteristic time (for example τ = L/v0), then the barred variables v, p, 

ϱ,φ, t, x, y and z are dimensionless. Define the barred gradient operator by 

∇ = ∂ 

∂x ê1 + ∂ 

∂y ê2 + ∂ 

∂z ê3 

where all derivatives are with respect to the barred variables. The above change of variables reduces the 

Navier-Stokes-Duhem equations 

to the form 

ϱ ∂v 

∂t + ϱ(v ·∇) v = −ϱ∇φ −∇p +(λ∗ + µ ∗ )∇ (∇·v)+µ ∗ ∇ 2 v, (2.5.39) 

 

ϱ0v0 

ϱ 

τ 

∂v 

∂t + 

 

ϱ0v2 

0 

ϱ 

L 

v · ∇ v = −ϱ0gϱ∇ φ − 

 

p0 

∇p 

L 

+ (λ∗ + µ ∗ ) 

L2 v0∇ ∇·v + 

∗ µ v0 

L2 

∇ 2 v. 

(2.5.40) 

Now if each term in the equation (2.5.40) is divided by the coefficient ϱ0v2 0 /L, we obtain the equation 

Sϱ ∂v 

∂t + ϱ v · ∇ v = −1 

F 

which has the dimensionless coefficients 

E = p0 

ϱ0v 2 0 

= Euler number 

F = v2 0 

= Froude number, g is acceleration of gravity 

gL 

∗ λ 

ϱ∇ φ − E∇p + 

µ ∗ +1 

 

1 

R ∇ ∇·v + 1 

R ∇2v (2.5.41) 

R = ϱ0V0L 

µ ∗ 

S = L 

τv0 

= Reynolds number 

= Strouhal number. 

Dropping the bars over the symbols, we write the dimensionless equation using the above coefficients. 

The scaled equation is found to have the form 

Sϱ ∂v 

∗ 

1 

λ 

+ ϱ(v ·∇)v = − ϱ∇φ − E∇p + 

∂t F µ ∗ +1 

 

1 

1 

∇ (∇·v)+ 

R R ∇2v (2.5.42) 

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296 

Boundary Conditions 

Fluids problems can be classified as internal flows or external flows. An example of an internal flow 

problem is that of fluid moving through a converging-diverging nozzle. An example of an external flow 

problem is fluid flow around the boundary of an aircraft. For both types of problems there is some sort of 

boundary which influences how the fluid behaves. In these types of problems the fluid is assumed to adhere 

to a boundary. Let rb denote the position vector to a point on a boundary associated with a moving fluid, 

and let r denote the position vector to a general point in the fluid. Define v(r) as the velocity of the fluid at 

the point r and define v(rb) as the known velocity of the boundary. The boundary might be moving within 

the fluid or it could be fixed in which case the velocity at all points on the boundary is zero. We define the 

boundary condition associated with a moving fluid as an adherence boundary condition. 

Definition: (Adherence Boundary Condition) 

An adherence boundary condition associated with a fluid in motion 

is defined as the limit lim v(r) =v(rb) whererbis the position 

r→rb 

vector to a point on the boundary. 

Sometimes, when no finite boundaries are present, it is necessary to impose conditions on the components 

of the velocity far from the origin. Such conditions are referred to as boundary conditions at infinity. 

Summary and Additional Considerations 

Throughout the development of the basic equations of continuum mechanics we have neglected thermodynamical 

and electromagnetic effects. The inclusion of thermodynamics and electromagnetic fields adds 

additional terms to the basic equations of a continua. These basic equations describing a continuum are: 

Conservation of mass 

The conservation of mass is a statement that the total mass of a body is unchanged during its motion. 

This is represented by the continuity equation 

where ϱ is the mass density and vk is the velocity. 

Conservation of linear momentum 

∂ϱ 

∂t +(ϱvk ),k =0 or Dϱ 

Dt + ϱ∇· V =0 

The conservation of linear momentum requires that the time rate of change of linear momentum equal 

the resultant of all forces acting on the body. In symbols, we write 

where Dvi 

Dt 

= ∂vi 

∂t 

 

D 

ϱv 

Dt V 

i 

dτ = F 

S 

i (s) ni 

 

dS + ϱF 

V 

i (b) 

dτ + 

n 

α=1 

F i (α) 

(2.5.43) 

∂vi + ∂xk vk is the material derivative, F i (s) are the surface forces per unit area, F i (b) are the 

represents isolated external forces. Here S represents the surface and 

body forces per unit mass and F i (α) 

V represents the volume of the control volume. The right-hand side of this conservation law represents the 

resultant force coming from the consideration of all surface forces and body forces acting on a control volume.

Surface forces acting upon the control volume involve such things as pressures and viscous forces, while body 

forces are due to such things as gravitational, magnetic and electric fields. 

Conservation of angular momentum 

The conservation of angular momentum states that the time rate of change of angular momentum 

(moment of linear momentum) must equal the total moment of all forces and couples acting upon the body. 

In symbols, 

 

D 

ϱeijkx 

Dt V 

j v k 

dτ = eijkx 

S 

j F k 

(s) dS + ϱeijkx 

V 

j F k (b) 

where M i (α) represents concentrated couples and F k (α) 

Conservation of energy 

dτ + 

n 

represents isolated forces. 

(eijkx 

α=1 

j 

(α) F k (α) + M i (α) 

) (2.5.44) 

The conservation of energy law requires that the time rate of change of kinetic energy plus internal 

energies is equal to the sum of the rate of work from all forces and couples plus a summation of all external 

energies that enter or leave a control volume per unit of time. The energy equation results from the first law 

of thermodynamics and can be written 

D 

Dt (E + K) = ˙W + ˙Qh 

(2.5.45) 

where E is the internal energy, K is the kinetic energy, W˙ is the rate of work associated with surface and 

body forces, and ˙ Qh is the input heat rate from surface and internal effects. 

 

Let e denote the internal specific energy density within a control volume, then E = ϱe dτ represents 

V 

the total internal energy of the control volume. The kinetic energy of the control volume is expressed as 

K = 1 

 

ϱgijv 

2 V 

i v j dτ where vi is the velocity, ϱ is the density and dτ is a volume element. The energy (rate 

of work) associated with the body and surface forces is represented 

 

˙W = gijF 

S 

i (s) vj 

dS + ϱgijF 

V 

i (b) vj n 

dτ + (gijF i (α) vj + gijM i (α) ωj ) 

where ωj is the angular velocity of the point xi (α) , F i (α) are isolated forces, and M i (α) are isolated couples. 

Two external energy sources due to thermal sources are heat flow qi and rate of internal heat production ∂Q 

∂t 

per unit volume. The conservation of energy can thus be represented 

 

D 

ϱ(e + 

Dt V 

1 

2 gijv i v j 

) dτ = (gijF 

S 

i (s) vj − qin i 

) dS + (ϱgijF 

V 

i (b) vj + ∂Q 

) dτ 

∂t 

+ 

n 

α=1 

(gijF 

α=1 

i (α) vj + gijM i (α) ωj + U (α)) 

(2.5.46) 

where U (α) represents all other energies resulting from thermal, mechanical, electric, magnetic or chemical 

sources which influx the control volume and D/Dt is the material derivative. 

In equation (2.5.46) the left hand side is the material derivative of an integral of the total energy 

et = ϱ(e + 1 

2 gijv i v j ) over the control volume. Material derivatives are not like ordinary derivatives and so 

297

298 

we cannot interchange the order of differentiation and integration in this term. Here we must use the result 

that 

 

D 

∂et 

et dτ = 

Dt V 

V ∂t + ∇·(et 

V ) dτ. 

To prove this result we consider a more general problem. Let A denote the amount of some quantity per 

unit mass. The quantity A can be a scalar, vector or tensor. The total amount of this quantity inside the 

control volume is A = 

ϱA dτ and therefore the rate of change of this quantity is 

V 

∂A 

∂t = 

 

∂(ϱA) 

dτ = 

V ∂t 

D 

 

ϱA dτ − ϱA 

Dt V 

S 

V · ˆndS, 

which represents the rate of change of material within the control volume plus the influx into the control 

volume. The minus sign is because ˆn is always a unit outward normal. By converting the surface integral to 

a volume integral, by the Gauss divergence theorem, and rearranging terms we find that 

 

D 

∂(ϱA) 

ϱA dτ = 

Dt V 

V ∂t + ∇·(ϱA 

V ) dτ. 

In equation (2.5.46) we neglect all isolated external forces and substitute F i (s) = σijnj, F i (b) = bi where 

σij = −pδij + τij. We then replace all surface integrals by volume integrals and find that the conservation of 

energy can be represented in the form 

∂et 

∂t + ∇·(et V )=∇(σ · V ) −∇·q + ϱb · V + ∂Q 

(2.5.47) 

∂t 

where et = ϱe + ϱ(v2 1 + v2 2 + v2 3 )/2 is the total energy and σ = 3 3 i=1 j=1 σij êi êj is the second order stress 

tensor. Here 

σ · V = −p 3 

3 

3 

V + τ1jvj ê1 + τ2jvj ê2 + τ3jvj ê3 = −p V + τ · V 

j=1 

j=1 

j=1 

and τij = µ ∗ (vi,j + vj,i)+λ∗δijvk,k is the viscous stress tensor. Using the identities 

ϱ D(et/ϱ) 

Dt 

∂et 

= 

∂t + ∇·(et V ) and ϱ D(et/ϱ) 

Dt 

together with the momentum equation (2.5.25) dotted with V as 

ϱ D V 

Dt · V = ϱ b · V −∇p · V +(∇·τ ) · V 

the energy equation (2.5.47) can then be represented in the form 

where Φ is the dissipation function and can be represented 

= ϱDe 

Dt + ϱD(V 2 /2) 

Dt 

ϱ De 

Dt + p(∇· V )=−∇ · q + ∂Q 

+Φ (2.5.48) 

∂t 

Φ=(τijvi) ,j − viτij,j = ∇·(τ · V ) − (∇·τ ) · V. 

As an exercise it can be shown that the dissipation function can also be represented as Φ = 2µ ∗ DijDij +λ ∗ Θ 2 

where Θ is the dilatation. The heat flow vector is determined from the Fourier law of heat conduction in

terms of the temperature T as q = −κ∇ T ,whereκis the thermal conductivity. Consequently, the energy 

equation can be written as 

ϱ De 

Dt + p(∇· V )= ∂Q 

+Φ+∇(k∇T ). (2.5.49) 

∂t 

In Cartesian coordinates (x, y, z) weuse 

D ∂ ∂ 

= + Vx 

Dt ∂t ∂x 

∇· V = ∂Vx ∂Vy 

+ 

∂x 

∇·(κ∇T )= ∂ 

∂x 

In cylindrical coordinates (r, θ, z) 

and in spherical coordinates (ρ, θ, φ) 

∂y 

 

κ ∂T 

∂x 

∂ ∂ 

+ Vy + Vz 

∂y ∂z 

∂Vz 

+ 

∂z 

 

+ ∂ 

 

κ 

∂y 

∂T 

 

+ 

∂y 

∂ 

 

κ 

∂z 

∂T 

 

∂z 

D ∂ ∂ Vθ ∂ ∂ 

= + Vr + + Vz 

Dt ∂t ∂r r ∂θ ∂z 

∇·V = 1 ∂ 1 

(rVr)+ 

r ∂r r2 ∂Vθ ∂Vz 

+ 

∂θ ∂z 

∇·(κ∇T )= 1 

 

∂ 

rκ 

r ∂r 

∂T 

 

+ 

∂r 

1 

r2 

∂ 

κ 

∂θ 

∂T 

 

+ 

∂θ 

∂ 

 

κ 

∂z 

∂T 

 

∂z 

∂ Vθ ∂ Vφ ∂ 

+ Vρ + 

∂ρ ρ ∂θ ρ sin θ ∂φ 

∂ 

1 ∂ 

(ρVρ)+ 

∂ρ ρ sin θ ∂θ (Vθ sin θ)+ 1 ∂Vφ 


∇·(κ∇T )= 1 

ρ2 

∂ 

ρ 

∂ρ 

2 κ ∂T 

 

1 

+ 

∂ρ ρ2 

∂ 

κ sin θ 

sin θ ∂θ 

∂T 

 

+ 

∂θ 

D ∂ 

= 

Dt ∂t 

∇· V = 1 

ρ2 1 

ρ 2 sin 2 θ 

 

∂ 

κ 

∂φ 

∂T 

 

∂φ 

The combination of terms h = e + p/ϱ is known as enthalpy and at times is used to express the energy 

equation in the form 

ϱ Dh Dp ∂Q 

= + −∇·q +Φ. 

Dt Dt ∂t 

The derivation of this equation is left as an exercise. 

Conservative Systems 

Let Q denote some physical quantity per unit volume. Here Q can be either a scalar, vector or tensor 

field. Place within this field an imaginary simple closed surface S which encloses a volume V. The total 

amount of Q within the surface is given by 

V Qdτ and the rate of change of this amount with respect 

to time is ∂ 

 

∂t Qdτ. The total amount of Q within S changes due to sources (or sinks) within the volume 

and by transport processes. Transport processes introduce a quantity J, called current, which represents a 

flow per unit area across the surface S. The inward flux of material into the volume is denoted 

S − J · ˆndσ 

(ˆn is a unit outward normal.) The sources (or sinks) SQ denotes a generation (or loss) of material per unit 

volume so that 

V SQ dτ denotes addition (or loss) of material to the volume. For a fixed volume we then 

have the material balance 

 

 

∂Q 

dτ = − J · ˆndσ+ SQ dτ. 

V ∂t S 

V 

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300 

Using the divergence theorem of Gauss one can derive the general conservation law 

∂Q 

∂t + ∇· J = SQ 

(2.5.50) 

The continuity equation and energy equations are examples of a scalar conservation law in the special case 

where SQ =0. In Cartesian coordinates, we can represent the continuity equation by letting 

Q = ϱ and J = ϱ V = ϱ(Vx ê1 + Vy ê2 + Vz ê3) (2.5.51) 

The energy equation conservation law is represented by selecting Q = et and neglecting the rate of internal 

heat energy we let 

 

3 

 

J = (et + p)v1 − viτxi + qx ê1+ 

i=1 

 

3 

 

(et + p)v2 − viτyi + qy ê2+ 

(2.5.52) 

i=1 

 

3 

(et + p)v3 − 

 

ê3. 

i=1 

viτzi + qz 

In a general orthogonal system of coordinates (x1,x2,x3) the equation (2.5.50) is written 

∂ 

∂t ((h1h2h3Q)) + ∂ 

((h2h3J1)) + 

∂x1 

∂ 

((h1h3J2)) + 

∂x2 

∂ 

((h1h2J3)) = 0, 

∂x3 

where h1,h2,h3 are scale factors obtained from the transformation equations to the general orthogonal 

coordinates. 

The momentum equations are examples of a vector conservation law having the form 

where a is a vector and T is a second order symmetric tensor T = 

∂a 

∂t + ∇·(T )=ϱ b (2.5.53) 

3 

k=1 j=1 

3 

Tjk êj êk. In Cartesian coordinates 

we let a = ϱ(Vx ê1 + Vy ê2 + Vz ê3) andTij = ϱvivj + pδij − τij. In general coordinates (x1,x2,x3) the 

momentum equations result by selecting a = ϱ V and Tij = ϱvivj + pδij − τij. In a general orthogonal system 

the conservation law (2.5.53) has the general form 

∂ 

∂t ((h1h2h3a)) + ∂ 

 

 

(h2h3T · ê1) + 

∂x1 

∂ 

 

 

(h1h3T · ê2) + 

∂x2 

∂ 

 

 

(h1h2T · ê3) = ϱ 

∂x3 

b. (2.5.54) 

Neglecting body forces and internal heat production, the continuity, momentum and energy equations 

can be expressed in the strong conservative form 

where 

∂U 

∂t 

+ ∂E 

∂x 

+ ∂F 

∂y 

+ ∂G 

∂z 

⎡ 

ρ 

⎤ 

⎢ ρVx ⎥ 

⎢ ⎥ 

U = ⎢ ρVy ⎥ 

⎣ ⎦ 

ρVz 

et 

=0 (2.5.55) 

(2.5.56)

⎡ 

⎢ 

E = ⎢ 

⎣ 

ρVx 

ρV 2 x + p − τxx 

ρVxVy − τxy 

ρVxVz − τxz 

⎡ 

(et + p)Vx − Vxτxx − Vyτxy − Vzτxz + qx 

ρVy 

⎢ 

F = ⎢ 

⎣ 

ρVxVy − τxy 

ρV 2 

⎤ 

y + p − τyy 

ρVyVz − τyz 

⎥ 

⎦ 

⎡ 

(et + p)Vy − Vxτyx − Vyτyy − Vzτyz + qy 

ρVz 

⎤ 

⎢ 

G = ⎢ 

⎣ 

ρVxVz − τxz 

ρVyVz − τyz 

+ p − τzz 

⎥ 

⎦ 

ρV 2 

z 

(et + p)Vz − Vxτzx − Vyτzy − Vzτzz + qz 

where the shear stresses are τij = µ ∗ (Vi,j + Vj,i)+δijλ ∗ Vk,k for i, j, k =1, 2, 3. 

Computational Coordinates 

⎤ 

⎥ 

⎦ 

(2.5.57) 

(2.5.58) 

(2.5.59) 

To transform the conservative system (2.5.55) from a physical (x, y, z) domain to a computational (ξ,η,ζ) 

domain requires that a general change of variables take place. Consider the following general transformation 

of the independent variables 

ξ = ξ(x, y, z) η = η(x, y, z) ζ = ζ(x, y, z) (2.5.60) 

with Jacobian different from zero. The chain rule for changing variables in equation (2.5.55) requires the 

operators 

∂( ) ∂( ) 

= 

∂x ∂ξ ξx + ∂() 

∂η ηx + ∂() 

∂ζ ζx 

∂( ) ∂( ) 

= 

∂y ∂ξ ξy + ∂() 

∂η ηy + ∂() 

∂ζ ζy 

∂( ) ∂( ) 

= 

∂z ∂ξ ξz + ∂() 

∂η ηz + ∂() 

∂ζ ζz 

(2.5.61) 

The partial derivatives in these equations occur in the differential expressions 

dξ =ξx dx + ξy dy + ξz dz 

dη =ηx dx + ηy dy + ηz dz 

dζ =ζx dx + ζy dy + ζz dz 

or 

⎡ 

⎣ dξ 

⎤ ⎡ 

dη ⎦ = 

dζ 

In a similar mannaer from the inverse transformation equations 

we can write the differentials 

⎣ ξx ξy ξz 

ηx ηy ηz 

ζx ζy ζz 

⎤ ⎡ 

⎦ ⎣ dx 

⎤ 

dy ⎦ (2.5.62) 

dz 

x = x(ξ,η,ζ) y = y(ξ,η,ζ) z = z(ξ,η,ζ) (2.5.63) 

dx =xξ dξ + xη dη + xζ dζ 

dy =yξ dξ + yη dη + yζ dζ 

dz =zξ dξ + zζ dζ + zζ dζ 

or 

⎡ 

⎣ dx 

⎤ ⎡ 

dy ⎦ = 

dz 

⎣ xξ xη xζ 

yξ yη yζ 

zξ zη zζ 

⎤ ⎡ 

⎦ ⎣ dξ 

⎤ 

dη ⎦ (2.5.64) 

dζ 

301

302 

The transformations (2.5.62) and (2.5.64) are inverses of each other and so we can write 

⎡ 

⎤ ⎡ 

⎤−1 

⎦ = 

⎦ 

⎣ ξx ξy ξz 

ηx ηy ηz 

ζx ζy ζz 

⎣ xξ xη xζ 

yξ yη yζ 

zξ zη zζ 

⎡ 

=J ⎣ yηzζ − yζzη 

−(yξzζ − yζzξ) 

−(xηzζ − xζzη) 

xξzζ − xζzξ 

⎤ 

xηyζ − xζyη 

−(xξyζ − xζyξ) ⎦ 

yξzη − yηzξ −(xξzη − xηzξ) xξyη − xηyξ 

By comparing like elements in equation (2.5.65) we obtain the relations 

ξx =J(yηzζ − yζzη) 

ξy = − J(xηzζ − xζzη) 

ξz =J(xηyζ − xζyη) 

ηx = − J(yξzζ − yζzξ) 

ηy =J(xξzζ − zζzξ) 

ηz = − J(xξyζ − xζyξ) 

ζx =J(yξzη − yηzξ) 

ζy = − J(xξzη − xηzξ) 

ζz =J(xξyη − xηyξ) 

The equations (2.5.55) can now be written in terms of the new variables (ξ,η,ζ) as 

∂U 

∂t 

(2.5.65) 

(2.5.66) 

+ ∂E 

∂ξ ξx + ∂E 

∂η ηx + ∂E 

∂ζ ζx + ∂F 

∂ξ ξy + ∂F 

∂η ηy + ∂F 

∂ζ ζy + ∂G 

∂ξ ξz + ∂G 

∂η ηz + ∂G 

∂ζ ζz =0 (2.5.67) 

Now divide each term by the Jacobian J and write the equation (2.5.67) in the form 

∂ 

∂t 

 

U 

+ 

J 

∂ 

∂ξ 

 

Eξx + Fξy + Gξz 

J 

+ ∂ 

 

Eηx + Fηy + Gηz 

∂η J 

+ ∂ 

 

Eζx + Fζy + Gζz 

∂ζ J 

 

∂ ξx 

− E 

+ 

∂ξ J 

∂ 

 

ηx 

+ 

∂η J 

∂ 

 

ζx 

∂ζ J 

 

∂ ξy 

− F 

+ 

∂ξ J 

∂ 

 

ηy 

+ 

∂η J 

∂ 

 

ζy 

∂ζ J 

 

∂ ξz 

− G 

+ 

∂ξ J 

∂ 

 

ηz 

+ 

∂η J 

∂ 

 

ζz 

=0 

∂ζ J 

(2.5.68) 

Using the relations given in equation (2.5.66) one can show that the curly bracketed terms above are all zero 

and so the transformed equations (2.5.55) can also be written in the conservative form 

where 

∂ U 

∂t + ∂ E 

∂ξ + ∂ F 

∂η + ∂ G 

=0 (2.5.69) 

∂ζ 

U = U 

J 

E = Eξx + Fξy + Gξz 

J 

F = Eηx + Fηy + Gηz 

J 

G = Eζx + Fζy + Gζz 

J 

(2.5.70)

Fourier law of heat conduction 

The Fourier law of heat conduction can be written qi = −κT,i for isotropic material and qi = −κijT,j 

for anisotropic material. The Prandtl number is a nondimensional constant defined as Pr = cpµ∗ 

κ 

the heat flow terms can be represented in Cartesian coordinates as 

∗ cpµ ∂T 

qx = − 

Pr ∂x 

∗ cpµ ∂T 

qy = − 

Pr ∂y 

∗ cpµ ∂T 

qz = − 

Pr ∂z 

Now one can employ the equation of state relations P = ϱe(γ − 1), cp = γR 

γ−1 , cpT = γRT 

γ−1 

so that 

and write the 

above equations in the alternate forms 

µ 

qx = − 

∗ 

∂ γP 

Pr(γ − 1) ∂x ϱ 

µ 

qy = − 

∗ 

∂ γP 

Pr(γ − 1) ∂y ϱ 

µ 

qz = − 

∗ 

∂ γP 

Pr(γ − 1) ∂z ϱ 

 

γP 

The speed of sound is given by a = 

ϱ = γRT and so one can substitute a2 in place of the ratio γP 

in the above equations. 

ϱ 

Equilibrium and Nonequilibrium Thermodynamics 

High temperature gas flows require special considerations. In particular, the specific heat for monotonic 

and diatomic gases are different and are in general a function of temperature. The energy of a gas can be 

written as e = et + er + ev + ee + en where et represents translational energy, er is rotational energy, ev is 

vibrational energy, ee is electronic energy, and en is nuclear energy. The gases follow a Boltzmann distribution 

for each degree of freedom and consequently at very high temperatures the rotational, translational and 

vibrational degrees of freedom can each have their own temperature. Under these conditions the gas is said 

to be in a state of nonequilibrium. In such a situation one needs additional energy equations. The energy 

equation developed in these notes is for equilibrium thermodynamics where the rotational, translational and 

vibrational temperatures are the same. 

Equation of state 

It is assumed that an equation of state such as the universal gas law or perfect gas law pV = nRT 

holds which relates pressure p [N/m 2 ], volume V [m 3 ], amount of gas n [mol],and temperature T [K] where 

R [J/mol − K] is the universal molar gas constant. If the ideal gas law is represented in the form p = ϱRT 

where ϱ [Kg/m3 ] is the gas density, then the universal gas constant must be expressed in units of [J/Kg−K] 

(See Appendix A). Many gases deviate from this ideal behavior. In order to account for the intermolecular 

forces associated with high density gases, an empirical equation of state of the form 

M1 

p = ρRT + 

n=1 

βnρ n+r1 + e −γ1ρ−γ2ρ2 

M2 

cnρ n+r2 

involving constants M1,M2,βn,cn,r1,r2,γ1,γ2 is often used. For a perfect gas the relations 

e = cvT γ = cp 

cv 

cv = R 

γ − 1 

n=1 

cp = γR 

γ − 1 

h = cpT 

hold, where R is the universal gas constant, cv is the specific heat at constant volume, cp is the specific 

heat at constant pressure, γ is the ratio of specific heats and h is the enthalpy. For cv and cp constants the 

relations p =(γ− 1)ϱe and RT =(γ− 1)e can be verified. 

303

304 

EXAMPLE 2.5-1. (One-dimensional fluid flow) 

Construct an x-axis running along the center line of a long cylinder with cross sectional area A. Consider 

the motion of a gas driven by a piston and moving with velocity v1 = u in the x-direction. From an Eulerian 

point of view we imagine a control volume fixed within the cylinder and assume zero body forces. We require 

the following equations be satisfied. 

Conservation of mass ∂ϱ 

∂t +div(ϱ V ) = 0 which in one-dimension reduces to ∂ϱ ∂ 

+ (ϱu) =0. 

∂t ∂x 

Conservation of momentum, equation (2.5.28) reduces to ∂ ∂ 2 

(ϱu)+ ϱu 

∂t ∂x 

+ ∂p 

∂x =0. 

Conservation of energy, equation (2.5.48) in the absence of heat flow and internal heat production, 

∂e ∂e 

becomes in one dimension ϱ + u + p 

∂t ∂x 

∂u 

=0. Using the conservation of mass relation this 

∂x 

equation can be written in the form ∂ ∂ 

(ϱe)+ 

∂t ∂x (ϱeu)+p∂u 

∂x =0. 

In contrast, from a Lagrangian point of view we let the control volume move with the flow and consider 

advection terms. This gives the following three equations which can then be compared with the above 

Eulerian equations of motion. 

Conservation of mass d 

Dϱ 

(ϱJ) = 0 which in one-dimension is equivalent to + ϱ∂u 

dt Dt ∂x =0. 

Conservation of momentum, equation (2.5.25) in one-dimension ϱ Du ∂p 

+ 

Dt ∂x =0. 

Conservation of energy, equation (2.5.48) in one-dimension ϱ De 

+ p∂u =0. In the above equations 

Dt ∂x 

D() ∂ ∂ 

Dt = ∂t ()+u ∂x (). The Lagrangian viewpoint gives three equations in the three unknowns ρ, u, e. 

In both the Eulerian and Lagrangian equations the pressure p represents the total pressure p = pg + pv 

where pg is the gas pressure and pv is the viscous pressure which causes loss of kinetic energy. The gas pressure 

is a function of ϱ, e and is determined from the ideal gas law pg = ϱRT = ϱ(cp − cv)T = ϱ( cp 

cv − 1)cvT or 

pg = ϱ(γ − 1)e. Some kind of assumption is usually made to represent the viscous pressure pv as a function 

of e, u. The above equations are then subjected to boundary and initial conditions and are usually solved 

numerically. 

Entropy inequality 

Energy transfer is not always reversible. Many energy transfer processes are irreversible. The second 

law of thermodynamics allows energy transfer to be reversible only in special circumstances. In general, 

the second law of thermodynamics can be written as an entropy inequality, known as the Clausius-Duhem 

inequality. This inequality states that the time rate of change of the total entropy is greater than or equal to 

the total entropy change occurring across the surface and within the body of a control volume. The Clausius- 

Duhem inequality places restrictions on the constitutive equations. This inequality can be expressed in the 

form 

 

 

D 

ϱs dτ ≥ s 

Dt V 

S 

 

Rate of entropy increase 

i 

n 

ni dS + ρb dτ + B (α) 

V 

α=1 

 

Entropy input rate into control volume 

where s is the specific entropy density, s i is an entropy flux, b is an entropy source and B (α) are isolated 

entropy sources. Irreversible processes are characterized by the use of the inequality sign while for reversible

Figure 2.5-3. Interaction of various fields. 

processes the equality sign holds. The Clausius-Duhem inequality is assumed to hold for all independent 

thermodynamical processes. 

If in addition there are electric and magnetic fields to consider, then these fields place additional forces 

upon the material continuum and we must add all forces and moments due to these effects. In particular we 

must add the following equations 

Gauss’s law for magnetism ∇· B =0 

Gauss’s law for electricity ∇· D = ϱe 

Faraday’s law ∇× E = − ∂ B 

∂t 

Ampere’s law ∇× H = J + ∂ D 

∂t 

1 

√ g 

1 

√ g 

∂ 

∂x i (√ gB i )=0. 

∂ 

∂x i (√ gD i )=ϱe. 

ɛ ijk Ek,j = − ∂Bi 

∂t . 

ɛ ijk Hk,j = J i + ∂Di 

∂t . 

where ϱe is the charge density, J i is the current density, Di = ɛ j 

i Ej + Pi is the electric displacement vector, 

Hi is the magnetic field, Bi = µ j 

i Hj + Mi is the magnetic induction, Ei is the electric field, Mi is the 

magnetization vector and Pi is the polarization vector. Taking the divergence of Ampere’s law produces the 

law of conservation of charge which requires that 

∂ϱe 

∂t + ∇· J =0 

∂ϱe 

∂t 

+ 1 

√ g 

∂ 

∂x i (√ gJ i )=0. 

The figure 2.5-3 is constructed to suggest some of the interactions that can occur between various 

variables which define the continuum. Pyroelectric effects occur when a change in temperature causes 

changes in the electrical properties of a material. Temperature changes can also change the mechanical 

properties of materials. Similarly, piezoelectric effects occur when a change in either stress or strain causes 

changes in the electrical properties of materials. Photoelectric effects are said to occur if changes in electric 

or mechanical properties effect the refractive index of a material. Such changes can be studied by modifying 

the constitutive equations to include the effects being considered. 

From figure 2.5-3 we see that there can exist a relationship between the displacement field Di and 

electric field Ei. When this relationship is linear we can write Di = ɛjiEj and Ej = βjnDn, whereɛji are 

305

306 

dielectric constants and βjn are dielectric impermabilities. Similarly, when linear piezoelectric effects exist 

we can write linear relations between stress and electric fields such as σij = −gkijEk and Ei = −eijkσjk, 

where gkij and eijk are called piezoelectric constants. If there is a linear relation between strain and an 

electric fields, this is another type of piezoelectric effect whereby eij = dijkEk and Ek = −hijkejk, where 

dijk and hijk are another set of piezoelectric constants. Similarly, entropy changes can cause pyroelectric 

effects. Piezooptical effects (photoelasticity) occurs when mechanical stresses change the optical properties of 

the material. Electrical and heat effects can also change the optical properties of materials. Piezoresistivity 

occurs when mechanical stresses change the electric resistivity of materials. Electric field changes can cause 

variations in temperature, another pyroelectric effect. When temperature effects the entropy of a material 

this is known as a heat capacity effect. When stresses effect the entropy in a material this is called a 

piezocaloric effect. Some examples of the representation of these additional effects are as follows. The 

piezoelectric effects are represented by equations of the form 

σij = −hmijDm Di = dijkσjk eij = gkijDk Di = eijkejk 

where hmij, dijk, gkij and eijk are piezoelectric constants. 

Knowledge of the material or electric interaction can be used to help modify the constitutive equations. 

For example, the constitutive equations can be modified to included temperature effects by expressing the 

constitutive equations in the form 

σij = cijklekl − βij∆T and eij = sijklσkl + αij∆T 

where for isotropic materials the coefficients αij and βij are constants. As another example, if the strain is 

modified by both temperature and an electric field, then the constitutive equations would take on the form 

eij = sijklσkl + αij∆T + dmijEm. 

Note that these additional effects are additive under conditions of small changes. That is, we may use the 

principal of superposition to calculate these additive effects. 

If the electric field and electric displacement are replaced by a magnetic field and magnetic flux, then 

piezomagnetic relations can be found to exist between the variables involved. One should consult a handbook 

to determine the order of magnitude of the various piezoelectric and piezomagnetic effects. For a large 

majority of materials these effects are small and can be neglected when the field strengths are weak. 

The Boltzmann Transport Equation 

The modeling of the transport of particle beams through matter, such as the motion of energetic protons 

or neutrons through bulk material, can be approached using ideas from the classical kinetic theory of gases. 

Kinetic theory is widely used to explain phenomena in such areas as: statistical mechanics, fluids, plasma 

physics, biological response to high-energy radiation, high-energy ion transport and various types of radiation 

shielding. The problem is basically one of describing the behavior of a system of interacting particles and their 

distribution in space, time and energy. The average particle behavior can be described by the Boltzmann 

equation which is essentially a continuity equation in a six-dimensional phase space (x, y, z, Vx,Vy,Vz). We

will be interested in examining how the particles in a volume element of phase space change with time. We 

introduce the following notation: 

(i) r the position vector of a typical particle of phase space and dτ = dxdydz the corresponding spatial 

volume element at this position. 

(ii) V the velocity vector associated with a typical particle of phase space and dτv 

corresponding velocity volume element. 

(iii) 

= dVxdVydVz the 

Ω a unit vector in the direction of the velocity V = v Ω. 

(iv) E = 1 

2 mv2 kinetic energy of particle. 

(v) d Ω is a solid angle about the direction Ωanddτ dE d Ω is a volume element of phase space involving the 

solid angle about the direction Ω. 

(vi) n = n(r, E, Ω,t) the number of particles in phase space per unit volume at position r per unit velocity 

at position V per unit energy in the solid angle d Ωattimetand N = N(r, E, Ω,t)=vn(r, E, Ω,t) 

the number of particles per unit volume per unit energy in the solid angle d Ωattimet. The quantity 

N(r, E, Ω,t)dτ dE d Ω represents the number of particles in a volume element around the position r with 

energy between E and E + dE having direction Ω in the solid angle d Ωattimet. 

(vii) φ(r, E, Ω,t)=vN(r, E, Ω,t) is the particle flux (number of particles/cm2 − Mev − sec). 

(viii) Σ(E ′ → E, Ω ′ → Ω) a scattering cross-section which represents the fraction of particles with energy E ′ 

and direction Ω ′ that scatter into the energy range between E and E + dE having direction Ωinthe 

solid angle d Ω per particle flux. 

(ix) Σs(E,r) fractional number of particles scattered out of volume element of phase space per unit volume 

per flux. 

(x) Σa(E,r) fractional number of particles absorbed in a unit volume of phase space per unit volume per 

flux. 

Consider a particle at time t having a position r in phase space as illustrated in the figure 2.5-4. This 

particle has a velocity V in a direction Ω and has an energy E. In terms of dτ = dx dy dz, ΩandEan element of volume of phase space can be denoted dτdEd Ω, where d Ω=d Ω(θ, ψ) =sinθdθdψ is a solid angle 

about the direction Ω. 

The Boltzmann transport equation represents the rate of change of particle density in a volume element 

dτ dE d Ω of phase space and is written 

d 

dt N(r, E, Ω,t) dτ dE d Ω=DCN(r, E, Ω,t) (2.5.71) 

where DC is a collision operator representing gains and losses of particles to the volume element of phase 

space due to scattering and absorption processes. The gains to the volume element are due to any sources 

S(r, E, Ω,t) per unit volume of phase space, with units of number of particles/sec per volume of phase space, 

together with any scattering of particles into the volume element of phase space. That is particles entering 

the volume element of phase space with energy E, which experience a collision, leave with some energy 

E − ∆E and thus will be lost from our volume element. Particles entering with energies E ′ >Emay, 

307

308 

Figure 2.5-4. Volume element and solid angle about position r. 

depending upon the cross-sections, exit with energy E ′ − ∆E = E and thus will contribute a gain to the 

volume element. In terms of the flux φ the gains due to scattering into the volume element are denoted by 

 

d Ω ′ 

 

dE ′ Σ(E ′ → E, Ω ′ → Ω)φ(r, E ′ , Ω,t) dτ dE d Ω 

and represents the particles at position r experiencing a scattering collision with a particle of energy E ′ and 

direction Ω ′ which causes the particle to end up with energy between E and E + dE and direction Ωind Ω. 

The summations are over all possible initial energies. 

In terms of φ the losses are due to those particles leaving the volume element because of scattering and 

are 

Σs(E,r)φ(r, E, Ω,t)dτ dE d Ω. 

The particles which are lost due to absorption processes are 

Σa(E,r)φ(r, E, Ω,t) dτ dE d Ω. 

The total change to the number of particles in an element of phase space per unit of time is obtained by 

summing all gains and losses. This total change is 

 

dN 

dτ dE dΩ = 

dt 

d Ω ′ 

 

dE ′ Σ(E ′ → E, Ω ′ → Ω)φ(r, E ′ , Ω,t) dτ dE d Ω 

− Σs(E,r)φ(r, E, Ω,t)dτ dE dΩ 

− Σa(E,r)φ(r, E, Ω,t) dτ dE d Ω 

(2.5.72) 

The rate of change dN 

dt 

+ S(r, E, Ω,t)dτ dE d Ω. 

on the left-hand side of equation (2.5.72) expands to 

dN 

dt 

∂N ∂N dx ∂N dy ∂N dz 

= + + + 

∂t ∂x dt ∂y dt ∂z dt 

+ ∂N dVx ∂N dVy ∂N dVz 

+ + 

∂Vx dt ∂Vy dt ∂Vz dt

which can be written as 

where d V 

dt = F 

m 

dN 

dt 

= ∂N 

∂t + V ·∇rN + F 

m ·∇ V N (2.5.73) 

represents any forces acting upon the particles. The Boltzmann equation can then be 

expressed as 

∂N 

∂t + V ·∇rN + F 

m ·∇ V N =Gains−Losses. (2.5.74) 

If the right-hand side of the equation (2.5.74) is zero, the equation is known as the Liouville equation. In 

the special case where the velocities are constant and do not change with time the above equation (2.5.74) 

can be written in terms of the flux φ and has the form 

 

1 ∂ 

v ∂t + 

Ω ·∇r +Σs(E,r)+Σa(E,r) φ(r, E, Ω,t)=DCφ (2.5.75) 

where 

 

DCφ = 

d Ω ′ 

 

dE ′ Σ(E ′ → E, Ω ′ → Ω)φ(r, E ′ , Ω ′ ,t)+S(r, E, Ω,t). 

The above equation represents the Boltzmann transport equation in the case where all the particles are 

the same. In the case of atomic collisions of particles one must take into consideration the generation of 

secondary particles resulting from the collisions. 

Let there be a number of particles of type j in a volume element of phase space. For example j = p 

(protons) and j = n (neutrons). We consider steady state conditions and define the quantities 

(i) φj(r, E, Ω) as the flux of the particles of type j. 

(ii) σjk( Ω, Ω ′ ,E,E ′ ) the collision cross-section representing processes where particles of type k moving in 

direction Ω ′ with energy E ′ produce a type j particle moving in the direction ΩwithenergyE. 

(iii) σj(E) =Σs(E,r)+Σa(E,r) the cross-section for type j particles. 

The steady state form of the equation (2.5.64) can then be written as 

Ω ·∇φj(r, E, Ω)+σj(E)φj(r, E, Ω) 

= 

 

k 

σjk( Ω, Ω ′ ,E,E ′ )φk(r, E ′ , Ω ′ )d Ω ′ dE ′ 

(2.5.76) 

where the summation is over all particles k = j. 

The Boltzmann transport equation can be represented in many different forms. These various forms 

are dependent upon the assumptions made during the derivation, the type of particles, and collision crosssections. 

In general the collision cross-sections are dependent upon three components. 

(1) Elastic collisions. Here the nucleus is not excited by the collision but energy is transferred by projectile 

recoil. 

(2) Inelastic collisions. Here some particles are raised to a higher energy state but the excitation energy is 

not sufficient to produce any particle emissions due to the collision. 

(3) Non-elastic collisions. Here the nucleus is left in an excited state due to the collision processes and 

some of its nucleons (protons or neutrons) are ejected. The remaining nucleons interact to form a stable 

structure and usually produce a distribution of low energy particles which is isotropic in character. 

309

310 

Various assumptions can be made concerning the particle flux. The resulting form of Boltzmann’s 

equation must be modified to reflect these additional assumptions. As an example, we consider modifications 

to Boltzmann’s equation in order to describe the motion of a massive ion moving into a region filled with a 

homogeneous material. Here it is assumed that the mean-free path for nuclear collisions is large in comparison 

with the mean-free path for ion interaction with electrons. In addition, the following assumptions are made 

(i) All collision interactions are non-elastic. 

(ii) The secondary particles produced have the same direction as the original particle. This is called the 

straight-ahead approximation. 

(iii) Secondary particles never have kinetic energies greater than the original projectile that produced them. 

(iv) A charged particle will eventually transfer all of its kinetic energy and stop in the media. This stopping 

distance is called the range of the projectile. The stopping power Sj(E) = dE 

dx represents the energy 

loss per unit length traveled in the media and determines the range by the relation dRj 1 

dE = Sj(E) or 

Rj(E) = E dE 

0 

′ 

Sj(E ′ ) . Using the above assumptions Wilson, et.al.1 show that the steady state linearized 

Boltzmann equation for homogeneous materials takes on the form 

Ω ·∇φj(r, E, Ω) − 1 ∂ 

Aj ∂E (Sj(E)φj(r, E, Ω)) + σj(E)φj(r, E, Ω) 

= 

 

k=j 

dE ′ d Ω ′ σjk( Ω, Ω ′ ,E,E ′ )φk(r, E ′ , Ω ′ ) 

(2.5.77) 

where Aj is the atomic mass of the ion of type j and φj(r, E, Ω) is the flux of ions of type j moving in 

the direction ΩwithenergyE. 

Observe that in most cases the left-hand side of the Boltzmann equation represents the time rate of 

change of a distribution type function in a phase space while the right-hand side of the Boltzmann equation 

represents the time rate of change of this distribution function within a volume element of phase space due 

to scattering and absorption collision processes. 

Boltzmann Equation for gases 

Consider the Boltzmann equation in terms of a particle distribution function f(r, V,t)whichcanbe 

written as 

∂ 

∂t + V ·∇r + F 

m ·∇ 

 

V f(r, V,t)=DCf(r, V,t) (2.5.78) 

for a single species of gas particles where there is only scattering and no absorption of the particles. An 

element of volume in phase space (x, y, z, Vx,Vy,Vz) can be thought of as a volume element dτ = dxdydz 

for the spatial elements together with a volume element dτv = dVxdVydVz for the velocity elements. These 

elements are centered at position r and velocity V at time t. In phase space a constant velocity V1 can be 

thought of as a sphere since V 2 

1 

= V 2 

x 

2 2 + Vy + Vz . The phase space volume element dτdτv changes with time 

since the position r and velocity V change with time. The position vector r changes because of velocity 

1John W. Wilson, Lawrence W. Townsend, Walter Schimmerling, Govind S. Khandelwal, Ferdous Kahn, 

John E. Nealy, Francis A. Cucinotta, Lisa C. Simonsen, Judy L. Shinn, and John W. Norbury, Transport 

Methods and Interactions for Space Radiations, NASA Reference Publication 1257, December 1991.

and the velocity vector changes because of the acceleration F 

m .Heref(r, V,t)dτdτv represents the expected 

number of particles in the phase space element dτdτv at time t. 

Assume there are no collisions, then each of the gas particles in a volume element of phase space centered 

at position r and velocity V1 move during a time interval dt to a phase space element centered at position 

r + V1dt and V1 + F 

mdt. If there were no loss or gains of particles, then the number of particles must be 

conserved and so these gas particles must move smoothly from one element of phase space to another without 

any gains or losses of particles. Because of scattering collisions in dτ many of the gas particles move into or 

out of the velocity range V1 to V1 + d V1. These collision scattering processes are denoted by the collision 

operator DCf(r, V,t) in the Boltzmann equation. 

Consider two identical gas particles which experience a binary collision. Imagine that particle 1with 

velocity V1 collides with particle 2 having velocity V2. Denote by σ(V1 → V ′ 

1 , V2 → V ′ 

2 ) dτV1dτV2 the 

conditional probability that particle 1is scattered from velocity V1 to between V ′ 

1 and V ′ 

1 + d V ′ 

1 and the 

struck particle 2 is scattered from velocity V2 to between V ′ 

2 and V ′ 

2 + d V ′ 

2 . We will be interested in collisions 

ofthetype( V ′ 

1 , V ′ 

2 ) → ( V1, V2) for a fixed value of V1 as this would represent the number of particles scattered 

into dτV1. Also of interest are collisions of the type ( V1, V2) → ( V ′ 

1, V ′ 

2) for a fixed value V1 as this represents 

particles scattered out of dτV1. Imagine a gas particle in dτ with velocity V ′ 

1 subjected to a beam of particles 

with velocities V ′ 

2. The incident flux on the element dτdτV ′ 

1 is | V ′ 

1 − V ′ 

2|f(r, V ′ 

2,t)dτV ′ 

2 

and hence 

σ( V1 → V ′ 

1, V2 → V ′ 

2) dτV1dτV2dt | V ′ 

1 − V ′ 

2|f(r, V ′ 

2,t) dτV ′ 

2 

(2.5.79) 

represents the number of collisions, in the time interval dt, which scatter from V ′ 

1 to between V1 and V1 + d V1 

as well as scattering V ′ 

2 to between V2 and V2 + d V2. Multiply equation (2.5.79) by the density of particles 

in the element dτdτV ′ 

1 and integrate over all possible initial velocities V ′ 

1 , V ′ 

2 and final velocities V2 not equal 

to V1. This gives the number of particles in dτ which are scattered into dτV1dt as 

 

Nsin = dτdτV1dt 

dτV2dτV ′ 

2 

dτV ′ 

1 σ( V ′ 

1 → V1, V ′ 

2 → V2)| V ′ 

1 − V ′ 

2 |f(r, V ′ 

1 ,t)f(r, V ′ 

2 

In a similar manner the number of particles in dτ which are scattered out of dτV1dt is 

Nsout = dτdτV1dtf(r, 

V1,t) dτV2 

Let 

dτV ′ 

2 

,t). (2.5.80) 

dτV ′ 

1 σ( V ′ 

1 → V1, V ′ 

2 → V2)| V2 − V1|f(r, V2,t). (2.5.81) 

W ( V ′ 

1 → V1, V ′ 

2 → V2) =| V1 − V2| σ( V ′ 

1 → V1, V ′ 

2 → V2) (2.5.82) 

define a symmetric scattering kernel and use the relation DCf(r, V,t)=Nsin − Nsout to represent the 

Boltzmann equation for gas particles in the form 

 

∂ 

∂t + V ·∇r + F 

m ·∇ 

V 

f(r, V1,t)= 

 

(2.5.83) 

dτ V ′ 1 

dτ V ′ 2 

dτV2 W ( V1 → V ′ 

1 , V2 → V ′ 

2 ) f(r, V ′ 

1 ,t)f(r, V ′ 

2 ,t) − f(r, V1,t)f(r, V2,t) . 

Take the moment of the Boltzmann equation (2.5.83) with respect to an arbitrary function φ( V1). That 

is, multiply equation (2.5.83) by φ( V1) and then integrate over all elements of velocity space dτV1. Define 

the following averages and terminology: 

311


• The particle density per unit volume 

 

n = n(r, t) = dτV f(r, V,t)= 

where ρ = nm is the mass density. 

• The mean velocity 

V1 = V = 1 

+∞ 

n 

−∞ 

For any quantity Q = Q(V1) define the barred quantity 

Q = Q(r, t) = 1 

 

n(r, t) 

+∞ 

−∞ 

f(r, V,t)dVxdVydVz 

V1f(r, V1,t)dV1xdV1ydV1z 

Q( V )f(r, V,t) dτV = 1 

+∞ 

n 

−∞ 

(2.5.84) 

Q( V )f(r, V,t)dVxdVydVz. (2.5.85) 

Further, assume that F 

m is independent of V , then the moment of equation (2.5.83) produces the result 

∂ 

nφ + 

∂t 

3 

i=1 

∂ 

∂xi 

nV1iφ − n 

3 

i=1 

Fi 

m 

∂φ 

∂V1i 

=0 (2.5.86) 

known as the Maxwell transfer equation. The first term in equation (2.5.86) follows from the integrals 

∂f(r, V1,t) 

∂t 

φ( V1)dτV1 = ∂ 

 

∂t 

f(r, V1,t)φ( V1) dτV1 = ∂ 

(nφ) (2.5.87) 

∂t 

where differentiation and integration have been interchanged. The second term in equation (2.5.86) follows 

from the integral 

 

V1∇rf φ( 

3 

∂f 

V1)dτV1 = V1i φdτV1 

∂xi i=1 

3 ∂ 

= 

∂xi 

 

V1iφf dτV1 

(2.5.88) 

= 

∂V1i 

i=1 

3 

i=1 

∂ 

∂xi 

nV1iφ . 

The third term in equation (2.5.86) is obtained from the following integral where integration by parts is 

employed 

 

F 

m ∇ 

 

fφdτV1 = 

V1 

3 

 

Fi ∂f 

φdτV1 

m ∂V1i 

i=1 

+∞ 3 

 

Fi ∂f 

= φ 

dV1xdV1ydV1y 

m ∂V1i 

−∞ i=1 

 

∂ Fi 

= − 

∂V1i m φ 

 

fdτV1 

= −n ∂ 

 

Fi 

m φ 

 

= − Fi ∂φ 

m 

(2.5.89) 

∂V1i

since Fi does not depend upon V1 and f(r, V,t) equals zero for Vi equal to ±∞. The right-hand side of 

equation (2.5.86) represents the integral of (DCf)φ over velocity space. This integral is zero because of 

the symmetries associated with the right-hand side of equation (2.5.83). Physically, the integral of (Dcf)φ 

over velocity space must be zero since collisions with only scattering terms cannot increase or decrease the 

number of particles per cubic centimeter in any element of phase space. 

In equation (2.5.86) we write the velocities V1i in terms of the mean velocities (u, v, w) and random 

velocities (Ur,Vr,Wr) with 

V11 = Ur + u, V12 = Vr + v, V13 = Wr + w (2.5.90) 

or V1 = Vr + V with V1 = Vr + V = V since Vr = 0 (i.e. the average random velocity is zero.) For 

future reference we write equation (2.5.86) in terms of these random velocities and the material derivative. 

Substitution of the velocities from equation (2.5.90) in equation (2.5.86) gives 

or 

∂(nφ) 

∂t 

Observe that 

∂ 

 

+ n(Ur + u)φ + 

∂x 

∂ 

 

n(Vr + v)φ + 

∂y 

∂ 

 

 

n(Wr + w)φ 

∂z 

∂(nφ) ∂ ∂ ∂ 

+ nuφ + nvφ + nwφ 

∂t ∂x ∂y ∂z 

+ ∂ 

nUrφ 

∂x 

+ ∂ 

nVrφ 

∂y 

+ ∂ 

nWrφ 

∂z 

− n 

nuφ = 

+∞ 

−∞ 

− n 

3 

i=1 

Fi 

m 

3 

i=1 

∂φ 

Fi 

m 

∂V1i 

∂φ 

∂V1i 

=0. 

=0 (2.5.91) 

(2.5.92) 

uφf(r, V,t)dVxdVydVz = nuφ (2.5.93) 

and similarly nvφ = nvφ, nwφ = nwφ. This enables the equation (2.5.92) to be written in the form 

n ∂φ 

∂t 

+ nu∂φ 

∂x 

 

∂n 

+ φ 

∂t 

∂φ ∂φ 

+ nv + nw 

∂y ∂z 

∂ ∂ ∂ 

+ (nu)+ (nv)+ 

∂x ∂y ∂z (nw) 

+ ∂ 

nUrφ 

∂x 

+ ∂ 

nVrφ 

∂y 

+ ∂ 

nWrφ 

∂z 

− n 

 

3 

i=1 

Fi 

m 

∂φ 

∂V1i 

=0. 

(2.5.94) 

The middle bracketed sum in equation (2.5.94) is recognized as the continuity equation when multiplied by 

m and hence is zero. The moment equation (2.5.86) now has the form 

n Dφ 

Dt 

∂ 

+ nUrφ 

∂x 

+ ∂ 

nVrφ 

∂y 

+ ∂ 

nWrφ 

∂z 

− n 

3 

i=1 

Fi 

m 

∂φ 

∂V1i 

=0. (2.5.95) 

Note that from the equations (2.5.86) or (2.5.95) one can derive the basic equations of fluid flow from 

continuum mechanics developed earlier. We consider the following special cases of the Maxwell transfer 

equation. 

313

314 

(i) In the special case φ = m the equation (2.5.86) reduces to the continuity equation for fluids. That is, 

equation (2.5.86) becomes 

∂ 

∂t (nm)+∇·(nm V1) =0 (2.5.96) 

which is the continuity equation 

∂ρ 

∂t + ∇·(ρ V )=0 (2.5.97) 

where ρ is the mass density and V is the mean velocity defined earlier. 

(ii) In the special case φ = m V1 is momentum, the equation (2.5.86) reduces to the momentum equation 

for fluids. To show this, we write equation (2.5.86) in terms of the dyadic V1 V1 in the form 

∂ 

 

nm 

∂t 

 

V1 + ∇·(nm V1 V1) − n F =0 (2.5.98) 

or 

∂ 

 

ρ( 

∂t 

Vr + 

V ) + ∇·(ρ( Vr + V )( Vr + V )) − n F =0. (2.5.99) 

Let σ = −ρ Vr Vr denote a stress tensor which is due to the random motions of the gas particles and 

write equation (2.5.99) in the form 

ρ ∂ V 

∂t + V ∂ρ 

∂t + ρ V (∇· V )+ V (∇·(ρ V )) −∇·σ − n F =0. (2.5.100) 

The term 

∂ρ 

V ∂t + ∇·(ρ 

V ) = 0 because of the continuity equation and so equation (2.5.100) reduces 

to the momentum equation 

∂ 

ρ 

V 

∂t + V ∇· 

V = n F + ∇·σ. (2.5.101) 

For F = q E + q V × B + mb,whereqischarge, E and B are electric and magnetic fields, and b is a 

body force per unit mass, together with 

σ = 

3 

i=1 j=1 

the equation (2.5.101) becomes the momentum equation 

3 

(−pδij + τij)eiej 

(2.5.102) 

ρ D V 

Dt = ρ b −∇p + ∇·τ + nq( E + V × B). (2.5.103) 

In the special case were E and B vanish, the equation (2.5.103) reduces to the previous momentum 

equation (2.5.25) . 

(iii) In the special case φ = m 

2 V1 · V1 = m 2 

2 (V11 +V 2 

12 +V 2 

13) is the particle kinetic energy, the equation (2.5.86) 

simplifies to the energy equation of fluid mechanics. To show this we substitute φ into equation (2.5.95) 

and simplify. Note that 

φ = m 

 

(Ur + u) 

2 

2 + (Vr + v) 2 + (Wr + w) 2 

 

φ = m 

 

U 

2 

2 r + V 2 

r + W 2 r + u2 + v 2 + w 2 

(2.5.104)

since uUr = vVr = wWr =0. Let V 2 = u 2 + v 2 + w 2 and C 2 r = U 2 r 

(2.5.104) in the form 

Also note that 

and that 

nUrφ = nm 

2 

= nm 

2 

nVrφ = nm 

2 

nWrφ = nm 

2 

φ = m 

2 

+ V 2 

r + W 2 r 

and write equation 

 

C 2 r + V 2 

. (2.5.105) 

 

Ur(Ur + u) 2 + Ur(Vr + v) 2 + Ur(Wr + w) 2 

 

 

UrC 2 r 

2 + uU 2 

r + vUrVr + wUrWr 

 

VrC 2 r + uVrUr + vV 2 

r + wVrWr 

 

WrC 2 r + uWrUr + vWrVr + wW 2 r 

 

 

(2.5.106) 

(2.5.107) 

(2.5.108) 

are similar results. 

We use ∂ 

∂V1i (φ) =mV1i together with the previous results substituted into the equation (2.5.95), and 

find that the Maxwell transport equation can be expressed in the form 

ρ D 

 

C 

Dt 

2 

2 

r V 

+ = − 

2 2 

∂ 

 

ρ[uU 

∂x 

2 

r + vUrVr + wUrWr] 

− ∂ 

 

ρ[uVrUr + vV 

∂y 

2 

 

r + wVrWr] 

− ∂ 

 

ρ[uWrUr + vWrVr + wW 

∂z 

2 r ] 

 

− ∂ 

 

∂x 

ρ UrC 2 

r 

2 

− ∂ 

 

∂y 

ρ VrC 2 

r 

2 

− ∂ 

 

∂z 

ρ WrC 2 

r 

2 

+ n F · (2.5.109) 

V. 

Compare the equation (2.5.109) with the energy equation (2.5.48) 

ρ De 

2 D V 

+ ρ = ∇(σ · 

Dt Dt 2 

V ) −∇·q + ρb · V (2.5.110) 

where the internal heat energy has been set equal to zero. Let e = C2 r 

2 

random motion of the gas particles, F = mb,andlet ∇·q = − ∂ 

∂x 

= − ∂ 

∂x 

 

ρ UrC 2 r 

2 

 

k ∂T 

∂x 

 

− ∂ 

 

∂y 

 

− ∂ 

 

k 

∂y 

∂T 

∂y 

 

− 

2 

∂ 

 

∂z 

 

− ∂ 

 

k 

∂z 

∂T 

 

∂z 

ρ VrC 2 r 

denote the internal energy due to 

ρ WrC 2 r 

2 

 

(2.5.111) 

represent the heat conduction terms due to the transport of particle energy mC2 r 

2 by way of the random 

particle motion. The remaining terms are related to the rate of change of work and surface stresses giving 

− ∂ 

 

ρ[uU 

∂x 

2 r + vUrVr 

 

+ wUrWr] 

− ∂ 

 

ρ[uVrUr + vV 

∂y 

2 

 

r + wVrWr] 

− ∂ 

 

ρ[uWrUr + vWrVr + wW 

∂z 

2 r ] 

 

= ∂ 

∂x (uσxx + vσxy + wσxz) 

= ∂ 

∂y (uσyx + vσyy + wσyz) 

= ∂ 

∂z (uσzx + vσzy + wσzz) . 

(2.5.112) 

315

316 

This gives the stress relations due to random particle motion 

σxx = − ρU 2 r 

σxy = − ρUrVr 

σxz = − ρUrWr 

σyx = − ρVrUr 

σyy = − ρV 2 

r 

σyz = − ρVrWr 

σzx = − ρWrUr 

σzy = − ρWrVr 

σzz = − ρW 2 r . 

(2.5.113) 

The Boltzmann equation is a basic macroscopic model used for the study of individual particle motion 

where one takes into account the distribution of particles in both space, time and energy. The Boltzmann 

equation for gases assumes only binary collisions as three-body or multi-body collisions are assumed to 

rarely occur. Another assumption used in the development of the Boltzmann equation is that the actual 

time of collision is thought to be small in comparison with the time between collisions. The basic problem 

associated with the Boltzmann equation is to find a velocity distribution, subject to either boundary and/or 

initial conditions, which describes a given gas flow. 

The continuum equations involve trying to obtain the macroscopic variables of density, mean velocity, 

stress, temperature and pressure which occur in the basic equations of continuum mechanics considered 

earlier. Note that the moments of the Boltzmann equation, derived for gases, also produced these same 

continuum equations and so they are valid for gases as well as liquids. 

In certain situations one can assume that the gases approximate a Maxwellian distribution 

f(r, 

m 

3/2 

V,t) ≈ n(r, t) 

exp − 

2πkT 

m 

2kT V · 

V 

(2.5.114) 

thereby enabling the calculation of the pressure tensor and temperature from statistical considerations. 

In general, one can say that the Boltzmann integral-differential equation and the Maxwell transfer 

equation are two important formulations in the kinetic theory of gases. The Maxwell transfer equation 

depends upon some gas-particle property φ which is assumed to be a function of the gas-particle velocity. 

The Boltzmann equation depends upon a gas-particle velocity distribution function f which depends upon 

position r, velocity V and time t. These formulations represent two distinct and important viewpoints 

considered in the kinetic theory of gases.

EXERCISE 2.5 

◮ 1. Let p = p(x, y, z), [dyne/cm2 ] denote the pressure at a point (x, y, z) inafluidmediumatrest 

(hydrostatics), and let ∆V denote an element of fluid volume situated at this point as illustrated in the 

figure 2.5-5. 

Figure 2.5-5. Pressure acting on a volume element. 

(a) Show that the force acting on the face ABCD is p(x, y, z)∆y∆z ê1. 

(b) Show that the force acting on the face EFGH is 

 

−p(x +∆x, y, z)∆y∆z ê1 = − p(x, y, z)+ ∂p 

∂x ∆x + ∂2p ∂x2 (∆x) 2 

+ ··· ∆y∆z ê1. 

2! 

(c) In part (b) neglect terms with powers of ∆x greater than or equal to 2 and show that the resultant force 

in the x-direction is − ∂p 

∆x∆y∆z ê1. 

∂x 

(d) What has been done in the x-direction can also be done in the y and z-directions. Show that the 

resultant forces in these directions are − ∂p 

∂y ∆x∆y∆z ê2 and − ∂p 

∂z ∆x∆y∆z ê3. (e) Show that −∇p = 

 

∂p 

− 

∂x ê1 + ∂p 

∂y ê2 + ∂p 

∂z ê3 

 

is the force per unit volume acting at the point (x, y, z) of the fluid medium. 

◮ 2. Follow the example of exercise 1above but use cylindrical coordinates and find the force per unit volume 

at a point (r, θ, z). Hint: An element of volume in cylindrical coordinates is given by ∆V = r∆r∆θ∆z. 

◮ 3. Follow the example of exercise 1above but use spherical coordinates and find the force per unit volume 

at a point (ρ, θ, φ). Hint: An element of volume in spherical coordinates is ∆V = ρ2 sin θ∆ρ∆θ∆φ. 

◮ 4. Show that if the density ϱ = ϱ(x, y, z, t) is a constant, then v r ,r =0. 

◮ 5. Assume that λ∗ and µ ∗ are zero. Such a fluid is called a nonviscous or perfect fluid. (a) Show the 

Cartesian equations describing conservation of linear momentum are 

∂u ∂u ∂u ∂u 

+ u + v + w 

∂t ∂x ∂y ∂z = bx − 1 ∂p 

ϱ ∂x 

∂v ∂v ∂v ∂v 

+ u + v + w 

∂t ∂x ∂y ∂z = by − 1 ∂p 

ϱ ∂y 

∂w ∂w ∂w ∂w 

+ u + v + w 

∂t ∂x ∂y ∂z = bz − 1 ∂p 

ϱ ∂z 

where (u, v, w) are the physical components of the fluid velocity. (b) Show that the continuity equation can 

be written 

∂ϱ ∂ ∂ ∂ 

+ (ϱu)+ (ϱv)+ (ϱw) =0 

∂t ∂x ∂y ∂z 

317

318 

◮ 6. Assume λ∗ = µ ∗ = 0 so that the fluid is ideal or nonviscous. Use the results given in problem 5 and 

make the following additional assumptions: 

• The density is constant and so the fluid is incompressible. 

• The body forces are zero. 

• Steady state flow exists. 

• Only two dimensional flow in the x-yplane is considered such that u = u(x, y), v = v(x, y) and 

w =0. (a) Employ the above assumptions and simplify the equations in problem 5 and verify the 

results 

u ∂u ∂u 1 ∂p 

+ v + 

∂x ∂y ϱ ∂x =0 

u ∂v ∂v 1 ∂p 

+ v + 

∂x ∂y ϱ ∂y =0 

∂u ∂v 

+ 

∂x ∂y =0 

(b) Make the additional assumption that the flow is irrotational and show that this assumption 

produces the results 

∂v ∂u 

− =0 and 

∂x ∂y 

1 

2 

2 2 

u + v + 1 

p = constant. 

ϱ 

(c) Point out the Cauchy-Riemann equations and Bernoulli’s equation in the above set of equations. 

◮ 7. Assume the body forces are derivable from a potential function φ such that bi = −φ,i. Show that for an 

ideal fluid with constant density the equations of fluid motion can be written in either of the forms 

∂v r 

∂t + vr ,sv s = − 1 

ϱ grm p,m − g rm φ,m or 

∂vr 

∂t + vr,sv s = − 1 

ϱ p,r − φ,r 

◮ 8. The vector identities ∇ 2 v = ∇ (∇·v) −∇×(∇×v) and (v ·∇) v = 1 

∇ (v · v) − v × (∇×v) are 

2 

used to express the Navier-Stokes-Duhem equations in alternate forms involving the vorticity Ω = ∇×v. 

(a) Use Cartesian tensor notation and derive the above identities. (b) Show the second identity can be written 

in generalized coordinates as v j v m ,j = g mj v k vk,j − ɛ mnp ɛ ijk gpivnvk,j. Hint: Show that ∂v2 

∂xj =2vkvk,j. ◮ 9. Use problem 8 and show that the results in problem 7 can be written 

or 

∂vr ∂t − ɛrnp rm ∂ 

Ωpvn = −g 

∂vi 

∂t − ɛijkv j Ω k = − ∂ 

∂x i 

∂x m 

 

p v2 

+ φ + 

ϱ 2 

 

p v2 

+ φ + 

ϱ 2 

◮ 10. In terms of physical components, show that in generalized orthogonal coordinates, for i = j, therate 

of deformation tensor Dij can be written D(ij) = 1 

 

hi ∂ 

2 hj ∂xj 

v(i) 

+ 

hi 

hj ∂ 

hi ∂xi 

v(j) 

, no summations 

hj 

and for i = j there results D(ii) = 1 ∂v(i) v(i) 

− 

hi ∂xi h2 3 ∂hi 1 

+ v(k) 

i ∂xi hihk 

k=1 

∂hi 

, no summations. (Hint: See 

∂xk Problem 17 Exercise 2.1.)

Figure 2.5-6. Plane Couette flow 

◮ 11. Find the physical components of the rate of deformation tensor Dij in Cartesian coordinates. (Hint: 

See problem 10.) 

◮ 12. Find the physical components of the rate of deformation tensor in cylindrical coordinates. (Hint: See 

problem 10.) 

◮ 13. (Plane Couette flow) 


Assume a viscous fluid with constant density is between two plates as illustrated 

(a) Define ν = µ∗ 

ϱ as the kinematic viscosity and show the equations of fluid motion can be written 

∂v i 

∂t + vi ,s vs = − 1 

ϱ gim p,m + νg jm v i ,mj + gij bj, i =1, 2, 3 

(b) Let v =(u, v, w) denote the physical components of the fluid flow and make the following assumptions 

• u = u(y), v= w =0 

• Steady state flow exists 

• The top plate, with area A, isadistanceℓabovethe bottom plate. The bottom plate is fixed and 

a constant force F is applied to the top plate to keep it moving with a velocity u0 = u(ℓ). 

• p and ϱ are constants 

• The body force components are zero. 

Find the velocity u = u(y) 

(c) Show the tangential stress exerted by the moving fluid is F 

A = σ21 

∗ u0 

= σxy = σyx = µ . This 

ℓ 

example illustrates that the stress is proportional to u0 and inversely proportional to ℓ. 

◮ 14. In the continuity equation make the change of variables 

t = t ϱ 

, ϱ = , v = 

τ ϱ0 

v 

v0 

, x = x 

L 

y z 

, y = , z = 

L L 

and write the continuity equation in terms of the barred variables and the Strouhal parameter. 

◮ 15. (Plane Poiseuille flow) Consider two flat plates parallel to one another as illustrated in the figure 

2.5-7. One plate is at y = 0 and the other plate is at y =2ℓ. Letv =(u, v, w) denote the physical components 

of the fluid velocity and make the following assumptions concerning the flow The body forces are zero. The 

derivative ∂p 

∂x = −p0 is a constant and ∂p ∂p 

= =0. The velocity in the x-direction is a function of y only 

∂y ∂z 

319

320 

Figure 2.5-7. Plane Poiseuille flow 

with u = u(y) andv = w = 0 with boundary values u(0) = u(2ℓ) =0. The density is constant and ν = µ ∗ /ϱ 

is the kinematic viscosity. 

(a) Show the equation of fluid motion is ν d2u p0 

+ =0, u(0) = u(2ℓ) =0 

dy2 ϱ 

(b) Find the velocity u = u(y) and find the maximum velocity in the x-direction. (c) Let M denote the 

mass flow rate across the plane x = x0 = constant, ,where0≤ y ≤ 2ℓ, and 0 ≤ z ≤ 1. 

Show that M = 2 

3µ ∗ ϱp0ℓ 3 . Note that as µ ∗ increases, M decreases. 

◮ 16. The heat equation (or diffusion equation) can be expressed div ( k grad u)+H = ∂(δcu) 

, where c is the 

∂t 

specific heat [cal/gm C], δ is the volume density [gm/cm 3 ], H is the rate of heat generation [cal/sec cm 3 ], u 

is the temperature [C], k is the thermal conductivity [cal/sec cm C]. Assume constant thermal conductivity, 

volume density and specific heat and express the boundary value problem 

k ∂2u = δc∂u, 

∂x2 ∂t 

0

Figure 2.5-8. Rayleigh impulsive flow 

where erf and erfc are the error function and complimentary error function respectively. Pick a point on the 

line y = y0 =2 √ ν and plot the velocity as a function of time. How does the viscosity effect the velocity of 

the fluid along the line y = y0? 

◮ 19. Simplify the Navier-Stokes-Duhem equations using the assumption that there is incompressible and 

irrotational flow. 

◮ 20. Let ζ = λ∗ + 2 

3 µ∗ and show the constitutive equations (2.5.21) for fluid motion can be written in the 

form 

σij = −pδij + µ ∗ 

 

vi,j + vj,i − 2 

3 δijvk,k 

 

+ ζδijvk,k. 

◮ 21. (a) Write out the Navier-Stokes-Duhem equation for two dimensional flow in the x-y direction under 

the assumptions that 

• λ ∗ + 2 

3 µ∗ = 0 (This condition is referred to as Stoke’s flow.) 

• The fluid is incompressible 

• There is a gravitational force b = −g∇ h Hint: Express your answer as two scalar equations 

involving the variables v1,v2,h,g,ϱ,p,t,µ ∗ plus the continuity equation. (b) In part (a) eliminate 

the pressure and body force terms by cross differentiation and subtraction. (i.e. take the derivative 

of one equation with respect to x and take the derivative of the other equation with respect to y 

and then eliminate any common terms.) (c) Assume that ω = ω ê3 where ω = 1 

 

∂v2 ∂v1 

− and 

2 ∂x ∂y 

derive the vorticity-transport equation 

dω 

dt = ν∇2 ω where 

dω 

dt 

= ∂ω 

∂t 

∂ω 

+ v1 

∂x 

∂ω 

+ v2 

∂y . 

Hint: The continuity equation makes certain terms zero. (d) Define a stream function ψ = ψ(x, y) 

satisfying v1 = ∂ψ 

∂y 

and v2 = − ∂ψ 

and show the continuity equation is identically satisfied. 

∂x 

Show also that ω = − 1 

2 ∇2 ψ and that 

∇ 4 ψ = 1 

ν 

If ν is very large, show that ∇ 4 ψ ≈ 0. 

∂∇ 2 ψ 

∂t 

∂ψ ∂∇ 

+ 

∂y 

2ψ ∂ψ 

− 

∂x ∂x 

∂∇2 

ψ 

. 

∂y 

321

322 

◮ 22. In generalized orthogonal coordinates, show that the physical components of the rate of deformation 

stress can be written, for i = j 

σ(ij) =µ ∗ 

 

hi ∂ 

hj ∂xj 

v(i) 

+ 

hi 

hj ∂ 

hi ∂xi 

v(j) 

, 

hj 

no summation, 

and for i = j = k 

σ(ii) =−p +2µ ∗ 

 

1 

hi 

∂v(i) 1 

+ 

∂xi hihj 

+ λ∗ 

 

∂ 

h1h2h3 ∂x1 {h2h3v(1)} + ∂ 

v(j) ∂hi 1 

+ v(k) 

∂xj hihk 

∂hi 

∂xk 

∂x2 {h1h3v(2)} + ∂ 

 

{h1h2v(3)} , no summation 

∂x3 ◮ 23. Find the physical components for the rate of deformation stress in Cartesian coordinates. Hint: See 

problem 22. 

◮ 24. Find the physical components for the rate of deformations stress in cylindrical coordinates. Hint: See 

problem 22. 

◮ 25. Verify the Navier-Stokes equations for an incompressible fluid can be written ˙vi = − 1 

ϱ p,i + νvi,mm + bi 

where ν = µ∗ 

ϱ is called the kinematic viscosity. 

◮ 26. Verify the Navier-Stokes equations for a compressible fluid with zero bulk viscosity can be written 

˙vi = − 1 

ϱ p,i + ν 

3 vm,mi + νvi,mm + bi with ν = µ∗ 

ϱ the kinematic viscosity. 

◮ 27. The constitutive equation for a certain non-Newtonian Stokesian fluid is σij = −pδij +βDij +γDikDkj. 

Assume that β and γ are constants (a) Verify that σij,j = −p,i + βDij,j + γ(DikDkj,j + Dik,jDkj) 

(b) Write out the Cauchy equations of motion in Cartesian coordinates. (See page 236). 

◮ 28. Let the constitutive equations relating stress and strain for a solid material take into account thermal 

stresses due to a temperature T . The constitutive equations have the form eij = 1+ν 

E σij − ν 

E σkk δij +αT δij 

where α is a coefficient of linear expansion for the material and T is the absolute temperature. Solve for the 

stress in terms of strains. 

◮ 29. Derive equation (2.5.53) and then show that when the bulk coefficient of viscosity is zero, the Navier- 

Stokes equations, in Cartesian coordinates, can be written in the conservation form 

∂(ϱu) 

∂t + ∂(ϱu2 + p − τxx) 

+ 

∂x 

∂(ϱuv − τxy) 

+ 

∂y 

∂(ϱuw − τxz) 

∂z 

∂(ϱv) ∂(ϱuv − τxy) 

+ + 

∂t ∂x 

∂(ϱv2 + p − τyy) 

+ 

∂y 

∂(ϱvw − τyz) 

∂z 

∂(ϱw) ∂(ϱuw − τxz) 

+ + 

∂t ∂x 

∂(ϱvw − τyz) 

+ 

∂y 

∂(ϱw2 + p − τzz) 

∂z 

= ϱbx 

= ϱby 

= ϱbz 

where v1 = u,v2 = v,v3 = w and τij = µ ∗ (vi,j + vj,i − 2 

3 δijvk,k). Hint: Alternatively, consider 2.5.29 and use 

the continuity equation.

◮ 30. Show that for a perfect gas, where λ ∗ = − 2 

3 µ∗ and η = µ ∗ is a function of position, the vector form 

of equation (2.5.25) is 

ϱ Dv 

Dt = ϱ b −∇p + 4 

3 ∇(η∇·v)+∇(v ·∇η) − v ∇2 η +(∇η) × (∇×v) − (∇·v)∇η −∇×(∇×(ηv)) 

◮ 31. Derive the energy equation ϱ Dh Dp ∂Q 

= + −∇·q +Φ. Hint: Use the continuity equation. 

Dt Dt ∂t 

◮ 32. Show that in Cartesian coordinates the Navier-Stokes equations of motion for a compressible fluid 


ρ Du 

Dt =ρbx − ∂p 

 

∂ ∗ ∂u 

+ 2µ 

∂x ∂x ∂x + λ∗∇· 

V + ∂ 

 

µ 

∂y 

∗ ( ∂u ∂v 

+ 

∂y ∂x ) 

 

+ ∂ 

 

µ 

∂z 

∗ ( ∂w ∂u 

+ 

∂x ∂z ) 

 

ρ Dv 

Dt =ρby − ∂p 

 

∂ ∗ ∂v 

+ 2µ 

∂y ∂y ∂y + λ∗∇· 

V + ∂ 

 

µ 

∂z 

∗ ( ∂v ∂w 

+ 

∂z ∂y ) 

 

+ ∂ 

 

µ 

∂x 

∗ ( ∂w ∂w 

+ 

∂y ∂x ) 

 

ρ Dv 

Dt =ρbz − ∂p 

 

∂ ∗ ∂w 

+ 2µ 

∂z ∂z ∂z + λ∗∇· 

V + ∂ 

 

µ 

∂x 

∗ ( ∂w ∂u 

+ 

∂x ∂z ) 

 

+ ∂ 

 

µ 

∂y 

∗ ( ∂v ∂w 

+ 

∂z ∂y ) 

 

where (Vx,Vy,Vz) =(u, v, w). 

◮ 33. Show that in cylindrical coordinates the Navier-Stokes equations of motion for a compressible fluid 


2 DVr Vθ ϱ − =ϱbr − 

Dt r 

∂p 

 

∂ ∗ ∂Vr 

+ 2µ 

∂r ∂r ∂r + λ∗∇· 

V + 1 

 

∂ 

µ 

r ∂θ 

∗ ( 1 ∂Vr ∂Vθ Vθ 

+ − 

r ∂θ ∂r r ) 

 

+ ∂ 

 

µ 

∂z 

∗ ( ∂Vr ∂Vz 

+ 

∂z ∂r ) 

 

+ 2µ∗ 

r (∂Vr 

1 ∂Vθ Vr 

− − 

∂r r ∂θ r ) 

 

DVθ VrVθ 

ϱ + =ϱbθ − 

Dt r 

1 

 

∂p 1 ∂ 

+ 2µ 

r ∂θ r ∂θ 

∗ ( 1 ∂Vθ Vr 

+ 

r ∂θ r )+λ∗∇· 

V + ∂ 

 

µ 

∂z 

∗ ( 1 ∂Vz ∂Vθ 

+ 

r ∂θ ∂z ) 

 

+ ∂ 

 

µ 

∂r 

∗ ( 1 ∂Vr ∂Vθ Vθ 

+ − 

r ∂θ ∂r r ) 

 

+ 2µ∗ 

r (1 

∂Vr ∂Vθ Vθ 

+ − 

r ∂θ ∂r r ) 

ϱ DVz 

Dt =ϱbz − ∂p 

 

∂ ∗ ∂Vz 

+ 2µ 

∂z ∂z ∂z + λ∗∇· 

V + 1 

 

∂ 

µ 

r ∂r 

∗ r( ∂Vr ∂Vz 

+ 

∂z ∂r ) 

 

+ 1 

 

∂ 

µ 

r ∂θ 

∗ ( 1 ∂Vz ∂Vθ 

+ 

r ∂θ ∂z ) 

 

◮ 34. Show that the dissipation function Φ can be written as Φ = 2µ ∗ DijDij + λ ∗ Θ 2 . 

◮ 35. Verify the identities: 

(a) ϱ D 

Dt (et/ϱ) = ∂et 

∂t + ∇·(et V ) (b) ϱ D 

Dt (et/ϱ) =ϱ De D 2 

+ ϱ V /2 . 

Dt Dt 

◮ 36. Show that the conservation law for heat flow is given by 

∂T 

∂t 

+ ∇·(Tv − κ∇T )=SQ 

where κ is the thermal conductivity of the material, T is the temperature, Jadvection = Tv, 

Jconduction = −κ∇T and SQ is a source term. Note that in a solid material there is no flow and so v =0and 

323

324 

the above equation reduces to the heat equation. Assign units of measurements to each term in the above 

equation and make sure the equation is dimensionally homogeneous. 

◮ 37. Show that in spherical coordinates the Navier-Stokes equations of motion for a compressible fluid can 

be written 

ϱ( DVρ 

2 Vθ − 

Dt + V 2 φ 

)=ϱbρ − 

ρ 

∂p 

 

∂ ∗ ∂Vρ 

+ 2µ 

∂ρ ∂ρ ∂ρ + λ∗∇· 

V + 1 

 

∂ 

µ 

ρ ∂θ 

∗ (ρ ∂ 1 ∂Vρ 

(Vθ/ρ)+ 

∂ρ ρ ∂θ ) 

 

+ 1 

 

∂ 

µ 


∗ 1 ∂Vρ ∂ 

( + ρ 

ρ sin θ ∂φ ∂ρ (Vφ/ρ)) 

 

+ µ∗ ∂Vρ 2 ∂Vθ 4Vρ 2 ∂Vφ 

(4 − − − 

ρ ∂ρ ρ ∂θ ρ ρ sin θ ∂φ − 2Vθ cot θ 

+ ρ cot θ 

ρ 

∂ cot θ ∂Vρ 

(Vθ/ρ)+ 

∂ρ ρ ∂θ ) 

ϱ( DVθ VρVθ 

+ 

Dt ρ − V 2 φ cot θ 

)=ϱbθ − 

ρ 

1 

∗ 

∂p 1 ∂ 2µ 

+ 

ρ ∂θ ρ ∂θ ρ (∂Vθ 

∂θ + Vρ)+λ ∗ ∇· 

V 

+ 1 

 

∂ 

µ 


∗ sin θ ∂ 

( 

ρ ∂θ (Vφ/ sin θ)+ 1 ∂Vθ 

ρ sin θ ∂φ ) 

 

+ ∂ 

 

µ 

∂ρ 

∗ (ρ ∂ 1 ∂Vρ 

(Vθ/ρ)+ 

∂ρ ρ ∂θ ) 

 

+ µ∗ 

 

1 ∂Vθ 1 ∂Vφ 

2 − 

ρ ρ ∂θ ρ sin θ ∂φ − Vθ 

 

cot θ 

cot θ +3 ρ 

ρ 

∂ 

 

1 ∂Vρ 

(Vθ/ρ)+ 

∂ρ ρ ∂θ 

 

DVφ VφVρ 

ϱ + 

Dt ρ + VθVφ 

 

cot θ 

= ϱbφ − 

ρ 

1 

 

∂p ∂ 

+ µ 

ρ sin θ ∂φ ∂ρ 

∗ 

 

1 ∂Vρ ∂ 

+ ρ 

ρ sin θ ∂φ ∂ρ (Vφ/ρ) 

 

+ 1 

∗ ∂ 2µ 1 ∂Vφ 

ρ sin θ ∂φ ρ sin θ ∂φ + Vρ 

 

+ Vθ cot θ + λ ∗ ∇· 

V 

+ 1 

 

∂ 

µ 

ρ ∂θ 

∗ 

 

sin θ ∂ 

ρ ∂θ (Vφ/ sin θ)+ 1 

 

∂Vθ 


+ µ∗ 

 

1 ∂Vρ ∂ 

3 

+ ρ 

ρ ρ sin θ ∂φ ∂ρ (Vφ/ρ) 

 

sin θ ∂ 

+2cotθ 

ρ ∂θ (Vφ/ sin θ)+ 1 

 

∂Vθ 


◮ 38. Verify all the equations (2.5.28). 

◮ 39. Use the conservation of energy equation (2.5.47) together with the momentum equation (2.5.25) to 

derive the equation (2.5.48). 

◮ 40. Verify the equation (2.5.55). 

◮ 41. Consider nonviscous flow and write the 3 linear momentum equations and the continuity equation 

and make the following assumptions: (i) The density ϱ is constant. (ii) Body forces are zero. (iii) Steady 

state flow only. (iv) Consider only two dimensional flow with non-zero velocity components u = u(x, y) and 

v = v(x, y). Show that there results the system of equations 

u ∂u ∂u 1 ∂P 

∂v 1 ∂P 

+ v + =0, u∂v + v + 

∂x ∂y ϱ ∂x ∂x ∂y ϱ ∂y =0, 

∂u ∂v 

+ 

∂x ∂y =0. 

Recognize that the last equation in the above set as one of the Cauchy-Riemann equations that f(z) =u−iv 

be an analytic function of a complex variable. Further assume that the fluid flow is irrotational so that 

∂v ∂u 

1 2 2 

− =0. Show that this implies that u + v 

∂x ∂y 2 

+ P 

= Constant. If in addition u and v are derivable 

ϱ 

from a potential function φ(x, y), such that u = ∂φ 

∂φ 

∂x and v = ∂y , then show that φ is a harmonic function. 

By constructing the conjugate harmonic function ψ(x, y) the complex potential F (z) =φ(x, y)+iψ(x, y) is 

such that F ′ (z) =u(x, y) − iv(x, y) andF ′ (z) gives the velocity. The family of curves φ(x, y) =constantare 

called equipotential curves and the family of curves ψ(x, y) = constant are called streamlines. Show that 

these families are an orthogonal family of curves.

§2.6 ELECTRIC AND MAGNETIC FIELDS 

Introduction 

In electromagnetic theory the mks system of units and the Gaussian system of units are the ones most 

often encountered. In this section the equations will be given in the mks system of units. If you want the 

equations in the Gaussian system of units make the replacements given in the column 3 of Table 1. 

MKS 

symbol 

Table 1. MKS AND GAUSSIAN UNITS 

MKS 

units 

Replacement 

symbol 

GAUSSIAN 

units 

E (Electric field) volt/m E statvolt/cm 

B (Magnetic field) weber/m 2 B 

c 

D (Displacement field) coulomb/m 2 D 

4π 

H (Auxiliary Magnetic field) ampere/m c H 

4π 

gauss 

statcoulomb/cm 2 

oersted 

J (Current density) ampere/m 2 J statampere/cm 2 

A (Vector potential) weber/m 

A 

c 

gauss-cm 

V (Electric potential) volt V statvolt 

ɛ (Dielectric constant) 

µ (Magnetic permeability) 

Electrostatics 

A basic problem in electrostatic theory is to determine the force F on a charge Q placed a distance r 

from another charge q. The solution to this problem is Coulomb’s law 

F = 1 qQ 

er 

(2.6.1) 

4πɛ0 r2 where q, Q aremeasuredincoulombs,ɛ0 =8.85 × 10 −12 coulomb 2 /N · m 2 is called the permittivity in a 

vacuum, r is in meters, [ F ] has units of Newtons and er is a unit vector pointing from q to Q if q, Q have 

thesamesignorpointingfromQto q if q, Q are of opposite sign. The quantity E = F/Q is called the 

electric field produced by the charges. In the special case Q =1,wehave E = F and so Q = 1 is called 

a test charge. This tells us that the electric field at a point P can be viewed as the force per unit charge 

exerted on a test charge Q placed at the point P. The test charge Q is always positive and so is repulsed if 

q is positive and attracted if q is negative. 

The electric field associated with many charges is obtained by the principal of superposition. For 

example, let q1,q2,...,qn denote n-charges having respectively the distances r1,r2,...,rn from a test charge 

Q placed at a point P. The force exerted on Q is 

F = F1 + F2 + ···+ Fn 

F = 1 

 

q1Q 

4πɛ0 r2 er1 + 

1 

q2Q 

r2 er2 + ···+ 

2 

qnQ 

r2 

ern 

n 

(2.6.2) 

n 

or E = E(P 

F 1 qi 

)= = eri 

Q 4πɛ0 

r 

i=1 

2 i 

ɛ 

4π 

4πµ 

c 2 

325

326 

where E = E(P ) is the electric field associated with the system of charges. The equation (2.6.2) can be generalized 

to other situations by defining other types of charge distributions. We introduce a line charge density 

λ ∗ ,(coulomb/m), a surface charge density µ ∗ ,(coulomb/m 2 ), a volume charge density ρ ∗ ,(coulomb/m 3 ), 

then we can calculate the electric field associated with these other types of charge distributions. For example, 

if there is a charge distribution λ ∗ = λ ∗ (s) along a curve C, wheres is an arc length parameter, then we 

would have 

E(P )= 1 

 

er 

4πɛ0 C r2 λ∗ds (2.6.3) 

as the electric field at a point P due to this charge distribution. The integral in equation (2.6.3) being a 

line integral along the curve C and where ds is an element of arc length. Here equation (2.6.3) represents a 

continuous summation of the charges along the curve C. For a continuous charge distribution over a surface 

S, the electric field at a point P is 

E(P )= 1 

 

er 

4πɛ0 S r2 µ∗dσ (2.6.4) 

where dσ represents an element of surface area on S. Similarly, if ρ∗ represents a continuous charge distribution 

throughout a volume V , then the electric field is represented 

E(P )= 1 

4πɛ0 

 

V 

er 

r 2 ρ∗ dτ (2.6.5) 

where dτ is an element of volume. In the equations (2.6.3), (2.6.4), (2.6.5) we let (x, y, z) denote the position 

of the test charge and let (x ′ ,y ′ ,z ′ ) denote a point on the line, on the surface or within the volume, then 

r =(x − x ′ ) e1 +(y − y ′ ) e2 +(z − z ′ ) e3 

(2.6.6) 

represents the distance from the point P to an element of charge λ ∗ ds, µ ∗ dσ or ρ ∗ dτ with r = |r| and er = r 

r . 

If the electric field is conservative, then ∇× E = 0, and so it is derivable from a potential function V 

by taking the negative of the gradient of V and 

E = −∇V. (2.6.7) 

For these conditions note that ∇V · dr = − E · dr is an exact differential so that the potential function can 

be represented by the line integral 

P 

V = V(P )=− E · dr 

α 

(2.6.8) 

where α is some reference point (usually infinity, where V(∞) = 0). For a conservative electric field the line 

integral will be independent of the path connecting any two points a and b so that 

b a b 

V(b) −V(a) =− E · dr − − E · dr = − E · dr = 

α 

α 

a 

b 

a 

∇V · dr. (2.6.9) 

Let α = ∞ in equation (2.6.8), then the potential function associated with a point charge moving in 

the radial direction er is 

r 

V(r) =− E · dr = 

∞ 

−q 

r 

1 q 1 

dr = 

4πɛ0 ∞ r2 4πɛ0 r |r q 

∞ = 

4πɛ0r .

By superposition, the potential at a point P for a continuous volume distribution of charges is given by 

V(P )= 1 

 

ρ 

4πɛ0 V 

∗ 

 

1 µ 

dτ and for a surface distribution of charges V(P )= 

r 4πɛ0 S 

∗ 

dσ and for a line 

r 

distribution of charges V(P )= 1 

 

λ 

4πɛ0 C 

∗ 

ds; and for a discrete distribution of point charges 

r 

V(P )= 1 

N qi 

. When the potential functions are defined from a common reference point, then the 

4πɛ0 ri 

i=1 

principal of superposition applies. 

The potential function V is related to the work done W in moving a charge within the electric field. 

The work done in moving a test charge Q from point a to point b is an integral of the force times distance 

moved. The electric force on a test charge Q is F = Q E and so the force F = −Q E is in opposition to this 

force as you move the test charge. The work done is 

W = 

b 

a 

F · dr = 

b 

a 

−Q E · dr = Q 

b 

a 

∇V · dr = Q[V(b) −V(a)]. (2.6.10) 

The work done is independent of the path joining the two points and depends only on the end points and 

the change in the potential. If one moves Q from infinity to point b, then the above becomes W = QV (b). 

An electric field E = E(P ) is a vector field which can be represented graphically by constructing vectors 

at various selected points in the space. Such a plot is called a vector field plot. A field line associated with 

a vector field is a curve such that the tangent vector to a point on the curve has the same direction as the 

vector field at that point. Field lines are used as an aid for visualization of an electric field and vector fields 

in general. The tangent to a field line at a point has the same direction as the vector field E at that point. 

For example, in two dimensions let r = x e1 + y e2 denote the position vector to a point on a field line. The 

tangent vector to this point has the direction dr = dx e1 + dy e2. If E = E(x, y) =−N(x, y) e1 + M(x, y) e2 

is the vector field constructed at the same point, then E and dr must be colinear. Thus, for each point (x, y) 

onafieldlinewerequirethatdr = K E for some constant K. Equating like components we find that the 

field lines must satisfy the differential relation. 

dx 

−N(x, y) = 

dy 

M(x, y) =K 

or M(x, y) dx + N(x, y) dy =0. 

(2.6.11) 

In two dimensions, the family of equipotential curves V(x, y) =C1 =constant, are orthogonal to the family 

of field lines and are described by solutions of the differential equation 

N(x, y) dx − M(x, y) dy =0 

obtained from equation (2.6.11) by taking the negative reciprocal of the slope. The field lines are perpendicular 

to the equipotential curves because at each point on the curve V = C1 we have ∇V being perpendicular 

to the curve V = C1 and so it is colinear with E at this same point. Field lines associated with electric 

fields are called electric lines of force. The density of the field lines drawn per unit cross sectional area are 

proportional to the magnitude of the vector field through that area. 

327

328 

Figure 2.6-1. Electric forces due to a positive charge at (−a, 0) and negative charge at (a, 0). 


Find the field lines and equipotential curves associated with a positive charge q located at the point 

(−a, 0) and a negative charge −q located at the point (a, 0). 

Solution: With reference to the figure 2.6-1, the total electric force E on a test charge Q =1place 

at a general point (x, y) is, by superposition, the sum of the forces from each of the isolated charges and is 

E = E1 + E2. The electric force vectors due to each individual charge are 

where k = 1 

4πɛ0 

E1 = kq(x + a) e1 + kqy e2 

r 3 1 

E2 = −kq(x − a) e1 − kqy e2 

r 3 2 

is a constant. This gives 

E = E1 + 

kq(x + a) 

E2 = 

− 

This determines the differential equation of the field lines 

r 3 1 

kq(x+a) 

r 3 1 

dx 

− kq(x−a) 

r 3 2 

To solve this differential equation we make the substitutions 

x + a 

cos θ1 = 

r1 

with r 2 1 =(x + a) 2 + y 2 

with r 2 2 =(x − a)2 + y 2 

kq(x − a) 

r3 

kqy 

e1 + 

2 

r3 − 

1 

kqy 

r3 

e2. 

2 

= 

kqy 

r 3 1 

dy 

− kqy 

r 3 2 

x − a 

and cos θ2 = 


. (2.6.13) 

r2 

(2.6.14)

Figure 2.6-2. Lines of electric force between two opposite sign charges. 

as suggested by the geometry from figure 2.6-1. From the equations (2.6.12) and (2.6.14) we obtain the 

relations 

which implies that 

− sin θ1 dθ1 = r1dx − (x + a) dr1 

r 2 1 

2r1dr1 =2(x + a) dx +2ydy 

− sin θ2 dθ2 = r2 dx − (x − a)dr2 

r 2 2 

2r2 dr2 =2(x − a) dx +2ydy 

− sin θ1 dθ1 = − 

− sin θ2 dθ2 = − 

(x + a)ydy 

r3 1 

(x − a)ydy 

r3 2 

+ y2 dx 

r 3 1 

+ y2 dx 

r 3 2 

(2.6.15) 

Now compare the results from equation (2.6.15) with the differential equation (2.6.13) and determine that 

y is an integrating factor of equation (2.6.13) . This shows that the differential equation (2.6.13) can be 

written in the much simpler form of the exact differential equation 

− sin θ1 dθ1 +sinθ2 dθ2 =0 (2.6.16) 

in terms of the variables θ1 and θ2. The equation (2.6.16) is easily integrated to obtain 

cos θ1 − cos θ2 = C (2.6.17) 

where C is a constant of integration. In terms of x, y the solution can be written 

These field lines are illustrated in the figure 2.6-2. 

x + a 

 

(x + a) 2 + y2 − 

x − a 

= C. (2.6.18) 

(x − a) 2 + y2 329

330 

The differential equation for the equipotential curves is obtained by taking the negative reciprocal of 

the slope of the field lines. This gives 

dy 

dx = 

r 3 1 

kq(x−a) 

r3 2 

kqy 

r3 1 

− kq(x+a) 

r 3 1 

− kqy 

r 3 2 

This result can be written in the form 

 

(x + a)dx + ydy (x − a)dx + ydy 

− 

+ 

=0 

which simplifies to the easily integrable form 

− dr1 

r 2 1 

+ dr2 

r 2 2 

in terms of the new variables r1 and r2. An integration produces the equipotential curves 

or 

1 

(x + a) 2 + y 2 − 

=0 

r 3 2 

1 

r1 

. 

− 1 

r2 

=C2 

1 

(x − a) 2 + y 2 =C2. 

The potential function for this problem can be interpreted as a superposition of the potential functions 

V1 = − kq 

and V2 = kq 

associated with the isolated point charges at the points (−a, 0) and (a, 0). 

r1 

r2 

Observe that the electric lines of force move from positive charges to negative charges and they do not 

cross one another. Where field lines are close together the field is strong and where the lines are far apart 

the field is weak. If the field lines are almost parallel and equidistant from one another the field is said to be 

uniform. The arrows on the field lines show the direction of the electric field E. If one moves along a field 

line in the direction of the arrows the electric potential is decreasing and they cross the equipotential curves 

at right angles. Also, when the electric field is conservative we will have ∇× E =0. 

In three dimensions the situation is analogous to what has been done in two dimensions. If the electric 

field is E = E(x, y, z) =P (x, y, z) e1 + Q(x, y, z) e2 + R(x, y, z) e3 and r = x e1 + y e2 + z e3 is the position 

vector to a variable point (x, y, z) on a field line, then at this point dr and E must be colinear so that 

dr = K E for some constant K. Equating like coefficients gives the system of equations 

dx 

P (x, y, z) = 

dy 

Q(x, y, z) = 

dz 

= K. (2.6.19) 

R(x, y, z) 

From this system of equations one must try to obtain two independent integrals, call them u1(x, y, z) =c1 

and u2(x, y, z) =c2. These integrals represent one-parameter families of surfaces. When any two of these 

surfaces intersect, the result is a curve which represents a field line associated with the vector field E. These 

type of field lines in three dimensions are more difficult to illustrate. 

The electric flux φE of an electric field E over a surface S is defined as the summation of the normal 

component of E over the surface and is represented 

 

φE = E · ˆn dσ with units of Nm2 

(2.6.20) 

C 

S

where ˆn is a unit normal to the surface. The flux φE can be thought of as being proportional to the number 

of electric field lines passing through an element of surface area. If the surface is a closed surface we have 

by the divergence theorem of Gauss 

 

φE = ∇· 

Edτ = E · ˆn dσ 

where V is the volume enclosed by S. 

V 

Gauss Law 

Let dσ denote an element of surface area on a surface S. A cone is formed if all points on the boundary 

of dσ are connected by straight lines to the origin. The cone need not be a right circular cone. The situation 

is illustrated in the figure 2.6-3. 

Figure 2.6-3. Solid angle subtended by element of area. 

We let r denote a position vector from the origin to a point on the boundary of dσ and let ˆn denote a 

unit outward normal to the surface at this point. We then have ˆn · r = r cos θ where r = |r| and θ is the 

angle between the vectors ˆn and r. Construct a sphere, centered at the origin, having radius r. This sphere 

intersects the cone in an element of area dΩ. The solid angle subtended by dσ is defined as dω = dΩ 

. Note 

r2 that this is equivalent to constructing a unit sphere at the origin which intersect the cone in an element of 

area dω. Solid angles are measured in steradians. The total solid angle about a point equals the area of the 

sphere divided by its radius squared or 4π steradians. The element of area dΩ is the projection of dσ on the 

ˆn · r 

ˆn · r dΩ 

constructed sphere and dΩ =dσ cos θ = dσ so that dω = dσ = . Observe that sometimes the 

r r3 r2 dot product ˆn · r is negative, the sign depending upon which of the normals to the surface is constructed. 

(i.e. the inner or outer normal.) 

The Gauss law for electrostatics in a vacuum states that the flux through any surface enclosing many 

charges is the total charge enclosed by the surface divided by ɛ0. The Gauss law is written 

 

Qe 

E 

for charges inside S 

ɛ0 

· ˆn dσ = 

(2.6.21) 

0 for charges outside S 

S 

S 

331

332 

where Qe represents the total charge enclosed by the surface S with ˆn the unit outward normal to the surface. 

The proof of Gauss’s theorem follows. Consider a single charge q within the closed surface S. The electric 

field at a point on the surface S due to the charge q within S is represented E = 1 q 

4πɛ0 r2 er and so the flux 

integral is 

 

 

φE = E 

q er · ˆn 

· ˆn dσ = 

S 

S 4πɛ0 r2 dσ = q 

 

dΩ q 

= (2.6.22) 

4πɛ0 S r2 ɛ0 

since er · ˆn 

r2 cos θdσ 

= 

r2 = dΩ 

 

= dω and dω =4π. By superposition of the charges, we obtain a similar 

r2 S 

n 

result for each of the charges within the surface. Adding these results gives Qe = qi. For a continuous 

 

i=1 

distribution of charge inside the volume we can write Qe = ρ 

V 

∗ dτ, whereρ ∗ is the charge distribution 

per unit volume. Note that charges outside of the closed surface do not contribute to the total flux across 

the surface. This is because the field lines go in one side of the surface and go out the other side. In this 

case E · ˆn dσ = 0 for charges outside the surface. Also the position of the charge or charges within the 

S 

volume does not effect the Gauss law. 

The equation (2.6.21) is the Gauss law in integral form. We can put this law in differential form as 

follows. Using the Gauss divergence theorem we can write for an arbitrary volume that 

 

 

E · ˆn dσ = ∇· 

S 

V 

 

ρ 

Edτ = 

V 

∗ 

dτ = 

ɛ0 

Qe 

= 

ɛ0 

1 

 

ρ 

ɛ0 V 

∗ dτ 

which for an arbitrary volume implies 

∇· E = ρ∗ 

. (2.6.23) 

The equations (2.6.23) and (2.6.7) can be combined so that the Gauss law can also be written in the form 

∇ 2 V = − ρ∗ 

ɛ0 

which is called Poisson’s equation. 

EXAMPLE 2.6-2 

Find the electric field associated with an infinite plane sheet of positive charge. 

Solution: Assume there exists a uniform surface charge µ ∗ and draw a circle at some point on the plane 

surface. Now move the circle perpendicular to the surface to form a small cylinder which extends equal 

distances above and below the plane surface. We calculate the electric flux over this small cylinder in the 

limit as the height of the cylinder goes to zero. The charge inside the cylinder is µ ∗ A where A is the area of 

the circle. We find that the Gauss law requires that 

 

E · ˆn dσ = Qe 

= 

ɛ0 

µ∗A (2.6.24) 

ɛ0 

S 

where ˆn is the outward normal to the cylinder as we move over the surface S. By the symmetry of the 

situation the electric force vector is uniform and must point away from both sides to the plane surface in the 

direction of the normals to both sides of the surface. Denote the plane surface normals by en and − en and 

assume that E = β en on one side of the surface and E = −β en on the other side of the surface for some 

constant β. Substituting this result into the equation (2.6.24) produces 

 

E · ˆn dσ =2βA (2.6.25) 

S 

ɛ0

since only the ends of the cylinder contribute to the above surface integral. On the sides of the cylinder we 

will have ˆn ·±en = 0 and so the surface integral over the sides of the cylinder is zero. By equating the 

results from equations (2.6.24) and (2.6.25) we obtain the result that β = µ∗ 

and consequently we can write 

2ɛ0 

E = µ∗ 

en where en represents one of the normals to the surface. 

2ɛ0 

Note an electric field will always undergo a jump discontinuity when crossing a surface charge µ ∗ . As in 

theaboveexamplewehave Eup = µ∗ 

en and 

2ɛ0 

Edown = − µ∗ 

2ɛ en so that the difference is 

Eup − Edown = µ∗ 

It is this difference which causes the jump discontinuity. 

ɛ0 

en or E i n (1) 

i + Ein (2) µ∗ 

i + =0. (2.6.26) 

ɛ0 


Calculate the electric field associated with a uniformly charged sphere of radius a. 

Solution: We proceed as in the previous example. Let µ ∗ denote the uniform charge distribution over the 

surface of the sphere and let er denote the unit normal to the sphere. The total charge then is written as 

q = µ 

Sa 

∗ dσ =4πa 2 µ ∗ . If we construct a sphere of radius r>aaround the charged sphere, then we have 

by the Gauss theorem 

E · er dσ = Qe 

ɛ0 

= q 

. 

ɛ0 

(2.6.27) 

Sr 

Again, we can assume symmetry for E and assume that it points radially outward in the direction of the 

surface normal er and has the form E = β er for some constant β. Substituting this value for E into the 

equation (2.6.27) we find that 

 

 

E · er dσ = β dσ =4πβr 2 = q 

. (2.6.28) 

Sr 

This gives E = 1 q 

4πɛ0 r2 er where er is the outward normal to the sphere. This shows that the electric field 

outside the sphere is the same as if all the charge were situated at the origin. 

Sr 

For S a piecewise closed surface enclosing a volume V and F i = F i (x 1 ,x 2 ,x 3 ) i =1, 2, 3, a continuous 

vector field with continuous derivatives the Gauss divergence theorem enables us to replace a flux integral 

of F i over S by a volume integral of the divergence of F i over the volume V such that 

 

S 

F i 

ni dσ = 

V 

F i ,i dτ or 

 

S 

ɛ0 

 

F · ˆn dσ = 

V 

div Fdτ. (2.6.29) 

If V contains a simple closed surface Σ where F i is discontinuous we must modify the above Gauss divergence 

theorem. 


We examine the modification of the Gauss divergence theorem for spheres in order to illustrate the 

concepts. Let V have surface area S which encloses a surface Σ. Consider the figure 2.6-4 where the volume 

V enclosed by S and containing Σ has been cut in half. 

333

334 

Figure 2.6-4. Sphere S containing sphere Σ. 

Applying the Gauss divergence theorem to the top half of figure 2.6-4 gives 

 

F i n T 

 

 

i dσ + 

dσ + 

dσ = 

ST 

Sb1 

F i n bT 

i 

ΣT 

F i n ΣT 

i 

VT 

F i ,i dτ (2.6.30) 

where the ni are the unit outward normals to the respective surfaces ST , Sb1 and ΣT . Applying the Gauss 

divergence theorem to the bottom half of the sphere in figure 2.6-4 gives 

 

 

 

 

dσ + 

dσ + 

dσ = 

F 

SB 

i n B i 

Sb2 

F i n bB 

i 

ΣB 

F i n ΣB 

i 

VB 

F i ,i dτ (2.6.31) 

Observe that the unit normals to the surfaces Sb1 and Sb2 are equal and opposite in sign so that adding the 

equations (2.6.30) and (2.6.31) we obtain 

 

F 

S 

i 

ni dσ + F 

Σ 

i n (1) 

 

i dσ = 

F 

VT +VB 

i ,i dτ (2.6.32)

where S = ST + SB is the total surface area of the outside sphere and Σ = ΣT +ΣB is the total surface area 

of the inside sphere, and n (1) 

i is the inward normal to the sphere Σ when the top and bottom volumes are 

combined. Applying the Gauss divergence theorem to just the isolated small sphere Σ we find 

 

F 

Σ 

i n (2) 

 

i dσ = F 

VΣ 

i ,i dτ (2.6.33) 

where n (2) 

i 

is the outward normal to Σ. By adding the equations (2.6.33) and (2.6.32) we find that 

 

F 

S 

i 

ni dσ + F 

Σ 

i n (1) 

i + F i n (2) 

 

i dσ = F 

V 

i ,i dτ (2.6.34) 

where V = VT + VB + VΣ. The equation (2.6.34) can also be written as 

 

F 

S 

i 

ni dσ = F 

V 

i 

,i dτ − F 

Σ 

i n (1) 

i + F i n (2) 

 

i dσ. (2.6.35) 

In the case that V contains a surface Σ the total electric charge inside S is 

 

Qe = 

V 

ρ ∗ 

dτ + µ 

Σ 

∗ dσ (2.6.36) 

where µ ∗ is the surface charge density on Σ and ρ ∗ is the volume charge density throughout V. The Gauss 

theorem requires that 

S 

E i ni dσ = Qe 

ɛ0 

= 1 

 

ρ 

ɛ0 V 

∗ dτ + 1 

 

µ 

ɛ0 Σ 

∗ dσ. (2.6.37) 

In the case of a jump discontinuity across the surface Σ we use the results of equation (2.6.34) and write 

 

E 

S 

i 

ni dσ = E 

V 

i 

,i dτ − E 

Σ 

i n (1) 

i + Ein (2) 

 

i dσ. (2.6.38) 

Subtracting the equation (2.6.37) from the equation (2.6.38) gives 

 

E 

V 

i ,i − ρ∗ 

 

dτ − E 

ɛ0 

Σ 

i n (1) 

i + Ein (2) 

 

µ∗ 

i + dσ =0. (2.6.39) 

ɛ0 

For arbitrary surfaces S and Σ, this equation implies the differential form of the Gauss law 

E i ,i = ρ∗ 

Further, on the surface Σ, where there is a surface charge distribution we have 

ɛ0 

. (2.6.40) 

E i n (1) 

i + Ein (2) µ∗ 

i + =0 (2.6.41) 

ɛ0 

which shows the electric field undergoes a discontinuity when you cross a surface charge µ ∗ . 

335

336 

Electrostatic Fields in Materials 

When charges are introduced into materials it spreads itself throughout the material. Materials in 

which the spreading occurs quickly are called conductors, while materials in which the spreading takes a 

long time are called nonconductors or dielectrics. Another electrical property of materials is the ability to 

hold local charges which do not come into contact with other charges. This property is called induction. 

For example, consider a single atom within the material. It has a positively charged nucleus and negatively 

charged electron cloud surrounding it. When this atom experiences an electric field E the negative cloud 

moves opposite to E while the positively charged nucleus moves in the direction of E.If E is large enough it 

can ionize the atom by pulling the electrons away from the nucleus. For moderately sized electric fields the 

atom achieves an equilibrium position where the positive and negative charges are offset. In this situation 

the atom is said to be polarized and have a dipole moment p. 

Definition: When a pair of charges +q and −q are separated by a distance 2 d the electric dipole 

moment is defined by p =2 dq, wherephasdimensionsof[Cm]. In the special case where d has the same direction as E and the material is symmetric we say that p 

is proportional to E and write p = α E,whereαis called the atomic polarizability. If in a material subject 

to an electric field their results many such dipoles throughout the material then the dielectric is said to be 

polarized. The vector quantity P is introduced to represent this effect. The vector P is called the polarization 

vector having units of [C/m2 ], and represents an average dipole moment per unit volume of material. The 

vectors Pi and Ei are related through the displacement vector Di such that 

For an anisotropic material (crystal) 

where ɛ j 

i 

is called the dielectric tensor and αj 

i 

Pi = Di − ɛ0Ei. (2.6.42) 

Di = ɛ j 

i Ej and Pi = α j 

i Ej 

is called the electric susceptibility tensor. Consequently, 

Pi = α j 

i Ej = ɛ j 

i Ej − ɛ0Ei =(ɛ j j 

i − ɛ0δi )Ej so that α j 

i = ɛj 

i 

(2.6.43) 

j 

− ɛ0δi . (2.6.44) 

A dielectric material is called homogeneous if the electric force and displacement vector are the same for any 

two points within the medium. This requires that the electric force and displacement vectors be constant 

parallel vector fields. It is left as an exercise to show that the condition for homogeneity is that ɛ j 

i,k =0. 

A dielectric material is called isotropic if the electric force vector and displacement vector have the same 

direction. This requires that ɛ j 

i = ɛδi j where δi j is the Kronecker delta. The term ɛ = ɛ0Ke is called the 

dielectric constant of the medium. The constant ɛ0 =8.85(10) −12 coul 2 /N · m2 is the permittivity of free 

space and the quantity ke = ɛ 

ɛ0 is called the relative dielectric constant (relative to ɛ0). For free space ke =1. 

Similarly for an isotropic material we have α j 

i 

linear medium the vectors P , D and E are related by 

= ɛ0αeδ j 

i where αe is called the electric susceptibility. For a 

Di = ɛ0Ei + Pi = ɛ0Ei + ɛ0αeEi = ɛ0(1 + αe)Ei = ɛ0KeEi = ɛEi 

(2.6.45)

where Ke =1+αe is the relative dielectric constant. The equation (2.6.45) are constitutive equations for 

dielectric materials. 

The effect of polarization is to produce regions of bound charges ρb within the material and bound 

surface charges µb together with free charges ρf which are not a result of the polarization. Within dielectrics 

we have ∇· P = ρb for bound volume charges and P · en = µb for bound surface charges, where en is a 

unit normal to the bounding surface of the volume. In these circumstances the expression for the potential 

function is written 

V = 1 

and the Gauss law becomes 

 

4πɛ0 V 

 

ρb 1 µb 

dτ + dσ 

r 4πɛ0 S r 

(2.6.46) 

ɛ0∇· E = ρ ∗ = ρb + ρf = −∇ · P + ρf or ∇(ɛ0 E + P )=ρf . (2.6.47) 

Since D = ɛ0 E + P the Gauss law can also be written in the form 

∇· D = ρf or D i ,i = ρf . (2.6.48) 

When no confusion arises we replace ρf by ρ. In integral form the Gauss law for dielectrics is written 

 

D · ˆn dσ = Qfe 

(2.6.49) 

where Qfe is the total free charge density within the enclosing surface. 

S 

Magnetostatics 

A stationary charge generates an electric field E while a moving charge generates a magnetic field B. 

Magnetic field lines associated with a steady current moving in a wire form closed loops as illustrated in the 

figure 2.6-5. 

Figure 2.6-5. Magnetic field lines. 

The direction of the magnetic force is determined by the right hand rule where the thumb of the right 

hand points in the direction of the current flow and the fingers of the right hand curl around in the direction 

of the magnetic field B. The force on a test charge Q moving with velocity V in a magnetic field is 

Fm = Q( V × B). (2.6.50) 

The total electromagnetic force acting on Q is the electric force plus the magnetic force and is 

 

F = Q E +( V × B) 

(2.6.51) 

337

338 

which is known as the Lorentz force law. The magnetic force due to a line charge density λ∗ moving along 

acurveCis the line integral 

 

Fmag = λ ∗ ds( V × 

B)= I × Bds. (2.6.52) 

C 

Similarly, for a moving surface charge density moving on a surface 

 

Fmag = µ ∗ dσ( V × 

B)= K × Bdσ (2.6.53) 

and for a moving volume charge density 

 

Fmag = 

S 

V 

C 

S 

ρ ∗ dτ( V × 

B)= 

V 

J × Bdτ (2.6.54) 

where the quantities I = λ∗V , K = µ ∗ V and J = ρ∗ V are respectively the current, the current per unit 

length, and current per unit area. 

A conductor is any material where the charge is free to move. The flow of charge is governed by Ohm’s 

law. Ohm’s law states that the current density vector Ji is a linear function of the electric intensity or 

Ji = σimEm, whereσim is the conductivity tensor of the material. For homogeneous, isotropic conductors 

σim = σδim so that Ji = σEi where σ is the conductivity and 1/σ is called the resistivity. 

Surround a charge density ρ∗ with an arbitrary simple closed surface S having volume V and calculate 

the flux of the current density across the surface. We find by the divergence theorem 

 

 

J · ˆn dσ = ∇· Jdτ. (2.6.55) 

S 

If charge is to be conserved, the current flow out of the volume through the surface must equal the loss due 

to the time rate of change of charge within the surface which implies 

 

 

J · ˆn dσ = ∇· 

S 

V 

Jdτ = − d 

 

ρ 

dt V 

∗ 

dτ = − 

V 

∂ρ∗ dτ 

∂t 

(2.6.56) 

or 

∇· 

V 

J + ∂ρ∗ 

 

dτ =0. 

∂t 

(2.6.57) 

This implies that for an arbitrary volume we must have 

V 

∇· J = − ∂ρ∗ 

. (2.6.58) 

∂t 

Note that equation (2.6.58) has the same form as the continuity equation (2.3.73) for mass conservation and 

so it is also called a continuity equation for charge conservation. For magnetostatics there exists steady line 

currents or stationary current so ∂ρ∗ 

∂t = 0. This requires that ∇· J =0.

Figure 2.6-6. Magnetic field around wire. 

Biot-Savart Law 

The Biot-Savart law for magnetostatics describes the magnetic field at a point P due to a steady line 

current moving along a curve C and is 

B(P )= µ0 

 

I × er 

4π C r2 ds (2.6.59) 

with units [N/amp · m] and where the integration is in the direction of the current flow. In the Biot-Savart 

law we have the constant µ0 =4π × 10−7 N/amp2 which is called the permeability of free space, I = I et is 

the current flowing in the direction of the unit tangent vector et to the curve C, er is a unit vector directed 

from a point on the curve C toward the point P and r is the distance from a point on the curve to the 

general point P. Note that for a steady current to exist along the curve the magnitude of I must be the 

same everywhere along the curve. Hence, this term can be brought out in front of the integral. For surface 

currents K and volume currents J the Biot-Savart law is written 

B(P )= µ0 

 

K × er 

4π S r2 dσ 

 

µ0 J 

and B(P 

× er 

)= 

4π r2 dτ. 


Calculate the magnetic field B adistancehperpendicular to a wire carrying a constant current I. 

Solution: The magnetic field circles around the wire. For the geometry of the figure 2.6-6, the magnetic 

field points out of the page. We can write 

I × er = I et × er = Iê sin α 

where ê is a unit vector tangent to the circle of radius h which encircles the wire and cuts the wire perpendicularly. 

V 

339

340 

For this problem the Biot-Savart law is 

B(P )= µ0I 

4π 

In terms of θ we find from the geometry of figure 2.6-6 

Therefore, 

 

ê 

ds. 

r2 tan θ = s 

h with ds = h sec2 θdθ and cos θ = h 

r . 

B(P )= µ0 

π 

θ2 

But, α = π/2+θ so that sin α =cosθ and consequently 

B(P )= µ0Iê 

4πh 

θ2 

θ1 

θ1 

Iê sin αhsec 2 θ 

h 2 / cos 2 θ 

dθ. 

cos θdθ= µ0Iê 

4πh (sin θ2 − sin θ1). 

For a long straight wire θ1 →−π/2 andθ2 → π/2 to give the magnetic field B(P )= µ0Iê 

2πh . 

For volume currents the Biot-Savart law is 

B(P )= µ0 

 

4π V 

and consequently (see exercises) 

J × er 

r 2 dτ (2.6.60) 

∇· B =0. (2.6.61) 

Recall the divergence of an electric field is ∇· E = ρ∗ 

is known as the Gauss’s law for electric fields and so 

ɛ0 

in analogy the divergence ∇· B = 0 is sometimes referred to as Gauss’s law for magnetic fields. If ∇· B =0, 

then there exists a vector field A such that B = ∇× A. The vector field A is called the vector potential of 

B. Note that ∇· B = ∇·(∇ × A)=0. Also the vector potential A is not unique since B is also derivable 

from the vector potential A + ∇φ where φ is an arbitrary continuous and differentiable scalar. 

Ampere’s Law 

Ampere’s law is associated with the work done in moving around a simple closed path. For example, 

consider the previous example 2.6-5. In this example the integral of B around a circular path of radius h 

which is centered at some point on the wire can be associated with the work done in moving around this 

path. The summation of force times distance is 

 

○ 

B · dr = ○ B · ê ds = µ0I 

 

○ ds = µ0I (2.6.62) 

2πh 

C 

C 

 

where now dr = ê ds is a tangent vector to the circle encircling the wire and ○ ds =2πh is the distance 

C 

around this circle. The equation (2.6.62) holds not only for circles, but for any simple closed curve around 

the wire. Using the Stoke’s theorem we have 

 

○ 

B · dr = (∇× 

B) · en dσ = µ0I = µ0 J · en dσ (2.6.63) 

C 

S 

C 

S

where J · en dσ is the total flux (current) passing through the surface which is created by encircling 

S 

some curve about the wire. Equating like terms in equation (2.6.63) gives the differential form of Ampere’s 

law 

∇× B = µ0 J. (2.6.64) 

Magnetostatics in Materials 

Similar to what happens when charges are introduced into materials we have magnetic fields whenever 

there are moving charges within materials. For example, when electrons move around an atom tiny current 

loops are formed. These current loops create what are called magnetic dipole moments m throughout the 

material. When a magnetic field B is applied to a material medium there is a net alignment of the magnetic 

dipoles. The quantity M, called the magnetization vector is introduced. Here M is associated with a 

dielectric medium and has the units [amp/m] and represents an average magnetic dipole moment per unit 

volume and is analogous to the polarization vector P used in electrostatics. The magnetization vector M 

acts a lot like the previous polarization vector in that it produces bound volume currents Jb and surface 

currents Kb where ∇× M = Jb is a volume current density throughout some volume and M × en = Kb is a 

surface current on the boundary of this volume. 

From electrostatics note that the time derivative of ɛ0 ∂ E 

∂t has the same units as current density. The 

total current in a magnetized material is then Jt = Jb + Jf + ɛ0 ∂ E 

∂t where Jb is the bound current, Jf is the 

free current and ɛ0 ∂ E 

∂t is the induced current. Ampere’s law, equation (2.6.64), in magnetized materials then 

becomes 

∇× B = µ0 Jt = µ0( Jb + ∂ 

Jf + ɛ0 

E 

∂t )=µ0 ∂ 

J + µ0ɛ0 

E 

∂t 

(2.6.65) 

where J = Jb + Jf . The term ɛ0 ∂ E 

∂t is referred to as a displacement current or as a Maxwell correction to 

the field equation. This term implies that a changing electric field induces a magnetic field. 

An auxiliary magnet field H defined by 

Hi = 1 

Bi − Mi 

µ0 

(2.6.66) 

is introduced which relates the magnetic force vector B and magnetization vector M. This is another constitutive 

equation which describes material properties. For an anisotropic material (crystal) 

Bi = µ j 

i Hj and Mi = χ j 

i Hj 

(2.6.67) 

where µ j 

i is called the magnetic permeability tensor and χji 

is called the magnetic permeability tensor. Both 

of these quantities are dimensionless. For an isotropic material 

µ j 

i = µδj i where µ = µ0km. (2.6.68) 

Here µ0 =4π × 10 −7 N/amp 2 is the permeability of free space and km = µ 

µ0 

coefficient. Similarly, for an isotropic material we have χ j 

i 

is the relative permeability 

= χmδ j 

i where χm is called the magnetic sus- 

ceptibility coefficient and is dimensionless. The magnetic susceptibility coefficient has positive values for 

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342 

materials called paramagnets and negative values for materials called diamagnets. For a linear medium the 

quantities B, M and H are related by 

Bi = µ0(Hi + Mi) =µ0Hi + µ0χmHi = µ0(1 + χm)Hi = µ0kmHi = µHi 

(2.6.69) 

where µ = µ0km = µ0(1 + χm) is called the permeability of the material. 

Note: The auxiliary magnetic vector H for magnetostatics in materials plays a role similar to the 

displacement vector D for electrostatics in materials. Be careful in using electromagnetic equations from 

different texts as many authors interchange the roles of B and H. Some authors call H the magnetic field. 

However, the quantity B should be the fundamental quantity. 1 

Electrodynamics 

In the nonstatic case of electrodynamics there is an additional quantity Jp = ∂ P 

∂t 

current which satisfies 

∇· Jp = ∇· ∂ P 

∂t 

and the current density has three parts 

∂ 

= 

∂t ∇· P = − ∂ρb 

∂t 

J = Jb + Jf + Jp = ∇× M + Jf + ∂ P 

∂t 

called the polarization 

(2.6.70) 

(2.6.71) 

consisting of bound, free and polarization currents. 

Faraday’s law states that a changing magnetic field creates an electric field. In particular, the electromagnetic 

force induced in a closed loop circuit C is proportional to the rate of change of flux of the magnetic 

field associated with any surface S connected with C. Faraday’s law states 

 

○ E · dr = − ∂ 

 

B · en dσ. 

∂t 

C 

Using the Stoke’s theorem, we find 

 

(∇× 

S 

 

∂ 

E) · en dσ = − 

S 

B 

∂t · en dσ. 

The above equation must hold for an arbitrary surface and loop. Equating like terms we obtain the differential 

form of Faraday’s law 

∇× E = − ∂ B 

. (2.6.72) 

∂t 

This is the first electromagnetic field equation of Maxwell. 

Ampere’s law, equation (2.6.65), written in terms of the total current from equation (2.6.71) , becomes 

which can also be written as 

∇× B = µ0(∇× M + Jf + ∂ P 

∂t )+µ0ɛ0 

∂ E 

∂t 

∇×( 1 

µ0 

S 

B − M)= Jf + ∂ 

∂t ( P + ɛ0 E) 

1 D.J. Griffiths, Introduction to Electrodynamics, Prentice Hall, 1981. P.232. 

(2.6.73)

or 

∇× H = Jf + ∂ D 

. (2.6.74) 

∂t 

This is Maxwell’s second electromagnetic field equation. 

To the equations (2.6.74) and (2.6.73) we add the Gauss’s law for magnetization, equation (2.6.61) and 

Gauss’s law for electrostatics, equation (2.6.48). These four equations produce the Maxwell’s equations of 

electrodynamics and are now summarized. The general form of Maxwell’s equations involve the quantities 

for i =1, 2, 3. Therearealsothequantities 

Ei, Electric force vector, [Ei] =Newton/coulomb 

Bi, Magnetic force vector, [Bi] =Weber/m 2 

Hi, Auxilary magnetic force vector, [Hi] =ampere/m 

Di, Displacement vector, [Di] =coulomb/m 2 

Ji, Free current density, [Ji] =ampere/m 2 

Pi, Polarization vector, [Pi] =coulomb/m 2 

Mi, Magnetization vector, [Mi] =ampere/m 

ϱ, representing the free charge density, with units [ϱ] =coulomb/m 3 

ɛ0, Permittivity of free space, [ɛ0] =farads/m or coulomb 2 /Newton · m 2 

µ0, Permeability of free space, [µ0] = henrys/m or kg· m/coulomb 2 

In addition, there arises the material parameters: 

µ i j, magnetic permeability tensor, which is dimensionless 

ɛ i j , dielectric tensor, which is dimensionless 

α i j , electric susceptibility tensor, which is dimensionless 

χ i j, magnetic susceptibility tensor, which is dimensionless 

These parameters are used to express variations in the electric field Ei and magnetic field Bi when 

acting in a material medium. In particular, Pi,Di,Mi and Hi are defined from the equations 

Di =ɛ j 

i Ej = ɛ0Ei + Pi 

The above quantities obey the following laws: 

ɛ i j = ɛ0δ i j + α j 

i 

Bi =µ j 

i Hj = µ0Hi + µ0Mi, µ i j = µ0(δ i j + χi j ) 

Pi =α j 

i Ej, and Mi = χ j 

i Hj for i =1, 2, 3. 

Faraday’s Law This law states the line integral of the electromagnetic force around a loop is proportional 

to the rate of flux of magnetic induction through the loop. This gives rise to the first electromagnetic field 

equation: 

∇× E = − ∂ B 

∂t 

or ɛ ijk Ek,j = − ∂Bi 

. 

∂t 

(2.6.75) 

. 

343

344 

Ampere’s Law This law states the line integral of the magnetic force vector around a closed loop is 

proportional to the sum of the current through the loop and the rate of flux of the displacement vector 

through the loop. This produces the second electromagnetic field equation: 

∇× H = Jf + ∂ D 

∂t 

or ɛ ijk Hk,j = J i f + ∂Di 

. (2.6.76) 

∂t 

Gauss’s Law for Electricity This law states that the flux of the electric force vector through a closed 

surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic 

field equation: 

∇· D = ρf or D i ,i = ρf or 

1 ∂ 

√ 

g ∂xi √ i 

gD = ρf . (2.6.77) 

Gauss’s Law for Magnetism This law states the magnetic flux through any closed volume is zero. This 

produces the fourth electromagnetic field equation: 

∇· B =0 or B i ,i =0 or 

1 ∂ 

√ 

g ∂xi √ i 

gB =0. (2.6.78) 

When no confusion arises it is convenient to drop the subscript f from the above Maxwell equations. 

Special expanded forms of the above Maxwell equations are given on the pages 176 to 179. 

Electromagnetic Stress and Energy 

Let V denote the volume of some simple closed surface S. Let us calculate the rate at which electromagnetic 

energy is lost from this volume. This represents the energy flow per unit volume. Begin with the 

first two Maxwell’s equations in Cartesian form 

ɛijkEk,j = − ∂Bi 

∂t 

(2.6.79) 

ɛijkHk,j =Ji + ∂Di 

. 

∂t 

(2.6.80) 

Now multiply equation (2.6.79) by Hi and equation (2.6.80) by Ei. This gives two terms with dimensions of 

energy per unit volume per unit of time which we write 

ɛijkEk,jHi = − ∂Bi 

∂t Hi 

Subtracting equation (2.6.82) from equation (2.6.81) we find 

(2.6.81) 

ɛijkHk,jEi =JiEi + ∂Di 

∂t Ei. (2.6.82) 

ɛijk(Ek,jHi − Hk,jEi) =− JiEi − ∂Di 

∂t Ei − ∂Bi 

∂t Hi 

ɛijk [(EkHi),j − EkHi,j + Hi,jEk] =− JiEi − ∂Di 


∂t Hi 

Observe that ɛjki(EkHi),j is the same as ɛijk(EjHk),i so that the above simplifies to 

ɛijk(EjHk),i + JiEi = − ∂Di 


∂t Hi. (2.6.83)

Now integrate equation (2.6.83) over a volume and apply Gauss’s divergence theorem to obtain 

 

 

 

ɛijkEjHkni dσ + JiEi dτ = − ( 

S 

V 

V 

∂Di 

∂t Ei + ∂Bi 

∂t Hi) dτ. (2.6.84) 

The first term in equation (2.6.84) represents the outward flow of energy across the surface enclosing the 

volume. The second term in equation (2.6.84) represents the loss by Joule heating and the right-hand side 

is the rate of decrease of stored electric and magnetic energy. The equation (2.6.84) is known as Poynting’s 

theorem and can be written in the vector form 

 

( E × 

H) · ˆn dσ = (− E · ∂ D 

∂t − H · ∂ B 

∂t − E · J) dτ. (2.6.85) 

For later use we define the quantity 

S 

V 

Si = ɛijkEjHk or S = E × H [Watts/m 2 ] (2.6.86) 

as Poynting’s energy flux vector and note that Si is perpendicular to both Ei and Hi and represents units 

of energy density per unit time which crosses a unit surface area within the electromagnetic field. 

Electromagnetic Stress Tensor 

Instead of calculating energy flow per unit volume, let us calculate force per unit volume. Consider a 

region containing charges and currents but is free from dielectrics and magnetic materials. To obtain terms 

with units of force per unit volume we take the cross product of equation (2.6.79) with Di and the cross 

product of equation (2.6.80) with Bi and subtract to obtain 

−ɛirsɛijk(Ek,jDs + Hk,jBs) =ɛrisJiBs + ɛris 

which simplifies using the e − δ identity to 

which further simplifies to 

−(δrjδsk − δrkδsj)(Ek,jDs + Hk,jBs) =ɛrisJiBs + ɛris 

 

∂Di 

∂t Bs + ∂Bs 

∂t Di 

 

∂ 

∂t (DiBs) 

−Es,rDs + Er,sDs − Hs,rBs + Hr,sBs = ɛrisJiBs + ∂ 

∂t (ɛrisDiBs). (2.6.87) 

Observe that the first two terms in the equation (2.6.87) can be written 

whichcanbeexpressedintheform 

Er,sDs − Es,rDs =Er,sDs − ɛ0Es,rEs 

=(ErDs),s − ErDs,s − ɛ0( 1 

2 EsEs),r 

=(ErDs),s − ρEr − 1 

2 (EjDjδsr),s 

=(ErDs − 1 

2 EjDjδrs),s − ρEr 

Er,sDs − Es,rDs = T E rs,s 

− ρEr 

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346 

where 

T E rs = ErDs − 1 

2 EjDjδrs 

is called the electric stress tensor. In matrix form the stress tensor is written 

⎡ 

⎣ E1D1 − 1 

2EjDj E1D2 E1D3 

T E rs = 

E2D1 E2D2 − 1 

2EjDj E2D3 

E3D1 E3D2 E3D3 − 1 

2EjDj (2.6.88) 

⎤ 

⎦ . (2.6.89) 

By performing similar calculations we can transform the third and fourth terms in the equation (2.6.87) and 

obtain 

where 

Hr,sBs − Hs,rBs = T M rs,s 

T M rs = HrBS − 1 

2 HjBjδrs 

is the magnetic stress tensor. In matrix form the magnetic stress tensor is written 

T M rs = 

⎡ 

⎣ B1H1 − 1 

The total electromagnetic stress tensor is 

Then the equation (2.6.87) can be written in the form 

or 

2BjHj B1H2 B1H3 

B2H1 B2H2 − 1 

2BjHj B2H3 

B3H1 B3H2 B3H3 − 1 

2BjHj (2.6.90) 

(2.6.91) 

⎤ 

⎦ . (2.6.92) 

Trs = T E rs + T M rs . (2.6.93) 

Trs,s − ρEr = ɛrisJiBs + ∂ 

∂t (ɛrisDiBs) 

ρEr + ɛrisJiBS = Trs,s − ∂ 

∂t (ɛrisDiBs). (2.6.94) 

For free space Di = ɛ0Ei and Bi = µ0Hi so that the last term of equation (2.6.94) can be written in terms 

of the Poynting vector as 

∂Sr 

µ0ɛ0 

∂t 

= ∂ 

∂t (ɛrisDiBs). (2.6.95) 

Now integrate the equation (2.6.94) over the volume to obtain the total electromagnetic force 

 

 

 

 

∂Sr 

ρEr dτ + ɛrisJiBs dτ = Trs,s dτ − µ0ɛ0 

V 

V 

V 

V ∂t dτ. 

Applying the divergence theorem of Gauss gives 

 

 

 

 

ρEr dτ + ɛrisJiBs dτ = Trsns dσ − µ0ɛ0 

V 

V 

S 

V 

∂Sr 

∂t 

dτ. (2.6.96) 

The left side of the equation (2.6.96) represents the forces acting on charges and currents contained within 

the volume element. If the electric and magnetic fields do not vary with time, then the last term on the 

right is zero. In this case the forces can be expressed as an integral of the electromagnetic stress tensor.

EXERCISE 2.6 

◮ 1. Find the field lines and equipotential curves associated with a positive charge q located at (−a, 0) and 

a positive charge q located at (a, 0). The field lines are illustrated in the figure 2.6-7. 

Figure 2.6-7. Lines of electric force between two charges of the same sign. 

◮ 2. Calculate the lines of force and equipotential curves associated with the electric field 

E = E(x, y) =2y e1 +2x e2. Sketch the lines of force and equipotential curves. Put arrows on the lines of 

force to show direction of the field lines. 

◮ 3. A right circular cone is defined by 

x = u sin θ0 cos φ, y = u sin θ0 sin φ, z = u cos θ0 

with 0 ≤ φ ≤ 2π and u ≥ 0. Show the solid angle subtended by this cone is Ω = A 

r2 =2π(1 − cos θ0). 

◮ 4. Acharge+qis located at the point (0,a) and a charge −q is located at the point (0, −a). Show that 

the electric force E at the position (x, 0), where x>ais E = 1 −2aq 

4πɛ0 (a2 + x2 e2. 

) 3/2 

◮ 5. Let the circle x2 + y2 = a2 carry a line charge λ∗ . Show the electric field at the point (0, 0,z)is 

E = 1 λ 

4πɛ0 

∗az(2π) e3 

(a2 + z2 . 

) 3/2 

◮ 6. Use superposition to find the electric field associated with two infinite parallel plane sheets each 

carrying an equal but opposite sign surface charge density µ ∗ . Find the field between the planes and outside 

of each plane. Hint: Fields are of magnitude ± µ∗ 

and perpendicular to plates. 

2ɛ0 

◮ 7. For a volume current J the Biot-Savart law gives B = µ0 

 

J × er 

4π V r2 dτ. Show that ∇· B =0. 

Hint: Let er = r 

r and consider ∇·( J × r 

). Then use numbers 13 and 10 of the appendix C. Also note that 

r3 ∇× J = 0 because J does not depend upon position. 

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348 

◮ 8. A homogeneous dielectric is defined by Di and Ei having parallel vector fields. Show that for a 

homogeneous dielectric ɛ j 

i,k =0. 

◮ 9. Show that for a homogeneous, isotropic dielectric medium that ɛ is a constant. 

◮ 10. Show that for a homogeneous, isotropic linear dielectric in Cartesian coordinates 

Pi,i = αe 

ρf . 

1+αe 

◮ 11. Verify the Maxwell’s equations in Gaussian units for a charge free isotropic homogeneous dielectric. 

∇· E = 1 

ɛ ∇· D =0 

∇· B =µ∇ H =0 

∇× E = − 1 ∂ 

c 

B µ ∂ 

= − 

∂t c 

H 

∂t 

∇× H = 1 ∂ 

c 

D 4π 

+ 

∂t c J = ɛ ∂ 

c 

E 

∂t 

+ 4π 

c σ E 

◮ 12. 

charge. 

Verify the Maxwell’s equations in Gaussian units for an isotropic homogeneous dielectric with a 

∇· D =4πρ 

∇· ∇× 

B =0 

E = − 1 ∂ 

c 

B 

∂t 

∇× H = 4π 

c J + 1 ∂ 

c 

D 

∂t 

◮ 13. For a volume charge ρ in an element of volume dτ located at a point (ξ,η,ζ) Coulombs law is 

E(x, y, z) = 1 

 

4πɛ0 

ρ 

erdτ 

r2 (a) Show that r 2 =(x− ξ) 2 +(y − η) 2 +(z − ζ) 2 . 

(b) Show that er = 1 

r ((x − ξ) e1 +(y − η) e2 +(z − ζ) e3) . 

(c) Show that 

E(x, y, z) = 1 

 

4πɛ0 V [(x − ξ) 2 +(y − η) 2 +(z − ζ) 2 ] 

(d) Show that the potential function for E is V = 1 

 

 

(x − ξ) e1 +(y − η) e2 +(z − ζ) e3 

1 

ρdξdηdζ = 

3/2 4πɛ0 

V 

V 

∇ 

 

er 

r2 

ρdξdηdζ 

ρ(ξ,η,ζ) 

[(x − ξ) 2 +(y − η) 2 +(z − ζ) 2 dξdηdζ 

] 1/2 

4πɛ0 V 

(e) Show that E = −∇V. 

(f) Show that ∇ 2 V = − ρ 

Hint: Note that the integrand is zero everywhere except at the point where 

ɛ 

(ξ,η,ζ) = (x, y, z). Consider the integral split into two regions. One region being a small sphere 

about the point (x, y, z) in the limit as the radius of this sphere approaches zero. Observe the identity 

er 

∇ (x,y,z) 

r2 

 

er 

= −∇(ξ,η,ζ) 

r2 

enables one to employ the Gauss divergence theorem to obtain a 

surface integral. Use a mean value theorem to show − ρ 

 

er ρ 

· ˆndS = 4π since ˆn = − er. 

4πɛ0 S r2 4πɛ0 

◮ 14. Show that for a point charge in space ρ∗ = qδ(x − x0)δ(y − y0)δ(z − z0), where δ is the Dirac delta 

function, the equation (2.6.5) can be reduced to the equation (2.6.1). 

◮ 15. 

(a) Show the electric field E = 1 

r2 er is irrotational. Here er = r 

r is a unit vector in the direction of r. 

(b) Find the potential function V such that E = −∇V which satisfies V(r0) =0forr0 > 0.

◮ 16. 

(a) If E is a conservative electric field such that E = −∇V, then show that E is irrotational and satisfies 

∇× E =curl E =0. 

(b) If ∇× E =curl E = 0, show that E is conservative. (i.e. Show E = −∇V.) 

Hint: The work done on a test charge Q = 1 along the straight line segments from (x0,y0,z0) to 

(x, y0,z0) andthenfrom(x, y0,z0) to(x, y, z0) and finally from (x, y, z0) to(x, y, z) can be written 

x 

y 

z 

V = V(x, y, z) =− E1(x, y0,z0) dx − E2(x, y, z0) dy − E3(x, y, z) dz. 

Now note that 

x0 

∂V 

∂y = −E2(x, 

z 

∂E3(x, y, z) 

y, z0) − 

dz 

z0 ∂y 

and from ∇× E = 0 we find ∂E3 ∂E2 

∂V 

= , which implies 

∂y ∂z ∂y = −E2(x, y, z). Similar results are obtained 

for ∂V ∂V 

and 

∂x ∂z . Hence show −∇V = E. 

◮ 17. 

(a) Show that if ∇· B = 0, then there exists some vector field A such that B = ∇× A. 

The vector field A is called the vector potential of B. 

Hint: Let 1 

A(x, y, z) = s B(sx, sy, sz) × rdswhere r = x e1 + y e2 + z e3 

1 

dBi 

and integrate 

0 ds s2 ds by parts. 

(b) Show that ∇·(∇× A)=0. 

◮ 18. Use Faraday’s law and Ampere’s law to show 

0 

g im (E j 

,j ),m − g jm E i ,mj 

y0 

∂ 

= −µ0 

∂t 

 

J i ∂E 

+ ɛ0 

i 

∂t 

◮ 19. Assume that J = σ E where σ is the conductivity. Show that for ρ = 0 Maxwell’s equations produce 

µ0σ ∂ E 

∂t 

and µ0σ ∂ B 

∂t 

∂ 

+ µ0ɛ0 

2E ∂t2 =∇2E + µ0ɛ0 

∂ 2 B 

∂t 2 =∇2 B. 

Here both E and B satisfy the same equation which is known as the telegrapher’s equation. 

◮ 20. Show that Maxwell’s equations (2.6.75) through (2.6.78) for the electric field under electrostatic 

conditions reduce to 

∇× E =0 

∇· D =ρf 

Now E is irrotational so that E = −∇V. Show that ∇ 2 V = − ρf 

ɛ . 

z0 

349

350 

◮ 21. Show that Maxwell’s equations (2.6.75) through (2.6.78) for the magnetic field under magnetostatic 

conditions reduce to ∇× H = J and ∇· B =0. The divergence of B being zero implies B can be derived 

from a vector potential function A such that B = ∇× A.Here A is not unique, see problem 24. If we select 

A such that ∇· A = 0 then show for a homogeneous, isotropic material, free of any permanent magnets, that 

∇ 2 A = −µ J. 

◮ 22. Show that under nonsteady state conditions of electrodynamics the Faraday law from Maxwell’s 

equations (2.6.75) through (2.6.78) does not allow one to set E = −∇V. Why is this? Observe that 

∇· B =0sowecanwrite B = ∇× A for some vector potential 

A. Using this vector potential show that 

Faraday’s law can be written ∇× E + ∂ 

A 

= 0. This shows that the quantity inside the parenthesis is 

∂t 

conservative and so we can write E + ∂ A 

= −∇V for some scalar potential V. The representation 

∂t 

E = −∇V − ∂ A 

∂t 

is a more general representation of the electric potential. Observe that for steady state conditions ∂ A 

∂t =0 

so that this potential representation reduces to the previous one for electrostatics. 

◮ 23. Using the potential formulation E = −∇V − ∂ A 

derived in problem 22, show that in a vacuum 

∂t 

(a) Gauss law can be written ∇ 2 V + ∂∇· A ρ 

= − 

∂t ɛ0 

(b) Ampere’s law can be written 

 

∇× ∇× 

A = µ0 

∂V ∂ 

J − µ0ɛ0∇ − µ0ɛ0 

∂t 

2A ∂t2 (c) Show the result in part (b) can also be expressed in the form 

 

∇ 2 A ∂ 

− µ0ɛ0 

 

A 

−∇ ∇· 

∂t 

 

∂V 

A + µ0ɛ0 = −µ0 

∂t 

J 

◮ 24. The Maxwell equations in a vacuum have the form 

∇× E = − ∂ B 

∂t 

∇× H = ∂ D 

∂t + ρ V ∇· D = ρ ∇· B =0 

where D = ɛ0 E, B = µ0 H with ɛ0 and µ0 constants satisfying ɛ0 µ0 =1/c2 where c is the speed of light. 

Introduce the vector potential A and scalar potential V defined by B = ∇× A and E 

∂ 

= − A 

∂t −∇V. 

Note that the vector potential is not unique. For example, given ψ as a scalar potential we can write 

B = ∇× A = ∇×( A + ∇ ψ), since the curl of a gradient is zero. Therefore, it is customary to impose some 

kind of additional requirement on the potentials. These additional conditions are such that E and B are 

not changed. One such condition is that A and V satisfy ∇· A + 1 

c2 ∂V 

=0. This relation is known as the 

∂t 

Lorentz relation or Lorentz gauge. Find the Maxwell’s equations in a vacuum in terms of A and V and show 

that 

∇ 2 − 1 

c2 ∂2 ∂t2 

V = − ρ 

ɛ0 

and 

 

∇ 2 − 1 

c2 ∂2 ∂t2 

A = −µ0ρ V.

◮ 25. In a vacuum show that E and B satisfy 

∇ 2 E 

1 

= 

c2 ∂2E ∂t2 ∇ 2 B 

1 

= 

c2 ∂2B ∂t2 ∇· E =0 ∇ B =0 

◮ 26. 

(a) Show that the wave equations in problem 25 have solutions in the form of waves traveling in the 

x- direction given by 

E = E(x, t) = E0e i(kx±ωt) 

and 

B = B(x, t) = B0e i(kx±ωt) 

where E0 and B0 are constants. Note that wave functions of the form u = Ae i(kx±ωt) are called plane 

harmonic waves. Sometimes they are called monochromatic waves. Here i2 = −1 is an imaginary unit. 

Euler’s identity shows that the real and imaginary parts of these type wave functions have the form 

A cos(kx ± ωt) and A sin(kx ± ωt). 

These represent plane waves. The constant A is the amplitude of the wave , ω is the angular frequency, 

and k/2π is called the wave number. The motion is a simple harmonic motion both in time and space. 

That is, at a fixed point x the motion is simple harmonic in time and at a fixed time t, themotionis 

harmonic in space. By examining each term in the sine and cosine terms we find that x has dimensions of 

length, k has dimension of reciprocal length, t hasdimensionsoftimeandωhas dimensions of reciprocal 

time or angular velocity. The quantity c = ω/k is the wave velocity. The value λ =2π/k has dimension 

of length and is called the wavelength and 1/λ is called the wave number. The wave number represents 

the number of waves per unit of distance along the x-axis. The period of the wave is T = λ/c =2π/ω 

and the frequency is f =1/T. The frequency represents the number of waves which pass a fixed point 

in a unit of time. 

(b) Show that ω =2πf 

(c) Show that c = fλ 

(d) Is the wave motion u =sin(kx − ωt)+sin(kx + ωt) a traveling wave? Explain. 

(e) Show that in general the wave equation ∇ 2 φ = 1 

c2 ∂2φ have solutions in the form of waves traveling in 

∂t2 either the +x or −x direction given by 

φ = φ(x, t) =f(x + ct)+g(x − ct) 

where f and g are arbitrary twice differentiable functions. 

(f) Assume a plane electromagnetic wave is moving in the +x direction. Show that the electric field is in 

the xy−plane and the magnetic field is in the xz−plane. 

Hint: Assume solutions Ex = g1(x − ct), Ey = g2(x − ct),Ez = g3(x − ct),Bx = g4(x − ct), 

By = g5(x − ct),Bz = g6(x − ct) wheregi,i =1, ..., 6 are arbitrary functions. Then show that Ex 

does not satisfy ∇· E = 0 which implies g1 must be independent of x and so not a wave function. Do 

the same for the components of B. Since both ∇· E = ∇· B =0thenEx = Bx = 0. Such waves 

are called transverse waves because the electric and magnetic fields are perpendicular to the direction 

of propagation. Faraday’s law implies that the E and B waves must be in phase and be mutually 

perpendicular to each other. 

351

352 

BIBLIOGRAPHY 

• Abramowitz, M. and Stegun, I.A., Handbook of Mathematical Functions, 10thed, 

New York:Dover, 1972. 

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• Aris, Rutherford, Vectors, Tensors, and the Basic Equations of Fluid Mechanics, 

Englewood Cliffs, N.J.:Prentice-Hall, 1962. 

• Atkin, R.J., Fox, N., An Introduction to the Theory of Elasticity, 

London:Longman Group Limited, 1980. 

• Bishop, R.L., Goldberg, S.I.,Tensor Analysis on Manifolds, New York:Dover, 1968. 

• Borisenko, A.I., Tarapov, I.E., Vector and Tensor Analysis with Applications, New York:Dover, 1968. 

• Chorlton, F., Vector and Tensor Methods, Chichester,England:Ellis Horwood Ltd, 1976. 

• Dodson, C.T.J., Poston, T., Tensor Geometry, London:Pittman Publishing Co., 1979. 

• Eisenhart, L.P., Riemannian Geometry, Princeton, N.J.:Univ. Princeton Press, 1960. 

• Eringen, A.C., Mechanics of Continua, Huntington, N.Y.:Robert E. Krieger, 1980. 

• D.J. Griffiths, Introduction to Electrodynamics, Prentice Hall, 1981. 

• Flügge, W., Tensor Analysis and Continuum Mechanics, New York:Springer-Verlag, 1972. 

• Fung, Y.C., A First Course in Continuum Mechanics, Englewood Cliffs,N.J.:Prentice-Hall, 1969. 

• Goodbody, A.M., Cartesian Tensors, Chichester, England:Ellis Horwood Ltd, 1982. 

• Hay, G.E., Vector and Tensor Analysis, New York:Dover, 1953. 

• Hughes, W.F., Gaylord, E.W., Basic Equations of Engineering Science, New York:McGraw-Hill, 1964. 

• Jeffreys, H., Cartesian Tensors, Cambridge, England:Cambridge Univ. Press, 1974. 

• Lass, H., Vector and Tensor Analysis, New York:McGraw-Hill, 1950. 

• Levi-Civita, T., The Absolute Differential Calculus, London:Blackie and Son Limited, 1954. 

• Lovelock, D., Rund, H. ,Tensors, Differential Forms, and Variational Principles, New York:Dover, 1989. 

• Malvern, L.E., Introduction to the Mechanics of a Continuous Media, 

Englewood Cliffs, N.J.:Prentice-Hall, 1969. 

• McConnell, A.J., Application of Tensor Analysis, New York:Dover, 1947. 

• Newell, H.E., Vector Analysis, New York:McGraw Hill, 1955. 

• Schouten, J.A., Tensor Analysis for Physicists,New York:Dover, 1989. 

• Scipio, L.A., Principles of Continua with Applications, New York:John Wiley and Sons, 1967. 

• Sokolnikoff, I.S., Tensor Analysis, New York:John Wiley and Sons, 1958. 

• Spiegel, M.R., Vector Analysis, New York:Schaum Outline Series, 1959. 

• Synge, J.L., Schild, A., Tensor Calculus, Toronto:Univ. Toronto Press, 1956. 

Bibliography

Prefixes. 

Basic Units. 

APPENDIX A 

UNITS OF MEASUREMENT 

The following units, abbreviations and prefixes are from the 

Système International d’Unitès (designated SI in all Languages.) 

Abreviations 

Prefix Multiplication factor Symbol 

tera 10 12 T 

giga 10 9 G 

mega 10 6 M 

kilo 10 3 K 

hecto 10 2 h 

deka 10 da 

deci 10 −1 d 

centi 10 −2 c 

milli 10 −3 m 

micro 10 −6 µ 

nano 10 −9 n 

pico 10 −12 p 

Basic units of measurement 

Unit Name Symbol 

Length meter m 

Mass kilogram kg 

Time second s 

Electric current ampere A 

Temperature degree Kelvin ◦ K 

Luminous intensity candela cd 

Supplementary units 

Unit Name Symbol 

Plane angle radian rad 

Solid angle steradian sr 

353

354 

DERIVED UNITS 

Name Units Symbol 

Area square meter m 2 

Volume cubic meter m 3 

Frequency hertz Hz (s −1 ) 

Density kilogram per cubic meter kg/m 3 

Velocity meter per second m/s 

Angular velocity radian per second rad/s 

Acceleration meter per second squared m/s 2 

Angular acceleration radian per second squared rad/s 2 

Force newton N (kg· m/s 2 ) 

Pressure newton per square meter N/m 2 

Kinematic viscosity square meter per second m 2 /s 

Dynamic viscosity newton second per square meter N · s/m 2 

Work, energy, quantity of heat joule J (N· m) 

Power watt W (J/s) 

Electric charge coulomb C (A· s) 

Voltage, Potential difference volt V (W/A) 

Electromotive force volt V (W/A) 

Electric force field volt per meter V/m 

Electric resistance ohm Ω (V/A) 

Electric capacitance farad F (A· s/V) 

Magnetic flux weber Wb (V · s) 

Inductance henry H (V· s/A) 

Magnetic flux density tesla T (Wb/m 2 ) 

Magnetic field strength ampere per meter A/m 

Magnetomotive force ampere A 

Physical constants. 

4arctan1=π =3.14159 26535 89793 23846 2643 ... 

 

lim 1+ 

n→∞ 

1 

n = e =2.71828 18284 59045 23536 0287 ... 

n 

Euler’s constant γ =0.5772156649 01532 86060 6512 ... 

 

γ = lim 1+ 

n→∞ 

1 

 

1 1 

+ + ···+ − log n 

2 3 n 

speed of light in vacuum = 2.997925(10) 8 ms −1 

electron charge = 1.60210(10) −19 C 

Avogadro’s constant = 6.02252(10) 23 mol −1 

Plank’s constant = 6.6256(10) −34 Js 

Universal gas constant = 8.3143 JK −1 mol −1 = 8314.3 JKg −1 K −1 

Boltzmann constant = 1.38054(10) −23 JK −1 

Stefan–Boltzmann constant = 5.6697(10) −8 Wm −2 K −4 

Gravitational constant = 6.67(10) −11 Nm 2 kg −2

APPENDIX B 

CHRISTOFFEL SYMBOLS OF SECOND KIND 

1. Cylindrical coordinates (r, θ, z) =(x 1 ,x 2 ,x 3 ) 

x = r cos θ 

y = r sin θ 

z = z 

r ≥ 0 

0 ≤ θ ≤ 2π 

−∞

356 

3. Parabolic cylindrical coordinates (ξ,η,z) =(x 1 ,x 2 ,x 3 ) 

x = ξη 

y = 1 

2 (ξ2 − η 2 ) 

z = z 

−∞

5. Elliptic cylindrical coordinates (ξ,η,z) =(x 1 ,x 2 ,x 3 ) 

x =coshξ cos η 

y =sinhξ sin η 

z = z 

ξ ≥ 0 

0 ≤ η ≤ 2π 

−∞

358 

7. Bipolar coordinates (u, v, z) =(x1 ,x2 ,x3 ) 

a sinh v 

x = 

, 


0 ≤ u

9. Prolate spheroidal coordinates (u, v, φ) =(x 1 ,x 2 ,x 3 ) 

x = a sinh u sin v cos φ, u ≥ 0 

y = a sinh u sin v sin φ, 0 ≤ v ≤ π 

h 2 1 = h2 2 

h 2 2 = a2 (sinh 2 u +sin 2 v) 

z = a cosh u cos v, 0 ≤ φ

360 

11. Toroidal coordinates (u, v, φ) =(x 1 ,x 2 ,x 3 ) 

a sinh v cos φ 

x = , 


0 ≤ u

12. Confocal ellipsoidal coordinates (u, v, w) =(x 1 ,x 2 ,x 3 ) 

x 2 = (a2 − u)(a2 − v)(a2 − w) 

(a2 − b2 )(a2 − c2 , u < c 

) 

2

362 

APPENDIX C 

VECTOR IDENTITIES 

The following identities assume that A, B, C, D are differentiable vector functions of position while 

f,f1,f2 are differentiable scalar functions of position. 

1. 

2. 

A · ( B × C)= B · ( C × A)= C · ( A × B) 

A × ( B × C)= B( A · C) − C( A · B) 

3. ( A × B) · ( C × D)=( A · C)( B · D) − ( A · D)( B · C) 

4. 

A × ( B × C)+ B × ( C × A)+ C × ( A × B)=0 

5. ( A × B) × ( C × D)= B( A · C × D) − A( B · C × D) 

= C( A · B × C) − D( A · B × C) 

6. ( A × B) · ( B × C) × ( C × A)=( A · B × C) 2 

7. ∇(f1 + f2) =∇f1 + ∇f2 

8. ∇·( A + B)=∇· A + ∇· B 

9. ∇×( A + B)=∇× A + ∇× B 

10. ∇(f A)=(∇f) · A + f∇· A 

11. ∇(f1f2) =f1∇f2 + f2∇f1 

12. ∇×(f A)=)∇f) × A + f(∇× A) 

13. ∇·( A × B)= B · (∇× A) − A · (∇× B) 

14. ( A ·∇) 

| 

A = ∇ 

A| 2 

 

− 

2 

A × (∇× A) 

15. ∇( A · B)=( B ·∇) A +( A ·∇) B + B × (∇× A)+ A × (∇× B) 

16. ∇×( A × B)=( B ·∇) A − B(∇· A) − ( A ·∇) B + A(∇· B) 

17. ∇·(∇f) =∇ 2 f 

18. ∇×(∇f) =0 

19. ∇·(∇× A)=0 

20. ∇×(∇× A)=∇(∇· A) −∇ 2 A

A 

Absolute differentiation 120 

Absolute scalar field 43 

Absolute tensor 45,46,47,48 

Acceleration 121, 190, 192 

Action integral 198 

Addition of systems 6, 51 

Addition of tensors 6, 51 

Adherence boundary condition 294 

Aelotropic material 245 

Affine transformation 86, 107 

Airy stress function 264 

Almansi strain tensor 229 

Alternating tensor 6,7 

Ampere’s law 176,301,337,341 

Angle between vectors 80, 82 

Angular momentum 218, 287 

Angular velocity 86,87,201,203 

Arc length60, 67, 133 

Associated tensors 79 

Auxiliary Magnetic field 338 

Axis of symmetry 247 

B 

Basic equations elasticity 236, 253, 270 

Basic equations for a continuum 236 

Basic equations of fluids 281, 287 

Basis vectors 1,2,37,48 

Beltrami 262 

Bernoulli’s Theorem 292 

Biharmonic equation 186, 265 

Bilinear form 97 

Binormal vector 130 

Biot-Savart law 336 

Bipolar coordinates 73 

Boltzmann equation 302,306 

Boundary conditions 257, 294 

Bulk modulus 251 

Bulk coefficient of viscosity 285 

C 

Cartesian coordinates 19,20,42, 67, 83 

Cartesian tensors 84, 87, 226 

INDEX 363 

Cauchy stress law 216 

Cauchy-Riemann equations 293,321 

Charge density 323 

Christoffel symbols 108,110,111 

Circulation 293 

Codazzi equations 139 

Coefficient of viscosity 285 

Cofactors 25, 26, 32 

Compatibility equations 259, 260, 262 

Completely skew symmetric system 31 

Compound pendulum 195,209 

Compressible material 231 

Conic sections 151 

Conical coordinates 74 

Conjugate dyad 49 

Conjugate metric tensor 36, 77 

Conservation of angular momentum 218, 295 

Conservation of energy 295 

Conservation of linear momentum 217, 295 

Conservation of mass 233, 295 

Conservative system 191, 298 

Conservative electric field 323 

Constitutive equations 242, 251,281, 287 

Continuity equation 106,234, 287, 335 

Contraction 6, 52 

Contravariant components 36, 44 

Contravariant tensor 45 

Coordinate curves 37, 67 

Coordinate surfaces 37, 67 

Coordinate transformations 37 

Coulomb law 322 

Covariant components 36, 47 

Covariant differentiation 113,114,117 

Covariant tensor 46 

Cross product 11 

Curl 21, 173 

Curvature 130, 131, 134, 149 

Curvature tensor 134, 145 

Curvilinear coordinates 66, 81 

Cylindrical coordinates 18, 42, 69

364 INDEX 

D 

Deformation 222 

Derivative of tensor 108 

Derivatives and indicial notation 18, 31 

Determinant 10, 25, 32, 33 

Dielectric tensor 333 

Differential geometry 129 

Diffusion equation 303 

Dilatation 232 

Direction cosines 85 

Displacement vector 333 

Dissipation function 297 

Distribution function 302 

Divergence 21, 172 

Divergence theorem 24 

Dot product 5 

Double dot product 50, 62 

Dual tensor 100 

Dummyindex4,5 

Dyads 48,62,63 

Dynamics 187 

E 

e Permutation symbol 6, 7, 12 

e-δ identity 12 

Eigenvalues 179,189 

Eigenvectors 179,186 

Einstein tensor 156 

Elastic constants 248 

Elastic stiffness 242 

Elasticity 211,213 

Electrostatic field 322,333 

Electric flux 327 

Electric units 322 

Electrodynamics 339 

Electromagnetic energy 341 

Electromagnetic stress 341,342 

Elliptic coordinates 72 

Elliptical cylindrical coordinates 71 

Enthalpy 298 

Entropy 300 

Epsilon permutation symbol 83 

Equation of state 300 

Equilibrium equations 273,300 

Elastic constants 243,248 

Equipotential curves 325 

Eulernumber294 

Euler-Lagrange equations 192 

Eulerian angles 201, 209 

Eulerian form 287 

Eulerian system 227 

Eulers equations of motion 204 

F 

Faraday’s law 176,301, 340 

Field lines 324, 327 

Field electric 322 

First fundamental form 133,143 

Fourier law 297, 299 

Free indices 3 

Frenet-Serret formulas 131, 188 

Froude number 294 

Fluids 281 

G 

Gas law 300 

Gauss divergence theorem 24, 330 

Gauss equations 138 

Gauss’s law for electricity 176,301,328 

Gauss’s law for magnetism 176,301,341 

Gaussian curvature 137,139, 149 

Geodesics 140, 146 

Geodesic curvature 135, 140 

General tensor 48 

Generalized e − δ identity 84, 104 

Generalized Hooke’s law 242 

Generalized Kronecker delta 13, 31 

Generalized stress strain 242 

Geometry in Riemannian Space 80 

Gradient 20, 171 

Gradient basis 37 

Green’s theorem 24 

Group properties 41, 54 

Generalized velocity 121 

Generalized acceleration 121

H 

Hamiltonian 208 

Heat equation 316 

Hexagonal material 247 

Higher order tensors 47, 93 

Hooke’s law 212, 242, 252 

Hydrodynamic equations 283 

I 

Ideal fluid 283 

Idemfactor 50 

Incompressible material 231 

Index notation 1, 2, 14 

Indicial notation 1, 2, 14,24 

Inner product 52 

Inertia 30 

Integral theorems 24 

Intrinsic derivative 120 

Invariant 43 

Inviscid fluid 283 

Isotropic material 248 

Isotropic tensor 104 

J 

Jacobian 17, 30, 40, 101, 127 

Jump discontinuity 330 

K 

Kronecker delta 3, 8, 13, 31, 76 

Kinetic energy 201 

Kinematic viscosity 302 

L 

Lagrange’s equation of motion 191, 196 

Lagrangian 209 

Laplacian 174 

Linear form 96 

Linear momentum 209, 287 

Linear transformation 86 

Linear viscous fluids 284 

Lorentz transformation 57 

Lame’s constants 251 

INDEX 365 

M 

Magnitude of vector 80 

Magnetostatics 334,338 

Magnetic field 334 

Magnetization vector 337 

Magnetic permeability 337 

Material derivative 234, 288 

Material symmetry 244, 246 

Maxwell equations 176, 339 

Maxwell transfer equation 308 

Maximum, minimum curvature 130, 140 

Mean curvature 137, 148 

Metric tensor 36, 65 

Meusnier’s Theorem 150 

Mixed tensor 49 

Mohr’s circle 185 

Moment of inertia 30, 184, 200 

Momentum 217, 218 

Multilinear forms 96, 98 

Multiplication of tensors 6, 51 

N 

Navier’s equations 254, 257 

Navier-Stokes equations 288, 290 

Newtonian fluids 286 

Nonviscous fluid 283 

Normal curvature 135, 136 

Normal plane 188 

Normal stress 214 

Normal vector 130, 132 

Notation for physical components 92 

O 

Oblate Spheroidal coordinates 75 

Oblique coordinates 60 

Oblique cylindrical coordinates 102 

Order 2 

Orthogonal coordinates 78, 86 

Orthotropic material 246 

Outer product 6, 51 

Osculating plane 188

366 INDEX 

P 

Parallel vector field 122 

Pappovich-Neuber solution 263 

Parabolic coordinates 70 

Parabolic cylindrical coordinates 69 

Particle motion 190 

Pendulum system 197, 210 

Perfect gas 283, 299 

Permutations 6 

Phase space 302 

Physical components 88, 91,93 

Piezoelectric 300 

Pitch,roll, Yaw 209 

Plane Couette flow 315 

Plane Poiseuille flow 316 

Plane strain 263 

Plane stress 264 

Poisson’s equation 329 

Poisson’s ratio 212 

Polar element 273 

Polarization vector 333 

Polyads 48 

Potential energy 191 

Potential function 323 

Poynting’s vector 341 

Pressure 283 

Principal axes 183 

Projection 35 

Prolated Spheroidal coordinates 74 

Pully system 194, 207 

Q 

Quotient law 53 

R 

Radius of curvature 130, 136 

Range convention 2, 3 

Rate of deformation 281, 286 

Rate of strain 281 

Rayleighimplusive flow 317 

Reciprocal basis 35, 38 

Relative scalar 127 

Relative tensor 50, 121 

Relative motion 202 

Relativity 151 

Relative motion 155 

Reynolds number 294 

Ricci’s theorem 119 

Riemann Christoffel tensor 116, 129,139, 147 

Riemann space 80 

Rectifying plane 188 

Rigid body rotation 199 

Rotation of axes 85, 87, 107 

Rules for indices 2 

S 

Scalar 40, 43 

Scalar invariant 43, 62, 105 

Scalar potential 191 

Scaled variables 293 

Second fundamental form 135, 145 

Second order tensor 47 

Shearing stresses 214 

Simple pulley system 193 

Simple pendulum 194 

Skew symmetric system 3, 31 

Skewed coordinates 60, 102 

Solid angle 328 

Space curves 130 

Special tensors 65 

Spherical coordinates 18, 43, 56, 69, 103,194 

Stokes flow 318 

Stokes hypothesis 285 

Stokes theorem 24 

Straight line 60 

Strain 218, 225, 228 

Strain deviator 279

Stress 214 

Stress deviator 279 

Strong conservative form 298 

Strouhal number 294 

St Venant 258 

Subscripts 2 

Subtraction of tensors 51, 62 

Summation convention 4, 9 

Superscripts 2 

Surface 62, 131 

Surface area 59 

Surface curvature 149 

Surface metric 125, 133 

Susceptibility tensor 333 

Sutherland formula 285 

Symmetric system 3, 31, 51, 101 

Symmetry 243 

System 2, 31 

T 

Tangential basis 37 

Tangent vector 130 

Tensor and vector forms 40, 150 

Tensor derivative 141 

Tensor general 48 

Tensor notation 92, 160 

Tensor operations 6, 51, 175 

Test charge 322 

Thermodynamics 299 

Third fundamental form 146 

Third order systems 31 

Toroidal coordinates 75, 103 

Torus 124 

Transformation equations 17, 37, 86 

Transitive property 45,46 

Translation of coordinates 84 

Transport equation 302 

Transposition 6 

Triad 50 

Trilinear form 98 

Triple scalar product 15 

INDEX 367 

U 

Unit binormal 131, 192 

Unit normal 131, 191 

Unit tangent 131, 191 

Unit vector 81, 105 

V 

Vector identities 15, 20, 315 

Vector transformation 45, 47 

Vector operators 20, 175 

Vector potential 188 

Velocity 95, 121, 190, 193 

Velocity strain tensor 281 

Viscosity 285 

Viscosity table 285 

Viscous fluid 283 

Viscous forces 288 

Viscous stress tensor 285 

Vorticity 107, 292 

W 

Wave equation 255, 269 

Weighted tensor 48, 127 

Weingarten’s equation 138, 153 

Work 191, 279 

Work done 324 

Y 

Young’s modulus 212

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