Tensor Analysis and Its Applications 9781482850673, 9781482850680

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Tensor Analysis and Its Applications

KHAN

Copyright © 2015 by Khan. ISBN:

Softcover eBook

978-1-4828-5067-3 978-1-4828-5068-0

All rights reserved. No part of this book may be used or reproduced by any means, graphic, electronic, or mechanical, including photocopying, recording, taping or by any information storage retrieval system without the written permission of the author except in the case of brief quotations embodied in critical articles and reviews. Because of the dynamic nature of the Internet, any web addresses or links contained in this book may have changed since publication and may no longer be valid. The views expressed in this work are solely those of the author and do not necessarily reflect the views of the publisher, and the publisher hereby disclaims any responsibility for them.

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Contents Preface

1 Preliminaries 1.1 n - Dimensional Space 1.2 Superscript and Subscript 1.3 Einstein’s Summation Convention 1.4 Dummy Suffix and Real Suffix 1.5 Transformation of Coordinates 1.6 Kronecker Delta 1.7 Matrices and Determinants

2 Tensor Algebra and its Calculus 2.1 Tensors 2.2 Operations on Tensors 2.2.1 Addition and Subtraction of Tensors 2.2.2 Multiplication of Tensors 2.2.3 Contraction 2.2.4 Inner Product or Scalar Product 2.2.5 Quotient Law 2.2.6 Extension of Rank of a Tensor 2.3 Types of Tensors 2.3.1 Invariant Tensors and Equality of Tensors 2.3.2 Fundamental Tensor 2.3.3 Symmetric Tensor and Anti - Symmetric (Skew - Symmetric) Tensor 2.3.4 Reciprocal Tensor 2.3.5 Reducible Tensor and Irreducible Tensor 2.3.6 Relative Tensor 2.3.7 Raising and Lowering Suffix: Associated Tensors

3 Christoffel’s Symbols and their Properties 3.1 Christoffel’s Symbols or Christoffel’s Brackets 3.2 Symmetric Properties of Christoffel’s Symbols

3.3 Tensor Laws of Transformation of Christoffel’s Symbols 3.4 Transitive Property for Laws of Transformation of Christoffel’s Symbols

4 Covariant Differentiation of Tensors and their Properties 4.1 Comma Notation 4.2 Covariant Differentiation of Vectors (Tensors of Rank one) 4.3 Covariant Derivative of a Covariant Tensor of Rank two 4.4 Covariant Derivative of a Contravariant Tensor of Rank two 4.5 Covariant Derivative of a Mixed Tensor of Rank two 4.6 Covariant Derivative of the Fundamental Tensors 4.7 Laws of Covariant Differentiation 4.8 Covariant Derivative of an Invariant (or a Scalar) 4.9 Tensor Form of Gradient, Divergence, Laplacian and Curl 4.10 Divergence of a Tensor 4.11 Intrinsic Derivatives 4.12 Parallel Displacement of Vectors

5 Riemannian Symbols and its Properties 5.1 Riemannian Symbols of Second Kind 5.2 Properties of the Riemann Curvature Tensor 5.3 Contraction of the Riemann Curvature Tensor 5.4 Riemannian Symbols of First Kind 5.5 Properties of Covariant Curvature Tensor 5.6 Contraction of Bianchi’s Identities (Einstein tensor) 5.7 Riemannian Curvature 5.8 Flat Space 5.9 Uniform Vector Field 5.10 Conditions for Flat Space - time 5.11 Space of Constant Curvature 5.12 Einstein space

6 Application of Tensor Calculus 6.1 Introduction 6.2 Application of Tensor Calculus

6.2.1 Tensor Calculus and Differential Geometry 6.2.2 Tensor Calculus and Riemannian Geometry 6.2.3 Tensor Calculus and Theory of Relativity 6.2.4 Tensor Calculus and Elasticity 6.2.5 Tensor Calculus and Physics 6.2.6 Tensor Calculus and Other Fields like Economics, Probability and Engineering Bibliography

Preface

T he notion of a tensor was introduced and studied by Professor Gregorio Ricci of the University of Padua (Italy) in 1887 primarily as the extension of vectors. The classical definitions of scalars (having magnitude only) and vectors (with magnitude and direction as well) do not cover completely many physical and geometrical quantities. For instance, stress in an elastic body and curl of a vector supposedly misunderstood as vectors are much more than vectors. Indeed, they are tensors. The object of this book is to provide a compact exposition of the fundamental results in the theory of tensors and also to illustrate the power of the tensor technique by applications to differential geometry, elasticity, and relativity. This book is intended to be a textbook, suitable for use in a undergraduate (honors) / postgraduate Mathematics course. This book is useful for those who are totally unfamiliar with tensor analysis. Those who are already acquainted with tensor analysis will fell interested in application of tensor analysis in Riemannian geometry, theory of relativity and many other disciplines of science and technology. This book consists of six chapters. The first chapter deals with brief concepts of n dimensional space, superscript and subscript, Einstein’s summation convention, dummy suffix and real suffix, transformation of coordinates, Kronecker delta and matrices and determinants etc. which are referred to in subsequent chapters. We begin second chapter by defining tensors and in sequence, we discuss in detail about the operations of the tensors and the different types of the tensors. We introduce the notion of the Christoffel’s symbols and discuss in detail about its properties in chapter third. The concept of covariant differentiation of tensors and its properties, tensor form of gradient, divergence, Laplacian and curl, divergence of a tensor, Intrinsic derivatives and parallel displacement of vectors etc. have been studied in detail in chapter fourth. In chapter five, we have studied about Riemann’s symbols (first kind and second kind) and its properties and have also discussed on a flat space, uniform vector field, conditions for flat space - time, space of constant curvature, Einstein space etc., which are mainly useful for postgraduate students and also for researchers. In last chapter, we have discussed the applications of tensor calculus which are mainly useful for researchers. I have freely consultanted the books on tensor analysis, Riemannian geometry by the authors given in the bibliography. I am indebted to these authors for providing a good and exhaustive literature on the subject due to which the writing of this book could be made possible. I wish to express my best thanks to the publisher for bringing out the book in such a nice form.

Any comment and suggestions for improving the text will be most welcome.

Quddus Khan

1 Preliminaries

W e begin this chapter by defining n - dimensional space and giving some other fundamental conceptual definitions to which we shall refer throughout this book.

1.1 n - Dimensional Space Consider an ordered set of n real variables These variables will be called the coordinates of a point. The space generated by all points corresponding to distinct values of the coordinates is called an n - dimensional space and is denoted by Vn. A curve in Vn is defined as the collection of the points which satisfy the n equations

where u is a parameter and are n functions of u, which obey certain continuity condition. A subspace Vm of Vn is defined for as the collection of points which satisfy the n equations

where the variables

are the coordinates of Vm. Also,

are n

functions of satisfying certain conditions of continuity. In particular, the subspace Vm is called a curve, a surface and a hypersurface according as and

respectively.

1.2 Superscript and Subscript The suffixes i and j in are called superscript and subscript, respetively. The upper position always denotes superscript and lower position denotes subscript. The suffix i in the coordinate do not have the character of power indices. Usually powers will be dented by bracket. Thus, means square of

1.3 Einstein’s Summation Convention If any index in a term is repeated, then a summation with respect to that index over the range 1, 2, …, n is implied. This convention is known as Einstein’s Summation. The expression

is represented by Summation convention means drop the sigma sign and adopt the convention. In view of summation convention, we have

(1.1) Hence by summation convention we mean that if a suffix occurs twice in a term, once in upper position and once in lower position, then that suffix implied sum over defined range. If the range is not given, then we assume that the range is from 1 to n.

1.4 Dummy Suffix and Real Suffix If a suffix occurs twice in a term, once in upper position and once in lower position, then that suffix is called a dummy suffix. For example, in

, i is a dummy suffix. Evedently,

(1.2) and (1.3) From (1.2) and (1.3), we have

(1.4) which shows that a dummy suffix can be replaced by another dummy suffix not already appearing in the expressin. Also, two or more than two dummy suffixes can be interchanged. For example,

(1.5) Dummy suffix is also known as umbral suffix or dextral index. A suffix which is not repeated is called a real or free suffix. For example,

is a real suffix in

A real suffix can not replaced by another real suffix. Thus (1.6)

1.5 Transformation of Coordinates

Let us consider a space Vn with the coordinate system

Then n equations

(1.7) where

are single - valued continuous differentiable functions of coordinates, define a new

coordinate system coordinates. The differentials

The equation (1.7) are said to define a transformation of are transformed according to the relations

that is (1.8)

1.6 Kronecker Delta Kronecker delta (introduced and studied by Leopold Kronecker) is denoted by the symbol and is defined by

(1.9)

Also,

and (1.10)

By summation convention,

which in view of the statement

for

instead of saying

gives

=1+1+…+1

(1.11) Also, by summation convention,

, using (1.9)

and , using (1.9)

(1.12) In particular, if

in (1.12), then

using (1.11).

1.7 Matrices and Determinants

A set of mn numbers (real or complex), which are arranged in a rectangular array in m rows and n columns, is called an matrix (to be read as ‘m by n’ matrix). An matrix is usually written as

where

We also use the brackets

or

or double bars

in place of

the square brackets to denote matrices. An matrix for which (that is, the number of rows and columns are not same) is called a rectangular matrix of order For examples: where is a rectangular matrix of order Also, an matrix for which (that is, the number of rows and columns are same) is called a square matrix of order n. It is also called an n - rowed square matrix. For example: is a square matrix of order n.

where

The elements of a square matrix for which that is, are called the diagonal elements and the line along which they lie is called the principal diagonal of the matrix. A square matrix (i)

is said to be

a symmetric matrix if

and (1.13)

(ii) a skew (or, anti) - symmetric matrix if (1.14) where

For example,

that is, the

element of

is a symmetric matrix and

is a skew - symmetric matrix respectively.

is the

element of A.

A square matrix

all zero, that is,

whose elements above and below of the principal diagonal are

for all

is called the diagonal matrix. For example,

and

are diagonal matrices. A diagonal matrix whose all elements are equal is called a scalar matrix. A scalar matrix whose all elements are equal to 1 is called a unit matrix or an identity matrix and is denoted by (unit matrix of order n). That is, a square matrix

is a unit matrix if

when

and

when For example, respectively. Two matrices (i)

and

are scalar matrix and unit marix of order 3,

and

are said to be equal if they are of the same size (that is, they have same order) and the

elements in the corresponding place of two matrices are same, that is, for each pair of subscripts i and j. (ii) are added or subtracted only when if they are of the same type (that is, they have same order). (iii) are multiplied together only when if the number of columns in A is same as the number of rows in B. Let

be a square matrix of order n. Then the determinant of A is denoted by For example, the determinant of a square matrix

or

is

(1.15)

Here

may be taken as the general element of this determinant. The suffix i and j denote the

row and column, respectively to which the element

belongs. The symbol

is called a determinant of order 2, and its value is equal to the product of the elements along the principal diagonal minus the product of the off diagonal elements. That is, Also, the symbol

is called a determinant of order 3, and its value is the number

which is called the expansion of the determinant along its first row of a determinant of order 3. If we leave a row and a column of a determinant of order 3, passing through the element then the second order determinant thus obtained is called the minor of the element shall denote it by example,

and we

In this way we can get 9 minors corresponding to the 9 elements of

For

the minor of the element

the minor of the element The minor

multiplied by

and so on. is called the cofactor of the element

and is denoted by

Thus, and (1.16) Let the elements have

be functions of independent variables

Now, from (1.15), we

(1.17) First determinant of right hand side

where

is the cofactor of the element

in determinant

Second determinant of right hand side

Similarly, the last determinant of right hand side Finally,

Thus, (1.18)

Similarly, (1.19) Let be a square matrix of order n and an indeterminate. Then the matrix is called the characteristic matrix of A, where is the unit matrix of order n. Also, the determinant

which is an ordinary polynomial in and the equation

(1.20) of degree n, is called the characteristic polynomial of A

(1.21) is called the characteristic equation of the matrix A and the roots of this equation are called the characteristic roots of the matrix A. If is a characteristic root of a square matrix A of order n, then a non - zero vector X such that (1.22) is called a characteristic vector of the matrix A corresponding to the characteristic root Noted that the characteristic root is also known as characteristic value or eigen value or latent root or proper value of a square matrix A. The set of the eigen values of A is called the spectrum of A.

Exercises 1.

What do you mean by the terms n - dimensional space and subspace?

2.

What do you mean by curve?

3.

Define superscript and subscript

4.

what do you mean by dummy suffix and real suffix?

5.

Show that the dummy suffix can be replaced by another dummy suffix not already appearing in the given expression.

6.

What do you mean by Kronecker delta? Show that

7.

Show that

8.

Define matrix and the principal diagonal of a matrix.

9.

Define symmetric and skew - symmetric matrices with examples.

10. Define diagonal matrix, scalar matrix and identity matrix with suitable examples. 11. Define determinant of a square matrix. 12. Define minor and cofactor of the element

of a given determinant.

13. Define eigen value and eigen vector of a given matrix A. 14. What is the difference between the matrix and the determinant? 15. What do you mean by spectrum of a square matrix

16. If

is a determinant of a square matrix

then prove that

where the symbols have their usual meaning. 17. Define characteristic equation of a square matrix A, characteristic roots of the matrix A and its characteristic vector of the matrix A corresponding to the characteristic root

2 Tensor Algebra and its Calculus

W e begin this chapter by introducing the notion of the tensors and giving in detail the fundamental concepts of the operations (that is, algebra) of tensors and their different types, which are mainly useful for those readers who are totally unfamiliar with tensors. Those who are already acquainted with tensors calculus will fell interested in application of tensors in Riemannian geometry, theory of relativity and many other disciplines of science and engineering.

2.1 Tensors A vector has only one direction and magnitude. A quantity having more than one direction gives the notion of a tensor. For example, if a balloon is pressed, pressure is in different directions. It represents a tensor. The tensor formulation was originated by G. Ricci and it become rather popular when Albert Einstein used it as a natural tool for the description of his general theory of relativity. The concept of a tensor has its origin in the developments of differential geometry by Gauss, Riemann and Christoffel. The emergence of tensor calculus, as a systematic branch of mathematics is due to Ricci and his pupil Levi - Civita. A tensor can be defined as follows: A tensor is a system of quantities or functions whose components obey a certain law of transformation of coordinates from one system to another. The rank or order of a tensor is defined as the total number of indices of the components of a tensor. A tensor of rank zero is a scalar. The examples of scalars are mass, time, distance, speed, temperature, energy etc., which are invariant under changes of the coordinate system. A tensor of rank one is a vector. The examples of vectors are displacement, velocity, acceleration, force, electric field etc. A tensor of the second rank is the next in order of complexity after scalars and vectors. By a second rank tensor is meant a quantity uniquely specified by nine real numbers in 3 dimensional space. The stress tensor, the moment of inertia tensor, the deformation tensor etc. are the examples of the second rank tensor. In n - dimensional space, it is specified by functions. Let

be n functions of coordinates

and if

are transformed to

in another

coordinate system

according to the rule

(2.1) then the functions rank one.

are called components of a contravariant vector or a contravariant tensor of be n functions of coordinates

Again, let another coordinate system

and if

are transformed to

in

according to the rule

(2.2) then the functions are called components of a covariant vector or a covariant tensor of rank one. The upper position of the suffix indicates the contravariant character whereas the lower position of the suffix indicates the covariant character of a tensor. That is, the superscript indices indicate the contravariant character whereas the subscript indices indicate the covariant character. Also, the tensors Let

be

another coordinate system

and

are of types

and

functions of coordinates

respectively. and if

are transformed to

in

according to the rule

(2.3) then the functions

are called components of a contravariant tensor of rank two.

Similarly, let in another coordinate system

be

functions of coordinates according to the rule

and if

are transformed to

(2.4) then the functions

are called components of a covariant tensor of rank two.

Again, let

be

in another coordinate system

functions of coordinates

and if

are transformed to

according to the rule

(2.5) then the functions

are called components of a mixed tensor of rank two. An example of

mixed tensor of rank two is Konecker delta It is noted that the tensors and are of types (2,0), (0,2) and (1,1) respectively. Also, the tensor of higher ranks are defined as follows: (i) Contravariant tensor of rank

(2.6) (ii) Covariant tensor of rank

(2.7) (iii) Mixed tensor of rank

(2.8) If a tensor is defined at all points of a curve or throughout the space itself, then we say that it forms a tensor field. However we will use interchangeably the terms tensor and tensor field. A tensor of rank m in n - dimensions has components. Tensors have their many applications in most of the branches of theoretical physics such as mechanics, fluid mechanics, elasticity, theory of relativity and electromagnetism etc. Tensors have their also applications to Riemannian geometry, differential geometry, general theory of relativity and many other disciplines of science and engineering.

Example 2.1

Show that the velocity of a fluid at any point is a contravariant tensor of rank one.

Solution Let

be the coordinate of a moving particle with the time t. We have

(i) as the velocity of the particle. In transformed coordinates, the components of the velocity are

using (i) which follows the tensor law of transformation and hence the velocity of a fluid at any point is a contravariant tensor of rank one.

Example 2.2 Show that the tangent vector to a curve is a contravariant tensor of rank one.

Solution Let

be a curve in

with two neighbouring points

limiting position of the chord PQ with components Applying the fact

where

for

and

The

determines the tangent vector to C at P.

treated as a function of

we obtain

form the components of the tangent vector to the deformed curve

(i) at P. The

relation (i) shows that satisfies the tensor law of transformation and hence contravariant tensor of rank one.

Example 2.3 Show that the normal to a surface (constant)

is a

is a covariant tensor of rank one.

Solution The normal to a surface is (constant) (i) being scalar function satisfying the relation

(ii) Using the fact

in (ii), we have

(iii) and

being components of the tangent vectors, (iii) imply the normal character of

and

For (i), the relations (iii) can be written as

(iv) as

are linearly independent. The transformation (iv) for the components of the normal vector

are in agreement with tensor of rank one.

That is, the normal to a surface is

Example 2.4 Show that the Kronecker delta is a mixed tensor of rank two.

Solution Let

be a Kronecker delta. Now,

is a covariant

which shows that

satisfies the tensor law of transformation and hence

one contravariant suffix and one covariant suffix. Thus,

is a tensor. But it has

is a mixed tensor of rank two.

Theorem 2.1 Show that the law of transformation for a contravariant vector is transitive.

Proof Let be a contravariant vector (or, a contravariant tensor of rank one). Consider coordinate transformations

In case of transformation

Again, in case of transformation

we have

we have

since

which shows that if we make the direct transformation from (i) to (iii), we get the same law of transformation. Which shows that the law of transformation for contravariant vector is transitive, that is, the transformations of a contravariant vector form a group.

Theorem 2.2 Prove that the transformation of a mixed tensor possesses the group property.

Proof

Let

be a mixed tensor of rank two. Consider coordinate transformations

In case of transformation

we have

Again, in case of transformation

we have

which shows that if we make the direct transformation from (i) to (iii), we get the same law of transformation. Which shows that the transformation of a mixed tensor possesses the group property, that is, the transformation of a tensor forms a group.

Theorem 2.3 Prove that if the components of a tensor vanishes in one coordinate system, they vanish identically in all coordinate systems.

Proof Let

be a mixed tensor of rank 5. Then by tensor law of transformation

(i) In relation (i), the components in

are the components of a tensor in

coordinate system, whereas

are

coordinate system.

Let (ii) From (i) and (ii), we have

Again, by tensor law of transformation,

If

then

Finally, we have

If we consider the transformation

since

if and only if then

(previously shown)

Thus, we have

which shows that the components of a tensor vanishes in every coordinate system.

2.2 Operations on Tensors The operations on tensors are given as follows: Addition and subtraction of tensors, multiplication of tensors or outer product or open product of two tensors, inner product or scalar product, contraction etc. The division, in the usal sense, of one tensor by another is not defined.

2.2.1 Addition and Subtraction of Tensors It is clear that we cannot expect to give any tensorial meaning to the expression because it cannot satisfy the transformation law (2.8). It does, however, follow from this equation that any linear combination of tensors of the same type whose coefficients are invariant is a tensor of the same type. For example, from the two tensors tensor

and

we can form the

which will satisfy the transformation law (2.8) provided that

and

are

invariants. In particular, are said to be the sum and difference of the two tensors respectively. Thus, the addition and subtraction of tensors are defined as follows: The addition or subtraction of two tensors is defined only in the case of tensors of same rank and same type. Same type means the same number of contravariant and covariant indices. In other words, two tensors can be added or subtracted provided they are of same rank and similar character. Similar character means the same number of contravariant and covariant indices.

The sum or difference of two tensors is a tensor of the same rank and similar character. This is proved in the following theorems:

Theorem 2.4 Prove tthat the sum or difference of two tensors is a tensor.

Proof Let

and

be two tensors. Then the tensor law of transformation of these tensors are

(i) and (ii)

(a) The sum of

and

is defined as

(iii) Now, adding (i) and (ii), we get

which in view of (iii) gives

which shows that is,

satisfies the tensor law of transformation and hence

is a tensor. That

is a tensor.

(b) The difference of

and

is defined as

(iv) Now, subtracting (ii) from (i), we get

which in view of (iv) gives

which shows that is,

satisfies the tensor law of transformation and hence

is a tensor. That

is a tensor.

Thus, if

and

are two tensors, then their sum or difference is also a tensor.

Theorem 2.5 Show that a linear combination of tensors of the same type and same rank is a tensor of same type and same rank.

Proof Let

and

be third rank mixed tensors and

be scalars, then

(i) is also a third rank mixed tensor. This can be seen as follows: Since

and

are third rank mixed tensors, therefore their tensor law of transformation are

(ii) and (iii)

Now,

using (ii) and (iii)

using (i) which shows that

satisfies the tensor law of transformation and hence

is a tensor. That

is, is a tensor. This is, of course, also true for contravariant, covariant and mixed tensors of any rank. Note 1.

Structures of the form

2.

Structures of the form

etc. are not allowed. etc. are allowed.

Example 2.5 Show that the sum of two tensors of type

is a tensor of type

Solution Let

and

be two tensors of type

and their sum be

(i) Since and transformation are

are mixed tensors of type

therefore their tensor law of

(ii) and (iii) Adding (ii) and (iii) and then using (i), we have

which confirms the tensor law of transformation for the tensor of type

Hence the sum of

two tensors of type

is a tensor of type

2.2.2 Multiplication of Tensors Let us select two tensors, one of contravariant order p and covariant order r, the other of contravariant order q and covariant order s. It then follow from (2.8) that the products of the components form a mixed tensor of contravariant order and covariant order This tensor is called the outer product of the two tensors. Thus, the product of two tensors are defined as follows: The product of two tensors or outer product or open product of two tensors is a tensor whose rank is the sum of the ranks of the two tensors. Thus, if r and s are the ranks of two tensors, then their product will be a tensor of rank More generally, if and are two tensors of types and and having ranks and respectively, then their product will also be a tensor of type whose rank is This is proved in the following theorem

Theorem 2.6 Show that the outer product of two tensors is a tensor whose order is the sum of the orders of the these two tensors.

Proof Let

and respectively.

be two tensors of types

and

and having ranks

and

Let (i) be the outer product of two tensors tensor of type and its rank is Since and law of transformation are

and

. If we show that then the result will follow.

are mixed tensors of types

(ii)

and

is a

therefore their tensor

and (iii) Now, multiplying (ii) and (iii), we get

(iv) Using (i) in (iv), we have

which confirms that is a tensor of type This completes the required result.

and its rank is

Theorem 2.7 Prove that the open product of two vectors is a tensor of order two, but the converse is not true.

Proof Let

be the open product of two vectors

and

then

(i) We have to show that Since

and

is a tensor of rank two.

are tensors of rank one, therefore their tensor law of transformation are

(ii) and (iii)

Now, multiplying (ii) and (iii), we get

which in view of (i) gives

which shows that two.

satisfies the tensor law of transformation and hence

is a tensor of order

The converse is not true. That is, the tensor product of two vectors is a tensor of order two. But every tensor of order two is not necessarily the tensor product of two vectors.

Example 2.6 Show that the product of two tensors of type

and

is a tensor of type

Solution Let and of transformation

be two tensors of type

and

respectively. Hence by tensor law

(i) and (ii)

Let (iii) be the outer product of two tensors we show that will follow. Multiplying

and

of type

is a tensor of type

(i)

and

(ii)

properly

and

respectively. If

having rank

and

then

using

which confirms the tensor law of transformation for the tensor of type

then the result

(iii),

we

get

Hence the

product of two tensors of type

and

is a tensor of type

having rank

2.2.3 Contraction An algebraic operation by which the rank of a mixed tensor is lowered by two is known as contraction. In the process of contraction one contravariant and one covariant suffixes of a mixed tensor are equal and the repeated index (suffix) summed over, the result is a tensor of rank lowered by two than the original tensor. The contraction process enables us to obtained a tensor of order from a mixed tensor of order It is noted that the contraction process can not applied with respect to two indices of same nature, that is, both contravariant or both covariant. With the help of contraction, we can obtained some other tensors which are lowered in rank than the original tensor. For example, Ricci tensor, zero tensor and Einstein tensor etc. are obtained from the tensor of type (1,3). These tensors will be studied later in sections 5.3 and 5.6 of chapter five. The contraction process reduces the rank of a tensor by two, which is proved in the following theorem:

Theorem 2.8 Show that contraction reduces the rank of a tensor by two.

Proof Let

be a mixed tensor of rank five, then by tensor law of transformation

(i) Now, taking

in (i), we get

But according to the law of transformation, it is a mixed tensor of rank three. Hence

is a

mixed tensor of rank three, while reduces the rank of a tensor by two.

is a mixed tensor of rank five. It means that contraction

2.2.4 Inner Product or Scalar Product We can also combine multiplication and contraction to produce new tensors. From the tensors and we may obtain such tensors as and many others. This process is called an inner multiplication of two tensors and the resulting tensor is called an inner product of two tensors. Thus, the inner product of two tensors are defined as follows: The process of a product of two tensors one contravariant and one covariant suffixes be equal, is called an inner product of two tensors. For example,

etc. all are inner

products of tensors and Also, the outer product of two tensors followed by a contraction results a new tensor is formed, called an inner product of two tensors and the process is called the inner multiplication of two tensors. The inner product is also called scalar product. Note carefully that we never contract two indices of the same type as the resulting sum is not necessarily a tensor. Also it should now be clear that with our index notation, the summation convention generally applies to two indices one of which is a superscript and the other a subscript.

Theorem 2.9 Show that the inner product of the tensors

and

is a tensor of rank three.

Proof Let

and

be two tensors and the inner product of these tensors be

If we show that

is a tensor of rank three, then the theorem will be proved. Since and are two tensors of rank two and three respectively, therefore their tensor law of transformation are

(i) and (ii)

From (i) and (ii), we have

But according to the law of transformation, it is a mixed tensor of rank three. Hence mixed tensor of rank three.

is a

2.2.5 Quotient Law In a tensor analysis it is often necessary to ascertain whether a given entity is a tensor or not. The direct method requires us to find out if the given entity obeys the tensor law of transformation or not. In practice this is trouble some and a simpler test is provided by a law known as quotient law. This law enables us to determine the tensor character of the expression. It is stated in the following theorem:

Theorem 2.10 An entity whose inner product with an arbitrary tensor of rank one (contravariant or covariant) is a tensor, is itself a tensor. In other words, a set of quantities, whose inner product with an arbitrary vector is a tensor, is itself a tensor.

Proof Let

and

be a set of quantities and an arbitrary vector (or, tensor of rank one)

respectively. Let the inner product of

and

be

(i) which is a tensor. We have to show that

is a tensor. From (i), we have

(ii)

and (iii)

Since

is a tensor, therefore by tensor law of transformation

which in view of (ii) and (iii) gives

which in view of

gives

which implies that

(iv) Since

is an arbitrary vector (or, tensor of rank one), hence

(v) From (iv) and (v), we have

which confirms the tensor law of transformation. Hence is a tensor. Thus, a set of quantities, whose inner product with an arbitrary vector (or, tensor of rank one) is a tensor, is itself a tensor.

Example 2.7 If a quantity

Solution

is such that

is a tensor for any vector

then show that

is a tensor.

Let (i) Since

is given to be a tensor, therefore, by tensor law of transformation

which in view of (i) gives

(ii) But

is a tensor of rank one, so by tensor law of transformation

(iii) From (ii) and (iii), we have

which must hold for all values of

showing that

and, therefore

is a tensor of rank three and hence completes the required result.

Example 2.8 Show that the Kronecker delta

is a mixed tensor of rank two.

Solution Let

and

product of

be a contravariant vector and a Kronecker delta respectively, then the inner and

be

which is a tensor of rank one. Thus, by quotient law, contravariant suffix and one covariant suffix. Hence,

is a tensor. Also,

is a mixed tensor of rank two.

has one

2.2.6 Extension of Rank of a Tensor The rank of a tensor can be extended by differentiating its each component with respect to variable

Let us consider as an example, a simple case in which the original tensor is of rank

zero, that is, a scalar of variables

whose derivatives with respect to variable

the scalar is

are

In other system

such that

which shows that transforms like the components of a tensor of rank one. Thus, the differentiation of a tensor of rank zero gives a tensor of rank one. In general we may say that the differentiation of a tensor with respect to variable yields a new tensor of rank one greater than the original tensor. The rank of a tensor can also be extended when a tensor depends upon another tensor and the differenttiation with respect to that tensor is performed. As an example consider a tensor S of rank zero (that is, a scalar) depending upon another tensor

then

Thus, the rank of the tensor of rank zero has been extended by two.

2.3 Types of Tensors There are so many types of tensors such as invariant tensors, equality of tensors, alternate tensor, fundamental tensor, symmetric tensor and anti - symmetric (or skew - symmetric) tensor, reciprocal tensor, reducible tensor, irreducible tensor, relative tensor, and associated tensors etc. which will have discussed one by one as follows:

2.3.1 Invariant Tensors and Equality of Tensors A tensor which have the same components in all coordinates system are said to be invariant tensors. For example: Kronecker delta symbol and Levi - Civita symbol (Epsilon tensor) are the important examples of such tensors. Two tensors of the same rank and same type are said to be equal if their components are one to one are equal. For example,

for all values of the indices.

Again, if two tensors are equal in one coordinate system, then they will also be equal in all

coordinate system.

Levi - Civita Symbol Levi - Civita symbol is also called an epsilon tensor or an alternating tensor or a permutation tensor. In 3 - dimensional space, Levi - Civita symbol is a tensor of rank three and is denoted by while in 4 - dimensional space, it is a tensor of rank four and is denoted by The Levi - Civita symbol is defined as the quantity

(in 3 - dimensional space which is anti

- symmetric in all its indices). Thus, only non - vanishing components of all the indices are different and they are equal to permutation of that is,

As

or

according as

are those for which is an even or odd

is anti - symmetric in every pair of indices, therefore

It can be shown that

for all values of indices

that is, Levi - Civita symbol or

alternate tensor is invariant.

Theorem 2.11 Prove that respectively.

is invariant, where

and

are contravariant vector and covariant vector

Proof Let and transformation

be any two vectors (or tensors having rank one). Then by tensor law of

(i) and (ii)

From (i) and (ii), we have

which on changing dummy

which shows that

is invariant.

Theorem 2.12 Show that the inner product of two tensors of rank one (one contravariant tensor of rank one and other covariant tensor of rank one) is a tensor of rank zero (or, a scalar, that is, invariant).

Proof Let

and

be two tensors having rank one. Then by tensor law of transformation

(i) and (ii) Let

be the inner product of

and

, then

(iii) From (i) and (ii), we have

which on changing dummy p by i gives

which in view of (iii) gives

showing that

is an invariant, that is, the product

transformation and hence

is a scalar, that is,

remains invariant under the coordinate is a tensor of rank zero.

Theorem 2.13 Show that Kronecker delta is a mixed tensor of rank two and also show that it is an invariant, that is, it has the same components in every coordinate system.

Proof Let

be an arbitrary vector (or, tensor of rank one) and

be a Kronecker delta. Evidently,

contravariant tensor of rank one. Thus, the inner product of tensor. Again,

with an arbitrary vector is a tensor. Hence, by Quotient law,

is a

has two distinct indices one of which is on upper position and the other is on

lower position. Hence

which shows that system.

is a mixed tensor of rank two. Then by tensor law of transformation,

is an invariant. That is,

has same components in every coordinate

Example 2.9 Show that the contraction of the outer product of the tensors

and

is an invariant.

Solution Let

and

be two tensors having rank one, then by tensor law of transformation

(i) and (ii) From (i) and (ii), we have

Contracting over i and j, we have

which on changing dummy a by i gives

which shows that

is an invariant and hence the result is completed.

2.3.2 Fundamental Tensor The distance ds between two neighboring points with coordinates

and

is given by (2.9)

where and are functions of the coordinates and may vary from point to point. Equation (2.9) is called the metric equation and ds is the interval or line - element. The space which satisfied the relation (2.9) is called the Riemannian space.

The function

are

in number, real and need not be positive but their determinant

(2.10) is non - zero. Since

and

in equation (2.9) are contravariant vectors and

is an invariant for any

arbitrary choice of and from quotient law, it follows that is a covariant tensor of rank two. This tensor is known as a fundamental tensor or metric tensor. In addition to the metric tensor

there are two more fundamental tensors

are known as the conjugate or reciprocal tensor of

and

which

and Kronecker delta, respectively.

Also, (2.11)

and (2.12) The function as follows: Let

in equation (2.11) is a contravariant tensor of rank two and this can be proved

be a contravariant vector; then its inner product with

gives an arbitrary covariant

vector

(2.13) Also,

using (2.13) using (2.12)

That is, the inner product of vector

with an arbitrary covariant vector

and hence by quotient law,

gives a contravariant

is a contravariant tensor of rank two. It is also

symmetric in k and i, that is, Since

and

are covariant and contravariant tensors of rank two, respectively, using

quotient law in equation (2.12), it is seen that is a mixed tensor of rank two. The three tensors defined through equations (2.9), (2.11) and (2.12) are called fundamental tensors and are of basic importance in general theory of relativity. While in mechanic, the metric

is related to the square of the speed through the relation

Also,

Hence, system.

that is, it has constant components which are independent of the coordinate

2.3.3 Symmetric Tensor and Anti - Symmetric (Skew - Symmetric) Tensor The tensor

is not necessarily the same as the tensor

(In the language of matrices

is the

transpose of ). If however two contravariant (or covariant) indices of a tensor can be interchanged without altering the value of the tensor, then the tensor is said to be symmetric with respect to these indices. For example, if

(2.14) then the conravariant tensor of second order

(or covariant tensor of second order

to be symmetric. A symmetric tensor of the second order has at most components, which will be shown later. For a tensor of higher order

if

) is said

independent

(2.15) then the tensor

is said to be symmetric with respect to the indices

In general, if a tensor

of type

satisfy the relation

(2.16) then the tensor

is symmetric in first two indices

The energy - momentum tensor of electromagnetic field, trace - free and the general energy momentum are the examples of symmetric tensors. Similarly, if two contravariant (or covariant) indices of a tensor can be interchanged with alter in sign but not in magnitude, then the tensor is said to be anti - symmetric (or skew - symmetric) with respect to these indices. For example, if

(2.17) then the conravariant tensor of second order

(or covariant tensor of second order

) is said

to be skew - symmetric. A skew - symmetric tensor of the second order has at most independent components, as will be shown later. For a tensor of higher order

if

(2.18) then the tensor

is said to be skew - symmetric with respect to the indices

In general, if a tensor

of type

satisfy the relation

(2.19) then the tensor

is skew - symmetric in first two indices

The electromagnetic field tensor is the example of anti - symmetric tensor. It is noted that the symmetric (or, anti - symmetric) property of a tensor is independent of

coordinate system and symmetric (or, anti - symmetric) property of a tensor is defined only when the indices are of the same type, however the Konecker delta, is a mixed tensor which possesses symmetric property with respect to its two indices, which will be shown later.

Theorem 2.14 Show that

is a covariant symmetric tensor.

Proof Since (i) Since

is an invariant and therefore from (i), it follows that

(ii) Since (iii) From (ii) and (iii), we have

Since

and

are arbitrary contravariant vectors, that is

which follows the tensor law of transformation and hence Also,

therefore

is a second rank covariant tensor.

(iv) where (v) Now,

using (v) (vi) which shows that

is symmetric tensor.

Again,

using (v) (vii) which shows that is skew - symmetric tensor. Now, changing dummy indices on right - hand side of (iv), we get

using (vi) and (vii) (viii) Adding the relations (iv) and (viii), we get

But is symmetric tensor, therefore symmet-ric tensor.

is also symmetric tensor. Hence

is a covariant

Theorem 2.15 Prove that symmetric (or, skew - symmetric) property of a tensor is independent of coordinate system.

Proof Let and j in

be a mixed tensor of rank four which is symmetric with respect to the first two indices i coordinate system, that is

(i) By tensor law of transformation,

(ii) using (i)

by (ii) which shows that the given tensor is again symmetric with respect to the first two indices i and j in new coordinate system. This result can also be proved for contravariant indices. Again, let

be a mixed tensor of rank four which is anti - symmetric with respect to the

first two indices i and j in

coordinate system, that is

(iii) Using (iii) in (ii), we have

by (ii). which shows that the given tensor is again anti - symmetric with respect to the first two indices i and j in new coordinate system. This result can also be proved for contravariant indices. Thus, the symmetric (or, skew - symmetric) property of a tensor is independent of coordinate system.

Theorem 2.16 Prove that a symmetric tensor

has

, whereas a skew - symmetric tensor

independent components in n - dimensional space has

independent components in

Proof Let be a symmetric tensor of rank two and hence it has given bellow:

components in

which are

Number of independent components corresponding to a repeated suffix is n. Also, the number of independent components of

corresponding to distinct suffixes is

Due to symmetric property this number is reduced to The total number of independent components of a symmetric tensor

Again, let be a skew - symmetric tensor of rank two and hence it has which are given bellow:

Now,

is skew - symmetric, that is,

Taking

components in

we get

But Therefore which shows that the number of independent components corresponding to a repeated suffix is zero. Also, the number of independent components of

corresponding to distinct suffixes is

Due to anti - symmetric property this number is reduced to The total number of independent components of a skew - symmetric tensor

Theorem 2.17

Show that a tensor of type (0, 2) is expressible as a sum of two tensors, one of which is symmetric and the other is anti - symmetric (skew - symmetric).

Proof Let

be a tensor of type

Evidently

(i) where (ii) and (iii) Now, interchanging i and j in (ii), we have

by (ii) which shows that

is symmetric.

Again, interchanging i and j in (iii), we have

by (iii) which shows that

is anti - symmetric.

Since the sum or difference of two tensors is a tensor of the same rank and similar character. It follows that

and

both are second rank covariant tensors.

Thus, from (i), we see that symmetric and

Theorem 2.18

is expressible as a sum of two tensors

is anti - symmetric (or skew - symmetric).

and

where

is

Prove that the symmetric (or skew - symmetric) property of a tensor is defined only when the indices are of the same type.

Proof Let be symmetric with respect to indices i and ℓ, that is, one contravariant and other covariant, respectively, then

(i) Since

is a tensor, therefore by tensor law of transformation

using (i) by interchanging dummy ℓ and i

(ii) But according to tensor law of transformation

(iii) From (ii) and (iii), we have which shows that the symmetric property of a tensor is defined only when the indices are of the same type. That is, both indices are either at covariant position or at contravariant position. This result can also be proved for skew - symmetric property.

Example 2.10 Show that the energy - momentum tensor

for the electromagnetic field is symmetric and

Solution The energy - momentum tensor is

which can be written as

(i) Interchanging i and j in (i),we have

by interchanging dummy i and j by (i) which shows that

is symmetric.

Moreover, the relation (i) gives

which gives

Thus, the energy - momentum tensor

is symmetric and

Example 2.11 Let

and

that

be a symmetric tensor and a skew - symmetric tensor respectively. Then show and also for any arbitrary tensor

Solution Since

is a symmetric tensor and

is an anti - symmetric tensor, so that

which on interchanging i and j keeping in mind i and j are dummy indices in right - hand side of

the above relation gives

But

therefore

Moreover, we know that any tensor

can be written as the sum of its symmetric part

and its anti - symmetric part

thus

and

Example 2.12 If

is a fourth rank tensor, then show that

Solution We have

which completes the required result.

2.3.4 Reciprocal Tensor A contravariant tensor

and covariant tensor

satisfying

and (2.20) are called the reciprocal tensors. For example,

is the reciprocal tensor of the tensor

and

is the reciprocal tensor of the tensor because The reciprocal tensor is also known as the conjugate tensor. It is important to note that a tensor of the second order has a conjugate only if its determinant is not zero.

2.3.5 Reducible Tensor and Irreducible Tensor A tensor quantity is said to be reducible if it can be decomposed into parts which transforms among themselves. If such a decomposition is not possible then the quantity is called irreducible. The irreducible tensors find their applications in the decomposition of the Riemannian curvature tensor. The Riemannian curvature tensor will have studied in detail in chapter five. Consider a second rank mixed tensor

and thus

trace of

then by tensor law of transformation

transforms as

which shows that the trace of Now consider

transforms into itself and therefore the tensor

is reducible.

(2.21) Then by tensor law of transformation, we have

using (2.21) which can not be further decomposed as it has no trace and accordingly Moreover, the trace

is also irreducible.

Thus, we have shown that the irreducible parts of tensor

is irreducible.

which is obtained from

are its trace

by

2.3.6 Relative Tensor If the components of a tensor

transforms according to the rule

and the trace - free

(2.22) where

is the Jacobian of the transformation, then the tensor

is called a relative

tensor of weight ω. The constant weight ω may be a positive or negative integer. If ω = 1, then the relative tensor is called the tensor density. If ω = 0 then the relative tensor is called an absolute tensor. Obviously we have been dealing above with absolute tensor. A relative tensor of rank one is called a relative vector. Thus, if

then

is called a relative vector of weight ω. A relative vector of weight one is called a vector

density and a relative vector of weight zero is called an absolute vector. A relative tensor of rank zero is called a relative scalar. Thus, if

Then I is called a relative scalar of weight ω. A relative scalar of weight one is called scalar density. A relative scalar of weight zero is called an absolute scalar. As an example of scalar density, consider the determinant of an ordinary covariant tensor of rank two. From the properties of the determinant of a product of matrices, we have

Therefore, the determinant of a covariant tensor of rank two is a scalar density of weight two. The algebraic operations (For example, addition, subtraction, multiplication, contraction, etc.) of relative tensors are same as that of absolute tensors (ordinary tensors) and we have 1.

A linear combination of tensor densities of the same type and same weigtht is a new tensor density of the same type and same weight.

2.

The outer product of two tensor densities of order p and q and weight ω1 and ω2, respectively, lead to a tensor density of order p = q and weight (ω1 + ω2).

3.

The contraction of indices on a mixed tensor density of order p and weight ω is a tensor density of order p – 2 and weght ω.

From these rules it follows that a tensor density of weight ω can be expressed as the product of an ordinary tensor and a scalar density of weight ω. Thus, the weight of the tensor density remains unchanged by raising and lowering of indices, through metric tensor, of tensor densities. Further, the symmetric properties of tensor densities follows the same procedure as that of

ordinary tensors (absolute tensors).

Theorem 2.19 Prove that the relative tensor law of transformation possess the group property.

Proof Let

be a relative tensor of weight ω. Consider the coordinate transformation that is

In case of the transformation

In case of the transformation

which in view of

we have

we have

gives

This shows that if we make the direct transformation from (i) to (iii), we get the same law of transformation. Which shows that the transformation of a relative tensor form a group, that is, the relative tensor law of transformation possess the group property.

Example 2.13 Prove that

Solution Since

is a tensor density.

Taking the determinant of both sides, we have

which leads to

showing that

is a tensor density.

Example 2.14 If is a second rank covariant symmetric tensor and density.

then show that

is a scalar

Solution Let

be a covariant tensor of rank two. Then by tensor law of transformation

which in view of

showing that

gives

is a scalar density.

2.3.7 Raising and Lowering Suffix: Associated Tensors Let us now define the operations of raising and lowering of the indices of a tensor with the help of contravariant and covariant fundamental tensors and respectively. By raising a suffix, a covariant index is changed to a contravariant index and by lowering a suffix, a contravariant index is changed to a covariant index. The operations of raising or lowering of the indices merely depend on the inner multiplication of a given tensor with the contravariant or covariant fundamental tensors and respectively. The fundamental tensors enable us to perform the operation of raising and lowering the suffixes of a tensor which changes a covariant suffix into a contravariant one and vice versa. The process really consists of forming the inner product of a given tensor with one of the

fundamental tensor. Raising a suffix for a vector is defined by

(2.23) and lowering by

(2.24) For a more general tensor raising is defined by

(2.25) and lowering by

(2.26) The process of raising and lowering an index can be applied for any tensor. Thus

It can be easily shown that the entities obtained as a result of raising and lowering operations with the metric (or, fundamental) tensor are themselves tensor. For example, let us consider equation (2.23). In

coordinate system, the relation (2.23) can be written as

which shows that transform like a contravariant tensor of rank one, that is, is a contravariant tensor of rank one. The tensors obtained as a result of the raising and lowering operations with fundamental

tensors are called associated tensors. The inner product of the fundamental tensor covariant vector

and the contravariant vector

which is called associate vector

gives a

We define

Similarly, we define

and say that the vector

is associate to the vector

The relation in between a vector and its associate is reciprocal, for the vector associate to

is

This process of association is often referred to as ‘lowering the superscript’ or ‘raising the subscript’ respectively. The process of raising and lowering the indices can performed on tensors. From the tensor we can form associate tensors like The dot notation is introduced to indicate which indices have been raised or lowered. The dots will be omitted when there is no possibility of confusion. For example, we shall write Note very carefully that although and are conjugate tensors, the tensors and are not as a rule conjugate. The process of obtaining associated tensors from a given tensor by raising and lowering operations may be summarize up as follows: (i)

Inner multiplication with contravariant fundamental tensor with raising.

(ii) Inner multiplication with covariant fundamental tensor lowering.

gives substitution

gives substitution with

(iii) Inner multiplication with mixed fundamental tensor gives substitution without raising or lowering. The process of raising and lowering the indices are reciprocal to each other, which is proved in the following theorem:

Theorem 2.20 Show that the process of raising and lowering the indices are reciprocal to each other.

Proof Let

be a mixed tensor of rank three. First raising the index k by taking its inner product with

contravariant fundamental tensor

we have

Now lowering the index p by taking its inner product with covariant fundamental tensor have

(i) we

(ii) The right - hand side is the same as the given tensor lowering the indices are reciprocal to each other.

Hence the process of raising and

Example 2.15 Prove that where the symbols have their usual meaning.

Solution (a) Since (i) and

using (i) (ii)

Also, (iii) and using (iii)

(iv) From (ii) and (iv), we have

(v) Now, changing dummy k by i and ℓ by j, respectively on left - hand side of (v), we get

(b) Since (vi) and (vii) From (vi) and (vii), we have

Example 2.16 Show that a linear combination of the tensors of type

is a tensor of type

Solution Let be tensors each of them is of type combination of these tensors is

Let

be scalars. Any linear

(i)

If we show is a tensor of type From (i), we have

In view of

then the required result will be completed.

are invariant, that is,

the above relation reduces to

which confirms the tensor law of transformation for the tensor of type tensor of type

Hence

is a

Example 2.17 If

is a skew - symmetric tensor, then prove that

Solution Let

be a skew - symmetric tensor. Then

(i) Now

using (i) which completes the required result.

Example 2.18

If

then show that the conjugate tensor

for

and

(no summation).

Solution Let

Since

Now,

Since

is the conjugate tensor of

and hence

Evidently,

Again, Generalizing this result, we have

Also, Similarly, Generalizing this result, we have

Example 2.19 Prove that the scalar product of a relative contravariant vector of weight ω and a relative covariant vector of weight

is a relative scalar of weight

Solution Let

be a relative contravariant vector of weight ω and

weight We have to prove that show that

be a relative covariant vector of

is a relative scalar of weight

For this we must

Now, (i)

and (ii) From (i) and (ii), we have

which completes the required result.

Example 2.20 If is an arbitrary contravariant vector and covariant tensor of rank two.

is an invariant, then show that

is a

Solution Let

be an arbitrary contravariant vector and

be an invariant, that is

(i)

which in view of (2.1) gives

(ii) Interchanging suffixes i and j in (ii), we have

(iii) Again, interchanging the dummy suffixes

in (iii), we get

(iv) Adding (ii) and (iv), we get

(v) But (vi) From (v) and (vi), we have

which on simplifying gives

Since

and

are arbitrary, that is,

Therefore

which confirms the tensor law of transformation for a covariant tensor of rank two. Hence is a covariant tensor of rank two.

Example 2.21 If

and

are contravariant vectors and

is a covariant tensor of rank two, find the nature

of

Solution Let and be contravariant vectors and law of transformation

which shows that

be a covariant tensor of rank two, then by tensor

is an invariant (scalar).

Example 2.22 If the tensors and are symmetric, and satisfying the equations

are components of contravariant vectors

prove that

Solution For

we have

(i) and (ii) Multiplying (i) and (ii) respectively by

and then subtracting, we get

Interchanging i and j in the second and fourth terms and noting that (iii)

we get

(iv) This proves the first required result. Multiplying equation (i) by

and using (iv), we have

This proves the second required result.

Exercises 1.

Define a tensor and its order (rank of a tensor). Also, show that its transformation law possesses a group property.

2.

Define a contravariant vector (or a contravariant tensor of rank one) and show that its transformation law form a group.

3.

Show that the law of transformation for a contravariant vector is transitive.

4.

Define a mixed tensor of rank two and show that its law of transformation possesses group property.

5.

Define a contravariant and a covariant tensor of rank one. Also, prove that the transformation of a covariant vector form a group.

6.

Prove that if the components of a tensor vanishes in one coordinate system, they vanish identically in all coordinate systems.

7.

Define addition and subtraction of two tensors and show that their sum or difference are also a tensor of the same rank and similar character.

8.

If are two mixed tensors of rank three, then show that their sum or difference are also a tensor of the same rank and similar character.

9.

Show that the difference of two tensors of type

is a tensor of type

10. Show that a linear combination of the tensors of type 11.

is a tensor of type

Show that the sum or difference of two tensors of the type

is a tensor of the type

12. Show that the tensor product of two contravariant vectors (or, two covariant vectors) is a tensor of rank two. But the converse is not true, that is, every contravariant (or, covariant) tensor of rank two is not necessarily the tens- or product of two contravariant vectors (or, two covariant vectors). 13. Define tensor product and show that the tensor product of two tensors of type is a tensor of type whose rank is

and

14. Prove that the open product of two vectors is a tensor of rank two. Is the converse is true? 15.

Define contraction and show that contraction reduces rank of a tensor by two.

16. Explain the process of contraction of tensor. Also, show that 17. Define inner product of two tensors. Also, prove that the inner product of the tensors and

is a tensor of rank 5.

18. Show that the inner product of

and

is a scalar (that is, a tensor of rank zero).

19. State and prove Quotient law of tensors. 20. Show that

is a tensor of rank zero.

21. Define an invariant tensor and show that

is an invariant.

22. Define fundamental tensors and show that tensor and mixed tensor of rank two respectively.

are covariant tensor, contravariant

23. Define symmetric tensor and skew - symmetric tensor and show that a tensor of type (2, 0) can be expressed as a sum of a symmetric tensor and a skew - symmetric tensor. 24. Show that every tensor of type (0, 2) can be expressed as the sum of a symmetric tensor and a skew - symmetric tensor. 25. Show that Kronecker delta is a symmetric mixed tensor of rank two. 26. If a tensor is skew-symmetric in suffixes i and j, then the tensor independent components.

has

27. Show that the symmetric property of a tensor is independent of coordinate system. 28. Show that a symmetric tensor dimensional space

has

independent components in n -

where as a skew - symmetric tensor

has

independent

components in 29. Prove that the symmetric property of a tensor is defined only when the indices are of the same type. 30. If

are components of a tensor of the type (0,2) and if the equation

holds with respect to a basis, then prove that either and

is symmetric.

and

is skew - symmetric or,

31. If

then prove that

32. Define reciprocal tensor and show that

is reciprocal of

33. Define conjugate tensor and show that

and

and vice versa.

are conjugate to each other.

34. Define reducible tensor and irreducible tensor. Show that irreducible, where

is reducible where as

is

is a mixed tensor of rank two and

35. Define a relative tensor, relative vector and relative scalar and prove that the relative tensor law of transformation possess group property. 36. If is a second rank covariant symmetric tensor and density. 37. Define associated tensor and show that tensor of rank one respectively.

then show that

is scalar

are contravariant and covariant

38. Show that the process of raising and lowering the indices are reciprocal to each other. 39. Let

and

be reciprocal symmetric tensors of rank two. Then show that

where 40. Show that

is an invariant.

41. Show that if the index is first raised and then lowered or vice versa, the original tensor is obtained. 42. Show that is a contravariant tensor of rank one whereas tensor of rank one.

is a covariant

43. Prove that the entities obtained as a result of raising and lowering operations with the fundamental metric tensor are themselves a tensor.

3 Christoffel’s Symbols and their Properties

W e begin this chapter by introducing a special type of symbols, known as Christoffel’s symbols having their many applications in most of branches of theoretical physics, Riemannian geometry, theory of relativity and many other disciplines of sciences and engineering and in sequence give some more elementary properties of Christoffel’s symbols, sufficient examples and theorems on it, which are mainly useful for graduate (honours) / postgraduate students and also for researchers.

3.1 Christoffel’s Symbols or Christoffel’s Brackets Although we found that

is a tensor of type (0, 2) having rank two because it satisfy the tensor

law of transformation. But the partial derivative of is not a tensor. We investigate an expression involving the partial derivatives of the fundamental tensors are tensor or not. In 1870, Christoffel and Lipschitz introduced and studied a special type of symbols, and called the Christoffel’s symbols of the first kind and second kind, respectively and are defined as

(3.1) and (3.2) The Christoffel’s symbol of first kind can also be denoted by

or,

and that of second

kind may be represented by The symbols and that is, the Christoffel’s symbols of the first kind and second kind are symmetric in the indices i and j and are not tensors, which will be shown later. Also, Christoffel’s symbols are certain functions involving the derivatives of the compo-nents of the fundamental tensors

From (3.2), it is evident that

and

Also,

which in view of

gives

(3.3) while (3.4) is a consequence of (3.1).

3.2 Symmetric Properties of Christoffel’s Symbols The symbols

and

are not necessarily the same as the symbols

and

then the symbols

are said to be

respectively. If symmetric in j and k.

Now, (3.5) Interchanging j and k in (3.5), we have

using (3.5) (3.6) which shows that

is symmetric in j and k.

Again, using (3.6) using (3.2) (3.7)

which shows that

is symmetric in j and k.

Thus, the Christoffel’s symbols

are symmetric in

Theorem 3.1 Prove that the necessary and sufficient condition that all the Christoffel’s symbols vanish at a point is that

are constant.

Proof Let

be a constant, then

(i) Now, the Christoffel’s symbols of first kind is

using (i). (ii) and the Christoffel’s symbols of second kind is

which shows that the Christoffel’s symbols vanish at a point is that Conversely, let the Christoffel’s symbols vanish at a point. Then

which implies that

which implies that constant for every i and j.

are constant.

Theorem 3. 2 If

is denoted by the symbol

Then prove that

Proof We know that

Therefore (i)

Similarly, (ii)

and (iii)

From (i), (ii) and (iii), and using the partial derivative property

we have

(iv) But it is given that

(v) From (iv) and (v), we have

which completes the required result.

Theorem 3. 3

Show that

where the symbols have their usual meaning.

Proof We know that (i) where

is the cofactor of the element

in the determinant

In view of (i), we have

which implies that

, using (3.2)

which implies that

which completes the required result.

Theorem 3. 4 Show that the Christoffel’s symbols have dimensional space

independent components in an n -

Proof We know that

is a covariant symmetric tensor of rank two. Hence the number of independent

components of

in

is

Consequently

components. Since each of symbols symbols

and

symbols have

has

and

has

independent

is a linear combination of

independent components in

Hence each of

that is, the Christoffel’s

independent components in

3.3 Tensor Laws of Transformation of Christoffel’s Symbols We know that

is a covariant tensor of order two and hence by tensor law of transformation

which on differentiating partially with respect to

gives

(3.8) Similarly, on differentiation of

partially with respect to

which on changing dummy

we have

in the

term on right hand side gives (3.9)

Also, on differentiation of

partially with respect to

which on changing dummy

we have

in the

term on right hand side gives

(3.10) Since

is symmetric, that is, (3.11)

Also, (3.12) In view of

and using

we have

(3.13) which shows that the Christoffel’s symbol of first kind does not follow the tensor law of transformation due the presence of Since

term on right - hand side and hence it is not a tensor.

is a contravariant tensor of order two, therefore by tensor law of transformation

(3.14) Multiplying (3.13) by (3.14) properly and noting that we have

using (3.2) and

(3.15) which shows that the Christoffel’s symbol of second kind does not follow the tensor law of transfor-mation due the presence of term on right - hand side and hence it is not a tensor. From (3.13) and (3.15), it follows that the Christoffel’s symbols are not tensor quantities. From a linear transformations of the type

(3.16) where

are constants.

From (3.16), we have

(3.17) Using (3.17) in (3.13) and (3.15), we have

(3.18) and (3.19) The relations (3.18) and (3.19) shows that Christoffel’s symbols are tensors relative to the linear transformation of the type (3.16). In such cases the Christoffel’s symbols are called Pseudo tensors.

3.4 Transitive Property for Laws of Transformation of Christoffel’s Symbols The laws of transformation for Christoffel’s symbols are

(3.20) Here we are considering coordinate transformations that is

The equation (3.20) is taken as in the case of In the case of the transformation is

(3.21)

In view of (3.20), the relation (3.21) gives

(3.22) But (3.23) Differentiating (3.23) with respect to

we have

Multiplying the above equation by

we get

Replacing dummy i by k in

term of left - hand side, we get (3.24)

Using (3.24) in (3.22), we get

which shows that if we make the direct transformation from (i) to (iii), we get the same law of transformation. Hence, the laws of transformations for Christoffel’s symbols possess group property.

Example 3.1 Prove that

where the symbols have their usual meaning.

Solution which on differentiation with respect to

gives

which in view of

gives

(i) Multiplying (i) by

and noting that

we get

which in view of (3.2) gives

Replacing i by h and j by h in

terms on right hand side in the above relation, we get

Replacing m by i and ℓ by j in the above relation and using the symmetric property of have

we

which completes the required result.

Example 3.2 Calculate the Christoffel’s symbols of both kinds for spaces, where

Solution (a) The Christoffel’s symbols of first kind is defined by

(i) There arises in all four cases:

Case - I: If

then (i) becomes

Case - II: If

then (i) gives

Case - III: If

then (i) gives

Case - I V: If

and noting that

(i) gives

(b) We know that if the coordinate system is orthogonal, then symbols of second kind is given by

The Christoffel’s

Clearly, if and if (ii) Now, applying four cases of (i), we have Case - I: If

Case - II: If

then (ii) gives

then (ii) gives

Case - III: If

then (ii) gives

Case - IV: If

then (ii) gives

Example 3.3 For the metric

where a, b, c, d are functions of coordinates

evaluate the Christoffel’s symbols

Solution The given metric is

Clearly,

Since

and the rest all are zero.

therefore the coordinate system is orthogonal, that is

Hence

Now,

Example 3.4 Calculate the non - vanishing Christoffel’s symbols of second kind corresponding to the metric

Solution The given metric is

Clearly, Since

therefore the coordinate system is orthogonal, that is

Hence

and the rest all are zero. Since is constant, therefore the non - vanishing Christoffel’s symbols of second kind are as follows:

Example 3.5 Calculate the non - vanishing Christoffel’s symbols corresponding to the metric

Solution

The given metric is

Let us take

then clearly, and the rest all are zero.

Since

therefore the coordinate system is orthogonal, that is

Hence

and the rest all are zero. Since follows:

are constants, therefore the non - vanishing Christoffel’s symbols are as

(i)

(ii)

(iii)

(iv)

Similarly, (v) (vi)

Example 3.6 Find the Christoffel’s symbols corresponding to the metric

Solution The given metric is

Let us take

Now,

Now, Similarly,

and the rest all are zero. Now, Similarly, In a similar manner, we have

then clearly,

and the rest all are zero.

Again, we have

Now,

and the rest all are zero.

Example 3.7 Find the Christoffel’s symbols corresponding to the metric

Solution The given metric is

Let us take Now,

then clearly,

and the rest all are zero.

Now,

and the rest all are zero. Now,

and the rest all are zero. Again, we have

Now,

and the rest all are zero.

Example 3.8 Find the Christoffel’s symbols corresponding to the metric

Solution The given metric is

Clearly,

and the rest all are zero.

Now,

Now, Now,

and the rest all are zero.

and the rest all are zero. Again, we have

Now,

and the rest all are zero.

Example 3.9 If

and

are the components of two symmetric covariant tensors and

corresponding Christoffel’s symbols of the second kind, then prove that

are the defined by

are components of a mixed tensor of the order indicated by the indices.

Solution We know that

(i) In view of (i), we have

(ii)

and (iii) Now, subtracting (ii) from (iii), we get

(iv) Using the fact

in (iv), we get

which confirms the tensor law transformation for a mixed tensor of rank three. Hence mixed tensor of rank three. Note 1. The metric tensor in spherical polar coordinate is given by

expanding along R1 In this case,

with

then clearly,

2. The metric tensor in spherical polar coordinate is given by

and the rest all are zero.

is a

expanding along R1 In this case,

with

then clearly,

and the rest all are zero.

Exercises 1.

Define Christoffel’s symbols of first kind and second kind, respectively.

2.

Show that

3.

Prove that

4.

Prove that

and also established the relation

where the symbols have their usual meaning. 5.

Show that

6.

Show that if

7.

Show that Christoffel’s symbols

8.

Determine the number of independent components of Christoffel’s symbols in

9.

Obtain tensor laws of transformation of Christoffel’s symbols and hence show that the Christoffel’s symbols are not tensor quantities.

are symmetric in i and j.

10. In which cases, Christoffel’s symbols are tensor quantities? 11. Define Pseudo tensors.

12. Prove that the transformation of Christoffel’s symbols form a group. 13. Prove that the necessary and sufficient condition that all the Christoffel’s symbols vanish at a point is that

are constant. and i, j, k are unequal, then

14. Show that if

15. Find the non - vanishing Christoffel’s symbols of second kind corresponding to the metric

16. Find the non - vanishing Christoffel’s symbols corresponding to the metric 17. For the metric

where a, b, c, d are functions of coordinates

evaluate the Christoffel’s symbols

18. Calculate the Christoffel’s symbols

corresponding to the metric

19. Determine the Christoffel’s symbols of the second kind in (i) rectangular (ii) cylindrical and (iii) spherical coordinates. (Hints: (i) in rectangular coordinates we have

so that

(ii) in cylindrical coordinates and (iii) in spherical coordinates

we have 20. Find the Christoffel’s symbols for a surface of a sphere in a 2 - dimensional Riemannian space. 21. Show that the Christoffel’s symbols hsve 18 independent components in 3 - dimensional space. 22. Show that the Christoffel’s symbols are not tensor quantities but they are tensors relative to the linear transformation of the type

where

are constants.

23. If and are the components of two symmetric covariant tensors and are the corresponding Christoffel’s symbols of the second kind, then prove that the quantities contravariant and

are the components of a mixed tensor, the indices ofcovariant.

being an index of

24. Let be two metric tensors, and be the corresponding Christoffel symbols. Then prove that the difference of these symbols is a tensor.

Answer 10. In a linear transformation of the type where are constants, the Christoffel’s symbols are tensor quantities. 15. The non - vanishing Christoffel’s symbols of second kind are

16. The non-vanishing Christoffel’s symbols are

17. The Christoffel’s symbols are

18.

. (i)

(ii)

and the rest all are zero and (iii)

and the rest all are zero. 20.

and the rest all are zero, and the rest all are zero.

4 Covariant Differentiation of Tensors and their Properties

G enerally we see that the covariant derivative of a tensor with respect to the variable

is extended the rank of the given tensor by one at covariant position. In this chapter, we have discussed in detail about the covariant differentiation of tensors and its properties, divergence of tensor, intrinsic derivatives and parallel displacement of vectors, which are mainly useful for graduate (honours) / postgraduate students and also for researchers.

4.1 Comma Notation We begin this section by introducing comma notation, which are much helpful for defining covariant derivative of vectors (contravariant or covariant), covariant derivative of tensors (contravariant or covariant or mixed) of order two and higher order and covariant derivative of the fundamental tensors. The comma notation are given as follows: Let

be a covariant tensor of rank one. The tensor

derivative of

with respect to

and

of order two is defined as covariant

is defined as

(4.1) Also, (4.2)

(4.3)

(4.4) and in general,

(4.5)

From the above we see that the notation of type (i) the partial derivative of

contains

terms. They are

with respect to

(ii) the terms containing with positive sign similar to that

and (iii) the terms containing with negative sign similar to that

4.2 Covariant Differentiation of Vectors (Tensors of Rank one) Let us investigate the tensor character, if any, of the partial derivatives of a covariant vector (or a covariant tensor of rank one). We start by differentiating partially the transformation law of covariant vector

(4.6) with respect to

gives

(4.7) But (4.8) The presence of the last term on the right - hand side of the relation (4.7) shows that the partial derivatives

do not form form a tensor. To obtain a tensor involving the partial derivatives,

put in term on on the right - hand side of the relation (4.7) and eliminate the partial derivatives of the second order by means of (4.8) and this gives us

which on rearranging the terms gives

which in view of (4.6) gives

(4.9) Using (4.1) in (4.9), we have

(4.10) which confirms the tensor law of transformation for a covariant tensor of rank two. Hence is a covariant tensor of rank two and it is called the covariant derivative of

with respect to

Some authors use the notation In order to set up the corresponding entity for contravariant vectors, we may conveniently start by differentiating the transformation law of contravariant vector (or a contravariant tensor of rank one)

(4.11) with respect to

gives

(4.12) The presence of the last term on the right - hand side of the relation (4.12) shows that the partial derivatives do not form a tensor. For the elimination of the partial derivatives of the second order, using (4.8) in (4.12), we have

which on rearranging the terms gives

which in view of (4.11) and

gives

which on changing dummy

Using (4.2) in the above relation, we have

(4.13) which confirms the tensor law of transformation for a mixed tensor of rank two. Hence mixed tensor of rank two and it is called the covariant derivative of with respect to

is a Some

authors use the notation As above we constructed tensors containing the partial derivatives of vectors (tensor having rank one). Can we now extend the process of covariant differentiation to tensors? The answer is in affirmative. In sequence, we discuss the covariant differentiation of a covariant tensor of rank two, a contravariant tensor of rank two and a mixed tensor of rank two one by one as follows:

4.3 Covariant Derivative of a Covariant Tensor of Rank two Let

be a covariant tensor of rank two. Then by tensor law of transformation (4.14)

Differentiating (4.14) partially with respect to

we get

(4.15) Now,

(4.16) and

(4.17) Using (4.16) and (4.17) in (4.15), we have

which on rearranging the terms gives

Using (4.3) in the above relation, we have

(4.18) which confirms the tensor law of transformation for a covariant tensor of rank three. Hence, is a covariant tensor of rank three and it is called the covariant derivative of respect to

Some authors use the notation

4.4 Covariant Derivative of a Contravariant Tensor of Rank two Let

be a contravariant tensor of rank two. Then by tensor law of transformation

with

(4.19) Differentiating (4.19) with respect to

we get

(4.20) Now,

(4.21) and

(4.22) Using (4.21) and (4.22) in (4.20), we get

which on changing dumm on the right hand side and rearranging the terms gives

Using (4.4) in the above relation, we have (4.23) which confirms the tensor law of transformation for a mixed tensor of rank three. Hence mixed tensor of rank three and it is called the covariant derivative of

is a

with respect to

Some authors use the notation

4.5 Covariant Derivative of a Mixed Tensor of Rank two Let

be a mixed tensor of rank two. Then by tensor law of transformation

(4.24) Differentiating (4.24) partially with respect to

we get

(4.25) which in view of dummy in (4.25) and then using (4.8), we get

which on rearranging the terms gives

term on the right hand side

which in view of (4.24) gives

which on interchanging i and h, term,

in

term on the left hand side and

term on the right hand side of the above relation gives

Using (4.5) in the above relation, we have

(4.26) which confirms the tensor law of transformation for a mixed tensor of rank three. Hence a mixed tensor of rank three and it is called the covariant derivative of

is

with respect to

Some authors use the notation In general,

is a mixed tensor of rank

and it is called the covariant

derivative of

with respect to

Also,

derivative of

with respect to

can be proved in the same way as the covariant

derivative of

The covariant

with respect to

4.6 Covariant Derivative of the Fundamental Tensors We shall now apply the process of covariant differentiation of tensors to fundamental tensors

(i) Covariant derivative of a covariant fundamental tensor

with respect to

is given by

(ii) Since

according as

which in view of

Therefore,

gives

Multiplying on both sides of the above relation by

and noting that

we obtain

which in view of (3.2) gives

which in view of (4.4) gives

(iii) Covariant derivative of mixed fundamental tensor

Thus, the covariant derivative of the fundamental tensors zero. That is,

with respect to

is given by

are all identically equal to

It means that the three fundamental tensors are treated

as constants. Due to this reason the fundamental tensors

are defined as covariant

constants.

4.7 Laws of Covariant Differentiation Covariant differentiation of tensors obeys the following laws: (i) The covariant derivative of the sum (or difference) of two tensors is the sum (or difference) of their covariant derivatives. This law can be immediantly deduced from the relation (4.5). (ii) The covariant derivative of an outer (or inner) product of two tensors is equal to the sum of two terms obtained by outer (or inner) multiplication of each tensor with the covariant derivative of the other tensor. This law can be proved as follows: Let

be a mixed tensor of rank three and

be a covariant tensor of rank one. Let

(4.27) Since open (or outer) product of two tensors is a tensor and hence

is a tensor, therefore,

(4.28) Using (4.27) in (4.28), we get

Using (4.1) and (4.5) in the above relation, we have

This type of proof is quite general and is applicable to any case of outer product of two tensors. The inner product of two tensors is a tensor formed by outer product and contraction. The contraction of i and k gives us

which makes it clear that the rule also applies to inner products. (iii) The fundamental tensors For example

are constants with respect to covariant differentiation.

It is obvious that the covariant differentiation of products, sums and differences of the tensors obeys the same rules as its ordinary differentiation, which is proved in the following theorem:

Theorem 4.1 Show that the covariant differentiation of sum and product of the tensors obeys the same rules as its ordinary differentiation.

Proof Let

and

be covariant tensors of rank two. Let

(i) Since sum of two tensors is a tensor of the same rank and similar character. Hence

is a

second rank covariant tensor. Since (ii) Using (i) in (ii), we get

using (ii) which shows that the covariant differentiation of sum of the tensors obeys the same rule as its ordinary differentiation. This result can also be proved for any types (contravariant or covariant or mixed) of tensors. Again, let

be a mixed tensor of rank two and

be a covariant tensor of rank one. Let

(iii) Since open or outer product of two tensors is a tensor and hence

is a tensor, therefore,

(iv) Using (iii) in (iv), we get

Using (4.1) and (4.5) in the above relation, we have

which shows that the tensors product obeys the same rules as its ordinary differentiation.

4.8 Covariant Derivative of an Invariant (or a Scalar) The covariant derivative of an invariant (or a scalar) is the same rule as its ordinary derivative. That is

where

is a scalar. Which will be prove it in the following theorem:

Theorem 4.2

Show that the covariant derivative of an invariant (or a scalar) is the same rule as its ordinary derivative.

Proof Let

be an invariant and

be a covariant vector. Then the product

The covariant derivative of the product

with respect to

is a covariant vector.

is written as

As the covariant differentiation of the product obeys the same rules as its ordinary differentiation, so we must have

(ii) Equating (i) and (ii), we have

As

is an arbitrary vector, that is

so that

which shows that the covariant derivative of an invariant (or a scalar) is the same rule as its ordinary derivative.

Theorem 4.3 Prove that the operation of covariant differentiation does not possess, in general, the commutative property of ordinary differentiation. But the covariant differentiation of an invariant is commutative.

Proof Let

be a covariant tensor of rank one. The tensor

covariant derivative of

with respect to

and

having of order two is defined as

(i) Also, (ii) and in general,

(iii) Interchanging i and j in (i) and (ii), we have

(iv) and (v) From (i) and (iv), we have

Similarly, from (ii) and (v), we have

Again, from (iii), we have

(vi) From (iii) and (vi), we have

which shows that the operation of covariant differentiation does not possess, in general, commutative property of ordinary differentiation.

Again, let respect to

Since

be an invariant and

is a covariant vector and its covariant derivative with

is

(vii) is a tensor of rank zero, so that its covariant derivative is its partial derivatives, that is

(viii) In view of equation (viii), the relation (vii) gives

which shows that the covariant differentiation of an invariant is commutative.

Theorem 4.4 Prove that the covariant derivative of a linear combination of tensors, with constant coefficients, equals to the linear combination of these tensors after the covariant differentiation was performed.

Proof Let

and

be two mixed tensors of rank two and a and b be scalars, then

which completes the required result.

4.9 Tensor Form of Gradient, Divergence, Laplacian and Curl The gradient of a scalar (or invariant) or that is

is defined as its ordinary derivative is denoted by

(4.29) Let

be a contravariant tensor of rank one. The divergence of

of its covariant derivative and is denoted by

is defined as the contraction

Thus

(4.30) The divergence of a covariant vector

is denoted by

and is defined as

(4.31) The gradient of a scalar function

is a covariant vector

(say) given by

(4.32) The contravariant vector associated with

is given by

The Laplacian is

Again, curl of a vector

denoted by

and is defined as

(4.33)

4.10 Divergence of a Tensor The divergence of a tensor is defined as its contracted covariant derivative with respect to the index of differentiation and any superscript.

Let

be a contravariant tensor of rank two. The covariant derivative of

with respect to

is defined as

The divergence of

is defined as

(4.34) If the tensor

is ant - symmetric, then

which on rearranging the terms gives

But

Therefore

(4.35) Using (4.35) the relation (4.34), we have

which completes the required result.

Theorem 4.5 Show that

where the symbols have their usual meaning.

Solution Let

be a contravariant vector. Now (i)

Also, (ii) Putting

in (ii), we get

(iii) From (i) and (iii), we have

4.11 Intrinsic Derivatives Let

be a tensor of type

whose components are functions of t along the curve

The intrinsic derivative is defined as

(4.36)

Accordingly, the intrinsic derivative is a tensor of the same order and type as the original tensor. The intrinsic derivative of an invariant is given by

(4.37) That is, the intrinsic derivative of an invariant coincides with its total derivative. Intrinsic derivative of higher order are easily defined. For example,

(4.38) In general, intrinsic differentiation is not commutative. From (4.1), (4.2),(4.3), (4.4) and (4.5), we calculate that

and From the definition of intrinsic derivatives, it follows that they obey the same three laws which apply to covariant derivatives.

4.12 Parallel Displacement of Vectors In Euclidean space which is referred to rectangular Cartesian coordinates, the concept of parallelism has the following significance: A parallel field of vector

is obtained throughout an Euclidean space if the

components of are constants. Thus if is a family of curves defined by the parameter s, this definition of parallelism is analytically expressed as

(4.39) or, (4.40) From this equation the change in the values of

for an infinitesimal displacement

along

the curve is given by

(4.41) Any of equations (4.39), (4.40) and (4.41) expresses the property of parallelism in Euclidean space referred to a rectangular coordinate system. Now the question is how the concept of parallelism may be generalized to a Riemannian space. Let us consider a general coordinate transformation from rectangular coordinate system

to

another coordinate system expressed as

is

then the transformation of contravariant vector

(4.42) Let us now see whether the conditions of parallelism (4.39), (4.40) and (4.41) satisfied by vector

in rectangular coordinate system

system that is, whether From (4.42), we have

are true for vector

in the new coordinate

is zero in new coordinate system.

which in view of (4.39) gives

(4.43) except when coordinate transformation is linear. Thus we see that the conditions of parallelism (4.39), (4.40) and (4.41) satisfied by vector in old coordinate system

are not generally true for vector

in the new coordinate system

We have

(4.44)

In view of (4.44), the relation (4.43) gives

(4.45) We have

which on differentiating with respect to

which on multiplication by

gives

gives

which gives

(4.46) Using (4.46) in (4.45), we have

(4.47) From the transformation law of christoofe’s sysbols of second kind

But in rectangular coordinate system

therefore

(4.48) Using (4.48) in (4.47), we have

(4.49) Interchanging primed and unprimed letters, we get

(4.50) This equation represents the condition of parallel displacement of contravariant vector the curve

along

in Riemannian space. Condition (4.50) may be written as

(4.51) which is the intrinsic derivative of

with respect to s.

Thus the contravariant vector is displaced parallelly along a curve derivative with respect to s is zero. From (4.50), the change in the values of due to the displacement Riemannian space is represented as

if its intrinsic along a curve in a

(4.52) Similarly the condition of parallel displacement of covariant vector given by

along the curve

is

(4.53) The change in the values written as

The angle

due to the displacement

between the two vectors

along a curve in a Riemannian space is

is given by

which on differentiation gives

When both

from fields of parallel vectors, this equation reduces to

that neither of the vectors

provided

is null. That is, the angle between two non-null vectors

remains constant whilst both undergo parallel propagation along the same curve.

Theorem 4.6 Let

be scalar functions of coordinates

Let A be an arbitrary vector. Then

Proof (a) We know that

(i) being an arbitrary contravariant vector. Taking

in (i), we get

or, (ii)

(b) (iii) (c) First prove the results (a) and (b). Taking divergence of both sides in (iii), we get

which gives

Finally, we have (d) First prove the result (a). Replacing A by

in (ii), we get

(iv) Interchanging

in (iv), we get

Theorem 4.7 Let

and

be a covariant vector and a scalar function respectively. Then

Proof (a) We know that

and

(i) (b) First prove the result (a). Putting

in (i), we get

(ii)

Interchanging

in (ii), we get

Theorem 4.8 Let contravariant and covariant components of the same vector A be

and

respectively.

Then show that

Proof By definition,

That is,

Theorem 4.9 Prove that the first covariant derivative of a covariant vector vector itself is gradient of a scalar

is symmetric, if and only if the

Proof Let

be covariant vector and given that or

which on integration gives

or,

which gives

Conversely, given that

then

which implies that

Example 4.1 Prove that

What form does the above equation assume if

is skew - symmetric?

Solution Since Putting

in the above relation, we get

(i) which completes the first part of the problem. Again, let

is skew - symmetric, so that

(ii) Now

which implies that

But

Therefore

(iii) Using (iii) in (i), we get

Thus if

is skew - symmetric, then

Example 4.2 For the metric

show that

if the vector

has the components

Solution The given metric is

Let us take

then clearly,

orthogonal coordinate system and so

therefore we have Therefore

Now, (i)

(ii) Thus,

which completes the required result.

Example 4.3 If

are scalars, then show that

Solution We have

which completes the required result.

Example 4.4 For a scalar

show that

Solution

which completes the required result.

Example 4.5 Evaluate

in spherical polar coordinates

Solution The given metric is

Let us take

then clearly, it means that the coordinate system is orthogonal

and so

Therefore

(i) and the rest all are zero.

Now, (ii)

Also,

which is the expression for the Laplacian in spherical polar coordinates. Thus, by a method of tensor analysis the well known formula of vector calculus has been obtained in a straight forward manner.

Example 4.6 If

is a skew - symmetric tensor, then show that

is a tensor of rank four.

Solution Let

is skew - symmetric tensor, so that

(i) Now,

(ii) Now,

on interchanging dummy suffixes a and k

which implies that

But

therefore

(iii) Similarly, we can show that

(iv) In view of (iii) and iv), the relation (ii) reduces to

But

is a tensor of rank four. Hence

is also a tensor of rank four.

Example 4.7 If at a specified point, the derivatives of with respect to are all zero, then prove that the components of covariant derivatives at that point are the same as its ordinary derivatives.

Solution Let (i) Let

be a mixed tensor of rank two. The covariant derivative

with respect to

is

(ii) Since

both contain terms of the type

and so, by virtue of (i), we have

(iii) In view of (iii), the relation (ii) gives

which shows that the components of covariant derivatives at that point are the same as its ordinary derivatives.

Example 4.8 If

is symmetric tensor, then show that

Solution Let

is symmetric tensor, so that

Since which in view of dummy

gives

(i) But

(ii) Now,

or, (iii) Using (iii) in (ii), we have

(iv) Using (iv) in (i), we have

which completes the required results.

Example 4.9 Show that

where the symbols have their usual meaning.

Solution Since which in view of dummy

gives

which completes the required results.

Example 4.10 If

is the curl of a covariant vector

then prove that

and that this is equivalent to

where the symbols have their usual meaning.

Solution Let

be the curl of a covariant vector

so that

(i) Interchanging i and j in (i), we get

using (i) (ii) which shows that

is anti - symmetric.

From (i), we have

(iii) Again, (iv) Using (iii) in (iv), we have

(v)

Similarly,

(vi) and

(vii) Adding (v), (vi) and (vii) and using symmetric property of Christoffel’s symbols, we have (viii) Again, we know that

(ix) Similarly,

(x) and (xi) Adding (ix), (x) and (xi), we get

using (ii)

using (viii).

Thus,

which completes the required results.

Example 4.11 If

is an anti - symmetric tensor of rank two, then show that

is a tensor.

Solution We have

(i) Since

is an anti - symmetric tensor, that is

(ii) Using (ii) in (i), we get

(iii) Since left - hand side of (iii), being the sum of the covariant tensors each of rank three, is a

covariant tensor of rank three. Hence right - hand side of (iii), that is, covariant tensor of rank three.

is a

Example 4.12 If

is a vector (or, a tensor of rank one), then show that, in general,

is not a tensor but

is a tensor.

Solution Let

be a covariant tensor of rank one. Then by tensor law of transformation

(i) Differentiating (i) partially with respect to

we get

(ii) From the tensor law of transformation, it is obvious that the second term on the right - hand side of (ii) is zero. Thus,

is also a tensor only when is, in general, not a tensor.

Again, we have

(iii) Since right - hand side of (iii), being the difference of two covariant tensor each of rank two, is a covariant tensor of rank two. Therefore left - hand side of (iii), that is covariant tensor of rank two.

is also a

Example 4.13 Prove that the necessary and sufficient condition that the curl of a vector field vanishes is that the vector field is gradient.

Solution Let (i) From (i), we have

which in view of (4.1) gives

which in view of

gives

which implies that

(ii) Integrating (ii), we get

(iii) Taking

Conversely, let

(a scalar) in (iii), we get

be a vector such that

Now, (iv) Also, (v)

Subtracting (v) from (iv) and using

, we have

which in view (4.1) gives

which completes the required result.

Example 4.14 Prove that a necessary and sufficient condition that the covariant derivative of a covariant vector be symmetric is that the vector must be gradient. Solution Let

be a covariant vector and covariant derivative of

Since

be symmetric if and only if

therefore

which in view of (4.1) gives

which in view of

gives

which implies that

which on integration gives

(i) Taking

(a scalar) in (i), we get

Conversely, let

be a vector such that

Now, (ii) Also, (iii)

Subtracting (iii) from (ii) and using

, we have

which in view of (4.1) gives

which completes the required result.

Exercises 1.

What do you mean by the covariant derivative of a covariant tensor of rank one? Show that it is a covariant tensor of rank two.

2.

Define the covariant derivative of a contravariant vector and show that it is a mixed tensor of rank two.

3.

Obtain covariant derivative of a covariant vector with respect to the fundamental tensor

4.

Show the covariant derivative of a covariant tensor of rank two is a covariant tensor of rank three.

5.

Show that the covariant derivative of a contravariant tensor of rank two is a mixed tensor of rank three.

6.

Define the covariant derivative of a mixed tensor of rank two and show that it is a mixed tensor of rank three.

7.

Show that the covariant differentiation of sum and product of tensors obeys the same rule

as its ordinary differentiation. 8.

Define the covariant derivative of a scalar and show that it is same as its ordinary differentiation.

9.

Prove that the covariant derivative of an invariant is same as its ordinary differentiation.

10. Define a covariant constant tensor and show that the covariant derivative of the tensor and

all vanish identically.

11. Prove that the operation of covariant differentiation does not possess, in general, the commutative property of ordinary differentiation. 12. Prove that the covariant differentiation of an invariant is commutative. 13. Define divergence of a vector and curl of a vector and show that the necessary and sufficient condition that thecurl of a vector field vanishes is that the vector field is gradient. 14. Let

be contravariant vector and covariant vector respectively, then show that

15. Show that

where the symbols have their usual meaning. 16. Prove that the necessary and sufficient condition for the first covariant derivative of a covariant vector be symmetric is that the vector be a gradient. 17. Prove that the fundamental tensors are covariant constants. 18. Define the curl of a covariant vector field and establish the identity

where denotes the covariant differentiation and covariant vector.

are the components of the curl of a

19. Prove that the tensors may be treated as constant in covariant differentiation with respect to the fundamental tensor. 20. Prove that the covariant derivatives of the tensors theorem). 21. Prove that

all vanish identically (Ricci’s

where the symbols have their usual meaning. 22. If

is symmetric tensor, then show that

where the symbols have their usual meaning. 23. If

is the curl of a covariant vector

then prove that

where the symbols have their usual meaning. 24. Prove that the metric tensors are covariant constant with respect to the Christooffel’s symbols (Ricci’s Lemma). 25. Write the laws of covariant differentiation and show that the covariant differentiation of sum and product of tensors obeys the same rule as its ordinary differentiation. 26. Define intrinsic derivative and show that 27. Define parallel displacement of vectors. 28. Prove that the intrinsic derivative of

is

where the symbols have their usual meaning. 29. If four.

is a skew - symmetric tensor, then show that

30. Show that

where the symbols have their usual meaning.

is a tensor of rank

5 Riemannian Symbols and its Properties W e begin this chapter by introducing the Riemann’s symbols (first kind and second kind) and in sequence, giving in detail its properties. We have also discussed in detail the fundamental concepts of a flat space, uniform vector field, conditions for flat space - time, space of constant curvature and Einstein space etc., which are mainly useful for postgraduate students and researchers.

5.1 Riemannian Symbols of Second Kind The covariant derivative of a tensor is a tensor; hence it can be differentiated covariantly again to obtain a new tensor. This tensor is called the second covariant derivative of the given tensor. With the help of second covariant derivative of the given tensor, we will now investigate the commutative problem with respect to covariant differentiation. Let us begin with the covariant derivative of a covariant tensor is given by

of rank one. The covariant derivative of

with respect to

(5.1) which is a covariant tensor of rank two. Taking covariant derivative of (5.1) with respect to

we get

After rearranging the terms in above relation, we get

(5.2)

Interchanging j and k in (5.2), we get

(5.3) Now, subtracting (5.3) from (5.2) and using the symmetry property of

and the fact of

we get

(5.4) Interchanging the dummy b and a in last two terms on the right hand side of (5.4), we get

(5.5) where (5.6) Since left - hand side of (5.5) is the difference of two tensors each of rank three. Therefore left hand side of (5.5) is a covariant tensor of rank three. Consequently right - hand side of (5.5) is also a covariant tensor of rank three. But (see in section 2.2.5 of chapter two) that

is an arbitrary vector. It follows from quotient law is a mixed tensor of the fourth order, of

contravariant order one and covariant order three. Also, determinant form as

defined by (5.6), can be written in

(5.7) The four rank tensor, given by equation (5.6), is known as the mixed Riemann - Christoffel tensor or, curvature tensor of second kind or, Riemann’s symbol of the second kind. In some

literature it is often referred to as Riemann- Christoffel tensor of second kind. This tensor was first discovered by Riemann (1826 - 1866) and after many years by Christoffel (1829 - 1900). The Riemann curvature tensor plays a central role in the geometric structure of a Riemann space. This tensor is not only important in describing the geometry of the curved space - time, but also form other tensor. We can construct other tensor which give a complete description of the gravitational field. It is clear from the equation (5.5), the necessary and sufficient conditions that the covariant differentiation of all vectors be commutative, is that Riemann - Christoffel tensor be identically zero.

Note We use some important notation in this book, which are given as follows:

5.2 Properties of the Riemann Curvature Tensor The curvature tensor satisfies its skew - symmetric property as well as its cyclic property and differential property, that is

(a) skew - symmetric property of

(b) cyclic property of

(c) differential property of

relative to the indices j and k, that is,

that is

that is,

Proof (a) Interchanging j and k in the equation (5.6), we get (5.8) Now, adding (5.6) and (5.8), we get

which shows that k. (b) Changing

is skew - symmetric (or, anti - symmetric) with respect to the indices j and

cyclically in (5.6), we get

(5.9)

Again, changing

cyclically in (5.9), we get

(5.10) Adding (5.6), (5.9) and (5.10) and using the symmetry property of

we get

(5.11) which shows that (c) At the pole

we have

quantities of types

It means that at

is cyclic with respect to Since the Christoffel’s symbol of both kinds contain the

Hence they vanish at

that is,

Thus, we have

the covariant derivative reduces to ordinary partial derivative.

Differentiating covariantly with respect to geodesic coordinates with the pole at

of (5.6) and then imposing the condition of

we get

(5.12) Changing

cyclically in (5.12), we get

(5.13)

Again, changing

cyclically in (5.13), we get

(5.14) Now, adding (5.12), (5.13) and (5.14) and using the fact

we get

(5.15) which is called the differential property of second identity.

and this property is also known as Bianch’s

5.3 Contraction of the Riemann Curvature Tensor It is obvious that Riemann - Christoffel tensor can be contracted in three ways: (a) Contraction with respect to the indices a and k, (b) Contraction with respect to the indices a and j, and (c) Contraction with respect to the indices a and i. But is skew - symmetric with respect to j and k, so that the process (a) and (b) gives only one independent tensor. Thus, there are only two ways of contracting the Riemann - Christoffel tensor

one way leads to a Ricci tensor while the other way leads to a zero tensor.

(i) Ricci Tensor The contraction of

with respect to the indices a and k, gives the second rank tensor called

the Ricci tensor and denoted by

We replace k by a in (5.6), we get

(5.16) Interchanging

in (5.16), we get

(5.17) Comparing (5.16) and (5.17) keeping in mind the symmetry property of we get

and dummy indices,

(5.18) which shows that the Ricci tensor Also, the contraction of with negative sign.

is a symmetric tensor.

with respect to the suffixes a and j gives the same Ricci tensor

For (5.19) If the Ricci tensor is further contracted, we obtained an invariant and is defined as

called the scalar curvature

(5.20) (ii) Zero Tensor The contraction of with respect to the suffixes i by a in (5.6), we get

gives the zero tensor. Now, we replace

(5.21)

Interchanging the dummy a and b in fourth term of (5.21) and using

we get

Thus, the contraction of

with respect to the suffixes

leads to a zero tensor.

5.4 Riemannian Symbols of First Kind The covariant curvature tensor of rank four is denoted by

and is defined by

(5.22) The symbol is also known as Riemann’s symbol of first kind or the covariant form of Riemann - Christoffel tensor or Riemann - Christoffel symbol of first kind or covariant curvature tensor. It is not difficult to verify that the defining formula (5.22) for convenient deterninantal form

can be written in the

which will be found useful in the listing proerties of this tensor in section 5.5. From (5.6) and (5.22), we have

(5.23)

Since

which in view of

gives

(5.24)

Similarly, (5.25)

Using (5.24) and (5.25) in (5.23), we have

(5.26) In last two terms of (5.26), the dummy b changed into a and expanded the equation, we get

(5.27) which is the required expression for Equation (5.27) is also expressible as

(5.28) which is also the required expression for

5.5 Properties of Covariant Curvature Tensor The curvature tensor

satisfies the following properties:

(a) Anti - symmetric in first two indices h and i, that is,

(b) Anti - symmetric in last two indices j and k, that is,

(c) Symmetric in two pairs of indices hi and jk, that is,

(d) Cyclic property of

that is, (Bianchi’s first identity), and

(e) Differential property, that is, (Bianchi’s second identity).

Proof (a) Anti - Symmetric in First Two Indices h and i of Interchanging h and i in (5.28), we get

which on interchanging the dummy suffixes a and b gives

(5.29) Now, adding (5.28) and (5.29) and using the symmetric property of

which shows that

we get

is skew - symmetric in first two indices h and i.

(b) Anti - Symmetric in Last two Indices j and k of Interchanging j and k in (5.28), we get

(5.30) Adding (5.28) and (5.30), we get

which shows that

is skew - symmetric in last two indices

(c) Symmetric in two pairs of indices Interchanging h and j in (5.28), we get

of

which on interchanging k and i gives

which on interchanging the dummy indices a and b in

term on the right - hand side gives

(5.31) Comparing (5.31) and (5.28), keeping in mind the symmetric property of

and the fact

we get

which shows that

is symmetric in two pairs of indices

(d) Cyclic Property of Changing

cyclically in (5.28), we get

(5. 32) Again, changing

cyclically in (5.32), we get

(5.33) Adding (5.28), (5.32) and (5.33), keeping in mind the symmetric property of fact

and the

we get

(5.34) which shows that

is cyclic with respect to i, j and k.

(e) Differential Property of As the part of (c) in section 5.2, we proceed upto the relation (5.15), that is

which on multiplying by

and summing for a gives

(5.35) which is called the differential property of and this property is also known as Bianchi’s second identity. Since each term of the equations (5.15) and (5.35) are tensor. Hence, (5.15) and (5.35) are tensorial equations. It means that (5.15) and (5.35) are true in every coordinate system and at every point. Hence, (5.15) and (5.35) are true in all coordinate system throughout and so (5.15) and (5.35) are identities, which are called Bianchi’s identities in honour of its discoverer Bianchi.

5.6 Contraction of Bianchi’s Identities (Einstein tensor) Now, applying anti - symmetric property in the second term with respect to k and ℓ of the relation (5.15), we get

which on contraction with respect to a and k gives

(5.36) But

by definition of Ricci tensor. Hence the relation (5.36) gives

which in view of (5.37) Now, changing the dummy indices ℓ and a into i in relation (5.37), we get

(5.38)

(5.39) From (5.38) and (5.39), we get

(5.40) where (5.41) is called an Einstein tensor. The covariant form of Einstein tensor is (5.42) By definition of divergence of a tensor,

using (5.40). Thus, the divergence of an Einstein tensor is identically zero.

Theorem 5.1 Show that the curvature tensor has

independent components in an n - dimensional

space.

Proof Let

be a covariant curvature tensor. We have to show that

has

independent

components in an n - dimensional space. Since is a tensor of rank four, so that it has components. All of them are not independent of each other due to the properties listed below:

The following four cases arise: Case – I: When Hence,

has one like suffix, that is, it is of type

By skew - symmetric property,

has no component.

Case – II: When has two unlike suffixes, that is, it is of type Since h can be had in n ways. After giving a particular value to h, the remaining values can be given to i. Hence h and i can be had in ways. By skew - symmetric property,

Which shows that i and h can be interchanged. Due to this property, the number

is

reduced to By cyclic property,

showing that cyclic property is itself satisfied. Hence, there is no reduction due to this property. Also, there is no reduction due to symmetric property. Therefore, the number of independent components of

is

Case – III: When suffixes

has three unlike suffixes, that is, it is of type can be had in

It can be easily shown that the

ways. Due to symmetric property, that is,

this number is reduced to

By cyclic property,

showing that cyclic property is itself satisfied. Hence, there is no reduction due to this property. Also, there is no reduction due to anti - symmetric property. Therefore, the number of independent components of

is

Case – IV: When has four unlike suffixes, that is, h, i, j and k all being unequal. Now, h can be had in n ways. After giving a particular value to h, the remaining values can be given to i. Thus h and i can be had in ways. After giving particular values to h and i, the remaining values can be given to j and k. Thus h, i, j and k can be had in Due to anti - symmetric property,

the number is reduced to Again, due the symmetric property,

the number is reduced to Further, due to cyclic property,

Out of three components only two are independent. It means that the number is reduced to

Thus, the number of independent components of

is

Finally, the total number of independent components of

is

ways.

Particular Cases: 1.

In 2 - dimensional space (that is, when n = 2), there exists only one independent component.

2.

In 3 - dimensional space (that is, when n = 3), there exists 6 independent components.

3.

In 4 - dimensional space (that is, when n = 4), there exists 20 independent components.

5.7 Riemannian Curvature From any two vectors

at a point of a Riemannian space, we can construct an invariant

Let us consider what happens if we replace the vectors

by the two linear

combinations

(5.43) where

are scalars (invariants). From (5.43), we have

which on simplification gives

Thus, the expression

which is an invariant with respect to coordinate transfor-

mations, is almost an invariant under linear transforma-tions of vectors. In order to obtain an expression which is also invariant under linear transformations of vectors. Let us evaluate

in view of (5.43), we have

(5.44) where θ is the angle between the vectors

Moreover,

(5.45) From equations (5.44) and (5.45), we have

Since

are scalars (invariants),

is also a scalar (invariants) and thus

and

are also invariant under linear

transformations of vectors. This means that the ratio of the two invariants and

is also an invariant denoted by

so that

(5.46) This invariant is called the Riemannian curvature or curvature associated with the vectors of the space. Noted that the denominator of is unity if the vectors are orthogonal unit vectors. At any point of a 2 - dimensional space there exist only two independent vectors. Hence, the Riemannian curvature of a space is uniquely determined at each point. Its value is easily found by choosing the two vectors whose components are of type respectively. Then

(5.47)

5.8 Flat Space A space is said to be a flat if its curvature vanishes at every point of the space. Also, the necessary and sufficient condition that a space to be a flat is that the Riemann - Christoffel tensor be identically zero, as will be shown later. A familiar example of a flat space is the Euclidean plane for which the metric is rectangular Cartesian coordinates and

in

in polar coordinates.

The examples of a flat space are the Euclidean space with line - element or, a pseudo - Euclidean space - time with line - element

It may be

noted that a flat space need not have the topology of a Euclidean space. For Example, if we consider a plane, a subspace of a Euclidean space, and roll it to make a cylinder then on the surface of the cylinder we have a flat space. Note 1. The following statements are equivalent: The curvature tensor, defined by equations (5.6) and (5.27), vanishes if and only if the (i) space is flat. (ii) parallel transport of vectors is independent of path (or, the connection is integrable). (iii) Covariant derivatives commute.

5.9 Uniform Vector Field Let

be a covariant tensor of rank one, then by definition

(5.48) Let

then (5.48) gives

(5.49) From (5.49), we have

(5.50) which represents the condition for the construction of uniform vector field by parallel displacement of vector. From (5.50), we have

From this it follows that

must be integrable and hence must be perfect differential.

By usual condition of perfect differential

we get

which implies that

which in view of (5.49) gives

Changing the dummy

in the first term, we get

which in view of (5.6) gives

Since

is an arbitrary, so that

Therefore

Thus, the construction of uniform vector field is possible only when the Riemann - Christoffel tensor vanishes throughout. It is obvious that when the curvature tensor differential equations (5.50) are integrable.

vanishes,

5.10 Conditions for Flat Space - time A region of the world in which it is possible to construct a Galilean frame of reference is called flat. When

are constant it is possible to construct Galilean coordinates. Thus, when

are

constant, the space - time is flat or homaloidal. Further when are constant, the 3 - index symbols (Christo-ffel’s symbols) all vanish. But these are not tensos and therefore they will not continue to vanish when other coordinates are substituted in the same flat region. But when are constant, the Riemann - Christoffel tensor being composed of products and derivatives of 3 index symbols will vanish and since it is a tensor, it will continue to vanish when other coordinate system is substituted in the same flat region. Thus the vanishing of Riemann Christoffel tensor is a necessary condition for flat space- time. This is also sufficient as proved below. When

it is possible to construct a uniform vector field by parallel displacement

of a vector all over the region. Let that

is not a tensor suffix). Then

be four uniform vector fields given by

(note

gives

(5.51)

Let

be another coordinate system related to the old by the equation

(5.52) Since

is an invariant, therefore

which in view of (5.52) gives (5.53) Differentiating (5.53) with respect to

we get

which in view of (5.51) gives

By changing dummy suffixes gives

which on integrating gives

Hence the space is a flat and the vanishing of Riemann - Christoffel tensor is necessary and sufficient condition for flat space - time.

Theorem 5.2 Prove that the necessary and sufficient condition for a space to be a flat is that its curvature vanishes.

Proof The Riemannian curvature or curvature associated with the vectors

of the space is

(i) Let

for all vectors

It follows from equation (i)

(ii) Also, (iii)

(iv) and (v) Since

are arbitrary vectors, addition of equations (ii), (iii), (iv) and (v), yields

which in view of

But

gives

Therefore

which in view of

gives

(vi) Changing

cyclically in (vi), we have

(vii) From (vi) and (vii), we have

(viii)

Since (ix) From (viii) and (ix), we have

But

therefore

which means that the space is a flat. Conversely, let the space be a flat, that is,

then from equation (i), we have

which completes the required result.

Example 5.1 Show that the 2 - dimensional space

is a flat space.

Solution The given space is

Let us take

then clearly,

and the rest all are zero, that is

for

Therefore

Now,

and the rest all are zero, that is,

The non - vanishing components of Christoffel’s symbol of second kind are

The only component of Riemann tensor in 2 - dimensions is

which shows that the 2 - dimensional space

is a flat space.

Example 5.2 Show that

is the metric of a flat space.

Solution The given metric is

Let us take that is

then clearly,

and the rest all are zero,

Therefore

Now,

and the rest all are zero, that is, The only non - vanishing derivative is Christoffel’s symbol of second kind are

while the non - vanishing components of

Also, By the definition of Riemannian tensor

and the space is a flat.

Example 5.3 Show that the space with the metric

is a flat space.

Solution The given metric is

Let us take

then clearly, and the

rest all are zero, that is,

Now,

Therefore

and

The only non - vanishing Christoffel’s symbols are

which is constant. Therefore

and hence the space is a flat.

5.11 Space of Constant Curvature The Riemannian curvature of the space is given by (5.46) associated with the vectors Let us now investigate spaces in which the Riemannian curvature at every point does not depend on the choice of the associated vectors

Thus from (5.46), we have for all vectors of

which implies that

(5.54) where

is now a function of the coordinates

Taking covariant derivative (5.54) with respect to

and using the fact

we have

(5.55) Similarly,

(5.56)

(5.57) Adding (5.55), (5.56) and (5.57) and using the Bianchi’s identity, we have

(5.58) Taking inner multiplication by

Hence, if

then

of the relation (5.58) and noting that

which yields

theorem; ‘if at each point of a space

we have

as a constant. Thus, we have proved Schur’s the Riemannian curvature is a function of the

coordinates only, then it is constant throughout the curvature. Thus, for a space of a constant curvature

Such a

is called a space of constant

A space which is homogeneous with respect to the Ricci tensor

is defined as Einstein space.

where

is a function of the coordinates

5.12 Einstein space

That is, to say, if at every point of a space, then that space is called Einstein space (introduced and studied by Albert Einstein). In other words, if the Ricci tensor satisfies

(5.59) where

is some scalar (or invariant), then it is called an Einstein space.

Taking inner multiplication of (5.59) with

we have

(5.60) where

and

are constant. Consequently, from (5.59) and (5.60), we have

(5.61)

Albert Einstein (March 14, 1879 - Apri 18, 1955)

Hence, a space is an Einstein space if at every point of a space. Also, a necessary and sufficient condition for a space to be an Einstein space is that

at every point of the space.

Example 5.4 For the metric

obtain the values of

and

Solution The given metric is

Now,

Therefore, (i) Also,

which leads to

(ii) Similarly,

(iii) From (ii) and (iii), we have

(iv) Since

which leads to

(v) From equations (iv) and (v), we have

Example 5.5 Show that a space of constant curvature is an Einstein space.

Solution For a space of constant curvature, we have

which on multiplying on both sides by

gives

which in view of

reduces to

(i) which on multiplying on both sides by

which in view of

gives

reduces to

(ii) From equations (i) and (ii), we have

which shows that a space of constant curvature is an Einstein space.

Example 5.6 Prove that a space

satisfying

is an Einstein space, where is a constant.

is constant. Also, show that for Einstein spaces, the scalar curvature

Solution Given that

which on contraction with respect to the indices ℓ and k gives

(i)

where

is a scalar. Thus the given space is an Einstein space.

Now, multiplying (i) by

, we get

which shows that the scalar curvature is a constant.

Example 5.7 Show that a space

of constant curvature is an Einstein space. Hence deduce that

Solution The Riemannian curvature unit vectors

Since

at any point P of the space

determined by the orientation of

is given by

is constant and therefore it is independent of the orientation. Hence

which on multiplying on both sides by

gives

which in view of

reduces to

(i) which on multiplying on both sides by

gives

which in view of

reduces to

(ii) From (i) and (ii), we have

which shows that the space result.

is an Einstein space. The equation (ii) proves the second required

Example 5.8 Prove that if then curvature of an Einstein space is constant.

and hence deduce that when

then the scalar

Solution The Bianchi’s identity is

which in view of

gives

which on contraction with respect to a and k gives

which in view of

which in view of

which in view of

reduces to

reduces to

gives

which on changing the dummy indices ℓ into a gives

(i) But (ii)

From (i) and (ii), we get

(iii) This completes the first part of the problem. Again, let the space

be an Einstein space, so that

which on multiplying on both sides by

gives

which in view of

gives

which in view of

gives

Using (iii) and

we have

which in view of (ii) gives

For

which shows that

the above relation gives

is constant (absolute constant), which completes the required result.

Example 5.9 Show that every space

is an Einstein space.

Solution For a 2 - dimensional space

(i) Also,

Thus, (ii)

Also, (iii)

(iv) Again,

That is, (v)

Now,

. (vi) Similarly,

(vii) and

(viii) From the relations (vi), (vii) and (viii), we have

which shows that the space

is an Einstein space.

Example 5.10 For a 2 - dimensional manifold, prove that

Solution Proceed upto the relation (v) of example 9.9, that is

(v) Thus,

Since,

is only non - vanishing component of

which completes the required result.

Exercises 1.

Prove that

where

is an arbitrary covariant tensor of rank one and deduce that

is a tensor.

2.

Obtain an expression for Riemann - Christoffel tensor of the second kind.

3.

Show that the Riemann’s symbol of the second kind the indices j and k and also show that

is skew - symmetric relative to

where the symbols have their usual meaning. 4.

Prove that the curvature tensor of second kind can be contracted in two different ways: one of these leads to a zero tensor and the other to a symmetric tensor.

5.

Define Ricci tensor and show that it is a symmetric.

6.

Define covariant curvature tensor and also find the expression for it.

7.

Show that the curvature tensor symmetric in h, i and j, k.

8.

Show that

is symmetric in pairs of indices and skew -

where the symbols have their usual meaning. 9.

Show that the curvature tensor differential property.

10. Show that

is satisfied its cyclic property as well as its

where the symbols have their usual meaning. 11. Show that the divergence of word tensor (that is, Einstein tensor) vanishes, that is, 12. Define an Einstein tensor of an Einstein tensor

Also, write its covariant form and show that the divergence is zero, that is,

13. Show that the curvature tensor dimensional space.

independent components in n –

has

14. Show that the number of independent components of curvature tensor does not exceed

15. State and prove Bianchi’s first and second identities for Riemannian - Christoffel tensor 16. Prove that

Hence or otherwise prove that

where the symbols have their usual meaning. 17. If

then show that

where the symbols have their usual meaning. 18. Show that

where the symbols have their usual meaning. 19. Show that the curvature tensor of a 4 - dimensional Riemannian space has at most 20 distinct non - vanishing components. 20. Define a flat space and show that the necessary and sufficient condition for a space to be a flat is that its curvature vanishes. 21. Show that the 2 - dimensional space 22. Show that

is a flat space.

is the metric of a flat space.

23. Show that the space with the metric

is a flat space.

24. Define Einstein space and show that a space of constant curvature is an Einstein space. 25. Prove that a space

is an Einstein space, where curvature is constant. 26. Show that every space

satisfying

is constant. Also, show that for Einstein spaces, the scalar is an Einstein space.

27. Define a space of constant curvature and Show that a space

of constant curvature is an

Einstein space. Hence deduce that 28. Prove that if then and hence deduce that when curvature of an Einstein space is constant. 29. Show that the space of constant curvature is an Einstein space. 30. Show that for a 2 - dimensional space

where the symbols have their usual meaning.

then the scalar

6 Application of Tensor Calculus

T ensor calculas have their many applications in most all the disciplines of science and engineer-ing. We begin this chapter by introducing the basic history of tensor calculus and in sequence, we study about the applications of tensor calculus in differential geometry, Riemannian geomet-ry, theory of relativity, elasticity, physics and other fields like economics, probability and engineering etc. which are mainly useful for researchers.

6.1 Introduction A vector quantity have only one direction and magnitude but a quantity having more than one direction gives the notion of tensors. It means that the notion of tensors are as the extension of vectors. The scalar and vector quantities do not cover completely many physical and geometrical quantities. For instance, stress in an elastic body and curl of a vector supposedly misunderstood as vectors are much more than vectors. Indeed, they are tensors. The great Mathematician Gregorio Ricci was first introduced and studied about the tensors, so G. Ricci is known as a founder of tensor calculus. The emergence of tensor calculus, otherwise known as the absolute differential calculus, as a systematic branch of mathematics is due to Ricci and his pupil Levi - Civita. In collaboration they published the first memoir on this subject: ‘Methodes de calcul differential absolu et leurs applications, Mathematische Annalen, vol. 54, (1901). The concept of a tensor has its origin in the develop-ments of differential geometry by Gauss, Riemann and Christoffel. The investigation of relations which remain valid when we change from one coordinate system to any other, is the chief aim of tensor calculus. It drew the attention of mathematician and physicist for its application, after the publication of a memoir by Ricci and Levi - Civita in 1901. Since then the tensor calculus has now become one of the essential techniques of modern theoretical physics. Coordinate transformation rules are framed to classify different geometric quantities such as scalars, vectors (contravariant and covariant) and tensors of different orders. This list of quantities is inexhaustive. There are quantities, which are neither of the above: Christoffel symbols are a few of them. The axiomatic approach to scalars (as elements of an algebraic field), vectors (as elements of a vector space) and tensors (as elements of vector spaces which are tensor products of two or more vector spaces) is the latest one.

6.2 Application of Tensor Calculus In the course of study, we find that the nature of a tensor quantity does not change by the change of coordinate system. Due to this property, it is very useful for describing the physical laws mathematically and so it is of great use in general Relativity theory, Differential geometry, Riemannian geometry, Mechanics, Elasticity, hydrodynamics, electromagnetic theory and many other disciplines of Science and engineering. The laws of physics can not depend on the frame of reference which the physicist chooses for the purpose of description. Accordingly it is aesthetically desirable and often convenient to utilise the tensor calculus as the mathematical background in which such laws can be formulated. In particular, Einstein found it an excellent tool for the presentation of general relativity theory. By use of tensor calculus we can observe that the space is flat or not. For example; if the Riemannian curvature is zero, then the space is flat. The application of tensor calculus are discussed here by taking one by one the following topics of application areas:

6.2.1 Tensor Calculus and Differential Geometry Differential geometry is that branch of Mathematics which is treated with the help of differential calculus and its study involves curve in spaces and surfaces. Monge (1746 - 1818) together with Gauss (1777 - 1855) are considered as the founder of differential geometry.

The Christoffel symbols have already studied in chapter of this book. The relations in between Christoffel symbols and fundamental coefficients are given as follows: The Christoffel’s symbol of first kind is denoted by

and defined by

, (6.1) where

denote differentiation with respect to

respectively.

From (6.1), we have

(6.2) For particular values of

we have the following

The Christoffel’s symbol of the second kind is denoted by

and defined by

(6.3) For particular values of

we have the following

From the above, we see that the Christoffel symbols exclusively depend upon the coefficients of the first fundamental form and their first parial derivatives. Also, In the theory of surfaces, the Gauss formulae (Gauss equations) are given by

(6.4) where the symbols have their usual meaning. The Gauss equations (6.4) can also be written in terms of tensor notation as

(6.5)

Also, the Mainardi - Codazzi equations

(6.6) can also be written in terms of Christoffel symbols of second kind as

(6.7) In tensor notation, the components of the geodesic curvature vector are

(6.8) respectively. The geodesic equarions

can be written in terms of the Christoffel’s symbol of first kind and second kind (that is, in tensor notations) as follows: (i) The geodesic equarions in terms of the Christoffel’s symbol of first kind

(6.9) (ii) The geodesic equarions in terms of the Christoffel’s symbol of second kind

(6.10) The Gaussian curvature in terms of tensor notation is given by

(6.11) From the above we see that the tensor calculus are used so much in Differential geometry. Thus the relations in between tensor calculus and Differential geometry are much wide and by use of

tensor calculus, the Differential geometry becomes so simple and interesting for readers.

6.2.2 Tensor Calculus and Riemannian Geometry Riemannian geometry is that branch of geometry which is based upon Riemannian metric. The German Mathematician Bernhard Riemann is known as a founder of Riemannian geometry because he was the first person who introduced and studied in detail about Riemannian geometry. The quadratic differential form

Bernhard Riemann (September 17, 1826 – July 20, 1866)

which expresses the distance (ds) between two neighbouring points with coordinates is called a Riemannian metric. is called a metric tensor and the space characterized by a Riemannian metric is called a Riemannian space. We can write

The contribution of

to

is zero, hence there is no loss of generality in

assuming that is symmetric. Thus is a covariant symmetric tensor of second order called the fundamental tensor of the Riemannian space. The quadratic differential form Since

is an invariant which can be prove as follows:

is a covariant tensor of rank two, therefore by tensor law of transformation

which in view of

gives

which on changing dummy

showing that The angle

between

gives

is an invariant. is given by

(6.12)

If the coordinate curves of the parameter

are orthogonal, then

which implies that (6.13) From (6.12) and (6.13), we have

which shows that are orthogonal if In a Riemannian geometry, a Riemannian connection on a Riemannian manifold M is a connection

satisfying the following two conditions:

(i) that is, the connection

is symmetric, or torsion - free.

that is, the connection is a metric compatible or, metric connection or, the covariant derivative of the metric tensor g is zero. The relations (i) and (ii) can be written in tensor form or, in local coordinate system as

By use of the local coordinate system, we can prove easily the fundamental theorem of Riemannian geometry “there exists a unique Riemannian connection on a Riemannian manifold” and other theorems of Riemannian geometry. In tensor notation, we write in place of the fundamental tensor of type (0,2). In a Riemannian geometry, the Riemannian curvature tensor K(X, Y, Z) of type (1,3) is defined as

(6.14) where

is a Lie bracket and

tensor notation),

is a Riemannian connection. In local coordinate system (or, in

is denoted by

and is defined as

(6.15) The relations (6.14) is in free indices form and (6.15) is in local coordinate system. The Bianchi’s first and second identities for the curvature tensor of type (1,3) are given in free indices form

whereas in tensor form or in local coordinate system

respectively. The covariant curvature tensor

of type (0,4) is defined in free indices form

(6.16) whereas in tensor form or in local coordinate system

(6.17) respectively. The Bianchi’s first and second identities for the curvature tensor of type (0,4) are given in free indices form

whereas in tensor form or in local coordinate system

respectively. The contraction of the curvature tensor

of type (1,3) with respect to X is denoted by

and defined by

(6.18) whereas in tensor form, the contraction of

with respect to the indices h and k, gives the

second rank tensor called the Ricci tensor and is denoted by (6.19)

Also,

We can easily show that the Ricci tensor is symmetric tensor. In a Riemannian geometry, a Riemannian manifold (M, g) of dimension n is said to be an Einstein manifold if its Ricci tensor of type (0,2) is of the form (6.20) where

is some constant.

In local coordinate system, if Ricci tensor

satisfies

(6.21) where is some scalar (or invariant), then it is called an Einstein space. From the above we see that the tensor calculus are used so much in Riemannian geometry. Thus the relations in between tensor calculus and Riemannian geometry are much wide and by use of tensor calculus, the Riemannian geometry becomes so simple and interesting for readers.

6.2.3 Tensor Calculus and Theory of Relativity The general theory of Relativity which was developed by Einstein in order to disuss gravitation. He postulated the principle of covariance, which asserts that the laws of physics must be independent of the space - time coordinates. This swept away the privileged role of the Lorentz transformation. As a result of Minkowski space was replaced by a 4 - dimensional Riemannian space

with the general metric

(6.22) Einstein also introduced the principle of equivalence, which in essence states that the fundamental tensor

can be chosen to account for the presence of a gravitational field. That is,

depends on the distribution of matter and energy in physical space. Matter and energy can be specified by the energy - momentum tensor theory satisfies the equation

which in the special

The only forces, namely those due to gravitation, are

however already taken into account by the choice of the fundamental tensor ignore

We therefore

and, in accordance with the principle of covariance, the energy - momentum tensor

must now satisfy the equation

We shall write this equation in the equivalent form

where

is the mixed energy - momentum tensor. The problem now is to determine

as a function of the fundamental tensor

and their derivatives up to the second order, bearing

in mind that The Einstein tensor plays a very fundamental role in general theory of relativity. The Einstein tensor is already studied in chapter five of section 5.6. The Einstein tensor is defined as

(6.23) satisfies the equation

The equations of motion require

but very remarkably

is an identity in Riemannian geometry. This led Einstein to propose the relation

(6.24) In effect these equations form the link between the physical energy - momentum tensor

and

the geometrical tensor of the of general relativity. In order that Newton’s theory of gravitation can be deduced as a first approximation from Einstein theory, it was found necessary to choose

where k is the usual gravitational constant

The value

of c is and in c. g. s. units. The motion of a particle which moves under the action of some force system can be represented in Minkowski space of curve, called the world - line of the particle. In special theory, the world - lines of free particles and of light rays are respectively the geodesics and the null geodesics of Minkowski space. The principle of equivalence demands that all particles be regarded as free particles when gravitation is the only force under consideration. Then it follows from the principle of covariance that the world - line of a particle under the action of gravitational forces is a geodesic of the with the metric (6.21). If no forces act on the particle, then the world - line of a free particle is a geodesic of the Minkowski space. Similarly, the world - line of a light ray is a null - geodesic. The velocity of a light ray is the constant c, and so we see from the relation that for such a ray Accordingly the world - line of a light ray is a null geodesic of the Minkowski space. Also, the study of planetary motion, Einstein’s field equations, poisson’s equations as an approximation of Einstein’s field equations, Einstein’s universe and De Sitter’s universe etc.

become so simple and interesting by use of tensor calculus. Thus, tensor calculus provide the tool for description of the general theory of Relativity.

6.2.4 Tensor Calculus and Elasticity The forces acting on a body are either external or internal. The external forces may consist either of body forces such as gravity which act on every particle of it, or of surface forces which act on the external surface of the body, for example the pressure between two bodies in contact. When an elastic body is subjected to an external force or stress, it becomes deformed or strained. In the formulation of the equations governing the equilibrium of an elastic solid the Cartesian tensor of second rank are used. A Cartesian tensor is defined as follows: A Cartesian tensor of the order in a 3 - dimensional Euclidean space is defined as a set of quantities which transform according to equation (2.8) when the coordinates undergo a positive orthogonal transformation. This is a less straingent condition than that imposed on a tensor. So we see that all tensors are Cartesian tensor but a Cartesian tensor is not necessarily a tensor in the usual sence. A tensor transformed components satisfy

is a Cartesian tensor of the

order if the

(6.25) on change of the coordinates by the positive orthogonal transformation

We see from

that both are Cartesian vectors. Also, the Koneker delta is a Cartesian tensor of the second order because

in virtue of order.

Similarly, a permutation tensor

is a Cartesian tensor of the third

The fundamental tensor of the Euclidean space is the Koneker delta Hence all the Christoffel symbols are zero and so the comma notation for covariant derivatives now denotes the familiar partial derivatives which are Cartesian tensors. The study of elasticity in terms of tensors falls into three parts: a description of the strain or deformation of the elastic substance, a description of the force or stress which produces the deformation, and a generalization of Hook’s law in terms of tensor form. Deformation of a body may be described by giving the change in the relative position of the parts of the body when the body is subjected to some external force. As the result of this force the configuration of the body changes and we say that the body is in a strained state.

Let us consider a point P of the body, at the position O. Let there is another point Q, at the position

relative to some fixed origin or,

In the

unstrained state the coordinates of Q relative to P are Let each point of the body is subjected to an external force which causes displacement u(r) varying from point to point. Thus the point P is shifted to P1 with the displacement u(r) and point Q is shifted to Q1 with the displacement The coordinates of Q1 relative to P1 are

(see in figure 6.1). Then by

Taylor’s expansion, we have

Neglecting the second and higher order differentials, the above equation reduces to

which shows that the displacement of the point Q consists of two parts, one of which, that is, u(r) is same for all the points of the body and therefore, corresponds to a translation of the body as a whole. Relative displacement Q1 and P1 (that is, deformation) is

(6.26)

The quantity

being the gradient of a vector, is a dyadic, that is, a tensor of second rank.

Now,

(6.27)

(6.28) are symmetric and ant - symmetric respectively. The term

of the equation (6.26) is due to the rigid body rotation of the element. Therefore

the quantity is known as rotation tensor. The remaining symmetric part is taken as a pure strain known as strain tensor. The stress is defined as the internal force per unit area acting on a deformed body. The force in the direction acting on the face whose normal in the direction is where are actually pressures in the sence of force / area. Whenever the term force is used, it is understood that the are to be multiplied by the appropriate area. There are the forces acting on the small parallelopiped see in figure 10.2. The stress may be directed normally or tangentially to the surfaces on which they act. In case deforming forces act normally to a given area of an elastic medium, they produce pure alongation and the stresses are called the normal stresses. If the deforming forces act tangentially to the surface they produce shearing stresses. Then obviously

are the normal stresses for total force F act normal to the face area

where the component

of the

and so on. The tangential stresses

are the shearing stresses. Thus the total stress may be completely specified by the array matrix

(6.29) where the diagonal elements are normal stresses and the non- diagonal elements are shearing stresses. Let us assume that the stresses are homogeneous. Then the forces on the opposite faces will be

reversed in sign as shown in figure 6.3.

We have (6.30) and hence the array of stresses, given equation (6.29) is symmetric. The total force along

is

(6.31) for static equilibrium. Thus in summation convention, we have

(6.32) which is the law of transformation of a tensor of second rank as defined by equation Therefore, the array (6.29), with the condition (6.30) represents a symmetric tensor of second rank. This tensor is defined as stress tensor. In the elementary theory of elasticity, Hooke’s law states that the tension of an elastic string is proportional to the extension. In other words, stress is proportional to strain. The corresponding assumption in the general theory of elasticity is that the stress tensor is a linear homogeneous function of the strain tensor. That is

It follows from the quotient that

is a Cartesian tensor of the fourth order, and it is called the

elasticity tensor. Further from the symmetry of

we find that

is symmetric not

only with respect to the indices i and j but also with respect to A body is said to be homogeneous if the elastic properties of the body are independent of the point under consideration. This means that the components of the elasticity tensor are all constants for a homogeneous body. We call a body isotropic if the elastic properties at a point are the same in all directions at that point. This means that the elasticity tensor

transforms to

itself under any rotation of axes. A Cartesian tensor which transforms into itself under a rotation of axes is called an isotropic tensor.

6.2.5 Tensor Calculus and Physics

Tensor analysis have developed as the natural generalization of vector analysis, and therefore, it may be expected that tensor analysis has relatively vider field of applications. As pointed out earlier, the general theory of relativity was developed by using metric tensor, Christoffel symbols, curvature tensor, Ricci tensor, and Einstein tensor. By use of electromagnetic field tensor, inertia tensor, stress tensor, fundamental tensors etc., the readers feel interest in the study of physics and its relative field like dynamics and electrodynamics. The classical theory of electrodynamics, according to Lorentz, is specified by the electric potential φ which is a scalar and the magnetic potential magnetic field strength vector

which is a vector. The electric field strength vector

and the

are derived from these potentials by the equations

Using electrostatic units, Maxwell’s equations are

(6.33) where

is the current density vector and

is the charge density.

We introduce the skew - symmetrical tensor

defined by

and we immediately calculate that its non - vanishing components in the given coordinate system are

The non - vanishing contravariant components

may now be obtained and which are

We now write Maxwell’s equations (6.33) in terms of be respectively

and the results are readily verified to

(6.34) The first and last equations of the relation (6.34) combine together into the form

(6.35) whilst the remaining two of the relation (6.34) are accounted for by the equations of the set

(6.36) which do not vanish identically. We have accordingly written Maxwell’s equations in tensor form in Minkowski space. Thus they are invariant under the Lorentz group of transformations. Let the position of a moving particle P be determine by a vector r. If the curvilinear coordinates of the terminal point of r are denoted by particle can be written in the form

then the equations of the path C of the

(6.37) and we call the curve C, the trajectory of the particle. The velocity of a particle P is a vector

whose components are

(6.38) as the velocity of the particle. In transformed coordinates the components of the velocity are

using (6.38)

which follows the tensor law of transformation and hence the velocity of a particle at any point is a contravariant tensor (or, a contravariant vector) of rank one. The acceleration of a particle P is a vector and in the general case, velocity is the function of of time t. Let us write its intrinsic derivative with respect to t according to the equation

Then we have

(6.39) where

is the intrinsic derivative and

are the Christoffel symbols calculated from metric

tensor

In the cartesian rectangular coordinates the equation (6.38) reduces to

which is obviously the acceleration of the particle. Therefore, the quantity given by the equation (6.39) is called acceleration vector in curvilinear coordinates system. If m is the mass of the particle P, then the force vector (contravariant) in curvilinear coordinate is defined as

(6.40) The covariant vector associated with

may be readily be written from equation (2.24), that is,

(6.41) and (6.41) The quantity

is scalar accordingly to quotient rule. This quantity is defined as work

done when the point of application of covariant force is moved through a small displacement In cartesian coordinate system there is no distinction between covariant and contravariant components and hence From the above we see that the tensor calculus are used so much in physics. Thus the relations in between tensor calculus and physics are much wide and by use tensor calculus, the physics becomes so simple and interesting for readers.

6.2.6 Tensor Calculus and Other Fields like Economics, Probability and

Engineering Since the relations in between tensor calculus and Differential geometry are much wide and by use tensor calculus, the Differential geometry becomes so simple and interesting for readers. In economics, differential geometry has applications to the field of econometrics and in engineering, differential geometry can be applied to solve problems in digital signal processing. Geometric modeling (including computer graphics) and computer - aided geometric design draw on ideas from differential geometry. In probability, statistics, and information theory, one can interpret various structures as Riemannian manifolds, which yields the field of information geometry, particularly via the Fisher information metric. In structural geology, differential geometry is used to analyze and describe geologic structures. Also, in computer vision, differential geometry is used to analyze shapes. In last we can say that due to the use of tensor analysis readers will have felt interest in the sudy of theory of relativity, differential geometry, Riemannian geometry, mechanics, elasticity, hydrodynamics, electromagnetic theory, physics and many other disciplines of science and engineering.

Exercises 31. What do you mean by differential geometry and Riemannian geometry? 32. Write the Bianchi’s first and second identities in free indices form and in local coordinate system (or, in tensor notation). 33. Define an Einstein manifold and an Einstein spaces. 34. Write the Mainardi - Codazzi equations in free indices form and in tensor notation from. 35. Write the Gaussian curvature in terms of tensor notation. 36.

Prove that the quadratic differential form

37. Shows that

is an invariant. are orthogonal if

38. Define an Einstein tensor. 39. Define Cartesian tensor and show that the Koneker delta is a Cartesian tensor of the second order. 40. Define rotation tensor, stress tensor and strain tensor. 41. Define an elasticity tensor and an isotropic tensor. 42. Show that the velocity of a particle at any point is a contravariant tensor of rank one.

43. Write in short the application of tensor analysis.

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