tensors

This page intentionally left blank

Copyright © 2006, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to [email protected]

ISBN (13) : 978-81-224-2700-4

PUBLISHING FOR ONE WORLD

NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at www.newagepublishers.com

To My parents


FOREWORD

It gives me great pleasure to write the foreword to Dr. Nazrul Islam’s book entitled “Tensors and Their Applications. I know the author as a research scholar who has worked with me for several years. This book is a humble step of efforts made by him to prove him to be a dedicated and striving teacher who has worked relentlessly in this field. This book fills the gap as methodology has been explained in a simple manner to enable students to understand easily. This book will prove to be a complete book for the students in this field.

Ram Nivas Professor, Department of Mathematics and Astronomy, Lucknow University, Lucknow


PREFACE

‘Tensors’ were introduced by Professor Gregorio Ricci of University of Padua (Italy) in 1887 primarily as extension of vectors. A quantity having magnitude only is called Scalar and a quantity with magnitude and direction both, called Vector. But certain quantities are associated with two or more directions, such a quantity is called Tensor. The stress at a point of an elastic solid is an example of a Tensor which depends on two directions one normal to the area and other that of the force on it. Tensors have their applications to Riemannian Geometry, Mechanics, Elasticity, Theory of Relativity, Electromagnetic Theory and many other disciplines of Science and Engineering. This book has been presented in such a clear and easy way that the students will have no difficulty in understanding it. The definitions, proofs of theorems, notes have been given in details. The subject is taught at graduate/postgraduate level in almost all universities. In the end, I wish to thank the publisher and the printer for their full co-operation in bringing out the book in the present nice form. Suggestions for further improvement of the book will be gratefully acknowledged.

Dr. Nazrul Islam


CONTENTS

Foreword ............................................................................................................. vii Preface ................................................................................................................ ix

Chapter–1 Preliminaries ........................................................................................... 1-5 1.1. 1.2. 1.3. 1.4. 1.5. 1.6.

n-dimensional Space ............................................................................................... 1 Superscript and Subscript ....................................................................................... 1 The Einstein's Summation Convention ...................................................................... 1 Dummy Index ....................................................................................................... 1 Free Index ............................................................................................................. 2 Krönecker Delta ..................................................................................................... 2 Exercises ............................................................................................................... 5

Chapter–2 Tensor Algebra ..................................................................................... 6-30 2.1. 2.2. 2.3. 2.4. 2.5. 2.6. 2.7. 2.8. 2.9. 2.10. 2.11. 2.12. 2.13. 2.14. 2.15.

Introduction .......................................................................................................... 6 Transformation of Coordinates ................................................................................ 6 Covariant and Contravariant Vectors ......................................................................... 7 Contravariant Tensor of Rank Two .......................................................................... 9 Covariant Tensor of Rank Two ................................................................................ 9 Mixed Tensor of Rank Two..................................................................................... 9 Tensor of Higher Order ......................................................................................... 14 Scalar or Invariant ................................................................................................ 15 Addition and Subtraction of Tensors ....................................................................... 15 Multiplication of Tensors (Outer Product of Tensors) ............................................... 16 Contraction of a Tensor ........................................................................................ 18 Inner Product of Two Tensors .............................................................................. 18 Symmetric Tensors .............................................................................................. 20 Skew-symmetric Tensor ....................................................................................... 20 Quotient Law ....................................................................................................... 24

xii

Tensors and Their Applications

2.16. 2.17.

Conjugate (or Reciprocal) Symmetric Tensor .......................................................... 25 Relative Tensor .................................................................................................... 26 Examples ............................................................................................................ 26 Exercises ............................................................................................................. 29

Chapter–3 Metric Tensor and Riemannian Metric ............................................ 31-54 3.1. 3.2. 3.3. 3.4. 3.5. 3.6. 3.7. 3.8. 3.9. 3.10. 3.11. 3.12. 3.13.

The Metric Tensor ............................................................................................... 31 Conjugate Metric Tensor (Contravariant Tensor) ...................................................... 34 Length of a Curve ................................................................................................ 42 Associated Tensor ................................................................................................ 43 Magnitude of Vector ............................................................................................. 43 Scalar Product of Two Vectors .............................................................................. 44 Angle Between Two Vectors .................................................................................. 45 Angle Between Two Coordinate Curves .................................................................. 47 Hypersurface ....................................................................................................... 48 Angle Between Two Coordinate Hyper surface ........................................................ 48 n-Ply Orthogonal System of Hypersurfaces ............................................................. 49 Congruence of Curves .......................................................................................... 49 Orthogonal Ennuple .............................................................................................. 49 Examples ............................................................................................................ 52 Exercises ............................................................................................................. 54

Chapter–4 4.1. 4.2. 4.3. 4.4. 4.5. 4.6. 4.7. 4.8.

Christoffel’s Symbols and Covariant Differentiation ................................ 55-84

Christoffel’s Symbol............................................................................................. 55 Transformtion of Christoffel’s Symbols .................................................................. 64 Covariant Differentiation of a Covariant Vector ........................................................ 67 Covariant Differentiation of a Contravariant Vector ................................................... 68 Covariant Differentiation of Tensors ....................................................................... 69 Ricci’s Theorem .................................................................................................. 71 Gradient, Divergence and Curl ............................................................................... 75 The Laplacian Operator ......................................................................................... 80 Exercises ............................................................................................................. 83

Chapter–5 Riemann-Christoffel Tensor ............................................................ 85-110 5.1. 5.2. 5.3.

Riemann-Christoffel Tensor ................................................................................... 85 Ricci Tensor ........................................................................................................ 88 Covariant Riemann-Christoffel Tensor .................................................................... 89

5.4. 5.5. 5.6. 5.7.

Properties of Riemann-Christoffel Tensors of First Kind Ri j k l ................................... 91 Bianchi Identity .................................................................................................... 94 Einstein Tensor .................................................................................................... 95 Riemannian Curvature of V n .................................................................................. 96

Contents

5.8. 5.9. 5.10. 5.11. 5.12. 5.13.

xiii

Formula For Riemannian Curvature in Terms of Covariant Curvature Tensor of V n ......................................................................................... 98 Schur’s Theorem ............................................................................................... 100 Mean Curvature ................................................................................................. 101 Ricci Principal Directions .................................................................................... 102 Einstein Space ................................................................................................... 103 Weyl Tensor or Projective Curvature Tensor ......................................................... 104 Examples .......................................................................................................... 106 Exercises ........................................................................................................... 109

Chapter–6 The e-systems and the Generalized Krönecker Deltas ................ 111-115 6.1. 6.2. 6.3. 6.4.

Completely Symmetric ......................................................................................... 111 Completely Skew-symmetric ................................................................................ 111 e-system ........................................................................................................... 112 Generalized Krönecker Delta ................................................................................ 112

6.5.

Contraction of ä iá jkâ ã ............................................................................................ 114 Exercises ........................................................................................................... 115

Chapter–7 Geometry ........................................................................................ 116-141 7.1.

Length of Arc .................................................................................................... 116

7.2.

Curvilinear Coordinates in E 3 .............................................................................. 120

7.3. 7.4. 7.5. 7.6. 7.7. 7.8. 7.9.

Reciprocal Base System Covariant and Contravariant Vectors .................................. 122 On The Meaning of Covariant Derivatives ............................................................. 127 Intrinsic Differentiation ....................................................................................... 131 Parallel Vector Fields ........................................................................................... 134 Geometry of Space Curves ................................................................................. 134 Serret-Frenet Formulae ....................................................................................... 138 Equations of A Straight Line ................................................................................ 140 Exercises ........................................................................................................... 141

Chapter–8 Analytical Mechanics ..................................................................... 142-169 8.1. 8.2. 8.3. 8.4. 8.5. 8.6. 8.7. 8.8. 8.9.

Introduction ...................................................................................................... 142 Newtonian Laws ................................................................................................ 142 Equations of Motion of Particle ............................................................................ 143 Conservative Force Field ..................................................................................... 144 Lagrangean Equation of Motion ........................................................................... 146 Applications of Lagrangean Equations ................................................................... 152 Hamilton’s Principle ............................................................................................ 153 Integral Energy .................................................................................................. 155 Principle of Least Action ..................................................................................... 156

xiv


8.10. 8.11. 8.12. 8.13. 8.14. 8.15.

Generalized Coordinates ...................................................................................... 157 Lagrangean Equation of Generalized Coordinates ................................................... 158 Divergence Theorem, Green’s Theorem, Laplacian Operator and Stoke’s Theorem in Tensor Notation ................................................................................ 161 Gauss’s Theorem ............................................................................................... 164 Poisson’s Equation ............................................................................................. 166 Solution of Poisson’s Equation ............................................................................. 167 Exercises ........................................................................................................... 169

Chapter–9 Curvature of a Curve, Geodesic .................................................... 170-187 9.1. 9.2. 9.3. 9.4. 9.5. 9.6. 9.7. 9.8.

Curvature of Curve, Principal Normal................................................................... 170 Geodesics ......................................................................................................... 171 Euler’s Condition ............................................................................................... 171 Differential Equations of Geodesics ...................................................................... 173 Geodesic Coordinates ......................................................................................... 175 Riemannian Coordinates ...................................................................................... 177 Geodesic Form of a Line Element ........................................................................ 178 Geodesics in Euclidean Space .............................................................................. 181 Examples .......................................................................................................... 182 Exercises ........................................................................................................... 186

Chapter–10 Parallelism of Vectors ................................................................. 188-204 10.1. 10.2. 10.3. 10.4. 10.5.

Parallelism of a Vector of Constant Magnitude (Levi-Civita’s Concept) ..................... 188 Parallelism of a Vector of Variable Magnitude ......................................................... 191 Subspace of Riemannian Manifold ........................................................................ 193 Parallelism in a Subspace .................................................................................... 196 Fundamental Theorem of Riemannian Geometry Statement ..................................... 199 Examples .......................................................................................................... 200 Exercises ........................................................................................................... 203

Chapter–11 11.1. 11.2. 11.3. 11.4. 11.5. 11.6. 11.7.

Ricci’s Coefficients of Rotation and Congruence ....................... 205-217

Ricci’s Coefficient of Rotation ............................................................................. 205 Reason for the Name “Coefficients of Rotation” .................................................... 206 Curvature of Congruence .................................................................................... 207 Geodesic Congruence ......................................................................................... 208 Normal Congruence ........................................................................................... 209 Curl of Congruence ............................................................................................ 211 Canonical Congruence ........................................................................................ 213 Examples .......................................................................................................... 215 Exercises ........................................................................................................... 217

Contents

xv

Chapter–12 Hypersurfaces .............................................................................. 218-242 12.1. 12.2. 12.3. 12.4. 12.5. 12.6. 12.7. 12.8. 12.9. 12.10. 12.11. 12.12. 12.13. 12.14. 12.15.

Introduction ...................................................................................................... 218 Generalized Covariant Differentiation .................................................................... 219 Laws of Tensor Differentiation ............................................................................ 220 Gauss’s Formula ................................................................................................ 222 Curvature of a Curve in a Hypersurface and Normal Curvature, Meunier’s Theorem, Dupin’s Theorem ............................................................................................... 224 Definitions ......................................................................................................... 227 Euler’s Theorem ................................................................................................ 228 Conjugate Directions and Asymptotic Directions in a Hypersurface.......................... 229 Tensor Derivative of Unit Normal......................................................................... 230 The Equation of Gauss and Codazzi ..................................................................... 233 Hypersurfaces with Indeterminate Lines of Curvature ............................................ 234 Central Quadratic Hypersurfaces .......................................................................... 235 Polar Hyperplane ................................................................................................ 236 Evolute of a Hypersurface in an Euclidean Space ................................................... 237 Hypersphere ......................................................................................................238 Exercises ........................................................................................................... 241

Index .................................................................................................................... 243-245


CHAPTER – 1

PRELIMINARIES 1.1 n-DIMENSIONAL SPACE In three dimensional rectangular space, the coordinates of a point are (x, y, z). It is convenient to write (x1, x2, x3) for (x, y, z). The coordinates of a point in four dimensional space are given by (x1, x2, x3, x4). In general, the coordinates of a point in n-dimensional space are given by (x1, x2, x3,...., xn) such ndimensional space is denoted by V n. 1.2 SUPERSCRIPT AND SUBSCRIPT In the symbol Aklij , the indices i, j written in the upper position are called superscripts and k, l written in the lower position are called subscripts. 1.3 THE EINSTEIN'S SUMMATION CONVENTION n

Consider the sum of the series S = a1 x1 + a2 x 2 + ... + a n x n = Σ a i x i . By using summation convention, i =1 drop the sigma sign and write convention as n

Σ ai x i = a i x i

i =1

This convention is called Einstein’s Summation Convention and stated as “If a suffix occurs twice in a term, once in the lower position and once in the upper position then that suffix implies sum over defined range.” If the range is not given, then assume that the range is from 1 to n. 1.4 DUMMY INDEX Any index which is repeated in a given term is called a dummy index or dummy suffix. This is also called Umbral or Dextral Index. e.g. Consider the expression ai xi where i is dummy index; then ai xi = a1 x1 + a 2 x 2 + ⋅ ⋅ ⋅ + an x n

2


a jx j = a1x1 + a2 x2 + ⋅ ⋅ ⋅ + an x n

and

These two equations prove that a ix i = a j x j So, any dummy index can be replaced by any other index ranging the same numbers. 1.5 FREE INDEX Any index occurring only once in a given term is called a Free Index. e.g. Consider the expression aij x i where j is free index. 1.6 KRÖNECKER DELTA The symbol δij , called Krönecker Delta (a German mathematician Leopold Krönecker, 1823-91 A.D.) is defined by 1 if i = j δij =  0 if i ≠ j Similarly δij and δij are defined as 1 if i = j δ ij = 0 if i ≠ j 1 if i = j δ ij =  0 if i ≠ j

and

Properties 1. If x1, x2, ... xn are independent coordinates, then ∂x i = 0 if i ≠ j ∂x j ∂x i = 1 if i = j ∂x j This implies that ∂x i = δij ∂x j It is also written as 2.

∂x i ∂x k = δ ij . ∂x k ∂x j

δ ii = δ11 + δ 22 + δ 33 + ⋅ ⋅ ⋅ + δ nn

δ ii

(by summation convention)

= 1 + 1+ 1 + ⋅ ⋅ ⋅ + 1

δ ii = n 3.

a ij δkj = a ik Since

31 1 32 2 33 3 3n n a 3 j δ 2j = a δ2 + a δ2 + a δ2 + ⋅ ⋅ ⋅ + a δ 2

(as j is dummy index)

Preliminaries

3

= a32

(as ä 12 = ä 32 = ⋅ ⋅ ⋅ = ä n2 = 0 and ä 22 = 1)

In general, a ijδ kj = a i 1δ1k + a i 2 δ2k + a i 3 δ 3k + ⋅ ⋅ ⋅ + aik δkk + ⋅ ⋅ ⋅ + a in δ nk ik a ij δkj = a (as δ1k = δ2k = ⋅ ⋅ ⋅ = δnk = 0 and δ kk = 1)

4.

i

j

δ j δk = δik δ ij δkj = δ1i δ1k + δi2 δ 2k + δi3 δ3k + ⋅ ⋅ ⋅ + δ ii δ ik + ⋅ ⋅ ⋅ + δ in δ nk = δik (as ä1i = δi2 = δi3 = ⋅ ⋅ ⋅ = δin = 0 and δ ii = 1)

EXAMPLE 1 Write

dφ ∂φ dx1 ∂φ dx 2 ∂φ dx n + 2 + ⋅⋅⋅ + n = 1 using summation convention. dt ∂x dt ∂x dt ∂x dt

Solution dφ ∂φ dx 1 ∂φ dx 2 ∂φ dx n + + ⋅ ⋅ ⋅ + = dt ∂x 1 dt ∂x 2 dt ∂x n dt dφ ∂φ dx i = i dt ∂x dt EXAMPLE 2 Expand: (i) aij xixj; (ii) glm gmp Solution (i)

1 j 2 j n j aij x i x j = a1 j x x + a2 j x x + ⋅ ⋅ ⋅ + a nj x x

= a11 x1 x1 + a22 x 2 x 2 + ⋅ ⋅ ⋅ + ann x n x n aij x i x j = a11 ( x1 )2 + a22 ( x 2 ) 2 + ⋅ ⋅ ⋅ + ann (x n )2 (as i and j are dummy indices) (ii)

g lm g mp = g l 1 g 1 p + g l 2 g 2 p + ⋅ ⋅ ⋅ + g ln g np , as m is dummy index.

EXAMPLE 3 If aij are constant and aij = aji, calculate: ∂ ∂ (i) ∂x (aij x i x j ) (ii) ∂x ∂x (aij xi x j ) k k l Solution ∂ ∂ (aij x i x j ) = a ij ( xi x j ) (i) ∂x k ∂xk

4


∂x

∂x

j i = a ij xi ∂ x + a ij x j ∂ x k k

= a ij xi δ jk + aij x j δ ik ,

as

∂x j ∂x k

= (aij δ jk ) xi + (aij δ ik ) x j = aik xi + akj xj = aik xi + aki xi

as aij δjk = aik as j is dummy index

∂(aij x i x j ) = 2aik xi

∂ xk ∂(aij x i x j )

(ii)

∂x k

as given aik = aki

= 2aikxi

Differentiating it w.r.t. xl : ∂ 2 (aij xi x j ) ∂xk ∂xl

= 2 aik

∂xi ∂xl

= 2 aik δ il ∂ 2 (aij xi x j ) ∂xk ∂xl

as a ik δil = alk.

= 2alk

EXAMPLE 4 If aij xi xj= 0 where aij are constant then show that aij + aji = 0 Solution Given aijx ix j= 0 ⇒ almxlxm= 0 Differentiating it w.r.t. xi partially, ∂ (alm x l x m ) = 0 ∂xi ∂ l m a lm (x x ) = 0 ∂xi a lm

Since

since i and j are dummy indices

∂x l m ∂x m l x + alm x =0 ∂xi ∂x i ∂x l = δ li ∂xi

and

∂x m = δmi ∂xi

= δ jk

Preliminaries

5

almδ il x m + alm δ im x l = 0 aim x m + ali x l = 0 as alm δli = aim and alm δmi = ali . Differentiating it w.r.t. xj partially aim

∂x m ∂x l + ali ∂x j ∂x j = 0 aim δmj + ali δ lj = 0 aij + aji= 0

Proved.

EXERCISES 1. Write the following using the summation convention. (i) (x 1 )2 + (x2 )2 + (x3 )2 + . . . + (xn )2 (ii) ds2 = g 11 (dx1 )2 + g 22 (dx2 )2 + . . . + g nn (dxn )2 (iii) a 1 x1 x3 + a 2 x2 x3 + . . . + a n xn x3 2. Expand the following: (i) a ij xj

(ii)

∂ ∂x

i

i

( ga )

(iii) Aik Bi

3. Evaluate: (i) x j δ ij

(ii) δ ij δkj δkl

(iii) δ ij δij

4. Express b i j y i y j in the terms of x variables where yi = cij xj and b ij cik = δ ik .

ANSWERS 1. 2.

(i) xi xj (ii) ds2 = g ij dxi dxj 1 (i) a i1 x + a i2 x2 + a i3 x3 + . . . + a in xn (ii)

∂

1

1

∂x

( ga ) +

∂ ∂x

2

2

( g a ) + ⋅⋅⋅ +

(iii) A1k B1 + A2k B2 + ... + Ank Bn 3.

(i) xi

4. Cij xi xj

(ii) δ il

(iii) n

(iii) a i x i x 3 . ∂ ∂x

n

n

( ga )

CHAPTER – 2

TENSOR ALGEBRA 2.1 INTRODUCTION A scalar (density, pressure, temperature, etc.) is a quantity whose specification (in any coordinate system) requires just one number. On the other hand, a vector (displacement, acceleration, force, etc.) is a quantity whose specification requires three numbers, namely its components with respect to some basis. Scalers and vectors are both special cases of a more general object called a tensor of order n whose specification in any coordinate system requires 3n numbers, called the components of tensor. In fact, scalars are tensors of order zero with 3° = 1 component. Vectors are tensors of order one with 31 = 3 components. 2.2 TRANSFORMATION OF COORDINATES In three dimensional rectangular space, the coordinates of a point are (x, y, z) where x, y, z are real numbers. It is convenient to write (x1, x2, x3) for ( x, y , z ) or simply xi where i = 1, 2, 3. Similarly in n- dimensional space, the coordinate of a point are n-independent variables (x1, x2,..., x n) in X-coordinate system. Let ( x 1 , x 2 ,..., x n ) be coordinate of the same point in Y-coordinate system. Let x 1 , x 2 , …, x n be independent single valued function of x1, x2,...., xn, so that, x 1 = x 1 (x 1 , x 2 , ...., x n ) x 2 = x 2 ( x1 , x 2 , ...., x n ) x 3 = x 3 (x 1 , x 2 , ...., x n ) M

M

x n = x n (x1 , x 2 , ..., x n ) or x i = x i ( x1 , x 2 , ..., x n ) ;

i = 1, 2, …, n

…(1)

Tensor Algebra

7

Solving these equations and expressing x i as functions of x 1 , x 2 ,..., x n , so that xi = x i ( x1 , x 2 ,..., x n );

i = 1, 2, ..., n

The equations (1) and (2) are said to be a transformation of the coordinates from one coordinate system to another 2.3 COVARIANT AND CONTRAVARIANT VECTORS (TENSOR OF RANK ONE) Let (x1, x2, ..., xn) or x i be coordinates of a point in X-coordinate system and ( x 1 , x 2 ,..., x n ) or x i be coordinates of the same point in the Y-coordinate system. Let A i, i = 1, 2, ..., n (or A 1, A 2, ..., A n) be n functions of coordinates x 1, x2, ..., x n in X-coordinate system. If the quantities A i are transformed to A i in Y-coordinate system then according to the law of transformation ∂x i j ∂x j j A i or A = ∂x j ∂x i A Then A i are called components of contravariant vector. Let Ai , i = 1, 2,..., n (or A 1, A 2, …, A n) be n functions of the coordinates x1, x2, ..., x n Ai =

in X-coordinate system. If the quantities Ai are transformed to Ai in Y-coordinate system then according to the law of transformation ∂x j ∂x i Aj Ai or Aj = i ∂x ∂x j Then A i are called components of covariant vector. The contravariant (or covariant) vector is also called a contravariant (or covariant) tensor of rank Ai =

one. Note: A superscript is always used to indicate contravariant component and a subscript is always used to indicate covariant component.

EXAMPLE 1 If xi be the coordinate of a point in n-dimensional space show that dxi are component of a contravariant vector. Solution 1 2 n Let x1, x2, ..., xn or xi are coordinates in X-coordinate system and x , x ,..., x or x i are coordinates in Y-coordinate system. If

x i = x i (x1 , x 2 ,..., x n ) i i i dx i = ∂x dx 1 + ∂ x dx 2 + ⋅ ⋅ ⋅ + ∂ x dx n 1 2 n

∂x

∂x

∂x

8


d xi =

∂x i ∂x j

dx j

It is law of transformation of contravariant vector. So, dx i are components of a contravariant vector. EXAMPLE 2 Show that

∂φ is a covariant vector where φ is a scalar function. ∂x i

Solution Let x1, x2, ..., xn or x i are coordinates in X-coordinate system and x 1 , x 2 , ..., x n or x i are coordinates in Y-coordinate system. Consider

φ( x 1 , x 2 ,..., x n ) = φ( x1 , x 2 ,..., x n ) ∂φ =

∂φ 1 ∂φ 2 ∂φ ∂x + 2 ∂x + ⋅ ⋅ ⋅ + n ∂x n 1 ∂x ∂x ∂x

∂φ ∂φ ∂x1 ∂φ ∂x 2 ∂φ ∂x n = + + ⋅ ⋅ ⋅ + i ∂x ∂x 1 ∂x i ∂x 2 ∂x i ∂x n ∂x i ∂φ ∂φ ∂x j = ∂x i ∂x j ∂x i ∂φ ∂x j ∂φ = ∂x i ∂x i ∂x j

or

It is law of transformation of component of covariant vector. So, vector.

∂φ is component of covariant ∂x i

EXAMPLE 3 Show that the velocity of fluid at any point is a component of contravariant vector or Show that the component of tangent vector on the curve in n-dimensional space are component of contravariant vector. Solution Let

dx1 dx 2 dx n , , …, be the component of the tangent vector of the point ( x1 , x 2 ,..., x n ) i.e., dt dt dt

dx i be the component of the tangent vector in X-coordinate system. Let the component of tangent dt

Tensor Algebra

9

vector of the point ( x 1 , x 2 ,..., x n ) in Y-coordinate system are

dx i . Then x 1 , x 2 , ..., x n or x i being a dt

function of x1 , x 2 , ..., x n which is a function of t. So, d xi = dt

∂x i dx1 ∂x i dx 2 ∂x i dx n + + ⋅ ⋅ ⋅ + dt1 dt dx 2 dt dx n dt

d xi = dt

∂x i dx j dx j dt

It is law of transformation of component of contravariant vector. So,

dx i is component of dt

contravariant vector. i.e. the component of tangent vector on the curve in n-dimensional space are component of contravariant vector. 2.4 CONTRAVARIANT TENSOR OF RANK TWO Let A ij (i, j = 1, 2, ..., n) be n2 functions of coordinates x1, x2, ..., xn in X-coordinate system. If the quantities A ij are transformed to A ij in Y-coordinate system having coordinates x 1 , x 2 , ..., x n . Then according to the law of transformation A ij =

∂x i ∂x j kl A ∂x k ∂x l

ij Then A are called components of Contravariant Tensor of rank two.

2.5 COVARIANT TENSOR OF RANK TWO Let A ij (i, j = 1, 2, ..., n) be n 2 functions of coordinates x1, x2, ..., xn in X-coordinate system. If the quantities Aij are transformed to Aij in Y-coordinate system having coordinates x 1 , x 2 ,...,x n , then according to the law of transformation, ∂x k ∂x l Akl ∂x i ∂x j Then A ij called components of covariant tensor of rank two. Aij =

2.6 MIXED TENSOR OF RANK TWO Let Aij (i, j = 1, 2, ..., n) be n2 functions of coordinates x1, x2, ..., xn in X-coordinate system. If the quantities Aij are transformed to A ji in Y-coordinate system having coordinates x 1 , x 2 ,..., x n , then according to the law of transformation ∂x i ∂x l k Al = ∂x k ∂x Then Aij are called components of mixed tensor of rank two. A ji

10


Note:

(i) The rank of the tensor is defined as the total number of indices per component. (ii) Instead of saying that “ A ij are the components of a tensor of rank two” we shall often say “ Aij is a tensor of rank two.”

THEOREM 2.1 To show that the Krönecker delta is a mixed tensor of rank two. Solution Let X and Y be two coordinate systems. Let the component of Kronecker delta in X-coordinate system δij and component of Krönecker delta in Y-coordinate be δij , then according to the law of transformation δij =

∂x i ∂x i ∂x l ∂x k = ∂x j ∂x k ∂x j ∂x l

δij =

∂x i ∂x l k δl ∂x k ∂x j

This shows that Krönecker δij is mixed tensor of rank two. EXAMPLE 4 If Ai is a covariant tensor, then prove that

∂Ai do not form a tensor.. ∂x j

Solution Let X and Y be two coordinate systems. As given Ai is a covariant tensor. Then Ai =

∂x k Ak ∂x i

Differentiating it w.r.t. x j ∂Ai ∂ j = ∂x ∂x j ∂Ai ∂x k = ∂x j ∂x i

 ∂x k     ∂x i Ak    ∂Ak ∂ 2 xk + A k ∂x j ∂x i ∂x j

It is not any law of transformation of tensor due to presence of second term. So,

…(1) ∂Ai is not a ∂x j

tensor. THEOREM 2.2 To show that δij is an invariant i.e., it has same components in every coordinate system. Proof: Since δij is a mixed tensor of rank two, then δij =

∂x i ∂x l k δl ∂x k ∂x j

Tensor Algebra

11

∂x i = ∂x k

 ∂x l k     ∂x j δ l   

∂x l k ∂x k δl = ∂x ∂ x ∂x j ∂x j ∂x i ∂x i i = δ = δij δij = , as j ∂x j ∂x j =

∂ x i ∂x k k

j

, as

So, δij is an invariant. THEOREM 2.3 Prove that the transformation of a contravariant vector is transitive. or Prove that the transformation of a contravariant vector form a group. Proof:

Let Ai be a contravariant vector in a coordinate system x i (i = 1, 2,...,n) . Let the coordinates

x i be transformed to the coordinate system x i and x i be transformed to x i . When coordinate x i be transformed to x i , the law of transformation of a contravariant vector is ∂x p q A ... (1) ∂x q When coordinate x i be transformed to x i , the law of transformation of contravariant vector is Ap =

∂x i p A ∂x p ∂x i ∂x p q A i = ∂x i ∂x q A from (1) ∂x i q i = A A ∂x q This shows that if we make direct transformation from x i to x i , we get same law of transformation. This property is called that transformation of contravariant vectors is transitive or form a group. Ai =

THEOREM 2.4 Prove that the transformation of a covariant vector is transitive. or Prove that the transformation of a covariant vector form a group. Proof: Let Ai be a covariant vector in a coordinate system x i (i = 1, 2, ..., n) . Let the coordinates x i be transformed to the coordinate system x i and x i be transformed to x i . When coordinate x i be transformed to x i , the law of transformation of a covariant vector is ∂x q Aq ... (1) ∂x p When coordinate x i be transformed to x i , the law of transformation of a covariant vector is Ap =

Ai =

∂x p Ap ∂x i

12


∂x p ∂x q Aq ∂x i ∂x p ∂x q Aq Ai = ∂x i This shows that if we make direct transformation from x i to x i , we get same law of transformation. This property is called that transformation of covariant vectors is transitive or form a group. Ai =

THEOREM 2.5 Prove that the transformations of tensors form a group or Prove that the equations of transformation a tensor (Mixed tensor) posses the group property. Proof: Let Aij be a mixed tensor of rank two in a coordinate system x i (i = 1, 2,...,n) . Let the coordinates x i be transformed to the coordinate system x i and x i be transformed to x i . When coordinate x i be transformed to x i , the transformation of a mixed tensor of rank two is ∂x p ∂x s r As ... (1) ∂x r ∂x q When coordinate x i be transformed to x i , the law of transformation of a mixed tensor of rank two is Aqp =

A ji

∂x i ∂x q p Aq = ∂x p ∂x j =

A ji =

i

q

p

s

p

j

r

q

∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x

r

As from (1)

∂x i ∂x s r As ∂x r ∂x j

This shows that if we make direct transformation from x i to x i , we get same law of transformation. This property is called that transformation of tensors form a group. THEOREM 2.6 There is no distinction between contravariant and covariant vectors when we restrict ourselves to rectangular Cartesian transformation of coordinates. Proof: Let P(x, y) be a point with respect to the rectangular Cartesian axes X and Y. Let ( x , y ) be the coordinate of the same point P in another rectangular cartesian axes X and Y , Let (l1, m1) and (l2, m2) be the direction cosines of the axes X , Y respectively. Then the transformation relations are given by x = l1 x + m1 y   ...(1) y = l2 x + m2 y  and solving these equations, we have x = l1 x + l 2 y   y = m1 x + m2 y  put x = x1 , y = x 2 , x = x 1 , y = x 2

...(2)

Tensor Algebra

13

Consider the contravariant transformation ∂x i j i = A ; j = 1,2 A ∂x j ∂x i 1 ∂x i 2 i = A + 2A A ∂x1 ∂x for i = 1, 2 . ∂x 1 1 ∂x 1 2 A + 2A = A ∂x1 ∂x 2 ∂x 1 ∂x 2 2 2 = A + 2 A A ∂x1 ∂x 1

From (1)

∂x 2 2 = l , but x = x1 , y = x , x = x 1 , y = x ∂x 1

Then ∂x ∂x 1 = =l . ∂x ∂x 1 1 Similarly,     2 2 ∂y ∂x ∂y ∂x = l2 = 1 ; = m2 = 2  ∂x ∂y ∂x ∂x  ∂x ∂x 1 = m1 = 2 ; ∂y ∂x

...(3)

So, we have A 1 = l1 A1 + m1 A 2   A 2 = l2 A1 + m 2 A 2  Consider the covariant transformation

..(4)

∂x j A j ; j = 1,2 ∂x i ∂x1 ∂x 2 A + A2 Ai = 1 ∂x i ∂x1 Ai =

for i = 1, 2 . ∂x1 ∂x 2 A + A2 1 ∂x 1 ∂x 1 ∂x 1 ∂x 2 A + A2 A2 = 1 ∂x 2 ∂x 2 A1 =

From (3)

A1 = l1 A1 + m1 A2   A2 = l2 A1 + m2 A2 

...(5)

14


So, from (4) and (5), we have A 1 = A1 and A 2 = A2 Hence the theorem is proved. 2.7 TENSORS OF HIGHER ORDER (a) Contravariant tensor of rank r i i ...i Let Ai1i 2 ...ir be nr function of coordinates x1, x2, ..., xn in X-coordinates system. If the quantities A i 2 r are transformed to A i1i 2 ...i r in Y-coordinate system having coordinates x 1 , x 2 , ..., x n . Then according to the law of transformation A

Then Ai ii 2 ...i r

i1i 2 ...i r

=

∂x i1 ∂x i2

K

∂x i r

p1 p 2 ...p r ∂x p1 ∂x p 2 ∂x p r A are called components of contravariant tensor of rank r.

(b) Covariant tensor of rank s Let A j1 j 2 ... j s be n s functions of coordinates x1, x2, ..., xn in X-coordinate system. If the quantities A j1 j 2 ... j s are transformed to A j1 j 2 ... j s in Y- coordinate system having coordinates x 1 , x 2 ,...,x n . Then according to the law of transformation A j1 j 2 ... j s =

∂x q1 ∂x q 2 ∂x q s ⋅ ⋅ ⋅ ∂x j1 ∂x j 2 ∂x j s

Aq1 ,q 2 ...,q s

Then A j1 j 2 ... j s are called the components of covariant tensor of rank s. (c) Mixed tensor of rank r + s i i ...i Let A j11 2j 2 ...rj s be nr+s functions of coordinates x1, x2, ..., xn in X-coordinate system. If the quantities

Aij11i 2j 2......i rj s are transformed to A ij11ij22......i rj s in Y-coordinate system having coordinates x 1 , x 2 , … , x n . Then according to the law of transformation A ij11ij22......i rj s =

∂x i1 ∂x i 2 ∂x ir ∂x p1 ∂x p2 ∂x p r

∂x q1 ∂x q 2 ∂x q s ⋅ ⋅ ⋅ Aqp11qp22......q ps r j1 j2 js ∂x ∂x ∂x

i i ...i Then A j11 2j 2 ...rj s are called component of mixed tensor of rank r + s . 2 ... j s A tensor of type Ai1j1i 2j... is known as tensor of type (r, s ) , In (r,s), the first component r ir indicates the rank of contravariant tensor and the second component s indicates the rank of covariant tensor.

ij Thus the tensors Aij and A are type (0, 2) and (2, 0) respectively while tensor Aij is type (1, 1).

Tensor Algebra

15

EXAMPLE ijk is a mixed tensor of type (3, 2) in which contravariant tensor of rank three and covariant Alm tensor of rank two. Then according to the law of transformation

Almijk =

∂x i ∂x j ∂x k ∂x a ∂x b ∂x α ∂x β ∂xγ ∂x l ∂x m

αβγ Aab

2.8 SCALAR OR INVARIANT A function φ ( x1 , x 2 , ..., x n ) is called Scalar or an invariant if its original value does not change upon transform ation of coordinates from x1, x2, ..., xn to x 1 , x 2 , ..., x n . i.e.

φ( x1 , x 2 ,..., x n ) = φ( x 1 , x 2 ,..., x n ) Scalar is also called tensor of rank zero. For example, Ai Bi is scalar.. 2.9 ADDITION AND SUBTRACTION OF TENSORS THEOREM 2.7 The sum (or difference) of two tensors which have same number of covariant and the same contravariant indices is again a tensor of the same rank and type as the given tensors. Proof: Consider two tensors Aij11i 2j 2......i rj s and B ij11i 2j 2......i rj s of the same rank and type (i.e., covariant tensor of rank s and contravariant tensor of rank r.). Then according to the law of transformation ∂x i1 ∂x i 2 ∂x ir ∂x q1 ∂x q 2 ∂x q s A ij11ij22......i rj s = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ Aqp11qp22......q ps r p1 p2 pr j1 j2 js ∂x ∂x ∂x ∂x ∂x ∂x and B ij11ij22......i rj s =

∂x i1 ∂x i 2 ∂x ir ∂x q1 ∂x q 2 ∂x q s ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ Bqp11qp22......q ps r ∂x p1 ∂x p2 ∂x p r ∂x j1 ∂x j2 ∂x j s

Then A ji11 ij22......i rj s ± B ij11ij22......i rj s =

∂x i1 ∂x i 2 ∂x i r ∂x q1 ∂x q 2 … ∂x q s ⋅ ⋅ ⋅ ∂x p1 ∂x p2 ∂x p r ∂x j1 ∂x j 2 ∂x j s

(A

p1 p2 ...p r q1q 2 ...q s

± B qp11qp2 2......qps r

)

If A ij11ij22......i rj s ± B ji11ij22......irj s = C ij11ij22......i rj s and Aqp11qp2 2......q ps r ± Bqp11qp2 2......qps r = C qp11qp2 2......qps r So, ∂x i1 ∂x i 2 ∂x ir ∂x q1 ∂x q 2 ∂x q s ,..., pr ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ Cqp11,,qp22,..., qs ∂x p1 ∂x p2 ∂x p r ∂x j1 ∂x j 2 ∂x j s i , i ,..., i This is law of transformation of a mixed tensor of rank r+s. So, C j11 , j22 ,..., jrs is a mixed tensor of C ij11ij22......i rj s =

rank r+s or of type (r, s).

16


EXAMPLE 5 If Akij and Bnlm are tensors then their sum and difference are tensors of the same rank and type. Solution As given Akij and Bkij are tensors. Then according to the law of transformation Akij =

∂x i ∂x j ∂x r pq Ar ∂x p ∂x q ∂x k

and Bkij

∂x i ∂x j ∂x r pq Br = ∂x p ∂x q ∂x k

then Akij ± Bkij =

(

∂x i ∂x j ∂x r pq Ar ± Brpq p q k ∂x ∂x ∂x

)

If Akij ± B kij = C kij and Arpq ± B rpq = C rpq So, ∂x i ∂x j ∂x r pq Cr ∂x p ∂x q ∂x k The shows that C kij is a tensor of same rank and type as Akij and Bkij . C kij =

2.10 MULTIPLICATION OF TENSORS (OUTER PRODUCT OF TENSOR) THEOREM 2.8 The multiplication of two tensors is a tensor whose rank is the sum of the ranks of two tensors. i i ...i Proof: Consider two tensors A j11 2j 2 ...rj s (which is covariant tensor of rank s and contravariant tensor of k k2 ...km rank r) and Bl11l 2 ... (which is covariant tensor of rank m and contravariant tensor of rank n). Then ln according to the law of transformation.

A ij11ij22......i rj s =

∂x i1 ∂x i 2 ∂x ir ∂x q1 ∂x q 2 ∂x q r ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∂x p1 ∂x p2 ∂x p r ∂x j1 ∂x j 2 ∂x j s

Aqp11qp22......q ps r

∂x k1 ∂x k 2 ∂x km ∂xβ1 ∂x β2 ∂x βn ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∂x α1 ∂x α 2 ∂x α m ∂x l1 ∂x l 2 ∂x l n

Bβα11βα2 2......βαn m

and km Bl1kl12k...2 ... = ln

Then their product is A ij11ij22......i rj s Blk1l12k...2 ...l nkm =

∂x i1 ∂x p1

⋅⋅⋅

∂x i r ∂x q1 ∂x pr ∂x j1

⋅⋅⋅

∂ x q s ∂ x k1 ∂ x km ∂ x β 1 ∂x βn ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∂x j s ∂x α1 ∂x α m ∂x l1 ∂x l m Aqp11qp22......q ps r Bβα11βα2 2......βαn m

Tensor Algebra

17

If ...km i i ...i k k ...k C ij1ij2 ......i rjk1l kl 2 ... = A j11 j22 ...rjs Bl1l12 ...2 ln n 1 2 s 1 2 ln

and ...,p r C qp11qp2 2......qpsβr α1β12α...2β...nαm = Aqp11qp22..., Bβα11βα2 2......βαn m qs

So, ∂x q s ∂x k1 ∂x i1 ∂x ir ∂x q 1 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∂x j s ∂x α1 ∂x p1 ∂x p r ∂x j1 ∂x k m ∂x β1 ∂ x βn ⋅ ⋅ ⋅ α ⋅ l ⋅ ⋅ ⋅ l C qp11qp2 2......q ps βrα1 β1α2 2......βαn h ∂x m ∂x 1 ∂x n

...km C ji11ij22......irj ksl11kl 22 ... ln =

i i ...i

...k

This is law of transformation of a mixed tensor of rank r + m + s + n. . So, C j11 j22 ...rj ks1lk12l 2 ...lmn is a mixed tensor of rank r + m + s + n . or of type (r + m, s + n ) . Such product is called outer product or open proudct of two tensors. THEOREM 2.9 If A i and B j are the components of a contravariant and covariant tensors of rank one then prove that A iB j are components of a mixed tensor of rank two. Proof: As Ai is contravariant tensor of rank one and B j is covariant tensor of rank one. Then according to the law of transformation Ai =

∂x i k A ∂x k

...(1)

Bj =

∂x l Bl ∂x j

...(2)

and

Multiply (1) and (2), we get Ai B j =

∂x i ∂x l k A Bl ∂x k ∂x j

This is law of transformation of tensor of rank two. So, Ai B j are mixed tensor of rank two. Such product is called outer product of two tensors. EXAMPLE 6 Show that the product of two tensors Aij and B mkl is a tensor of rank five. Solution As Aij and B mkl are tensors. Then by law of transformation A ji =

∂x i ∂x q p Aq ∂x p ∂x j

and

Bmkl =

∂x k ∂x l ∂x t rs Bt ∂x r ∂x s ∂x m

18


Multiplying these, we get A ji Bmkl =

∂x i ∂x q ∂x k ∂x l ∂x t p rs A B ∂x p ∂x j ∂x r ∂x s ∂x m q t

This is law of transformation of tensor of rank five. So, Aij B mkl is a tensor of rank five. 2.11 CONTRACTION OF A TENSOR The process of getting a tensor of lower order (reduced by 2) by putting a covariant index equal to a contravariant index and performing the summation indicated is known as Contraction. In other words, if in a tensor we put one contravariant and one covariant indices equal, the process is called contraction of a tensor. ijk For example, consider a mixed tensor Alm of order five. Then by law of transformation,

∂x i ∂x j ∂x k ∂x s ∂x t pqr Ast ∂x p ∂x q ∂x r ∂x l ∂x m Put the covariant index l = contravariant index i, so that Almijk =

Aimijk =

∂x i ∂x j ∂x k ∂x s ∂x t pqr Ast ∂x p ∂x q ∂x r ∂x i ∂x m

=

∂x j ∂x k ∂x s ∂xt pqr Ast ∂x q ∂x r ∂x p ∂x m

=

∂x j ∂x k ∂x t s pqr δ p Ast ∂x q ∂x r ∂x m

Aimijk =

Since

∂x s = δ sp ∂x p

∂x j ∂x k ∂x t pqr A pt ∂x q ∂x r ∂x m

ijk This is law of transformation of tensor of rank 3. So, Aim is a tensor of rank 3 and type (1, 2) ijk while Alm is a tensor of rank 5 and type (2, 3). It means that contraction reduces rank of tensor by two.

2.12 INNER PRODUCT OF TWO TENSORS l l Consider the tensors Akij and Bmn if we first form their outer product Akij B mn and contract this by k putting l = k then the result is Akij B mn which is also a tensor, called the inner product of the given tensors. Hence the inner product of two tensors is obtained by first taking outer product and then contracting it.

EXAMPLE 7 If A i and B i are the components of a contravariant and covariant tensors of rank are respectively then prove that A iB i is scalar or invariant.

Tensor Algebra

19

Solution As A i and B i are the components of a contravariant and covariant tensor of rank one respectively, then according to the law of the transformation ∂x i p ∂x q A Bq = and = Bi A ∂x p ∂x i i

Multiplying these, we get ∂x i ∂x q p A Bq = AB ∂x p ∂x i ∂x q p ∂x q A B , = δ qp = since q ∂x p ∂x p i

i

= δ qp A p Bq p A i Bi = A B p

This shows that A iB i is scalar or Invariant. EXAMPLE 8 If Aij is mixed tensor of rank 2 and B mkl is mixed tensor of rank 3. Prove that Aij B mjl is a mixed tensor of rank 3. Solution As Aij is mixed tensor of rank 2 and B mkl is mixed tensor of rank 3. Then by law of transformation A ji =

∂x i ∂x q p ∂x k ∂x l ∂x t rs kl A Bt q and Bm = ∂x p ∂x j ∂x r ∂x s ∂x m

...(1)

Put k = j then Bmjl

=

∂x j ∂x l ∂x t rs Bt ∂x r ∂x s ∂x m

...(2)

Multiplying (1) & (2) we get A ij B mjl = = A ji Bmjl

∂x i ∂x q ∂x j ∂x l ∂x t p rs Aq Bt ∂x p ∂x j ∂x r ∂x s ∂x m ∂x i ∂x l ∂x t q p rs δr Aq Bt ∂x p ∂x s ∂x m

∂x i ∂x l ∂x t p qs Aq B t = ∂x p ∂x s ∂x m

since

∂x q ∂x j ∂x q = = δ qr ∂x j ∂x r ∂x r

since

δ qr Btrs = Btqs

This is the law of transformation of a mixed tensor of rank three. Hence Aij B mjl is a mixed tensor of rank three.

20


2.13 SYMMETRIC TENSORS A tensor is said to be symmetric with respect to two contravariant (or two covariant) indices if its components remain unchanged on an interchange of the two indices. EXAMPLE (1) The tensor A ij is symmetric if A ij = A ji ijk ijk = Almjik (2) The tensor Alm is symmetric if Alm

THEOREM 2.10 A symmetric tensor of rank two has only

1 n(n + 1) different components in n2

dimensional space. Proof: Let A ij be a symmetric tensor of rank two. So that Aij = A ji .  A11  21 A The component of A ij are  A31   M  A n1 

A12 A 22 A32 M An2

A13 A 23 A33 M An3

L L L L L

A1n   A2n  A 3n   M  A nn 

i.e., A ij will have n2 components. Out of these n2 components, n components A 11, A 22, A 33, ..., A nn are different. Thus remaining components are (n2– n). In which A 12 = A 21, A 23 = A 32 etc. due to symmetry. So, the remaining different components are

1 2 (n − n) . Hence the total number of different 2

components = n+

1 2 1 (n − n) = n(n + 1) 2 2

2.14 SKEW-SYMMETRIC TENSOR A tensor is said to be skew-symmetric with respect to two contravariant (or two covariant) indices if its components change sign on interchange of the two indices. EXAMPLE (i) The tensor A ij is Skew-symmetric of A ij = − A ji ijk ijk (ii) The tensor Alm is Skew-symmetric if Alm = − Almjik

THEOREM 2.11

A Skew symmetric tensor of second order has only

1 n(n − 1) different non-zeroo 2

components. Proof: Let A ij be a skew-symmetric tensor of order two. Then Aij = − A ji .

Tensor Algebra

21

 0  21 A ij The components of A are  A 31   M  A n1 

[Since A

ii

A12 0 A 32 M An 2

A13 A 23 0 M

L L L L

An3 L

A1n   A2n  A3n   M  0 

]

= − Aii ⇒ 2 A ii = 0 ⇒ Aii = 0 ⇒ A11 = A 22 = ⋅ ⋅ ⋅ A nn = 0

i.e., A ij will have n2 components. Out of these n2 components, n components A 11, A 22, A 33, ..., A nn are zero. Omitting there, then the remaining components are n2 –n. In which A 12 = – A 21, A 13 = – A 31 1 2 etc. Ignoring the sign. Their remaining the different components are (n − n) . 2 1 Hence the total number of different non-zero components = n(n − 1) 2 Note: Skew-symmetric tensor is also called anti-symmetric tensor.

THEOREM 2.12 A covariant or contravariant tensor of rank two say Aij can always be written as the sum of a symmetric and skew-symmetric tensor. Proof: Consider a covariant tensor A ij. We can write A ij as Aij =

1 1 ( A + A ji ) + ( Aij − A ji ) 2 ij 2

Aij = Sij + Tij where

Sij =

1 1 ( Aij + A ji ) and Tij = ( Aij − A ji ) 2 2

S ji =

1 ( A + Aij ) 2 ji

Now,

S ji = Sij So, Sij is symmetric tensor.. and Tij =

1 ( A + A ji ) 2 ij

1 ( A − Aij ) 2 ji 1 = − ( Aij − A ji ) 2

T ji =

T ji = −Tij or

Tij = −T ji So, Tij is Skew-symmetric Tensor..

22


EXAMPLE 9 j k If φ = a jk A A . Show that we can always write φ = b jk A j Ak where b jk is symmetric.

Solution As given φ = a jk A j A k Interchange the indices i and j

...(1)

φ = ak j A k A j

...(2)

Adding (1) and (2), 2φ = (a jk + akj ) A j A k φ =

1 (a jk + akj ) A j A k 2

φ = b jk A j A k where b jk =

1 (a + akj ) 2 jk

To show that b jk is symmetric. Since 1 (a + akj ) 2 jk 1 = (a kj + a jk ) 2 1 = (a jk + akj ) 2

b jk = bkj

bkj = b jk So, b jk is Symmetric. EXAMPLE 10  ∂Ti ∂T j If Ti be the component of a covariant vector show that  j − i ∂x  ∂x symmetric covariant tensor of rank two. Solution As Ti is covariant vector. Then by the law of transformation Ti =

∂x k Tk ∂x i

  are component of a Skew

Tensor Algebra

23

Differentiating it w.r.t. to x j partially,, ∂ ∂Ti j = ∂x j ∂x

 ∂x k     ∂x i Tk   

∂2 x k ∂x k ∂Tk T + = k ∂x j ∂x i ∂x i ∂x j ∂T j ∂x j

=

∂ 2 xk ∂x k ∂x l ∂Tk T + k ∂x j ∂x i ∂x i ∂x j ∂x l

...(1)

Similarly, ∂T j

∂ 2 xk ∂x k ∂x l ∂Tk T + k ∂x i ∂x i ∂x j ∂x j ∂x i ∂x l Interchanging the dummy indices k & l ∂T j ∂x i

=

=

∂2 x k ∂x k ∂x l ∂Tl T + k ∂x i ∂x j ∂x i ∂x j ∂x k

...(2)

Substituting (1) and (2), we get ∂x k ∂x l  ∂Tk ∂Tl  ∂Ti ∂T j  −  − = ∂x i ∂x j  ∂x l ∂x k  ∂x j ∂x i This is law of transformation of covariant tensor of rank two. So,

∂Ti ∂T j − are component of ∂x j ∂x i

a covariant tensor of rank two. ∂Ti ∂T j − To show that is Skew-symmetric tensor.. ∂x j ∂x i Let ∂Ti ∂T j − Aij = ∂x j ∂x i ∂T j ∂Ti − A ji = ∂xi ∂x j  ∂T

∂T 

j i = −  ∂ x j − ∂x i   

A ji = − Aij or

Aij = − A ji

∂Ti ∂T j − is Skew-symmetric. ∂x j ∂x i ∂Ti ∂T j − So, are component of a Skew-symmetric covariant tensor of rank two. ∂x j ∂x i So, Aij =

24


2.15 QUOTIENT LAW By this law, we can test a given quantity is a tensor or not. Suppose given quantity be A and we do not know that A is a tensor or not. To test A, we take inner product of A with an arbitrary tensor, if this inner product is a tensor then A is also a tensor. Statement If the inner product of a set of functions with an atbitrary tensor is a tensor then these set of functions are the components of a tensor. The proof of this law is given by the following examples. EXAMPLE 11 Show that the expression A(i,j,k) is a covariant tensor of rank three if A(i,j,k)B k is covariant tensor of rank two and B k is contravariant vector Solution Let X and Y be two coordinate systems. As given A (i, j, k)B k is covariant tensor of rank two then A ( i, j, k ) B k =

∂x p ∂x q ∂x i ∂x j

A( p, q, r )B r

...(1)

Since B k is contravariant vector. Then ∂x k ∂x r Bk = r Br Br = k B k or ∂x ∂x So, from (1) A ( i, j, k ) B k = A ( i, j, k ) B k = A ( i, j , k ) =


A ( p , q, r )

∂x p ∂x q ∂x r ∂x i ∂x j ∂x k ∂x p ∂x q ∂ x r

∂x r ∂x k

Bk

A( p, q , r )B k

A( p , q , r )

∂ x i ∂x i ∂x k

As B k is arbitrary.. So, A(i , j , k ) is covariant tensor of rank three. EXAMPLE 12 If A (i, j, k)A iB jCk is a scalar for arbitrary vectors A i, B j, Ck. Show that A(i, j, k) is a tensor of type (1, 2). Solution Let X and Y be two coordinate systems. As given A(i , j, k ) A i B j C k is scalar. Then i

j

p

q

A (i, j, k ) A B C k = A( p, q, r ) A B C r

...(1)

Tensor Algebra

25

Since Ai , B i and C k are vectors. Then ∂x i p A Ai = or ∂x p ∂x j q B or Bj = ∂x q ∂x k C k = r Cr or ∂x So, from (1) i

j

A (i, j , k ) A B C k = A( p, q, r)

∂x p i A ∂x i ∂x q j B = ∂x j ∂x r k C = ∂x k

Ap = Bq Cr

∂x p ∂xq ∂x k ∂x i ∂x j ∂xr

Ai B jCk

As A i , B j , C k are arbitrary.. Then

∂x p ∂xq ∂x k A( p, q, r ) A ( i, j , k ) = ∂xi ∂x j ∂xr So, A(i, j, k) is tensor of type (1, 2). 2.16

CONJUGATE (OR RECIPROCAL) SYMMETRIC TENSOR

Consider a covariant symmetric tensor Aij of rank two. Let d denote the determinant Aij with the elements Aij i.e., d = Aij and d ≠ 0 . Now, define A ij by A ij =

Cofactor of Aij is the determinan t Aij

d A is a contravariant symmetric tensor of rank two which is called conjugate (or Reciprocal) tensor of Aij . ij

THEOREM 2.13 If Bij is the cofactor of Aij in the determinant d = |Aij| ≠ 0 and Aij defined as A ij =

Bij d

Then prove that Aij A = . Proof: From the properties of the determinants, we have two results. kj

(i) ⇒

δ ki

Aij Bij = d Aij

Bij d

=1

Aij Aij = 1,

given

Aij =

Bij d

26


(ii)

Aij B kj = 0 Aij

B kj d

= 0,

Aij Akj = 0

d ≠0 if i ≠ k

from (i) & (ii) 1 if i = k Aij Akj = 0 if i ≠ k  Aij Akj = δik

i.e., 2.17 RELATIVE TENSOR

i i ...i If the components of a tensor A j11 2j 2 ...rj s transform according to the equation 2 ...k r Alk11l 2k... ls

∂x = ∂x

ω

Aij11i 2j 2......i rj s

∂x k1 ∂x k2 ∂x k r ∂x j1 ∂x j2 ∂x js ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∂x i1 ∂x i 2 ∂xi r ∂x l r ∂x l 2 ∂x l s

∂x is the Jacobian of transformation. If ∂x ω = 1, the relative tensor is called a tensor density. If w = 0 then tensor is said to be absolute.

i i ...i Hence A j11 2j 2 ...rj r is called a relative tensor of weight ω, where

MISCELLANEOUS EXAMPLES 1. Show that there is no distinction between contravariant and covariant vectors when we restrict ourselves to transformation of the type i m i x i = am x + b ; where a's and b's are constants such that

a ir ami = δ rm Solution Given that or

i m i x i = am x + b a im x m = x i − bi

...(1) ...(2)

Multiplying both sides (2) by a ir , we get a ir ami x m = a ir x i − bi a ir δ rm x m = a ir x i − bi a ir as given a ir ami = δ rm i i i i x r = a r x − b a r as δ rm x m = x r

or

i i i i xs = as x − b as

Differentiating Partially it w.r.t. to x i ∂x s = a is ∂x i

..(3)

Tensor Algebra

27

Now, from (1) i s i x i = as x + b

∂x i a is s = ∂x The contravariant transformation is

...(4)

∂x i s A = a is A s = A ∂x s The covariant transformation is

...(5)

i

∂x s As = a is As ...(6) ∂x i Thus from (5) and (6), it shows that there is no distinction between contravariant and covariant tensor law of transformation Ai =

2. If the tensors aij and g ij are symmetric and u i , v i are components of contravariant vectors satisfying the equations (aij − kgij )u i = 0, i, j = 1, 2,...,n ( aij − k'gij )v i = 0, k ≠ k ′ . Prove that gij u i v j = 0, a ij u i v j = 0. Solution The equations are (aij − kgij )u i = 0

...(1)

(aij − k ′gij )v i = 0 j

...(2) j

Multiplying (1) and (2) by u and v respectively and subtracting, we get aij u i v j − aij v i u j − kgij u i v j + k ′gij u j v i = 0 Interchanging i and j in the second and fourth terms, a ij u i v j − a ji v j u i − kgij u i v j + k ′g ji u i v j = 0 As aij and g ij is symmetric i.e., a ij = a ji & g ij = g ji − kgij v j u i + k ′gij u i v j = 0 i

j

(k ′ − k ) g ij u v = 0 g ij u i v j = 0 since k ≠ k ′ ⇒ k − k ′ ≠ 0 Multiplying (1) by v j , we get a ij v j u i − kgij u i v j = 0 a ij u i v j = 0 as g ij u i v j = 0.

Proved.

28


3. If Aij is a Skew-Symmetric tensor prove that i

k

i k

(δ j δl + δl δ j ) Aik = 0 Solution Given Aij is a Skew-symmetric tensor then Aij = − A ji . Now, i

k

i k

(δ j δl + δl δ j ) Aik = δ ij δ kl Aik + δil δ kj Aik = δ ij Ail + δil Aij = A jl + Alj i

k

i k

(δ j δl + δl δ j ) Aik = 0

as A jl = − Alj

4. If aij is symmetric tensor and bi is a vector and a ij bk + a jk bi + a kib j = 0 then prove that a ij = 0 or bk = 0 . Solution The equation is aij bk + a jk bi + akib j = 0 aij bk + a jk bi + akib j = 0 ⇒ By tensor law of transformation, we have a pq

∂x p ∂x q ∂x r ∂x p ∂x q ∂x r ∂x p ∂x q ∂x r b + a b + a b =0 r pq r pq r ∂x i ∂x j ∂x k ∂x j ∂x k ∂x i ∂x k ∂x i ∂x j

 ∂x p ∂x q ∂x r ∂x p ∂x q ∂x r ∂x p ∂x q ∂x r  + j + a pq br  i  j k ∂x ∂x k ∂x i ∂x k ∂x i ∂x j   ∂x ∂x ∂x

= 0

a pq br = 0 ⇒ a pq = 0 or br = 0

⇒

⇒ a ij = 0 or bk = 0 5. If a mn x x = bmn x x for arbitrary values of x r , show that a(mn) = b(mn) i.e., a mn + a nm = bmn + bnm m n

m n

If a mn and bmn are symmetric tensors then further show the a mn = bmn . Solution Given a mn x m x n = bmn x m x n (a mn − bmn )x m x n = 0

Tensor Algebra

29

Differentiating w.r.t. x i partially (ain − bin )x n + (ami − bmi ) x m = 0 Differentiating again w.r.t. x j partially (aij − bij ) + (a ji − b ji ) = 0 a ij + a ji = bij + b ji or

a mn + anm = bmn + bnm or a (mn ) = b( mn )

Also, since a mn and bmn are symmetric then a mn = a nm , bmn = bnm . So, 2 amn = 2bmn a mn = bmn

EXERCISES 1. Write down the law of transformation for the tensors (i) Aij (ii) Bkij ijk (iii) Clm

2. If Arpq and Bts are tensors then prove that Arpq Bts is also a tensor.. 3. If Aij is a contravariant tensor and Bi is covariant vector then prove that Aij Bk is a tensor of rank three and Aij Bj is a tensor of rank one. 4. If Ai is an arbitrary contravariant vector and Cij Ai Aj is an invariant show that Cij + Cji is a covariant tensor of the second order. 5. Show that every tensor can be expressed in the terms of symmetric and skew-symmetric tensor. 6. Prove that in n-dimensional space, symmetric and skew-symmetric tensor have

n n (n + 1) and (n − 1) 2 2

independent components respectively. 7. If U ij ≠ 0 are components of a tensor of the type (0, 2) and if the equation fUij + gU

ji

= 0 holds

w.r.t to a basis then prove that either f = g and U ij is skew-symmetric or f = –g and U ij is symmetric. i k i k 8. If Aij is skew-symmetric then ( B j Bl + B l B j ) Aik = 0 . ij

i

9. Explain the process of contraction of tensors. Show that aij a = δ j .

30


10. If Arpq is a tensor of rank three. Show that Arpr is a contravariant tensor of rank one. ij k k 11. If a k λi µ j γ is a scalar or invariant, λi , µ j , γ are vectors then akij is a mixed tensor of type (2, 1). h i h k 12. Show that if a hijkλ µ λ µ = 0 where λi and µ i are components of two arbitrary vectors then

ahijk + ahkji + a jihk + a jkhi = 0

13. Prove that Aij Bi C j is invariant if Bi and C j are vector and Aij is tensor of rank two. 14. If A(r, s, t) be a function of the coordinates in n-dimensional space such that for an arbitrary vector Br of the type indicated by the index a A(r, s, t)Br is equal to the component Cst of a contravariant tensor of order two. Prove that A(r, s, t) are the components of a tensor of the form Arst . 15. If Aij and Aij are components of symmetric relative tensors of weight w. show that A

ij

ij

= A

∂x ∂x

w− 2

and

Aij = Aij

∂x ∂x

w+ 2

16. Prove that the scalar product of a relative covariant vector of weight w and a relative contravariant vector of weight w′ is a relative scalar of weight w + w ′ .

CHAPTER – 3

METRIC TENSOR AND RIEMANNIAN METRIC 3.1 THE METRIC TENSOR In rectangular cartesian coordinates, the distance between two neighbouring point are (x, y, z) and ( x + dx, y + dy, z + dz ) is given by ds 2 = dx 2 + dy 2 + dz 2 . In n-dimensional space, Riemann defined the distance ds between two neighbouring points x i and x i + dx i (i = 1, 2,...n) by quadratic differential form ds 2 = g11 (dx 1 ) 2 + g12 dx 1dx 2 + ⋅ ⋅ ⋅ + g1n dx1 dx n + g12 (dx 2 )dx1 + g 22 (dx 2 ) 2 + ⋅ ⋅ ⋅ + g 2 n dx 2 dx n +

. . . . . . . . .. . . . . . . . . . . . . . . . .

+

+ g n1 dx n dx1 + g n 2 dx n dx 2 + ⋅ ⋅ ⋅ ⋅ + g nn (dx n )2 ds 2 = g ij dx i dx j (i, j = 1,2,...n)

...(1)

using summation convention. Where g ij are the functions of the coordinates x i such that g = g ij ≠ 0 The quadratic differential form (1) is called the Riemannian Metric or Metric or line element for ndimensional space and such n-dimensional space is called Riemannian space and denoted by Vn and g ij is called Metric Tensor or Fundamental tensor.. The geometry based on Riemannian Metric is called the Riemannian Geometry. THEOREM 3.1 The Metric tensor g ij is a covariant symmetry tensor of rank two. Proof: The metric is given by i j ds 2 = g ij dx dx

...(1)

32


Let x i be the coordinates in X-coordinate system and x i be the coordinates in Y-coordinate system. Then metric ds2 = gij dxidxj transforms to ds 2 = g ij d x i dx j . Since distance being scalar quantity. i j i j ds 2 = g ij dx dx = gij dx dx

So,

The theorem will be proved in three steps. (i) To show that dxi is a contravariant vector. x i = x i (x1, x2 ,...x n )

If

dxi =

∂x i 1 ∂x i 2 ∂x i n dx + dx + ⋅ ⋅ ⋅ + dx ∂x i ∂x 2 ∂x n

dxi =

∂x i k dx ∂x k

It is law of transformation of contravariant vector. So, dx i is contravariant vector.. (ii) To show that g ij is a covariant tensor of rank two. Since dxi =

∂x i k ∂x j l j dx ∂x and d x = ∂x k ∂x l

from equation (2) ∂x i ∂x k ∂x i g ij dx i dx j = g ij k ∂x ∂x i k l g g kl dx dx = ij k ∂x g ij dx i dx j = g ij

dxk

∂x j l dx ∂x l

∂x j k l ∂x dx ∂x l ∂x j k l dx dx ∂x l

Since g ij dx i dx j = g kl dx k dx l (i, j are dummy indices). i j   g kl − g ij ∂x ∂x  ∂x k ∂x l 

or

or

g kl − gij

 k l dx dx = 0   ∂x i ∂x j = 0 as dx k and dx l are arbitrary.. ∂x k ∂x l g kl = g ij

∂x j ∂x j ∂x k ∂x l

g ij = g kl

∂x k ∂x l ∂x i ∂x j

So, g ij is covariant tensor of rank two.

...(2)

Metric Tensor and Riemannian Metric

33

(iii) To show that g ij is symmetric. Then g ij can be written as g ij =

1 1 (g + g ji ) + ( gij − g ji ) 2 ij 2

g ij = Aij + Bij 1 g +g ( ij ji ) = symmetric 2 1 Bij = (g ij − g ji ) = Skew-symmetric 2 Aij =

where

g ij dx i dx j = ( Aij + Bij )dxi dx j from (3)

Now,

( gij − Aij )dx i dx j = Bij dxi dx j

(4)

Interchanging the dummy indices in Bij dx i dx j , we have Bij dxi dx j = B ji dx i dx j Bij dxi dx j = − B ij dx i dx j Since Bij is Skew-symmetric i.e., Bij = −B ji Bij dx i dx j + Bij dx i dx j = 0 2 Bij dx i dx j = 0 ⇒

Bij dxi dx j = 0

So, from (4), ( gij − Aij )dx i dx j = 0 ⇒

g ij = Aij as dx i , dx j are arbitrary..

So, g ij is symmetric since Aij is symmetric. Hence g ij is a covariant symmetric tensor of rank two. This is called fundamental Covariant Tensor. THEOREM 3.2 To show that g ij dx i dx j is an invariant. Proof: Let x i be coordinates of a point in X-coordinate system and x i be coordinates of a same point in Y-coordinate system. Since g ij is a Covariant tensor of rank two. Then,

g ij = g kl

∂x k ∂x 1 ∂x i ∂x j

34


∂x k ∂x l =0 ∂x i ∂x j

g ij − g kl

⇒

k l   gij − g kl ∂x ∂x  ∂x i ∂x j 

 i j dx dx = 0  

( g ij dx dx ) = g kl

∂x k ∂x l i j dx dx ∂x i ∂x j

= g kl

∂x k i ∂x l dx dx j ∂x i ∂x j

i

j

g ij dx i dx j = g kl dx k dx l So, g ij dx i dx j is an ivariant. 3.2 CONJUGATE METRIC TENSOR: (CONTRAVARIANT TENSOR) The conjugate Metric Tensor to g ij , which is written as g ij , is defined by Bij g ij = g (by Art.2.16, Chapter 2) where Bij is the cofactor of g ij in the determinant g = g ij ≠ 0 . By theorem on page 26 Aij Akj = δik g ij g kj = δik

So, Note

(i) (ii)

Tensors g ij and g ij are Metric Tensor or Fundamental Tensors. g ij is called first fundamental Tensor and g ij second fundamental Tensors.

EXAMPLE 1 Find the Metric and component of first and second fundamental tensor is cylindrical coordinates. Solution Let (x1, x2, x3) be the Cartesian coordinates and (x1, x2 , x3 ) be the cylindrical coordinates of a point. The cylindrical coordinates are given by x = r cosθ , y = r sin θ, z = z So that x1 = x, x 2 = y, x 3 = z and x 1 = r , x 2 = θ, x 3 = z

...(1)

Let g ij and g ij be the metric tensors in Cartesian coordinates and cylindrical coordinates respectively.


35

The metric in Cartesian coordinate is given by 2 2 2 ds 2 = dx + dy + dz

12 2 2 3 2 ds 2 = (dx ) + (dx ) +(dx )

...(2)

i j ds 2 = g ij dx dx

But

= g11(dx1 )2 + g12 dx1dx2 + g13 dx 1 dx 3 + g 21 dx 2 dx 1 + g22(dx2)2 + g 23 dx 2 dx 3 + g 31dx 3 dx 1 + g 32 dx 3 dx 2 + g33(dx3)3

...(3)

Comparing (2) and (3), we have g11 = g 22 = g 33 = 1 and g12 = g13 = g 21 = g 23 = g31 = g 32 = 0 On transformation g ij = g ij

∂x i ∂x j , since g ij is Covariant Tensor of rank two. (i, j = 1, 2, 3) ∂x i ∂x j

for i = j = 1. 2

g11

 ∂x1   ∂x 2   ∂x 3  = g11 1  + g 22  1  + g 33  1   ∂x   ∂x   ∂x 

2

since g12 = g13 = ⋅ ⋅ ⋅ = g32 = 0 . 2

2

 ∂x   ∂y   ∂z  g11 = g11  + g 22   + g 33    ∂r   ∂r   ∂r 

and g11 = g 22 = g 33

2

Since x = r cos θ, y = r sin θ, z = z ∂x ∂y ∂z = cosθ, = sin θ, =0 ∂r ∂r ∂r = 1. g11 = cos 2 θ + sin 2 θ + 0 g11 = 1

Put i = j = 2.  ∂x 1 g 22 = g11  2  ∂x

2

  2  + g 22  ∂x   ∂x 2   2

g 22

2

  3  + g 33  ∂x    ∂x 2     2

 ∂x   ∂y   ∂z   + g 22   + g 33   = g11   ∂θ   ∂θ   ∂θ 

2

2

36


Since g11 = g 22 = g 33 = 1 ∂z ∂x ∂y =0 = −r sin θ, = r cosθ, ∂θ ∂θ ∂θ g 22 = (− r sin θ )2 + (r cos θ)2 + 0 = r 2 sin 2 θ + r 2 cos 2 θ g 22 = r 2 Put i = j = 3. 2

g 33

 ∂x1   ∂x 2   ∂x 3        g + g + g = 11  3  22  33  3 3  ∂x   ∂x   ∂x  2

 ∂x   ∂y   ∂z  = g11   + g 22   + g 33    ∂z   ∂z   ∂z  Since

∂x ∂y ∂z = 0, = 0, = 1 . So, g 33 = 1 . ∂z ∂z ∂z

So,

g11 = 1, g 22 = r 2 , g 33 = 1

and

g12 = g13 = g 21 = g 23 = g31 = g 32 = 0

(i) The metric in cylindrical coordinates i j ds 2 = g ij dx dx i, j = 1, 2,3.

( )

1 ds 2 = g11 dx

2

( )

+ g 22 dx 2

2

( )

+ g 33 dx 3

since g12 = g13 = ⋅ ⋅ ⋅ = g32 = 0 2 2 2 2 ds 2 = dr + r (dθ ) + dφ

(ii) The first fundamental tensor is

g ij

since

 g11 g12 g13  1 0 0     2  =  g 21 g 22 g 23  = 0 r 0  g 31 g32 g 33  0 0 1   

1 g = g ij = 0 0 2 g= r

0 r2 0

0 0 1

2


37

(iii) The cofactor of g are given by B11 = r 2 , B22 = 1, B33 = r 2 and B12 = B21 = B13 = B 23 = B31 = B32 = 0 ij The second fundamental tensor or conjugate tensor is g =

and

g 11 =

cofactor of g11 in g g

g 11 =

B11 r 2 = 2 =1 g r

g 22 =

B12 1 = 2 g r

g 33 =

B33 r 2 = 2 =1 g r

Bij g

.

g 12 = g13 = g 21 = g 23 = g 31 = g 32 = 0

1  Hence the second fundamental tensor in matrix form is 0  0 

0 0 1  0 . r2  0 1

EXAMPLE 2 Find the matrix and component of first and second fundamental tensors in spherical coordinates. Solution Let ( x1 , x 2 , x 3 ) be the cartesian coordinates and ( x 1 , x 2 , x 3 ) be the spherical coordinates of a point. The spherical coordinates are given by x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ So that x1 = x, x 2 = y , x 3 = z and x 1 = r , x 2 = θ , x 3 = φ Let g ij and g ij be the metric tensors in cartesian and spherical coordinates respectively.. The metric in cartesian coordinates is given by 2 2 2 ds 2 = dx + dy + dz

( ) + (dx ) + (dx )

ds 2 = dx1 But

2

2 2

3 2

i j ds 2 = g ij dx dx ; (i, j = 1, 2, 3)

38


⇒ g11 = g 22 = g 33 = 1 and g12 = g 23 = g13 = g 21 = g31 = g 32 = 0 On transformation g ij = g ij

∂x i ∂x j ∂x i ∂x j

(since g ij is covariant tensor of rank two) (where i, j = 1,2,3). ∂x1 ∂x 1 ∂x 2 ∂x 2 ∂x 3 ∂x 3 + g + g 22 33 ∂x i ∂x j ∂x i ∂x j ∂x i ∂x 1 since i, j are dummy indices. Put i = j = 1 g ij = g11

2

2

g11

 ∂x 1   ∂x 2   ∂x 3      g + g + g = 11  1  22  33  1 1  ∂x   ∂x   ∂x

g11

 ∂x   ∂y   ∂z  = g11   + g 22   + g 33    ∂r   ∂r   ∂r 

2

2

   

2

2

Since x = r sin θ cos φ, y = r sin θ sin φ , z = r cos θ ∂x ∂z ∂y = cos θ = sin θ sin φ, = sin θ cos φ, ∂r ∂r ∂r and g11 = g 22 = g 33 = 1 . So, g11 = (sin θ cos φ )2 + (sin θ sin φ )2 + cos2 θ g11 = 1 put i = j = 2 g 22

 ∂x 1 = g11  2  ∂x

2

  2  + g 22  ∂x   ∂x 2   2

2

  3  + g 33  ∂x   ∂x 2  

 ∂x   ∂y   ∂z  g 22 = g11   + g 22   + g33    ∂θ   ∂θ   ∂θ 

   

2

2

since g11 = g 22 = g 33 = 1 ∂x ∂y ∂z = r cos θ cos φ, = r cos θ sin φ, = −r sin θ ∂θ ∂θ ∂θ g 22 = (r cos θ cos φ )2 + (r cos θ sin φ)2 + (− r sin θ)2 g 22 = r 2


39

Put i = j = 3 2

2

g 33

 ∂x1   ∂x 2   ∂x 3      g + g + g = 11  3  22  33  3 3  ∂x   ∂x   ∂x

g 33

 ∂x   ∂y   ∂z  = g11   + g 22   + g33    ∂φ   ∂φ   ∂φ 

2

2

   

2

2

since g11 = g 22 = g 33 = 1 ∂x ∂y ∂z = −r sin θ sin φ, = r sin θ cos φ, =0 ∂φ ∂φ ∂φ

and

g 33 = (− r sin θ sin φ)2 + (r sin θ cos φ )2 + 0 g 33 = r 2 sin 2 θ So, we have g11 = 1, g 22 = r 2 , g 33 = r 2 sin 2 θ g12 = g13 = g 21 = g 23 = g31 = g 32 = 0

and

(i) The Metric in spherical coordinates is i j ds 2 = g ij dx dx ; i, j = 1, 2, 3

( )

1 ds 2 = g11 dx

2

( )

+ g 22 dx 2

2

( )

+ g 33 dx 3

2

ds 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 (ii) The Metric tensor or first fundamental tensor is

g ij and

0   g11 g12 g13  1 0   g g g  = 0 r 2 0 =  21 22 23     g 31 g32 g 33  0 0 r 2 sin 2 θ  

g = g ij

1 =0 0

0 r2 0

0 0 = r 4 sin 2 θ r 2 sin 2 θ

(iii) The cofactor of g are given by B11 = 1, B22 = r 2 , B33 = r 2 sin 2 θ and B12 = B21 = B31 = B13 = B23 = B32 = 0 The second fundamental tensor or conjugate tensor is g ij =

Bij g

.

40


g 11 =

cofactor of g11 in g B11 = g g

r 4 sin 2 θ r 4 sin 2 θ =1 =

g 11 g

22

B 22 r 2 sin 2 θ = 4 2 = g r sin θ

g 22 = g

33

1 r2

B33 r2 = = g r 4 sin 2 θ

g 33 =

1 r sin 2 θ 2

12 13 21 31 32 and g = g = g = g = g = 0 Hence the fundamental tensor in matrix form is

g  21 = g  g 31  11

g

ij

g

12

g 22 g 32

 0 g   0 23  g  =   g 33  0  

0 1 r2

13

0

   0   1  r 2 sin 2 θ   0

EXAMPLE 3 If the metric is given by

( ) ( ) 2

ds 2 = 5 dx 1 + 3 dx 2

2

( )

+ 4 dx 3

2

− 6dx 1dx 2 + 4dx 2 dx 3

Evaluate (i) g and (ii) g ij . Solution i j The metric is ds 2 = g ij dx dx ;

(i, j = 1, 2, 3)

( )

1 ds 2 = g11 dx

2

+ g12 dx 1dx 2 + g13 dx1 dx 3 + g 21dx 2 dx1

+ g 22 (dx 2 ) 2 + g 23 dx 2 dx 3 + g31 dx 3 dx1 + g 32 dx 3 dx 2 + g 33 (dx 3 )2 Since g ij is symmetric ⇒ g ij = g ji i.e.,

g12 = g 21, g 23 = g32 , g13 = g 31


So,

41

1 2 2 2 3 2 1 2 ds 2 = g11 (dx ) + g 22 (dx ) + g33 (dx ) + 2 g12 dx dx

+ 2 g 23 dx 2 dx 3 + 2 g13 dx1 dx 3

...(1)

Now, the given metric is 1 2 2 2 3 2 1 2 2 3 ds 2 = 5(dx ) + 3(dx ) + 4(dx ) − 6dx dx + 4dx dx Comparing (1) and (2) we have

g11 = 5, g 22 = 3, g 33 = 4, 2g12 = −6 ⇒ g12 = −3 = g 21 2 g 23 = 4 ⇒ g 23 = 2 = g 32 , g13 = 0 = g 31 5 −3 0

g11 g12 g13

g = g ij = g 21 g 22 g 23 = − 3 3 2 = 4 g 31 g 32 g33 0 2 4 (ii) Let Bij be the cofactor of g ij in g. Then B11 = Cofactor of g11 =

3 2 =8 2 4

B22 = Cofactor of g 22 =

5 0 = 20 04

B33 = Cofactor of g 33 =

5 −3 =6 −3 3 −3 2 = 12 = B 21 0 4

B12 = Cofactor of g12 = − B13 = Cofactor of g13 =

B 23

Since

−3 3 = −6 = B31 0 2

= Cofactor of g 23 = –

g ij =

5 −3 0

2

= −10 = B 32

Bij g

We have B 8 g 11 = 11 = = 2; g 22 = 5, g 33 = 3 , g 12 = g 21 = 3, g 13 = g 31 = − 3 , g 23 = g 32 = − 5 g 4 2 2 2

...(2)

42


Hence,

g ij

   =   − 

3 3 −  2 5 3 5 −  2  3 5 3 − 2 2 2  2

3.3 LENGTH OF A CURVE Consider a continuous curve in a Riemannian Vn i.e., a curve such that the coordinate x i of any current point on it are expressible as functions of some parameter, say t. The equation of such curve can be expressed as x i = xi (t) The length ds of the arc between the points whose coordinate s are x i and x i + dx i given by i j ds 2 = g ij dx dx

If s be arc length of the curve between the points P1 and P2 on the curve which correspond to the two values t1 and t 2 of the parameter t. s =

∫

P2 P1

ds =

∫

t2 

t1

12

i j   gij dx dx  dt  dt dt  

NULL CURVE dx i dx j = 0 along a curve. Then s = 0. Then the points P1 and P2 are at zero distance, despite dt dt of the fact that they are not coincident. Such a curve is called minimal curve or null curve. If g ij

EXAMPLE 4 A curve is in spherical coordinate xi is given by

Find length of arc 1 ≤ t ≤ 2.

2 −1  1  3 2 x1 = t, x = sin   and x = 2 t − 1 t  

Solution In spherical coordinate, the metric is given by 1 2 1 2 2 2 1 2 2 3 2 ds 2 = (dx ) + ( x ) (dx ) + ( x sin x ) (dx )


43

2 −1 1 , x3 = 2 t 2 − 1 x1 = t, x = sin t

given

dx 2 = −

dt t t −1 2

(

2

, dx 3 =

2t t 2 −1

dt

2

2   2t  dt    −1 1   +t − + t sin sin dt     t t 2 − 1    t   t 2 − 1   

ds = (dt )

2

2

2

2

)

−1 2  1 1 2t dt  − 2  dt , dx 3 = 2 ⋅ t 2 − 1 2  1  t  1−  t

1

dx 1 = dt , dx 2 =

dt 2 4t 2 2 dt + + (dt )2 = ds 2 2 t −1 t −1 2

5t 2 2 ds 2 = t 2 − 1 dt ds =

t

5

t −1 2

dt

Now, the length of arc, 1 ≤ t ≤ 2, is 2

∫

t2

t1

3.4

ds =

5

∫

2

1

5  t2 −1   = dt = 2  12  t2 −1  1 t

15 units

ASSOCIATED TENSOR

i i ... i A tensor obtained by the process of inner product of any tensor A j11 2j 2 ... rj s with either of the fundamental

tensor g ij or g ij is called associated tensor of given tensor.. e.g. Consider a tensor Aijk and form the following inner product g αi Aijk = Aαjk ; g αj Aijk = Aikα ; g αk Aijk = Aijα All these tensors are called Associated tensor of Aijk . Associated Vector Consider a covariant vector Ai . Then g ik Ai = A k is called associated vector of Ai . Consider a contravariant vector A j . Then g jk A j = Ak is called associated vector of A j .

44


3.5 MAGNITUDE OF VECTOR i The magnitude or length A of contravariant vector A . Then A is defined by

A= or

g ij Ai A j

i j A 2 = g ij A A

i Also, A2 = A j A j as g ij A = A j i.e., square of the magnitude is equal to scalar product of the vector and its associate.

The magnitude or length A of covariant vector Ai . Then A is defined by A=

g ij Ai A j

A 2 = g ij Ai A j or A vector of magnitude one is called Unit vector. A vector of magnitude zero is called zero vector or Null vector. 3.6 rSCALAR r r r PRODUCT OF TWO VECTORS Let A and B be two vectors. Their scalar product is written as A ⋅ B and defined by r r A ⋅ B = Ai Bi r r i i j j Also, A ⋅ B = A B i = g ij A B since Bi = g ij B r r i ij i ij A ⋅ B = Ai B = g Ai B j since B = g B j Thus

i.e.,

r r i i j 2 A ⋅ A = A Ai = g ij A A = A r A = A = g ij Ai A j

Angle two vectors r between r Let A and B be two vectors. Then r r r r A ⋅ B = A B cos θ r r g ij A i B j A⋅B cos θ = r r = ⇒ AB g ij A i A j g ij B i B j since

r A =

r g ij Ai A j ; B =

g ij B i B j

This is required formula for cos θ . Definition

r r The inner product of two contravariant vectors A (or A i ) and B (or Bi ) associated with a symmetric tensor g ij is defined as g ij A i B j . It is denoted by r r i j g ( A , B ) = g ij A B


45

r r THEOREM 3.3 The necessary and sufficient condition that the two vectors A and B at 0 be orthogonal r r if g ( A, B ) = 0 r r Proof: Let θ be angle between the vectors A and B then r r r r A ⋅ B = A B cos θ r r or A ⋅ B = AB cos θ g ij A i B j = AB cos θ ⇒

cos θ =

g ij Ai B j

...(1)

AB

r r If A and B are orthogonal then θ = π ⇒ cos θ = 0 then from (1) 2 i j g ij A B = 0 ⇒

r r r r i j g A, B = 0 since g A, B = g ij A B

( )

( )

i j Conversely if g ij A B = 0 then from (1)

cos θ = 0 ⇒ θ = π . 2 r r So, two vectors A & B are orthogonal.

r r r r Note: (i) If A and B be unit vectors. Then A = B = 1. Then

Proved.

r r cos θ = A ⋅ B = gij A i B j

r r π π (ii) Two vectors A and B will be orthogonal if angle between them is i.e., θ = then 2 2 π cos θ = cos θ = = 0 2

3.7 ANGLE BETWEEN TWO VECTORS THEOREM 3.4 To show that the definition of the angle between two vectors is consistent with the requirement cos 2θ ≤ 1. OR To justify the definition of the angle between two vectors. OR To show that the angle between the contravariant vectors is real when the Riemannian Metric is positive definition. r r Proof: Let θ be the angle between unit vectors A and B then i j j ij ij i cos θ = g ij A B = A j B = A j B Bi = g A j Bi = A Bi

46


To show that θ is real i.e., |cosθ| ≤ 1. Consider the vector lAi + mB i when l and m are scalars. The square of the magnitude of i i j j lAi + mB i = g ij (lA + mB ) (lA + mB ) 2 i j i j i j 2 i j = g ij l A A + g ij lmA B + m lg ij B A + m g ij B B

= l 2 + 2lm cos θ + m 2 Since i j 2 g ij A i A j = A 2 = 1; g ij B B = B = 1.

and

r r r g ij A i B j = cosθ; as A & B are unit vector i.e., A = 1 ⇒ A 2 = 1 .

Since square of magnitude of any vector ≥ 0. So, the square of the magnitude of lAi + mB i ≥ 0. or l 2 + 2lm cos θ + m 2 ≥ 0

(l + m cos θ )2 + m 2 − m 2 cos 2 θ

≥0

(l + m cosθ) 2 + m 2 (1 − cos2 θ) ≥ 0 This inequality holds for the real values of l & m. if 1 − cos2 θ ≥ 0 ⇒

cos 2 θ ≤ 1 cos θ ≤ 1 Proved.

THEOREM 3.5 The magnitude of two associated vectors are equal. Proof: Let A and B be magnitudes of associate vectors A i and A i respectively. Then i j A 2 = g ij A A

...(1)

ij B 2 = g Ai A j

...(2)

and From equation (1) A 2 = (gij Ai ) A j A2 = Aj A j

...(3)

since g ij A i = A j (Associate vector) From equation (2) ij B 2 = (g Ai )A j j B2 = A Aj

...(4)


47

since g ij Ai = A j from (3) and (4) 2 A2 = B A=B ⇒ i So, magnitude of Ai and A are equal.

3.8 ANGLE BETWEEN TWO COORDINATE CURVES Let a Vn referred to coordinate x i , (i = 1, 2, ... n) . For a coordinate curve of parameter x l, the coordinate x l alone varies. Thus the coordinate curve of parameter x l is defined as x i = c i , ∀i except i = l ...(1) i, C s where are constants. Differentiating it, we get dx i = 0, ∀i , except i = l and dx l ≠ 0 Let Ai and B i be the tangent vectors to a coordinate curve of parameters x p and x q respectively.. Then Ai = dx i = (0,...0, x p , 0...0)

...(2)

B i = dx i = (0,...0, x q , 0...0)

...(3)

If θ is required angle then cos θ =

=

=

cos θ =

g ij A i B j g ij A i A j

g ij B i B j

g pq A p B q g pp A p A p g qq B q B q g pq A p B q g pp g qq A p B q g pq g pp g qq

which is required formula for θ . The angle wij between the coordinate curves of parameters x i and x j is given by cos wij =

g ij g ii g jj

...(4)

48


If these curves are orthogonal then cos wij = cos ⇒

π =0 2

g ij = 0

Hence the x i coordinate curve and x j coordinate curve are orthogonal if g ij = 0 . 3.9 HYPERSURFACE The n equations xi = xi (u1) represent a subspace of Vn . If we eliminate the parameter u1, we get (n –1) equations in x j, s which represent one dimensional curve. Similarly the n equations xi = xi (u1,u2) represent two dimensional subspace of V n. If we eliminating the parameters u1, u2, we get n –2 equations in xi,s which represent two dimensional curve V n. This two dimensional curve define a subspace, denoted by V 2 of V n. Then n equations xi = xi (u1, u2, ... un–1) represent n – 1 dimensional subspace V n–1 of V n. If we eliminating the parameters u1, u2, ...un–1, we get only one equation in x i, s which represent n –1 dimensional curve in V n. This particular curve is called hypersurface of V n. i

Let φ be a scalar function of coordinates x i . Then φ (x ) = constant determines a family of hypersurface of V n. 3.10 ANGLE BETWEEN TWO COORDINATE HYPERSURFACE Let

and

φ (x i ) = constant

...(1)

ψ( x i ) = constant

...(2)

represents two families of hypersurfaces. Differentiating equation (1), we get ∂φ i dx = 0 ∂x i

...(3)

∂φ ∂φ is orthogonal to dx i . Hence is normal to φ = constant, since dx i is i ∂x ∂x i tangential to hypersurface (1). This shows that

∂ψ is normal to the hypersurface (2). If ω is the angle between the hypersurface (1) ∂x i and (2) then ω is also defined as the angle between their respective normals. Hence required angle ω is given by Similarly

∂φ ∂ψ ∂x i ∂x j ∂φ ∂φ ∂ψ ∂ψ g ij i g ij i j ∂x ∂x ∂x ∂x j g ij

cos ω =

...(4)


49

If we take

and

φ = x p = constant

...(5)

ψ = x q = constant The angle ω between (5) and (6) is given by

...(6)

∂x p ∂x q ∂x i ∂x j q q ∂x p ∂x p ij ∂x ∂x g ij i g ∂x ∂x j ∂x i ∂x j g ij

cos ω =

g ij δip δ qj

=

g ij δip δ qj g ij δqi δ qj g pq

cos ω =

g pp g qq

...(7)

The angle ωij between the coordinate hypersurfaces of parameters x i and x j is given by g ij

cos ωij =

g ii g jj

...(8)

If the coordinate hypersurfaces of parameters x i and x j are orthogonal then ωij =

π 2

cos ωij = 0

⇒

from (8), we have g = 0 . ij

Hence the coordinate hypersurfaces of parameters x i and x j are orthogonal if g ij = 0 . 3.11 n-PLY ORTHOGONAL SYSTEM OF HYPERSURFACES If in a Vn there are n families of hypersurfaces such that, at every point, each hypersurface is orthogonal to the n − 1 hypersurface of the other families which pass through that point, they are said to form as n-ply orthogonal system of hypersurfaces. 3.12 CONGRUENCE OF CURVES A family of curves one of which passes through each point of Vn is called a congruence of curves. 3.13 ORTHOGONAL ENNUPLE An orthogonal ennuple in a Riemannian Vn consists of n mutually orthogonal congruence of curves.

50


THEOREM 3.6 To find the fundamental tensors g ij and g ij in terms of the components of the unit tangent e h ( h = 1, 2,... n ) to an orthogonal ennuple. Proof: Consider n unit tangents e ih (h = 1, 2,...n) to conguence e h ( h = 1, 2,... n ) of an orthogonal ennuple in a Riemannian Vn . The subscript h followed by an upright bar simply distinguishes one congruence from other. It does not denote tensor suffix. The contravariant and covariant components of eh | are denoted by eh | and eh |i respectively.. Suppose any two congruences of orthogonal ennuple are eh | and ek | so that g ij ehi |ekj| = δ hk

...(1)

ehi |ek |i = δ hk from (1), g ij ehi | ekj| = 0 g ij ehi | ehj| = 1

and We define

ehi | =

cofactorof eh |i in determinant eh |i e h|i

Also, from the determinant property, we get n

∑e e

i h| h | j

h =1

= δij

...(2)

Multiplying by e jk n

∑e e

i h| h| j

h =1

g j k = δ ij g jk

n

∑e e

i k h | h|

or

h =1

= g ik

...(3)

Again multiplying (2) by g ik . n

∑e e

i h | h | j g ik

h =1

or

i = δ j g ik

g jk =

∑e

h |k eh | j

...(4)

from (3) and (4) n

g ij =

∑e

h |i

h =1

e h| j

...(5)


51

n

g ij =

∑e

i h|

e hj|

...(6)

h =1

This is the required results. Corollary: To find the magnitude of any vector u is zero if the projections of u on eh | are all zero. Proof: Let n

i

u =

∑C e

i h h|

...(7)

h =1

Then n

u i ek |i =

∑C e e

i h h | k |i

=

h =1

or

n

∑C δ h

h k

= Ck

h =1

C k = u i ek |i

...(8)

C k = projection of u i on ek |i i.e., Using (8), equation (7) becomes n

ui =

∑u e j

i h| j ek |

h =1

Now,  u 2 = u iu i =    =



∑ C e   ∑ C e i h h|

h



k k |i

k

  from (7)  

∑C C e e

i k h | k| i

h

h ,k

=

∑C C δ h

k

h k

h ,k

=

∑C C h

h

∑ (C

)2

h n

u2 =

h

h =1

This implies that u = 0 iff u 2 = 0 iff C h = 0 . Hence the magnitude of a vector u is zero iff all the projections of u (i.e. of ui) on n mutually orthogonal directions ehi | are zero.

52


Miscellaneous Examples 1. If p and q are orthogonal unit vectors, show that ( g hj g ik − g hk gij ) p h q i p j q k = 1 Solution Since p and q are orthogonal unit vectors. Then g ij pi q j = 0, p 2 = q 2 = 1 . Now, h j i k h k i j ( g hj g ik − g hk gij ) p h q i p j q k = g hj g ik p p q q − g hk g ij p q q p

= ( g hi p h p j ) ( gik q i q k ) − (g hk p j q k ) ( gij q i p j ) = p2.q2 – 0.0 =1.1 = 1 (since g hi p h p j = 1 & g hk p h q k = 0 ) 2. If θ is the inclination of two vectors A and B show that sin θ = 2

(g hi g ik − g hk g ij ) A h A j B i B k g hj gik A h A j B j B k

Solution If θ be the angle between the vectors A and B then g ij A j B i

cos θ =

gij A i A j g ik B i B k

But sin 2 θ = 1 − cos 2 θ

sin 2 θ = 1 −

=

( gij B i A j ) (g hk A h B k ) ( g hj A h A j ) ( g ik B i B k )

(g hj g ik − g hk g ij ) A h A j B i B k g hj g ik A h A j B i B k

3. If X ij are components of a symmetric covariant tensor and u, v are unit orthogonal to w and satisfying the relations ( X ij − αg ij )u i + γw j = 0 ( X ij − βg ij ) v i + δw j = 0 where α ≠ β prove that u and v are orthogonal and that


53

X ij u i v j = 0 Solution Suppose X ij is a symmetric tensor. Since u i , v j are orthogonal to wi then u i wi = 0

...(1)

v i wi = 0

...(2)

( X ij − αgij )u i + γw j = 0

...(3)

( X ij − βgij )vi + δw j = 0 where α ≠ β.

...(4)

given

Multiply (3) & (4) by v j , u j respectively and using (1) and (2), we have ( X ij − αg ij )u iv j = 0

...(5)

( X ij − βg ij )v iu j = 0

...(6)

Interchanging the suffixes i & j in the equation (6) and since g ij , X ij are symmetric, we get ( X ij − αg ij )u iv j = 0 Subtract (6) & (7) we get

... (7)

(β − α)g iju i v j = 0 Since β ≠ α and β − α ≠ 0. Hence, g ij u i v j = 0 So, u and v are orthogonal. Using (8) in equation (5) & (6), we get

...(8)

X ij u i v i = 0

Proved.

4. Prove the invariance of the expression

g dx1 dx 2 ...dx n for the element volume.

Solution Since g ij is a symmetric tensor of rank two. Then g ij =

∂x k ∂x l g kl ∂x i ∂x j

Taking determinant of both sides g ij Since

∂x k = ∂x i

∂x l ∂x j

g kl

∂x = J (Jacobian) ∂x g kl = g

&

g ij = g

54


So, g = gJ 2 or J=

g g

Now, the transformation of coordinates from x l to x i , we get ∂x 1 2 n dx 1 dx 2 ... dx n = ∂x d x d x ... dx = Jd x 1d x 2 ... d x n dx 1dx 2 ... dx n =

g 1 2 d x d x ... d x n g

g dx1 dx2 ... dx n =

g d x1d x 2... d x n

So, the volume element dv = g dx1 dx 2 ... dx n is invariant.

EXERCISES 1. For the Metric tensor g ij defined g kl and prove that it is a contravariant tensor. 2. Calculate the quantities g i j for a V3 whose fundamental form in coordinates u, v, w, is 2

2

2

adu + bdv + cdw + 2 fdvdw + 2 gdwdu + 2 hdudv 3. Show that for an orthogonal coordinate system g 11 =

1 1 1 , g 33 = , g 22 = g11 g 22 g 33

4. For a V2 in which g11 = E , g12 = F , g 21 = G prove that g = EG − F 2 , g 11 = G g , g 12 = − F g , g 22 = E g 1 n( n + 1) . 2 6. If vectors u i , vi are defined by u i = g ij u j , v i = g ij v j show that u i = g ij u j , u i vi = u i v i and u i g ij u j = u i g ij u j 7. Define magnitude of a unit vector. prove that the relation of a vector and its associate vector is reciprocal. 5. Prove that the number of independent components of the metric g ij cannot exceed

8. If θ is the angle between the two vectors Ai and B i at a point, prove that 2

sin θ =

(g hi gik − g hk gij ) A h Ai B j B k ghi g jk Ah Ai B j B k

9. Show that the angle between two contravariant vectors is real when the Riemannian metric is positive definite.

CHAPTER – 4

CHRISTOFFEL'S SYMBOLS AND COVARIANT DIFFERENTIATION 4.1 CHRISTOFFEL'S SYMBOLS The German Mathematician Elwin Bruno Christoffel defined symbols

[ij, k ] =

1  ∂gi k ∂g j k ∂g i j   + − k  , (i, j, k = 1, 2,...n ) 2  ∂x j ∂x i ∂x  called Christoffel 3-index symbols of the first kind.

...(1)

k   = g k l [ij, l ] i j 

...(2)

and

called Christoffel 3-index symbols of second kind, where g i j are the components of the metric Tensor or fundamental Tensor. There are n distinct Christoffel symbols of each kind for each independent g i j . Since g i j is 1 n (n + 1) independent components. So, the number of 2 1 1 independent components of Christoffel’s symbols are n ⋅ n(n + 1) = n 2 (n + 1) . 2 2 symmetric tensor of rank two and has

k THEOREM 4.1 The Christoffel's symbols [ij, k ] and   are symmetric with respect to the indices i i j  and j. Proof: By Christoffel’s symbols of first kind

[ij, k ] =

∂g ij  1  ∂g ik ∂g jk  j + ∂ − k  i 2  ∂x ∂x ∂x 

Interchanging i and j, we get

[ ji, k ] =

1  ∂g jk ∂gik ∂g ji   + − k  2  ∂x i ∂x j ∂x 

56


=

1  ∂g i k ∂g j k ∂g i j   + − k  since g ij = g ji 2  ∂x j ∂x i ∂x 

[ ji, k ] = [ ij , k ] Also, by Christoffel symbol of second kind  k   = g k l [ij, l ] i j = g k l [ ji, l ] since [ij , l ] = [ ji, l ] k k   =  j i i j   

Proved.

THEOREM 4.2 To prove that (i)

[ij, m] =

(ii)

[ik , j ] + [ jk , i] =

 k  g km   i j 

∂g ij ∂x k

∂g ij

 i   j  = − g jl   − g im   l k ∂x   m k  Proof: (i) By Christoffel’s symbol of second kind (iii)

k

k   = g k l [ij, l ] i j  Multiplying this equation by g k m , we get  k  g k m   = g k m g k l [ij, l ] i j 

= δ lm [ij , l ] as g k m g k l = δ lm

k g km   = [ij, m] i j  (ii) By Christoffel’s symbol of first kind

and

[ik , j ] =

1  ∂g k j ∂g i j ∂g i k    + − 2  ∂x i ∂x k ∂x j 

...(1)

[ jk ,i ] =

1  ∂g ki ∂g ji ∂g jk  + − 2  ∂x j ∂x k ∂x i

...(2)

  

adding (1) and (2),

[ik , j ] + [ jk , i ] =

1  ∂ g i j ∂g ji    + g = g ji 2  ∂ x k ∂x k  since ij

Christoffel's Symbols and Covariant Differentiation

[ik , j ] + [ jk , i] =

57

1 ∂g ij ∂g ij ⋅2 = 2 ∂x k ∂x k

(iii) Since g ij g lj = δ li . Differentiating it w.r.t. to x k , we get g ij

∂glj ∂x k

+ glj

∂g ij =0 ∂x k

Multiplying this equation by g lm , we get g ij g lm

∂g lj ∂x k

+ g lm g lj g lm glj δ mj

∂g ij =0 ∂x k

∂g ∂g ij − g ij g lm ljk k = ∂x ∂x ∂g lj ∂g ij ij lm − g g { [ lk , j ] + [ jk , l ] } = [lk , j ] + [ jk , l ] . = since ∂x k ∂x k

{

}

{

∂g im lm ij ij lm = − g g [lk , j ] − g g [ jk , l ] ∂x k

}

∂g im  i   m = − g lm   − g ij   k ∂x l k   j k Interchanging m and j, we get  i   j  ∂g ij lj  im  − g − g =     ∂x k  l k   m k 

or

∂g ij  i   j  − g ij   − g im   as g lj = g jl k = ∂x l k  m k  Proved.

( )

 i  ∂ log g THEOREM 4.3 To show that i j  =   ∂x j Proof: The matrix form of g ik is

g ik

 g11 g12 ... g1n  g  21 g 22 ... G2 n   =  M     g n1 g n 2 ... g nn 

58


and

g = gik

g11 g 21 = M

g12 ... g1n g 22 ... g 2 n

g n1

g n 2 ... gnn

But

g ik g = δlk

Take

l = k.

il

g ik g ik = δ kk = 1 ⇒

g ik = [g ik ]−1 =

Gik (Theorem 2.13 , Pg 25) g

where Gik is cofactor of g ik in the determinant g ik ⇒

g = g ik Gik

Differentiating w.r.t. g ik partially ∂g ∂g ik =1 = Gik since ∂g ik ∂g ik Now, ∂g ∂g ∂g ik j = ∂ g ik ∂x j ∂x = Gik

∂gik ∂x j

But Gik = gg ik ∂g ik ∂gik = gg j ∂x j ∂x ∂g 1 ∂g g ik ikj j = g ∂x ∂x 1 ∂g ∂g ik = g ik { [ jk , i] + [ij, k ] } as = [ jk , i] + [ij, k ] g ∂x j ∂x j ik ik = g [ jk , i] + g [ij , k ]

 k  i =  +   j k  i j 

...(1)


59

 i   i  1 ∂g  +   as k is dummy indices j =  g ∂x  j i  j i  i  1 ∂g = 2  j g ∂x  j i  i  1 ∂g =   j 2 g ∂x  j i ∂ log( g ) ∂x

j

 i    j i 

= 

Proved.

EXAMPLE 1 If g ij ≠ 0 show that

g αβ

∂ ∂x j

∂  β   β   = ∂x j [ik , α] − i k    i k 

( [βj, α] + [αj , β] )

Solution By Christoffel’s symbol of second kind  β βα   = g [ik , α] i k   Multiplying it by g αβ , we get β  βα g αβ   = g αβ g [ik , α ] i k    β  βα g αβ   = [ik , α ] as g αβ g = 1 i k 

Differentiating it w.r.t. to x j partially ∂  β  ∂  [ik , α] g = j  αβ   ∂x j ∂x  i k  g αβ

since g αβ

∂  β   + ∂ x j i k 

 β  ∂g α β ∂ ik,α   j =  i k  ∂ x ∂x j

[

∂g αβ ∂x j

]

= [αj, β] + [βj, α]

∂  β  β  ∂  +   ([αj, β] + [βj , α]) = [ik , α] j  i k i k ∂x     ∂x j

60


g αβ

∂ ∂x j

∂ β   β   = ∂x j [ik , α] − i k  ( [αi, β] + [βj, α ] )   i k 

Solved.

EXAMPLE 2  k Show that if g ij = 0 for i ≠ j then (i)   = 0 whenever i, j and k are distinct. i j   i  1 ∂ log gii  i  1 ∂ log g ii  i  1 ∂g jj = = − (ii)   = (iii) (iv)     i j 2 gii ∂x i i i  2 ∂x i j  2 ∂x  j j Solution The Christoffel’s symbols of first kind

[ij, k ] = (a) If i = j = k The equation (1) becomes

[ii, i] = (b) If i = j ≠ k The equation (1) becomes

[ii, k ] =

1  ∂g jk ∂g ik ∂g ij  + − 2  ∂x i ∂x j ∂x k

  

1 ∂gii 2 ∂x i

1  ∂g i k ∂gi k ∂g i i  + i – k 2  ∂x i ∂x ∂x 

Since g ik = 0 as i ≠ k (given) 1 ∂g ii 1 ∂g jj or [ jj, i] = − k 2 ∂x 2 ∂xi

[ii, k ] =

−

[ij,i ] =

1  ∂g ji ∂g ii ∂g ij  + j − i  2  ∂x i ∂x ∂x 

(c) i = k ≠ j

=

1 ∂g ii , as g ij = 0, i ≠ j 2 ∂x j

(d) i ≠ j ≠ k

[ij, k ] = 0

as g ij = 0 , g j k = 0 , i ≠ j ≠ k

(i) as i, j, k are distinct i.e., i ≠ j ≠ k  k    = g kl [ij, l ] since g k l = 0, k ≠ l i j 

...(1)


61

k    =0 i j  i   = g ii [ii, i ] i i 

(ii)

= g ii ⋅ =

1 ∂gii from (a) 2 ∂x i

1 1 ∂g ii g ii = i as gii 2 gii ∂x

i 1 ∂ log g ii   = 2 ∂x i i i  (iii)

i=k≠ j i  ii   = g [ij, i] i j  1 1 ii = g [ij, i] as g = ii gii  i  1 ∂ log gii   = 2 ∂x j i j 

(iv)

j=k≠i  i  ii   = g [ jj, i] j j   1 1 ∂g j j = − from (b) g ii 2 ∂x i − 1 ∂g j j  i    = 2g i i i ∂x  j j

EXAMPLE 3 If ds 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 , find the values of  1  3  [22, 1] and [13, 3], (ii)   and   2 2 1 3 Solution (i)

The given metric is metric in spherical coordinates, x1 = r , x 2 = θ, x 3 = φ. Clearly, g11 = 1, g 22 = r 2 , g 33 = r 2 sin 2 θ and gi j = 0 for i ≠ j

Solved.

62


Also,

22 g 11 = 1, g =

1 1 g 33 = , 2 r sin 2 θ r2 (See Ex. 2, Pg. 39, and g ij = 0, for i ≠ j. )

(i) Christoffel Symbols of first kind are given by

[ij, k ] =

1  ∂g jk ∂g ik ∂g ij  + −  , i, j, k = 1, 2,3 2  ∂x i ∂x j ∂x k 

Taking i = j = 2 and k = 1 in (1)

[22, 1] = =

1  ∂g 21 ∂g 21 ∂g 22  + 2 − 1  Since g21 = 0. 2  ∂x 2 ∂x ∂x  1  ∂0 ∂0 ∂r 2   2 + 2 − 1 2  ∂x ∂x ∂x 

= −

1 ∂r 2 = −r 2 ∂r

Taking i = 1, j = k = 3 in (1)

[13,3] = =

1  ∂g33 ∂g13 ∂g13  + 3 − 3 2  ∂x1 ∂x ∂x  1 ∂r 2 sin 2 θ since g13 = 0 2 ∂r

[13, 3] =

r sin 2 θ (ii) Christoffel symbols of the second kind are given by k   = g kl [ij, l] = g k 1 [ij ,1] + g k 2 [ ij ,2] + g k 3 [ ij,3] i j  Taking k = 1, i = j = 2.  1    = g 11 [22, 1] + g 12 [22, 2 ] + g13 [22, 3] 2 2    1    = 1[22, 1] + 0[22, 2 ]+ 0[22, 3] Since g 12 = g 13 = 0 2 2    1    = −r 2 2

...(1)


63

and 3  31 32 33   = g [13, 1] + g [13, 2] + g [13, 3] 1 3   =

1 [13, 3] Since g 31 = g 32 = 0 2 r sin θ 2

 3  1 1 2   = 2 2 ⋅ r sin θ = 1 3 r r sin θ   4.2 TRANSFORMATION OF CHRISTOFFEL'S SYMBOLS The fundamental tensors g ij and g ij are functions of coordinates x i and [ij, k ] is also function of coordinates x i . Let g ij , g − ij and [ ij , k ] in another coordinate system x i . (i ) Law of Transformation of Christoffel's Symbol for First Kind Let [i j , k ] is a function of coordinate x i and [ ij , k ] in another coordinate system x i . Then [ ij , k ] =

1  ∂gik ∂g ik ∂g ij  + −   2  ∂x i ∂x j ∂x k 

...(1)

Since g ij is a covariant tensor of rank two. Then g ij =

∂x p ∂x 2 g pq ∂x i ∂x j

...(2)

Differentiating it w.r.t. to x k , we get ∂g ij ∂x k

=

∂ ∂x k

 ∂x p ∂x q     ∂x i ∂x j g pq   

=

∂ ∂x k

 ∂x p ∂x q   ∂x i ∂x j 

p q ∂g  pq  g pq + ∂x ∂x  i j ∂x ∂x ∂x k 

∂g i j

 ∂ 2 x p ∂x q ∂x p ∂ 2 x q  =  ∂x k ∂x i ∂x j + ∂x i ∂x k ∂x j ∂xk 

p q ∂g r  pq ∂x  g pq + ∂x ∂x  ∂x i ∂x j ∂x r ∂x k 

...(3)

Interchanging i, k and also interchanging p, r in the last term in equation (3) ∂g kj ∂x i

 ∂ 2 x p ∂x q ∂x p ∂ 2 x q + k =  i k j ∂x ∂x i ∂x j  ∂x ∂x ∂x

r q ∂g p  rq ∂x  g pq + ∂x ∂x  ∂x k ∂x j ∂x p ∂x i 

...(4)

64


and interchanging j, k and also interchange q, r in the last term of equation (3)  ∂ 2 x p ∂x q ∂x p ∂ 2 x q ∂ g ik + i =  j i i ∂x j ∂x ∂x k ∂x j  ∂x ∂x ∂x

p r q ∂g  pr  g pq + ∂x ∂x ∂x i k j q  ∂ x ∂ x ∂ x ∂ x 

...(5)

Substituting the values of equations (3), (4) and (5) in equation (1), we get [ij, k ] =

1  ∂ 2 x p ∂x q ∂x p ∂x q ∂x r  2 i j k g pq + i 2  ∂x ∂x ∂x ∂x ∂x j ∂x k

[ij, k ] =

∂ 2 x p ∂x q ∂x p ∂x q ∂x r 1  ∂g rp ∂g qr ∂g pq  g + + p − pq ∂x i ∂x j ∂x k ∂x i ∂x i ∂x k 2  ∂x q ∂x ∂x r

 ∂g rp ∂g qr ∂g pq  q + − ∂x p ∂x r  ∂x

  

  

∂ 2 x p ∂x q ∂x p ∂x q ∂x r g + [ pq, r] ...(6) pq ∂x i ∂x j ∂x k ∂x i ∂x i ∂x k It is law of transformation of Christoffel's symbol of the first kind. But it is not the law of transformation of any tensor due to presence of the first term of equation (6). So, Christoffel's symbol of first kind is not a tensor. [ij, k ] =

(ii) Law of Transformation of Christoffel's Symbol of the Second Kind  k  kl kl g [ ij , l ] =   is function of coordinates x i and g [ij, l ] = Let i j  i x . Then

 k    in another coordinate system i j 

∂ 2 x p ∂x q ∂x p ∂x q ∂x r g pq + i [ pq, r ] from (6) i j l ∂x ∂x ∂x ∂x ∂x j ∂x l is contravariant tensor of rank two. [ij, l ] =

kl As g

g kl =

∂x k ∂x l st g ∂x s ∂x t

Now g kl [ij, l ] =

∂x k ∂x l st ∂ 2 x p ∂x q ∂x k ∂x l st ∂x p ∂x q ∂x r g g + g [pq, r] pq ∂x s ∂x t ∂x i x j ∂x l ∂x s ∂x t ∂x i ∂x j ∂x l  ∂x l ∂x q  st ∂ 2 x p ∂x k  g g + pq  ∂x t ∂x l  ∂x i x j ∂x s  

 ∂x l ∂x r  ∂x p ∂x q st    ∂x t ∂x l  ∂x i ∂x j g [ pq, r ]  

=

∂x k ∂x s

=

∂x k 2 st ∂ 2 x p ∂x k r ∂x p ∂x q st ∂x l ∂x q δ g g + δ g [ pq , r ] = δqt as t pq t s i j s i j t l ∂x ∂x x ∂x ∂x ∂x ∂x ∂x

=

∂x k ∂ 2 x p sq ∂x k ∂x p ∂x q sr g g + g [ pq, r ] as δ rt g st = g sr pq ∂x s ∂x i ∂x j ∂x s ∂x i ∂x j


65

∂x p ∂ 2 x p s ∂x k ∂x p ∂x 2  s  δp + s   ∂x s ∂x i ∂x j ∂x ∂x i ∂x j  p q   s  sr = δ sp and g [ pq, r ] =  p q   

g kl [ij, l ] = Since g sq g pq

 k    = i j 

∂x k ∂ 2 x s ∂x k ∂x p ∂x q + ∂x s ∂x i ∂x j ∂x s ∂x i ∂x j

 s     p q

....(7)

It is law of transformation of Christoffel’s symbol of the second kind. But it is not the law of transformation of any tensor. So, Christoffel’s symbol of the second kind is not a tensor. Also, multiply (7) by

∂x s , we get ∂x k ∂x s dx k

Since

 k  ∂x s ∂x k ∂ 2 x s ∂x s ∂x k ∂x p ∂x q +   = ∂x k ∂x s ∂x i ∂x j ∂x k ∂x s ∂x i ∂x j i j 

 s    p q

∂x s ∂x k = δ ss = 1 ∂x k ∂x s

∂x s ∂x k

 k  ∂ 2 xs ∂x p ∂x q +   = ∂x i ∂x j ∂x i ∂x j i j 

∂x s ∂ 2 xs = ∂x k ∂x i ∂x j

 s     p q

 k  ∂x p ∂x q  s  i j  − i  j    ∂x ∂x  p q 

...(8)

It is second derivative of x s with respect to x ’s in the terms of Christoffel’s symbol of second kind and first derivatives. THEOREM 4.4 Prove that the transformation of Christoffel’s Symbols form a group i.e., possess the transitive property. Proof: Let the coordinates x i be transformed to the coordinate system x i and x j be transformed to x i . When coordinate x i be transformed to x i , the law of transformation of Christoffel’s symbols of second kind (equation (7)) is  k  ∂x k ∂ 2 x s ∂x k ∂ x p ∂ x q  s  +   =   ∂ x s ∂ x i ∂x j ∂x s ∂x i ∂ x j  p q  i j  When coordinate x i be transformed to x i . Then  k  ∂x i ∂x j ∂x r  r  ∂ 2 x k ∂x r + =     u v k ∂x u ∂x v ∂x k i j  ∂x ∂x ∂x u v

...(1)

66


 s  ∂x k ∂x p ∂x q ∂x i ∂x j ∂x r ∂ 2 x s ∂x k ∂x i ∂x j ∂x r +   = p q   ∂x s ∂x i ∂x j ∂x u ∂x v ∂x k ∂x i ∂x j ∂x s ∂x u ∂x v ∂x k +

∂ 2 x k ∂x r ∂x u ∂x v ∂x k

2 k r  r   s  ∂x p ∂x q ∂x r ∂ x ∂x  =  + +  u v s u v k u v  p q  ∂x ∂x ∂x ∂x ∂x ∂x

∂ 2 x s ∂x r ∂x i ∂x j ∂x i ∂x j ∂x s ∂x u ∂x v as

∂x

p i

∂x

∂x ∂x

i

u

=

∂x

p

∂x

u

...(2)

Since we know that ∂x s ∂x i ∂x s = ∂x i ∂x u ∂x u

...(3)

Differentiating (3) w.r.t. to x v , we get ∂ ∂x v

 ∂x s   ∂x i 

 ∂x i ∂x s ∂  ∂x i    ∂x u + ∂x i ∂x v  ∂x u  

2 s  = ∂ x   ∂x u ∂x v

∂ 2 x s ∂x j ∂x i ∂x s ∂ 2 x i ∂ 2 xs + = ∂x i ∂x j ∂x v ∂x u ∂x i ∂x u ∂x v ∂x u ∂x v Mutiply (5) by

...(4)

∂x r . ∂x s

∂ 2 x s ∂x j ∂x i ∂x r ∂ 2 x i ∂x s ∂x r ∂ 2 x s ∂x r + = ∂x i ∂x j ∂x v ∂x u ∂x s ∂x u ∂x v ∂x s ∂x s ∂x u ∂x v ∂x s Replace dummy index i by k in second term on L.H.S. ∂ 2 x s ∂x j ∂x i ∂x r ∂ 2 x k ∂x r ∂ 2 x s ∂x r + = ∂x i ∂x j ∂x v ∂x u ∂x s ∂x u ∂x v ∂x k ∂x u ∂x v ∂x s Using (5) in equation (2), we get  r   s  ∂x p ∂x q ∂x r ∂ 2 x s ∂x r +   =   u v s ∂x u ∂x v ∂x s u v  p q ∂x ∂x ∂x

...(5)

...(6)


67

The equation (6) is same as the equation (1). This shows that if we make direct transformation from x i to x i we get same law of transformation. This property is called that transformation of Christoffel's symbols form a group. 4.3 COVARIANT DIFFERENTIATION OF A COVARIANT VECTOR Let Ai and Ai be the components of a covariant vector in coordinate systems x i and x i respectively.. Then ∂x p Ap Ai = ∂x i

...(1)

Differentiating (1) partially w.r.t. to x j , ∂Ai ∂2 x p ∂x p ∂A p A + = p ∂x j ∂x j ∂x i ∂x i ∂x j ∂Ai ∂2 x p ∂x p ∂A p ∂x q A + = p ∂x j ∂x j ∂x i ∂x i ∂x q ∂x j It is not a tensor due to presence of the first term on the R.H.S. of equation (2). Now, replace dum m y index p by s in the first term on R.H.S. of (2), we have ∂Ai ∂ 2 xs ∂x p ∂A p ∂x q As + i j = j i ∂x ∂x ∂x ∂x ∂x q ∂x j Since we know that from equation (8), page 65, ∂ 2 xs ∂x s = ∂x i ∂x j ∂x k Substituting the value of

 k  ∂x p ∂x q  s   − i  j  i j  ∂x ∂x  p q 

∂ 2 xs in equation (3), we have ∂x i ∂x j  ∂x s ∂Ai =  k ∂x j  ∂x

 k  ∂x p ∂x q  s   ∂x p ∂A p ∂x q  − A +     s ∂x i ∂x q ∂x j i j  ∂x i ∂x j  p q  

 k  ∂x s ∂x p ∂x q =   k As − i ∂x ∂x j i j  ∂x  k  ∂Ai ∂x p ∂x q A + =   k ∂x j ∂x i ∂x j i j 

 s  ∂x p ∂x q ∂A p   As + i ∂x ∂x j ∂x q  p q

 ∂A p  s   q − As     ∂x  p q 

...(2)

...(3)

68


 k  ∂x p ∂x q ∂Ai − A =  k ∂x i ∂x j ∂x j i j 

 ∂A p  s     ∂x q − As  p q    

...(4)

Now, we introduce the comma notation Ai , j =

∂Ai k  − Ak   j ∂x i j

...(5)

Using (5), the equation (4) can be expressed as Ai , j =

∂x p ∂x q Ap, q ∂x i ∂x j

..(6)

It is law of transformation of a covariant tensor of rank two. Thus, Ai , j is a covariant tensor of rank two. So, Ai , j is called covariant derivative of Ai with respect to x j . 4.4 COVARIANT DIFFERENTIATION OF A CONTRAVARIANT VECT OR Let Ai and A i be the component of contravariant vector in coordinate systems x i and x i respectively.. Then ∂x i s A ∂x s ∂x s i A As = ∂x i Ai =

or

Differentiating it partially w.r.t. to x j , we get ∂A s ∂ 2 xs ∂x s ∂A i i A + = ∂x j ∂x j ∂x i ∂x i ∂x j Since from equation (8) on page 65,

...(1)

∂ 2 xs ∂ x s  k  i ∂ x p ∂x q  s  =  A −   ∂ x k i j  ∂ x i ∂x j  p q  ∂x j ∂x i substituting the value of

∂ 2 xs in the equation (1), we get ∂x j ∂x i ∂A s ∂x s = ∂A j ∂x k

∂A s ∂A q ∂x s = ∂A q ∂x j ∂x k

 k  i ∂x p ∂x q  s  i ∂x s ∂A i  A − i  A + i ∂x ∂x j  p q  ∂x ∂x j i j   k  i ∂x p i ∂x q  A − i A ∂x ∂x j i j

 s  ∂x s ∂A i  + i j  p q  ∂x ∂x


69

Interchanging the dummy indices i and k in the first term on R.H.S. and put ∂A s ∂x q ∂x s = ∂x q ∂x j ∂x i ∂x q ∂x j

 ∂A s  ∂x s p  s    ∂x q + A  p q  = ∂x i    i ∂A i + Ak  j ∂x k

∂x p i A = A p we get ∂x i

 i  k ∂x q p  s  ∂x s ∂A i  A − j A  + i j ∂x k j  p q  ∂x ∂x   i   k 

 ∂x i ∂x q  = j  ∂x s ∂x j

 k ∂A i A + j j  ∂x

   

 ∂A s  p  s   + A   q  ∂x  p q  

...(2)

Now, we introduce the comma notation ∂Ai  i  + Ak   j ∂x k j  Using (3), the equation (2) can be expressed as A,i j =

...(3)

∂x p ∂x q Ap, q ...(4) ∂x i ∂x j It is law of transformation of a mixed tensor of rank two. Thus, Ai , j is a mixed tensor of rank two. Ai , j is called covariant derivative of Ai with respect to x j . A,ij =

4.5 COVARIANT DIFFERENTIATION OF TENSORS Covariant derivative of a covariant tensor of rank two. Let Ai j and Ai j be the components of a covariant tensor of rank two in coordinate system x i and x i respectively then Aij =

∂x p ∂x q A pq ∂x i ∂x j

...(1)

Differentiating (1) partially w.r.t. to x k ∂Aij

∂A pq ∂x k

=

 ∂x p ∂x q   ∂x i ∂x j 

  A pq  

∂x k ∂Aij

∂x p ∂x q ∂Apq ∂x r ∂ 2 x p ∂x q ∂x p ∂ 2 x q + A + A pq ...(2) = pq ∂x i ∂x j ∂x r ∂x k ∂x k ∂x i ∂x j ∂x i ∂x k ∂x j

∂x k as

∂x p ∂x q ∂A pq ∂ + k i j k ∂x ∂x ∂x ∂x

=

∂A pq ∂x r (since A pq components in x i coordinate) ∂x r ∂x k

70


A pq

∂ 2 x p ∂x q ∂ 2 x l ∂x q A = lq ∂x i ∂x k ∂x j ∂x i ∂x k ∂x j

A pq

∂ 2 x p ∂x q ∂x q A = lq ∂x i ∂x k ∂x j ∂x j

 h  ∂x l  l  ∂x p ∂x r    h −    i k i k  ∂x  p r  ∂x ∂x 

Since we know that from equation (8) on page 65.  h  ∂x l  l  ∂x p ∂x r ∂ 2 xl  h −  i i k =  k ∂x ∂x i k  ∂x  p r  ∂x ∂x A pq

 h  ∂x l ∂x 2  l  ∂ x p ∂ x q ∂x r ∂ 2 x p ∂x q A − A     lq lq = i k ∂x h ∂ x j  p r  ∂x i ∂x j ∂x k   ∂x i ∂x k ∂x j

A pq

 h  ∂ 2 x p ∂x q  l Ahj −   i k j =  ∂x ∂x ∂x i k  p as Ahj = Alq

 ∂x p ∂x q ∂x r  Alq i r ∂x ∂x j ∂x k

...(3)

∂x l ∂x q by equation (1) ∂x h ∂x j

and Apq

∂x p ∂ 2 x q ∂x p ∂ 2 x l A = pl ∂x i ∂x j ∂x k ∂x i ∂x j ∂x k = Apl

∂x p ∂x i

 h  ∂x l  l  ∂x q ∂x r    h −    j k  j k  ∂x q r  ∂x ∂x 

 h  ∂x p ∂x l  l  ∂x q ∂x r ∂x p =  A − Apl  pl  ∂x i ∂x h q r  ∂x j ∂x k ∂x i  j k  Apq

 h  ∂x p ∂ 2 x q  l  ∂x p ∂x q ∂x r A − A pl =     i ih j k ∂x i ∂x j ∂x k  j k  q r  ∂x ∂x ∂x

...(4)

Substituting the value of equations (3) and (4) in equation (2) we get,  ∂A pq  h   h   l   l  ∂x p ∂x q ∂x r  r − Alq  p r  − A pl q r  +   Ahj +  =  Aih i j k k     ∂x ∂x ∂x  ∂x i k   j k  ∂x ∂Aij

 h   h   ∂A pq  l   l  ∂x p ∂x q ∂x r − A − A − A − A     =     ih hj lq pl  r i j k ∂x k  j k  i k   p r q r  ∂x ∂x ∂x  ∂x ∂Aij


Aij , k =

71

∂Aij

 h h  −   Aih −   Ahj , then ∂x j k i k  k

Aij , k = Apq , r

∂x p ∂x q ∂x r ∂x i ∂x j ∂x k

It is law of transformation of a covariant tensor of rank three. Thus, A ij,k is a covariant tensor of rank three. So, Aij , k is called covariant derivative of Aij w.r.t. to x k . Similarly we define the covariant derivation xk of a tensors A ij and Aij by the formula ∂A ij  i   j + A lj   + A il   A ,k = k ∂x l k  l k  ∂A ij  i   l  + A lj   − Ali  Aij ,k =  k ∂x l k   j k ij

and

ij ... l In general, we define the covariant deriavative xk of a mixed tensor Aab by the formula ... c ij ... l Aab ... c ,k

Note:

ij ...l ∂Aab pj ... l  i  ip... l  j  ij... p  l  ...c + Aab =  + Aab... c   + ⋅ ⋅ ⋅ + Aab... c   ...c  k ∂x p k  p k p k  p  ij... l  p  ij ... l  p  − A ijpb......l c   − Aap... c   − ⋅ ⋅ ⋅ − Aab ... p   a k  b k  c k 

Ai , k is also written as Ai , k = ∇ k Ai .

4.6 RICCI'S THEOREM The covariant derivative of Kronecker delta and the fundamental tensors gij and gij is zero. Proof: The covariant derivative xp of Kronecker delta is δ ij,k =

∂δ ij

 i   l  + δ lj   − δ il   ∂x l k   j k k

 i   i  = 0+  −   j k  j k  δ ij,k = 0 as

∂δ ij

 i   i  = 0; δlj   =   ∂x l k   j k  k

Also, consider first the tensor gij and the covariant derivative of gij is g ij , k =

∂g ij

m  m − g mj   − gim   ∂x i k  j k k

72


g ij , k = ∂g ij

But

∂x k

∂g ij

m − [ik , j ]− [ jk , i] as g mj   = [ik , j ] ∂x i k  k

= [ik , j ]+ [ jk, i]

g ij , k = ∂g ij − ∂g ij ∂x k ∂x k g ij , k = 0

So,

W e can perform a sim ilar calculation for the tensor gij.

Since we know that g im g mj = δ ij . Similarly taking covariant derivative, we get im i g ,im k g mj + g g mj , k = δ j , k

But

im gmj,k = 0 and δ ij , k = 0. So, g , k = 0 as g mj ≠ 0

EXAMPLE 4 Prove that if A ij is a symmetric tensor then Ai j, j =

(

)

∂g jk ∂ 1 Ai j g − A jk j 2 ∂x i g ∂x

1

Solution Given that A ij be a symmetric tensor. Then A ij = A ji

...(1)

We know that Ai j, k

∂Ai j  j  l  + Ail   − Al j   = k ∂x l k  i k 

Ai j, j

∂Ai j  j  l  + Ail   − Al j   = j ∂x l j  i j 

Put k = j, we get

(

)

=

∂ log g ∂Ai j + Ail − Al j g hl [ ij, h] j ∂x ∂x l

=

∂Ai j Ai j ∂ g + − A jh [ij, h] since A ij is symmetric. ∂x j g ∂x j

Ai j, j =

(

...(2)

)

1 ∂ Ai j g − A jk [ij, k ] g ∂x j

...(3)


73

But A jk [ij , k ] =

1 jk  ∂g jk ∂g ki ∂g ij A  i + − 2 ∂x j ∂x k  ∂x

A jk [ij , k ] =

∂g ij 1  jk ∂g jk jk ∂g ki jk  A + A − A 2 ∂x i ∂x j ∂x k

     

∂g ∂g ki A kj kji j = ∂x ∂x On Interchanging the dummy indices j & k. A jk

A jk

...(4)

∂g ji ∂g ki jk A = since Aij = A ji ∂x k ∂x j

∂g ∂g ki − A jk ijk = 0 as g ij = g ji j ∂x ∂x Using (5), equation (4) becomes ⇒

A jk

A jk [ij , k ] =

...(5)

1 jk ∂g jk A 2 ∂x i

Put the value of A jk [ij , k ] in equation (3), we get Ai j, j =

(

)

∂g jk 1 ∂ Ai j g 1 − A jk j 2 g ∂x ∂x i

Proved.

EXAMPLE 5  k  k Prove that   −   are components of a tensor of rank three where i j a i j b are the Christoffel symbols formed from the symmetric tensors aij and bij . Solution Since we know that from equation (8), page 65. ∂ 2 xs ∂x s = ∂x i ∂x j ∂x k ∂x s ∂x k or

 k  ∂x p ∂x q  s   − i  j  i j  ∂x ∂x  p q 

 k  ∂ 2 xs ∂x p ∂x q  s  +   =   ∂x i ∂x j ∂x i ∂x j  p q i j   k   ∂ 2 xs ∂x p ∂x q +   =  i j ∂x i ∂x j i j   ∂x ∂x

 s  ∂x k   s  p q  ∂x

 k  k   and   i j a i j b

74


Using this equation, we can write

and

 ∂ 2 xs  k  ∂x p ∂x q +  = i j   ∂x i ∂x j  ∂x ∂x i j  a

 s   ∂x k    s  p q  a  ∂x

 k   ∂ 2 xs ∂x p ∂x q   =  i j+ i i j b ∂x ∂x j  ∂x ∂x

 s   ∂x k    s  p q b  ∂x

Subtracting, we obtain  s   k   k   s   ∂x p ∂x q ∂x k −  −      i     = j s  p q a  p q b  ∂x ∂x ∂x i j a i j b Put  s   s  s   −  = A pq  p q a  p q b Then above equation can written as ∂x p ∂x q ∂x k ∂x i ∂x j ∂x s It is law of transformation of tensor of rank three. s Aijk = Apq

 k  k So,   −   are components of a tensor of rank three. i j a i j b EXAMPLE 6 If a specified point, the derivatives of gij w.r.t. to xk are all zero. Prove that the components of covariant derivatives at that point are the same as ordinary derivatives. Solution Given that ∂g ii = 0, ∀ i , j, k at P 0 ...(1) ∂x k Let Aij be tensor.. Now, we have to prove that Aij , k = Aij ,k =

∂Aij ∂x k

∂A ij

at P0 .

 i  i  α  + A αj   − Aα   ∂x α k  j k k

...(2)


75

∂g ij  k Since   and [ij,.k ] both contain terms of the type and using equation (1) we get ∂x k i j   k   = 0 = [ij, k] at P0 . i j  So, equation (2) becomes ∂A ij Aij ,k = at P0 ∂x k 4.7 GRADIENT, DIVERGENCE AND CURL (a) Gradient If φ be a scalar function of the coordinates, then the gradient of φ is denoted by grad φ =

∂φ ∂x i

which is a covariant vector. (b) Divergence The divergence of the contravariant vector Ai is defined by div A i =

∂A i  i  + Ak   i ∂x  k i

It is also written as A,ii The divergence of the covariant vector Ai is defined by div Ai = g ik Aik EXAMPLE 7 i Prove that div A =

(

1 ∂ g Ak g ∂x k

)

Solution: If Ai be components of contravariant vector then i div A i = A, i =

∂A i  i  + Ak   i ∂x k i 

Since

(

)

 i  ∂ 1 ∂ g log g =   = k ∂x g ∂x k k i  So, div A i =

∂A i 1 ∂ g k + A i ∂x g ∂x k

76


Since i is dummy index. Then put i = k , we get ∂A k 1 ∂ g k + A k ∂x g ∂x k

div A i =

(

1 ∂ g Ak g ∂x k

div A i =

)

...(1) Proved

(c) Curl Let Ai be a covariant vector then ∂Ai  k − Ak   j ∂x i j 

Ai , j =

∂A j

 k − Ak   ∂x  j i

Aj, i =

and

i

are covariant tensor. So, Ai , j − A j , i =

∂Ai ∂A j − is covariant tensor of second order, which is called curl of Ai . ∂x j ∂x i

Thus curl Ai = Ai , j − A j ,i Note: curl A i is a skew-symmetric tensor. Since A j, i –A i,j = – (A i,j –A j,i)

EXAMPLE 8 If Aij be a skew-symmetric tensor of rank two. Show that Aij , k + A jk ,i + Aki, j =

∂Aij ∂x

k

+

∂A jk ∂x

i

+

∂Aki ∂x j

Solution Since we know that Aij , k =

A jk ,i =

Aki, j =

∂Aij

 l   l  − Alj   − Ail   i k ∂x    j k k

∂A jk

h  l  − Alk   − A jl   ∂x  j i  j k i

∂Aki  l − Ali  j ∂x k

  l   − Akl   j i j 


77

Adding these, we get Aij , k + A jk ,i + Aki, j =

∂Aij ∂x k

+

∂A jk

+

∂x i

 l  ∂Aki   l  − A + A      lj jl ∂x j  i k  k i 

  l   l    l   l  −  Ali   + Ail    −  Akl   + Alk     j k    i j   j i  k j   l Since   is symmetric i.e., i k 

 l   l    =   etc. i k  k i  =

∂Aij ∂x

k

+

 l − k Since Aij is skew-symmetric. Then Alj

∂A jk ∂x

+

i

∂Aki  l  −   ( Alj + A jl ) ∂x j i k 

  l   ( Ali + Ail ) −   ( Akl + Alk ) j i j = − A jl ⇒ Alj + A jl = 0 . Similarly,,

Ali + Ail = 0 and Akl + Alk = 0 So, Aij , k + A jk ,i + Aki, j =

∂Aij ∂x

k

+

∂A jk ∂x

∂Aki ∂x j

+

i

THEOREM 4.5 A necessary and sufficient condition that the curl of a vector field vanishes is that the vector field be gradient. Proof: Suppose that the curl of a vector Ai vanish so that curl Ai = Ai , j − A j , i = 0

...(1)

To prove that Ai = ∇φ, φ is scalar.. Since from (1), Ai , j − A j ,i = 0 ⇒ ⇒ ⇒ ⇒

∂Ai ∂A j − =0 ∂x j ∂xi ∂A j ∂Ai = j ∂x i ∂x ∂A j j ∂Ai j dx dx = ∂xi ∂x j ∂ A j dx j dAi = ∂x i

(

)

Integrating it we get Ai =

∫

(

∂ A j dx j i ∂x

)

78


∂ A j dx j i ∂x ∂φ Ai = , where φ = ∂x i or Ai = ∇φ. Conversely suppose that a vector Ai is such that =

∫

∫

A j dx j

Ai = ∇φ, φ is scalar.. To prove curl Ai = 0 Now, Ai = ∇φ =

and So,

∂φ ∂x i

∂ 2φ ∂Ai = ∂x j ∂xi ∂x j ∂A j ∂ 2φ = ∂x i ∂x i ∂x j ∂Ai ∂A j − =0 ∂x j ∂xi

So,

curl Ai = Ai , j − A j ,i =

So,

curl Ai = 0

∂Ai ∂A j − =0 ∂x j ∂x i Proved.

THEOREM 4.6 Let φ and ψ be scalar functions of coordinates xi. Let A be an arbitrary vector then (i)

div (φA) = φ div A + A ⋅ ∇φ

(ii)

∇(φψ) = φ∇ψ + ψ∇φ

(iii)

∇ 2 (φψ) = φ∇ 2 ψ + ψ∇ 2 ψ + 2∇φ ⋅ ∇ψ

(iv)

div (ψ∇φ) = ψ∇ 2 φ + ∇φ ⋅ ∇ψ

Proof: (i) Since we know that div A i =

(

1 ∂ g Ai g ∂x i

)

...(1)

replace A i by φA i , we get div( φA ) = i

(

1 ∂ g φAi g ∂x i

)


1 g

=

=

∂φ ∂x

i

79

 ∂ ( g Ai )  i ∂φ  φ + g A ⋅  i  ∂x i ∂x  i

i

⋅A +φ

1 ∂( g A ) g

∂x

i

= ∇φ ⋅ A i + φ div A i Thus div( φA) = ∇φ ⋅ A + φ div A

...(2)

(ii) By definition of gradient, ∇(φψ) =

∂(φψ ) ∂x i

= φ

∂ψ ∂φ +ψ i i ∂x ∂x

∇(φψ) = φ∇ψ + ψ∇φ Thus (iii) Taking divergence of both sides in equation (3), we get div (∇φψ) = div [φ∇ψ + ψ∇φ ] ∇ 2 (φψ) = div (φ∇ψ) + div (ψ∇φ) = ∇φ ⋅ ∇ψ + φ div (∇ψ) + ∇ψ ⋅ ∇φ + ψdiv (∇φ) ∇ 2 (φψ) = φdiv (∇ψ) + ψdiv (∇φ) + 2∇φ ⋅ ∇ψ Thus, ∇ 2 (φψ) = φ∇ 2 ψ + ψ∇ 2 φ + 2∇φ ⋅ ∇ψ (iv) Replace A by ∇ψ in equation (3), we get div (φ∇ψ) = ∇φ ⋅ ∇ψ + φ div (∇ψ) div (φ∇ψ) = ∇φ ⋅ ∇ψ + φ∇ 2 ψ THEOREM 4.7 Let Ai be a covariant vector and φ a scalar function. Then (i)

curl (φA) = A × ∇φ + φ curlA

(ii)

curl (ψ∇φ) = ∇φ × ∇ψ

Proof: (i) Let Ai be a covariant vector then curl A = curl Ai = Ai , j − A j ,i Replacing Ai by φAi , we get

( )

curl (φAi ) = (φAi ), j − φA j , i

...(3)

80


= φ, jAi + φAi , j − φ, i A j − φA j , i = ( Ai φ, j − A j φ, i) + φ( Ai , j − A j ,i ) = Ai × ∇φ + φcurl Ai So, curl (φA) = A × ∇φ + φcurlA

...(1)

(ii) Replacing A by ∇ψ in equation (1), we get curl (φ∇ψ) = φcurl (∇ψ) + ∇ψ × ∇φ Interchange of φ and ψ , we get Since curl (∇φ ) = 0. So,

curl (ψ∇φ) = ψcurl (∇φ) + ∇φ × ∇ψ.

curl (ψ∇φ) = ∇φ × ∇ψ.

Proved.

4.8 THE LAPLACIAN OPERATOR The operator ∇ 2 is called Laplacian operator read as "del square". THEOREM 4.8 If φ is a scalar function of coordinates x i then ∇ 2φ =

∂  ∂φ   g g kr r  k ∂x  g ∂x 

1

Proof: Since ∇ 2 φ = div grad φ

...(1)

and grad φ =

∂φ , ∂x r

which is covariant vector. But we know that any contravariant vector A k associated with Ar (covariant vector) is A k = g kr Ar (Sec Art. 3.4, Pg 43) Now, the contravariant vector A k associated with A k = g kr Since div A i =

∂φ (Covariant vector) is ∂x r

∂φ ∂x r

(

)

1 ∂ g Ak , (Sec Ex. 7, Pg 75) g ∂x k


81

So, from (1) ∂φ  1  ∇ 2 φ = div  g kr r  = ∂x   g

∂φ   ∂  g g kr r  ∂ x   k ∂x

Proved.

EXAMPLE 9 Show that, in the cylindrical coordinates, ∇ 2V =

1 ∂  ∂V  1 ∂ 2V ∂ 2V r + + r ∂r  ∂r  r 2 ∂θ 2 ∂z 2

by Tensor method Solution The cylindrical coordinates are (r , θ, z ). If V is a scalar function of (r , θ, z ). Now, ∇ 2V = ∇ ⋅ ∇V Since ∇V = i

∂V ∂V ∂V +j +k ∂r ∂θ ∂z

Let ∂V ∂V ∂V , A2 = , A3 = ...(1) ∂r ∂θ ∂z Then ∇V = iA1i + jA2 j + kA3 k , since ∇V is covariant tensor. The metric in cylindrical coordinates A1 =

is ds 2 = dr 2 + r 2 dθ2 + dz 2 here, x1 = r , x 2 = θ, x 3 = z . Since ds 2 = g ij dx i dx j g11 = 1, g 22 = r 2 , g 33 = 1 and others are zero. g11 g12 g13

1 0 0

g = g ij = g 21 g 22 g 23 = 0 r 2 0 = r 2 g 31 g 32 g33 0 0 1 g =r

(See Pg. 34, Example 1)

Now,

div (∇V ) = div Ai =

(

k 1 ∂ gA ∂x k g

)

82


1

1 ∂( g A ) =

div (∇V ) =

g

1

∂x

2

+

1 ∂( g A ) g

∂x

2

3

+

∂( g A ) ∂x

3

1  ∂ ( rA1 ) ∂ ( rA 2 ) ∂ ( rA3 )  + +   r  ∂r ∂θ ∂z 

We can write A k = g kq Aq (Associated tensor) A k = g k1 A1 + g k 2 A2 + g k 3 A3 Put k = 1 11 12 13 A1 = g A1 + g A2 + g A3

A1 = g 11 A1 as g 12 = g 13 = 0 Similarly, A 2 = g 22 A2 A3 = g 33 A3 and g 11 =

Cofactor of g11 in g r 2 = 2 =1 g r

g 22 =

Cofactor of g 22 in g 1 = 2 g r

g 33 =

Cofactor of g 33 in g r 2 = 2 = 1 (See. Pg. 34, Ex.1) g r

So, 11 A1 = g A1 = A1 1 22 A 2 = g A2 = 2 A2 r 33 A3 = g A3 = A3 .

or A1 = A1, A 2 = 1 A2 , A3 = A3 . r2 from (1), we get ∂V 1 ∂V ∂V 2 3 A1 = ∂r , A = r 2 ∂θ , A = ∂z from (2),

...(2)


83

So, 1  ∂  ∂V  ∂  1 ∂V  ∂  ∂V i ∇ 2V = div A = r  ∂r  r ∂r  + ∂θ  r r 2 ∂θ  + ∂ z  r ∂z       div (∇V ) =

=

1  ∂  ∂V r r  ∂r  ∂r

 ∂  1 ∂V +   ∂θ  r ∂θ

  

 ∂  ∂V    + r   ∂z  ∂z  

1  ∂  ∂V  1 ∂ 2 V ∂ 2V  + r +  r  r  ∂r  ∂r  r ∂θ 2 ∂z 2 

div (∇V ) =

1 ∂  ∂V  1 ∂ 2V ∂ 2V r + + r ∂r  ∂r  r 2 ∂θ 2 ∂z 2

∇ 2V =

1 ∂  ∂V  1 ∂ 2V ∂ 2V r + + r ∂r  ∂r  r 2 ∂θ 2 ∂z 2

EXERCISES 1. Prove that the expressions are tensors (a) Aij , l =

(b)

Air jk ,l

α   α  −   Aαj −   Ai α  jl  ∂x i l

∂Aij l

=

∂Air jk ∂x

l

−  i

 r  Aα α r α α A −   Ar −   Ar   l  αjk  j l  i αk k l  ijα + α l  ijk

2. Prove that

Ai j, j

=

1 ∂ ( Ai g

j

∂x

g) j

3. If A ijk is a skew-symmetric tensor show that

j k  − Ak   i j

∂ ( g Ai jk ) is a tensor.. g ∂xk

1

4. Prove that the necessary and sufficient condition that all the Christoffel symbols vanish at a point is that g ij are constant. 5. Evaluate the Christoffel symbols in cylindrical coordinates. 6. Define covariant differentiation of a tensor w.r. to the fundamental tensor gij . Show that the covariant differentiation of sums and products of tensors obey the same result as ordinary differentiation. 7. Let contravariant and covariant components of the same vector A be Ai and Ai respectively then prove that i div A = div Ai

84


8. If Aij is the curl of a covariant vector. Prove that Aij ,k + A jk, i + Aki, j = 0

9. To prove that u ⋅∇u = – ucurl u if u a vector of constant magnitude. 10. A necessary and sufficient condition that the covariant derivative vector be symmetric is that the vector must be gradient. 11. Show that, in spherical coordinates 2 ∇ V =

by tensor method.

1 ∂  2 ∂V  1 ∂  ∂V  1 ∂ 2V r + 2  sin θ + 2 2 2 ∂ θ  r sin θ ∂ φ2 r ∂r  ∂ r  r sin θ ∂ θ

CHAPTER – 5

RIEMANN-CHRISTOFFEL TENSOR 5.1 RIEMANN-CHRISTOFFEL TENSOR If A i is a covariant tensor then the covariant x j derivative of A i is given by A i,j =

∂Ai  α  −   Aα ∂x j i j 

...(1)

Differentiating covariantly the equation (1) w.r. to x k , we get  α  ∂Ai , j  α −   Aα, j −   Ai , α Ai , jk = k ∂x i k   j k  =

 ∂A  α    α   ∂A  β    i −   A  −    α −   Aβ  α j j  ∂x   ∂ x      i j   i k   α j     α   ∂A  γ   −    αi −   Aγ  i k   ∂x i k   ∂ ∂x k

∂ Ai 2

Ai , jk =

∂x k ∂x

j

 α  +  i k  Interchanging j and k in equation (2), we get

 β  α

  α  ∂Ai  α   γ   Aβ −   α +   Aγ j   j k  ∂x  j k  i α 

α  ∂  i k 

∂ Ai − ∂x j ∂x k ∂x j 2

Ai , kj =

 α  ∂   α  ∂ Aα  α  ∂ Aα i j  − Aα −   −  k k j ∂x i k  ∂x  i k  ∂ x

 α  ∂A  α  ∂A Aα −   αj −   αk i k  ∂x i j  ∂x

...(2)

86


 α  +  i j  Subtract equation (3) from (2), we get `

Ai , jk − Ai , kj

 β   α  ∂Ai  α    Aβ −   α +   α k  k i  ∂x k j 

α  =   i k 

 γ    Aγ i α

 α  ∂   β   α   β  i j  Aα −      Aβ −  Aβ + k ∂x α j  i j  α k 

...(3)

α ∂  i k  A α ∂x j

Interchanging of α and β in the first and third term of above equation, we get

Ai , jk − Ai , kj

   α  ∂    i k  − =  ∂x j  

 α  ∂  i j  ∂x k

 β   α +   i k  β

  β  −   j  i j 

    α  Aα   β k  

Ai , jk − Ai , kj = Aα Riαjk

...(4)

...(5)

where

Riαjk

 α   α  ∂  ∂  i k   i j   β  − + =  ∂x j ∂xk  i k 

 α   β   −  β j   i j 

 α    β k 

...(6)

Since A i is an arbitrary covariant tensor of rank one and difference of two tensors A i, jk – Ai, kj is a covariant tensor of rank three. Hence it follows from quotient law that Riαjk is a mixed tensor of rank four. The tensor Riαjk is called Riemann Christoffel tensor or Curvature tensor for the metric g ij dx i dx j . The symbol Riαjk is called Riemann’s symbol of second kind. Now, if the left hand side of equation (4) is to vanish i.e., if the order of covariant differentiation is to be immaterial then Riαjk = 0 Since A α is arbitrary. In general Riαjk ≠ 0, so that the order of covariant differentiation is not immaterial, It is clear from the equation (4) that “a necessary and sufficient condition for the validity of inversion of the order of covariant differentiation is that the tensor Riαjk vanishes identically.. Remark The tensor

i Rikl

∂ ∂x k =  i    j k

∂  i    l ∂x α k  +  i   α     j l   j k

 i    α l  α     j l

...(7)

Riemann-Christoffel Tensor

87

THEOREM 5.1 Curvature tensor Riαjk is anti symmetric w.r.t. indices j and k. Proof: We know that from (7) curvature tensor ∂ ∂x j

 α   α      β j  β k 

∂ ∂ xk

Riαjk =  α   α  +      β      i

Riαjk

j 

   i j 

i k 

 α  ∂  i k  = − ∂x j

 β     i k 

 α  ∂  i j   α  +  ∂x k β j 

 β   α   −  i k  β k 

 β    i j 

Interchanging j and k, we get

R iαkj

α α ∂  ∂   i j  − i k  +  α   β  −  α   β  = β k  i j  β j i k  ∂x k ∂x j         α  ∂    i k  − = − j  ∂x  

Rikα

j

 α  ∂  i j  ∂x k

 α   +  β j  i

  β  α   β  −    k  β k  i j   

= – Riαjk

So, Riαjk antisymmetric w.r.t. indices j and k. Theorem 5.2 To prove that Riαjk + R αjk i + R kiα j = 0 Proof: Since we know that ∂ ∂x

∂ ∂x k

j

α Rijk =  α   α 

  i j 

Rijαk

  i k 

 α  ∂    = i k  − ∂x j

+

 α   α      β j  β k   β    i j 

 α  ∂  i j  α  +  ∂x k β j

 β    i k 

 β   β   −  i k  i j

 α    β k 

...(1)

88


Similarly R αj ki

α α ∂  ∂   j i j k α   β   β   α =  k  −  i  +     −   ∂x ∂x β k   j i  j k  β i

...(2)

and

Rkiα

j

 α  ∂  k j  − = ∂ xi

 α  ∂  k i   α  +  ∂x j β i 

 β   β   −  k j k i 

 α    β j

...(3)

On adding (1), (2) and (3), we get Riαj k + R αj k i + Rkαi j = 0 This is called cyclic property. 5.2 RICCI TENSOR The curvature tensor Riαjk can be contracted in three ways with respect to the index α and any one of its lower indices Rααj k , Riαα k , Riαj α Now, from equation (7), art. 5.1, ∂ ∂x j α Rαjk =  α    α j 

∂ ∂x k  α    α k 

 α  ∂  α k  = − ∂x j =

∂ 2 log g ∂x j ∂ x k

+

 α    β j 

 α    β k 

 β    α j 

 β    α k 

 α  ∂  α j  α  +  ∂x k β j –

 β   α   −  α k  β k 

∂ 2 log g ∂ xk∂ x j

 α  ∂ log g Since α k  = and α and β are free indices Rααj k = 0 . ∂x k   Also for Riαj α . Write R i j for R αi j α

 β    α j 


89

Rij = Rijαα

∂ ∂  α   j α ∂x ∂x β j  = +  α  α  β      i j  i α i j 

α ∂  i α Rij =  j  − ∂x

Rij =

∂ 2 log g ∂x j ∂x i

 α    β α  β   i α

α ∂  i  j +  α   β  −  α   β  β j  i α β α i j  ∂x α       α ∂   i j α  β   α  β  −  α  +   −    β ∂x  j i α β αi j 

...(1)

Interchanging the indices i and j we get α ∂ j α R ji =  i  − ∂x

R ji =

∂ 2 log g ∂x i ∂x j

α ∂  j  i +  α  β  −  α   β  β i  j α  β α  j i  ∂x α       α ∂   i j β α  α β  −  α  +    −   ∂x i α β j  β α  i j 

...(2)

(Since α and β are dummy indices in third term). Comparing (1) and (2), we get Rij = R ji Thus Rij is a symmetric Tensor and is called Ricci Tensor.. For Riααk : Riααk = − Rikαα = −R ik 5.3 COVARIANT RIEMANN-CHRISTOFFEL TENSOR The associated tensor Ri j k l = g iα R αjkl

...(1)

is known as the covariant Riemann-Christoffel tensor or the Riemann-Christoffel tensor of the first kind. Expression for Rijkl Rijkl = gi α Rαjkl

90


 ∂  α  ∂  α   β   α   β   α  = g iα  k  j l  − l  j k  +  j l  β k  −  j k  β l            ∂x   ∂x 

…(2)

Now, g iα

∂  α ∂   = ∂x k  j l  ∂x k =

  α   α  ∂g i α  gi α  j l  −  j l  ∂x k  

∂[ jl , i ]  α  ∂g i α −  k ∂x k  j l  ∂x

...(3)

Similarly, g iα

∂  α  ∂[ jk , i ]  α  ∂gi α −  =  l l  ∂x  j k  ∂x l  j k  ∂x

...(4) udv duv vdu   By the formula dx = dx − dx   

Using (3) and (4) in equation (2), we get Ri j k l =

=

Ri j k l =

∂[ jl , i]  α  ∂gi α ∂[ jk , i ] −  k − + ∂x k ∂ xl  j l  ∂ x

 α  ∂g i α  β    l + gi α    j k  ∂x  j l 

 α   β    − g iα   β k   j k 

 α    β l 

∂[ jl , i ] ∂[ jk , i ]  α  ∂g iα  α  ∂g iα  βα   β  α − +  l −  j l  k + g i α  j l  β k  − g iα  j k  β l  k l j k ∂x ∂x   ∂x   ∂x       ∂[ jl , i ] ∂ [ jk , i]  α   α − + ( [il, α ] + [αl , i ] ) −  ( [ik , α] + [αk , i] ) k l ∂x ∂x  j k  j l  β  β  +  [βk , i] −  [βl , i ]  j l  j k

Ri j k l =

∂[ jl , i ] ∂ [ jk , i]  α   α − + [il, α] −  [ik , α ] k l ∂x ∂x  j k  j l

...(5)

It is also written as

Ri j k l

 α ∂ ∂   k k +  j k ∂x = ∂x [ jk , i ] [ jl, i] [ik ,α ]

 α    j l [il, α ]

...(6)

But we know that

[ jl, i] =

1  ∂gli ∂g ji ∂g jl   + − i  2  ∂x j ∂x l ∂x 

...(7)


91

[ jk ,i ] =

and

1  ∂g ki ∂g ji ∂g ik  + − i 2  ∂x j ∂x k ∂x

  

...(8)

Using (7) and (8), equation (5) becomes 1 ∂ Rijkl = 2 ∂x k

 ∂g li ∂g ji ∂g jl  j + − ∂ xl ∂ xi  ∂x

 1 ∂  − l  2 ∂x

 ∂g ki ∂g ji ∂g jk  j + k − ∂x ∂x i  ∂x

  

 α α  + [il, α ] −  [ik , α ]  j k  j l 2 2 ∂ g jk ∂ g jl 1  ∂ g il ∂ g ik + − − = 2 j k i l j l i k ∂x ∂x ∂x ∂x ∂x ∂x  ∂x ∂x 2

Rijkl

2

 +  

α  α   [il , α] −  [ik , α ]  j k  j l

...(9)

Since  α  α αβ   = g αβ [ jk , β] and  j l  = g [ jl , β] j k     Ri j k l =

∂ 2 g jk ∂ 2 g jl 1  ∂ 2 g il ∂ 2 g ik + − − 2  ∂x j ∂x k ∂x i ∂x l ∂x j ∂x l ∂x i ∂x l

   

+ g αβ [ jk , β][il, α ] − g αβ [ jl , β][ik , α ]

...(10)

This is expression for Ri j k l . The equation (9) can also be written as Ri j k l

∂ 2 g jk ∂ 2 g jl 1  ∂ 2 gil ∂ 2 gik + − − = 2  ∂x j ∂x k ∂x i ∂x l ∂x j ∂x l ∂x i ∂x k   α  + g αβ    j k 

   

 β   α   β  αβ    − g    i l   j l  i k 

5.4 PROPERTIES OF RIEMANN-CHRISTOFFEL TENSORS OF FIRST KIND Ri j k l (i)

R jikl = − Rijkl

(ii)

Rijlk = −Rijkl

(iii)

Rklij = Rijkl

(iv)

Rijkl + Riklj + Riljk = 0

92


Proof: We know that from equation (9), Pg. 91. Rijkl =

∂ 2 g jk ∂ 2 g jl 1  ∂ 2 gil ∂ 2 g ik + − − 2  ∂x j ∂x k ∂x i ∂x l ∂ 2 x j ∂x l ∂x i ∂x k 

   

 α α  + [il, α ] −  [ik , α ]  j k  j l

...(1)

(i) Interchanging of i and j in (1), we get R jikl

2 ∂ 2 g jk ∂ 2 gik ∂ 2 gil 1  ∂ g jl = − − =  i k 2 ∂x ∂x ∂x j ∂x l ∂x i ∂x l ∂x j ∂x k 

= −

 α  + [ jl, α]− iαl  [ jk , α]  i k    

∂ 2 g jk ∂ 2 g jl 1  ∂ 2 gil ∂ 2 gik + − − 2  ∂x j ∂x k ∂x i ∂x l ∂x i ∂x k ∂x j ∂x l

  α   +   [ jl , α ] −   i k  

 α    [ jk , α ] i l 

R jikl = − Rijkl or

Rijkl = − R jikl

(ii) Interchange l and k in equation (1) and Proceed as in (i) (iii) Interchange i and k in (1), we get ∂ 2 g ji ∂ 2 g jl 1  ∂ 2 g kl ∂ 2 g ki + − − R kjil = 2  ∂x j ∂x i ∂x k ∂x l ∂x j ∂x l ∂x k ∂x i Now interchange j and l, we get R klij

2 ∂ 2 g lj 1  ∂ g kj ∂ 2 g li ∂ 2 g ki =  l i+ k j − l j − k i 2  ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x

For  α  αβ  β   α  [kj, α ] =   g    l i   k j  l i 

 β  αβ  α  =  g   k j  l i   β  =  [li, β] k j α α  [li, β] = k j [li, α]   l i 

 α   +  [kl, α] −  α [ki, α]   j i  j l   α  +  [kj, α] −  α [ki, α ]  l i l j  


93

So, R klij =

2 ∂ 2 g lj 1  ∂ g kj ∂ 2 gli ∂ 2 g ki + − − 2  ∂x l ∂xi ∂x k ∂x j ∂x l ∂x j ∂x k ∂x i

  α  α  +  [li, α] −  [ki, α]  k j  l j 

R klij = Rijkl , from (1) from equation (1) Rijkl

2 ∂ 2 g jk ∂ 2 g jk ∂ 2 g jl 1  ∂ g il = 2 j k + i l − j l − j k ∂x ∂x ∂x ∂x ∂x ∂x  ∂x ∂x

  α  + [il , α] −  jα l [ik , α ]   j k    

Riklj

2 ∂ 2 g kj 1  ∂ g ij ∂ 2 g kl ∂ 2 g il =  k l + i j − k j − i l 2  ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x

 α   +  [ij, α] −  α [il , α]  k l  k j  

2 ∂ 2 g lj ∂ 2 gij ∂ 2 g lk 1  ∂ g ik + − − 2  ∂x l ∂x j ∂x i ∂x k ∂x l ∂x k ∂x i ∂x j 

 α  + [ik , α] − lαk [ij, α ]  l j    

Riljk =

On adding these equations, we get Rijkl + Riklj + R iljk = 0 This property of Rijkl is called cyclic property.. Theorem 5.3 Show that the number of not necessarily independent components of curvature tensor 1 2 2 n (n − 1) . does not exceed 12 Or Show that number of distinct non-vanishing components of curvature tensor does not exceed 1 2 2 n (n − 1) . 12 Proof: The distinct non-vanishing components of Rijkl of three types. (i) Symbols with two distinct indices i.e., Rijij . In this case total number of distinct non-vanishing 1 components of Rijkl are n(n − 1) . 2 (ii) Symbols with three distinct indices i.e., Rijik . In this case, total number of distinct non1 vanishing components Rijkl are n(n − 1) (n − 2) . 2 (iii) Symbols Rijkl with four distinct indices. In this case, total number of distinct non-vanishing 2

components of Rijkl is

2

n (n − 1) . 12

94


Hence the number of distinct non-vanishing components of the curvature tensor Rijkl does not 1 2 2 n (n − 1) . exceed 12 Remark When Rijkl is of the form Riiii i.e., all indices are same In this case, Riiii has no components. 5.5 BIANCHI IDENTITY It states that R ijkl, m + R ijlm, k + R ijmk ,l = 0 and

Rhjkl, m + R hjlm, k + Rhjmk ,l = 0

Proof: Introducing geodesic coordinate1 in which Christoffel symbols are constant with the pole at P0 . Since we know that

R ijkl

R ijkl

R ijkl, m

∂ ∂  i    k l ∂x ∂x mk  + =  ii   m      jk jl     jk  i ∂   jl  − = ∂x k

i    ml   m    jl 

i  ∂   jk  +  i  m  −  m  i        ∂x l mk  jl   jk ml 

 i  i  ∂2   ∂2    j l −  j k  = ∂x m ∂x k ∂x m ∂x l

...(1)

 i   m Since  j k ,  j l  etc. are constant at pole.     So, their derivatives are zero. ∂ ∂x l R ijlm =  i     j l 

∂ ∂x m  i     j m

1 Details of geodesic coordinate given in chapter curvature

+

  α

i   l 

 α     j l 

 i    α m  α     j m

in curve . Geodesic.


95

 i   i  ∂  ∂       i  =  j m −  j l  +   l ∂x ∂ xm α l 

R ijlm, k

 α   i   −   j m α m 

 α     j l 

 i  2 i  ∂2   ∂    j m −  j l  = ∂x k ∂x l ∂x k ∂x m

...(2)

and

R ijmk

R ijmk

R ijmk

∂ ∂  i    m k ∂x ∂x β m + =  i   i   β         j m  j k   j m

 i    β k   β     j k

 i   i  ∂  ∂   j k  −  j m +  i   β  −  i   α  =       ∂x m ∂x k β m   j k  β k   j m   i  2 i  ∂2   ∂   j k j m   = − ∂x m ∂x l ∂x k ∂x l

...(3)

On adding (1), (2) and (3), we get R ijkl, m + R ijlm, k + R ijmk ,l = 0

...(4)

Multiplying R ijkl, m by g hi i.e., g hi R ijkl, m = Rhjkl, m Then equation (4) becomes Rhjkl, m + R hjlm, k + Rhjmk ,l = 0 ...(5) Since every term of equation (4) and (5) is a tensor. So, equation (4) and (5) are tensor equations and therefore hold in every coordinate system. Further, P0 is an arbitrary point of Vn . Thus there hold throughout Vn . Hence equation (4) or (5) is called Bianchi identity.. 5.6 EINSTEIN TENSOR 1 i i Theorem 5.4 To prove the tensor R j − δ j R is divergence free. 2 Proof: We know that from equation (5)

96


Rhjkl, m + R hjlm, k + R hjmk,l = 0 Multiply it by g hl g jk , we get g hl g jk R hjkl,m + g hl g jk R hjlm, k + g hl g jk Rhjmk, l = 0 g jk R jk, m + g hl g jk (−R hjml, k ) + g hl g jk (− R jhmk, l ) = 0 Since Rhjlm = −R hjml & Rhjmk = − R jhmk g jk R jk, m − g jk R jm , k − g hl R hm, l = 0 R , m − Rmk , k − R ml ,l = 0 Since g jk R jk = R R , m − R mk , k − Rmk , k = 0 R , m − 2 Rmk , k = 0 Rmk , k −

1 R, m = 0 2

1 Rmk , k − δ km R , k = 0 2

since R , m = δkm R, k

 k 1 k   Rm − δ m R  = 0 2  , k 1 Rmk − δ km R is divergence free. 2 1 i 1 i The tensor Rmk − δ km R = Gmk or R j − δ j R is known as Einstein Tensor.. 2 2 5.7 RIEMANN CURVATURE OF A V n Consider two unit vectors pi and qi at a point P 0 of V n. These vectors at P 0 determine a pencil of directions deferred by ti = αpi + βqi. α and β being parameters. One and only one geodesic will pass through P 0 in the direction of p i . Similarly one and only one geodesic will pass through in the direction qi. These two geodesics through P 0 determined by the orientation of the unit vectors pi and qi. Let this surface is denoted by S. The Gaussian curvature of S at P 0 is defined to be the Riemannian Curvature of Vn at P0 for the orientation determined by pi and qi. Let the coordinates y i of Vn are Riemannian coordinates with origin at P 0. The equation of surfaces S in given by

i.e.,

y i = ( p i α + q iβ)s

...(1)

y i = p iu 1 + q i u 2

...(2)


97

where αs = u1 and βs = u 2 , three parameters namely α, β, s can be reduced to two parameters u and u 2 . Here u1 and u 2 are coordinates of any current point on S. 1

Let ds 2 = bαβ du α du β be the metric for the surface S. where bαβ = g ij

∂y i ∂ y j (α, β = 1, 2) ∂uα ∂uβ

 k  γ  Let   and   be the Christoffel symbols corresponding to the coordinates y i and u α . i j α β  g   R Let αβγδ and R hijk be curvature tensor corresponding to the metrices b αβ du αdu β and g ij dy i dy j . Since the Greek letters α, β, γ, δ take values 1,2 and so that the number of independent non1 1 2 2 vanishing components of Rαβγδ is n2 (n 2 − 1) for n = 2, i.e., they are ⋅ 2 2 − 1 = 1. Let us 12 12 transform the coordinate system u α to u ′ α and suppose that the corresponding value of R1212 are

(

)

' . R1212 Then

' = Rαβγδ R1212

∂u α ∂u β ∂u γ ∂u δ ∂u ′1 ∂u ′ 2 ∂u ′1 ∂u ′ 2

= R1βγδ

∂u1 ∂u β ∂u γ ∂u δ ∂u 2 ∂u β ∂u γ ∂u δ + R 2 βγδ ∂u ′1 ∂u′ 2 ∂u ′1 ∂u′ 2 ∂u ′1 ∂u ′ 2 ∂u ′1 ∂u ′ 2

= R12 γδ

∂u 1 ∂u 2 ∂u γ ∂u δ ∂u 2 ∂u 1 ∂u γ ∂u δ + R 21γδ ∂u′1 ∂u′ 2 ∂u′1 ∂u′ 2 ∂u′1 ∂u′ 2 ∂u′1 ∂u′ 2

= R1212

∂u1 ∂u 2 ∂u1 ∂u 2 ∂u1 ∂u 2 ∂u 2 ∂u1 + R 1221 ∂u ′1 ∂u ′ 2 ∂u ′1 ∂u ′ 2 ∂u ′1 ∂u ′ 2 ∂u ′1 ∂u ′ 2

+ R 2112

∂u 2 ∂u1 ∂u1 ∂u 2 ∂u 2 ∂u1 ∂u 2 ∂u 1 + R 2121 ∂u11 ∂u ′ 2 ∂u′ 2 ∂u ′ 2 ∂u ′1 ∂u′ 2 ∂u ′1 ∂u′ 2

2  ∂u 1 ∂u 2 ∂u1 ∂u 2 ∂u 2 ∂u ′  ∂u 2 ∂u1    R − 2 + = 2112  ∂u′1 ∂u′ 2 ∂u ′1 ∂u ′ 2 ∂u′ 2 ∂u′ 2  ∂u ′1 ∂u ′ 2    

= R1212 J 2 where J = so, ′ = R1212 J 2 R1212

∂u1 ∂u ′1

∂u1 ∂u ′ 2

∂u 2 ∂u ′1

∂u 2 ∂u ′ 2

=

∂u ∂u′

...(3)

98


Again b′αβ = bαβ b′αβ = bαβ

⇒ or from (3) and (4), we get

∂u γ ∂u δ ∂u ′ α ∂u ′β ∂u γ ∂u′ α

∂u δ ∂u′ β

b′ = bJ 2

...(4)

R1212 ′ R = 1212 = K (say) b′ b

...(5)

This shows that the quantity K is an invariant for transformation of coordinates. The invariant K is defined to be the Guasian curvature of S. Hence K is the Riemannian Curvature of S at P0 . Since Riemannian Coordinates yi with the origin at P0 . We have as geodesic coordinates with the pole at P0 . Therefore

Then

 k  γ    ,   = 0 at P0 i j  g α βb

and

 ∂[ij , h]g ∂[ik , h] g Rhijk =  − + ∂y k ∂y j 

  at P0 .  

 ∂[βγ, α] b ∂[βδ, α] b  +  at P0 . Rαβγδ =  − ∂u δ ∂u γ   R1212 = −

∂[21,1] b ∂u

2

+

∂[22,1] b ∂u1

at P0

...(6)

from (5) we get 1  ∂[21, 1] b ∂[22, 1]b  +  K = b  − ∂u 2 ∂u 1  

...(7)

This is required expression for Riemannian curvature at P0 . 5.8 FORMULA FOR RIEMANNIAN CURVATURE IN THE TERMS OF COVARIANT CURVATURE TENSOR OF V n  k γ  Let   and   be the Christoffel symbols of second kind relative to the metrices bαβdu αdu β i j  g α β 


99

and g ij dy i dy j respectively. We have

[αβ, γ]b

=

∂y i ∂y j ∂y k [ij, k ]g ∂u α ∂u β ∂u γ

[21, 1]b

=

∂y i ∂y j ∂y k [ij, k ]g ∂u 2 ∂u1 ∂u1

...(8)

= q i p j p k [ij , k ] g using (2) Now, ∂[21, 1] b ∂u

2

=

∂[ij, k ]g

qi p j p k

∂u 2

i j k = q p p

∂[ij, k ]g ∂y h ∂y h ∂u 2

i j k h = q p p q

∂[ij, k ]g ∂y h

Interchanging h and k, we get ∂[21, 1] b ∂u 2

i k j h = qq p p

∂[ij, h ]g

...(9)

∂y k

Similarly, ∂[22, 1] b ∂u

1

i k j h = qq p p

∂[ij, h ]g

...(10)

∂y k

Using (9) and (10), equation (6) becomes  ∂[ij, h ]g ∂[ik , h]g h i j k R1212 = p q p q  − ∂y k + ∂y j  h i j k R1212 = p q p q Rhijk at P0

Since bαβ = g ij

∂y i ∂y j ∂u α ∂u β

b11 = g ij

∂y i ∂y j = g ij p i p j 1 1 ∂u ∂u

b11 = g hj p h p j

  at P0  ...(11)

100


Similarly b22 = g ij q i q j = g ik q i q k b12 = g ij p i q j = g ji p j q i = g hk p h q k b11 b12

b = b b = b11b22 − b12 b21 21 22

[

b = p h qi p j qk g hig ik − gij ghk

]

...(12)

Dividing (11) and (12), we get p h q i q k Rhijk R1212 = K= b p h qi p j q k ( gik g hj − g ij g hk )

...(13)

This is formula for Riemannian Curvature of Vn at P0 determined by the orientation of Unit vectors p i and q i at P0 . 5.9 SCHUR’S THEOREM If at each point, the Riemannian curvature of a space is independent of the orientation choosen then it is constant throughout the space. Proof: If K is the Riemannian curvature of Vn at P for the orientation determined by unit vectors p i and q i then it is given by p h q i p j q k R hijk

K=

(g ik g hj − g ij g hk ) p h qi p j q k Let K be independent of the orientation choosen. Then equation (1) becomes K=

...(1)

R hijk g ik g hj − gij g hk

Rhijk = K (g ik g hj − g ij g hk )

...(2)

We have to prove that K is constant throughout the space Vn . If N = 2, the orientation is the same at every point. So, consider the case of Vn when n > 2. Since g ij are constants with respect to covariant differentation, therefore covariant differentation of (2) gives Rhijk ,l = ( g ik g hj − g ij g hk ) K ,l

...(3)

where K, l is the partial derivative of K.

Taking the sum of (3) and two similar equations obtained by cyclic permutation of the suffices, j, k and l.


101

( g hj g ik − g hk g ij ) K ,i +( g hk g il − g hl g ik )K , j +( g hl g ij − g hj g il )K , k = Rhijk, l + Rhikl, j + Rhilj,k

...(4)

Here n > 2 therefore three or more distinct values to indices j, k, m can be given . Multiplying (4) by g hj and using g hj g hl = δ lj , we get (ng ik − δ ih g hk )k , l +(g il δ kj − gik δ l j ) K , j +(δ ih g h l − ngil )K , k = 0 or, (n − 1) g ik K , l + 0 + (1 − n) gil K , k = 0 for δ ij = 0, i ≠ j g ik K ,l − g il K , k = 0 Multiplying by g ik and using g ik δik = n, g il g ik = δ kl , we get

nK,l −δ kl K ,k = 0 or (n − 1)K ,l = 0 or

K, k = 0 as (n − 1)K ,l = 0

∂K = 0. ∂x l integrating it, we get K = constant. This proves that the partial derivatives of K w.r.t. to x’s are all zero. or

Consequently K is constant at P. But P is an arbitrary point of Vn . Hence K is constant throughout Vn . 5.10 MEAN CURVATURE The sum of mean curvatures of a Vn for a mutually orthogonal directions at a point, is independent of the ennuple choosen. Obtain the value of this sum. Or Prove that the mean curvature (or Riccian Curvature) in the direction ei at a point of a Vn is the sum of n – 1 Riemmanian curvatures along the direction pairs consisting of the direction and n – 1 other directions forming with this directions an orthogonal frame. i Proof: Let e hi be the components of unit vector in a given direction at a point P of a Vn . Let e k be the components of unit vector forming an orthogonal ennuple.

Let the Riemannian curvature at P of Vn for the orientation determined by l hi and e ik (h ≠ k ) be denoted by K hk and given by K hk =

ehp e qk λ rk λ sk R pqrs ehp e qk e rh eks ( g pr q qs − g ps g qr ) e hp1 e qk1 e rh1 e ks1 R pqrs

= (e p e r g ) (e q e s g ) − (e p e s g ) (e q e r g ) pr ps h h k k qs h k k h qr

...(1)

102


Since unit vectors e hi , λ ik are orthogonal. Therefore e hp e hr g pr = 1 p s

e h eh g ps = 0 etc. and Using these in equation (1), then equation (1) becomes

K hk = K hk = n

∑K

=

∑e k =1

n

∑K

1×1 − 0 × 0 e hp e kq ehr

e ks R pqrs

...(2)

n

hk

k =1

Put

e hp ekq e rh e ks R pqrs

hk

k =1

p q r s e e e h k h k

R pqrs

= M h . Then n

p r M h = eh eh

∑e k =1

q s e k k

R pqrs

p r qs = e h e h g R pqrs p r qs = − e h e h g R qprs p r M h = − eh e h R pr

...(3)

This shows that M h is independent of (n – 1) orthogonal direction choosen to complete an orthogonal ennuple. Here M h is defined as mean curvature or Riccian curvature of Vn for the direction e hp1 . Summing the equation (1) from h = 1 to h = n, we get n

∑M

h

h =1

p r = − eh |e h| R pr

= − g pr R pr = –R n

or

∑M h =1

h

= − g pr R pr == −R

This proves that the sum of mean curvatures for n mutual orthogonal directions is independent of the directions chosen to complete an orthogonal ennuple and has the value − R. 5.11 RICCI’S PRINCIPAL DIRECTIONS Let e ih1 is not a unit vector and the mean curvature M h is given by Mh =

Rij e hi e hj g ij e hi ehj


⇒

103

( Rij + M h g ij ) e ih ehj = 0

Differentiating it w.r. to e ih1 , we get ∂M h g jk ehj e hk + 2 ( Rij + Mg ij ) e hj = 0 ∂ ehi

...(1)

For maximum and minimum value of M h . ∂M h =0 ∂e ih

Then equation (1) becomes ( Rij + Mg ij )ehi = 0 These are called Ricci's Principal direction of the space as they are principal directions of Ricci tensor Rij . 5.12 EINSTEIN SPACE A space, which is homogeneous relative to the Ricci tensor Rij is called Einstein space. If space is homogeneous then we have Rij = λg ij

...(1)

Inner multiplication by g ij , we get R = λn since Rij g ij = R and g ij gij = n ⇒

1 λ = nR

from (1) Rij =

R g n ij

Hence a space is an Einstein space if Rij =

R g at every point of the space. n ij

Theorem 5.5 To show that a space of constant Curvature is an Einstein space. Proof: Let the Riemannian curvature K at P of Vn for the orientation determined by p i and q i , is given by p h q i p j q k R hijk K = p hq i p j q k ( g g − g g ) ik hj ij hk Since K is constant and independent of the orientaion.

104


Rhijk K = (g g − g g ) ik hj ij hk Rhijk = K (g ik g hj − g ij g hk ) Multiplying by g hk Kg hk ( g ik g hj − g ij g hk ) = g hk Rhijk K (δhi g hj − ng ij ) = Rij Since g hk g ik = δ hi ; g hk g hk = n and g hk Rhi jk = Rij K (g ij − ng ij ) = Rij K (1 − n) g ij = Rij

...(1)

ij

Multiplying by g , we get Kn(1 − n) = R as g ij Rij = R

...(2)

from (1) & (2) Rij = (1 − n) gij ⋅

R 1 = Rgij n(1 − n ) n

R g n ij This is necessary and sufficient condition for the space Vn to be Einstein space. ⇒

Rij =

5.13 WEYL TENSOR OR PROJECTIVE CURVATURE TENSOR Weyl Tensor denoted as Whijk and defined by Whijk = Rhijk +

1 ( g R − g kh g ij ) 1 − n ki hj

Theorem 5.6 A necessary and sufficient condition for a Riemannian V n (n > 3) to be of constant curvature to that the Weyl tensor vanishes identically throughout Vn . Proof: Necessary Condition: Let K be Riemannian Curvature of Vn . Let K = constant. We have to prove that Whijk = 0 Since we know that K=

p h q i p j q k R hijk (g hj g ik − g ij g hk ) p h qi p j q k

= constant


105

Since K be independent of the orientation determined by the vector p i and q i . Then K=

R hijk g hj gik − gij g hk

...(1)

Multiplying by g hk , we get g hk Rhijk = g hk K (g hj g ik − g ij g hk ) Rij = K (δkj gik − ng ij ) Rij = K (1 − n) g ij

...(2)

Multiplying by g ij again, we get g ij Rij = K (1 − n) g ij g ij R = K (1 − n )n

...(3)

Putting the value of K from (3) in (2), we get Rij =

R g n ij

...(4)

The equation (3) shows that R is constant since K is constant. Now, the W tensor is given by Whijk = Rhijk +

1 [ g R − g hk g ij ] 1 − n ik hj

Whijk = Rhijk +

1  R R  g ik g hj − g hk g ij   1−n  n n 

from (5), we get

R = Rhijk + n(1 − n) [ gik g hj − g hk gij ] = Rhijk +

Rhijk R ⋅ , by. eqn. (1) n(1 − n) K

Whijk = Rhijk + K

Rhijk K

, by equation (3)

Whijk = 2 R hijk Since K is constant. The equation (5) shows that Whijk = 0 . This proves necessary condition.

106


Sufficient Condition Let Whijk = 0. Then we have to prove that K is constant. Now, Whijk = 0. Rhijk +

⇒

1 [ g R − g hk Rij ] = 0 1 − n ik hj

Multiplying by g hk , we get Rij +

⇒

1 [g hk g ik Rhj − g hk g hk Rij ] 1− n 1 Rij + [δh R − nRij ] 1 − n i hj R Rij + ij (1 − n) 1− n 2 R ij

⇒

=0 =0 =0 =0

Rij = 0

Since Rij = 0 ⇒ g hk Rhijk = 0. hk ⇒ g = 0 or Rhijk = 0

If Rhijk = 0, then clearly K = 0. So, K is constant. If g hk = 0 then K=

K=

R hijk p h qi p j q k (g hj g ik − 0 g ij ) p h qi p j q k R hijk p h q i p j q k ( p h p j g hj ) (q i q k g ik )

=

R hijk p h q i p j q k p2 ⋅ q 2

h i j k K = Rhijk p q p q since p 2 = 1, q 2 = 1

K = constant as Rij = 0 This proves sufficient condition. EXAMPLE 1 For a V2 referred to an orthogonal system of parametric curves (g 12 = 0) show that R12 = 0, R11 g 22 = g 22 g11 = R1221 ij R = g Rij =

Consequently Rij =

1 Rg . 2 ij

2R1221 g11 g 22


107

Solution Given that g12 = 0 so that g 12 = 0. Also, g ij =

1 1 1 22 ⇒ g 11 = & g =g . g ij g11 22

The metric of V2 is given by i j ds 2 = g ij dx dx ; (i, j = 1, 2) 1 2 2 2 ds 2 = g11 (dx ) + g 22 (dx ) Since g12 = 0.

We know that g1hk Rhijk = Rij . and

g = g ij =

g11 g12 g 0 = 11 = g11 g 22 g 21 g 22 0 g 22

(i) To prove R12 = 0 hk h1 R12 = g Rh12 k = g Rh 2 as Rh 122 = 0

R12

= g 21 R2121 as R1121 = 0 =0

(ii) To prove R11 g 22 = R22 g11 = R1221 R11 = g hk Rh 11k = g 2 k R 211k R 2112 22 R11 = g R 2112 = g 22 So, R11 = and

R 2112 g 22

R22 = g hk Rh 22k = g 1k R122k 11 = g R1221 =

So,

...(1)

R22 =

R1221 g11

R1221 g11

...(2)

from (1) and (2) R 11g22 = R 1221=R 22g11 (iii) To prove R=

2R1221 g11 g 22

ij i1 i2 R = g Rij = g Ri 1 + g Ri 2

… (3)

108


= g 11 R11 + g 22 R22 for g 12 = 0 =

R 11 R 22 + g 11 g 22

R1221 R1221 [Q by eqn. (3)] = g g +g g 22 11 11 22 2R1221 R= g11 g 22 (iv) To prove Rij =

1 Rg 2 ij R=

2R1221 g11 g 22

R=

2R1221 as g = g11 g 22 g

R1221 =

1 Rg 2

The eqn (3) expressed as R11 g 22 =

1 Rg = R 22 g11 . 2

it becomes R11 =

Rg Rg g Rg = 11 22 = 11 2 g 22 2g 22 2

Rg Rg11 g 22 Rg22 = = R22 = 2 g11 2g11 2 So, 1 1 Rg11 & R22 = g 22 2 2 1 = g12 as R12 = 0 = g12 2

R11 = R12 This prove that Rij =

1 R . 2 ij

EXAMPLE 2 The metric of the V 2 formed by the surface of a sphere of radius r is ds 2 = r 2 dθ 2 + r 2 sin 2 θdφ 2 1 in spherical polar coordinates. Show that the surface of a sphere is a surface of constant curvature 2 . r Solution Given that ds 2 = r 2 d θ 2 + r 2 sin 2 θd φ 2 Since r is radius of curvature then r is constant.


g11 = r 2 ,

109

g 22 = r 2 sin 2 θ, g12 = 0,

g = g11 g 22 = r 4 sin 2 θ.

We can prove R 1221 = r2 sin 2 θ Now, the Riemannian curvature K of V n is given by p hq i p jq k . K= h i j k p q p q ( g hj gik − g hk gij ) At any point of V2 there exists only two independent vectors. Consider two vectors whose components are (1, 0) and (0, 1) respectively in V2 . Then K=

R 1212 R = 1212 g11 g 22 g

K=

r 2 sin 2 θ 1 = = constant. r 4 sin 2 θ r 2

EXERCISES 1. Show that Rijkl =

 α  ∂[ jl, i] ∂[ jk , i]  α  − + [ il , α ] −     [ik , α] ∂xk ∂xl  j k   j l

2. Show that 2 ∂ g jk ∂ 2 g ik ∂ g jl 1  ∂ g il + − − 2  ∂x j ∂xk ∂xi ∂xl ∂ x j ∂xl ∂xi ∂ xk  3. Using the formula of the problem 2. Show that 2

Rijkl =

2

  αβ  + g ( [ jk , β][il , α] − [ jl , β][ik , α ] ) . 

Rijkl = − R jikl = − Rijlk = Rklij and Rijkl + Riklj + R iljk = 0

4. Show that the curvature tensor of a four dimensional Riemannian space has at the most 20 distinct non-vanishing components. h

5. (a) If prove that the process of contraction applied to the tensor Rijk generates only one new tensor

Rij which is symmetric in i and j. 1 Rihhj , (h, i, j being unequal). g hh 6. Show that when in aV3 the coordinates can be chosen so that the components of a tensor g ij are zero when i, j, k are unequal then (b)

If ds2 = g11 ( dx1 ) 2 + g 22 (dx 2 )2 + g 33 ( dx3 )2 . Prove that Rij =

(i) Rhj =

1 Rhiij g ii

110


(ii) Rhh =

1 1 Rhiih + Rhjjh g ii g jj

7. Prove that if 1 ∂R α αj α Ri = g Rij then Ri , α = 2 i ∂x and hence deduce that when n > 2 then scalar curvature of an Einstein space is constant. 8. If the Riemannian curvature K of Vn at every point of a neighbourhood U of Vn is independent of the direction chosen, show that K is constant throughout the neighbourhood U. Provided n > 2. 9. Show that a space of constant curvature K0 is an Einstein space and that R = K 0 n (1 –n ). 10. Show that the necessary and sufficient condition that Vn be locally flat in the neighbourhood of 0 is that Riemannian Christoffel tensor is zero. 11. Show that every V2 is an Einstein space. 12. For two dimensional manifold prove that K= −

R 2

13. Show that if Riemann-Christoffel curvature tensor vanishes then order of covariant differentiation is commutative.

CHAPTER – 6

THE e-SYSTEMS AND THE GENERALIZED KRÖNECKER DELTAS The concept of symmetry and skew-symmetry with respect to pairs of indices can be extended to cover to pairs of indices can be extended to cover the sets of quantities that are symmetric or skewsymmetric with respect to more than two indices. Now, consider the sets of quantities A i ... i k or Ai l ... i k depending on k indices written as subscripts or superscripts, although the quantities A may not represent tensor. 6.1 COMPLETELY SYMMETRIC The system of quantities A i1 ... i k (or Ai1... i k ) depending on k indices, is said to be completely symmetric if the value of the symbol A is unchanged by any permutation of the indices. 6.2 COMPLETELY SKEW-SYMMETRIC The systems Ai1 ... ik or ( Ai1... i k ) depending on k indices, is said to be completely skew-symmetric if the value of the symbol A is unchanged by any even permutation of the indices and A merely changes the sign after an odd permutation of the indices. Any permutation of n distinct objects say a permutation of n distinct integers, can be accomplished by a finite number of interchanges of pairs of these objects and that the number of interchanges required to bring about a given permutation form a perscribed order is always even or always odd. In any skew-symmetric system, the term containing two like indices is necessarily zero. Thus if one has a skew-symmetric system of quantities Aijk where i, j, k assume value 1, 2, 3. Then A122 = A112 = 0 A123 = − A213 , A312 = A123 etc. In general, the components Aijk of a skew-symmetric system satisfy the relations. Aijk = − Aikj = − A jik Aijk = A jki = Akij

112


6.3 e-SYSTEM Consider a skew-symmetric system of quantities ei1 ... i n ( or ei1 ,..., i n ) in which the indices i1 ... in assume values 1,2,...n. The system ei1 ... i n ( or ei1... i n ) is said to be the e-system if

ei1 ... in (or ei1 ... in )

= +1; when i1 , i2 ,..., in   an even permutation of number1, 2, ..., n   = −1; when i1 , i2 ,..., in   an odd permutation of number1, 2, ..., n  = 0 in all other cases 

EXAMPLE 1 Find the components of system eij when i, j takes the value 1,2. Solution The components of system eij are e11 , e12 , e21 , e 22 . By definition of e-system, we have e11 = 0, e12 = 1,

indices are same since i j has even permutation of 12

e 21 = − e12 = −1

since i j has odd permutation of 12

e22 = 0,

indices are same

EXAMPLE 2 Find the components of the system ei jk . Solution By the definition of e-system, e123 = e 231 = e321 = 1 e213 = e132 = e321 = −1 eijk = 0 if any two indices are same. 6.4 GENERALISED KRÖNECKER DELTA A symbol δi1j1...... ikjk depending on k superscripts and k subscripts each of which take values from 1 to n, is called a generalised Krönecker delta provided that (a) it is completely skew-symmetric in superscripts and subscripts

The e-Systems and the Generalized Krönecker Deltas

113

(b) if the superscripts are distinct from each other and the subscripts are the same set of numbers as the superscripts. The value of symbol  = 1; an even number of transposition is required to arrange the  superscripts in the same order as subscripts.  i1 ... i k δ j ... j  = −1; where odd number of transpositions arrange the superscripts 1 k  in the same order as subscripts   = 0, in all other cases the value of the symbolis zero EXAMPLE 3 Find the values of δ ijkl . Solution By definition of generalised Kronecker Delta, δ ijkl = 0 if i = j or k = l or if the set. ij is not the set kl. 22 23 δ11 pq = δ pq = δ13 = ⋅ ⋅ ⋅ = 0

i.e.,

δ ijkl = 1 if kl is an even permutation of ij = δ 21 = δ13 = δ31 = δ 23 = ⋅⋅⋅ =1 δ12 12 21 13 31 23

i.e.,

and δ ijkl = −1 if kl is an odd permutation of ij. 31 21 = δ13 δ12 = δ13 21 31 = δ12 = ⋅ ⋅ ⋅ = − 1

i.e.,

i i ... i Theorem 6.1 To prove that the direct product e 1 2 n e j1 j2 ... jn of two systems e i1 ... in and e j1 j 2 ... jn is the

generalized Krönecker delta. i i ... i n

Proof: By definition of generalized Krönecker delta, the product e 1 2

ej

1 j2 ... jn

has the following values.

(i) Zero if two or more subscripts or superscripts are same. (ii) +1, if the difference in the number of transpositions of i1 , i2 ,...,in and j1 , j2 ,..., jn from 1,2,...n is an even number. (iii) –1, if the difference in the number of transpositions of i1, i2,...,in and j1, j2, ...jn from 1, 2,..n an odd number. Thus we can write e i 1i 2 ... i n e j1 j 2 ... THEOREM 6.2 To prove that (i) e i 1 i 2 ... i n = δ 1j1 2j 2 ... nj n i i ... i

i i ... i (ii) ei1i 2 ... i n = δ 1j12j 2 ...n jn

jn

i i ... i = δ 1j1 2j 2 ... nj n

114


Proof: By Definition of e-system, e i 1 i 2 ... i n ( or e i1 i 2 ... i n ) has the following values. (i) +1; if i1 , i2 ,...,in is an even permutation of numbers 1,2,...n. (ii) -1; if i1 , i2 ,...,in is an odd permutation of numbers 1,2,...n (iii) 0; in all other cases Hence by Definition of generalized krönecker delta, we can write i i ... i i i ... i (1) e 1 2 n = δ11 22... nn

and 1 2 ... n (2) ei1i2 ... in = δi1i 2 ... in

6.5 CONTRACTION OF ä iáâãjk jk Let us contract δiαβγ on k and γ . For n = 3, the result is ij1 ij 2 ij 3 ij δ ijk αβγ = δ αβ1 + δ αβ2 + δαβ 3 = δαβ

This expression vanishes if i and j are equal or if α and β are equal. If i = 1, and j = 2, we get δ123 α β 3. Hence

δ12 αβ

+ 1; if αβis an even permutatio n of 12  − 1; if αβ is an odd permutatio n of 12 =   0; if αβ is not permutatio n of 12

Similarly results hold for all values of α and β selected from the set of numbers 1, 2, 3. Hence + 1; if i j is an even permutatio n of αβ  − 1; if i j is an odd permutatio n of αβ  ij δ αβ =  0; if two of the subscripts or superscripts are equal or when the  subscripts and superscripts are not formed from the same numbers.  1 If we contract δ ijαβ . To contract δ ijαβ first contract it and the multiply the result by . We obtain 2 a system depending on two indices 1 ij 1 i 1 i2 i3 δ iα = 2 δαj = 2 (δ α1 + δ α2 + δα 3 )

The e-Systems and the Generalized Krönecker Deltas

(

1 12 1 13 It i = 1 in δiα then we get δ α = δα 2 + δ α3 2

115

)

This vanishes unless α = 1 and if α = 1 then δ11 = 1. Similar result can be obtained by setting i = 2 or i = 3. Thus δ iα has the values. (i) 0 if i ≠ α, (α, i = 1, 2, 3, ) (ii) 1 if i = α . By counting the number of terms appearing in the sums. In general we have 1 ij ij δ iα = n −1 δ αj and δ ij = n(n − 1) We can also deduce that (n − k )! i1i2 ... ir i r −1... i k δi1j1i2j...2 ...i r jr = (n − r )! δ j1 j2 ... j r j r −1... j k

...(1)

...(2)

and δi1j1i2j...2 ...i r jr = n (n −1) (n − 2) ⋅ ⋅ ⋅ (n − r + 1) =

n! n − r!

...(3)

or e i1 i 2 ... i n e i1 i 2 ... i n = n!

...(4)

and from (2) we deduce the relation e i1i 2 ... iri r+1... i n e j1 j2 ... j r jr +1 ... i n = n!

...(5)

EXERCISE 1. Expand for n = 3 i

α

(a) δ αδ j

αβ i

12 i j

(c) δ ij x y

ij 2 1

ij i j ij (c) e aα aβ = e a .

(b) δ ij x x

j

ij

(d) δij

2. Expand for n = 2 ij 1 2

(a) e ai a j

(b) e ai a j

ijk 3. Show that δ ijk = 3 ! if i, j, k = 1, 2, 3.

4. If a set of quantities Ai1i 2 ... ik is skew-symmetric in the subscripts (K in number) then i ... i

δ 1j1... jk Ai1 ... ik = k! A j1 ... j k k

5. Prove that εijk g is a covariant tensor of rank three where where ε ijk is the usual permutation symbol.

CHAPTER – 7

GEOMETRY 7.1 LENGTH OF ARC Consider the n-dimensional space R be covered by a coordinate system X and a curve C so that i ...(1) C : x i = x (t ), (i = 1,2..., n) which is one-dimensional subspace of R. Where t is a real parameter varying continuously in the interval t1 ≤ t ≤ t 2 . The one dimensional manifold C is called arc of a curve. 1 2  1 2 dx n  n dx dx Let F  x , x ,..., x , dt dt ,..., dt  be a continuous function in the interval t1 ≤ t ≤ t 2 . Wee   i  dx  assume that F  x,  > 0, unless every dx = 0 and that for every positive number k  dt  dt 1 2 1 2 n    1 2 dx dx dx n   x , x ,..., x n , dx , dx ,... dx  . F  x1 , x 2 ,..., x n , k ,k ,..., k kF =  dt dt dt  dt dt dt   

The integral  dx  F  x,  dt t1  dt  is called the length of C and the space R is said to be metrized by equation (2). s=

∫

t2

...(2)

dx Different choices of functions F  x,  lead to different metric geometrices.  dt  If one chooses to define the length of arc by the formula dx p dx q dt, (p,q = 1, 2, ..., n) ...(3) t1 dt dt dx p dx p dxq , then the resulting where g pq ( x) is a positive definite quadratic form in the variable dt dt dt geometry is the Riemannian geometry and space R metrized in this way is the Riemannian n-dimensional space R n. s=

∫

t2

g pq ( x)

Geometry

117

Consider the coordinate transformation T : x i = x i ( x1 ,..., x n ) such that the square of the element of arc ds, p q ds 2 = g pq dx dx

...(4)

ds 2 = dx i dx i

...(5)

can be reduced to the form Then the Riemannian manifold R n is said to reduce to an n-dimensional Euclidean manifold E n. The Y-coordinate system in which the element of arc of C in E n is given by the equation (5) is called an orthogonal cartesian coordinate system. Obviously, E n is a generalization of the Euclidean plane determined by the totality of pairs of real values ( x 1 , x 2 ). If these values ( x 1 , x 2 ) are associated with the points of the plane referred to a pair of orthogonal Cartesian axes then the square of the element of arc ds assumes the familiar form ds 2 = (d x1)2 + (d x 2 )2 . dx   dx    dx  THEOREM 7.1 A function F  x,  satisfying the condition F  x, k  = kF  x ,  for every dt    dt   dt  k > 0. This condition is both necessary and sufficient to ensure independence of the value of the t2  dx  integral s = t F  x,  dt of a particular mode of parametrization of C. Thus if t in C : x i = xi (t) 1  dt  is replaced by some function t = φ(s ) and we denote x i [φ(s )] by ξ i (λ) . so that x i (t ) ξ i (s ) we have

∫

equality

∫

t2 t1

 dx  F  x ,  dt =  dt 

∫

s2 s1

F (ξ, ξ′) ds

ds i and t1 = φ (s1 ) and t2 = φ (s 2 ). ds Proof: Suppose that k is an arbitary positive number and put t = ks so that t1=ks 1 and t2=ks 2. Then i C : x i = x (t ) becomes i where ξ′ =

C : x i (ks) = ξ i (t ) and

Substituting these values in

or

dx i (ks ) dx i (ks) =k ds dt t2  dx  s= F  x ,  dt we get t1  dt 

ξ′ i (s ) =

∫

s=

∫

s=

∫

s2 s1 s2 s1

dx (ks )  F  x (ks), kds dt   F [ξ(s ),ξ′( s )]ds

dx    dx  We must have the relation F (ξ, ξ′ ) = F  x, k  = kF  x, . Conversely, if this relation is true dt    dt 

118


for every line element of C and each k > 0 then the equality of integrals is assumed for every choice of parameter t = φ ( s), φ1 (s ) > 0, s1 ≤ s ≤ s 2 with and t2 = φ (s 2 ). i

Note: (i) (ii)

dx Here take those curves for which x i (t ) and are continuous functions in t1 ≤ t ≤ t 2. dt

 

A function F  x,

dx  dx    dx   satisfying the condition F  x, k  = kF  x,  for every k > 0 is called dt  dt    dt  i

positively homogeneous of degree 1 in the

dx . dt

EXAMPLE 1 What is meant, consider a sphere S of radius a, immersed in a three-dimensional Euclidean manifold E 3 , with centre at the origin (0, 0, 0) of the set of orthogonal cartesian axes O − X 1 X 2 X 3 . Solution Let T be a plane tangent to S at (0, 0,–a) and the points of this plane be referred to a set of orthogonal cartesian axes O′ –Y 1Y 2 as shown in figure. If we draw from O (0, 0, 0) a radial line OP , interesting the sphere S at P ( x1 , x 2 , x 3 ) and plane T at Q( x 1 , x 2 , −a) then the points P on the lower half of the sphere S are in one-to-one correspondence with points ( x 1 , x 2 ) of the tangent plane T.. 3

X

2

X O P1 1

O´

P2 C

Y

2

X

Q2 K

1

Y

Q1 Fig. 7.1

If P ( x1 , x 2 , x 3 ) is any point on the radial line OP,, then symmetric equations of this line is x1 − 0 x2 − 0 x3 − 0 = =λ = 2 x1 − 0 x − 0 −a− 0 or x1 = λx 1 , x 2 = λx 2 , x 3 = −λa Since the images Q of points P lying on S, the variables x i satisfy the equation of S, ( x1 )2 + (x 2 )2 + (x 3 ) 2 = a 2

( ) ( )

λ2  x 1 + x 2  2

or

2

+ a2  = a2 

...(6)

Geometry

119

Solving for λ and substituting in equation (6), we get ax1

x = 1

and

( x 1 ) 2 + ( x 2 )2 + a 2

ax 2

, x2 =

( x 1 )2 + (x 2 )2 + (x 3 ) 2

− a2

x = 3

...(7)

( x 1) 2 + ( x 2 ) 2 + ( x 3 ) 2

These are the equations giving the analytical one-to-one correspondence of the points Q on T and points P on the portion of S under consideration. Let P1 ( x1 , x 2 , x 3 ) and P2 ( x1 + dx1 , x 2 + dx 2 , x 3 + dx 3 ) be two close points on some curve C lying on S. The Euclidean distance P1 P2 along C, is given by the formula ds 2 = dx i dx i , (i = 1, 2, 3)

...(8)

Since ∂x i p dx = ∂x p d x , (P =1, 2) i

Thus equation (8) becomes ds 2 =

∂ xi ∂ xi d x pd x q p q ∂x ∂ x

= g pq ( x )d x pd x q , (p,q = 1, 2) ∂ xi ∂ x i . ∂ x p ∂x q If the image K of C on T is given by the equations where g pq (x ) are functions of x i and g pq =

 x 1 = x 1 ( t ) K:  x 2 = x 2 (t ), t1 ≤ t ≤ t2

then the length of C can be computed from the integral s=

∫

t2 t1

g pq

dx p dxq dt dt dt

A straight forward calculation gives 1 1 2 (x d x − x 2 d x 1)2 a2 2 1  1 2 2 2  1 + ( x ) + ( x )  a 2 

(d x 1)2 + (d x 2) 2 + ds2 = and

{

}

...(9)

120

Tensors and Their Applications 2

2

2

2 1  d x1   d x 2      +  + 12  x1 d x − x 2 d x       t 2  dt  dt dt  a   dt  dt s= t 1 1 2 2 2 1 1 + 2 (x ) + ( x ) a So, the resulting formulas refer to a two-dimensional manifold determined by the variables ( x 1 , x 2 ) in the cartesian plane T and that the geometry of the surface of the sphere imbeded in a three-dimensional Euclidean manifold can be visualized on a two-dimensional manifold R 2 with metric given by equation (9). 1 If the radius of S is very large then in equation (9) the terms involving 2 can be neglected. Then a equation (9) becomes

∫

{

}

1 2 2 2 ...(10) ds 2 = (d x ) + (d x ) . Thus for large values of a, metric properties of the sphere S are indistinguishable from those of the Euclidean plane. The chief point of this example is to indicate that the geometry of sphere imbedded in a Euclidean 3-space, with the element of arc in the form equation (8), is indistinguishable from the Riemannian geometry of a two-dimensional manifold R 2 with metric (9). the latter manifold, although referred to a cartesian coordinate system Y, is not Euclidean since equation (9) cannot be reduced by an admissible transformation to equation (10).

7.2 CURVILINEAR COORDINATES IN E 3 Let P (x ) be the point, in an Euclidean 3-space E 3 , referred to a set of orthogonal Cartesian coordinates Y.. Consider a coordinate transformation T : x i = x i (x 1 , x 2 , x 3 ), (i = 1, 2, 3) ∂ xi Such that J = ∂ x j ≠ 0 in some region R of E 3 . The inverse coordinate transformation i 1 2 3 T −1 : x i = x ( x , x , x ), (i = 1, 2, 3) will be single values and the transformations T and T −1 establish one-to-one correspondence between the sets of values ( x1 , x 2 , x 3 ) and ( x 1 , x 2 , x 3 ).

The triplets of numbers ( x1 , x 2 , x 3 ) is called curvilinear coordinates of the points P in R. 1 2 3 If one of the coordinates x , x , x is held fixed and the other two allowed to vary then the point P traces out a surface, called coordinate surface. If we set x1 = constant in T then ...(1) x1 ( x 1 , x 2 , x 3 ) = constant defines a surface. If constant is allowed to assume different values, we get a one-parameter family of surfaces. Similarly, x 2 (x 1 , x 2 , x 3 ) = constant and x 3 ( x 1 , x 2 , x 3 ) = constant define two families of surfaces. The surfaces

Geometry

121 2 3 x1 = c1 , x = c2 , x = c3

3 Y

X

… (2)

3 2 X

1 X

P

2 Y

1 Y

Fig. 7.2

intersect in one and only one point. The surfaces defined by equation (2) the coordinate surfaces and intersection of coordinate surface pair-by pair are the coordinate lines. Thus the line of intersection of x1 = c1 and x 2 = c2 is the x 3 − coordinate line because along this the line the variable x 3 is the only one that is changing. EXAMPLE 2 Consider a coordinate system defined by the transformation 3

Y

X

X2

1

Y

O X

2

3

Y1 Fig. 7.3

x 1 = x1 sin x 2 cos x 3

...(3)

x 2 = x1 sin x 2 sin x 3

...(4)

x 3 = x1 cos x 2

...(5)

122


The surfaces x1 = constant are spheres, x 2 = constant are circluar cones and x3 = constant are planes passing through the Y 3-axis (Fig. 7.3). The squaring and adding equations (3), (4) and (5) we get, ( x 1 )2 + (x 2 )2 + (x 3 ) 2 = ( x1 sin x 2 cos x 3 ) 2 + (x1 sin x 2 sin x 3 )2 + (x 1 cos x 2 ) 2 On solving ( x 1 )2 + (x 2 )2 + (x 3 ) 2 = (x1 )2 x1 =

(x 1 ) 2 + ( x 2 ) 2 + ( x 3 )2

...(6)

Now, squaring and adding equations (3) and (4), we get ( x 1 )2 + (x 2 )2 = ( x1 sin x 2 sin x 3 )2 + (x1 sin x 2 sin x 3 )2 ( x 1 )2 + (x 2 )2 = ( x1 )2 (sin x 2 ) 2 x1 sin x 2 =

(x 1 ) 2 + ( x 2 ) 2

...(7)

Divide (7) and (5), we get tan x 2 =

or

(x1 )2 + (x 2 )2 x3

1 2 2 2  −1  ( x ) + ( x )  tan x =   x3   2

...(8)

Divide (3) and (4), we get x2 tan x 3 1 = x ⇒

2 −1  x x 3 = tan  1 x

   

...(9)

So, the inverse transformation is given by the equations (6), (8) and (9). If x1 > 0, 0 < x 2 < π, 0 ≤ x 3 < 2π. This is the familiar spherical coordinate system. 7.3 RECIPROCAL BASE SYSTEMS Covariant and Contravariant Vectors

r r r Let a cartesian coordinate system be determined by a set of orthogonal base vectors b1 , b2 ,b3 then the position vector rr of any point P (x 1 , x 2 , x 3 ) can be expressed as r r ...(1) r = bi x i (i = 1, 2, 3)

Geometry

123

r Since the base vectors bi are independent of the position of the point P (x 1 , x 2 , x 3 ) . Then from (1), r r dr = bi dx i ...(2) If P (x 1 , x 2 , x 3 ) and Q (x 1 + dx 1 , x 2 + d x 2, x 3 + d x 3 ) be two closed point. The square of the element of arc ds between two points is r r ds 2 = dr ⋅ dr from equation (2), r i r j ds 2 = bi dx ⋅ b j dx r r = bi ⋅ b j d x id x j r r i j ds 2 = δij d x d x ; since bi ⋅ b j = δ ij

3 Y

X a3

3 2 X

a2

a1

1 X

P b3

r O

b2

b1

2 Y

1 Y Fig. 7.4.

r r r r r r r (b1 , b2 , b3 areorthogonal base vector i.e.,b1 ⋅ b1 = 1 & b1 ⋅ b2 = 0) . ds 2 = d x id x i;

as äij = 1, i = j

= 0, i ≠ j a familiar expression for the square of element of arc in orthogonal cartesian coordinates. Consider the coordinate transformation i 1 2 3 x i = x (x , x , x ), (i = 1, 2, 3)

r define a curvilinear coordinate system X. The position vector r is a function of coordinates x i . i.e., r r r = r (x i ), (i = 1, 2, 3) Then r r ∂r i dr = dx ...(3) ∂x i

124


and

ds 2 = d rr ⋅ d rr r r ∂r ∂r ⋅ d x id x j = ∂ xi ∂ x j ds 2 = g ij dx i dx j

where g ij

r r ∂r ∂r ⋅ = ∂ xi ∂ x j

r ∂r The vector is a base vector directed tangentially to X i - coordinate curve. ∂x i Put r ∂r r = ai i ∂x Then from (3) and (4) r r r r dr = ai d x i and g ij = a i ⋅ a j

...(4)

...(5)

...(6)

Now, from equations (2) and (6), we get r r a j dx j = bi d x i r ∂x i j r dx a j dx j = bi ∂x j ⇒

r ∂ xi r , as d x j are arbitrary a j = bi ∂x j

r So, the base vectors a j transform according to the law for transformation of components of covariant vectors. r The components of base vectors a i , when referred to X-coordinate system, are r r r a1 : (a1 , 0,0), a 2 : (0, a2 , 0), a3 : (0,0, a3 ). ...(7) and they are not necessarily unit vectors. In general, r r r r g11 = ar1 ⋅ ar1 ≠ 1, g 22 = a 2 ⋅ a2 ≠ 1, g 33 = a3 ⋅ a3 ≠ 1. If the curvilinear coordinate system X is orthogonal. Then r r r r g ij = a i ⋅ a j = ai a j cos θij = 0, if i ≠ j. r r Any vector A are can be written in the form A = k d rr where k is a scalar..

...(8)

...(9)

Geometry

125

r r ∂r i d r = d x we have Since ∂ xi

r ∂r r i A = ∂x i (kdx ) r r i A = ai A

r where Ai = kdxi . The numbers Ai are the contravariant components of the vector A Consider three non-coplanar vectors r r r r r r a 2 × a3 ar 2 = a3 × a1 , ar 3 = a1 × a2 r1 ...(10) a = rr r , [ar1ar2 ar3 ] [ar1ar2ar3 ] [a1a 2a3 ] r r r r rr r where a 2 × a3 , etc. denote the vector product of a 2 and a3 and [a1 a2 a3 ] is the triple scalar r r r prduct a1 ⋅ a2 × a3 . Now, r r r rr r a2 × a3 ⋅ a1 [a1 a2 a3 ] r1 r = = 1. a ⋅ a1 = [ar ar ar ] [ar1ar2ar3 ] 1 2 3 r r r r r r a2 × a3 ⋅ a2 [a2 a2 a3 ] r1 r = = 0. a ⋅ a 2 = [ar ar ar ] [ar1ar2 ar3 ] 1 2 3

r r r Since [a2 a2 a3 ] = 0. Similarly,

r r r r a 1 ⋅ a3 = a 2 ⋅ a1 = ⋅ ⋅ ⋅ = 0 r r r r a 2 ⋅ a2 = 1, a 3 ⋅ a 3 = 1

Then we can write

r r a i ⋅ a j = δij

EXAMPLE 3 rr r To show that [a1 a2 a3 ] = Solution

[

]

g and ar1ar 2 ar 3 = 1 where g = g ij . g

The components of base vectors ai are r r r a1 : (a1 , 0,0), a 2 : (0, a2 , 0) and a3 : (0, 0, a3 ) Then a1 0 0 rr r [a1a2a3 ] = 0 a2 0 = a1a2a3 0 0 a3

...(11)

126


and

g= g ij

g11 g12 g13 = g 21 g 22 g 23 g 31 g 32 g33

from equations (8) and (9), we have g11 = ar1 ⋅ ar1 ⇒ a12 = g11 Similarly a22 = g 22 , a 32 = g 33 and g12 = ar1 ⋅ ar2 = 0, g13 = ar1 ⋅ ar3 = 0 etc. So, a12 0 g= 0 0

0

a22

0 = a12 a22 a32

0

a32

g = a1 a2 a3 from eqn. (11) and (12), we have [ar1ar2ar3 ] = g

...(12)

r r r 1 Since the triple products [a 1 a 2 a 3 ] = . Moreover,, g r r r r r r a2 × a3 r a 3 × a1 r a1 × a 2 r , a = , a = a1 = r1 r 2 r 3 r r r r r r 2 3 [a a a ] [a1 a 2 a 3 ] [a1 a 2 a 3 ] The system of vectors ar1 , ar 2 , ar3 is called the reciprocal base system. r r r Hence if the vectors a 1 , a 2 , a 3 are unit vectors associated with an orthogonal cartesian coordinates then the reciprocal system of vector defines the same system of coordinates. Solved. r r ri The differential of a vector r in the reciprocal base system is d r = a d xi . where dxi are the components of drr. Then r r ds 2 = dr ⋅ dr r r = (a i dxi ) ⋅ (a j dx j ) r r = a i ⋅ a j dxi dx j

where

ij ds 2 = g dxi dx j r r g ij = a i ⋅ a j = g ji

...(13)

Geometry

127

The system of base vectors determined by equation (10) can be used to represent an arbitrary r r r vector A in the form A = a i Ai , where Ai are the covariant components of A. r r Taking scalar product of vector Ai a i with the base vector a j , we get r r r r Ai a i ⋅ a j = Ai δij = A j as a i ⋅ a j = δ ij . 7.4 ON THE MEANING OF COVARIANT DERIVATIVES

r r ∂A r THEOREM 7.2 If A is a vector along the curve in E 3 . Prove that = A,αj aα j ∂x i r ∂A Also, prove that A,i j = j . Where Ai are component of A. ∂x r r Proof: A vector A can be expressed in the terms of base vectors ai as r r A = Ai ai r r r ∂r where a i = i and Ai are components of A. ∂x r The partial derivative of A with respect to x j is r r ∂A ∂Ai r i ∂ai a + A = i ∂x j ∂x j ∂x j r r Since g ij = a i ⋅ a j . Differentiating partially it w.r.t. x k , we have r r ∂g ij ∂a j ∂a i r ⋅a j + k = ∂x k ∂x k ∂x Similarly, r r ∂g jk ∂a j r ∂ak ⋅ a + = k ∂x i ∂xri ∂xri ∂g ik ∂ai r ∂a ⋅ ak + kj and j = j ∂x ∂x ∂x r Since A can be written as r r i ri A = a i A = a Ai r Taking scalar product with a j , we have r r r r a i ⋅ a j A i = a i ⋅ a j Ai ⇒

...(1)

r ⋅ ai r ⋅ aj r ⋅ ai

g ij A i = δ ij Ai = A j

r r r r As ai ⋅ a j = g ij , a i ⋅ a j = δ ij and δ ij Ai = A j . We see that the vector obtained by lowering the index in Ai is precisely the covariant vector Ai .

128


r The two sets of quantities Ai and Ai are represent the same vector A referred to two different base systems. EXAMPLE 4 Show that g iα g jα = δ ij . Solution Since we know that Then

g i α = ari ⋅ arα and g jα = ar j ⋅ ar α r r r r g i α g jα = (ai ⋅ aα ) (a j ⋅ a α )

r r r r = (ai ⋅ a j ) (aα ⋅ a α )

r r = δ ij δ αα as ai ⋅ a j = δ ij g i α g jα = δij as δ αα = 1. But

r r ∂r ai = ∂x i r r r r ∂a j ∂ai ∂ 2r ∂ 2r = = = ∂x j ∂x j dx i ∂x i ∂x j ∂xi

So,

r r ∂a j ∂ai = ∂x j ∂x i

Now, [ij, k ] = Substituting the value of

or Hence

1  ∂g ik ∂g jk ∂g ij  + − k 2  ∂x j ∂x i ∂x

∂g ∂g ik ∂g jk , i and ijk , we get j ∂x ∂x ∂x r 1 ∂a i r [ij,k] = ⋅ 2 j ⋅ a k 2 ∂x r ∂ai r ⋅ ak [ij,k] = ∂x j r ∂ai rk 1 rk j = [ ij, k ]a , as r = a ∂x ak

r ∂a i r α r r ⋅ a = [ij, k ]a k ⋅ a α ∂x j

  , Christoffel’s symbol 

Geometry

129

r r = [ij, k ]g kα , Since g kα = a k ⋅ a α

r α  α  ∂ai r α [ i j , k ] g kα =   , . a = as   i j  ∂x j i j  r  α  r ∂ai =   aα i j  ∂x j r ∂ai Substituting the values of in equation (1), we get ∂x j

...(2)

r ∂Ai r  α  i r ∂A ai +   A a α = ∂x j ∂x j i j  α ∂A r α  r a +   A i aα = j α ∂x i j  r  ∂A α  α  i  r ∂A =  j +   A  aα ∂x j i j    ∂x r ∂A ∂A α  α  i α r α = A , a since A = +  A j α ,j ∂x j ∂x j i j  Thus, the covariant derivative A,αj of the vector A α is a vector whose components are the r ∂A r components of j referred to the base system ai . ∂x r α  ∂ai = 0 If the Christoffel symbols vanish identically i.e.,   the = 0, from (2). ∂x j i j  Substituting this value in equation (1), we get r ∂A ∂Ai r ai = ∂x j ∂x j But

∂Ai  i  β + A ∂x j β j   i  ∂Ai   = 0. A,i j = as ∂x j β j  A,i j =

Proved. r r THEOREM 7.3 If A is a vector along the curve in E 3 . Prove that A j , k a j wheree A j are components r of A. . r Proof: If A can be expressed in the form r r A = Ai a i

130


r where Ai are components of A. r The partial derivative of A with respect to x k is r r ∂A ∂Ai ri ∂a i a + Ai k = ∂x k ∂x k ∂x r r Since a i ⋅ a j = δij , we have,

...(1)

Differentiating it partially w.r. to x k , we get r r ∂a i r r i ∂a j ⋅ a + a ⋅ =0 j ∂x k ∂x k r r r i ∂a j ∂a i r ⋅a j = − a ⋅ k ∂x ∂x k

r r But a i ⋅ aα = δiα . Then

r r α  = − ai ⋅ aα  , j k r ∂a i ∂x k r ∂a i ∂x k

Since

r ∂a j

 α r =  a α ∂x  j k k

r α  ⋅ a j = − δiα    j k r  i  ⋅a j = −    j k r  i r ∂a i −  a j , = ∂x k  j k

1 rj as ar = a j

r ∂a i substituting the value of in equation (1), we get ∂x k r ∂Ai ri  i r j ∂A a − Ai  a k = ∂x k ∂x  j k =

∂Ai r j  i rj a − Ai  a k ∂x  j k

r  ∂Ai  i  r ∂A − Ai   a j k =  k ∂x  j k   ∂x r ∂A j  i  ∂A rj A = − Ai   = A a , Since j, k k k j, k ∂x ∂x  j k Proved.

Geometry

131

7.5 INTRINSIC DIFFERENTIATION r Let a vector field A(x ) and i C : x i = x (t ), t1 ≤ t ≤ t 2 r be a curve in some region of E 3 . The vector A(x ) depend on the parameter t and if A(x ) is a differentiable vector then r r dA ∂A dx j ⋅ = dt ∂x j dt r j dA α r dx = A, j aα dt dt Since we know

r  ∂A α  α  r α r ∂A A , a =  j +   aα j α =  ∂x i j ∂x j

(See Pg. 127, Theo. 7.2)

So, r  ∂A α  α  i  r dx j dA =  j + i j  A  aα dt dt     ∂x r  dAα  α  i dx j  r dA +  A =   aα dt dt  i j  dt dAα  α  i dx j +  A is called the absolute or Intrinsic derivative of A α with respect dt i j dt   δA α . to parameter t and denoted by δt The formula

δA α dAα  α  i dx j δA δA = . So, δt = dt + i j A dt is contravariant vector. If A is a scalar then, obviously, δt δt   Some Results (i) If Ai be covariant vector dAi  α  dx β δAi = dt − i β Aα dt δt   (ii) (iii)

δA ij δAij  i  αj dx β  j  iα dx β = + + A A δt dt α β dt α β dt δA ij δt

=

δA ij

 i  α dx β  α  i dx β + −  A A dt α β j dt i β α dt

132


(iv)

δA ijk δt

=

δA ijk

 i  α dx β  α  i dx β  α  i dx β + − −  A jk A A dt dt  j β αk dt k β j α dt α β 

EXAMPLE 5 If g ij be components of metric tensor, show that

δg ij δt

= 0.

Solution The intrinsic derivative of g ij is δg ij δt

=

=

α dx β  α  dx β −   g αj −  g iα dt i β dt β j  dt

dg ij

∂g ij dx β  α  dx β  α  dx β − g − g     α j i α dt β j  dt ∂x β dt i β

 ∂g ij  α   α   dx β  − g –    g iα  α j =  ∂x β   i β β j   dt    δg ij δt as

 ∂g ij  dx β =  β − [iβ, j ] − [βj, i]  ∂x  dt

α   α    gαj = [ i β, j ] and   g i α = [β j , i ] . β j  i β  ∂g ij

But

∂x β δg ij

So,

δt

= [iβ, j ] + [ jβ, i]. = 0.

EXAMPLE 6 Prove that d ( g ij A i A j ) dt

= 2 g ij A i

δA i δt

Solution Since g ij Ai a j is scalar.. Then d ( g ij A i A j ) dt

=

δ(g ij A i A j ) δt

Geometry

133

δ( A i A j ) , since g ij is independent of t. δt  δA i j δA j  A + Ai = g ij   δt   δt Interchange i and j in first term, we get = g ij

d ( g ij A i A j ) dt d ( g ij A i A j ) dt

 i δA j δA j  + Ai = g ij  A  as g ij is symmetric. δt δt   = 2 g ij Ai

δA j δt Proved.

EXAMPLE 7 Prove that if A is the magnitude of Ai then A, j =

Ai , j Ai A

Solution Given that A is magnitude of Ai . Then Since g ik A i A k = Ai A i g ik A i A k = A 2 Taking covariant derivative w.r. to x j , we get g ik A,i j A k + g ik A i A,kj = 2 AA, j Interchange the dummy index in first term, we get g ki A,kj A i + g ik A i A,kj = 2 AA, j 2 g ik Ai A,kj = 2 AA, j g ik A i A,kj = AA, j

(

)

Ai g ik A,kj = AA, j Ai Ai , j = AA, j since g ik A,kj = Ai , j A, j =

Ai , j Ai A

Proved.

134


7.6 PARALLEL VECTOR FIELDS Consider a curve i C : x i = x (t ), t1 ≤ t ≤ t 2 , (i = 1, 2, 3) r in some region of E 3 and a vector A localized at point P of C. If we construct at every point of C a vector equal to A in magnitude and parallel to it in direction, we obtain a parallel field of vector along the curve C.

3 X

3 Y

2 X C 1 X

P

O

2 Y

1 Y

r r if A is a parallel field along C then the vector A do not change along the curve and we can write

r r dA = 0. It follows that the components Ai of A satisfy a set of simultaneous differential equations dt ∂A i = 0 or ∂t dAi  i  α dxβ + =0 A dt α β dt This is required condition for the vector field Ai is parallel. 7.7 GEOMETRY OF SPACE CURVES Let the parametric equations of the curve C in E 3 be C : x i = x i (t ), t1 ≤ t ≤ t 2

(i = 1, 2, 3).

The square of the length of an element of C is given by i j ds 2 = g ij dx dx and the length of arc s of C is defined by the integral

s=

∫

t2 t1

gij

dxi dx j dt dt dt

...(1)

...(2)

from (1), we have g ij

dx i dx j =1 ds ds

...(3)

Geometry

Put

135

dx i = λi . Then equation (3) becomes ds g ij λi λj = 1

...(4) r r The vector λ , with components λi , is a unit vector. Moreover,, λ is tangent to C, since its dx i components λi , when the curve C is referred to a rectangular Cartesian coordinate Y,, becomes λi = . ds These are precisely the direction cosines tangent vector to the curve C. r of the r Consider a pair of unit vectors and µ (with components λi and µ i respectively) at any point λ r P of C. Let λ is tangent to C at P Fig. (7.6). λ λ + dλ P(x)

(Q(x + dx) µ

C r + dr

r

O Fig. 7.6

r r The cosine of the angle θ between λ and µ is given by the formula i j cos θ = g ij λ λ

r r and if λ and µ are orthogonal, then equation (5) becomes g ij λi µ j = 0

...(6)

δλi j δλj i λ + gij λ =0 δs δs

…(7)

r Any vector µ satisfying equation (6) is said to be normal to C at P.. Now, differentiating intrinsically, with respect to the are parameter s, equation (4), we get g ij

as g ij is constant with respect to s. Interchange indices i and j in second term of equation (7) we get g ij

...(5)

δλi j δλi j λ + gij λ =0 δs δs

Since g ij is symmetric. Then 2 g ij λi

δλj =0 δs

136


g ij λi

⇒

δλ j =0 δs

δλj either vanishes or is normal to C and if does not vanish δs δλj we denote the unit vector co-directional with by µ j and write δs we see that the vector

µj =

j 1 δλ j K = δλ , δs K δs

...(8)

where K > 0 is so chosen as to make µ j a unit vector.. The vector µ j is called the Principal normal vector to the curve C at the point P and K is the curvature of C. r r The plane determined by the tangent vector λ and the principal normal vector µ is called the osculating plane to the curve C at P. r Since µ is unit vector g ij µi µ j = 1

...(9)

Also, differentiating intrinsically with respect to s to equation (6), we get g ij

δλj j δλ j µ + g ij λi =0 δs δs

or g ij λi

δµ j δλi j µ = − gij δs δs = − Kg ij µ i µ j Since

g ij λi g ij λi

δλi = K µi δs

δµ j = − K , since g ij µ i µ j = 1. δs

δµ j +K =0 δs

δµ j + g ij λi λ j K = 0 as g ij λi λj = 1 δs  δµ j  g ij λi  + Kλ j  = 0  δs 

g ij λi

This shows that the vector

δµ j + Kλ j is orthogonal to λi . δs

Geometry

137

r Now, we define a unit vector v , with components v j , by the formula  1  δµ j j v i = τ  δs + Kλ   

...(10)

r r δµi r + Kλi the vector v will be orthogonal to both λ and µ . δs To choose the sign of τ in such a way that where τ =

g eijk λi µ j v k = 1 r r so that the triad of unit vectors λ , µ and νr forms a right handed system of axes. Since eijk

∂ xi g = is a relative tensor of weight −1 and ∂xj

...(11)

2

it follows that ε ijk = g eijk is an

obsolute tensor and hence left hand side of equation (11) is an invariant v k in equation (11) is determined by the formula ijk v k = ε λi µ j

...(12)

1 ijk α where λ i and µ j are the associated vectors g iα λα and g iα µ and ε ijk = e is an absolute tensor.. g The number τ appearing in equation (10) is called the torsion of C at P and the vector vr is the binormal. We have already proved that in Theorem 7.2, Pg. 127. r ∂A r = A,αi a α ∂x i r if the vector field A is defined along C, we can write r i ∂A ∂x i α ∂x r A a = ...(13) , i ∂s α ∂x i ∂s Using definition of intrinsic derivative, i δA α α dx = A,i δs ds Then equation (13) becomes r r r dA δA α r ∂A dx i dA a ⋅ ⋅ ⋅ ; as = = ...(14) ds δs α ds ∂x i ds r r Let r be the position vector of the point P on C then the tangent vector λ is determined by r dr r r = λi ai = λ ds from equation (14), we get r r dλ δλα r r d 2r = aα = c ...(15) ⇒ 2 = ds δs ds

138


where

r r c is a vector perpendicular to λ.

r

r

With each point P of C we can associate a constant K , such that c K = µ is a unit vector.. Since

r c r = µ K 1 δλα r r a , from (15) µ = K δs α

from equation (8), we get α r r µ = µα aα , since µ α = 1 δλ K δs

7.8 SERRET-FRENET FORMULA The serret-frenet formulas are given by

δλ 1 δλi δλi i or = Kµ , K > 0 whereK = K δs δs δs i  1  δµ δµi i i δµi + K λ i i i (ii) ν = τ  δs = τν – Kλ where τ = δs + K λ  or δs   k δν (iii) = – τµ k δs First two formulas have already been derived in article (7.7), equation (8) and (10). Proof of (iii) From equation (12), article (7.7), we have i

(i)

µi =

ε ijk λ i µ j = ν k where λi , µ i , ν k are mutually orthogonal. Taking intrinsic derivative with respect to s, we get δµ j δλi δνk µ j + ε ijk λ i = δs δs δs From formulas (i) and (ii), we get ε ijk

ε ijk Kµ iµ j + εijk λ i (τν j – Kλ j ) = ε ijk λ i (τν j – Kλ j ) = ε ijk λ iν j τ – Kε ijk λ j λi =

δνk δs δνk , δs δνk δs

Since ε ijk µ i µ j = 0

Geometry

139

ε ijk λ i λ j = 0

Since

δνk δs

ε ijk λi ν j τ =

Since ε ijk λ i ν j = µ k , but ε ijk are skew-symmetric. Then ε ijk λ i ν j = – µ k So, – µkτ =

δνk δs

δνk k = – τµ δs

or

δν i = – τµi δs This is the proof of third Serret-Frenet Formula Expanded form of Serret-Frenet Formula. ⇒

(i)

(ii)

(iii)

dλi  i  j dx k i + λ ds  j k  ds = Kµ

or

d 2 xi  i  dx j dx k +  = Kµ i ds 2  j k  ds ds

dµi  i  j dx k +  µ i i ds  jk  ds = τν – Kλ dv i  i  j dx k i +  ν ds  jk  ds = – τµ

EXAMPLE 8 Consider a curve defined in cylindrical coordinates by equation  x1 = a  2  x = θ( s )  x3 = 0  This curve is a circle of radius a. The square of the element of arc in cylindrical coordinates is 1 2 1 2 2 2 3 2 ds 2 = (dx ) + ( x ) (dx ) + (dx )

so that g11 = 1,

g22 = (x1 )2 ,

g33 = 1,

gij = 0,

i≠ j

It is easy to verify that the non-vanishing Christoffel symbols are (see Example 3, Page 61)  1 2 2 1   = x1 ,   =   = 1 . 22 12 21 x      

140


dθ dx i The components of the tangent vector λ to the circle C are λi = so that λ1 = 0, λ2 = , ds ds λ3 = 0. Since λ is a unit vector, gijλiλj= 1 at all points of C and this requires that

(x1 )2  ddsθ 

2

2

2  dθ  = a   =1  ds 

2

dθ 1 So,   = 2 and by Serret-Frenet first formula (expanded form), we get  ds  a  1  2 dx 2 dλ1  1  j dx k 1 + λ Kµ = ds  j k  ds =  2 2λ ds = – a     1

 2  2 dx1 dλ2  2  j dx k Kµ2 = ds +  j k λ ds =  2 1λ ds = 0     dλ3  3  j dx k Kµ = ds +  j k λ ds = 0   3

Since µ is unit vector, g ij µi µ j = 1 and it follows that K =

1 , 1 = –1, µ 2 = 0, µ 3 = 0 a µ

Similarly we can shows that τ = 0 and ν1 = 0, ν 2 = 0, ν 3 = 1. 7.9 EQUATIONS OF A STRAIGHT LINE Let Ai be a vector field defined along a curve C in E 3 such that i C : x i = x (s ) .

s1 ≤ s ≤ s 2,

(i = 1, 2, 3).

s being the arc parameter. If the vector field Ai is parallel then from article 7.6 we have δA i =0 δs or

dAi  i  α dx β + A ds α β ds = 0

(1)

We shall make use of equation (1) to obtain the equations of a straight line in general curvilinear r coordinates. The characteristic property of straight lines is the tangent vector λ to a straight line is r directed along the straight line. So that the totality of the tangent vectors λ forms a parallel vector field.

Geometry

141

i Thus the field of tangent vector λ =

dxi must satisfy equation (1), we have ds

d 2 xi  i  dx α dx β δλi + = = 0  δs ds 2 α β ds ds The equation

d 2 xi  i  d x α d x β +  = 0 is the differential equation of the straight line. ds 2 α β ds d s

EXERCISE 1. Show that

d (gij AiB j ) dt

2. Show that Ai , j – A j , i =

= g ij

δA i j δB i B + g ij A i δt δt

∂ Ai ∂A j – ∂x j ∂ xi

3. If Ai = g ij A j show that Ai , k = giα Aα,k 4. Show that

d ( gij Ai B j ) dxk

i i = Ai,k B + Bi,k A

5. Show that δ2 λi dK i µ – K ( τνi – Kλi ) = ds ds 2 δ 2µi δs2 δ 2νi

=

dτ i dK i ν – ( K 2 + τ2 )µi – λ ds ds dτ

= τ( Kλ – τv ) – µ ds δs2 i

i

i

6. Find the curvature and tension at any point of the circular helix C whose equations in cylindrical coordinates are C : x1 = a, x2 = θ, x 3 = θ Show that the tangent vector λ at every point of C makes a constant angle with the direction of X 3 axis. Consider C also in the form y1 = a cosθ, y2 = a sinθ, y3 = θ. Where the coordinates yi are rectangular Cartesion.

CHAPTER – 8

ANALYTICAL MECHANICS 8.1 INTRODUCTION Analytical mechanics is concerned with a mathematical description of motion of material bodies subjected to the action of forces. A material body is assumed to consist of a large number of minute bits of matter connected in some way with one another. The attention is first focused on a single particle, which is assumed to be free of constraints and its behaviour is analyzed when it is subjected to the action of external forces. The resulting body of knowledge constitutes the mechanics of a particle. To pass from mechanics of a single particle to mechanics of aggregates of particles composing a material body, one introduces the principle of superposition of effects and makes specific assumptions concerning the nature of constraining forces, depending on whether the body under consideration is rigid, elastic, plastic, fluid and so on. 8.2 NEWTONIAN LAWS 1. Every body continues in its state of rest or of uniform motion in a straight line, except in so far as it is compelled by impressed forces to change that state. 2. The change of motion is proportional to the impressed motive force and takes place in the direction of the straight line in which that force is impressed. 3. To every action there is always an equal and contrary reaction; or the mutual actions of two bodies are always and oppositely directed along the same straight line. The first law depends for its meaning upon the dynamical concept of force and on the kinematical idea of uniform rectilinear motion. The second law of motion intorduces the kinematical concept of motion and the dynamical idea of force. To understand its meaning it should be noted that Newton uses the term motion in the sense of momentum, i.e., the product of mass by velocity, this, "change of motion" means the time of change of momentum. In vector notation, the second law can be stated as r d (mv ) r … (1) F = dt

Analytical Mechanics

143

If we postulate the invariance of mass then equation (1) can be written as r r … (2) F = ma r d (mv ) r from (1) if = 0. F = 0 then dt So that r m v = constant. hence vr is constant vector.. Thus the first law is a consequence of the second. The third law of motion states that accelerations always occur in pairs. In term of force we may say that if a force acts on a given body, the body itself exerts an equal and oppositely directed force on some other body. Newton called the two aspects of the force of action and reaction. 8.3 EQUATIONS OF MOTION OF A PARTICLE THEOREM 8.1 The work done in displacing a particle along its trajectory is equal to the change in the kinetic energy of particle. Proof: Let the equation of path C of the particle in E 3 be i C : x i = x (t )

… (1)

and the curve C the trajectory of the particle. Let at time t, particle is at P {x i (t )}. If v i be the component of velocity of moving particle then dx i dt and if ai be the component of acceleration of moving particle then vi =

ai

δv i dv i  i  j d x k = δt = dt +  j k v d t  

ai

d 2 x i  i  dx j dx k +  = dt 2  j k  dt dt

… (2)

… (3)

δv i  i  is the intrinsic derivative and the   are the Christoffel symbols calculated from the δt  jk metric tensor gij If m be the mass of particle. Then by Newton’s second law of motion where

Fi = m

δv i = ma i δt

… (4)

r We define the element of work done by the force F in producing a displacement drr by invariant r dW = F .drr . r Since the components of F and drr are F i and dxi respectively..

144


Then dW = gij F i dxj … (5) j i = F j dx where F j = gij F The work done in displacing a particle along the trajectory C, joining a pair of points P 1 and P 2, is line integral

∫

W=

P2

P1

Fi dx i

… (6)

using equation (4) then equation (6) becomes W=

∫

=

∫

W=

∫

m g ij

δv i dx j δt

m g ij

δv i dx j dt δt dt

P2

P1

P2

P1

P2

P1

δv i j v dt δt

m g ij

… (7)

Since g ij v i v j is an invariant then δ (g ij vi v j )

=

d ( g v iv j ) dt ij

δt d δvi j ( g ij v i v j ) = 2 g ij v or dt δt δvi j 1 d (g ij v i v j ) g ij v = ⇒ 2 dt δt using this result in equation (7), we get W=

∫

P2 m

P1

d ( g v i v j ) dt 2 dt ij

m [g v i v j ]PP12 2 ij Let T2 and T1 is kinetic energy at P 2 and P 1 respectively. W = T2 – T1 W=

m m v2 gij v i v j = is kinetic energy of particle. 2 2 We have the result that the work done by force F i in displacing the particle from the point P 1 to 1 the point P 2 is equal to the difference of the values of the quantity T = mv 2 at the end and the 2 beginning of the displacement. where T =

8.4 CONSERVATIVE FORCE FIELD The force field F i is such that the integral W =

∫

P2

P1

Fi dx i is independent of the path.

Therefore the integrand F i dxi is an exact differential dW = F i dxi

… (8)


145

of the work function W. The negative of the work function W is called the force potential or potential energy. We conclude from equation (8) that ∂V Fi = – i … (9) ∂x i where potential energy V is a function of coordinates x . Hence, the fields of force are called conservative ∂V . ∂x i THEOREM 8.2 A necessary and sufficient condition that a force field Fi, defined in a simply connected region, be conservative is that Fi,j = Fj,i. if F i = –

Proof: Suppose that F i conservative. Then F i = –

∂V ∂x i

Now, ∂Fi  k  −   Fk ∂x j i j  ∂V ∂  − i   ∂x  −  k F F i,j =   k ∂x j i j  F i,j =

=−

k ∂ 2V −  Fk j i ∂x ∂x  j i 

… (1)

and F j,i =

∂F j

k −   Fk ∂x  j i i

Similarly, F j,i = –

∂ 2V k −  Fk i j ∂x ∂x  ji

From equation (1) and (2), we get F i,j = F j,i conversely, suppose that F i,j = F j,i Then ∂F j  k  ∂Fi  k  −   Fk = −   Fk j ∂x ∂x i  j i i j  ⇒ ∂F j  k  ∂Fi as   due to symmetry.. j = ∂x i i j ∂x

… (2)

146


Take F i = −

∂V ∂x i

Then ∂Fi ∂ 2V − j i j = ∂x ∂x ∂x ∂ 2V = − i j ∂x ∂x ∂  ∂V  −  = ∂x i  ∂x j  ∂F j ∂Fi j = ∂x ∂x i Fi = −

So, we can take

∂V . ∂x i

Hence, F i is conservative. 8.5 LAGRANGEAN EQUATIONS OF MOTION Consider a particle moving on the curve C : x i = x i (t ) At time t, let particle is at point P (xi). The kinetic energy T =

Since x& i = v i.

1 mv 2 can be written as 2 1 i j T = m g ij x& x& 2

1 m g jk x& j x& k 2 Differentiating it with respect to x& i , we get or

T=

… (1)

 ∂x& j k &k  1 ∂T j ∂x   & & m g x + x jk = 2  ∂x& i i  & ∂ x ∂x& i  

(

)

=

1 m g jk δij x& k + δki x& j 2

=

1 1 m g jk δ ij x& k + m g jk δ ki x& j 2 2

(

1 m gik x& k + g ji x& j 2 1 j j = m (gij x& + gij x& ) 2

=

) as

gij = g ji


147

∂T = m g ij x& j ∂x& i ∂T = m g ik x& k ∂x& i

or

…(2)

Differentiating equation (2) with respect to t, we get

(

d  ∂T  d g x& k   = m dt  ∂x& i  dt ik

)

d k k = m  g ik x& + g ik &x&   dt   ∂g ik dx j k  k m =  ∂x j dt x& + g ik &x&    d ∂T ∂g j k k = m ikj x& x& + m g ik &x& dt ∂x& i ∂x 1 m g jk x& j x&k 2 Differentiating it with respect to xi, we get ∂T 1 ∂g jk j k m x& x& i = ∂x 2 ∂x i Now,

… (3)

Since T =

d  ∂T  dt  ∂x& i

… (4)

 ∂T ∂g 1 ∂g  − i = m ikj x& j x&k + m g ik&x&k − m jki x& j x& k  ∂x 2 ∂x ∂x k = m g ik &x& +

1 gik j k 1 1 ∂g m j x& x& + m g ik x& j x& k − m jki x& j x& k 2 ∂x 2 2 ∂x

k = m g ik &x& +

1 g ik j k 1 ∂g ij k j 1 ∂g jk j k m x& x& + m k x& x& − m i x& x& 2 ∂x j 2 ∂x 2 ∂x

1  ∂gik ∂g ij ∂g jk  j k k  x& x& = m g ik &x& + 2 m  j + k − ∂x ∂x i   ∂x = m g ik &x&k + m [ jk , i ] x& j x& k = m g ik &x&k + m g il gil [ jk , i ]x& j x& k = m g il &x&l + m g il gil [ jk , i]x& j x& k

148


[

= m g il x&&l + g il [ jk , i ]x& j x& k

]

   l  = m g il  &x&l +  x& j x& k  ,  jk   d  ∂T  ∂T  − = m g il a l , dt  ∂x& i  ∂x i where al is component of acceleration or

l  as g il [ jk , i] =    j k

 l  j k l l Since a = x&& +  j k x& x&  

d  ∂T  ∂T  − = m ai dt  ∂x& i  ∂x i

d  ∂T  ∂T = Fi … (5)  − dt  ∂x& i  ∂x i where F i = m ai, component of force field. The equation (5) is Lagrangean equation of Motion. For a conservative system, F i = – d  ∂T  dt  ∂x& i

∂V . Then equation (5) becomes ∂x i

∂V  ∂T − i = − i ∂x  ∂x

d  ∂T  ∂(T − V )  − =0 … (6) dt  ∂x& i  ∂x i Since the potential V is a function of the coordinates xi alone. If we introduce the Lagrangean function L=T–V Then equation (6) becomes or

d  ∂L  ∂L  − =0 dt  ∂x& i  ∂x i

… (7)

EXAMPLE 1 Show that the covariant components of the acceleration vector in a spherical coordinate system with ds 2 = (dx1 ) 2 + ( x1 dx 2 )2 + (x1 ) 2 sin 2 x 2 (dx 3 ) 2 are a1 = &x&1 − x1 (x& 2 )2 − x1 (x& 3 sin x 2 ) 2

and

[

]

a2 =

d ( x1 )2 x& 2 − (x1 ) 2 sin x 2 cos x 2 ( x& 3 )2 dt

a3 =

d (x1 sin x 2 ) 2 x& 3 dt

[

]


149

Solution In spherical coordinate system, the metric is given by ds2 = (dx1 ) 2 + ( x1 dx 2 )2 + (x1 ) 2 sin 2 x 2 (dx 3 ) 2 If v is velocity of the paiticle then 2

2

2  2  3  ds   dx1  1 2  dx  1 2 2  dx  2 = + ( x ) + ( x sin x )   v =  dt   dt   dt   dt     

2

v 2 = ( x&1 )2 + (x1 ) 2 ( x& 2 ) 2 + ( x1 sin x 2 )2 (x& 3 ) 2 If T be kinetic energy then 1 2 mv 2 1 1 2 1 2 2 2 1 2 2 3 2 T = m (x& ) + ( x ) (x& ) + ( x sin x ) (x& ) 2 By Lagrangean equation of motion d  ∂T  ∂T  − = F i and m ai = F i dt  ∂x& i  ∂x i where F i and ai are covariant component of force field and acceleration vector respectively. So, T=

[

]

d  ∂T  ∂T  − = m ai dt  ∂x& i  ∂x i

… (2)

Take i = 1, m a1 =

d  ∂T  ∂T  − dt  ∂x&1  ∂x 1

m a1 =

1 d m m (2x&1 ) − 2x 1 ( x& 2 )2 + 2 x1 (sin x 2 ) 2 ( x& 3 )2 2 dt 2

from (1), we get

a1 =

[

[

dx&1 − x1 ( x& 2 )2 + x1 (sin x 2 ) 2 ( x& 3 )2 dt

]

]

a1 = &x&1 − x1 (x& 2 )2 − x1 (x& 3 sin x 2 ) 2 Take i = 2, m a2 =

d  ∂T  dt  ∂x& 2

m a2 =

1 d 1 2 2 1 m ( x ) 2 x& − m 2( x1 )2 sin x 2 cos x 2 (x&3 ) 2 2 dt 2

a2 =

[

[

… (1)

 ∂T − 2  ∂x

]

]

d ( x1 )2 x& 2 − (x1 ) 2 sin x 2 cos x 2 ( x& 3 )2 dt

150


Take i = 3 m a3 =

d  ∂T  ∂T  − dt  ∂x&3  ∂ x 3

m a3 =

1 d m 2( x& 3 ) ( x1 sin x 2 )2 − 0 2 dt

a3 =

[

[

d 3 1 2 x& ( x ) (sin x 2 ) 2 dt

]

]

EXAMPLE 2 Use Lagrangean equations to show that, if a particle is not subjected to the action of forces then its trajectory is given by yi = ait + bi where ai and bi are constants and the yi are orthogonal cartesian coordinates. Solution If v is the velocity of particle. Then we know that, v 2 = g ij y& i y& j where yi are orthogonal cartesian coordinates. Since gij = 0, i ≠ j gij = 1, i = j So, v 2 = ( y& i ) 2 But, 1 2 mv , T is kinetic energy.. 2 1 i 2 T = m( y& ) 2 The Lagrangean equation of motion is T=

d  ∂T  ∂T  − = Fi dt  ∂y& i  ∂yi Since particle is not subjected to the action of forces. So, F i = 0 Then

d 1   m 2 y& i  – 0 = 0 dt  2  m

dy& i =0 dt


151

dy& i =0 dt

or

y& i = ai yi = ait + bi ⇒ i i where a and b are constant. ⇒

EXAMPLE 3 Prove that if a particle moves so that its velocity is constant in magnitude then its acceleration vector is either orthogonal to the velocity or it is zero. Solution If v i be the component of velocity of moving particle then vi

dx i = dt

or

v i = x& i

given |v| = constant. Since g ij vi v j = | v |2 = constant Taking intrinsic derivative with respect to t, we get δ (g v i v j ) = 0 δt ij  δv i j δv j g ij  v + vi δt  δt

  =0  

g ij

δv i j δv j v + g ij v i =0 δt δt

g ij

δv i j δv i v + g ji v j = 0, (Interchange dummy index i and j in second term) δt δt 2 g ij g ij

δv i j v = 0 as gij = gji δt δvi j v =0 δt

This shows that acceleration vector

δv i δv i = 0. is either orthogonal to v i or zero i.e., δt δt

152


8.6 APPLICATIONS OF LAGRANGEAN EQUATIONS (i) Free-Moving Particle If a particle is not subjected to the action of forces, the right hand side of equation (5), 148, vanishes. Then we have d  ∂T  ∂T  − =0 dt  ∂x& i  ∂x i

… (1)

1 m y& i y& i . 2 Hence, the equation (1) becomes m &y&i = 0. Integrating it we get yi = ait + bi, which represents a straight line. (ii) Simple Pendulum Let a pendulum bob of mass m be supported by an extensible string. In spherical coordinates, the metric is given by If xi be rectangular coordinate system, then T =

ds2 = dr 2 + r 2 d φ2 + r 2 sin 2 φ d θ 2 If T be the kinetic energy, then T=

1 2 1 mv = m (r&2 + r 2 φ& 2 + r 2 sin 2 φ θ& 2 ) 2 2

O

Y2 r φ R mg cos φ

Y1

θ mg sin φ Y3

P mg

Fig. 8.1.

from Lagrangean equation of motion d  ∂T  ∂T  − = Fi dt  ∂x& i  ∂x i x1 = r , x 2 = φ, x 3 = θ. So, take

x1 = r

i = 1, 2, 3

… (1)


153

d  ∂T  ∂T  − = mg cos φ − R dt  ∂r&  ∂r from (1), we have R &r& − rφ&2 − r sin 2 φ θ&2 = g cos φ − m

… (2)

Take x2 = φ, we have r &φ& + 2r&φ& − r sin φ cos φ θ& 2 = – g sin φ

… (3)

and take x3 = θ, we have d 2& 2 (r θ sin φ) = 0 dt

… (4)

If the motion is in one plane, we obtain from equations (2), (3), and (4), by taking θ& = 0. R r&& − r φ& 2 = g cos φ − m r &φ& + 2r&φ& = – g sin φ g If r& = 0, we get, &φ& = −  sinφ which is equation of simple pendulum supported by an r inextensible string. For small angles of oscillation the vibration is simple harmonic. For large vibration the solution is given in the term of elliptic functions. 8.7 HAMILTON’S PRINCIPLE If a particle is at the point P 1 at the time t1 and at the point P 2 at the time t2, then the motion of the particle takes place in such a away that

∫

t2 t1

(δT + Fi δxi ) dt = 0

where xi = xi (t) are the coordinates of the particle along the trajectory and xi + δx i are the coordinates along a varied path beginning at P 1 at time t1 and ending at P 2 at time t2. Proof: Consider a particle moving on the curve t1 ≤ t ≤ t 2 C : x i = x (t ), At time t, let particle is at P(xi). If T is kinetic energy. Then i

T=

1 m g ij x& i x& j 2

or T = T ( x i , x& i ) i.e, T is a function of xi and x& i . Let C ' be another curve, joining t1 and t2 close to be C is C' : x i (ε, t ) = x i (t ) + δx i (t)

154


At t1 and t2 xi = x i = x i + ε δ x i ⇒

δx i (t1 ) = 0 and δx i (t 2 ) = 0

But T = T ( x i , x& i ). If δT be small variation in T.. Then ∂T i ∂T i δT = ∂x&i δx& + ∂x i δx Now,

∫ {(δT + F )δx }dt = ∫ t2

t1

∫

=

t1

∂T i δx dt + ∂x i

t2 t1

∫

t2 t1

∂T i δx& dt + ∂x&i

∫

∫

t2 t1

∂T i δx dt + ∂x i

t

 ∂T i  2  ∂x& i δx  −   t1

∫

t2 t1

t

 ∂T i  2 then  & i δx  = 0.  ∂x  t1 So,

∫ (δT + F δx )dt = ∫

∫

∫ (δT + F δx )dt = ∫ t1

i

i

∂x

t1

+ t2

∂T

t2

i

i

t1

t2 t1

i

i

δ x dt −

∫t

t2 1

d dt

 ∂T  i  ∂ x&

 i δ x dt 

Fi δx i dt

t2  ∂T

t1

 i d  ∂T   i −  i  + Fi δx dt  ∂x dt  ∂x&  

since particle satisfies the Lagrangean equation of motion. Then d  ∂T  ∂T  − = Fi dt  ∂x& i  ∂x i or

t1

Fi δx i dt

d  ∂T  i  δx dt + dt  ∂x&i 

Since δx i (t1 ) = 0, δx i (t 2 ) = 0.

t2

t2

∂T as 1st term ∂x& i

Integrating second term by taking

=

∂T ∂T   i δx i + i δx& i + Fi δx i  dt ∂x&  ∂x 

t2 

i

i

∂T d  ∂T  −   = – Fi ∂x i dt  ∂x& i 

∫

t2 t1

Fi δxi dt


155

So,

∫ ∫

t2 t1 t2

t1

(δT + Fi δx i ) dt =

∫

t2 t1

[−Fi + Fi ] δxi dt

(δT + Fi δx i ) dt = 0

Proved.

8.8 INTEGRAL OF ENERGY THEOREM 8.3 The motion of a particle in a conservative field of force is such that the sum of its kinetic and potential energies is a constant. Proof: Consider a particle moving on the curve C : x i = x i (t ), t1 ≤ t ≤ t 2 i At time t, let particle is at P (x ). If T is kinetic energy. Then

or

T=

1 m g ij x& i x& j 2

T=

1 m g ij v i v j 2

As T is invariant. Then Taking intrinsic derivative with respect to t, we get dT δT = dt δt =

δ 1   m gij v i v j  δt  2 

=

 δv i j 1 δv j  m g ij  v + vi 2 δt   δt

=

1  δv i j δv j i  m  gij v + g ij v  2  δt δt 

1  δvi j δvi j  = 2 m  gij δt v + g ji δt v  , Since i and j are dummy indices.   1 δv i j m 2 g v = ij 2 δt dT δv i j v = m gij dt δt or

dT δv j i m g v = ij dt δt

as g ij = g ji

156


j i = m g ij a v as

dT = m ai v i, dt

δv j = aj δt

since gij aj = ai

dT = F i v i, dt Since F i = m ai is a covariant component of force field. But given F i is conservative, then ∂V , where V is potential energy.. ∂x i

Fi = – So,

dT ∂V i v =– dt ∂x i ∂V dx i ∂x i dt

= −

dT dV = − dt dt dT dV d + = 0 ⇒ (T + V ) = 0 dt dt dt ⇒

T + V = h, where h is constant. Proved.

8.9 PRINCIPLE OF LEAST ACTION Let us consider the integral A=

∫

P2

mv.ds

(1)

p1

evaluated over the path i C : x i = x (t ), t1 ≤ t ≤ t 2 where C is the trajectory of the particle of mass m moving in a conservative field of force. In the three dimensional space with curvilinear coordinates, the integral (1) can be written as

A=

∫

=

∫

P2 p1

m g ij

t ( P2 ) t ( P1 )

dx i dx j dt

m gij

dx i dx j dt dt dt


Since T =

157

1 dx i dx j m g ij , we have 2 dt dt A=

∫

t ( P2 )

2T dt

t ( P1 )

This integral has a physical meaning only when evaluated over the trajectory C, but its value can be computed along any varied path joining the points P 1 and P 2. Let us consider a particular set of admissible paths C' along which the function T + V, for each value of parameter t, has the same constant value h. The integral A is called the action integral. The principle of least action stated as “of all curves C' passing through P 1 and P 2 in the neighbourhood of the trajectory C, which are traversed at a rate such that, for each C', for every value of t, T + V = h, that one for which the action integral A is stationary is the trajectory of the particle.” 8.10 GENERALIZED COORDINATES In the solution of most of the mechanical problems it is more convenient to use some other set of coordinates instead of cartesian coordinates. For example, in the case of a particle moving on the surface of a sphere, the correct coordinates are spherical coordinates r, θ, φ where θ and φ are only two variable quantities. Let there be a particle or system of n particles moving under possible constraints. For example, a point mass of the simple pendulum or a rigid body moving along an inclined plane. Then there will be a minimum number of independent coordinates required to specify the motion of particle or system of particles. The set of independent coordinates sufficient in number to specify unambiguously the system configuration is called generalized coordinates and are denoted by q 1 , q 2 , ... q n where n is the total number of generalized coordinates or degree of freedom. Let there be N particles composing a system and let x(i α) , (i = 1, 2,3), (α = 1, 2,...N ) be the positional coordinates of these particles referred to some convenient reference frame in E 3. The system of N free particles is described by 3N parameters. If the particles are constrained in some way, there will be certain relations among the coordinates x(i α) and suppose that there are r such independent relations, f i ( x1(1) , x(21) , x 3(1 ) ; x1( 2 ) , x(22 ) , x(32 ) ;...x1( N ) x(2N ) x(3N ) ) = 0, (i = 1, 2, ..., r)

… (1)

By using these r equations of constraints (1), we can solve for some r coordinates in terms of the remaining 3N – r coordinates and regard the latter as the independent generalized coordinates qi. It is more convenient to assume that each of the 3N coordinates is expressed in terms of 3N – r = n independent variables qi and write 3N equations. x(i α) = x(i α) (q1 , q 2 ,...,q n , t )

… (2)

where we introduced the time parameter t which may enter in the problem explicitly if one deals with moving constraints. If t does not enter explicitly in equation (2), the dynamical system is called a natural system. The velocity of the particles are given by differentiating equations (2) with respect to time. Thus x&(i α) =

∂x i(α ) ∂q j

q& j +

∂x i(α ) ∂t

… (3)

158


The time derivatives q& i of generalized coordinates qi the generalized velocities. For symmetry reasons, it is desirable to introduced a number of superfluous coordinates qi and describe the system with the aid of k > n coordinates q1, q2,..., qk. In this event there will exist certain relations of the form f j (q1 ....,q k , t ) = 0

… (4)

Differentiating it we get ∂f j i ∂f j q& + … (5) ∂t = 0 ∂q i It is clear that they are integrable, so that one can deduce from them equations (4) and use them to eliminate the superfluous coordinates . In some problems, functional relations of the type … (6) F j (q1 , q 2 ,...,q k ; q&1 ,...,q& k , t ) = 0, ( j = 1, 2, 3, ..., m) arise which are non-integrable. If non-integrable relations (6) occurs in the problems we shall say that the given system has k – m degrees of freedom, where m is the number or independent non-integrable relations (6) and k is the number of independent coordinates. The dynamical systems involving nonintegrable relations (6) are called non-holonomic to distinguish them from holonomic systems in which the number of degrees of freedom is equal to the number of independent generalized coordinates. In other words, a holonomic system is one in which there are no non-integrable relations involving the generalized velocities. 8.11 LAGRANGEAN EQUATIONS IN GENERALIZED COORDINATES Let there be a system of particle which requires n independent generalized coordinates or degree of freedom to specify the states of its particle. The position vectors xr are expressed as the function of generalized coordinates q i , (i = 1, 2,...,n) and the time t i.e., 1 2 n (r = 1, 2, 3, ) xr = xr (q , q ,...,q , t ); The velocity x& r of any point of the body is given by ∂x r dq j ∂x r x& = ∂q j dt + ∂t r

=

∂x r j ∂x r q& + , ∂t ( j = 1, 2, ..., n) ∂q j

where q& j are the generalized velocities. Consider the relation, with n degree of freedom, 1 2 n xr = xr (q , q ,...,q )

… (1)

involve n independent parameters qi. The velocities x& r in this case are given by ∂x r j q& , = x& ∂q j r

(r = 1, 2, 3; j = 1, 2, ..., n)

…(2)


159

where q& j transform under any admissible transformation, q k = q k (q1 ,...,q n ),

(k = 1, 2, ..., n)

… (3)

in accordance with the contravariant law. The kinetic energy of the system is given by the expression of the form T=

1 Σ m g rs x& r x& s , 2

(r,s = 1,2,3,)

… (4)

where m is the mass of the particle located at the point xr. The grs are the components of the metric tensor. Substituting the value of x& r from equation (2), then equation (4) becomes 1 ∂x r ∂x s i j q& q& T = 2 Σ m g rs i ∂q ∂q j T=

1 a q&i q& j 2 ij

aij = Σ m g rs

where

∂x r ∂r s , ∂q i ∂q j

… (5) (r, s = 1, 2, 3), (i, j = 1, ..., n)

1 a q&i q& j is an invariant and the quantities aij are symmetric, we conclude that the aij 2 ij are components of a covariant tensor of rank two with respect to the transformations (3) of generalized coordinates. Since T =

Since the kinetic energy T is a positive definite form in the velocities q& i , | aij |> 0. Then we construct the reciprocal tensor aij . Now, from art. 8.5, Pg. 146, by using the expression for the kinetic energy in the form (5), we obtain the formula, d  ∂T dt  ∂q& i

 ∂T  l  l  j k −    ∂qi = a il  &q& +  j k q& q&      

(6)

 l  where the Christoffel symbol   are constructed from the tensor akl.  jk Put l  q&&l +  q& j q& k = Q l jk so, the equation (6), becomes d  ∂T dt  ∂q& i

 ∂T l −  ∂qi = ail Q  = Qi (i = 1, 2, ..., n)

… (7)

160


∂x& r ∂x r ∂x& r ∂ 2 xr j ∂x&r d  ∂x r  & = , = q = Now, from the realtions & j and qi dt  q i  and using equations ∂q ∂q j ∂qi ∂x i ∂q j ∂ ∂  (2) and (4). Then by straightforward calculation, left hand member of equation (7) becomes d  ∂T dt  ∂q& i

 ∂T ∂x r − m a = r  ∂qi ∂q i  in which a j = g ij a i is acceleration of the point P. Also, Newton's second law gives m ar = F r

∑

… (8)

… (9)

Fr's

where are the components of force F acting on the particle located at the point P.. From the equation (9), we have

∑

m ar

∂x r = ∂q i

∑

Fr

∂x r ∂q i

∑

Fr

∂x r ∂q i

and equation (8) can be written as d  ∂T dt  ∂q& i

 ∂T −  ∂qi = 

… (10)

comparing (7) with (8), we conclude that Qi =

∑

Fr

∂x r ∂q i

where vector Qi is called generalized force. The equations d  ∂T dt  ∂q& i

 ∂T −  ∂qi = Qi 

… (11)

are known as Lagrangean equations in generalized coordinates. They give a system of n second order ordinary differential equations for the generalized coordinates qi. The solutions of these equations in the form C : q i = qi (t) Represent the dynamical trajectory of the system. If there exists a functions V (q1 , q 2 ,...,q" ) such that the system is said to be conservative and for such systems, equation (11) assume the form d  ∂L dt  ∂q& i

 ∂L −  ∂qi = 0  where L = T – V is the kinetic potential.

… (12)


161

Since L (q, q& ) is a function of both the generalized coordinates and velocities. ∂L i ∂L i dL = & i q&& + i q& dt ∂q ∂q from (12), we have

∂L d  ∂L i = ∂q dt  ∂q& i

… (13)

 .  

Then equation (13), becomes ∂L i d  ∂L dL q&& +  i = dt dt  ∂q& ∂q& i =

 i  q&  

d  ∂L i  q& dt  ∂q& i 

… (14)

since L = T – V but the potential energy V is not a function of the q& i ∂L i ∂T i q& = q& = 2T i & ∂q ∂q& i since

T=

1 a q& i q& j . 2 ij

Thus, the equation (14) can be written in the form d ( L − 2T ) d (T + V ) =0 = dt dt which implies that T + V = h (constant). Thus, along the dynamical trajectory, the sum of the kinetic and potential energies is a constant. 8.12 DIVERGENCE THEOREM, GREEN'S THEOREM, LAPLACIAN OPERATOR AND STOKE'S THEOREM IN TENSOR NOTATION (i) Divergence Theorem r Let F be a vector point function in a closed region V bounded by the regular surface S. Then r r div F = F ⋅ nˆ ds … (1)

∫

V

∫ S

where nˆ is outward unit normal to S. Briefly the theorem states that the integral with subscript V is evaluated over the volume V while r the integral in the right hand side of (1) measures the flux of the vector quantity F over the surface S. r In orthogonal cartesian coordinates, the divergence of F is given by the formula ∂F 1 ∂F 2 ∂F 3 r + + div F = ∂x 1 ∂x 2 ∂x 3

… (2)

162


r If the components of F relative to an arbitrary curvilinear coordinate system X are denoted by F i then the covariant derivative of F i is F ,ij =

∂F i  i  k +  F ∂x j k j 

r The invariant F, ij in cartesian coordinates represents the divergence of the vector field F . Also, r i j i j F ⋅ nˆ = g ij F n = F ni since g ij n = ni Hence we can rewrite equation (1) in the form

∫ F dV i ,i

∫ F n dS i

=

V

i

… (3)

S

(ii) Symmetrical form of Green's Theorem Let φ ( x1 , x 2 , x3 ) and ψ (x1 , x 2 , x3 ) be two scalar function in V.. Let φ i and ψ i be the gradients of φ and ψ respectively, so that ∂φ and ∂xi Put F i = φψi and from the divergence of we get ∇φ = φ i =

∇ψ = ψ i =

∂ψ ∂x i

F,i j = g ij Fi , j = g ij (φψi , j + ψi φ j ) Substituting this in equation (3), we get

∫g

ij

(φψi , j + ψ i φ j )dV

∫ φψ n dS i

=

V

i

… (4)

S

Since ∇ψ = ψi , then g ij ψi , j = ∇ 2 ψ

… (5)

Also, the inner product g ij ψi φ j can be written as g ij ψi φ j = ∇φ.∇ψ where ∇ denote the gradient and ∇ 2 denote the Laplacian operator.. Hence the formula (4) can be written in the form

∫(g

ij

φψi , j + g ij ψi .φ J ) dV

V

=

∫ φ nˆ ⋅ ∇ψ dS S

∫ (φ∇ ψ + ∇φ ⋅ ∇ψ) dV = ∫ φ nˆ ⋅ ∇ψ dS 2

V

S

∫ φ ∇ ψ dV = ∫ φ nˆ ⋅ ∇ ψ − ∫ ∇φ ⋅ ∇ψ dV 2

V i where nˆ ⋅ ∇ψ = ψi n =

∂ψ . ∂n

S

V

… (6)


163

Interchanging φ and ψ in equation (5), we get

∫ ψ∇ φ dV 2

=

V

∫ ψnˆ ⋅ ∇φ − ∫ ∇ψ ⋅ ∇φ dV S

… (7)

V

Subtracting equation (5) from equation (6), we get

∫ (φ∇ ψ − ψ∇ φ) dV 2

 ∂ψ

2

=

V

∂φ 

∫  φ ∂n − ψ ∂n  dS

… (8)

S

This result is called a symmetric form of Green's theorem. (iii ) Expansion form of the Laplacian Operator The Laplacian of ψ is given by ij ∇ 2 ψ = g ψi , j from (5) when written in the terms of the christoffel symbols associated with the curvilinear coordinates i x covering E 3,

 ∂2 ψ  k  ∂ψ ij ∇ 2 ψ = g  i j −   k i j  ∂x  ∂x ∂x i and the divergence of the vector F is F,ii =

∂F i  i  +  F ∂x i  j i 

   

j

… (9)

…(10)

i ∂ log g   = j i ∂ xj   The equation (10) becomes But we know that

F,ii

or ij If putting F i = g

F,ii

∂F i  ∂  +  j log g  F j = i ∂x  ∂x  =

1 ∂( g F i ) ∂x i g

∂ψ = g ij ψ j in equation (11), we get ∂x j ∂ψ   ∂  g g ij j  1  ∂x  g ij ψ j , i = i ∂x g

But from equation (5), we know that ij ∇ 2 ψ = g ψ j, i

… (11)

… (12)

164


Hence equation (12) becomes ∂ψ   ∂ g g ij j  1  ∂x  ∇ 2 ψ = g ij ψ j ,i = i g ∂x It is expansion form of Laplacian operator. (iv) Stoke's Theorem

r Let a portion of regular surface S be bounded by a closed regular curve C and let F be any vector point function defined on S and on C. The theorem of Stokes states that r r nˆ. curl F ds = F .λˆ ds … (13)

∫

∫

S

C

r where λ is the unit tangent vector to C and curl F is the vector whose components in orthogonal cartesian coordinates are determined from e1 ∂ r curl F = ∂ x1 F1

e2 ∂ ∂x 2 F2

e3 ∂ r ∂x 3 = ∇ × F F3

… (14)

where ei being the unit base vectors in a cartesian frame. We consider the covariant derivative F i,j of the vector F i and form a contravariant vector Gi = – ε ijk F j , k … (15) r we define the vector G to be the curl of F . dx i r i ijk Since nˆ . curl F = ni G = −ε F j ,k ni and the components of the unit tangent vector λ and . ds Then equation (13) may be written as

∫

− εijk F j , k ni ds S

The integral

∫ F dx i

=

∫

C i

Fi

dx i ds ds

… (16)

r is called the circulation of F along the contour C.

c

8.13 GAUSS'S THEOREM The integral of the normal component of the gravitational flux computed over a regular surface S containing gravitating masses within it is equal to 4 πm where m is the total mass enclosed by S. Proof: According to Newton's Law of gravitation, a particle P of mass m exerts on a particle Q of unit m mass located at a distancer r from P. Then a force of magnitude F = 2 . r Consider a closed regular surface S drawn around the point P and let θ be the angle between the unit outward normal to nˆ to S and the axis of a cone with its vertex at P.. This cone subtends an element of surface dS.


165

The flux of the gravitational field produced by m is

∫

r F .nˆ dS =

S

∫ S

m cos θ r 2 dw r 2 cos θ

2

r dw and dw is the solid angle subtended by dS. cos θ Thus, we have, r F .nˆ dS m dw = 4 πm =

where dS =

∫

∫

S

… (1)

S

n

F θ dS

r dω P m S

Fig. 8.2.

If there are n discrete particles of masses mi located within S, then r F .nˆ =

n

∑ i =1

mi cos θ i ri 2

and total flux is

∫

r F .nˆ dS = 4 π

n

∑m

S

… (2)

i

i =1

The result (2) can be easily generalized to continuous distributions of matter whenever such distribution no where melt the surface S. The contribution to the flux integration from the mass element ρ dV contained within V, is cos θρ dV r dS F .nˆ dS = r2 S

∫

∫ S

and the contribution from all masses contained easily within S is

∫ S

r F .nˆ dS =

 cos θ ρ dV   r2 V

∫ ∫ S

  dS  

… (3)

166


where

∫

denotes the volume integral over all bodies interior to S. Since all masses are assumed to be

V

interior to S,r never vanishes. So that the integrand in equation (3) is continuous and one can interchange to order of integration to obtain

∫

r F .nˆ dS =

V

S

 cos θ dS   dV s r2 

∫ ρ  ∫

… (4)

cos θ dS = 4π. Since it represents the flux due to a unit mass contained within S. r2 S Hence But

∫

r

∫ F.nˆ dS

=

∫

4 π ρ dV = 4 πm

… (5)

V

S

where m denotes the total mass contained within S. Proved. Gauss's theorem may be extended to cases where the regular surface S cuts the masses, provided that the density S is piecewise continuous. Let S cut some masses. Let S' and S" be two nearby surfaces, the first of which lies wholly within S and the other envelopes S. Now apply Gauss's theorem to calculate the total flux over S" produced by the distribution of masses enclosed by S since S" does not intersect them. We have r ( F .nˆ )i dS = 4 πm

∫

S"

r where the subscript i on F ⋅ nˆ refers to the flux due to the masses located inside S and m is the total mass within S. On the other hand, the net flux over S' due to the masses outside S, by Gauss's theorem is r ( F ⋅ nˆ)o ds = 0

∫

S' r where the subscript o on F ⋅ nˆ refers to the flux due to the masses located outside S. Now if we S' and S" approach S, we obtain the same formula (5) because the contribution to the r total flux from the integral (F ⋅ nˆ)o dS is zero.

∫

S'

8.14 POISSON'S EQUATION By divergence theorem, we have

∫

r F .nˆ ds =

V

S

and by Gauss's Theorem,

∫

r div F dV

r

∫ F.nˆ ds = 4 π∫ ρ dV from these, we have

S

V


∫

167

r (div F − 4πρ ) dV = 0

v

Since this relation is true for an arbitrary V and the integrand is piecewise continuous, then r div F = 4 πρ By the definition of potential function V, we have r F = −∇V and

div ∇V = ∇ 2V

So,

r div F = 4 πρ div ( −∇V ) = 4 πρ ∇ 2V = – 4 πρ

which is equation of poisson. If the point P is not occupied by the mass, then ρ = 0. Hence at all points of space free of matter the potential function V satisfies Laplace's equation ∇ 2V = 0 8.15 SOLUTION OF POISSON’S EQUATION We find the solution of Poisson’s Equation by using Green’s symmetrical formula. We know that Green's symmetrical formula ∂φ   ∂ψ (φ∇ 2 ψ − ψ∇ 2 φ) dV =  φ ∂n − ψ ∂n  dS … (1) S

∫

∫

V

where V is volume enclosed by S and φ and ψ are scalar point functions. 1 where r is the distance between the points P ( x1 , x 2 , x3 ) and Q ( y1 , y 2 , y 3 ) and V is r the gravitational potential. Put φ =

n

Q (y )

r P(x) n

Fig. 8.3.

S

168


1 has a discontinuity at x i = y i , delete the point P(x) from region of integration by r surrounding it with a sphere of radius ε and volume V'. Apply Green’s symmetrical formula to the 1 region V – V' within which and V possess the desired properties of continuity.. r 2 2 1 = 0. In region V − V', ∇ φ = ∇ r Then equation (1) becomes Since

1 2 ∇ ψ dV = r V −V'

∫

∫ S

 1 ∂ψ ∂ 1r    dS + − ψ  r ∂n ∂n   S'

∫

 1 ∂ψ ∂ 1r    ds − ψ  r ∂n ∂n  

… (2)

where nˆ is the unit outward normal to the surface S + S' bounding V – V' . S' being the surface of the ∂ ∂ =− . sphere of radius ε and ∂n ∂r Now

∫

S'

 1 ∂ψ ∂ 1r    dS = − ψ  r ∂n  ∂ n  

∫

S'

 1 ∂ψ ∂ 1r − − ψ  r ∂r ∂r   1 ∂ψ

=

  dS  

ψ 2 r dw 2  

∫  − r ∂r − r S'

= −

 ∂ψ



∫  r ∂r + ψ  dw S'

∫

S'

 1 ∂ψ ∂( 1r )  ∂ψ    = − ε  − ψ  dw − 4πψ  r ∂n  ∂n   ∂r  r = ε  S'

∫

… (3)

where ψ is the mean value of V over the sphere S' and w denote the solid angle. 1 2 3 Let ψ( x , x , x ) = ψ (P ) as r → 0 then as ε → 0 from (3), we have

∫

S'

 1 ∂ψ ∂ 1r    − ψ  r ∂n ∂n  = −4πψ (P ) 

Then equation (2) becomes

∫

V

Since ε → 0 then

1

1 2 ∇ ψ dV = r

∫ r ∇ ψ dV

V'

2

= 0.

∫ S

 1 ∂ψ ∂ 1r    dS − 4πψ (P ) − ψ  r ∂n ∂n  


169

∂ 1r 1 1 2 1 1 ∂ψ 1 − ∇ ψ dV + dS − ψ dS ψ(P ) = 4 π r 4π S r ∂n 4π S ∂n V

∫

∫

∫

… (4)

This gives the solution of Poisson's equation at the origin. If ψ is regular at infinity, i.e., for sufficiently large value of r, ψ is such that (ψ) ≤

∂ψ m m ≤ and ∂r r 2 r

… (5)

where m is constant. If integration in equation (4) is extended over all space, so that r → ∞ . Then, using equation (5), equation (4) becomes ψ(P ) = −

1 ∇ 2ψ dV 4π ∞ r

∫

… (6)

But ψ is a potential function satisfying the Poisson's equation i.e. ∇ 2 ψ = −4πρ. Hence, from (6), we get ψ(P ) =

∫ρ ∞

dV r

This solution is Unique.

EXERCISES 1. Find, with aid of Lagrangian equations, the trajectory of a particle moving in a uniform gravitational field. 2. A particle is constrained to move under gravity along the line yi = ci s (i = 1, 2, 3). Discuss the motion. 3. Deduce from Newtonian equations the equation of energy T + V = h, where h is constant. 4. Prove that

∫ ψ n dS i

i

S

∫ ∇ ψ dV 2

=

V

where ψ i =

∂ψ . ∂x i

5. Prove that the curl of a gradient vector vanishes identically.

CHAPTER – 9

CURVATURE OF CURVE, GEODESIC 9.1 CURVATURE OF CURVE: PRINCIPAL NORMAL Let C be a curve in a given Vn and let the coordinates x i of the current point on the curve expressed as functions of the arc length s. Then the unit tangent t to the curve the contravariant components dxi t i = ds

...(1)

The intrinsic derivative (or desired vector) of t i along the curve C is called the first curvaturee r r vector of curve C relative to Vn and is denoted by p . The magnitude of curvature vector p is called first curvature of C relative Vn and is denoted by K: So, K=

g ij p i p j

r

where P i are contravariant components of p so that dx j

i P = t , j ds

i

 ∂t i  i α  i   dx  + t   =  j   α j   ds  ∂x

=

j ∂t i dx j α dx  i + t  ds  α ∂x j ds

  j

dxα dx j  i 

dt i

= ds + ds ds α j   =

d 2 xi ds

2

+

dx j dx k  i  ds ds  k

 , Replacing dummy index α by k j

Curvature of Curve, Geodesic

171

d 2 xi

dx j dx k  i 

 i 

 i 

pi = ds 2 + ds ds  j k  as  j k  = k j        r

If nˆ is a unit vector in the direction of p , then we have r p = k nˆ The vector nˆ is called the Unit principal normal. 9.2 GEODESICS Geodesics on a surface in Euclidean three dimensional space may be defined as the curve along which lies the shortest distance measured along the surface between any two points in its plane. But when the problem of find the shortest distance between any two given points on a surface is treated properly, it becomes very complicated and therefore we define the geodesics in V3 as follows: (i) Geodesic in a surface is defined as the curve of stationary length on a surface between any two points in its plane. (ii) In V3 geodesic is also defined as the curve whose curvature relative to the surface is everywhere zero. By generalising these definitions we can define geodesic in Riemannian Vn as (i) Geodesic in a Riemannian Vn is defined as the curve of minimum (or maximum) length joining two points on it. (ii) Geodesic is the curve whose first curvature relative to Vn is zero at all points. 9.3 EULER'S CONDITION THEOREM 9.1 The Euler condition for the integral

∫

t1

to

f (x i , x&i ) dt

to be staionary are ∂f ∂x

i

−

d  ∂f    =0 dt  ∂ x& i  dxi

x& = dt i = 1, 2, 3,... Proof: Let C be a curve in a Vn and A, B two fixed points on it. The coordinates x i of the current point P on C are functions of a single parameter t. Let t0 and t1 be the values of the parameter for the points A and B respectively. To find the condition for the integral

where

i

∫

t1

t0

f (x i , x&i ) dt

...(1)

to be stationary. Let the curve suffer an infinitesimal deformation to C ′, the points A and B remaining fixed while the current points P(xi) is displaced to P' (xi + ηi) such that ηi = 0 at A and B both.

172


P'

C’

A

B

P

C

Fig. 9.1

In this case the value of integral (1) becomes I ′ So, I′ =

∫ F (x t1

i

t0

)

& i dt + ηi , x& i + η

By Taylor's theorem  ∂f ∂f  + k  + ⋅ ⋅ ⋅ ⋅ ∂y   ∂x

F ( x + h , y + k ) = f ( x, y ) +  h

Then I′ =

∫

I′ =

∫

t1 

  ∂F i ∂F i  i i  F ( x , x& ) +  i η + i η&  + ⋅ ⋅ ⋅ dt ∂x  ∂x 

t0  t1 t0

F ( x i , x& i ) dt +

∫

t1  ∂F



t 0  ∂x i

ηi +

∂F i  η&  dt ∂x& i 

(Neglecting higher order terms in small quantities ηi ) I′ = I +

∫

∂F ∂f   i η i + i η& i  dt t0  ∂ x ∂ x&  t1 

δI = I ′ − I =

∫

t1 



∂F

t0  ∂ x i

ηi +

∂F i  η&  dt ∂ x&i 

...(2)

∂ zi & j η& i = x ∂x j

where Now,

∫

t

 ∂F i  1 & η dt =  ∂ x&i η  – t0 ∂ x &i t0 t1 ∂ F

i

= −

∫

t1 t0

∫

t1 t0

d  ∂F  i  η dt dt  ∂ x&i 

d  ∂F  i   η dt dt  ∂ x&i 

...(3)


173 t    ∂F 1 since  i ηi (t ) = 0, η i (t1 ) = ηi (t0 ) = 0   ∂x& t0 

Then equation (2) becomes δI =

∫

ti  ∂F



t0  ∂ x i

The integral I is stationary if δI = 0 . i.e., if

∫

t1  ∂F



t0  ∂ x i

−

−

d  ∂F   i   η dt dt  ∂ x&i  

...(4)

d  ∂F   i   η dt = 0 dt  ∂ x&i  

Since ηi are arbitrary and hence the integrand of the last integral vanishes, so that ∂F d  ∂F  −   = 0, (i = 1, 2,..., n) ∂ xi dt  ∂ x&i  Hence the necessary and sufficient condition for the integral (1) to be stationary are

… (5)

∂F d  ∂F  –   = 0, (i = 1, 2, …, n) ∂x i dt  dx i  These are called Euler's conditions for the integral I to be stationary. 9.4 DIFFERENTIAL EQUATIONS OF GEODESICS To obtain the differential equations of a geodesic in a Vn , using the property that it is a path of minimum (or maximum) length joining two points A and B on it. Proof: Consider a curve C in Vn joining two fixed points A and B on it and x i (t ) be the coordinates of point P on it. The length of curve C is s=

∫

ds = dt

B

gij

A

gij

dxi d x j dt dt dt

...(1)

dxi d x j dt dt

Put ds = dt

or

gij

d xi d x j = F (say) dt dt

...(2)

g ij x& i x& j = F

s& =

Then equation (1) becomes s=

∫

B A

F dt

...(3)

174


Since curve C is geodesic, then the integral (3) should be stationary, we have from Euler's condition ∂F ∂x

i

−

d  ∂F    =0 dt  ∂ x& i 

...(4)

Differentiating equation (2) with respect to x k and x& k we get, ∂F ∂x

and

=

k

1 ∂g ij i j x& x& 2 s& ∂ x k

∂F  1  1 2 gik x& i  = g ik x&i k = ∂x&  2s&  s& d  ∂F  1 1 ∂ g ik j i 1 i x& x& + g ik &x& i   = − 2 &s& g ik x& + dt  ∂ x& k  s& ∂x j s& s&

Putting these values in equation (4), we get 1 ∂g ij i j  1 1 ∂ g ik j i 1  x& x& − − 2 &s&g ik x& i + x& x& + g ik &x&i  = 0 j & & 2 s& ∂ x k s s s ∂ x    ∂g &s& 1 ∂ g ij  i j  x& x& = 0 g ik x& i +  ikj − s& 2 ∂x k   ∂x

g ik &x&i −

g ik &x&i −

&s& g x& i + [k , ij ] x&i x& j = 0 &s ik

multiplying it by g km , we get g km gik &x&i −

&s& km g g ik x&i + g km [k , ij ] x& i x& j = 0 s& m

km g km gik = δim and g [k , ij] = i j  

But

x&&m −

d 2xm dt

2

−

&s& m  m  i j x& +   x& x& = 0 s& i j 

&s& dx m  m  dx j dxk  Replacing dummy +  = 0  index i by k s& dt  j k  dt dt 

This is the differential equation for the geodesic in parameter t. Taking s = t , s& = 1, &s& = 0. Then equation (5) becomes d 2xm ds 2

 m  dx j dx k +  = 0 ...(6)  j k  ds ds

  

...(5)


175

which may also written as dx k ds

 dx m   ds 

   =0  ,k

Then the intrinsic derivative (or derived vector) of the unit tangent to a geodesic in the direction of the curve is everywhere zero. In otherwords, a geodesic of Vn is a line whose first curvature relative to Vn is identically zero. THEOREM 9.2 To prove that one and only one geodesic passes through two specified points lying in a neighbourhood of a point O of a Vn . OR To prove that one and only one geodesic passes through a specified point O of Vn in a prescribed direction. Proof: The differential equations of a geodesic curve in a Vn are d 2xm ds 2

 m  dx j dx k +  =0  j k  ds ds

These equations are n differential equations of the second order. Their complete integral involves 2n arbitrary constants. These may be determined by the n coordinates of a point P on the curve and the n components of the unit vector in the direction of the curve at P. Thus, in general, one and only one geodesic passes through a given point in a given direction. 9.5 GEODESIC COORDINATES A cartesian coordinate system is one relative to which the coefficients of the fundamental form are constants. Coordinates of this nature do not exists for an arbitrary Riemannian V n.. It is, however, possible to choose a coordinate system relative to which the quantities gij are locally constant in the neighbourhood of an arbitrary point P 0 of V n. Such a cartesian coordinate system is known as geodesic coordinate system with the pole at P 0. The quantities g ij are said to be locally constants in the neighbourhood of a point P0 if ∂g ij

= 0 at P0

∂ xk

and

∂g ij ∂ xk

≠ 0 elsewhere

k 

This shows that [ij, k ] = 0 , i j = 0 at P0 .   Since the covariant derivative of A ij with respect to xk is written as  h  ∂Aij  h   Aih −   A h j , see pg. 71 Aij , k = ∂x k −   j k  i k 

176


The covariant derivative of Aij at P0 with respect to x k reduces to the corresponding ordinary derivatives. Hence Aij , k =

∂Aij ∂x k

at P0

THEOREM 9.3 The necessary and sufficient condition that a system of coordintes be geodesic with pole at P0 are that their second covariant derivatives with respect to the metric of the space all vanish at P0. Proof: We know that (equation 8, Pg. 65) ∂2xs ∂x j ∂ x j ∂  ∂x s  ∂x j  ∂x i

or

=

∂x s  k  ∂x p ∂x q  s   −   ∂x k i j  ∂x i ∂x j  p q 

 ∂x s  k  ∂x p ∂x q  s   =  −    ∂x k i j ∂x i ∂x j  p q 

...(1)

Interchanging the coordinate system x i and x i in equation (1), we get ∂  ∂x s  ∂x j  ∂x i

−

 ∂x s  =  ∂x k 

∂x p ∂x q  s  ∂  = i j  ∂x ∂x  p q ∂x j =

∂ ∂x

j

 k  ∂x p ∂x q  s   − i  j  i j ∂x ∂ x  p q   ∂x s   ∂x i 

 ∂x s −  ∂x k 

(x )− x s ,i

= ( x ,is ), j

k  i

  j

k  ∂x s s   since k = x, k at P0 i j ∂ x   k  since  i j = 0 at P0   s ,k

Thus, x,sij

= −


 s     p q

Necessary Condition Let x s be a geodesic coordinate system with the pole at P0 so that  s    = 0 at P0  p q

Hence from (2), we have

x,sij = 0 at P0

Sufficient Condition Conversely suppose that x,sij = 0 at P0 .

...(2)


177

Then equation (2) becomes  s  ∂x p ∂x q   i j = 0  p q ∂x ∂x ⇒

 s  ∂x p ∂x q ≠ 0 ≠ 0 at P0   = 0 at P0 , as and ∂x i ∂x j  p q

So, x s is a geodesic coordinate system with the pole at P 0. 9.6 RIEMANNIAN COORDINATES A particular type of geodesic coordinates introduced by Riemann and known as Riemannian coordinates. Let C be any geodesic through a given point P0 , s the length of the curve measured from P0 and ξ i the quantities defined by  dxi  ξ =  ds  i

   o

...(1)

the subscript zero indicating as usual that the function is to be evaluated at P0 . The quantities ξi represents that only one geodesic will pass through P 0 in the direction of ξi in V n. Let yi be the coordinates of a point P on the geodesic C such that yi = sξi ...(2) where s is the arc length of the curve from P0 to P . The coordinates y i are called Riemannian coordinates. The differential equation of geodesic C in terms of coordinates y i relative to Vn is given by d 2 y i  i  dy i dy j +  =0 ds 2  j k  ds ds

...(3)

 i   is a christoffel symbol relative to the coordinates y i .  j k

where 

The differential equation (3) will be satisfied by (2), we have,  i  i j i 0+  ξ ξ = 0 since dy = ξ i  j k  ds or

 i  i j  ξ ξ = 0  j k

...(4)

using equation (2), equation (4) becomes  i  yi y j yi = ξi   = 0 as s  j k s s or

 i  i j  y y = 0  j k

The equation (5) hold throughout the Riemannian Vn .

...(5)

178


Since y i ≠ 0, y j ≠ 0, from (5) we get  i    = 0 at P0  j k

Hence the Riemannian coordinates are geodesic coordinate with the pole at P0 . THEOREM 9.4 The necessary and sufficient condition that the coordinates y i be Riemannian coordinates  i  i j  y y = 0 hold throughout the Riemannian Vn .  j k

is that 

 i  i j  y y = 0 (from equation 5) throughout Proof: If y are Riemannian coordinates then the condition   j k  i

the Riemannian Vn . d 2 y  i  dy i dy j  i  i j  Conversely if  j k  y y = 0 hold then ds 2 +  = 0 are saitsfied by y i = s ξi .  j k  ds ds   Hence yi are Riemannian coordinates. 9.7 GEODESIC FORM OF A LINE ELEMENT Let φ be a scalar invariant whose gradiant is not zero. Let the hypersurface φ = 0 be taken as coordinates hypersurface x1 = 0 and the geodesics which cut this hypersurface orthogonally as the coordinate lines of parameter x1 , this parameter measuring the length of arc along a geodesic from the hypersurface x1 = 0 . Since dx1 is the length of the vector µi is given by i j u 2 = g ij u u

i.e.,

(dx )

1 2

1 1 = g11dx dx

g 11 = 1 ⇒ Now, if v i is the tangent vector to the hypersurface x1 = 0 then we have

...(1)

2 3 n v i = ( 0, dx , dx ,..., dx )

since the vectors ui and v i are orthogonal vectors. Then, g ij u i v j = 0

⇒ ⇒

g 1 j u 1 v j = 0, [u i = 0, i = 2, 3, ..., n ] g 1 j v j = 0, as u 1 ≠ 0.

g1 j = 0, for j = 2, 3, …, n. ⇒ Again the coordinate curves of parameter x1 are geodesics. Then s = x1. If ti is unit tangent vector to a geodesic at any point then t1 = 1 and t i = 0, for i = 2, 3, …, n.

...(2)


179

Now, ti =

⇒ and

dx1 dx1 d 2xi ds 2

d xi d x i = 1 ds dx

= 1 and =

d 2xi

dxi dx1

= 0 for i ≠ 1

= 0 for i = 1, 2,..., n

dx12

Also, the differential equation of geodesic is d 2 x i  i  dx j dx k +  =0 ds 2  j k  ds ds using above results, we have  i  dx 1 dx1   =0 11 ds ds ⇒

i   =0 1 1

⇒

g ij [11, j ] = 0

⇒

[11, j ] = 0 1  2 ∂g 1 j ∂ g 11 − 2  ∂ x i ∂x j

ij as g ≠ 0

  = 0, since g 11 = i ⇒ ∂g 11 = 0  ∂x j 

So, ∂g 1 j ∂x1

= 0 for j ≠ 1

...(3)

from equations (1), (2), and (3), we have g 11 = 1, g1j = 0; ( j = 2, 3, …, n),

∂ g1 j ∂ x1

= 0, ( j = 2, 3, …, n)

The line element is given by i

ds 2 = gij dx dx

j

1 1 j k ds 2 = g 11dx dx + g jk dx dx 1 2 j k ds 2 = ( dx ) + g jk dx dx ; ( j = 2, 3, …, n, k = 2, 3, …, n)

...(4)

The line element (4) is called geodesic form of the line element. We note that the coordinate curves with parameter x 1 are orthogonal to the coordinate curve x i = c i (i = 1, 2, ..., n) at all points and hence to the hypersurfaces x 1 = c at each point. Note 2: The existence of geodesic form of the line element proves that the hypersurfaces φ= x1 = constant form a system of parallels i.e., the hypersurfaces φ = x 1 = constant are geodesically parallel hypersurfaces.

Note 1:

180


THEOREM 9.5 The necessary and sufficient condition that the hypersurfaces φ = constant form a system of parallel is that (∇φ)2 = 1. Proof: Necessary Condition Suppose that hypersurface φ = constant form a system of parallels then prove that (∇φ 2) = 1. Let us take the hypersurface φ = 0 as the coordinate hypersurface x1 = 0. Let the geodesics cutting this hypersurface orthogonally, be taken as coordinate lines of parameter x1. Then the parameters x1 measures are length along these geodesics from the hypersurface x1 = 0. This implies the existence of geodesic form of the line element namely 1 2 i j ds 2 = ( dx ) + g ij dx dx

...(1)

where i, j = 2, 3,..., n. From (1), we have g 11 = 1, g 1i = 0 for i ≠ 1.

from these values, it follows that g11 = 1, g1 i = 0, for i ≠ 1 Now, ij (∇φ)2 = ∇φ ⋅ ∇φ = g

ij = g

∂φ ∂φ ∂x i dx j

∂ x1 ∂x1 = g ij δ1i δ1j i j ∂ x ∂x

(∇φ)2 = g11 = 1

so

(∇φ)2 = 1 Sufficient Condition Suppose that (∇φ)2 = 1 then prove that the hypersurface φ = constant from a system of parallels. Let us taken φ = x1 and orthogonal trajectories of the hypersurfaces φ = x1 constant as the coordinate lines of parameter x1. Then the hypersurfaces x1 = constant xi = constant (i ≠ 1) are orthogonal to each other. The condition for this g1i = 0 for i ≠ 1 . Now, given that (∇φ)2 = 1 ⇒

g ij

∂φ ∂φ =1 ∂x i ∂x j

⇒

g ij

∂x1 ∂x1 =1 ∂x i ∂x j

⇒

g ij δ1i δ1j = 1

⇒

g11 = 1


181

Thus g11 = 1 and g1 i = 0 for i ≠ 1. Consequently g 11 = 1, g = 0 , for i ≠ 1. 1i

Therefore, the line element i j ds 2 = g ij dx dx

is given by ds 2 = (dx1 ) 2 + g ik dx i dx k ; (i, k = 2, 3, …, n)

which is geodesic form of the line element. It means that the hypersurfaces φ = x1 = constant form a system of parallels. 9.8 GEODESICS IN EUCLIDEAN SPACE Consider an Euclidean space S n for n-dimensions. Let y i be the Euclidean coordinates. The differential equation of geodesics in Euclidean space is given by d 2 y i  i  dy j dy k +  =0 ds 2  j k  ds ds

...(1)

In case of Euclidean coordinates the fundamental tensor g ij is denoted by aij and g ij = a ij = δij = 1, if i = j 0, if i ≠ j ∂g ij ∂x

k

=

∂a ij ∂x k

=0

k This implies that   = 0, [ij, k ] = 0 relative to S n . i  j Then equation (1) becomes d 2 yi ds 2

=0

Integrating it, we get dyi = ai, where ai is constant of integration. ds

Again Integrating, we get y i = ais + bi, where bi is constant of integration

...(2)

The equation (2) is of the form y = mx + c. Hence equation (2) represents a straight line. Since equation (2) is a solution of equation (1) and therefore the geodesic relative to S n are given by equation (2). Hence geodesic curves in Euclidean space S n are straight lines.

182


THEOREM 9.6 Prove that the distance l between two points P (y i) and Q (y' i ) in Sn is given by

∑ (y ′ − y ) n

i 2

i

l=

i =1

Proof: We know that geodesics in S n are straight line. Then equation of straight line in S n may be taken as yi = ais + bi ...(1) Let P ( y i ) and Q ( y' i) lie on equation (1). Then yi = ais + bi, y′ i = a i s′ + bi y′ i − y i = a i (s ′ − s ) Then equation (2) becomes y′ i − y i = a i l n

l2

∑ i =1

n

(ai ) 2 =

∑ ( y′

i

− y i )2

i =1

But a i is the unit tangent vector to the geodesics. Then n

∑ (a )

i 2

=1

i =1

So, n

l

2

=

∑ ( y′

i

− y i )2

i =1

n

l=

∑( y′

i

− y i )2

i =1

EXAMPLE 1 Prove that Pythagoras theorem holds in S n . Solution Consider a triangle ABC right angled at A i.e., ∠BAC = 90 o . C (y i3)

A ( y1i )

B ( y2i ) Fig. 9.2

...(2)


183

Then the lines AB and AC are orthogonal to each other. So, AB ⋅ AC = 0 a ij ( y i2 − y1i ) ( y i3 − y1i ) = 0

or

n

or

∑ (y i =1

i 2

− y1i ) ( y 3i − y1i ) = 0

...(1)

By distance formula, we have n

2

(AB) =

∑ (y

i 2

− y1i ) 2

...(2)

i 3

− y1i )2

...(3)

i 3

− y i2 )2

...(4)

i =1 n

2

(AC) =

∑ (y i =1 n

2

(BC) =

∑ (y i =1

Now, equation (4) can be written as n

2

(BC) =

∑ [( y

i 3

i =1 n

=

∑ [( y

i 3

− y1i )2 + ( y1i − y i2 ) 2 + 2( y3i − y1i ) ( y1i − y i2 ) ]

i =1 n

=

− y1i ) + ( y1i − y i2 )]2

∑

( y i3 − y1i )2 +

i =1 n

=

∑

n

∑ (y

i 1

− y i2 ) 2 + 2 × 0, [from (1)]

i =1

( y i3 − y1i )2 +

i =1

n

∑( y

i 1

− y2i )2

i =1

(BC)2 = ( AC)2 + ( AB)2 Hence Pythagoras theorem holds in S n . EXAMPLE 2 Prove that if θ is any solution of the differential equation (∇θ)2 = f (θ) then the hypersurfaces θ = constant constitute a system of parallels. Solution Given that (∇θ)2 = f (θ)

...(1)

184


Then prove that the hypersurfaces θ = constant form a system of parallel. Suppose φ=

or

dφ = dθ

Now,

∇φ =

f ( θ)

dθ f (θ)

, Then, dφ =

1 f ( θ)

∂φ ∂φ ∂θ = i = ∂θ ∂x i ∂x

 (∇φ)2 =  

=

dθ

∫

 ∇θ   f ( θ) 

1 f ( θ)

∇θ

2

1

1 1 (∇ θ) 2 = f (θ); from (1) f ( θ) f ( θ)

(∇φ)2 = 1 This proves that the hypersurfaces φ = constant form a system of parallels and therefore the hypersurfaces θ = constant. EXAMPLE 3 Show that it is always possible to choose a geodesic coordinates system for any Vn with an arbitrary pole P 0. Solution Let P0 be an arbitrary pole in a Vn . Let us consider general coordinate system x i . suppose the value of x i and P0 are denoted by x0i . Now consider a new coordinate system x j defined by the equation. x

j

1 2

 h  i l m m  ( x − x0 ) ( x − x0 ) l m  

j m m j = a m ( x − x0 ) + a h 

...(1)

The coefficients amj being constants and as such that their determinant do not vanish. Now we shall prove that this new system of coordinated x j defined by equation (1) is a geodesic coordinate system with pole at P0 i.e., second covariant derivative of x j vanishes at P.. Differentiating equation (1) with respect to x m , we get ∂x j ∂x  ∂x j   ∂x m 

m

1 2

  = a j at P m 0  0

Now, the Jacobian determinant

h  i l  2 ( x − x0 ) l m  

j j = a m + ah 

...(2)

...(3)


185

 ∂x j   m = a j ≠ 0 ∂x  m  0 and therefore the transformation given by equation (1) is permissible in the neighbourhood of P0 . Differentiating equation (2) with respect to x j , we get  ∂2 x j   j m  = a j  h   ∂x ∂x  h  0 l m  0

...(4)

But we know that

(x )

j ,lm 0

 ∂2x =  l

 h   −    ∂x ∂x  0  l m  0 j

m

 ∂x j   ∂x h 

   0

h  h  j  −   a h , (from (3) and (4)) l m  0 lm 

j = ah 

(x , )

j lm 0

=0

Hence equation (1) is a geodesic coordinate system with pole at P0 . EXAMPLE 4 If the coordinates xi of points on a geodesic are functions of arc lengths s and φ is any scalar function of the x's show that d pφ ds p

= φ, ij... l

d xi d x j dxl ⋅⋅ ⋅ ds ds ds

...(1)

Solution Since the coordinates x i lie on a geodesic. Then d 2 xi ds 2

 i  dx j dx k + =0   j k  ds ds

...(2)

Here the number of suffices i j...l is p. We shall prove the theorem by mathematical induction method. Since x′s are functions of s and φ is a scalar function of x′ s , we have dφ ∂φ dxi = ds ∂x i ds d 2φ ds2

=

or

dφ dx i = φ, i ds ds

∂φ, i dx j d 2xi + φ ,i ∂ x j ds ds 2

∂φ ,i dx i dx j = ∂x j ds ds − φ ,i

 i  d x j d x k , from (2)    j k  ds ds

...(3)

186


∂φ ,i dx i dx j = ∂x i ds ds − φ, m =

 m  dx j dx k    j k  ds ds

∂φ, i dx i dx j  m  dxi dx j − φ (adjusting the dummy index.)   , m ∂ x j ds ds i j  ds ds

 ∂φ, i  m   dx i dx j  − φ  ,m  =  j i j  ds ds  ∂x Equations (3) and (4) imply that the equation (1) holds for p = 1 and p = 2. Suppose that the equation (1) holds for p indices r1 , r2 ,...,r p so that d pφ

r

dx r1 dx r1 dx p ⋅ ⋅ ⋅ ds p ds ds ds Differentiating the equation (5) with respect to s, we get d p +1 φ ds p +1

=

= φ, r1 , r2 , ...,

rp

...(5)

r r r ∂φ, r1 r2 ... r p dx r1 dx p dx p +1 d 2 x r1 dx r2 dx p ⋅ ⋅ ⋅ + φ , ⋅ ⋅ ⋅ + ⋅ ⋅⋅ r r1 r2 ... rp ds ds ds ds ∂ x p +1 ds 2 ds

+ ⋅ ⋅ ⋅φ, r1 r2 ... rp substituting value of

...(4)

r

dx r1 d 2x p ⋅ ⋅⋅ ds ds 2

...(6)

d 2 x r1 etc. from (2) in (6) and adjusting dummy indices, we have ds 2  ∂ φ, r r ... r m  m   1 2 p   d p +1 φ − φ ⋅ ⋅ ⋅ φ     , m r ... r , r r ... m r 2 p 1 2 p +1 =  ∂x p +1 r r r r     ds  1 p +1   p p +1    r

dx r1 dx r2 dx p +1 ⋅⋅⋅ ds ds ds r

d x r1 d x r2 d x p +1 ⋅⋅ ⋅ ds ds ds This shows that the equation (1) holds for next values of p. But equation (1) holds for p = 1, 2, ... Hence equation (1) holds for all values of p. = φ ,r1 r2 .. r p rp +1

EXERCISES 1. Prove that at the pole of a geodesic coordinate system, the components of first covariant derivatives are ordinary derivatives. 2. If x i are geodesic coordinates in the neighbourhood of a point if they are subjected to the transformation i xi = x +

1 i c x j xk xl 6 jkl


187

where C′s are constants then show that xi are geodesic coordinates in the neighbourhood of O. 3. Show that the principal normal vector vanishes identically when the given curve is geodesic. 4. Show that the coordinate system x i defined by 1 i  j k i x i = x + 2  j k x x   is geodesic coordinate system with the pole at the origin. 5. Obtain the equations of geodesics for the metric ds2 = e−2 kt ( dx2 + dy2 + dz2 ) + dt 2 6. Obtain the differential equations of geodesics for the metric 1 2 2 2 2 ds2 = f ( x) dx + dy + dz + f ( x) dt 2   d2x 1 d  dx  1 d dt d2y d2z d 2t d (log f ) dx dt Ans : 2 −  (log f )   + (log f ) = 0 ; = 0 ; = 0 ; − = 0 2 dx ds dx ds ds   ds  ds  2 f 2 dx ds2 ds2 ds2

7. Find the differential equations for the geodesics in a cylindrical and spherical coordinates. 8. Find the rate of divergence of a given curve C from the geodesic which touches it at a given point.

CHAPTER – 10

PARALLELISM OF VECTORS

10.1 PARALLELISM OF A VECTOR OF CONSTANT MAGNITUDE (LEVI-CIVITA’S CONCEPT) Consider a vector field whose direction at any point is that of the Unit Vector ti. In ordinary space, the field is said to be parallel if the derivative of ti vanishes for all directions ui (say) and at every point of the field i.e., ∂t i j u =0 ∂x j Similarly in a Riemannian V n the field is said to be parallel if the derived vector of ti vanishes at each point for every direction ui at each point of V n. i.e., t,i j = u j = 0 It can be shown that it is not possible for an arbitrary V n. Consequently we define parallelism of vectors with respect to a given curve C in a V n. A vector ui of constant magnitude is parallel with respect to V n along the curve C if its derived vector in the direction of the curve is zero at all points of C i.e., dx j i u =0 … (1) ds , j where s is arc-length of curve C. The equation (1) can be written in expansion form as j  ∂ui  m  i  dx  j + u m j =0   ds ∂ x j ∂ui d x j m i  dx + u   =0 ∂ x j ds m j ds

du i  i + um  ds m

 dx j  j  ds = 0

… (2)

Parallelism of Vectors

189

This concept of parallelism is due to Levi-Civita. The vector ui is satisfying the equation (1) is said to a parallel displacement along the curve Now, multiplying equation (1) by gil, we get  dx j   g il  u,i j ds  = 0 

or

( gil u,i j )

dx j =0 ds

( g il u i ) , j

dx j =0 ds

ul , j

dx j ds = 0

u i, j

dx j =0 ds

 ∂ui  m  dx j − u  j m   =0 i j  ds  ∂x dui  m  dx j − um   =0 ds i j  ds The equation (2) and (3) can be also written as m i  j dui = − u m j  d x  

… (3)

… (4)

m  j dui = um i j  d x … (5)   The equation (4) and (5) give the increment in the components ui and ui respectively due to displacement dxj along C. and

THEOREM 10.1 If two vectors of constant magnitudes undergo parallel displacements along a given curve then they are inclined at a constant angle. Proof: Let the vectors ui and v i be of constant magnitudes and undergo parallel displacement along a curve C, we have (from equation (1), Pg. 188.)  dx j = 0  ds  j d x vi, j = 0 ds 

u,i j

at each point of C. Multiplying (1) by gil, we get dx j ( g il u,i j ) =0 ds

… (1)

190


or

ul , j

dx j =0 ds

u i, j

dx j =0 ds

… (2)

vi , j

dx j =0 ds

… (3)

Similarly,

Let φ be the angle between ui and v i then ui.v i = uv cos θ Differentiating it with respect to arc length s, we get d d (u i vi ) (uv cos θ) = ds ds i = (u v i ), j

− uv sin θ

dx j ds

j dθ dx j i dx vi + u i vi , j = u, j ds ds ds

Using equation (1) and (3), then equation (4) becomes dθ − uv sin θ =0 ds dθ sin θ = 0, as u ≠ 0, v ≠ 0 ⇒ ds dθ Either sinθ = 0 or =0 ⇒ ds Either θ = 0

⇒

or

θ = constant.

⇒ θ is constant. Since 0 is also a constant. THEOREM 10.2 A geodesic is an auto-parallel curve. Proof: The differential equation of the geodesic is given by (See Pg. 174, eqn. 6) d 2 x m  m  dx j dx k +  ds 2  j k  ds ds

=0

d  d x m   m  d x j d x k +  ds  ds   j k  ds ds = 0 ∂ ∂x j

 dx m   ds 

 d x j  m  d x j d xk   ds +  j k  ds ds = 0   

… (4)


 ∂  j  ∂x

 dx m   ds 

191

  m  dx k  dx j +   =0    j k  ds  ds

 d xm  dx j    ds  ds = 0  , j

or

t,mj

dx j =0 ds

dx m suffer a parallel displacement along a geodesic curve.This ds confirms that geodesic is an auto-parallel curve. Proved. This shows that the unit tangent vector

10.2 PARALLELISM OF A VECTOR OF VARIABLE MAGNITUDE Two vectors at a point are said to be parallel or to have the same direction if their corresponding components are proportional. Consequently the vector v i will be parallel to ui at each point of curve C provided v i = φui … (1) where φ is a function of arc length s. If ui is parallel with respect to Riemannian V n along the curve C. Then, u ,ij

dx j =0 ds

… (2)

The equation (1) shows that v i is of variable constant and parallel with respect to Riemannian V n so that v ,ij

dx j dxj i = (φ u ), j ds ds i i = (φ, j u + φu, j )

dx j ds

= φ, j

dx j i dxj u + φ u,i j ds ds

= φ, j

dx j i u Since ds

∂φ d x j i u = ∂x j ds v,i j

dφ i dx j u = ds ds d φ vi = ds φ

from (1)

u,i j

dx j =0 ds

192


i = v

d (log φ) ds

dx j d (log φ ) = v i f (s) where f (s) = … (3) ds ds Hence a vector v i of variable magnitude will be parallel with respect to V n if equation (3) is satisfied. Conversely suppose that a vector v i of variable magnitude such that dx j = v i f (s ) v,ij = ds to show that v i is parallel, with respect to V n . Take v,i j

ui = v i ψ (s)

… (5)

Then dx j dx j i = (v ψ), j ds ds j dx j i i dx ψ + ψ, j v = v, j ds ds ∂ψ dx j i i v f ( s ) ψ + v = ∂x j ds dψ  dx j i u,i j = v  ψf (s ) + … (6) ds  ds  dψ = 0. Select ψ such that ψf ( s ) + ds Then equation (6) becomes dx j u,i j =0 ds This equation shows that the vector u i is of constant magnitude and suffers a parallel displacement along curve C. The equation (5) shows that v i is parallel along C. Hence necessary and sufficient condition that a vector v i of variable magnitude suffers a parallel displacement along a curve C is that u ,ij

v,i j

dx j = v i f (s ). ds

EXAMPLE 1 Show that the vector v i of variable magnitude suffers a parallel displacement along a curve C if and only if dx k (v i v i,k − v i v l,k ) = 0, i = 1, 2, ..., n. ds Solution From equation (4), we have dx j v ,ij = v i f (s ) … (1) ds


193

Multiplying by v l, we get dx j = v l v i f (s ) ds Interchange the indices l and i, we get v l v,i j

dx j = v i v l f (s) ds Subtract (1) and (2), we get v i v,l j

(v l u,ik − v i v l,k )

… (2)

dx k = 0 by interchanging dummy indices j and k. ds

10.3 SUBSPACES OF A RIEMANNIAN MANIFOLD Let V n be Riemannian space of n dimensions referred to coordinates xi and having the metric ds2 = gij dxi dxj. Let V m be Riemannian space of m dimensions referred to coordinates y α and having the metric ds2 = axβ dyα dyβ, where m > n. Let Greek letters α , β, γ take the values 1, 2, ..., m and Latin indices i, j, k ... take the values 1, 2, … n. If the n independent variables xi are such that the coordinates ( y α ) of points in V m are expressed as a function of xi then V n is immersed in V m i.e. V n is a subspace of V m. Also V m is called enveloping space of V n. Since the length ds of the element of arc connecting the two points is the same with respect to V n or V m. it follows that gij dxi dxj = a αβdy α dy β ∂y α ∂x i ∂y α a gij = αβ i ∂x

gij dxi dxj = a αβ ⇒

∂y β i j dx dx ∂x j ∂y β ∂x j

… (1)

As dxi and dxj are arbitrary. This gives relation between gij and aαβ . THEOREM 10.3 To show that the angle between any two vectors is the same whether it is calculated with respect to Vm or Vn. Proof: Consider two vectors dx i and δx j defined at any point of V n and suppose that the same vectors in V m are represented by dy α and δy α respectively. If θ is the angle between dxi and δx i then cos θ =

g ij dx i δx j g ij dx i dx j g ij δx i δx j

If φ is the angle between the vectors dyα and δyα then cos φ =

aαβ dy αδy β aαβ dy α dy β aαβ δy αδy β

… (1)

194


∂ y α i ∂ yβ dx dx j ∂x i ∂x j ∂ yα i ∂ yβ j ∂ yα i ∂ yβ j d x d x a δx δx . αβ ∂x i ∂x j ∂ xi ∂x j aαβ

= aαβ

∂y α ∂y β i j dx dx ∂x i ∂x j ∂y α ∂y β i j ∂y α ∂y β i j dx dx a δx δx . αβ ∂xi ∂x j ∂x i ∂x j aαβ

= aαβ

cos φ =

Since g ij = aαβ

gij dx i dx j g ij dx i dx j g ij δx i δx j

… (2)

∂y α ∂y β (from equation (1), art. 10.3) ∂x i ∂x j

from (1) & (2) cos θ = cos φ ⇒ θ = φ.

Proved.

THEOREM 10.4 If Uα and ui denote the components of the same vector in Riemannian Vm and Vn respectively then to show that α i ∂y α = u U ∂x i Proof: Let the given vector be unit vector at any point P of V n. Let the component of the same vector x's and y's be a i and A α respectively. Let C be curve passing through s in the direction of the given vector then ∂y α ∂y α dx i ∂y α i = i = i a = … (1) A ∂s ∂x ds ∂x But the components of a vector of magnitude a are a times the corresponding components of the Unit vector in the same direction. Then α i i … (2) U α = aA , u = aa Multiplying (1) by a, we get α

∂y α (aa i ) = aA ∂x i ∂y α i α = u , using (2) U ∂x i α

Or

Proved.

THEOREM 10.5 To show that there are m-n linearly vector fields normal to a surface V n immersed in a Riemannian V m. Proof: Since V n is immersed in V m, the coordinates y α of points in V m are expressible as functions of coordinates xi in V n.


195

Now, dy α ∂y α dx i = ds ∂x i ds i for the curve x = s, we have

...(1)

dy α ∂y α = ds ∂x i

...(2)

dy α ∂ yα is a vector tangential to the curve in V m. Then from equation (2) it follows that ds ∂ xi i α in V m is tangential to the coordinate curve of transmeter x in V n. Let the Unit vectors N in V m be normal to each of the above vector fields of V n then Since

∂y α β N =0 ∂xi ∂y α (aαβ N β ) i = 0 ∂x ∂y α N α i = 0, ∂x a αβ

⇒ ⇒

⇒

a αβ N β y ,αi = 0

...(3)

i = 1, 2, …, n & α = 1, 2, …, m

The equation (3) are n equations in m unknowns N

α

 ∂y α (m > n). The coefficient matrix  i  ∂x

  is  

of order m × n and the rank of this matrix is n. It means that there will be only m – n linearly independent solution of N α . This shows that there are m – n linearly independent normals to V n to V m. EXAMPLE 2 Show that [ij, k] = [αβ, γ ]

∂y α ∂y β ∂y γ ∂ 2 y α ∂x β + a αβ ∂x i ∂x j ∂x k ∂xi ∂x j ∂x k

where [αβ, γ] and [i j, k ] are the Christoffel’s symbols of first kind relative to metrics a αβ dy a dy β and g ij dx i dx j . Solution Since relation between a αβ and g ij is given by ∂y α ∂y β ∂x i ∂x j Differentiating it with respect to xk, we get g ij = a αβ

∂g ij ∂x k

∂aαβ ∂y α ∂y β ∂y γ ∂ 2 y α ∂y γ ∂y α ∂ 2 y β + a + a = αβ αβ ∂x γ ∂x i ∂x j ∂x k ∂x i ∂x k ∂x j ∂x i ∂x j ∂x k

… (1)

196


Similarly ∂g jk ∂x i

∂aβγ ∂y α ∂y β ∂y γ ∂ 2 y α ∂y β ∂y α ∂ 2 yβ + a + a = αβ αβ ∂y α ∂x i ∂x j ∂x k ∂x j ∂x i ∂x k ∂x j ∂x k ∂x i

… (2)

and ∂a γα ∂y α ∂y β ∂y γ ∂g ki ∂ 2 y α ∂y β ∂y α ∂ 2 y β + a + a = αβ αβ ∂x j ∂x β ∂x i ∂x j ∂x k ∂x k ∂x j ∂x i ∂x k ∂x i ∂x j But we know that [ij, k ] =

and,

1  ∂g jk ∂g ki ∂g ij  + − 2  ∂x i ∂x j ∂x k

  

1  ∂gβγ ∂g γα ∂g αβ  − [αβ, γ] = 2  α + ∂y β ∂y γ   ∂y

… (3)

… (4)

… (5)

Substituting the value of (1), (2), (3) in equation (4) and using (5) we get, [ij, k ] = [αβ, γ ]

∂y α ∂y β ∂ y γ ∂ 2 yα ∂ yβ + a . αβ ∂x i ∂x j ∂ x k ∂ xi ∂ x j ∂ x k

10.4 PARALLELISM IN A SUBSPACE THEOREM 10.6 Let T α and t i be the components of the same vector t relative to Vm and Vn respectively. Let the vector t be defined along a curve C in V n. If p i and qα are derived vectors of t along C relative to Vn and Vm respectively. Then ∂y α = pi ∂xi Proof: Since from equation (1), theorem, (10.4), Pg. 194 we have qα

T

α

dT α ds

∂y α ∂x i 2 α dt i ∂y α dx j i ∂ y + t = ds ∂x i ∂x i ∂x j ds i = t

… (1) … (2)

Now, dx j ds β α dy = T,β ds  ∂T α  β γ  α   dy  + T   =  β γβ  ds  ∂x β ∂T α dy β γ  α  dy + T   = ∂xβ ds γβ ds

i p i = t, j

qα

q

α

β dT α γ  α  dy + T   = ds  γβ ds

… (3)

… (4)


Putting the value of

197

dT α and T γ from (2) and (1) in equation (4), we get ds 2 α γ β dt i ∂y α dx j i ∂ y i ∂y dy  α  + t + t qα =   ds ∂x i ∂x i ∂x j ds dx i ds γβ

q

α

=

2 α γ β ∂t i ∂y α dx j dx j i ∂ y i ∂y dy  α  + t + t   ∂x j ∂x i ds ∂x i ∂x j ds ∂x i ds  γβ

=

2 α γ β j ∂t i ∂y α dx j dx j i ∂ y i ∂y ∂y dx  α  + t + t   ∂x j ∂x i ds ∂x i ∂x j ds ∂x i ∂ x. j ds γβ

=

∂t i ∂ y α d x j i d x j  ∂ 2 y α ∂y β ∂y γ  α  + t + +   ds  ∂ x i ∂x j ∂x j ∂x i γβ ∂x j ∂x i ds

… (5)

But we know that [ij, k ] = [αβ, γ]

∂y α ∂y β ∂y γ ∂ 2 y α ∂y γ + a αγ ∂x i ∂x j ∂x k ∂x i ∂x j ∂x k

[ ji, k ] = [βγ, δ]

∂y β ∂y γ ∂y δ ∂ 2 y α ∂y δ + a αδ ∂x j ∂x i ∂x k ∂x i ∂x j ∂x k

 α ∂y β ∂y γ ∂y δ ∂ 2 y α ∂y δ a + a = βγ αδ j αδ ∂x ∂x i ∂x k ∂x i ∂x j ∂x k   ∂y δ a = αδ ∂x k Multiplying (5) by a αδ

   

… (6)

∂y δ , we get, using (6), ∂x k

aαδ

Or

2 α   α  ∂y β ∂y γ   j i + ∂ iy j  βγ ∂x ∂x ∂x ∂x  

j ∂y δ α ∂t i dx j ∂y α ∂y δ i dx q a + t [ij, k ] = αδ ds ∂x k ∂x j ds ∂x i ∂x k

qδ

 dx j  ∂t i  p ∂y δ g + t i  g pk  , j ik k =  ds  ∂x ∂x i j   dx j = ds

 ∂t i  i  j gik + t a   g ik   ∂x  a j   

= g ik

dx j i t, ds j

since g ij = aαβ

∂y α ∂y β ∂x i ∂x j

198


i = ( gik t ), j

= tk, j

dx j ds

dx j ds

∂y α dx j t = k, j ds ∂x k α ∂y q α k = pk ∂x qα

or

…(7) Proved.

Properties of Vm

(i) If a curve C lies in a subspace V n of V m and a vector field in V n is parallel along the curve C with regard to V m, then to show that it is also a parallel with regard to V n. Proof: If a vector field t is a parallel along C with respect to V m. then its derived vector qα vanishes i.e., qα = 0, α = 1, 2,... m. Hence equation (7) becomes pk = 0, k = 1, 2, ..., n. This shows that the vector field t is also parallel along C with respect to V n. (ii) To show that if a curve C is a geodesic in a V n it is a geodesic in any subspace V n of V m. α ∂y α Proff : Science, q = pi ∂x i Let t be unit tangent vector to the curve C them T α is unit tangent vector to C relative to V m and ti is unit tangent vector to C relative to V n. Now, pi = 0 ⇒ curve C is a geodesic in V n α q =0 ⇒ curve C is a geodesic in V m Also, qα = 0, ∀α ⇒ pi = 0, ∀i This shows that if a curve C in V n is a geodesic relative to V m then the same curve is also a geodesic relative to V n. (iii) A necessary and sufficient condition that a vector of constant magnitude be parallel with respect to V n along C in that subspace, is that its derived vector relative to V m for the direction of the curve be normal to V n. Proof: Let t be a vector of constant magnitude. This vector t is parallel along C relative to Vn iff pi = 0, ∀i iff q α But qα For

∂y α =0 ∂x i

∂ yα = 0 implies that qα is normal to V n. ∂x i

∂ yα = y,αi lying in V m is tangential to a coordinate curve of parameter xi in V n. ∂ xi


199

These statements prove that a necessary and sufficient condition that a vector of constant magnitude be parallel along C relative to V n is that its derived vector i.e., qα along C relative to V m be normal to V n. (iv) A necessary and sufficient condition that a curve be a geodesic in V n is that its principal subnormal relative to V m the enveloping space be normal to V n at all points of the curve. Proof: In particular let the vector t be unit tangent vector to the curve C. In this case qα is called principal normal the curve C. Also pi = 0, ∀i implies that the curve C is a geodesic relative to V n. Using result (iii), we get at once the result (iv). (v) To prove that the tendency of a vector is the same whether it is calculated with respect to V m. Proof: Since, we have ∂y α q α i = pi ∂x qα

∂y α dx i dxi p = i ds ∂x i ds

dy α dxi = pi ds ds β α dy dy dx j dx i Tα, β = ti, j ds ds ds ds i.e., tendency of Tα along C = tendency of ti along C. i.e., tendency of t along C relative to V m = tendency of t along C relative to V n. qα

10.5 THE FUNDAMENTAL THEOREM OF RIEMANNIAN GEOMETRY STATEMENT With a given Riemannian metric (or fundamental tensor) of a Riemannian manifold there is associated a symmetric affine connection with the property that parallel displacement (or transport) preserves scalar product. Proof: Let C be a curve in V n. Let pi and qi be two unit vectors defined along C. Suppose that the unit vectors pi and qi suffer parallel displacement along the curve C in V n, then we have dx j =0 … (1) ds dx j q ,ij and =0 … (2) ds Let gij be the given fundamental tensor of a Riemannian manifold. Hence, the scalar product of vectors p i and q i is g ij p i q j. Now, p,i j

( gij p i q j ),k

dx k =0 ds

200

Tensors and Their Applications k     dx k  j dx k  i j i  j dx   g ij  p,ik q + g p q + g ij  ,k ds   ij ,k ds  P q = 0 ds      

… (3)

Using equation (1) and (2), the equation (3) becomes, k    gij , k dx  p i q j = 0  ds   gij, k = 0

dx k ≠ 0) (Since and are unit vectors and ds ∂g ij m  m − g mj   − gim   = 0 k ∂x i k   j k ∂g ij − [ik , j ] − [ jk , i ] = 0 ∂x k ∂g ij = [ik , j] + [ jk , i ] ∂x k Now, using equation (4), we have ∂g jk ∂g ki ∂g ij + − = [ ji, k ] + [ki, j ] + [kj, i ] + [ij, k ] – ([ik , i] + [ jk , i ]) ∂x i ∂x j ∂x k since [ij, k ] = [ ji, k ]. So, ∂g jk ∂g ki ∂g ij + − = 2[ij, k ] ∂x i ∂x j ∂x k pi

qi

[ij,k] = k lk But we know that i j = g [ij, l ]   from (5), we have

1  ∂g jk ∂g ki ∂g ij  + − 2  ∂x i ∂x j ∂x k

  

k 1 lk  ∂g jl ∂g li ∂gij   = g  i + j − l 2 ∂x ∂x i j  ∂x

… (4)

… (5)

  

Proved.

EXAMPLE 3 If t i and T α are contravariant components in x's and y's respectively, of n vector field in V n immersed in V m. Show that T, αj = y,αij t i + y ,αi t,i j Solution Since we know that T

α

i = t

∂y α ∂x i


201

i α T α = t y, j . Taking covariant differentiation of both sides, we get

T, αj = (t i y α,i ), j = t,i j y,αi + t i ( y ,αi ), j T, αj = t i y ,αij + t,i j y ,αi EXAMPLE 4 Show that g ij

dx i dx j remains constant along a geodesic. ds ds

Solution i Let t =

dx i . Then ds

dx i dx j = g ij t i t j = t 2 ds ds Since we know that geodesics are autoparallel curves. Then gij

t,i j

dx j =0 ds t,i j t j = 0

or

… (1)

Now, dt 2 d d ( g ij t i t j ) = (ti t i ) = ds ds ds dx j i = (t i t i ), j t j = ( ti t ) , j ds dt 2 = (ti , j t i )t i + (t ,i j t i )ti = 0, ds

from (1)

Integrating it we get t2 = constant. So, g ij

dx i dx j remains constant along a geodesic. ds ds

EXAMPLE 5 If ti are the contravariant components of the unit tangent vector to a congruence of geodesics. Show that t i (ti , j + t j ,i ) = 0 and also show that | ti , j + t j ,i |= 0.

202


Solution Let t i denote unit tangent vector to a congruence of geodesic so that t,i j t j = 0 Since geodesics are auto-parallel curves. Then to prove that (i)

… (1)

t i (ti , j + t j ,i ) = 0

(ii) | ti , j + t j ,i |= 0 Since t i ti = t 2 = 1 t i ti = 1 Taking covariant derivative of both sides, we get (t i t i ), j = 0 or

t,i j t i + t i ti , j = 0

Since t is a free index. Then we have t,i j t i + ti , j t i = 0 2t i, j t i = 0 ⇒

ti , j t i = 0

… (2)

from (1) g ik t,i j t j = 0 or

tk, j t j = 0

⇒

t k ,i t i = 0

Thus

t i t j, i = 0

⇒

t j ,i t i = 0

⇒

t i t j, i = 0 … (3)

Adding (2) and (3), we get t i (ti , j + t j ,i ) = 0 Also, since ti ≠ 0, ∀i. Taking determinants of both sides we get | t i (ti , j + t j , i ) | = 0 ⇒

| ti , j + t j ,i | = 0 Solved.

EXAMPLE 6 When the coordinates of a V 2 are chosen so that the fundamental forms is (dx&' ) + 2 g12dx1 dx 2 + (dx 2 )2 , prove that the tangent to either family of coordinate curves suffers a parallel displacement along a curve of the other family.


203

Solution The metric is given by i j ds 2 = g ij dx dx

i,j = 1,2.

Comparing it with 1 2 1 2 2 2 ds 2 = (dx ) + 2 g12 dx dx + (dx )

(1)

We have, g11 = 1, g22 = 1. In this case, we have coordinate curves of parameters x1 and x2 respectively. The coordinate x1 curve is defined by xi = ci, ∀i, except i = 1. (2) 2 and the coordinate x curve is defined by x i = d i, ∀i, except i = 2 (3) i i where c and d are constants. Let p i and q i be the components of tangents vectors to the curves (2) and (3) respectively. Then we have pi = dxi = 0, ∀i, except i = 1 and qi = dxi = 0, ∀i, except i = 2 So, pi = (dx i, 0) and qi = (0,dx i) Let t be the unit tangent vector to the curve (2). Hence pi dx i i t = = (1,0) = p ds

where

p = dxi

So, we have dx j =0 ds Hence the tangent to the family of coordinates curves (2) suffers a parallel displacement along a curve of the family of curves (3). p,i j

EXERCISES 1. Explain Levi-Civita's concept of parallelism of vectors and prove that any vector which undergoes a parallel displacement along a geodesic is inclined at a constant angle to the curve. 2. Show that the geodesics is a Riemannian space are given by

d 2 x m  m  dx i dx k +  =0 ds 2 i k  ds ds Hence prove that geodesics are auto-parallel curves.

204


3. Establish the equivalence of the following definitions of a geodesic. (i) It is an auto–parallel curve. (ii) It is a line whose first curvature relative to Vn identically zero. (iii) It is the path extremum length between two points on it. 4. If u and v are orthogonal vector fields in a V n , prove that the projection on υ of the desired vector of u in its own direction is equal to minus the tendency of v in the direction of u. 5. If the derived vector of a vector u i is zero then to show that vector u i has a constant magnitude along curve. 6. Prove that any vector which undergoes a parallel displacement along a geodesic is inclined at a constant angle to the curve. 7. Prove that there are m – n linearly independent vector fields normal to a surface Vn immersed in a Riemannian Vm and they may be chosen in a multiply infinite number of ways. But there is only one vector field normal to the hyperface. 8. Show that the principal normal vector vanishes identically when the given curve in geodesic. 9. Show that if a curve is a geodesic of a space it is a geodesic of any subspace in which it lies. 10. Define parallelism in a subspace of Riemannian manifold. If a curve C lies in a subspace Vn of Vm and a vector field in Vn is parallel along C with respect to V m . Then show that it is also parallel with respect to V n .

CHAPTER – 11

RICCI'S COEFFICIENTS OF ROTATION AND CONGRUENCE 11.1 RICCI'S COEFFICIENTS OF ROTATION Let e ih| (h=1, 2, …n) be the unit tangents to the n congruences eh , of an orthogonal ennuple in a e Riemannian V n, . The desired vector of el|i in the direction of e k | has components el|i , j ekj| and the

projection of this vector on e ih| is a scalar invariant, denoted by γlhk, so that γlhk = el|i, j eih|ekj|

…(1)

The invariants γlhk are Ricci's Coefficients of Rotation. Since i being a dummy index, has freedom of movement. Then equation (1) may be written as γlhk = el |i , j eh|i ekj|

…(2)

The indices l, h and k are not tensor indices. But these indices in γlhk are arranged in proper way, the first index l indicates the congruence whose unit tangent is considered, the second h indicates the direction of projection and the third k is used for differentiation. THEOREM 11.1 To prove thast the Riccie's coefficients of rotation are skew-symmetric in the first two indices i.e., γlhk = –γlhk Proof: If e ih| (h = 1, 2, …, n) be n unit tangents to n congruences e h| of an orthgonal ennuple in a Vn then eh|i eli| = 0 convariant differentiation with respect to xj, we get eh|i eli| ,j = 0 eh |i , j eli| + eli|, j eh|i = 0

206


multiplying by ekj| and summing for j, we get eh |i , j eli|eki | + eii|, j eh|i e kj| = 0 γhlk + γlhk = 0 γhlk = –γlhk

or

…(1)

Note: Put l = h in equation (1), we get γllk = –γllk 2γllk = 0 γllk = 0

or or

THEOREM 11.2 To prove that

∑ γ lhk eh|i h

= el|i , j ekj|

Proof: since we know that γlhk = el|i, j eih|ekj| Multiplying by eh|m and summing for h.

∑ γ lhk eh|m = ∑ el|i , j eih|ekj|eh|m h h

j = el |i , j e k|

∑ (e e ) i

h|

h| m

h

= el |i , j ekj|δim since

∑ ehi | eh|m = δ h

i m

= (e l |i , j δ m ) e kj | i

∑γ

lhk eh |m

h

= el |m , j ekj |

Replacing m by i, we get

∑ γ lhkei|m h

= el |i , j e kj|

11.2 REASON FOR THE NAME "COEFFICIENTS OF ROTATION" Let Cm be a definite curve of the congruence whose unit tangent is emi | and P 0 a fidxed point on it. Let u1 be a unit vector which coincides with the vector eli| at P 0 and undergoes a parallel displayment along the curve Cm. Thus u i = eli| at P 0 and

u,ij ehj| = 0

… (1) … (2)

Ricci's Coefficients of Rotation and Congruence

207

If θ is the angle between the vectors ui and e ih| , we have cos θ = ui eh|i Differentiating it with respect to arc length sm along Cm, we get – sin θ

dθ i j ds m = (u e h|i ), j em | = (u i e h|i , j + u i, j eh |i ) emj |

– sin θ at the point P 0,θ =

dθ u i e h |i , j e mj | + u ,i j e mj | ds m =

…(3)

π , we have 2 –

dθ u i eh|i , j emj | ds m = = u i eh|i , j emj | = eli|eh|i, j emj | ;

from (1)

= e h|i , j eli| emj | – ⇒

dθ ds m = γhlm dθ ds m = – γhlm

… (4)

dθ In Eucliden space of three dimensions, ds is the arc-rate of rotation of the vector e ih| about the m curve Cm. Hence the quantities γhlmare called coefficients of rotation of the ennuple. Since it was discovered by Ricci and hence it is called Ricci's coefficient of rotation. 11.3 CURVATURE OF CONGRUENCE The first curvature vector pl| of curve of the congruence is the derived vector of e ih| in its own direction. Where e ih| (h = 1, 2, …, n) be unit tangents to n congruence eh| of an orthogonal ennuple in a Vn . If phi | be the contravariant component of first curvature vector pl|. Then by definition, we have phi | = ehi , j ehj| from theorem (11.2), we have phi | =

∑ γ hlh eil| l

… (1)

208


The magnitude of p ih| is called curvature of the curve of congruence e h| and denoted by K h | . Now, K h2| = g ij phi | p hj|  = g ij  

∑γ l

     

i  hlh e l | 

∑γ

i hmh em |

m

=

∑∑ g ij eli| emi |γ hlh γ hmh

=

∑∑ δ

=

∑ γ hmhγ hmh

l

l

   

m

l m γ hlm

m

γ hmh , Since g ei e i = l δm ij l| m|

m

K h2| =

∑ (γ

hmh )

2

m

This is the required formula for Kh|. 11.4 GEODESIC CONGRUENCE If all the curves of a congruence are geodesics then the congruence is called a geodesic congruence. THEOREM 11.3 A necessary and sufficient condition that congruence C of an orthogonal ennuple be a geodesic congruence is that the tendencies of all the other congruences of the ennuple in the direction of C vanish indentically. Or To obtain necessary and sufficient conditions that a congruence be a geodesic congruence. Proof: From equation (1), Pg. 207, we have phi | =

∑ γ hlh eil| l

Since

eli|

≠ 0 , ∀h and hence

phi |

= 0 iff γhlh = 0, ∀h . But pli| = 0 iff the congruence C is geodesic

congruence. Hence C is a geodesic congruence iff γhlh = 0, ∀h . But γhlh = – γlhh So, C is a geodesic congruence iff – γlhh = 0, –∀h . or C is a geodesic congruence iff γlhh = 0. Hence γlhh = 0 are the necessary and sufficient conditions that congruence with unit tangents ehi | be geodesic congruence. Again γlhh is the tendency of the vector eli| in the direction vector e ih| .


209

Thus a congruence C of an orthogonal ennuple is a geodesic congruence iff the tendency of all other congruences in the direction of C vanish identically. 11.5 NORMAL CONGRUENCE A normal congruence is one which intersects orthogonally a family of hypersurfaces. THEOREM 11.4 Necessary and sufficient conditions that the congruence eh | of an orthogonal ennuple be normal is that γnqp = γnpq Proof: Consider a congruence C of curves in a V n. Let φ( x1 , x 2 , …, x n ) = constant be a family of hypersurfaces. To determines a normal congruence whose tangent vector is grad φ or ∇ φ. Let ti be the convariant components of unit tangent vector to C. The congruence C is a normal congruence to a family of hypersurface φ( x1 , x 2 , …, x n ) = constant if φ, φ,1 φ, 2 = … n = y (say) = … (1) t1 t2 tn In order that (n – 1) differential equations given by equation (1) admit a solution which is not constant, these must constitute a complete system. From (1) φ,i = y or φ, i = yti ti ∂φ yti = ∂x i

⇒

Differentiating it with respect to x j , we get ∂y ∂x j

ti + y

∂t i ∂x j

=

∂ 2φ

∂x i ∂x j Interchanging indices i and j, we get ∂t j ∂y ∂ 2φ t + y = j ∂x i ∂x i ∂x j ∂x i Subtracting (2) and (3) we get ∂t   ∂y ∂t   ∂y  j ti + y i j  –  i t j + y i j  = 0 ∂x   ∂x dx   dx ∂t j   ∂y  ∂t ∂y  y ij – i  +  t i j – t j i  = 0 ∂x   ∂x ∂x   ∂x

… (2)

… (3)

210


Multiplying by tk,  ∂ti ∂t j  ∂y ∂y – ik ti i = 0 ytk  j – i  + ti tk j ∂x  ∂x dx  dx By cyclic permutation of i, j, k in (4), we get  ∂t j ∂t yti  k – kj ∂x  ∂x

…(4)

 ∂y ∂y  + t j t i k – i k ti =0 ∂x ∂x j 

… (5)

and ∂t   ∂t ∂y ∂y yti  ki – ik  + t k t j i – ti t j k = 0  ∂x ∂x  ∂x ∂x

… (6)

On adding (4) , (5) and (6) we get   ∂t ∂t j   ∂t j ∂t y t k  ij – i  + ti  k – kj ∂x  ∂x  ∂x   ∂x

  + t j  ∂tk – ∂ti   ∂x i ∂x k 

t k (ti , j – t j ,i ) + ti (t j , k – t k , j ) + t j (t k , i – ti ,k ) = 0

or

  = 0   … (7)

as y ≠ 0 , where i, j, k = 1, 2, …, n. These are the necessary and sufficient conditions that the given congruence be a normal congruence. Now suppose that the congruence is one of an orthogonal ennuple in V n. Let en|i be the unit tangents of given congruence C so that ti = en|i Then equation (7) becomes en|k (en| i,j – en| j,i) + en| i (en| j,k – en| k,j) + en| j (en|k,i – en|i,k) = 0 … (8) Now, multiplying equation (8) by e ip| eqR| , we get e n|k (en |i , j – en | j ,i ) e ip| eqk| + en |i (en | j , k − en| k , j ) e ip| eqk + en | j (en |k ,i – en |i , k ) e ip |eqk| where p and q are two new indices chosen from 1, 2, …, n – 1. i.e., p,q and n are unequal. e n |k eqk| = δ q = 0, q ≠ n h

But

en |i e kp| = δ hp = 0, p ≠ n so, we have e n| j (en |k ,i – en | j , k ) eip | eqk| = 0 e n| j (en |k , i e iq |e kp | – en | j , k eip |e kq| ) = 0 en|j (γnqp – γnpq) = 0 Since

en |k , i eip | eqk| = γnpq


211

γnqp – γnpq = 0, en| j ≠ 0 ⇒ γnpq = γnqp, … (9) ⇒ Conversely if equation (9) in true then we get equation (8). Which implies that equation (7) are satisfied bty en|i Hence en| is a normal congruence. Thus necessary and sufficient conditions that the congruence en| of an orthogonal ennuple be a normal congruence are that γnqp = γnpq (p, q = 1, 2, …, n – 1 such that p ≠ q) THEOREM 11.5 Necessary and sufficient conditions that all the congruences of an orthogonal ennuple be normal. Proof: If all the congruences of a orthogonal ennuple are normal. Then γnqp = γnpq (p, q = 1, 2, …, n – 1 such that p ≠ q) If the indices h, k, l and unequal then γhkl = γhlk … (1) But due to skew-symmetric property i.e., γhlk = – γlhk. So, γhkl = γhlk = – γlhk = – γlkh, from (1) = γklh (skew-symmetric property). = γkhl, from (1) γhkl = – γhkl , (skew-symmetric property). γhkl + γhkl = 0 2 γhkl = 0 ⇒ γhkl = 0 ⇒ where (l, h, k = 1, 2, …, n such that h, k, kl are unequal). 11.6 CURL OF CONGRUENCE The curl of the unit tangent to a congruence of curves is called the curl of congruence. If en| is a given congruence of curves then Curl en|i = en|i, j – en| j,i If curl en| i = 0 then congruence is irrotational. THEOREM 11.6 If a congruence of curves satisfy two of the following conditions it will also satisfy the third (a) that it be a satisfy the third (b) that it be a geodesic congruence (c) that it be irrotational. Proof: Consider an orthogonal ennuple and e ih| (h = 1, 2, …, n) be n unit tangent to n congruences of this orthogonal ennnuple. From theorem (11.2), we have

∑ γ lhkeh|i = el|i , j ekj| h

212


Putting l = n and j = m, we have n

∑γ

nhk eh |i

h =1

m = e n |i , m e k |

Now, multiplying by ek| j and summing with respect to k from 1 to n, we have n

∑ γ nhk eh|i ek | j = en|i ,me mk| ek| j

h,k =1

= en |i ,mδ mj

Since ekm| ek| j = δmj

n

∑γ

h, k =1

nhk eh|i e k | j

= e n|i, j

… (2)

By definition of curl of congruence, we have curl en|i = en| i, j – en| j,i =

n

n

h,k =1

n ,k =1

∑ γ nhke h|i ek| j – ∑ γ nhkeh| j ek |i ; n

curl en| i =

∑

from (2)

n

γ nhk eh |i ek | j –

h ,k =1

∑γ

h , k =1

nkh ek | j e h|i

n

curl en|i =

∑

(γnhk – γnkh) eh| iek| j

h ,k =1

… (3)

This double sum may be separated into two sums as follows. (i) Let h and k take the values 1, 2, …, n – 1. (ii) Either h = n or k = n or h = k = n. Now, the equation (3) becomes n −1

curl en| i =

∑

h ,k =1

n –1

(γ nhk – γ nkh ) eh |i ek | j +

n

+

∑

∑ h =1

(γ nhn – γ nnh ) e h|i en | j

( γ nnk – γ nkn ) en |i e n| j + (γ nnn – γ nnn ) en |i en | j

k =1

Since we know that γnnk = γnnh = γnnn= 0. So, n –1

curl en|i

=

∑

h , k =1 n–1

curl en|i

=

(γ nhk – γ nkh ) e h |i e k | j +

∑ h ,k

n –1

∑

γ nhn e h|i en | j –

h =1

(γ nhk – γ nkh ) eh|i ek | j +

n–1

∑γ

nkn en |i

ek| j

k =1

n–1

∑γ h =1

nhn ( eh |i en | j

– en |i ek | j )

… (4)


213

The first term on R.H.S. of equation (4) vanishes if γnhk – γnkh = 0 i.e., if γnhk = γnkh. i.e., if the congruence e n| is normal. Again the second term of R.H.S of equation (4) vanishes γ nhn = 0

if

i.e., if γ hnn = 0

i.e., if the congruence e n| is a geodesic congruence. Further, if first and second term on right hand side of equation (4) both vanishes then curl en|i = 0 Hence we have proved that if the congruence e n| satisfies any two of the following conditions then it will also satisfy the third. (a) e n| is a normal congruence (b)

e n| is irrotational

(c)

e n| is a geodesic congruence

11.7 CANONICAL CONGRUENCE It has been shown that given a congruence of curves, it is possible to choose, in a multiply infinite number of ways, n – 1 other congruences forming with the given congruence an orthogonal ennuple. Consider the system of n – 1 congruence discovered by Ricci, and known as the system canonical with respect to the given congruence. THEOREM 11.7 Necessary and sufficient conditions that the n – 1 congruences eh| of an orthogonal ennuple be canonical with respect to en| are γnhk + γnkh = 0; (h, k = 1, 2, …, n – 1, h ≠ k ). Proof: Let the given congruence en| be regarded as nth of the required ennuple. Let en|i be unit tangent to given congruence. X ij = ½ (en| i, j + en| j,i)

… (1)

Let us find a quantity ρ and n quantities ei satisfying the n + 1 equations e n|i ei = 0  i, j = 1, 2, …, n ( X ij – ωgij e i + ρE n| j = 0

… (2)

where ω is a scalar invariant. Writing equation (2) in expansion form, we have e n|1e1 + en |2 e 2 + …… + en| n e n = 0 and ( X 1 j – ωg1 j ) ei + ( X 2 j – ωg 2 j ) e 2 + … ( X nj – ωg nj ) e n + ρ e n| j = 0

… (3)

214


for j = 1, 2, …, n. 1

2

n

( X 11 – ωg11 ) e + ( X 21 – ωg 21 ) e + … + ( X n1 – ωg n 1 ) e + ρ en |1 = 0 ( X 12 – ωg12 ) e1 + ( X 22 – ωg 22 ) e 2 + … + ( X n 2 – ωg n 2 )e n + ρ en |2 = 0 M ( X 1n – ωg1 n ) e1 + ( X 2 n – ωg 2 n ) e 2 + … + ( X nn – ωg nn ) e n + ρ en |n = 0 from (3) e n|1 e1 + en |2 e 2 + … en |n e n + ρ.0 = 0 eliminating ρ and the quantities e1, e2, …en, we have the equation

(X 11 – ωg11 ) (X 21 – ωg 21 ) ( X12 – ωg12 ) ( X 22 – ωg 22 )

… …

(X n1 – ωg n1 ) ( X n2 – ωg n2 )

(X 2n – ωg 2n )

… …

(X nn – ωg nn )

M ( X1n – ωg1n ) en |1

en |2

en| n

en |1 e n|2 en |n 0

which is of degree n – 1 in ω. Hence there will be n – 1 roots of ω and these roots be ω1, ω2, …, ωn – 1. All roots are real. Let ωh be one of these roots and let the corresponding values of ρ and ei be denoted by ρh and e ih| respectively. Then ρh and e ih| will satisfy the equation (2), we have en |i e ih| = 0 and ( X ij – ωg ij ) eih | + ρh en | j = 0 Similarly ωk be another root of these roots, we have

… (4) … (5)

en |i e ik| = 0

… (6)

( X ij – ωg ij ) eik | + ρh en | j = 0

… (7)

Multiplying (5) by e kj| and (7) ehj| and using (4) and (6), we get ( X ij – ωh g ij ) ehi |ekj| = 0

… (8)

( X ij – ωk g ij ) eki |e hj| = 0

… (9)

Since X ij and g ij are symmetric tensor in i and j. Now, interchanging i and j in equation (9), we get ( X ij – ωk g ij ) ehi |e kj| = 0 Subtracting (10) and (8), we get

… (10)

(ωh – ωk ) g ij e ih| ekj | = 0

… (11)

Since ωk – ωk ≠ 0 as ωh ≠ ωh. ⇒

g ij eih| ekj| = 0

… (12)


215

This shows that e ih| and e kj| unit vectors are orthogonal to each other and hence the congruence e h| and e k| (h ≠ k) are orthogonal to each other. Hence the n – 1 congruence eh| (h = 1, 2, …, n) thus determined form an orthogonal ennuple with en|. Using equation (12), equation (10) becomes X ij e ih| ekj| = 0 Since from (1), X ij =

(

(

)

1 e + e n| j , i . Then we have, 2 n |i , j

)

1 e + en | j ,i ehi |ekj| = 0 2 n |i , j en |i , j ehi | ekj| + en| j ,i eik ehj = 0 γnhk + γnkh = 0; (h, k = 1, 2, …, n – 1 such that h ≠ k ) … (13) Conversely if equation (13) is true then (n – 1) congruences eh| of the orthogonal ennuple and canonical with respect to e n| . Hence necessary and sufficient condition that the n – 1 congruences e h| of an orthogonal ennuple be canonical with respect to en| are γnhk + γnkh = 0; (h, k = 1, 2, …, n such that h ≠ k ) THEOREM 11.8 Necessary and sufficient conditions that n –1 mutually orthogonal congruences e h| orthogonal to a normal congruence e n| , be canconical with respect to the later are γnhk = 0 where k, h = 1, 2,…, n – 1 such that h ≠ k. Proof: By theorem (11.7), Necessary and sufficient conditions that (n – 1) congruences e h| of an orthogonal ennuple be canonical with respect to en| are γnhk + γnkh = 0 (h, k = 1, 2, …, n – 1 such that h ≠ k). If the congruence en| is normal. Then γnhk = γnkh The given condition γ nhk + γ nkh = 0 becomes γnhk + γnhk = 0 2γnhk = 0 γnhk = 0

Proved.

EXAMPLE 1 1 If en| are the congruences canonical with respect to en| prove that (i) ωh = γnhk (ii) ρh = γhnn, 2 (iii) If en| is a geodesic congruence, the congruences canonical with respect to it are given by ( X ij − ωg ij )ei = 0 = 0 Solution Suppose (n – 1) congruences eh| of an orthogonal ennable in a V n are canonical with respect to the cougruence eh| then en |iehi | = 0

… (1)

216


and where

( X ij – ωh g ij ) ehi | + ρ h en | j = 0

X ij =

… (2)

1 (e + en| j,i) 2 n| i,j

… (3)

Since e in| unit tangents then g ij ehi |ehj| = 0

… (4)

(i) Multiplying equation (2) by e ih| , we get ( X ij – ωh g ij ) ehi | ehj| + ρh e n| j e hj| = 0 X ij eih| ehj| – ω h g ij eih |ehj| = 0

since en| j ehj| = 0

(from (1))

X ijeih| ehj| – ωh = 0, since g ijehi |ehj| = 1 1 (e n|i , j + en| j , i ) e ih|e hj| – ωh = 0; 2

from (3)

1 (e e i e j + en | j ,i eih| ehj | ) – ωh = 0 2 n|i , j h | h| 1 (γ + γ ) – ω = 0 nhh nhh h 2 ωh = γnhh (ii) Multiplying (2) by enj| , we get ( X ij – ωh g ij ) ehi | enj| + ρh e n| j e nj| = 0 X ij eih| enj| – ω h g ij ehi |enj| + ρ h × 1 = 0

1 (e n|i , j + en| j , i ) e ih|e hj| + ρh = 0, 2

from (1) since e ih| e hj| = 0

1 (e ei e j + en | j ,i ehi | ehj | ) + ρh = 0 2 n|i , j h | n| 1 (γ + γnnh) + ρh = 0 2 nhn ρh = – or

ρh =

1 γ , since γnnh = 0 2 nhn

1 γ 2 hnn

(iii) If en| is a geodesic congruence, then γhnn = 0 from the result (ii), we have 1 ρh = × 0 2

ρh = 0


217

from equation (2), we get

( X ij – ω h gij ) ehi | + 0en| j = 0 or

( X ij – ωh g ij ) e ih| = 0

this gives ( X ij – ωg ij ) e i = 0 . EXAMPLE 2 Prove that when a manifold admits an orthogonal systyem of n normal congruences then any of these in canonical with respect to each other congruence of the system. solution Let e ih| (h = 1, 2, …, n) be unit tangents to n normal congruences of an orthogonal ennuple in a V n. So that γlhk = 0, where l, h, k = 1, 2,...n such that l, h, k being unequal. It is required to show that a congruence eh| is canonical with respect to the congruence ek| (h, k = 1, 2, …, h ≠ k). We know that the n – 1 congruence en| of an orthogonal ennuple be canonical to en| iff γnhk + γnkh = 0 This condition is satisfied by virtue of equation (1). Hence (n – 1) congruences eh| of an orthogonal ennuple are canonical to en|. Similarly we can show that any n – 1 congruences are canonical to the remaining congruence. It follows from the above results that any one congruence is canonical with respect to each other congruence of the system.

EXERCISE 1. If φ is a scalar invariant n

∑ φ,ij eih|ehi | = ∆ φ 2

h =1

2. The coefficient of ρ n – 1 in the expansion of the determinant |φ,ij – ρg ij | is equal to ∇ 2 φ. 3. If eh| are the unit tangents to n mutually orthogonal normal congruences and e11 + be21 is also a normal congruence then ae11 – be21 is a normal congruence. 4. Show that ∂Q ∂ ∂φ ∂ ∂Q ( γ lhk – γ lkh ) – = ∂sl ∂sh ∂sk ∂sk ∂sh l

∑

where S h denotes the arc length of a curve through a point P of an ennuple and Q is a scalar invariant. 5. If the congruence eh| (h = 1, 2, … n – 1) of an orthogonal ennuple are normal, prove that they are canonical with respect to other congruence eh| .

CHAPTER – 12

HYPERSURFACES 12.1 INTRODUCTION We have already studied (Art 10.3 chapter 10) that if m > n then we call V n to be a subspace of V m and consequently V m is called enveloping space of V n. We also know that there are m – n linearly independent normals Nα to V n. (Art 10.3 Theorem 10.5, chapter 10). If we take m = n + 1 then V n is said to be hypersurface of the enveloping space V n + 1 Let V n be Riemannian space of n dimensions referred to cordinates xi and having the metric ds2 = i gij dx dxj. Let V n be Riemannian space of m dimensions referred to coordinates yα and having the metric ds2 = aαβ dyα dyβ. Where m > n. Let Greek letters α, β, γ … take the values 1, 2, …, m and latin indices i, j, k, … take the values 1, 2, …n. Then we have, the relation between aαβ and gij α β gij = aαβ ∂y ∂y

∂xi ∂x j

… (1)

Since the function yα are invariants for transformations of the coordinates xi in V n, their first covariant derivatives with respect to the metric of V n are the same as their ordinary derivatives with respect to the variables xi. yα, i =

i.e.,

∂yα ∂xi

Then equation (1) can be written as gij = aαβ y,αi yβ, j The vector of V n

+ i

… (2)

whose contravariant components are y,αi is tangential to the curve of

parameter xi in V n. Consequently if N α are the contravariant components of the unit vector normal to V n. Then we have a αβ N β y,αi = 0, and

a αβ

N αN β =

1

(i = 1, 2, …, n)

… (3) … (4)

Hypersurface

219

12.2GENERALISED COVARIANT DIFFERENTIATION Let C be any curve in V n and s its arc length. The along this curve the x's and the y's may be expressed as function of s only. Let uα and ν β be the components in the y's of two unit vector fields which are parallel along C with respect to V m. Similarly w i the components in x's of a unit vector field which is parallel along C with respect to V n. Now, uα is parallel along C relative to V m then we have, uα,β

dyβ =0 ds

 ∂u α  γ  dy β  β – uγ   =0 α β ds  ∂y ∂u α dy β  γ  dy β – u  =0 γ ∂y β ds α β ds duα  γ  dy β – uγ   =0 ds α β ds  α  dy duα = u γ β γ ds ds  

β

… (1)

similarly v β is parallel along C relative to V m then α

γ  γ  dy dvβ = – v α β ds ds  

… (2)

and wi is parallel along C relative to V n then j  i  dx dwi = – w  j k  ds ds  

k

… (3)

The Christoffel symbol with Greek indices being formed with respect to the aαβ and the y's and christoffel symbol with Latin indices with respect to the gij and the x's. Let Aβαi be a tensor field, defined along C, which is mixed tenser of the second order in the y's and α a covariant vector in the x's. Then the product uα v β wi Aβi is scalar invariant and it is a function of s along c. Its derivative with respect to s is also a scalar invariant. α Differentiating uα v β wi Aβi with respect to s, we have

dAα d dv β dw i i α (uα v β wi Aβαi ) = u v β w i βi + du α v β w i A α + u w A + uα v β Aβαi α βi α βi dt ds ds ds ds β

= uα v w

i

dAβαi ds

+

duδ β i δ dv δ dw j v w Aβi + uα w i Aδαi + u αvβ Aβαj ds ds ds

220


β i = u αv w

γ  α  dy γ α β i  δ  dy + Aβδi .u αv βwi   – Aδi .u αv w βγ  ds ds   δγ  ds

dAβαi

k α β i  j  dx – Aβj uα v w ik  ds , by equation (1), (2) & (3).   γ γ k   α β i dAβi δ  α  dy α  δ  dy α  j  dx u v w  + A . – A – A .    βi  δi  βj   = α δ γ  ds β γ  ds i k  ds   ds

= Scalar, along C Since the outer product uαv βwi is a tensor and hence from quotient law that the expression within

α α the bracket is tenser of the type Aβi and this tensor is called intrinsic derivative of Aβi with respect to s. The expression within the bracket can also be expressed as

dxk ds

Since α Aβi

 ∂Aβαi α δ   j   + Aβδi   y,γk – Aδαi   y,γk – Aβαj     ∂xk δγ  βγ  ik  

dxk is arbitrary.. ds

So, by Quotient law the expression within the bracket is a tensor and is called tensor derivative of with respect to xk. It is denoted by Aβαi;k . Then we have Aβαi;k =

∂Aβαi

α δ   j + Aβδi   y,γk – Aδαi   y,γk – Aβαj  . δγ βγ dx     ik  k

α Aβαi;k is also defined as generalised covariant derivative Aβi with respect to xk .

Note:– Semi-colon(;) is used to denote tensor differentiation. 12.3 LAWS OF TENSOR DIFFERENTIATION THEOREM 12.1 Tensor differentiation of sums and products obeys the ordinary rules of differentiation. Proof: Suppose Aβα , Bβα and B γ are tensors in V m. (i) To prove that ( Aβα + Bβα ); k = Aβα;k + Bβα; k Let the sum Aβα + Bβα be denoted by the tensor Cβα . Now, Cβα;k

=

∂Cβα

α α  + Cβα   y, ck – Caα   y, ck ∂x a c βc k

Hypersurface

∴

221

( Aβα

+

Bβα ); k

=

∂( Aβα + Bβα ) dx

k

a  α + ( Aβa + Aβα )   y,ck – ( Aaα + Baα )  y ,ck a c  β c 

 ∂Aβα  aα c αa  c   + A y – A y     β , k a , k =  ∂x k + ac β c    ∂Bβα α   a   + Bβa   y,ck – Baα   y,ck   ∂xk  ac βc  

( Aβα + Bβα ); k = Aβa ;α + Bβa ;k Hence the result (i) (ii) Prove that ( Aβα + Bγ ); k = Aαβ; k Bγ + Aαβ Bγ ; k α Aβα Bγ = Dβγ

Let

α Then Dβγ is a tensor

we have D αβγ ; k =

∴

( A αβ

Bγ ) ; k =

α ∂Dβγ

∂x

k

α  a a  + D;ak   yc – Dαaγ   y,ck – Dαβa   y,ck ac   βc  γc 

∂( Aβα B γ ) dx

k

 α a a a + Aβγ Bγ   y ,ck – Aaα Bγ   y,ck – Aβα B a   y ,ck ac  βc  βc 

 ∂Aβα   ∂B  a α α c αa  c   + A y – A y Bγ + Aβα  γ – Ba   y,ck      β , k a , k =  dxk k ac β c  γc    dx 

( Aβα B γ ); k = Aβα; k Bγ + Aaα Bγ ; k Hence the result (ii). Note:– α, β, γ, a, c take values from 1 to m while k take values from 1 to n. THEOREM 12.2 To show that aαβ; j = 0 or To prove that the metric tensor of the enveloping space is generalised covariant constant with respect to the Christoffel symbol of the subspace. Proof: We have (see pg. 220) aαβ;i =

=

∂aαβ dxi

∂aαβ dx i

γ γ  – aγβ   y,δi – aαγ   y,δi αβ   βδ 

– ( [αδ, β] + [βδ, α ] ) y,δi

222


or

aαβ;i =

or

aαβ;i =

∂aαβ dx

∂aαβ dx

–

∂aαβ δ y,i ∂ yδ

–

∂aαβ ∂yδ ∂ aαβ ∂ aαβ ⋅ i = – δ ∂y dx ∂x i ∂x i

i

i

= 0

Proved.

12.4 GAUSS'S FORMULA At a point of a hypersurface V n of a Riemannian space V n + 1, the formula of Gauss are given by y;αij = Ωij N α

Proof: Since yα is an invariant for transformation of the x's and its tensor derivative is the same as its covariant derivative with respect to the x’s, so that yα; i = yα, i =

dyα

… (1)

dxi

Again tenser derivative of equation (1) with respect to x’s is α y;αij = ( y;i ); j = ( y,αi )

y;αij

=

l  α  ∂ ( y,αi ) – y,αl   + yβ,i y,γj   j dx ij  βγ 

=

∂  ∂y α  α  l  β γ  α  – y,l   + y,i y, j   ∂x j  ∂xi  ij βγ 

=

∂ 2 yα

l  α  – y,αl   + yβ,i y,γj   ∂x ∂x ij βγ  i

… (2)

j

Interchanging j and i in (2), we have y;αji

∂ 2yα l  α  – y ,αl   + y β, j y ,γi   = j i ∂x ∂x  ji  βγ 

y;αji

∂ 2yα l  – y ,αl   + y β,l y ,γj = i j ∂x ∂x ij 

α    βγ

… (3) α 

α 

[On interchanging β and γ in third term of R.H.S. of (3) and using βγ  = γβ  ].     On comparing equation (2) and (3), we have y;αij = y;αji

So, y;αij is symmetrical with respect to indices i & j.

Hypersurface

223

Let g ij dx idx j and aαβdyα dyβ be fundamental forms corresponding to V n and V n + 1 respectively.. Then gij = aαβ

∂yα ∂yβ ∂xi ∂x j

gij = aαβ y;αi y;βj

or

Taking tensor derivative of both sides with respect to xk gij;k = aαβ;k y;αi y;βj + aαβ y;αi y;βjk + aαβ y;αik y;βj But

gij;k = 0

and

aαβ;k = 0.

we have, 0 = a αβ y ;αi y;βjk + aαβ y;αik y;βj or

aαβ y;αik y,βj + aαβ y,αi y;βjk = 0, using (1), Art. 12.4

… (4)

By cyclic permutation on i, j, k in (4), we have

and

aαβ yα; ji y,βk + aαβ y α, j y;βki = 0

… (5)

a αβ y;αkj y,βi + aαβ y α,k y β;ij = 0

… (6)

subtracting equation (4) from the sum of (5) and (6), we get. 2 aαβ y;αij y ,βk = 0 a αβ y ;αij y β,k = 0

or

… (7)

This shows that y;αij is normal (orthogonal) to y,βk . Since y,βk is tangential to V n and hence y;αij is normal to V n. Then we can write y;αij = N αΩij where

Nα

(8)

is unit vector normal to V n and Ω ij is a symmetric covariant tensor of rank two. Since

y;αij is a function of x's the tenser Ω ij is also a function of x's.

The equation (8) are called Gauss’s formula. From equation (8) y;αij = Ωij N α

y;αij aαβ N β = Ω ij aαβ N β N α = Ω ij N 2

224


y;αij aαβ N β = Ωij , or

Ω ij

=

N=1

y;αija αβN β.

The quadratic differential form Ωij dxi dx j

is called the second fundamental form for the hypersurface V n of V n + 1. The components of tenser Ω ij are said to be coefficient of second fundamental form. Note: The quadratic differential form g ij dx idx j is called first fundamental form. 12.5 CURVATURE OF A CURVE IN A HYPERSURFACE AND NORMAL CURVATURE If U α and u i be the contravariant components of the vector u relative to V n and V n + 1 respectively then we have (from chapter 10, Theorem 10.4) ∂y α i u = y,αi ui … (1) U ∂ xi Let the derived vector to vector u along C with respect to metric of V n and Vn + 1 are denoted by p and q respectively. Then

and

α

=

pi

=

qα

u,i j

dxj ds

α = U,β

d yβ ds

= U ;αj

dx j , (from equation 1, Art 12.4) ds

… (2)

Taking the tensor derivative of each side of equation (1) with respect to x's, we have U ;αj = y; ij u + y, i u , j α

α

i

i

By Gauss's formula, we have y;αij = Ωij N α Then

U ;αj = Ω ij N u + y ,i u , j α

i

α i

Putting the value of u;αj in equation (2), we have dx j ds

qα

α i α i = (Ωij N u + y, i u, j )

or

qα

j   α i i dx  α  Ω u =  ij  N + y,i p ds  

where

i pi = u, j

dx j ds

… (3)

Hypersurface

225

Now suppose that vector u is a the unit tangent t to the curve C. Then the derived vectors q and p are the curvature vectors of a C relatively to V n + 1 and V n respectively. Then equation (3) becomes. 

dx i dx j  α N + y ,αi p i ds 

 qα =  Ωij ds  Kn = Ω ij

Taking

dx i dx j , ds ds

… (4)

we get

α qα = Kn Nα + y,i pi

...(5)

Kn is called Normal curvature of V n at any point P of the curve C and Kn Nα is called normal curvature vector of V n + 1 in the direction of C. Meaunier's Theorem If Ka and Kn are the first curvature of C relative to V n + 1 and normal curvature of V n respectively r

r

and w is the angle between N and C (C being the unit vector of V n + 1 then the relation between Ka, Kn and w is given by Kn = Ka cos ω Proof: We know that qα = Kn N α + p i y,αi

...(1)

dxi dx j ds ds Let Ka and Kg be the first curvatures of C with respect V n + 1 and V n respectively then where

Kn = Ωij

Ka = r

a αβ q α q β , Kg =

gij pi p j

r

Let ω be the angle between N and C . Then

r

r r N ⋅ C = N ⋅ C cos ω

= cos ω r r N .C = cos ω

If b is a unit vector of V n + equation (1) becomes

1

as |N| = |C| = 1 ...(2) r

in the direction principal normal of C with respect to V n then

r r r Ka C = K g b + K n N

r Taking scalar product of equation (3) with N , we have r r r r r r K a N .C = K g N. b + K n N. N Ka cos ω = Kg ⋅ 0 + K n ⋅1;

Kn = Ka cos ω

… (3)

from (2) … (4) Proved.

226


EXAMPLE 1 Show that the normal curvature is the difference of squares of geodesic curvatures. Solution We know that (from Meurier’s Theorem, equation 3) r r r KaC = Kg b + Kn N

Taking modulus of both sides, we get 2 2 Ka2 = Kg + Kn 2 2 Kn2 = Ka – K g .

or

Theorem 12.3 To show that the first curvature in Vn + 1 of a geodesic of the hypersurface Vn is the normal curvature of the hypersurface in the direction of the geodesic. Proof: From, example 1, we have 2 2 Ka2 = Kg + Kn

If C is a geodesic of V n then p i = 0 ⇒ Kg = 0 Then we have Ka2 = Kn2 ⇒

Ka = K n.

Proved.

Dupin's Theorem The sum of normal curvatures of a hypersurface V n for n mutually orthogonal directions is an invariant and equal to Ω ij g ij . Proof: Let ehi | (h = 1, 2, …, n) be unit tangents to n congruences of an orthogonal ennuple in a V n. Let Knh be normal curvature of the hypersurface V n in the direction of the congruence eh | . Then Knh = Ωij eih | ehj| The sum of normal curvatures for n mutually orthogonal directions of an orthogonal ennuple is a V n is n

∑ Knh = ∑ Ω h =1

n

ij

eih| ehj|

h =1

= Ωij

n

∑e

i h|

ehj|

h =1

= Ω ij g ij = Scalar invariant

Proved.

Hypersurface

227

12.6 DEFINITIONS (a) First curvature (or mean curvature) of the hypersurface Vn at point P. It is defined as the sum of normal curvatures of a hypersurface V n form mutually orthogonal directions at P and is denoted by M. Then M = Ω ij g ij (b) Minimal Hypersurface The hypersurface V n is said to be minimal if M = 0 i.e., Ω ij g ij = 0 (c) Principle normal curvatures The maximum and minimum values of Kn are said to be the principle normal curvatures of V n at P. Since these maximum and minimum values of Kn correspond to the principal directions of the symmetric tensor Ωij . (d) Principal directions of the hypersurface at a point P. The principal directions determined by the symmetric tensor Ωij at P are said to be principal directions of the hypersurface at P. (e) Line of curvature in V n A line of curvature in a hypersurface V n is a curve such that its direction at any point is a principal direction. Hence we have n congruences of lines of curvature of a V n. THEOREM 12.4 To show that the mean curvature of a hypersurface is equal to the negative of the divergence of the unit normal. or To show that the first curvature of a hypersurface is equal to the negative of the divergence of the unit normal. or To show that the normal curvature of a hypersurface for any direction is the negative of the tendency of the unit normal in that direction. Proof: Let N be the unit normal vector to the hypersurface V n in y's and let Nα be its covariant components. Let t be the unit tangent vector to N congruences eh| (h = 1, 2, …, n) of an orthogonal ennuple in V n and let Thα| be the contravariant components in V n + 1 of t. Since t is orthogonal to N, therefore Th|α N α = 0

… (1)

Taking covariant derivative of equation (1) with regard to yβ provides Thα| ,β N α + Thα| N α, β = 0 Multiplying equation (2) by Thβ , we get Thα| ,β Thβ| N α + Thα| Thβ| N α, β = 0

… (2)

228


(Thα| ,βThβ| ) N α = – N α ,β Thβ| Thα|

… (3)

Now Thα|,βThβ| is the first curvature of the curve eh| and N α,β . Thβ| . Thα| is the tendency of Nα is the direction of eh| . Hence equation (3) implies that the normal component of the first curvature of the eh| relative to V n + 1 or the normal curvature of V n in the direction of the curve eh| = – tendency of N in the direction of the curve eh|

… (4)

Taking summation of both of (3) and (4) for h = 1, 2, …, n we have i.e., mean curvature or first curvature of a hypersurface = – divergence of the unit normal. Corollary: To prove that M = – divn + 1N Proof: Since N is a vector of unit magnitude i.e., constant magnitude, its tendency is zero. also by definition, the div n + 1 N and div n N differ only by the tendency of the vector N in its direction. But the tendency of N is zero hence it follows that divn N = – div n + 1 N Hence M = – div n N = – div n + 1 N. 12.7 EULER'S THEOREM Statement The normal curvature Kn of V n for any direction of ar in V n is given by n

Kn =

∑ Kh cos2 α h h =1

where Kh are the principal curvature and α h are the angles between direction of ar and the congruence eh| Proof: The principal directions in V n determined by the symmetric covariant tensor teser Ω ij are given by (Ωij – K h g ij ) p ih| = 0

… (1)

where Kh are the roots of the equation Ωij – Kgij = 0

… (2)

and phi | are the unit tangents to n congruences of lines of curvature. The roots of the equation (2) are the maximum and minimum values of the quality Kn defined by Ωij p i p j

Kn = g p i p i ij

… (3)

Hypersurface

229

Multiplying equation (1) by p ki | , (K ≠ h), we get (Ωij – K n gij ) pih| p ik| = 0 The principal directions satisfy the equation

(h ≠ k )

Ωij pih | pkj| = 0

… (4)

Let eh| be the unit tangents to the n congruences of lines of curvature. Then the principal curvatures are given by Kh = Ωij ehi | ehi |

(h = 1, 2, …, n)

Any other unit vector ar in V n is expressible in the form r a =

∑ eh| cosα h h n

or where

ai

=

… (5)

∑e

i

h=1

h|

cosα h

… (6)

r r cos α h| = a ⋅ e h | = ai eh|i

r

ai being the contravariant components of a and α h being the inclination of vector a to eh|. The r normal curvature of V n for the direction of a is given by Kn = Ωij a ia j from (6), we get  Kn = Ω ij   n

=

∑

h ,k =1 n

=

∑

∑e

i h|

h

 cos α h 

   

∑e

j k|

h

 cos α h 

(Ωij e ih| ekj | ) cosα h cosαk (Ωij ehi |ehj| ) cos 2 α h

h =1

n

2 Kn = ∑ Kh cos α h h =1

… (7)

This is a generalisation of Euler’s Theorem. 12.8 CONJUGATE DIRECTIONS AND ASYMPTOTIC DIRECTIONS IN A HYPERSURFACE r The directions of two vector, ar and b , at a point in V n are said to be conjugate if Ωij a ib j = 0

… (1)

and two congruences of curves in the hypersurface are said to be conjugate if the directions of the two curves through any point are conjugate.

230


A direction in V n which is self-conjugate is said to be asymptotic and the curves whose direction are along asymptotic directions are called asymptotic lines. Therefore the direction the vector ar at a point of V n be asymptotic if Ωij a i a j = 0

… (2)

The asymptotic lines at a point of a hypersurface satisfies the differential equation Ω ij dxi dx j = 0

… (3)

THEOREM 12.5 If a curve C in a hypersurface V n has any two of the following properties it has the third (i) it is a geodesic in the hypersurface Vn (ii) it is a geodesic in the enveloping space Vn + 1 (iii) it is an asymptotic line in the hypersurface V n. Proof: Let C be a curve in the hypersurface V n. The normal curvature Kn of the hypersurface V n in the direction of C is given by 2 2 Ka2 = Kg + Kn

… (1)

where Ka and Kg are the first curvatures of C relative to enveloping space V n + 1 and hypersurface V n respectively. Suppose C is a geodesic in the hypersurface V n [i.e., (i) holds] then Kg = 0. If C is also a geodesic in the enveloping space V n + 1 [i.e., (ii) holds] then Ka = 0. Now, using these values in equation (1), we have Kn2 = 0

⇒ Kn = 0

Implies that C is an asymptotic line is the hypersurface V n i.e., (iii) holds. Hence we have proved that (i) and (ii) ⇒ (iii) Similarly we have proved that (ii) and (iii) ⇒ (i) and (i) and (iii) ⇒ (ii) 12.9 TENSOR DERIVATIVE OF THE UNIT NORMAL The function y α are invariants for transformations of the coordinates x i in V n their first covariant derivatives with respect to the metric of V n are the same as their ordinary derivatives with respect to the variables x i. i.e.,

α y;αi = y,i =

∂ yα ∂ xi

… (1)

The unit Normal N α be the contravariant vector in the y’s whose tensor derivative with respect to the x’s is N;αi =

∂N α  α  β δ +   N y, i ∂x i β δ

… (2)

Hypersurface

231

aαβN β N α = 1

Since

… (3)

Tensor derivative of this equation with respect to

xi

gives

aαβN α N;βi + aαβ N;αi Nβ = 0

Interchanging α and β in Ist term, we get or

aβα N β N ;αi + a αβ N ;αi N β = 0 a αβ N β N ;αi + aαβ N ;αi N β = 0 2 aαβ N β N ;αi = 0 a αβ N β N ;αi = 0

… (4)

which shows that N;αi is orthogonal to the normal and therefore tangential to the hypersurface. Thus N;αi can be expressed in terms to tangential vectors y,αk to V n so that N;αi = Aiα y,αk

where Aik is a mixed tensor of second order in V n to be determined. Since unit normal N α is orthogonal to tangential vector y,αi in V n. Then aαβN α y,βi = 0

Taking tensor derivative with respect to x j, we get a αβ N ;αj y,βi + aαβ N α y,βij = 0 or

a αβ N ;αj y β,i + a αβ N α Ωij N β = 0

or

a αβ N ;αj y β,i + Ω ij (aαβ N α N β ) = 0

from equation (3),

since y,βij = Ωij N β

a αβ N α N β = 1

or

a αβ N ;αj y,βi + Ωij = 0

or

g ki A kj + Ωij = 0   ∂ y α ∂ yβ α α = g ki  since aαβ y, k y, i = a αβ k i ∂x ∂x  

Multiplying this equation by gim, we get g ki g im A kj + Ωij g im = 0

… (5)

232

Tensors and Their Applications k im δm =0 k A j + Ωij g

Aim + Ωij g im = 0 im Aim = – Ωij g

Substituting this value in equation (5), we get N;αi = Ω ij g y, k = – Ω ji g α

ik

N;αi = – Ωij g

jk

jk

y,αk

y ,αk

… (6)

This is the required expression for the tensor derivative of N α . Theorem 12.6 The derived vector of the unit normal with respect to the enveloping space, along a curve provided it be a line of curvature of the hypersurface. Proof: Since the tensor derivative of N α is N;αi = – Ωij g

jk

y ,αk

… (1)

Consider a unit vector e i tangential to the curve C. N;αi ei = – Ωij g

Then

jk

y,αk ei

… (2)

The direction of N;αi ei is identical with that of e i N;αi ei = – λy,αi ei ,

Then

(λ is scalar constant)

from (2), we have Ωij g jk y,αk ei = λy,αi e i Multiplying both sides of this equation by a αβ yβ,l Ω ij g jk e i (a αβ y ,αk y,βl ) = λe i (aαβ y,αi y ,βl ) Ωij g jkei gkl = λei gil

since g il = aαβ y α,i y β,l

Ωijδilei = λ ei g il Ωil ei – λei gil = 0

(Ω il – λg il ) ei = 0

(l = 1, 2, …, n)

This equation implies that the direction of e i is a principal direction for the symmetric tensor Ω il i.e., e i is a principal direction for the hypersurface V n. Hence by definition the curve C is a line of curvature V n.

Hypersurface

233

12.10 THE EQUATION OF GAUSS AND CODAZZI since we know that (pg. 86, equation 5) p Ai, jk – Ai, kj = A p Rijk

… (1)

where A i is a covariant tenser of rank one and difference of two tensers Ai, jk – Ai,kj is a covariant tenser of rank three. It y,αi are components of a covariant tensor of rank one in x 's. Then replacing A i by y,αi in equation (1), we get p y,αijk – y,αikj = y,αp Rijk = y,αp g ph Rhijk ,

p Since Rijk = g phRhijk

… (2)

where Rhijk are Riemann symbols for the tensor g ij We know that

and

y,αij = Ω ij N α

… (3)

N ,αi = – Ωij g ik y,αk

… (4)

α are Riemann symbols for the tensor a Let Rγδε αβ and evaluated at points of the hypersurface using equation (3) and (4), equation (2) becomes α y,αp g ph [ Rhijk – (Ωhj Ω ik – Ω hk Ω ij )] – N α (Ωij , k – Ωik , j ) – Rγδε y,γi y δ, j y,εk

… (5)

Multiplying equation (5) by a αβ yβ,l and summed with respect to α. Using the relations a αβ y ,αi y,βl = 0 and

g αβ = N β y,αl = 0,

we get Rlijk = (Ωlj Ωik – Ωlk Ωij ) + R βγδε y β,l y,γi y ,δj y,εk

… (6)

Multiplying (6) by a αβ N β and summing with respect to α. Using relations a αβ N β y α,l = 0 and

aαβ N β B β = 0

we get Ω ij ,k – Ωik , j + Rβγδε N β y ,γi y,δj y ,εk = 0

… (7)

Hence, The equation (6) are generalisation of the Gauss Characteristic equation and equation (7) of the Mainardi-Codazzi equations.

234


12.11 HYPERSURFACES WITH INDETERMINATE LINES OF CURVATURE A point of a hypersurface at which the lines of curvature are indeterminate is called an Umbilical Point. The lines of curvature may be indeterminate at every point of the hypersurface iff Ω ij = ωg ij

… (8)

where ω is an invariant The mean curvature M of such a hypersurface is given by M = Ωij g ij = ω gij g ij = ωn ⇒

ω=

M . n

So that the conditions for indeterminate lines of curvature are expressible as Ω ij =

M g , n ij

from (i)

If all the geodesics of a hypersurface V n are also geodesics of an enveloping V n hypersurface V n is called a totally geodesic hypersurface of the hypersurface V n + 1.

… (9) + 1.

They

THEOREM 12.7 A totally geodesic hypersurface is a minimal hypersurface and its-lines of curvature are indeterminate. Proof: We know that 2 2 Ka2 = Kn + K g

and a hypersurface is said to be minimal if M=0 and the lines of curvature are indeterminate if Ω ij =

M g ij n

… (1) … (2)

… (3)

If a hypersurface V n is totally geodesic then geodesics of V n are also geodesics of V n + 1. i.e., Ka = 0 = k g Now, from (1), we have Kn = 0 But normal curvature Kn is zero for an asymptotic direction. Hence a hypersurface V n is totally geodesic hypersurface iff the normal curvature Kn zero for all directions in V n and hence Ω ij = 0

M = Ω ij g ij = 0 i.e., equation (2) is satisfied. Hence, the totally geodesic hypersurface is minimal hypersurface. In this case equation (3) are satisfied hence the lines of curvature are indeterminate.

Hypersurface

235

12.12 CENTRAL QUADRATIC HYPERSURFACE Let xi be the cartesian in Euclidean space S n, so that the components gij of the fundamental tensor are constants. Let yi be the Riemmannian coordinates. If a fixed point O is taken as a pole and s the distance of any point P then Riemannian coordinates y i of P with pole O are given by y i = sξ i … (1) i ξ is unit tangent in the direction of OP. Let aij be the components in the x ′s of a symmetric tensor of the rank two and evaluated at the pole O. Then the equation y i aij y

j

=1

… (2)

represents a central quadratic hypersurface Substituting the value of equation (1) in equation (2), we get sξi aij sξi = 1 ξiaijξi =

1

… (3)

s2

The equation (3) showing that the two values of s are equal in magnitude but opposite in sign. The positive value of s given by equation (3) is the length of the radius of the quadric (2) for the direction ξ i . THEOREM 12.8 The sum of the inverse squares of the radii of the quadric for n mutually orthogonal directions at O is an invariant equal to a ij g ij . Proof: If e ih| , (h = 1, 2, …n) are the contravariant components of the unit tangents at O to the curves of an orthogonal ennuple in Sn. The radius Sh relative to the direction e ih| is given by aij ehi | ehj | = n

or

∑

h =1

(s h )–2

1 s 2h h

=

∑

aij ehi | ehj|

h =1

n

= a ij n

∑

h =1

(s h )–2

∑

eih | e hj|

h =1

= a ij g ij

Proved.

236


THEOREM 12.9 The equation of hyperplane of contact of the tangent hypercone with vertex at the point Q (y ) . Proof: Given (from equation 2, pg. 235) aij y i y j = 1 Differentiating it aij dy i y j + aij y i dy j = 0 or

aij dy i y j + aij y j dy i = 0 2 aij dy i y j = 0 aij dy i y j = 0

This shows that dy i is tangential to the quadric. Hence yj is normal to the quadric. The tangent hyperplane at the point P ( y j) is given by (Y i – y i ) aij y j = 0 aij Y i y j = aij y i y j aij Y i y j = 1

since aij y i y j = 1

… (4)

This equation represents the equation of tangent hyperplane at P ( y i) . If the tangent hyperplane P ( y j) passes through the point Q (y–i). Then we have y i aij y j = 1

… (5)

Thus all points of the hyperquadric, the tangent hyperplanes at which pass through Q lie on the hyperplane (5) on which yi is the current point. This is the hyperplane of contact of the tangent hypercone whose vertex is Q ( y i ) . 12.13 POLAR HYPERPLANE The polar hyperplane of the point R (~y i ) with respect to quadric (2) is the locus of the vertices of the hypercones which touch the hyperquadric along its intersections with hyperplanes through R. If Q ( y i ) is the vertex of such tangent hypercone, then R lies on the hyperplane of contact of Q so that yi a ~ yj =1 ij

Consequently for all positions of the hyperplane through R, Q lies on the hyperplane yi a ~ yj =1 ij

… (6)

This is required equations of the polar hyperplane of R and R is the pole of this hyperplane.

Hypersurface

237

12.14 EVOLUTE OF A HYPERSURFACE IN EUCLIDEAN SPACE Consider a hypersurface V n of Euclidean space Sn + 1 and let x i (i = 1, 2, … n) be coordinates of an arbitrary point P of V n whose components relative to Sn + 1 are

y α (α = 1, 2,L n + 1) Let N α be a unit normal vector at P relative to Sn + 1 so that tensor derivate N α becomes covariant derivative. So,

N ;αi = N ,αi = – Ωij g jk y,αk

… (1)

y;αij = y,αij = Ω ij N α

… (2)

n +1

and

gij =

∑

α =1

y,αi y,αj

… (3)

Let P ( y α ) be a point on the unit normal N α such that distance of P from P is ρ in the direction of N α such that y α = y α + ρN α … (4) Suppose P undergoes a displacement dx i in V n then the corresponding displacement d y α of P is given by d y α = ( y,αi + ρN ,αi ) dx i + N αdρ

… (5)

The vector ( y,αi + ρN ,αi ) dx i is tangential to V n whereas N αdρ is a normal vector. Therefore if the displacement of P ( y α ) be along the normal to the hypersurface then we have ( y,αi – ρN ,αi 0) dx i = 0

… (6)

Using equation (1) in equation (6), we get ( y,αi – ρg jk y,αk ) dx i = 0 Multiplying it by y,αi and summing with respect to α, we get, using equation (3), as ( gil – ρ Ωij g jk glk ) d xi = 0 ( gil – ρ Ω ij δlj ) d xi = 0 ( g il – ρΩil ) dx i = 0 ( gij – ρ Ωij ) d xi = 0

… (7)

This shows that the directions dxi given by equation (7) are principal directions of the hypersurface where the roots ρ of the equation |gij – ρΩ ij| = 0 are called principal radii of normal curvature. The locus of P ( y α ) satisfying the condition (4) is called evolute of the hypersurface V n of Sn + 1 where ρ is a root of (7). The evolute is also a hypersurface of Sn + 1.

238


12.15 HYPERSPHERE The locus of a point in Sn which moves in such way that it is always at a fixed distance R from a fixed point C (b α ) is called a hypersphere of radius R and centre C. Therefore the equation of such a hypersphere is given by n

∑ (y

α

α =1

– b α )2 = R 2

… (1)

THEOREM 12.10 The Riemannian curvatrure of a hypersphere of radius R is constant and equal to 1 . R2 Proof: Let the hypersurface be a V n of Sn + 1 and let its centre be taken as origin of Euclidean coordinates in Sn + 1. Then the hypersphere is given by

∑ (y

α 2

)

α

= R 2,

(α = 1, 2, …, n + 1)

… (2)

For the point in V n the y's are functions of the coordinates xi on the hypersphere. Differentiating equation (2) with respect to x i, we get

∑ y α y,αi α

=0

and again differentiating it with respect to

∑y

α ,j

y,αi +

α

∑y

α

α

… (3) x j,

we get.

y,αij = 0

… (4)

By Gauss formula, y,αij = Ω ij N α Using (5), equation (4) becomes g ij +

∑ y α Ωij N α α

=0

… (5)

… (6)

From equation (3) it follows that y,αi is perpendicular to y α . But y,αi is tangential to V m. Hence y α is normal to V n.. The equation (2) implies that the components of the unit vector N α are given by N

α

=

yα ⇒ y α = RN α R

Using (7), equation (6) becomes g ij +

∑RN

α

α

or or

g ij + RΩ ij

N α Ωij = 0

∑ (N α

α 2

) =0

g ij + R Ωij = 0

since ( N α )2 = 1

… (7)

240

Tensors and Their Applications 2.

To prove that for a space Vn with positive constant Riemannian curvature K these exists sets of n + 1 real coordinate y α satisfying the condition n +1

∑( y α =1

α

) = 1 K

where R 2 =

1 K

Proof: Using (9) in equation (1), we get

∑ (y α ) 2 α

=

1 K

Proved.

EXAMPLE 2 Show that the directions of two lines of curvature at a point of a hypersurface are conjugate. Solution The principal direction e ih| are given by Ωij ehi | ehj| = 0

… (1)

The shows that principal directions at a point of a hypersurface are conjugate. Thus we say that two congruences of lines of curvature are conjugate. EXAMPLE 3 Show that the normal curvature of hyper surface V n in an asymptotic direction vanishes. Solution Let us consider a curve C in a V n. If C is an asymptotic line then it satisfies the differential equation Ω ij dxi dx j = 0 dxi dx j =0 … (1) ds ds Now the normal curvature Kn of the hypersurface V n in an asymptotic direction in a V n is given i.e.,

Ωij

by Kn = Ωij i.e.,

Kn = 0

dxi dx j ds ds from (1)

EXAMPLE 4 To prove that if the polar hyperplane of the point R passes through a point P then that of P passes through R. Solution Let P( y i ) and R ( y i ) be two points. The polar hyperplane of R ( y i ) is

242


11. Prove that the necessary and sufficient condition that system of hypersurface with unit normal N be isothermic is that

ρ ( N div N – N .∇N ) = 0. 12. Show that the normal curvature of a subspace in an asymptotic direction is zero. 13. If straight line through a point P in S n meets a hyperquadric in A and B and the polar hyperplane of P in Q prove that P, Q are harmonic conjugates to A, B. 14. What are evolutes of a hypersurface in an Euclidean space. Show that the varieties ρ1 = constant in evolute are parallel, having the curves of parameter ρ1 as orthogonal geodesics of V n .

INDEX A Absolute 131 Addition and Subtraction of Tensors 15 Associated tensor 43 Asymptotic directions 229

B Bianchi identity 94 Binormal 137

C Canonical congruence 213 Christoffel’s symbols 55 Completely skew-symmetric 111 Completely symmetric 111 Concept 188 Congruence of curves 49 Conjugate (or Reciprocal) Symmetric Tensor 25 Conjugate directions 229 Conjugate metric tensor 34 Conservative force field 144 Contraction 18 Contravariant tensor of rank r 14 Contravariant Tensor of rank two 9 Contravariant vector 7 Covariant tensor of rank s 14

Covariant tensor of rank two 9 Covariant vector 7 Curl 76 Curl of congruence 211 Curvature 136 Curvature of Congruence 207 Curvature tensor 86

D Degree of freedom 157 Dextral Index 1 Divergence 75 Divergence Theorem 161 Dummy index 1 Dupin’s Theorem 226

E Einstein space 103 Einstein tensor 95 Einstein’s Summation Convention 1 Euler’s condition 171 Euler’s Theorem 228 Evolute 237

F First curvature 227 First curvature vector of curve 170

244

First fundamental tensor 34 Free Index 2 Fundamental tensor 31 Fundamental theorem 199

G Gauss Characteristic equation 233 Gauss’s formula 222 Gauss’s theorem 164 Generalised KrÖnecker delta 112 Generalized coordinates 157 Geodesic congruence 208 Geodesic coordinate system 175 Geodesics 171 Gradient 75 Green’s Theorem 162

H Hamilton’s principle 153 Hypersphere 238 Hypersurface 48, 218

I Inner product of two tensors 18 Integral of energy 155 Intrinsic derivative 131

K Kinetic energy 144 Krönecker Delta 2

L Lagrangean equation 148 Lagrangean function 148 Laplace’s equation 167 Laplacian operator 80, 163 Length of a curve 42 Levi-civita’s concept 188 Line element 31 Line of curvature 227


M Magnitude of vector 44 Mainardi-Codazzi equations 233 Mean curvature 101 Meaunier’s Theorem 225 Metric Tensor 31 Minimal curve 42 Minimal Hypersurface 227, 234 mixed tensor of rank r + s 14 mixed tensor of rank two 9

N Newtonian Laws 142 Normal congruence 209 n-ply orthogonal system of hypersurfaces 49 Null curve 42

O Orthogonal Cartesian coordinates 120 Orthogonal ennuple 49 Osculating plane 136 Outer Product of Tensor 16

P Parallel vector fields 134 Parallelism 188 Poisson’s equation 166 Polar hyperplane 236 Potential energy 145 Principal directions of the hypersurface 227 Principle normal vector 136 Principle normal curvatures 227 Principle of least action 156 Projective curvature tensor 104

Q Quotient Law 24

R Reciprocal base systems 122

Index

Relative Tensor 26 Ricci tensor 88 Ricci’s principal directions 102 Ricci’s coefficients of rotation 205 Ricci’s Theorem 71 Riemann curvature 96 Riemann’s symbol 86 Riemann-Christoffel Tensor 85 Riemannian coordinates 177 Riemannian Geometry 31, 116 Riemannian Metric 31 Riemannian space 31

245

Stoke’s Theorem 164 Straight line 140 Subscripts 1 Superscripts 1 Symmetric Tensors 20

T Tensor density 26 Torsion 137 Totally geodesic hypersurface 234 Transformation of Coordinates 6

U S Scalar product of two vectors 44 Schur’s theorem 100 Second fundamental tensors 34 Serret-Frenet formula 138 Simple Pendulum 152 Skew-Symmetric Tensor 20

Umbral 1 Unit principal normal 171

W Weyl tensor 104 Work function W 145

tensors

Recommend Documents