386
S

Continuum Mechanics

Embed Size (px)

DESCRIPTION

BY George Backus 1997

Citation preview

Continuum Mechanics

George Backus

Insitute for Geophysics and Planetary Physics

University of California� San Diego

S Samizdat

Press

Published by the Samizdat Press

Center for Wave Phenomena

Department of Geophysics

Colorado School of Mines

Golden� Colorado �����

and

New England Research

�� Olcott Drive

White River Junction� Vermont �����

c�Samizdat Press� ����

Samizdat Press publications are available via FTP

from landau�mines�edu or �����������

Or via the WWW from http���landau�mines�edu� samizdat

Permission is given to freely copy these documents�

Contents

Dictionary D��

Five Prerequisite Proofs PP��

I Tensors over Euclidean Vector Spaces �

� Multilinear Mappings �

��� De�nition of multilinear mappings � � � � � � � � � � � � � � � � � � � � � �

�� Examples of multilinear mappings � � � � � � � � � � � � � � � � � � � � � � �

�� Elementary properties of multilinear mappings � � � � � � � � � � � � � � � �

��� Permuting a multilinear mapping � � � � � � � � � � � � � � � � � � � � � � �

� De�nition of Tensors over Euclidean Vector Spaces ��

� Alternating Tensors� Determinants� Orientation� and n�dimensional Right�

handedness ��

�� Structure of �nV � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

� Determinants of linear operators � � � � � � � � � � � � � � � � � � � � � � � ��

� Orientation � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

� Tensor Products ��

��� De�nition of a tensor product � � � � � � � � � � � � � � � � � � � � � � � � �

�� Properties of tensor products � � � � � � � � � � � � � � � � � � � � � � � � � �

i

ii CONTENTS

�� Polyads � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

� Polyad Bases and Tensor Components ��

��� Polyad bases � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

�� Components of a tensor relative to a basis sequence� De�nition � � � � � �

�� Changing bases � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

��� Properties of component arrays � � � � � � � � � � � � � � � � � � � � � � � �

��� Symmetries of component arrays � � � � � � � � � � � � � � � � � � � � � � �

��� Examples of component arrays � � � � � � � � � � � � � � � � � � � � � � � � �

� The Lifting Theorem ��

Generalized Dot Products �

��� Motivation and de�nition � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Components of P hqiR � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Properties of the generalized dot product � � � � � � � � � � � � � � � � � � ��

��� Applications of the generalized dot product � � � � � � � � � � � � � � � � � ��

How to Rotate Tensors �and why� �

��� Tensor products of linear mappings � � � � � � � � � � � � � � � � � � � � � ��

�� Applying rotations and re�ections to tensors � � � � � � � � � � � � � � � � ��

�� Physical applications � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

��� Invariance groups � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Isotropic and skew isotropic tensors � � � � � � � � � � � � � � � � � � � � � ��

Di�erential Calculus of Tensors

��� Limits in Euclidean vector spaces � � � � � � � � � � � � � � � � � � � � � � ��

�� Gradients� De�nition and simple properties � � � � � � � � � � � � � � � � � ��

�� Components of gradients � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Gradients of dot products � � � � � � � � � � � � � � � � � � � � � � � � � � ���

��� Curvilinear coordinates � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

CONTENTS iii

��� Multiple gradients and Taylor�s formula � � � � � � � � � � � � � � � � � � � ��

��� Di�erential identities � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

�� Integral Calculus of Tensors ��

���� De�nition of the Integral � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

������ Mass distributions or measures � � � � � � � � � � � � � � � � � � � ���

����� Integrals � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Integrals in terms of components � � � � � � � � � � � � � � � � � � � � � � � �

�� Integral Identities ��

���� Linear mapping of integrals � � � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Line integral of a gradient � � � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Gauss�s theorem � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

���� Stokes�s theorem � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

���� Vanishing integral theorem � � � � � � � � � � � � � � � � � � � � � � � � � � �

���� Time derivative of an integral over a moving volume � � � � � � � � � � � � ��

���� Change of variables of integration � � � � � � � � � � � � � � � � � � � � � � ��

II Elementary Continuum Mechanics ���

�� Eulerian and Lagrangian Descriptions of a Continuum ���

��� Discrete systems � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

�� Continua � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Physical quantities � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

��� Derivatives of physical quantities � � � � � � � � � � � � � � � � � � � � � � ���

��� Rigid body motion � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

��� Relating continuum models to real materials � � � � � � � � � � � � � � � � ���

�� Conservation Laws in a Continuum ���

��� Mass conservation � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

����� Lagrangian form � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

iv CONTENTS

���� Eulerian form of mass conservation � � � � � � � � � � � � � � � � � ���

�� Conservation of momentum � � � � � � � � � � � � � � � � � � � � � � � � � ���

���� Eulerian form � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

��� Shear stress� pressure and the stress deviator � � � � � � � � � � � � ��

�� Lagrangian form of conservation of momentum � � � � � � � � � � � � � � � ���

��� Conservation of angular momentum � � � � � � � � � � � � � � � � � � � � � ���

����� Eulerian version � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

���� Consequences of�

ST��

S � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Lagrangian Form of Angular Momentum Conservation � � � � � � � � � � � ��

��� Conservation of Energy � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

����� Eulerian form � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

���� Lagrangian form of energy conservation � � � � � � � � � � � � � � � ��

��� Stu� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

����� Boundary conditions � � � � � � � � � � � � � � � � � � � � � � � � �

���� Mass conservation � � � � � � � � � � � � � � � � � � � � � � � � � � � �

���� Momentum conservation � � � � � � � � � � � � � � � � � � � � � � � �

����� Energy conservation � � � � � � � � � � � � � � � � � � � � � � � � � �

�� Strain and Deformation ���

���� Finite deformation � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

��� In�nitesimal displacements � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Constitutive Relations ��

���� Introduction � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Elastic materials � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

��� Small disturbances of an HTES � � � � � � � � � � � � � � � � � � � � � � � ��

���� Small disturbances in a perfectly elastic earth � � � � � � � � � � � � � � � ��

���� Viscous �uids � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

III Exercises ���

DICTIONARY D��

Logic

� �implies� �e�g� A� B where A and B are sentences��

� �implies and is implied by� �e�g� A� B where A and B are sentences��

i� �if and only if� � same as �

�� �is de�ned as�

� �for every�

� �there exists at least one�

�� �there exists exactly one�

� �such that�

Sets �same as �Classes��

�a�� � � � � a�� is an �ordered� sequence of objects� �a�� � � � � an� is also called an ordered

n�tuple� Order is important� and �a�� a�� a�� �� �a�� a�� a��� However� duplication�

e�g� a� � a�� is permitted�

fa�� � � � � a�g is a set of objects� Order is irrelevant� and fa�� a�� a�g � fa�� a�� a�g �

fa�� a�� a�g� etc� Duplication is not permitted� All a are di�erent�

�a�� � � � � �a�� � � � � a�� �� �a�� a�� a�� a�� a���

fa�� � � � � �a�� � � � � a�g �� fa�� a�� a�� a�� a�g�

� is the empty set� the set with no objects in it�

R �� set of all real numbers�

R�� �� set of all non�negative real numbers�

R� �� set of all positive real numbers�

D�� DICTIONARY

C �� set of all complex numbers�

f�� � � � � ng �� set of integers from � to n inclusive�

fx � p�x�g is the set of all objects x for which the statement p�x� is true� For example�

fx � x � y for some integer yg is the set of even integers� It is the same as

fx � x is an integerg�

� �is a member of�� Thus� �a � A� is read �the object a is a member of the set A� The

phrase �a � A� can also stand for �the object a which is a member of A�

�� �is not a member of�

� �is a subset of�� A � B means that A and B are sets and every member of A is a

member of B�

�is a proper subset of�� A B means that A � B and A �� B �i�e�� B has at least one

member not in A��

A B �� the set of all objects in both A and B� �Read �A meet B���

A � B �� the set of all objects in A or B or both� �Read �A join B���

AnB �� the set of all objects in A and not B� �Read �A minus B���

A� B �� the set of all ordered pairs of �a� b� with a � A and b � B�Cartesian

products�������������������

A� � � � �� An �� the set of all ordered n�tuples �a�� � � � � an� with a� � A�� az � A�� � � � � an �An�

�nA �� the set of all ordered n tuples �a�� � � � � an� with a�� a�� � � � � an all members of A�

Same as A� � � �� A �n times��

DICTIONARY D��

Functions

Function is an ordered pair �Df � f�� Df is a set� called the �domain� of the function�

and f is a rule which assigns to each d � Df an object f�d�� This object is called

the �value� of f at d� The function �Df � f� is usually abbreviated simply as f � In

the expression f�d�� d is the �argument� of f �

Range of function is a set Rf consisting of all objects which are values of f � Two

equivalent de�nitions of Rf are

Rf �� fx � �d � Df � x � f�d�gor

Rf �� ff�d� � d � Dfg �

f � g for functions means Df � Dg and for each d � D� f�d� � g�d��

f�W � for a function f and a set W is de�ned as ff�w� � w � Wg� If W Df � � � then

f�W � �� ��

d j f�d� is a shorthand way of describing a function� For example� if Df � R and

f�x� � x� for all x � R� we can describe f by saying simply x j x� under f � In

particular� j � under f �

f jW is the �restriction� of f to W � It is de�ned whenever W is a subset of Df � It is the

function �W� g� such that for each w � W � g�w� � f�w�� Its domain is W �

�Extension�� If g is a restriction of f � then f is an extension of g to Df �

f��� v��� f�u�� ��� Suppose U and V are sets and f is a function whose domain is U � V �

Suppose u� � U and v� � V � De�ne two functions g and h as follows�

Dg � U and� �u � U� g�u� � f�u� v��

Dh � V and� �v � V� h�v� � f�u�� v��

D�� DICTIONARY

= f ( )

R f

f

f

Df

-1

d

r

r d= f( ), d r-1

Figure D���

Then g is often written f��� v��� and h is often written f�u�� ��� The dot shows whereto put the argument of the function�

Pa�A f�a� is de�ned when A � Df and when f�A� consists of objects which can be

added �e�g�� real numbers or vectors�� The symbol stands for the sum of all the

values f�A� for a � A� The number of terms in the sum is the number of objects in

A� so A must be �nite unless some kind of convergence is assumed�

IU � the identity function on the set U � has domain U and e�ect u j u� That is�

IU�u� � u� �u � U �

�Invertible�� A function f is invertible if f�d� � f�d��� d � d�� �d� d� � Df � That is�

if r � Rf � there is exactly one d � Df such that r � f�d�� In other words� for each

r � Rf � the equation f�d� � r has exactly one solution d � Df �

�Inverse�� f��� If f is invertible� its inverse f�� is de�ned to be the function whose

domain is Rf and such that for each r � Rf � f���r� is that unique d � Df such that

DICTIONARY D��

f�d� � r� That is� d � f���r� is the unique solution of r � f�d�� Note Df�� � Rf �

Rf�� � Df � See Fig� D���

For example� if the domain is R�� both the functions x j x� and x j x� are

invertible� and their inverses are x j x��� and xj x���� However� if the domain

is R� the function x j x� is not invertible�

Note that if f is invertible� so is f��� and �f����� � f �

Mappings

A Mapping is an ordered triple �U� V� f� in which U and V are sets� f is a function�

Df � U � and Rf � V � We say that �the function f maps U into V �� The mapping

is often abbreviated simply as f if the context makes clear what U and V are�

f � U V is the usual way of writing the mapping �U� V� f� if one wants to note explicitly

what U and V are� The symbol �f � U V � is read �the mapping f of U into V ��

It can also stand for the sentence �f is a function which maps U into V �� In this

latter usage it is equivalent to �f�U� � V ��

F �U V � �� the set of all functions mapping U into V �

Injective� If f is invertible� f � U V is an �injective� mapping or an �injection��

Surjective� If Rf � V � f � U V is a �surjective� mapping or a �surjection��

Bijection� If f � U V is both injective and surjective� it is �bijective�� or a �bijection��

Note� If f � U V is a bijection� so is f�� � V U �

Composition g � f � Suppose f � U V and g � V W � Then g � f � U W is

the mapping de�ned by requiring for each u � U that �g � f��u� � g�f�u��� The

function g � f is called the �composition� of g with f � Note that the order in which

the functions are applied or evaluated runs backward� from right to left�

D�� DICTIONARY

Note� If f � U V and g � V W and h � W X then

h � �g � f� � �h � g� � f� �D���

Note� If f � U V and g � V W are bijections� so is g � f � U W � and

�g � f��� � f�� � g���

Note� If f � U V is a bijection� f�� � f � IU and f � f�� � IV �

Note� Suppose f � U V and g � V U � Then

i� if g � f � IU � then f is an injection�

ii� if f � g � IV � then f is an surjection�

iii� if g � f � IU � and f � g � IV then f is a bijection and g � f���

Note� If f � U V then f � IU � IV � f � f �

Permutations

A permutation is a bijection � � �� � � � � n f�� � � � � ng� It is an �n�permutation� or a

�permutation of degree n��

Sn �� the set of all n�permutations� It has n� members� The product �� of two per�

mutations � and � is de�ned to be their composition� � � �� If e � If������ng then

e� � �e � � and ���� � ���� � e for all � � Sn� These facts� and the above notes

on pg� iv� make Sn a group under the multiplication de�ned by �� � � � �� It is

called the symmetric group of degree n� Its identity is e�

�i� j� transposition or interchange� If i �� j and � � i� j � n �i�e� fi� jg � f�� � � � � ng� then�i� j� stands for the permutation � � Sn such that ��i� � j� ��j� � i� and ��k� � k

if k �� fi� jg� This permutation is called the �transposition� or �interchange� of i

and j� Note that �ij� � �ij� � e� the identity permutation�

Theorem � �See� e�g�� Birkho� and MacLane� A Survey of Modern Algebra�� If

� � Sn� � is the product of n� � or fewer transpositions� There are many ways to

DICTIONARY D�

write � as a product of transpositions� but all have the same parity �i�e�� all involve

an even number of transpositions or all involve an odd number of transpositions��

even� odd� If � � Sn� � is even or odd according as it can be written as the product of

an even or an odd number of transpositions�

sgn � �� �� if � is even� �� �� if � is odd�

Note� sgn ��� � ��� � �sgn ����sgn ���� and sgn e � �� sgn � is read �signum of ���

Note� sgn � � sgn ��� because � � ��� � e� and � � sgn e � sgn � � ��� �

�sgn ���sgn ����� But sgn � � ���

Arrays

An array of order q is a function f whose domain is

Df � f�� � � � � n�g � � � �� f�� � � � � nqg�

The array is said to have dimension n� � n� � � � �� nq�

fi����iq If i� � f�� � � � � n�g� i� � f�� � � � � n�g� � � � � iq � f�� � � � � xqg� then f�i�� � � � � iq� is

often written fi��i�����iq or f i����iq or fi�i�i�

i����iq etc� There are q ways to write it�

The object f�i�� � � � � iq� is called the entry in the array at location �i�� � � � � iq� or

address �i�� � � � � iq�� The integers i�� � � � � iq are the �indices�� and each can be either

a subscript or a superscript� What counts is their right�left order� We will never

use symbols like f ijkl �

�n� is the n�dimensional Kronecker delta� It is an n � n array with �n�ij � � if i �� j�

�n�ij � � if i � j� Usually the n� is omitted� and it is written simply �ij� or �

ij� or

�ij� or �i j�

�n� is the n�dimensional alternating symbol� It is an

n factorsz �� �n� n� � � �� n array� with nn

entries� all � or �� or ��� It is de�ned as follows�

D�� DICTIONARY

�n�i����in � � if any of i�� � � � � in are equal� i�e�� if fi�� � � � � ing �� f�� � � � � ng�

If fi�� � � � � ing � f�� � � � � ng� then there is a unique � � Sn � i� � ����� � � � � in �

��n�� In that case� �n�i����in � �

n��������n� � sgn��

Usually the �n� is written without the n as �i����in � or �i�

i����in � or �i����in � etc�

Example� ��� � �� ��� � �� ��� � ��� ��� � �� �n � �

Example �n � ��

���� � ���� � ���� � �

���� � ���� � ���� � ��

all other �ijk � �� Thus� ���� � �� ���� � �� etc�

Einstein index conventions� i� If an index appears once� without parentheses� in

each term of an equation� the equation is true for all possible values of the

index� Thus� ��ij � �ji� says that ��ij � �ji� for all i� j � f�� � � � � ng� This is

a true statement� ��ijk � �ijk� says that ��ijk � �jik for i� j� k�f�� � � g� Thisis a false statement� The correct statement is �ijk � ��jik�

ii� If an index appears twice in any expression but a sum or di�erence� a sum over

all possible values of that index is understood� Thus� if Aij is an n� n array�

Aii stands forPn

i �Aii� If ui and vj are �rst order arrays of dimension n� uivi

stands forPn

i � uivi� and ui�vi stands forPn

i � ui�vi� but ui � vi and ui � vi

stand for the i�th entries in two arrays� not forP

i�ui � vi� orP

i�ui � vi��

iii� If an index appears three or more times in an expression other than a sum or

di�erence� a mistake has been made� Thus� if Aijkl is an n� n� n� n array�

Aiiii is never written alone� The sum is written out asPn

i �Aiiii�

iv� Parentheses �protect� an index from the conventions� Thus� Aii� refers to a

particular element in the array� the one at location �ii�� And vi� � wi� means

equality only for the particular value of i under discussion� not for all possible

values of i� The element Aiiii in an n � n � n � n array is written Aiiii� to

DICTIONARY D��

make clear that no sum is intended� Strictly speaking� the de�nition of the

Kronecker delta� using these conventions� reads �ij� � � if i �� j� and �ii� � �

for all i�

An array A of order r is symmetric �antisymmetric� in its p�th and q�th indices

if they have the same dimension� np � nq� and

Ai� � � � ip��ipip�� � � � iq��iqiq�� � � � ir

� �Ai� � � � ip��iqip�� � � � iq��ipiq�� � � � ir�

�The � is for symmetric A� the � is for antisymmetric A�� The array is totally

symmetric �antisymmetric� if it is symmetric �antisymmetric� in every pair of

indices�

Some properties of �ij and �i����in are listed below�

�� �ij � �ji

� �n�ii � n

� �ijAjk����kp � Aik����kp for any array A of suitable dimension� �ijAk�jk����kp �

Ak�ik����kp� etc� In particular� �ij�jk � �ik�

�� �i����in is totally antisymmetric�

�� If Ai����in is an

n factorsz �� �n� n� � � �� n array which is totally antisymmetric� there

is a constant � such that

Ai����in � ��i����in�

�� �i����in�j����jn �P

��Sn �sgn�� �i�j���� � � � �i�j��n� �The Polar Identity�� The

two special cases of interest to us are

n � � �ij�kl � �ik�jl � �il�jk which implies �ij�kj � �jk and �ij�ij � �

n � �

�ijk�lmn � �il�jm�kn � �im�jn�kl � �in�jl�km

��im�jl�kn � �il�jn�km � �in�jm�kl

D��� DICTIONARY

which implies �ijk�lmk � �il�jm��im�jl and �ijk�ljk � �il and �ijk�ijk �

��

Some applications of �i����ln are these� If Aij is a real or complex n�n

array whose determinant is detA� then��

Ai�j�Ai�j� � � � Ainjn�j����jn � �detA� �i����in �

In real three�space� suppose x�� x�� x� is a right�handed triple of mutually

perpendicular unit vectors �that is� x� � � x� � x�� � � �� Suppose �u � ui xi

and �v � vj xj� Then

�u � �v � uivi

�u� �v � xi�ijkujvk� or

��u� �v�i �� xi � ��u� �v� � �ijkujvk�

If �r � ri xi is the position vector in real three�space� R�� and �w � R� R�

is a vector �eld� with

�w��r � � wi��r � xi� then

�r � �w � divergence of �w �wi

ri

�r� �w � curl of �w � xi�ijkwk

rj�

A very important general property of arrays is the following�

Remark � Suppose Aijk����kp is antisymmetric in i and j� while Sijl����lq is symmetric in

i and j� Then

Aijk����kP Sijl����lq � ��

Proof�

The subscripts k and l are irrelevant� so we omit them� We have AijSij �

�AjiSji because Aij � �Aji and Sij � Sji� Now replace the summation index

DICTIONARY D���

j by i and the index i by j� Then

AjiSji � AijSij�

Putting these two results together gives AijSij � �AijSij� so AijSij � ��

Vector Space Facts

Fields� A �eld is a set F of objects which can be added� subtracted� multiplied and

divided according to the ordinary rules of arithmetic� Examples are R� C� the set

of rational numbers� and the set of numbers a � bp� where a and b are rational�

We will use only the �elds R and C� and usually F � R�

De�nition of vector spaces� A vector space is an ordered quadruple �V � F � ad� sc�

with these properties�

i� V is a non�empty set� Its members are called vectors�

ii� F is a �eld� Its members are called scalars�

iii� ad � V � V V � This function is called vector addition�

iv� sc � F � V V � This function is called multiplication by scalars�

The remaining properties make more sense if we introduce a new notation� For any

�u��v � V and c � F we de�ne

�u� �v � � ad��u��v�

c�v � �vc �� sc�c� �v�

The additional properties which make �V� F� ad� sc� a vector space are properties of

ad and sc as follows �letters with are vectors� those without are scalars��

v� �u� �v � �v � �u

vi� �u� ��v � �w� � ��u� �v� � �w

D��� DICTIONARY

vii� �a � b��u � a�u� b�u

viii� a��u� �v� � a�u� a�v

ix� a�b�u� � �ab��u

x� ��u � ��v for any �u��v � V

xi� ��u � �u

The vector ��u is called the zero vector� written ��� The vectors �����v and �u������vare written ��v and �u� �v respectively�

Example �� F is a �eld� V is the set of all ordered n�tuples of members of F � so a

typical vector is

�u � �u�� � � � un� with u�� � � � � un � F�

The addition and multiplication functions are

�u� �v �� ad��u��v� �� �u� � v�� � � � � un � vn�

c�v �� sc�c� �v� �� �cv�� � � � � cvn�

The zero vector is �� � ��� � � � � ��� The space V is usually written F n�

Example �� F � R� V is the set of all ordered n tuples of members of R�� so a typical

vector is

�u � �u�� � � � � un� with u�� � � � � un � R��

The addition and multiplication functions are

�u� �v �� ad��u��v� �� �u�v�� � � � � unvn�

c�v �� sc�c� �v� �� �vc�� � � � � vcn��

The zero vector is �� � ��� �� � � � � ���

Example �� Let �V� F� ad� sc� be any vector space and let S be any non�empty set�

De�ne a new vector space �VS� F� adS� scS� as follows�

VS � F�S V�� the set of all functionsf � S V�

DICTIONARY D���

To de�ne adS and scS we note that if f and g are functions in VS and c � F then

adS�f� g� and scS�c� f� are supposed to be functions in VS� i�e�� adS�f� g� � S V

and scS�c� f� � S V � To de�ne these functions� we must say what values they

assign to each s � S� The de�nitions we adopt are these�

�adS�f� g�� �s� � f�s� � g�s� �D��

�scS�c� f�� �s� � cf�s�� �D��

Following the convention for vector spaces� we write adS�f� g� as f �g and scS�c� f�

as cf � Thus� f � g and cf are functions whose domains are S and whose ranges are

subsets of V � The de�nitions ��� and �� read thus� for any s � S

�f � g��s� � f�s� � g�s� �D���

�cf��s� � cf�s�� �D���

These are not �obvious facts�� They are de�nitions of the vector�valued functions

f � g and cf �

In VS� the zero vector is the function which assigns to each s � S the zero vector in

V �

Example �� F � real numbers � R� V � set of all continuous real�valued functions

on the closed unit interval � � x � �� ad and sc are de�ned as in Example �

page D��� �This is a vector space� because if f and g are continuous functions so

is f � g and so is cf for any real c��

Example �� F � R� V � set of all continuous positive real valued functions on

� � x � �� ad and sc as de�ned in Examples and �� This is not a vector space�

There are scalars c � F and vectors f � V such that cf � �V � For example� if f � V

then ����f � �V � In other words� the function sc in this case is not a mapping from

F � V to V �

Notation� Usually the vector space �V� F� ad� sc� will be called simply the vector space

V � We will almost never use the notations ad��u��v� or sc�c� �v�� but will use �u � �v

D��� DICTIONARY

and c�v and �vc� V will be called a vector space over F � or a real or complex vector

space if F � R or C�

Because of vector space rules v� and vi�� there is no ambiguity aboutPn

i � �ui when

�u�� � � � � �un � V � For example� if n � �� all of �u���u����u��u���� ��u���u�����u���u���

��u� � �u�� � �u�� � �u�� ��u� � ��u� � �u��� � �u�� etc�� are the same�

When �u�� � � � � �un � V and a�� � � � � an � F � we do not even need theP� We can use

the index conventions to writePn

i � ai�ui as ai�ui�

Facts about ��� If a � F � then a�� � �� because a�� � a����� � �a���� � ��� � ��� Also� if

a � F � �v � V � and a �� � � �v �� ��� then a�v �� ��� For suppose a�v � �� and a �� ��

Then a�� � F � and �v � ��v � �a��a��v � a���a�v� � a���� � ��� Finally� �� � �u � �u

because �� � �u � ��u� ��u � �� � ���u � ��u � �u�

Linear mappings� Suppose V and W are both vector spaces over the same �eld F � and

L � F�V W �� We call L a �linear mapping� if for any �u� �v � V and c � F it is

true that

L��u� �v� � L��u� � L��v�

L�c�u� � cL��u��

L�V W � is the set of all linear mappings from V to W � �

It is a vector space over F if it�s ad and sc are de�ned as in example � page D���

If L � L�V W � is a bijection� then L�� � L�W V �� And then L is called an

�isomorphism� and V and W are said to be isomorphic� Intuitively speaking� two

isomorphic vector spaces are really the same space� Each member of V corresponds

to exactly one member of W � and vice versa� and ad and sc can be applied in either

V or W with the same result� For example� if F � R� then the two spaces in

examples � and on page D�� are isomorphic� If V is the space in Example �

�Note that if L � L�V �W � and M � L�U � V �� then L �M � L�U � W ��

DICTIONARY D���

then the isomorphism L � Rn V is given by

L�u�� � � � � un� � �eu� � � � � � eun�

and

L���v�� � � � � vn� � �ln v�� � � � ln vn��

Subspaces� A subspace of vector space �V� F� ad� sc� is a vector space � V � F� ad� sc� with

these properties�

i� V � V

ii� fad � ad j eV � eViii� fsc � sc jF � eV �

If V is any subset of V � then � V � F� ad��� eV � eV � sc ���F� eV � is a subspace of �V� F� ad� sc�

if and only if

a� ad� eV � eV � � eV and also

b� sc�F � �V � � eV �

Intuitively� subspaces are lines� planes and hyperplanes in V passing through the

origin� �If eV is a subspace of V � and �v � eV � then ��v � eV by iii�� but ��v � ��� Thus

the zero vector in V is the zero vector in every subspace of V ��

Finite dimensional vector spaces� V is a vector space over �eld F � U is an arbitrary

non�empty subset of V �

Linear combination� If �u�� � � � � �un � V and a�� � � � � an � F � the vector ai�ui is

called a linear combination of f�u�� � � � �ung�

Span� The set of all linear combinations of �nite subsets of U is written sp U � It is

called the span of U � It is a subspace of V � so it is called the subspace spanned

or generated by U � If U is a subspace� U � sp U �

D��� DICTIONARY

Finite dimensional� A vector space V is ��nite dimensional� if V � sp U for

some �nite subset U V �

Linear dependence� A set U � V is �linearly dependent� if there are �nitely

many vectors �u�� � � � � �un � U and scalars a�� � � � � an such that

i� at least one of a�� � � � � an is not �� and also

ii� ai�ui � ���

It is an important theorem that if ��u�� � � � � �un� is a sequence such that f�u�� � � � � �ungis linearly dependent� then there is anm � f�� � � � � ng such that �um � spf�u�� � � � � �um��g�

Linear independence� A set U � V is linearly independent if it is not linearly

dependent� Linear independence of U means that whenever f�u�� � � � �ung � U

and ai�ui � ��� then a� � � � � � an � ��

Basis� Any linearly independent subset of vector space V which spans V is called

a basis for V � The following are important theorems about bases�

Theorem � If V is �nite dimensional� it has a basis� and all its bases are �nite

and contain the same number of vectors� This number is called the dimension

of V � written dimV �

Theorem � If V is �nite dimensional and W is a subspace of V � W is �nite

dimensional and dimW � dimV � Every basis for W is a subset of a basis for

V �

Theorem � If f�u�� � � � � �ung spans V then dimV � n� If dimV � n� then

f�u�� � � � � �ung is linearly independent� hence a basis for V �

Theorem � If V is �nite dimensional and U is a linearly independent subset

of V � U is �nite� say U � f�u�� � � � � �ung� and n � dimV � If n � dimV � U spans

V and so is a basis for V �

Theorem � Suppose B � f�b�� � � ��bng is a basis for V � Then there are func�

tions c�B� � � � � cnB � L�V F � such that for any �v � V

�v � cjB��v��bj�

DICTIONARY D��

The scalars v� � c�B��v�� � � � � vn � cnB��v�� which are uniquely determined by

�v and B� are called the coordinates of �v relative to the basis B� The linear

functions c�B� � � � � cnB are called the coordinate functionals for the basis B� �A

function whose values are scalars is called a �functional��� Clearly �bi � �ij�bj�

Since the coordinates of any vector relative to B are unique� it follows that

cjB��bi� � �i

j �D���

Theorem Suppose V andW are vector spaces over F � and B � f�b�� � � � ��bngis a basis for V � and f�w�� � � � � �wng�W � Then ��L � L�V W � � L��bi� � �wi�

�In other words� L is completely determined by LjB for one basis B� and

LjB can be chosen arbitrarily�� �The L whose existence and uniqueness are

asserted by the theorem is easy to �nd� For any �v � V � L��v� � ciB��v��wi��

This L is an injection if f�w�� � � � � �wng is linearly independent and a surjection

if W � spf�w�� � � � � �wng�

Linear operators� If L � L�V V �� L is called a �linear operator on V �� If B �

f�b�� � � � ��bng is any basis for V � then the array Lij � cjB�L�

�bi�� is called the matrix

of L relative to B� Clearly� L��bi� � Lij�bj� �The matrix of L relative to B is

often de�ned as the transpose of our Lij� It is that matrix � L��bi� � �bjL

ji� Our

de�nition is the more convenient one for continuum mechanics��

The determinant of the matrix of L relative to B depends only on L� not on B� so

it is called the determinant of L� written detL� A linear operator L is invertible

i� detL �� �� If it is invertible it is an isomorphism �a surjection as well as an

injection��

If L � L�V� V �� �v � V � � F � and �v �� ��� and if

L��v� � �v �D���

then is an �eigenvalue� of L� and �v is an �eigenvector�of L belonging to eigenvalue

� The eigenvalues of L are the solution � F of the polynomial equation

det�L� IV � � �� �D���

D��� DICTIONARY

Linear operators on a �nite dimensional complex vector space always have at least

one eigenvalue and eigenvector� On a real vector space this need not be so�

If L and M � L�V� V �� then L �M � L�V� V � and det�L �M� � �detL��detM��

Euclidean Vector Spaces

De�nition� A Euclidean vector space is an ordered quintuple �V�R� ad� sc� dp� with these

properties�

i� �V�R� ad� sc� is a �nite�dimensional �real� vector space

ii� dp is a functional on V � V which is symmetric� positive�de�nite and bilinear�

In more detail� the requirements on dp are these�

a� dp � V � V R� �dp is a functional�

b� dp��u��v� � dp��v� �u� �dp is symmetric�

c� dp��u� �u� � � if �u �� ��� �dp is positive de�nite�

d� for any �u� �v � V � dp��� �v � and dp��u� �� � L�V R�� �dp is bilinear��

dp is called the dot�product functional� and dp��u��v� is usually written �u � �v� In this

notation� the requirements on the dot product are

a�� �u � �v � R �dp is a functional�

b�� �u � �v � �v � �u �dp is symmetric�

c�� �u � �u � � if �u �� �� �dp is positive de�nite�

d�� for any �u� �u�� � � � � �un� �v � V and a�� � � � � an � F

�ai�ui� � �v � ai��ui � �v��v � �ai�ui� � ai��v � �ui�

��� dp is bilinear

Example �� �V�R� ad� sc� is as in Example �� page D��� and �u � �v � uivi� For n �

this is ordinary three�space�

DICTIONARY D���

Example �� �V�R� ad� sc� is as in Example �� page D��� If f and g are two functions in

V � their dot product is f � g �R �� dxf�x�g�x�� Note that without continuity� f �� �

need not imply f � f � �� For example� let f�x� � � in � � x � � except at x � ���

There let f���� � �� Then f �� � but f � f � ��

Length and angle� Schwarz and triangle inequalities� The length of a vector �v in

a Euclidean space is de�ned as k�v k ��p�v � �v� All nonzero�vectors have positive

length� From the de�nitions on page D����

j�u � �vj � k�uk k�vk �Schwarz inequality�

j�u� �vj � k�uk� k�uk �triangle inequality�

If �u �� �� and �v �� �� then �u � �v�k�u kk�v k is a real number between �� and �� so it

is the cosine of exactly one angle between � and �� This angle is called the angle

between �u and �v� and is written � ��u��v�� Thus� by the de�nition of � ��u��v��

�u � �v � k�ukk�vk cos � ��u��v��

In particular� �u � �v � �� � ��u��v� � ��� If k�uk � �� we will write u for �u�

Orthogonality� If V is a Euclidean vector space� and �u��v � V � P � V � Q � V � then

�u��v means �u � �v � ��

�u�Q means �u � �q � � for all �q � Q�

P�Q means �p � �q � � for all �p � P and �q � Q�

Q� �� f�x � �x � V and �x�Qg� This set is called �the orthogonal complement of Q��

It is a subspace of V � Obviously Q�Q�� Slightly less obvious� but true in �nite

dimensional spaces� is �Q��� � spQ�

Theorem Q spans V � Q� �n��o� In particular� if

n�b�� � � � ��bn

ois a basis for

V and �v ��bi � �� then �v � ��� Also� if �v � �x � � for all �x � V � then �v � ���

Orthonormal sets and bases� An orthonormal set Q in V is any set of mutually per�

pendicular unit vectors� I�e�� �u � Q � k�uk � �� and �u��v � Q� �u �� �v � �u � �v � ��

D��� DICTIONARY

An orthonormal set is linearly independent� so in a �nite dimensional Euclidean

space� any orthonormal set is �nite� and can have no more than dimV mem�

bers� An orthonormal basis in V is any basis which is an orthonormal set� If

B � f x�� � � � � xng is an orthonormal basis for V � the coordinate functionals ciB

are given simply by

ciB��v � � �v � xi� ��v � V� �D���

That is� for any �v� �v � ��v � xi� xi�

Theorem Every Euclidean vector space has an orthonormal basis� If U is a

subspace of Euclidean vector space V � U is itself a Euclidean vector space� and

every orthonormal basis for U is a subset of an orthonormal basis for V �

Linear functionals and dual bases� Let V be a Euclidean vector space� For any �v �V we can de�ne a �v � L�V R� by requiring simply

�v��y� � �v � �y� ��y � V� �D����

Theorem �� Let V be a Euclidean vector space� The mapping �v j �v de�ned

by equation ��� is an isomorphism between V and L�V R�� �Therefore we can

think of the linear functionals on V � the members of L�V R�� simply as vectors

in V ��

Proof�

We must show that the mapping �v j �v is an injection� a surjection�

and linear� Linearity is easy to prove� We want to show ai�vi � ai�vi�

But for any �y � V �

ai�vi��y� � �ai�vi� � �y � ai��vi � �y� � ai ��vi��y ��

� �ai�vi���y ��

�See de�nition of ai�vi in example � page D��� Therefore the two func�

tions ai�vi and ai�vi are equal�

DICTIONARY D���

To prove that �v j �v is injective� suppose �u � �v� We want to prove

�u � �v� For any �y � V � we know �u��y� � �v��y�� so �u � �y � �v � �y� or

��u��v� ��y � �� In particular� setting �y � �u��v we see ��u��v� � ��u��v� � ��

so ��u� �v� � ��� �u � �v�

To prove that �v j �v is surjective� let � L�V R�� We want to �nd

a �v � V � � �v� Let f x�� � � � � xng be an orthonormal basis for V and

de�ne �v � � xi� xi� Then for any �y � V � �v��y� � �v � �y � � xi�� xi � �y� �� xi�yi � �yi xi� � ��y�� Thus� �v � � QED�

Even in a Euclidean space� it is sometimes convenient to work with bases which are

not orthonormal� The advantage of an orthonormal basis x�� � � � � xn is that for any

�v � V � �v � ��v � xi� xi� An equation almost as convenient as this can be obtained for

an arbitrary basis� We need some de�nitions� All vectors are in a Euclidean vector

space V �

De�nition� Two sequences of vectors in V � ��u�� � � � � �um� and ��v�� � � � � �vm�� are dual to

each other if �ui � �vj � �ij�

Remark � If ��u�� � � � � �um� and ��v�� � � � � �vm� are dual to each other� each is linearly

independent�

Proof�

If ai�ui � ��� then �ai�ui� � �vj � �� so ai��ui � �vj� � �� so ai�ij � �� so aj � ��

De�nition� An ordered basis for V is a sequence of vectors ��b�� � � � ��bn� such that f�b�� � � � ��bngis a basis for V �

Remark � If two sequences of vectors are dual to each other and one is an ordered

basis� so is the other� �Proof� If ��v�� � � � � �vm� is a basis� m � dimV � Then� since

��u�� � � � � �um� is linearly independent� it is also a basis�

D��� DICTIONARY

Remark � If ��b�� � � � ��bn� is an ordered basis for V � it has at most one dual sequence

��b�� � � � ��bn��

Proof�

If ��b�� � � � ��bn� and ��v�� � � � � �vn� are both dual sequences to ��b�� � � � ��bn�� then

�bi ��bj � �ij� and �vi � bj � �ij� so ��bi��vi� ��bj � �� Then ��bi��vi� � �aj�bj� � �

for any ai� � � � � an � R� But any �u � V can be written �u � aj�bj� so

��bi � �vi� � �u � � for all �u � V � Hence� �bi � �vi � �� �bi � �vi�

Remark � If ��b�� � � � ��bn� is an ordered basis for V � it has at least one dual sequence

��b�� � � � ��bn��

Proof�

Let B � f�b�� � � ��bng� and let ciB be the coordinate functions for this basis�

They are linear functionals on V � so according to theorem ���� there is

exactly one �bi � V such that �bi � ciB� Then �bi��y� � ciB��y �� ��y � V � so

�bi ��y � ciB��y �� ��y � V � In particular� �bi ��bj � ciB��bj� � �i j �see ����� Thus

��b�� � � � ��bn� is dual to ��b�� � � � ��bn��

From the foregoing� it is clear that each ordered basis B � ��bi� � � � ��bn� for V has

exactly one dual sequence BD � ��b�� � � � ��bn� � and that this dual sequence is also an

ordered basis for V � It is called the dual basis for ��b�� � � � ��bn�� It is characterized

and uniquely determined by

�bi ��bj � �i j� �D����

There is an obvious symmetry� each of B and BD is the dual basis for the other�

If �v � V � and B � ��b�� � � � ��bn� is any ordered basis� we can write

�v � vj�bj� �D���

DICTIONARY D���

where vj � cjB��v� are the coordinates of �v relative to B� Then �bi � �v � vj�bi ��bj �

vj�ij � vi� Thus

vi � �bi � �v� �D���

Since ��b�� � � � ��bn� � BD is also an ordered basis for V � we can write

�v � vj�bj� �D����

Because duality is symmetrical� it follows immediately that

vj � �v ��bj� �D����

An orthonormal basis is its own dual basis� it is self�dual� If �bi � xi then �bi � xi�

and ����� ��� are the same as ��� and ����� We have

vj � vj � xj � �v�

In order to keep indices always up or down� it is sometimes convenient to write an

orthonormal basis � x�� � � � � xn� as � x�� � � � � xn�� with xi � xi� Then

vj � vj � xj � �v � xj � �v�

If ��b�� � � � ��bn� � B is an ordered basis for V � and BD � ��b�� � � � ��bn� is its dual basis�

and �v � V � we can write

�v � vi�bi � vi�bi� �D����

The vi are called the contravariant components of �v relative to B� and the vi are the

covariant components of �v relative to B �and also the contravariant components of �v

relative to BD�� The contravariant components of �v relative to B are the coordinates

of �v relative to B� The covariant components of �v relative to B are the coordinates

of �v relative to BD�

The covariant metric matrix of BD is

gij � �bi ��bj� �D����

D��� DICTIONARY

The contravariant metric matrix of BD is

gij � �bi ��bj� �D����

Clearly gij is the contravariant and gij the covariant metric matrix of BD�

We have vi � �bi � �v � �bi � ��bjvj� � ��bi ��bj�vj so

vi � gijvj� �D����

Similarly� or by the symmetry of duality�

vi � gijvj� �D���

The metric matrices can be used to raise or lower the indices on the components of

�v�

Transposes� Suppose V and W are Euclidean vector spaces and L � L�V W �� Then

there is exactly one LT � F�W V � such that

L��v� � �w � �v � LT ��w�� ���v� �w� � V �W� �D���

This LT is linear� i�e� LT � L�W V �� Furthermore� �LT �T � L� and if

a�� � � � � an � R and L�� � � � � Ln � L�V W �� then

�aiLi�T � ai�L

Ti � �D��

The linear mapping LT is called the �transpose� of L �also sometimes the adjoint

of L� but adjoint has a second� quite di�erent� meaning for matrices� so we avoid

the term��

If L � L�V W � and M � L�U V �� then L �M � L�U W � and it is easy to

verify that

�L �M�T � MT � LT � �D��

If L � L�V W � is bijective� then L�� � L�W V �� It is easy to verify that

LT � W V is also bijective and

�L���T � �LT ���� �D���

DICTIONARY D���

Orthogonal operators or isometries� If V is a vector space� the members of L�V V � are called �linear operators on V �� If V is Euclidean and L � L�V V �� then

L is called �orthogonal� or an �isometry� if kL��u�� L��v�k � k�u� �vk for all �u and

�v in V � If L � L�V V �� then each of the following conditions is necessary and

su!cient for L to be an isometry�

a� kL�vk � k�vk� ��v � V

b� L��u� � L��v� � �u � �v� ��u� �v � V

c� LT � L � IV

d� L � LT � IV

e� LT � L��

f� For every orthonormal basis B � V � L�B� is orthonormal� That is� if B �

f x�� � � � � xng is any orthonormal basis� then fL� x��� � � � � L� x�g is also an or�

thonormal basis�

g� There is at least one orthonormal basis B � L�B� is orthonormal�

h� For every orthonormal basis B � V � the matrix of L relative to B has or�

thonormal rows and columns �i�e�� LijL

ik � �jk� and Li

jLkj � �ik��

i� There is at least one orthonormal basis B � V relative to which the matrix

of L has either orthonormal rows �LijL

kj � �ik� or orthonormal columns

�LijL

ik � �ij��

If L is orthogonal� det L � ��� If detL � ��� L is called �proper�� Otherwise L is

�improper��

"�V � �� set of all orthonormal operators on V �

"��V � �� set of all proper orthonormal operators on V �

If L and M � "�V � or "��V �� the same is true of L �M and L��� Also IV � "�V �

and "��V �� Therefore� "�V � and "��V � are both groups if multiplication is de�ned

D��� DICTIONARY

to mean composition� "�V � is called the orthogonal group on V � while "��V � is

the proper orthogonal group�

If L � "��V �� L is a �rotation�� This implies that there is an orthonormal basis

x�� � � � � x�m� x�m�� and angles "�� � � � �"m between � and � such that

L� x�i��� � x�i�� cos �i � x�i sin �i

L� x�i� � � x�i�� sin �i � x�i cos �i

L� x�n��� � x�n���

������������i � f�� � � � � mg

The vector x�n�� is omitted in the above de�nition if dimV is even� and included

if dimV is odd� If included� it is the �axis� of the rotation�

If L � "�V � is improper� L is the product of a rotation and a �re�ection�� A

re�ection is an L for which there is an orthonormal basis x�� � � � � xn such that

L� x�� � � x�� L� xn� � xk� if k � �

Symmetric operators� A symmetric operator on Euclidean space V is an L � L�V� V � �LT � L� �Antisymmetric means LT � �L�� Each of the following conditions is ne�

cessary and su!cient for a linear operator L to be symmetric�

a� LT � L

b� L��u� � �v � L��v� � �u� ��u� �v � V

c� The matrix of L relative to every orthonormal basis for V is symmetric

d� The matrix of L relative to at least one orthonormal basis for V is symmetric�

e� There is an orthonormal basis for V consisting entirely of eigenvectors of L�

That is� there is an orthonormal ordered basis � x�� � � � � xn� for V � and a se�

quence of real numbers ��� � � � � n� �not necessarily di�erent� such that

L� xi�� � i� xi�� �i � f�� � � � � ng� �D���

Facts� The eigenvalues f�� � � � � ng are uniquely determined by L� but the eigenvectors

are not� For any real let ��L��� denote the set of all �v � V such that L��v� � �v�

DICTIONARY D��

Then ��L��� is a subspace of V � and dim ��L��� � � � is an eigenvalue of L� In

that case� ��L��� is called the eigenspace of L with eigenvalue � The eigenspaces

are uniquely determined by L� If � �� �� ��L���� � ��L�����

Positive�de�nite �semide�nite� symmetric operators� If L � L�V V � is sym�

metric� L is said to be positive de�nite when

L��v� � �v � � for all nonzero �v � V �D���

and positive semide�nite when

L��v� � �v � � for all �v � V� �D���

A symmetric linear operator is positive de�nite �semide�nite� i� all its eigenvalues

are positive �non�negative��

Every symmetric positive �semi� de�nite operator L has a unique positive �semi�

de�nite symmetric square root� i�e�� a symmetric positive �semi� de�nite operator

L��� such that

L��� � L��� � L �D���

Polar Decomposition Theorem� �See Halmos� p� ��� ��� Suppose V is Euclidean

and L � L�V V �� Then there exist orthogonal operators O� and O� and positive

semide�nite symmetric operators S� and S� such that

L � O� � S� � S� � O�� �D���

Both S� and S� are uniquely determined by L �in fact S� � �LTL���� and S� �

�LLT ������ If L�� exists� O� and O� are uniquely determined by L and are equal�

and S� and S� are positive de�nite� If detL � � then O� is proper�

The existence part of the polar decomposition theorem can be restated as follows�

for any L � L�V V � there are orthonormal bases � x�� � � � � xn� and � y�� � � � � yn� in

V and non�negative numbers �� � � � � n such that

D��� DICTIONARY

L� xi�� � i� yi�� i � �� � � � � n� �D���

If L is invertible� all i� � ��

Stated this way� the theorem remains true for any L�V W � if V and W are

Euclidean spaces and dimV � dimW � If dimV � dimW � then ��� holds for

i � �� � � � � m� with m � dimW � while for i � m��� � � � � n �with n � dimV � ��� is

replaced by

L� xi� � ��� �D���

Five Prerequisite Proofs

PP��

The Polar Identity

Let Sn �� group all permutations on f�� � � � � ng�

�i������in �� n dimensional alternating symbol

�� � if fi�� � � � � ing �� f�� � � � � ng�� sgn � if i� � ����� � � � � in � ��n�� � � Sn�

Then the polar identity is

X��Sn

� sgn ���i�j���� � � � �inj��n� � �i����in�j����jn� �PP���

Lemma PP�� Suppose Ai����in is an n � � � � � n dimensional n�th order array which is

totally antisymmetric �i�e� changes sign where any two indices are interchanged�� Then

Ai����in � �A�����n� �i����in � �PP��

Proof�

If any two of fi�� � � � � ing are equal� Ai�����in� � �� If they are all di�erent�

there is a � � Sn such that i� � ����� � � � � in � ��n�� This � is a product

of transpositions �interchanges�� and by carrying out these interchanges in

succession on A�����n one obtains A�������n�� Each interchange produces a sign

change� so the �nal sign is � or � according as � is even or odd� Thus�

A�������n� � � sgn��A����n � ��������n�A����n� Therefore� equation PP�� holds

both when fi�� � � � � ing contains a repeated index and when it does not�

PP�

PP�� PROOFS

Lemma PP�� Suppose W is a vector space and f � Sn W and � � Sn� Then

X��Sn

f��� �X��Sn

f����� �X��Sn

f����� �PP��

Proof�

Both the mapping � j ��� and the mapping � j �� are bijections of Sn to

itself� Therefore� each of the sums in �PP�� contains exactly the same terms�

These terms are vectors in W � so they can be added in any order�

Lemma PP�� De�ne

Ai����inj����jn �X��Sn

� sgn���i�j���� � � � �inj��n� �PP���

Then

Ai����inj����jn � Aj����jni����in �PP���

and for any � � Sn�

Ai����inj�������j��n� � � sgn ��Ai����inj����jn �PP���

Proof�

To prove �PP��� note that by lemma � since sgn ��� � sgn ��

Ai����inj����jn �X��Sn

� sgn���i�j������ � � � �inj��� �n��

The set of n pairs ������B� �

������

CA � � � � �

�B� n

����n�

CA���

is the same as the set ������B� ����

CA � � � � �

�B� ��n�

n

CA���

because a pair �B� i

j

CA

PROOFS PP��

is in the �rst set of pairs i� j � ����i� and in the second set of pairs i�

i � ��j�� Therefore

�i�j������ � � � �inj����n� � �i����j� � � � �i��n�jn

and

Ai����inj����jn �X��Sn

� sgn ���i����j� � � � �i��n�jn

�X��Sn

� sgn ���j�i���� � � � �jni��n�

� Aj����jni����in �

This proves equation �PP����

To prove equation �PP���� let � � Sn and let j� � k���� � � � � jn � k�n�� Then

from equation �PP����

Ai����ink�������k��n� �X��Sn

� sgn ���i�k����� � � � �ink���n��

Now � sgn ��� � � and so � sgn �� � � sgn���� sgn�� � � sgn �� �� sgn ��� sgn ��� �

� sgn��� sgn ���� and

Ai����ink�������k��n� � � sgn ��X��Sn

� sgn����i�k����� � � � �ink���n�

� � sgn ��X��Sn

� sgn���i�k���� � � � �ink��n� by Lemma PP�

� � sgn ��Ai����ink����kn�

This proves equation �PP����

Now we can prove the polar identity� Let Ai����inj����jn be as in �PP���� Then

�x i�� � � � � in� By lemma � and equation �PP����

Ai����inj����jn � Ai����in�����n�j����jn�

By equation �PP���� interchanging the i�s and j�s gives

Ai����inj����jn � A�����n j����jn�i����in � �PP���

PP�� PROOFS

Set i� � �� � � � � in � n� and equate these two� The result is

A�����n j����jn������n � A�����n�����n�j����jn

so

A�����n j����jn � A�����n �����n�j����jn� �PP���

Substituting �PP��� in �PP��� gives

Ai����inj����jn � A�����n �����n�i����in�j����jn� �PP���

Then from the de�nition �PP���� with ir � r� js � s

A�����n�����n �X��Sn

� sgn������� � � � �n�n��

There are n� terms in this sum� and a term vanishes unless ���� � �� ��� �

� � � � ��n� � n� that is� unless � is the identity permutation e� But sgn e � ��

so only one of the n� terms in the sum is nonzero� and this term is �� Therefore�

A�����n�����n � �� and equation �PP��� becomes the polar identity �PP����

Spectral Decomposition of a

Symmetric Linear Operator S on a

Euclidean Space V �

De�nition PP�� S � V V is symmetric i� S � ST � that is i� �u � S��v� � S��u� � �v for

all �u� �v � V �

Theorem PP��� Let S be a symmetric linear operator on Euclidean space V � Then V

has an orthonormal basis consisting of eigenvectors of S�

Lemma PP�� Suppose S is symmetric� U is a subspace of V � and S�U� � U � Then

S�U�� � U� and SjU� is symmetric as a linear operator on U��

Proof�

Suppose �w � U�� We want to show S��w� � U�� That is� for any �u � U � we

want to show �u �S��w� � �� But �u �S��w� � S��u� � �w� By hypothesis� S��u� � U

and �w � U�� so S��u� � �w � �� Thus� S��w� � U�� Therefore S�U�� � U��

And if the de�nition �PP��� above is satis�ed for all �u� �v � V � it certainly

holds for all �u� �v � U�� Hence� SjU� is symmetric�

Lemma PP�� Suppose S � V V is a symmetric linear operator on Euclidean space

V � Then S has at least one eigenvalue � such that S��v� � ��v� for some �v �� ���

Proof�

PP��

PP�� PROOFS

Let x�� � � � � xn be an orthonormal basis for V � and let Sij be the matrix of

S relative to this basis� so S� xi� � Sij xj� Then Sij � S� xi� � xj � xi �S� xj� � S� xj� � xi � Sji� so Sij is a symmetric matrix� A real number � is

an eigenvalue of S i� det�S � �I� � �� where I is the identity operator on

V � �If the determinant is �� then S � �I is singular� so there is a nonzero

�v � �S � �I���v� � ��� or S��v�� ��v � ��� or S��v� � ��v�� The matrix of S � �I

relative to the basis � x�� � � � � xn� is Sij���ij� so we want to prove that the n�th

degree polynomial in �� det�Sij���ij�� has at least one real zero� It certainlyhas a complex zero� �� so there is a complex n�tuple �r�� � � � � rn� such that

�Sij � ��ij�rj � �� or Sijrj � �ri�

Let � denote complex conjugation� Then

r�iSijrj � ��rir�i �� �PP����

Taking the complex conjugate of this equation gives �Sij is real�

riSijr�j � ���r�i ri��

Changing the summation indices on the left gives

rjSjir�i � ���r�i ri�� �PP����

But Sij � Sji� so the left hand sides of �PP���� and �PP���� are the same�

Also� r�i ri � rir�i �� �� Hence� �� � �� Thus � is real� and we have proved

lemma �PP����

Now we can prove theorem PP��� Let � be any �real� eigenvalue of S� Let v be

a unit eigenvector with this eigenvalue� Then S�spf vg� � spf�vg so by lemma

PP��� S�f vg�� � f vg� and Sjf vg� is symmetric� We proceed by induction on

dimV � Since dimf vg� dimV � we may assume theorem PP�� true on f vg��Adjoining v to the orthonormal basis for f vg� which consists of eigenvectors

of Sjf vg� gives an orthonormal basis for V consisting of eigenvectors of S�

PROOFS PP��

De�nition PP�� Let S be a linear operator on Euclidean space V � A �spectral decom�

position of S� is a collection of real numbers �� � � � �m and a collection of subspaces

of V � U�� � � � � Um� such that

i� SjUi� � �i� IjUi� for i � �� � � � � m

ii� V � U� � � � �� Um� �This means

a� Ui�Uj if i �� j� That is� if �ui � Ui and �uj � Uj then �ui � �uj � ��

b� every �v � V can be written in exactly one way as �v � �u� � � � � � �um with

�ui � Ui� �

Corollary PP�� If S is a symmetric linear operator on Euclidean space V � S has at

least one spectral decomposition�

Proof�

Order the di�erent eigenvalues of S which theorem PP�� produces as ��

� � � �m� Let Ui be the span of all the orthonormal basis vectors turned up

by that theorem which have eigenvalue �i�

Corollary PP�� S has at most one spectral decomposition�

Proof�

Let ���� � � � ��u and U ��� � � � U

�u be a second spectral decomposition� Let ��

be one of ���� � � � � ��u� and let �v � be a corresponding nonzero eigenvector� Then

S��v �� � ���v �� But �v � � �u� � � � � � �um where �ui � Ui� and S��v� � � ���v � �

���u� � � � �� ���um� Hence S��v�� � S��u�� � � � �� S��um� � ���u� � � � �� �m�um�

Hence�

��� � ����u� � � � �� ��� � �m��um � ��� �PP���

Since �v � �� ��� there is at least one �uj �� ��� Then� dotting this �uj into �PP���

gives ��� � �j���uj� � �uj� � �� so �� � �j � �� or �� � �j� Thus every ��i is

PP��� PROOFS

included in f��� � � � � �mg� By symmetry� every �i is included in f���� � � � � ��ug�Therefore the two sets of eigenvalues are equal� and u � m� Since the ei�

genvalues are ordered by size� ��i � �i for i � �� � � � � m� Now suppose

�u � � U �i � Then �u �i � �u� � � � �� �um and S��u �i � � �i��u

�i� � �i��u� � � � �� �um� �

S��u���� � ��S��um� � ����u��� � ���m�um so��i�����u��� � ����i��m��um � ���

Dot in �uj and we get ��i� �j��uj� � �uj� � �� Since �i �� �j� �uj� � �uj� � �� and

�uj � ��� Thus �u�i � �ui� Thus� �u�i � Ui� Thus U

�i � Ui� By symmetry� Ui � U �

i �

Thus Ui � U �i � QED�

Corollary PP�� If a linear operator on a Euclidean space has a spectral decomposition�

it is symmetric�

Proof�

Suppose S has spectral decomposition ���� U��� � � � � ��m� Um�� Let �u��v � V �

We can write �u �Pm

i � �ui� �v �Pm

i � �vi with �ui� �vi � Ui� Then

S��u� �mXi �

S��ui� �mXi �

�i�ui

S��v� �mXi �

S��vi� �mXi �

�i�vi

�u � S��v� �

�mXi �

�ui

���� mXj �

�j�vj

A �mX

i�j �

�j��ui � �vj�

�mXi �

�i��ui � �vi�

S��u� � �v �

�mXi �

�i�ui

���� mXj �

�vj

A �mX

i�j �

�i��ui � �vj�

�mXi �

�i��ui � �vi��

Hence �u � S��v� � S��u� � �v�

Corollary PP�� Two linear operators with the same spectral decomposition are equal�

Proof�

PROOFS PP���

Suppose S and S � have the same spectral decomposition ���� U��� � � � � ��m� Um��

Let �v � V � Then �v �Pm

i � �ui with �ui � Ui� Then

S��v� �mXi �

S��ui� �mXi �

�i�ui �mXi �

S ���ui�

� S ��

mXi �

�ui

�� S ���v��

Since this is true for all �v � V � S � S ��

PP��� PROOFS

Positive De�nite Operators and Their

Square Roots

De�nition PP�� A symmetric linear operator S is positive de�nite if �v � S��v� � � for

all �v �� ��� �S is positive semi�de�nite if �v � S��v� � � for all �v � V ��

Corollary PP�� A symmetric linear operator S is positive �semi� de�nite i� all its

eigenvalues are �non�negative� positive�

Proof�

�� Suppose S has an eigenvalue � � �� Let �v be a nonzero vector with

S��v� � ��v� Then �v � S��v� � ���v � �v� � �� so S is not positive de�nite�

�� Let ���� U��� � � � � ��m� Um� be the spectral decomposition of S� Let ��v � V �

�v �� ��� Then �v �Pm

i � �ui with �ui � Ui and at least one �ui �� ��� Then

S��v� �Pm

i � S��ui� �Pm

i � �i�ui� so

�v � S��v� �

�mXi �

�ui

���� mXj �

�juj

A �mX

i�j �

�j�ui � �uj

�mXi �

�i ��ui � �ui� � ��

Note� The proofs require an obvious modi�cation if S is positive semi�de�nite rather than

positive de�nite�

Theorem PP��� Suppose S is a symmetric� positive �semi� de�nite linear operator in

Euclidean space V � Then there is exactly one symmetric positive �semi� de�nite linear

PP��

PP��� PROOFS

operator Q � V V such that Q� � S� �Q is called the positive square root of S� and is

written S�����

Proof�

Existence� Let ���� U��� � � � � ��m� Um� be the spectral decomposition of S�

Since �i � �� let ����i be the positive square root of �i� Let Q be the linear op�

erator on V whose spectral decomposition is ������ � U��� � � � � ��

���m � Um�� Then

Q is symmetric by corollary PP��� and positive de�nite by corollary PP���

Let �v � V � Then �v �Pm

i � �ui with �u � Ui� Then

Q��v� �mXi �

Q��ui� �mXi �

����i �ui

Q���v� �mXi �

����i Q��ui� �

mXi �

����i �

���i �ui

�mXi �

�i�ui �mXi �

S��ui� � S��v��

Hence� Q� � S�

Uniqueness� Suppose Q is symmetric� positive de�nite� and Q� � S� Let

�q�� U��� � � � � �qu� Uu� be the spectral decomposition of Q� Then if �ui � Ui

we have S��ui� � Q��ui � QQ��ui� � Q�qi��ui�� � q�i��ui�� Thus q�i is an

eigenvalue of S� Therefore �q��� U��� � � � � �q�u� Uu� is a spectral decomposition

for S� Therefore u � m� U � Ui� q�i � �i� Since qi � �� qi � �

���i � Therefore Q

has the spectral decomposition ������ � U��� � � � � ��

���m � Um�� and is the operator

found in the existence part of the proof�

The Polar Decomposition Theorem

Theorem PP��� Suppose V is a Euclidean vector space� L � V V is a linear operator�

and detL � �� Then there are unique linear mappingsR�� S�� R�� S� with these properties�

i� L � R�S�

ii� L � S�R�

iii� R� and R� are proper orthogonal �i�e� R�� � RT and detR � ���

iv� S� and S� are positive de�nite and symmetric�

Further� R� � R�� so S� � RT� S�R��

Before giving the proof we need some lemmas�

Lemma PP�� If det �� �� LTL and LLT are symmetric and positive de�nite�

Proof�

It su!ces to consider LTL� because LLT � �LT �T �LT � and detLT � detL�

First� �LTL�T � LT �LT �T � LTL� so LTL is symmetric� Second� if �v �� ���

L��v� �� �� because L�� exists� But then L��v� � L��v� � �� And L��v� � L��v� ��v � LT �L��v�� � �v � ��LTL���v��� Thus� �v � ��LTL���v�� � �� so LTL is positive

de�nite�

Lemma PP� If S is positive de�nite and symmetric� detS � ��

Proof�

PP���

PP��� PROOFS

Let x�� � � � xn be an orthonormal basis for V consisting of eigenvectors for

S� namely s�� � � � � sn� �Some of these may be the same�� Relative to basis

x�� � � � � xn� the matrix of S is diagonal� and its diagonal entries are s�� � � � � sn�

Therefore detS � s�s� � � � sn� But since S is positive de�nite� all its eigenval�

ues are positive� Therefore detS � ��

Lemma PP� If A and B are linear operators on V and AB � I then B�� exists and

B�� � A� Hence� BA � I and A�� exists and A�� � B�

Proof�

If AB � I then �detA��detB� � detAB � det I � �� so detB �� �� Hence�

B�� exists� By its de�nition� B��B � BB�� � I� And if we multiply

AB � I on the right by B�� we have �AB�B�� � IB��� or A�BB��� � B���

or AI � B��� or A � B��� QED�

Lemma PP� If L is a linear operator on Euclidean space and L�� exists so does �LT ����

and �LT ��� � �L���T �

Proof�

Take transposes in the equation LL�� � I� The result is �L���TLT � I� By

lemma PP��� �LT ��� exists and equals �L���T �

Proof of Theorem PP���

First we deal with R� and S�� To prove these unique� we assume they exist and have

the asserted properties� Then L � R�S� implies LT � �R�S��T � ST

� RT� � S�R

��� so

LTL � S�R��� R�S� � S�IS� � S�

� � Now LTL is a symmetric positive de�nite operator�

and so is S�� Therefore� S� must be the unique symmetric positive de�nite square root of

LTL� i�e�

S� � �LTL����� �PP���

PROOFS PP��

Since detS� � �� S��� exists� Then L � R�S� implies

R� � LS��� � �PP����

Thus S� and R� are given explicitly and uniquely in terms of L� To prove the existence

of an R� and S� with the required properties� we de�ne them by �PP��� and �PP����

Then S� is symmetric positive de�nite as required� and we need show only that R� is

proper orthogonal� i�e� that detR� � �� and RT� � R��

� � But detR� � detL det�S��� � �

detL� detS� � � since detL � � and detS� � �� Therefore� if we can show RT� � R��

� � we

automatically have detR� � �� By lemma PP�� it su!ces to prove RT�R� � I� But RT

� �

�LS��� �T � �S��� �TLT � �ST� �

��LT � S��� LT so RT�R� � S��� LTLS��� � S��� S�

�S��� � I�

Next we deal with R� and S�� To prove them unique� we suppose they exist� Then

L � S�R� so LT � �S�R��

T � RT� S

T� � R��

� S�� But if detL � � then detLT � detL � ��

so there is only one way to write LT � R� S� with R� proper orthogonal and S� symmetric�

If R� is proper orthogonal� so is R��� � and we must have R��

� � R� so R� � R��� �

Then R� and S� are uniquely determined by L� In fact� S� � S� � �LT �TLT � LLT �

and R� � � R���� � �LTS��� ��� � S��L

T ���� To prove existence of R� and S� with

the required properties� we observe that LT � R� S� with R� proper orthogonal and S�

symmetric� Then �LT �T � L � � R� S��

T � S�T RT

� � S�R��� � Therefore we can take

S� � S�� R� � R��� �

Finally� to prove R� � R�� note that if L � S�R� then L � R�R��� S�R� � R��R

T� S�R���

We claim that RT� S�R� is symmetric and positive de�nite� If we can prove that� then

L � R��RT� S�R�� is of the form L � R�S�� and uniqueness of R� and S� requires R� � R��

S� � RT� S�R�� Symmetry is easy� We have �RT

� S�R��T � RT

� ST� �R

T� �

T � RT� S�R�� For

positive de�niteness let �v � V � �v �� ��� Then R���v� �� � because R��� exists� Then

R���v� � S��R���v�� � � because S� is positive de�nite� Then �v � RT� S��R���v�� � �� or

�v � ��RT� S�R����v�� � �� Thus RT

� S�R� is positive de�nite�

We shall not need the PDT when detL � or L�� fails to exist� For these cases� see

Halmos� p� ����

PP��� PROOFS

Representation Theorem for

Orthogonal Operators

De�nition PP�� A linear operator R on Euclidean vector space V is �orthogonal� i� it

preserves length� that is� kR��v�k � k�vk for every �v � V �

There are several other properties each of which completely characterizes orthogonal

operators �i�e�� R is orthogonal i� it is linear and has one of these properties�� We list

them in lemmas�

Lemma PP��� R is orthogonal i� R is linear and preserves all inner products� that is�

R��u� �R��v� � �u � �v for all �u��v � V �

Proof�

Obviously if R preserves all inner products it preserves lengths� The inter�

esting half is that if R preserves lengths then it preserves all inner products�

To see this� we note that for any �u and �v � V � k�u� �vk� � ��u� �v� � ��u� �v� �

�u � �u� �u � �v � �v � �u� �v � �v� That is

k�u� �vk� � k�uk� � �u � �v � k�vk�� �PP����

Applying this result to R��u� and R��v�� and using R��u � �v� � R��u� � R��v��

gives

kR��u� �v�k� � kR��u�k� � R��u� �R��v� � kR��v�k�� �PP����

PP���

PP��� PROOFS

If R preserves all lengths� then kR��u��v�k � k�u��vk� kR��u�k � k�uk� kR��v�k �k�vk� so subtracting �PP���� from �PP���� gives R��u� �R��v� � �u � �v�

Lemma PP��� R is orthogonal i� RT � R��� �Therefore if R is orthogonal R�� exists��

Proof�

If RT � R�� then RTR � I� Then for any �u��v � V � �u � RTR��v� � �u � �v so

R��u� �R��v� � �u ��v� Hence R is orthogonal� Conversely� suppose R orthogonal�

Then for any �u��v � V � R��u� � R��v� � �u � �v� so �u � RTR��v� � �u � �v� so �u ���RTR���v� � �v � � �� Fix �v� Then we see that RTR��v� � �v is orthogonal to

every �u � V � hence to itself� Hence it is ��� so RTR��v� � �v� Since this is true

for every �v � V � RTR � I� Then by lemma PP��� RT � R���

Corollary PP�� If either RTR � I or RRT � I� R is orthogonal�

Proof�

By lemma PP��� either equation implies RT � R���

Corollary PP� If R is orthogonal� so is R���

Proof�

By lemma PP���� R�� exists and is RT � But RT �RT �T � RTR � I� so RT

satis�es the second equation in corollary PP��� Hence RT is orthogonal�

The easiest way to discover whether a linear R � V V is orthogonal is to study its

matrix Rij relative to some orthonormal basis for V � Recall that we de�ne this matrix by

R� x�i � Rij xj �PP����

where � x�� � � � � xn� is the orthonormal basis for V which we are using� The matrix Lij of

an arbitrary linear operator L is de�ned in the same way�

L� x�i � Lij xj� �PP����

PROOFS PP���

The matrix of LT relative to � x�� � � � � xn� is the transposed matrix Lji� That is

LT � xi� � Lij xj or �LT �ij � Lji� �PP����

To see this� we note that xk � L� xi� � Lij xk � xj � Lij�kj � Lik� so

Lij � L� xi� � xj� �PP���

Therefore �LT �ij � LT � xi� � xj � xi � L� xj� � Lji�

Now we have

Remark � If R is orthogonal� then its matrix relative to every orthonormal basis satis�es

RijRik � �jk ��orthonormal columns�� �PP���

and

RjiRki � �jk ��orthonormal rows�� �PP��

Proof�

If R is orthogonal so is RT � Therefore� it su!ces to prove �PP���� We have

RT � xj� � �RT �ji xi � Rij xi� Then RRT � xj� � RijR� xi� � RijRik xk� But if

R is orthogonal� RRT � I� so RRT � xj� � �jk xk� Thus RijRik xk � �jk xk� or

�RijRik � �jk� xk � �� Fix j� Since x�� � � � � xn are linearly independent� we

must have RijRik � �jk � �� This is �PP����

Remark If R � L�V V � and there is one orthonormal basis for V relative to which

the matrix of R satis�es either �PP��� or �PP��� then R is orthogonal�

Proof�

Suppose that relative to the particular basis x�� � � � � xn� the matrix ofR satis�es

�PP���� Multiply by xk and add over k� using Rik xk � R� xi�� The result is

RijR� xi� � xj�

PP��� PROOFS

By linearity ofR� this meansR�Rij xi� � xj orRhRT � xj�

i� xj� or �RR

T �� xj� �

xj� But then for any scalars v�� � � � � �vn we have �RRT ��vj xj� � vjRRT � xj� �

vj xj� Hence RRT ��v� � �v for all �v � V � Hence RRT � I� Hence R is

orthogonal�

Next suppose R satis�es �PP�� rather than �PP���� Then �RT �ij�RT �ik �

�jk� so the matrix of RT satis�es �PP���� Hence� RT � R�� is orthogonal�

Hence so is �RT �T � �R����� � R�

Corollary PP� If the rows of a square matrix are orthonormal� so are the columns�

A matrix with this property is called orthogonal� What we have seen is that if R is

orthogonal� so is its matrix relative to every orthonormal basis� And if the matrix of R

relative to one orthonormal basis is orthogonal� R is orthogonal� so its matrix relative to

every orthonormal basis is orthogonal�

Now we discuss some properties of orthogonal operators which will eventually tell us

what those operators look like�

First� we introduce the notation in

De�nition PP�� The set of all orthogonal operators on Euclidean space V is written

"�V �� The set of all proper orthogonal operators �those with determinant � �� is written

"��V ��

Lemma PP��� "�V � and "��V � are groups if multiplication is de�ned as operator

composition�

Proof�

All we need to prove is that both sets are closed under the operations of taking

inverses and of multiplying� But ifR � "�V �� we know R�� � "�V � by remark

PP�� And det�R��� � �detR��� � � if detR � � so if R � "��V � then

R�� � "��V �� If R� and R� � "�V �� then �R�R��T �R�R�� � RT

�RT�R�R� �

RT� IR� � RT

�R� � I� so R�R� � "�V �� And if R�� R� � "��V � then

det�R�R�� � �detR���detR�� � �� so R�R� � "��V ��

PROOFS PP���

Lemma PP��� If R � "�V �� detR � �� or ���

Proof�

RTR � I so �detRT ��detR� � det I� But detRT � detR and det I � ��

Thus �detR�� � ��

Lemma PP��� Suppose R � "�V � and U is a subspace of V and R�U� � U � Then

RjU � "�U��

Proof�

By hypothesis� RjU is a linear operator on U � Since kR��v�k � k�vk for all

�v � V � this is certainly true for all �v � U � Hence� RjU satis�es de�nition

PP��

Lemma PP��� Under the hypothesis of lemma PP���� RT �U� � U �

Proof�

Suppose �u � U � Then there is a unique �u � � U such that �u � �RjU���u �� �be�

cause �RjU��� exists as a linear operator on U�� But �u � � U � so �RjU���u �� �

R��u ��� Thus �u � R��u ��� and R����u� � �u �� Then RT ��u� � �u �� Thus

RT ��u� � U if �u � U � QED

Lemma PP��� Under the hypotheses of lemma PP���� R�U�� � U��

Proof�

Suppose �w � U�� We want to show R��w� � U�� That is� �u �R��w� � � for all

�u � U � But �u �R��w� � RT ��u� � �w� By lemma PP��� RT ��u� � U � By hypothesis

�w�U � Hence� RT ��u� � �w � �� as required�

Lemma PP�� If R � "�V � and is an eigenvalue of R� � ���

Proof�

PP��� PROOFS

Suppose R��v� � �v and �v �� ��� Then k�vk � kR��v�k � k�vk �j j k�vk so

� �j j� QED�

Lemma PP�� Let R � "�V �� Let x�� � � � xn be an orthonormal basis for V � Let Rij

be the matrix of R relative to this basis� Suppose is a complex zero of the n�th degree

polynomial det�Rij � �ij�� Then j j�� ��

Proof�

There is a complex n�tuple �r�� � � � � rn� �� ��� � � � � �� such that Rijrj � ri�

Taking complex conjugates gives Rijr�j � �r�i � or Rikr

�k � �r�i � Multiplying

and summing over i gives

RijRikrjr�k � �rir

�i �

But RijRik � �jk� so rjr�j �j j� rir�i � Since rir�i � �� we have j j�� ��

Lemma PP�� In lemma PP���� � is also a zero of det�Rij � �ij� � ��

Proof�

Since �Rij � ��ij�r�j � � and �r��� � � � � r

�n� �� ��� � � � ��� det�Rij � ��ij� � ��

Lemma PP��� Suppose R� Rij as in lemma PP���� Suppose is not real and �r�� � � � � rn� ����� � � � � �� is a sequence of n complex numbers such that Rijrj � ri� Then riri � ��

Proof�

Rikrk � ri so �Rijrj��Rikrk� � �riri� or RijRikrjrk � �riri� or �jkrjrk �

�riri� or rjrj � �riri� or ��� ��riri � �� Since is not real� �� � �� �� so

riri � ��

Remark Suppose R � "�V � and V contains no eigenvectors of R� Then V contains

two mutually orthogonal unit vectors �x and �y such that for some � in � � ��

PROOFS PP���

�������������R� x� � x cos" � y sin"

R� y� � x sin" � y cos"�

�PP��

Proof�

Choose an orthonormal basis z�� � � � � zn for V � and let Rij be the matrix of

R relative to that basis� If � � det�Rij � �ij� has a real root� � it is an

eigenvalue of R� so R has an eigenvector� Hence no root of det�Rij � �ij�

is real� By lemma PP���� we can write any root as ei� where � � � or

� � �� By lemma PP���� we can �nd a root e�i� with � � �� Then

there is an n�tuple of complex numbers �r�� � � � � rn� �� ��� � � � � �� such that

Rjkrj � e�i�rk� �PP���

We may normalize �r�� � � � � rn� so that

r�jrj � �PP���

and by lemma PP��

rjrj � �� �PP���

Now write rj � xj � iyj where xj and yj are real�

Then separating the real and imaginary parts of �PP��� gives

Rjkxj � xk cos � � yk sin �

Rjkyj � �xk sin � � yk cos ��

Separating the real and imaginary parts of �PP��� gives

xjxj � yjyj � �

xjyj � �

PP��� PROOFS

and �PP��� gives

xjxj � yjyj � �

From these equations� clearly

xjxj � yjyj � � and xjyj � ��

Now take x � xj zj and y � yj zj� and we have the x� y� � whose existence is asserted in

the theorem�

Corollary PP� dimV � �

Remark Suppose R � "�V � and V contains no eigenvectors of R� Then V has an

orthonormal basis x�� y�� � � � xm� ym such that for each j � f� � � � � mg there is an "j in

� �j � with

�������������R� xj� � xj� cos �j� � yj� sin �j�

R� yj� � � xj� sin �j� � yj� cos �j�

� �PP���

Proof�

By remark PP�� we can �nd mutually orthogonal unit vectors x�� y� and ��

in � �� � such that �PP��� holds for j � �� Then R�spf x�� y�g� �spf x�� y�g� Hence� by lemma PP���� R�f x�� y�g�� � f x�� y�g�� Hence� by

lemma PP���� Rjf x�� y�g� � "�f x�� y�g��� Since R has no eigenvectors in V �

it has none in f x�� y�g�� Thus Rjf x�� y�g� satis�es the hypotheses of remark

PP�� and we can repeat the argument� �nding �� in � �� � an mutually

orthogonal unit vectors x�� y� in f x�� y�g� such that �PP��� holds for j � �

We can proceed in this way as long as the dimension of the remaining space

is � �� �The argument is by induction on dimV �� When the dimension of the

remaining space is �� we have an orthonormal basis for V with the required

properties�

PROOFS PP��

Corollary PP��� dimV is even� That is� if R � "�V � and dimV is odd� R has an

eigenvector� �This is obvious from another point of view� If dimV is odd� so is the degree

of the real polynomial det�R� I�� Hence� it has at least one real zero��

Theorem PP��� Suppose V is Euclidean vector space and R � "�V �� Then V has an

orthonormal basis x�� y�� � � � � xm� ym� z�� � � � zn� w� � � � � wp with these properties�

i� R� wj� � � wj

ii� R� zj� � zj

iii� for j � f�� � � � � mg� there is a �j in � �j � such that equation �PP��� holds�

Proof�

Let Z and W be the eigenspaces of V with eigenvalues �� and ��� Let

z�� � � � � zn be an orthonormal basis for Z and w�� � � � � wp an orthonormal basis

for W � Then R�Z �W � � Z �W � so R��Z �W ��� � �Z �W ��� Moreover�

�Z �W �� contains no eigenvectors of R� since all their eigenvalues are �� or

��� and these are already in Z orW � Thus remark PP�� applied to �Z�W ���

QED�

Corollary PP��� detR � ����p�

Proof�

Relative to the basis in theorem PP���� the matrix of R looks thus�

Each � diagonal block contributes a factor cos� �j � sin� �j � � to the

determinant� whence the result�

PP��� PROOFS

. . .

cosθ sin θ

θ-sin cosθ

1

1

. . .

-1 . . .{

cos sin

-sin cos

cosθ sin θ

θ-sin cosθ

1 1

1 1

θ θ

θ θ

2 2

2 2

m m

m m

n

p

terms

terms {-1

Figure PP��� Matrix of R�

PROOFS PP���

Corollary PP��� IfR � "��V �� then V has an orthonormal basis x�� y�� � � � � xm� ym� z�� � � � � zn

such that�����������������������������������������

i� R� zj� � zj for j � f�� � � � � ng�

ii� For j � f�� � � � � mg there is a �j in � �j � � such that

R� xj� � xj� cos �j� � yj� sin �j�

R� yj� � � xj� sin �j� � yj� cos �j��

� �PP���

Proof�

Since detR � ��� in theorem PP���� p is even� Therefore� we may divide

w�� � � � � wp into pairs� For any such pair� we have �for example�

R� w�� � � w� � w� cos � � w� sin �

R� w�� � � w� � � w� sin � � w� cos ��

If we adjoin these pairs to the pairs � xj� yj� given by theorem PP���� and take the

corresponding angles as � � �� we have the result of the corollary�

For obvious reasons� a linear operator on V which looks like �PP��� in some or�

thonormal basis is called a �rotation�� Corollary PP�� says that every proper orthogonal

operator is a rotation� If dimV � � the eigenvector z� is called the axis of the rotation�

and the angle #�� in � �� � � is the angle of the rotation� This terminology breaks down

only when m � � and n � in �PP���� In that case� R � I� The angle of rotation of I

is �� but its axis of rotation obviously cannot be de�ned�

For dimV � � corollary PP�� is called Euler�s theorem� If R � L�V V � and an

orthonormal basis x�� � � � � xn for V can be found such that R� x�� � � x�� R� xj� � xj for

j � then R is said to be a re�ection along x�� Obviously detR � �� for a re�ection�

PP��� PROOFS

so a re�ection is an �improper orthogonal� operator� Also� clearly a re�ection has the

property R�� � R� or R� � I�

If R� is a re�ection and R� is any other improper orthogonal operator� then R � R�R�

is orthogonal and detR � �det�R���detR�� � ����� � �� so R is proper orthogonal� But

then R�R � R��R� � R�� so R� � R�R� That is� every improper orthogonal operator is

the product of a rotation and a re�ection�

Part I

Tensors over Euclidean Vector Spaces

Chapter �

Multilinear Mappings

��� De�nition of multilinear mappings

M�V� � � � �� Vq W ��

V�� � � � � Vq�W are vector spaces over a single �eld F � and M � F �V�� � � ��Vq W ��

If M satis�es any of the following three equivalent conditions� M is called a multilinear

mapping from V� � � � �� Vq to W �

De�nition ����� For any ��v�� � � � � �vq� � V� � � � �� Vq and any p � f�� � � � � qg�

M��v�� � � � � �vp��� �� �vp��� � � � � �vq� � L�Vp W ��

�That is� M��v�� � � � � �vq� depends linearly on each of the separate vectors �v�� � � � � �vq if the

others are held �xed��

De�nition ���� If a � F � p � f�� � � � � qg� �xp and �yp � Vp� and ��v�� � � � � �vq� � V��� � ��Vqthen

i� M��v�� � � � � �vp��� a�vp� �vp��� � � � � �vq� � aM��v�� � � � � �vp��� �vp� �vp�n� � � � � �vq�� Also

ii� M��v�� � � � � �vp��� �xp � �yp� �vp��� � � � � �vq� � M��v�� � � � � �vp��� �xp� �vp��� � � � � �vq�

�M��v�� � � � � �vp��� �yp� �vP��� � � � � �vq� �

� CHAPTER �� MULTILINEAR MAPPINGS

De�nition ���� For each p � f�� � � � � qg� suppose a�p�� � � � � anpp� � F and �vp�� � � � � � �v p�

np �Vp� Then M

�ai����v

��i� � � � � � a

iqq��v

q�iq

�� ai��� � � � a

iqq�M

��v��i� � � � � � �v

q�iq

��

De�nition ���� M�V� � � � � � Vq W � denotes the set of all multilinear mappings

from V� � � � �� Vq to W �

��� Examples of multilinear mappings

In these examples� V�� � � � � Vq are Euclidean vector spaces� and the �eld F is R� The

vector space W need not be Euclidean �i�e� need not have a dot product� but it must be

real �i�e� its �eld is R��

Example ����� For each p between � and q� let �up be a �xed vector in Vp� Let �w� � W �

For any ��v�� � � � � �vq� � V� � � � �� Vq� de�ne

M��v�� � � � � �vq� � ��u� � �v�� � � � ��uq � �vq��w�� ������

To give the next example� some terminology is useful� Suppose that for each p �f�� � � � � qg Bp �

��bp�� � � � � ��b p�np

�is an ordered basis for Vp� Then the sequence of ordered

bases �B�� � � � � Bq� is called a �basis sequence� for V�� � � ��Vq� Let BDp � ��b �p�� � � � �

�bnpp��

be the ordered basis for Vp which is dual to Bp� Then �BD� � � � � � B

Dq � is �the basis sequence

dual to �B�� � � � � Bq���

Example ����� Let �B�� � � � � Bq� be a basis sequence for V�� � � ��Vq� Let �wi����iq be any

q�th order� n� � � � � � nq array of vectors from W � For any ��V�� � � � � �Vq� � V� � � � �� Vq

de�ne

M��v��� � � � � �vq�

����b i��� � �v ��

�� � �

��biqq� � �vq�

��wi����iq � �����

Then

M � M �V� � � � �� Vq W � �����

and also

M��b��j� � � � ��b

��jq

�� �wj����jq � ������

���� EXAMPLES OF MULTILINEAR MAPPINGS �

The multilinearity of M is almost obvious� ����� is a sum of n�n� � � � nq terms like

������� Each of these terms is linear in each �vp if �v�� � � � � �vp��� �vp��� � � � �vq are �xed�

To prove ������ we let �vp � �bp�jp in the de�nition ������ obtaining M��b

��j� � � � � �

�bq�jq � �

��b i�����b ��j� � � � � ��biqq���b q�jq ��wi����iq � �i� j� � � � �

iqjq �wi����iq � �wj����jq � There are no other examples�

Every member ofM�V� � � � �� Vq W � is like example ������ We have

Remark ������ Suppose V�� � � � � Vq are Euclidean vector spaces and W is a real vector

space� Suppose dimV� � n�� � � � � dimVq � nq� Suppose �B�� � � � � Bq� is a basis sequence

for V� � � � �� Vq� and �wi����iq is an n� � n� � � � �� nq array of vectors from W � Then ��M � M�V� � � � �� Vq W � �M��b

��i� � � � � ��b

q�iq � � �wi����iq �

Proof�

Example ��� shows at least one such M � To establish uniqueness suppose

M does satisfy ����� and ������� Let ��b�p�� � � � ��bnpp�� � BD

p be the ordered

basis for Vp which is dual to Bp � ��bp�� � � � � ��b p�np �� For any �vp� � Vp� we have

�vp� � vipp��bp�ip where �v

ipp� � vp� ��bipp� � Then

M��v��� � � � � �v q�� � M�vi����b��i� � � � � � v

iqq��bq�iq � �by de�nition ������

� bi��� � � � viqq�M��b

��i� � � � � ��b

q�iq � �by hypothesis�

� vi��� � � � viqq� �wi����iq

���bi��� � �v��

�� � �

��biqq� � �vq�

��wi����iq �

In other words� M is the function de�ned by ������

It follows that every M � M�V� � � � �� Vq W � has the form ������

� CHAPTER �� MULTILINEAR MAPPINGS

��� Elementary properties of multilinear mappings

Remark ������ If one of �v�� � � � � �vq is �� and if M � M�V� � � � � � Vq W � then

M��v�� � � � � �vq� � ��� the zero in W �

Proof�

We suppose �v� � ��� ThenM���� �v�� � � � � �vq� � M����� �v�� � � � � �vq� � �M���� �v�� � � � � �vq� �

��� where we have used de�nition ��������

Remark ������ M�V� � � � �� Vq W � is a vector space over F if we use the ad and

sc de�ned in example � page D��� or in equations ������� ������

Proof�

Refer to �subspaces� on page D���� We know from example � page D���

that F �V� � � � � � Vq W � is a vector space over F � Obviously� M�V� �� � �� Vq W �� is a subset of F �V� � � � �� V q� W �� Therefore� it su!ces

to prove that if a � F and M � N � M�V� � � � � � V q W �� then aM and

M � N � M�V� � � � � � Vq W �� That is� if M and N are multilinear� so

are aM and M �N � This is immediate from de�nition ����� because we know

it for linear functions� Alternatively� we can use de�nition ����� and write

everything out� We will do this for M �N � We have

�M �N��a i����v

��i� � � � � � a

iqq��v

q�iq

�� M

�ai����v

��i� � � � � a

iqq��v

q�iq

��N

�a i����v

��i� � � � � � a

iqq��v

q�iq

�� by de�nition of M �N�

� a i��� � � � a

iqq�M

��v��i� � � � � � �v

q�iq

�� ai��� � � � a

iqq�N

��v��i� � � � � � �v

q�iq

��because M and N are multilinear�

� a i��� � � � aiqa�

hM��v��i� � � � � � �v

q�iq

��N

��v��i� � � � � � �vq�iq

�ibecause W is a vector space over F

���� PERMUTING A MULTILINEAR MAPPING

� ai��� � � � aiqq�

h�M �N�

��v��i� � � � � � �v

q�iq

�iby de�nition of M �N�

Comparing the two ends of this chain of equalities� we have a proof thatM�N

is multilinear if M and N are� and if M �N is de�ned by �������

�� Permuting a multilinear mapping

De�nition ������ Suppose ��v�� � � � � �vq� � V� � � � � � Vq and � � r s � q� The q�

tuple ��v�� � � � � �vr��� �vs� �vr�� � � � � �vs��� �vr� �vs��� � � ��vq� will be abbreviated ��v�� � � ��vq�rs�� It is

simply ��v�� � � � � �vq� with �vr and �vs interchanged� Note that if Vr � Vs� then ��v�� � � � � �vq�rs� �V� � � � �� Vq�

De�nition ������ Suppose M � M�V� � � � � � Vq W � and � � r s � q� Suppose

Vr � Vs� Then we write �rs�M for the member of F �V� � � � � � Vq W � de�ned by

requiring for any ��v�� � � � � �vq� � V� � � � �� Vq that

��rs�M � ��v�� � � � � �vq� � M ��v�� � � � � �vq�rs� � �������

That is

��rs�M � ��v�� � � � � �vr��� �vr� �vr��� � � � � �vs��� �vs� �ss��� � � � � �vq�

� M ��v�� � � � � �vr��� �vs� �vr��� � � � � �vs��� �vr� �ss��� � � � � �vq� � ������

Remark ������ �rs�M � M�V� � � � �� Vq W � if Vr � Vs�

Proof�

Obvious from ������ and de�nition ��������

Remark ������ �rs� is a linear operator onM�V� � � � �� Vq W � if Vr � Vs�

� CHAPTER �� MULTILINEAR MAPPINGS

Proof�

By remark �������� �rs� � M�V�� � � ��Vq W � M�V�� � � ��Vq W ��

It remains to prove that �rs� is linear� Suppose a�� � � � � aN are scalars and

M�� � � �MN � M�V� � � � �� Vq W �� We want to prove that �rs��ajMj� �

aj��rs�Mj�� Let ��v�� � � � � �vq � V� � � � �� Vq�� Then

��rs��ajMj�� ��v�� � � � � �vq� � �ajMj���v�� � � � � �vq�rs� � ajhMj��v�� � � � � �vq�rs�

i

� aj

�h�rs�Mj

i��v�� � � � � �vq�

���ajh�rs�Mj

i���v�� � � � � �vq��

Since this is true for any ��v�� � � � � �vq� � V� � � � � � Vq� we have the required

result�

De�nition ������ Suppose M � r� s as in de�nition ������� If �rs�M � M � M is �sym�

metric� under �rs�� If �rs�M � �M � M is �antisymmetric� under �rs��

Remark ������ If Vr � Vs� the members ofM�V�� � � ��Vq W � which are symmetric

under �rs� constitute a subspace of M�V� � � � � � Vq W �� The same holds for the

members antisymmetric under �rs��

Proof�

It su!ces to show that if a is any scalar and M � N are symmetric �or antisym�

metric� under �rs�� so are aM andM�N � We give the proof for antisymmetry�

In that case� by hypothesis�

�rs�M � �M � �rs�N � �N�

But �rs� is linear� so �rs��aM � � a��rs�M � � a��M� � �aM � and �rs��M �

N� � �rs�M��rs�N � �M�N � ��M�N�� �Note that the set of members

ofM�V� � � � �� Vq W � which are symmetric �anti�symmetric� under �rs�

is simply the eigenspace of �rs� with eigenvalue ������ Of course� it is a

subspace ofM�V� � � � �� Vq W ���

���� PERMUTING A MULTILINEAR MAPPING �

De�nition ������ Suppose V� � � � � � Vq � V � Suppose M � M��qV W � and

� � Sq� Then �M is de�ned as that member of F��qV W � such that for any

�v�� � � ��vq � V

��M���v�� � � � � �vq� � M��v���� � � � � �v�q�

��

Remark ������ If V� � � � � � Vq � V then �M � M��qV W �� and in fact � is a

linear operator onM��qV W �� Moreover if �� � � Sq and M � M��qV W � then

���M� � ����M� ������

Proof�

Except for ������� the proof is like that of remark ��������� To prove equation

������� let �v�� � � � � �vq � V and let �w� � �v���� � � � � �wq � �v�q�� Then����M�

���v�� � � � � �vq� � ��M�

��v���� � � � � �v�q�

� ��M� ��w�� � � � � �wq� � M��w���� � � � � �w�q�

� M��v������� � � � � �v���q��

� M��v������ � � � � �v���q�

� �����M � ��v�� � � � � �vq�

Since this is true for all �v�� � � � � �vq � V � we have �������

De�nition ������ Suppose V� � � � � � Vq � V and M � M � ��qV W �� M is

totally symmetric �antisymmetric� if �M � M ��M � � sgn��M � for every � � Sq�

�� CHAPTER �� MULTILINEAR MAPPINGS

Remark ����� M is totally symmetric �antisymmetric� i� �rs�M � M�� �M� for

every transposition �rs��

Proof�

�� is obvious� For��� recall that any � � Sq is a product of transpositions�

By remark ������ the e�ects of these transpositions on M can be calculated

one after the other� If all leave M unchanged� so does �� If all multiply M by

��� then �M is M multiplied by sgn � �� �� for even�odd ���

Remark ����� The totally symmetric �antisymmetric� members of M��qV W �

constitute a subspace ofM��qV W ��

Proof�

We give the proof for the totally symmetric case� Suppose M�� � � � �MN are

totally antisymmetric members ofM��qV W � and a�� � � � � aM are scalars�

We want to show that aiMi is totally antisymmetric� But for any � � Sq� � is

linear operator on M��qV W �� so ��aiMi� � ai��Mi� � ai�� sgn��Mi� �

� sgn���aiMi�� That is� aiMi is totally antisymmetric�

Chapter �

De�nition of Tensors over Euclidean

Vector Spaces

If V�� � � � � Vq are Euclidean vector spaces� thenM�V�� � � ��Vq� R is called the �tensor

product of V�� � � � � Vq�� It is written V� � � � �� Vq� Thus

V� � � � �� Vq ��M�V� � � � �� Vq� R� ������

The multilinear functionals which are the members of V�� � � �� Vq are called the tensors

of order q over V� � � � �� Vq�

If V� � � � � � Vq � V � then V�� � � �� Vq �

q timesz �� �V � � � �� V � and this is abbreviated �qV �

Its members are called the tensors of order q over V � By convention� tensors of order zero

are scalars� that is

��V � R� �����

For tensors of order �� the notation requires some comment� When q � �� multilinearity

is simply linearity� That is M�V R� � L�V R�� Thus the tensors of order � over

a vector space V are simply the linear functional on V � This would not be a problem�

except that for q � �� ������ reduces to V� ��M�V� R�� which says

V � L�V R�� �����

��

�� CHAPTER �� DEFINITION OF TENSORS OVER EUCLIDEAN VECTOR SPACES

It is at this point that we require V to be a Euclidean vector space� If it is� then �v � V can

be safely confused with the linear functional �v � L�V R� de�ned by ����� Tensors

of order � over V are simply the vectors in V �

When V is not a Euclidean vector space� ����� is false� �The situation is very

subtle� If V is not Euclidean but is �nite dimensional� every basis B for V generates an

isomorphism between V and L�V R�� namely�bij ciB� But each of these isomorphisms

depends on the basis B chosen to produce it� There is no �natural�� basis�independent

isomorphism which can be used to identify vectors with linear functionals� See MacLane

and Birkho�� Algebra� MacMillan ����� p� � for a more detailed discussion�� Tensors

over vector spaces which are not Euclidean are constructed in a more complicated fashion

than ������� We will not need this generality� and tensors over Euclidean spaces have

some very useful properties for continuum mechanics� It is for this reason that we consider

only tensors over Euclidean spaces� See M� Marcus�s book� on reserve� for a treatment of

tensors over arbitrary vector spaces�

Chapter �

Alternating Tensors� Determinants�

Orientation� and n�dimensional

Right�handedness

In this section we discuss an application of tensors which some of you may have seen in

your linear algebra courses�

Let V be an n�dimensional Euclidean vector space� The set of totally antisymmetric

tensors of order q over V is written �qV � It is a subspace of �qV � �See remark ��������

We want to study �nV � The members of �nV are called �alternating tensors over V ��

��� Structure of �nV

We want to understand what the alternating tensors over V are like� It will turn out that

there are not very many of them� In fact dim�nV � �� We want to prove this and to

�nd a basis �vector� in the vector space �nV �

First we note a technique for verifying that A � �nV � We need to prove only that

A � �nV and that �rs�A � �A for every transposition �rs� � Sn� If this latter fact is

true� then clearly remark ���� tells us A � �nV �

A second property of an A � �nV which is equivalent to membership in �nV is that

��CHAPTER �� ALTERNATING TENSORS� DETERMINANTS� ORIENTATION� ANDN�DIMENSIONAL RIGHT�HANDEDNESS

for any ��v�� � � � � �vn� � �nV and any � � Sn�

A��v���� � � � � �v�n�

�� � sgn ��A ��v�� � � � �vn� � ������

This is obvious from the de�nition of �A and �nV � It implies the �rst step in our

attempt to understand alternating tensors�

Lemma ������ If A � �nV and two of �v�� � � � �vn are equal� then A��v�� � � � � �vn� � ��

Proof�

If �vr � �vs� interchanging them will not change A��v�� � � � � �vn�� But sgn �rs� �

��� so ������ says interchanging them changes the sign of A��v�� � � � � �vn��

Hence A��v�� � � � � �vn� � ��

Equation ������ and lemma ���� are equivalent to the single equation

A ��vi� � � � � � �vin� � �i����inA �v�� � � � � �vn� � �����

For if two of i�� � � � � in are equal� both sides of ����� vanish� If i�� � � � � in are all di�erent�

there is a permutation � � Sn such that i��� ����� � � � � in � ��n�� Then ����� reduces

to equation �������

Lemma ���� can be extended to

Remark ����� If A � �nV and �v�� � � � � �vn are linearly dependent� then A��v�� � � � � �vn� �

��

Proof�

By hypothesis� there are scalars c�� � � � � cn� not all zero� such that ci�vi � ���

If �v� � ��� then A��v�� � � � � �vn� � � by remark ������ If �v� �� � then at least

one of c�� � � � � cn is nonzero� Let cp be the last scalar which is nonzero� Then

c��v� � � � �� cp�vp � �� and we can divide by cp to write

�vp �p��Xi �

ai�vi

���� STRUCTURE OF �NV ��

where

ai � � cicp�

Then

A ��v�� � � � � �vn� � A

���v�� � � � � �vp��� p��Xi �

ai�vi� �vp��� � � � � �vn

A

�p��Xi �

aiA ��v�� � � � �vp��� �vi� �vp��� � � � � �vn�

� � by lemma �����

Remark ����� would be rather dull if A � �nV implied A � �� Therefore at this

point we pause to construct a nonzero member of �nV � Let B � ��b�� � � � ��bn� be any basis

for V � Let BD � ��b�� � � � ��bn� be its dual basis� De�ne AB � F��nV R� by requiring

AB ��v�� � � � � �vn� ���v� ��bi�

�� � �

��vn ��bin

��i����in �����

for any ��v�� � � � �vn� � �nV � From de�nition ������� it is clear that AB is multilinear�

so AB � �nV � because the values of AB are scalars� Furthermore� AB

��b�� � � � ��bn

���

�b� ��bi��� � �

��bn ��bin

��i����in � ��

i� � � � �nin�i����in � ������n � �� That is� if B �

��b�� � � � ��bn

��

AB

��b�� � � ��bn

�� � ������

when AB is de�ned by ������ Thus AB �� �� We claim also that AB � �nV � To show

this� is su!ces to show that �rs�AB � �AB for any transposition �rs�� The argument is

the same for all �rs�� so we consider ���� We have

����AB� ��v�� � � � � �vn� � AB ��v�� �v�� �v�� � � � � �vn�

���v� ��bi�

� ��v� ��bi�

� ��v� ��bi�

�� � �

��vn ��bin

��i����in�

If we relabel the summation index i� as i�� and relabel i� as i�� this is

����AB� ��v�� � � � � �vn� ���v� ��bi�

� ��v� ��bi�

� ��v� ��bi�

�� � �

��vn ��bin

��i� i� i����in

��CHAPTER �� ALTERNATING TENSORS� DETERMINANTS� ORIENTATION� ANDN�DIMENSIONAL RIGHT�HANDEDNESS

���v� ��bi�

�� � �

��vn ��bin

��i� i� i����in

� ���v� ��bi�

�� � �

��vn ��bin

��i� i� i����in

� �AB ��v�� � � � � �vn� �

Since these equations hold for any ��v�� � � � � �vn� � �nV � we have ���AB � �AB � QED

Now for every ordered basis B in V we have managed to construct a nonzero member

of �nV � namely AB� It looks as if �nV must be rather large� In fact it is not� We can

now easily prove

dim�nV � � if n dimV� ������

To see this� let B � ��b�� � � � ��bn� be a �xed ordered basis for V � We will prove that if

A � �nV then A is a scalar multiple of AB� Thus� fABg is a basis for �nV � The argument

goes thus� Let A � �nV � For any ��v�� � � � � �vn� � �nV we can write �vp � vpi�bi where vp

i �

�vp ��bi� BD � ��b�� � � � ��bn� being the basis dual to B� Then

A ��v�� � � � � �vn� � A�v�

i��bi� � � � � � vnin�bin

� v�i� � � � vn

inA��bi� � � � � ��bin

� v�i� � � � vn

in�i����inA��b�� � � � ��bn

�by �����

� A��b�� � � � ��bn

� ��v� ��bi�

�� � �

��vn ��bin

��i����in

� A��b�� � � � ��bn

�AB ��v�� � � � � �vn� �

Since this is true for all �v�� � � � � �vn � V � we have for ordered basis B � ��b�� � � ��bn� in V

���� DETERMINANTS OF LINEAR OPERATORS �

and for any A � �nV

A � A��b�� � � � ��bn

�AB� ������

Of course� A��b�� � � � ��bn

�� R� so ������ shows A � spfABg� This proves �������

Equation ���� permits us to prove the following test for linear independence of a

sequence of n vectors in an n�dimensional space�

Remark ������ Suppose A � �nV and A �� �� and dimV � n� Suppose �v�� � � � � �vn � V �

Then �v�� � � � � �vn are linearly dependent if and only if A��v�� � � � � �vn� � ��

Proof�

�� is remark ���� To prove ��� we want to prove that if A��v�� � � � � �vn� � �

then �v�� � � � � �vn are linearly dependent� If they are not� they are a basis B for

V � Then ������ says A � A��v�� � � � � �vn�AB� and since A �� �� we must have

A��v�� � � � � �vn� �� �� contrary to our hypothesis�

��� Determinants of linear operators

The fact that dim�nV � � makes possible a simple and elegant de�nition of the determ�

inant detL of a linear operator L � L�V V �� Choose any nonzero A � �nV � De�ne

A�L� � F ��nV R� by

A�L� ��v�� � � � � �vn� � A �L��v��� � � � � L��vn�� � � ��v�� � � � � �vn� � �nV� �����

Since L is linear and A is multilinear� de�nition ������� shows that A�L� is multilinear� It

is obvious that A�L� is totally antisymmetric if A is� Thus A�L� � �nV � Since A �� �� A is

a basis for �nV � and there is a scalar kA�L such that

A�L� � kA�LA� ����

From the de�nition ������ if a � R� �aA��L� � aA�L�� Thus by ����� �aA��L� � kA�LaA�

But� applying the general result ���� to aA instead of A� we have �aA��L� � kaA�L�aA��

Thus

kaA�L�aA� � �aA��L� � aA�L� � akA�LA � kA�L�aA��

��CHAPTER �� ALTERNATING TENSORS� DETERMINANTS� ORIENTATION� ANDN�DIMENSIONAL RIGHT�HANDEDNESS

Since A �� �� kaA�L � kA�L� But every member of �nV can be written aA for some a � R�

Therefore kA�L has the same value for all nonzero A � �nV � It is independent of A� and

depends only on L� Thus for any L � L�V V � there is a real number kL such that for

any nonzero A � �nV �

A�L� � kLA� ����

This equation is obviously true also for A � �� and hence for all A in �nV � The determ�

inant of L� detL� is de�ned to be kL� Thus� by the de�nition of detL�

A�L� � �detL�A �����

for any A � �nV � Returning to the de�nition ����� of A�L�� we have

A �L ��v�� � � � � � L ��vn�� � �detL�A ��v�� � � � � �vn� �����

for any A � �nV and any ��v�� � � � � �vn� � �nV �

Now suppose ��v�� � � � � �vn� is any ordered basis for V � and Lij is the matrix of L relative

to this basis� Then L��vi� � Lij�vj so

A �L ��v�� � � � � � L ��vn�� � AhL�

j��vj� � � � � � Lnjn�vjn

i

� L�j� � � � Ln

jnA ��v�� � � � � �vn� because A is multilinear

� L�j� � � � Ln

jn�j� � � � jnA ��vj�� � � � � �vjn� because

A is totally antisymmetric ������

��detLi

j�A ��v�� � � � � �vn�

where detLij is the determinant of the n � n matrix Li

j� Comparing this with ������

we see that

detL � detLij� �����

���� DETERMINANTS OF LINEAR OPERATORS ��

The determinant of a linear operator is the determinant of its matrix relative to any basis�

In deducing ����� from ����� we have used the fact that A��v�� � � � � �vn� �� �� This fact

follows from remark �������

Some properties of detL are easy to deduce from ������ For example� if K and L

are both linear operators on V �

det �K � L� � �detK� �detL� � �����

The proof is simple� From ������

A ��K � L� ��v�� � � � � � �K � L� �vn�� � det �K � L�A ��v�� � � � � �vn�

� A �K �L��v��� � � � � � K �L��vn���

� �detK�A �L ��v�� � � � � � L ��vn��

� �detK� �detL�A ��v�� � � � � �vn� �

These equations are true for any A � �nV and any f�v�� � � � � �vng � V � If we choose A �� �

and ��v�� � � � � �vn� an ordered basis� then A��v�� � � � � �vn� �� �� and we can cancel it� obtaining

������

This choice of A and ��v�� � � � � �vn� also shows that if we set L � IV in ����� �IV ��

identity mapping of V onto V � then

det IV � �� �����

As another application of ������ let A �� � and f�v�� � � � � �vng be a basis for V � Then�

as remarked in theorem ������ L is an isomorphism �� fL��v��� � � � � L��vn�g is linearly

independent� From remark ���� this is true�� A �L��v��� � � � � L��vn�� �� �� From ������

since A��v�� � � ��vn� �� �� A�L��v��� � � � � L��vr�� �� � � detL �� �� Thus a linear operator is

invertible� and hence an isomorphism� i� its determinant is nonzero�

��CHAPTER �� ALTERNATING TENSORS� DETERMINANTS� ORIENTATION� ANDN�DIMENSIONAL RIGHT�HANDEDNESS

If detL �� �� then L�� exists� and L �L�� � IV � so �detL��detL��� � det�L �L��� �det IV � �� Thus

det�L��

�� �detL��� � �����

Finally we claim

detLT � detL� ������

Recall that LT is the unique member of L�V V � �

L ��u� � �w � LT ��w� � �u� ��u� �w � V�

Let � x�� � � � � xn� be an ordered orthonormal basis for V � Relative to this basis B� the

matrix of LT is

�LT�i

j � cjBhLT

��bi�i

� xj �hLT � xi�

i� xi � L � xj�

� ciB �L �xj�� � Lji� and

detLT � det�LT�i

j ��LT��

j� � � ��LT�n

jn�j����jn

� Lj�� � � � Ljn

n�j����jn

�X��Sn

L���� � � � L�n�

n � sgn �� �

Now for any � � Sn�

Lij �

nL���

�� � � � � L�n�no

�� i � ��j��� j � ����i��� Lij �

nL�

������ � � � � Ln���n�

o�

Hence� nL���

�� � � � � L�n�no�nL������ � � � � � L���n�

n

o

���� ORIENTATION ��

so�

L���� � � � L�n�

n � L������ � � � Ln

���n��

Moreover� sgn � � sgn ���� Hence

detLT �X��Sn

L������� � � � � Ln

���n��sgn � ��

�� ������

Now suppose f � Sn W where W is any vector space� We claim

X��Sn

f����

��

X��Sn

f ��� � �����

The reason is that as � runs over Sn so does ���� The mapping � ��� is a bijection of

Sn to itself� But from ������ and ������

detLT �X��Sn

L���� � � � Ln

�n� sgn �

� L�i� � � � Ln

in�i����in � detL�

If L � "�V � �see page D��� then L�� � LT so L � LT � IV � Then �detL��detLT � �

det IV � �� so �detL�� � �� Thus

detL � �� if V � "�V ��

��� Orientation

Two ordered orthonormal bases for V � � x�� � � � � xn� and � x��� � � � � x�n�� are said to have

the same orientation i� one basis can be obtained from the other by a continuous rigid

motion� That is� there must be for each t in � � t � � an ordered orthonormal basis

� x��t�� � � � � xn�t�� with these properties�

i� xi��� � xi

ii� xi��� � x�i

iii� xi�t� depends continuously on t�

��CHAPTER �� ALTERNATING TENSORS� DETERMINANTS� ORIENTATION� ANDN�DIMENSIONAL RIGHT�HANDEDNESS

Since � x��t�� � � � � xn�t� is orthonormal� there is an Lt � "�V � � xi�t� � Lt� xi�� From

i� and iii� above�

i�� L� � IV

iii�� Lt depends continuously on t �i�e�� its components relative to any basis depend

continously on t� For example� its components relative to � x�� � � � � xn� are Lij �

xj � Lt� xi� � xj � xi�t�� and these are continuous in t��

Then detL� � � and detLt depends continuously on t� But Lt � "�V � for all t� so

detLt � �� for all t� Hence detLt � � and detL� � �� Thus�

Remark ������ If two ordered o�n� bases for V have the same orientation� the orthogonal

operators L which maps one basis onto the other has detL � ��

The converse of remark ��� is also true� as is clear from the comment on page D��

and from Theorem PP���� We will not prove that comment here� It is obvious when

dimV � � and for dimV � it is Euler�s theorem�

If two ordered orthonormal bases do not have the same orientation� we say they have

opposite orientations� The orthogonal operator L which maps one basis onto the other has

detL � ��� Given any ordered orthonormal basis B � � x�� � � � � xn�� if two other ordered

orthonormal bases are oriented oppositely to B� they have the same orientation� Proof�

Let the bases be B� and B��� Let L� and L�� be the orthogonal operators which

mapB ontoB� and B��� Then L����L���� is the orthogonal operator which maps

B� onto B��� and det�L����L����� � �detL��� det ��L����� � �detL����detL���� �

�������� � ���

Thus� we can divide the ordered orthonormal bases for V into two oppositely oriented

classes� Within each class� all bases have the same orientation� We call these two classes

the �orientation classes� for V � Choosing an orientation for V amounts to choosing one

of the two orientation classes for V �

It turns out that orientations are easily described in terms of alternating tensors�

���� ORIENTATION ��

De�nition ������ We call A � �nV �unimodular� if

jA � x�� � � � � xn�j � � �����

for every ordered orthonormal basis � x�� � � � � xn� of V �

If A and A� are both unimodular� neither is �� so there is a real nonzero a such that

A� � aA� Applying ����� to both A and A� for one particular � x�� � � � � xn� shows a � ���Thus

A� � �A ����

if A and A� are unimodular� Hence

Remark ������ �nV contains at most two unimodular members� Does it contain any$

To see that the answer is yes� we prove

Remark ������ Let B � � x�� � � � xn� and B� � � x��� � � � x�n� be any two ordered orthonor�

mal bases for V � Let AB be de�ned by ������� Then AB�B�� �an abbreviation for

AB� x��� � � � � x

�n�� is �� or �� according as B and B� have this same or opposite orienta�

tions� Therefore� AB and �AB are unimodular�

Proof�

There is an L � "�V � such that x�i � L� xi�� Then AB� x��� � � � � x

�n� � AB�L� x���� � � � L� xn��

� �detL�AB� x�� � � � � xn�� But detL � ��� and AB�B� � �� by ������ QED�

If A and �A are the two unimodular members of �nV � we see from remark �� that

A is positive on one of the two orientation classes of V � and �A is positive on the other

orintation class� Choosing an orientation class for V amounts simply to choosing one of

the two unimodular members of �nV � Therefore we accept

De�nition ������ An oriented Euclidean vector space is an ordered pair �V�A� in which

V is Euclidean vector space and A is one of the two unimodular alternating tensors over

V � A is called �the� alternating tensor of the oriented space� An ordered orthonormal

basis � x�� � � � � xn� is �positively� or �negatively� oriented according as A� x� � � � � xn� is ��

or ���

��CHAPTER �� ALTERNATING TENSORS� DETERMINANTS� ORIENTATION� ANDN�DIMENSIONAL RIGHT�HANDEDNESS

Chapter �

Tensor Products

�� De�nition of a tensor product

De�nition ����� Suppose U�� � � � � Up� V�� � � � � Vq are Euclidean vector spaces and P �U� � � � � � Up and Q � V� � � � � � Vq� De�ne PQ as the member of F�U� � � � � �Up � V� � � � � � Vq R� which assigns to any �p � q��tuple �u�� � � � � �up� �v�� � � � �vq� �U� � � � �� Up � V� � � � �� Vq the real number

PQ ��u�� � � � � �up� �v�� � � � � �vq� � P ��u�� � � � � �up�Q ��v�� � � � � �vq� � �������

By de�nition ������� it is clear that PQ is multilinear� so

PQ � U� � � � �� Up � V� � � � �� Vq� ������

PQ is called the �tensor product� of P andQ� Note that if P were inM�U��� � ��Up W � and Q were in M�V� � � � � � Vq W �� with W �� R� W � �� R� then PQ could not

be de�ned� because the product on the right of ������� would not be de�ned� Tensors can

be multiplied by one another� but in general multilinear mappings cannot� The tensor

product PQ is sometimes written P �Q�

�� Properties of tensor products

�� CHAPTER �� TENSOR PRODUCTS

Remark ������ Commutativity can fail� PQ need not equal QP � In fact� these two

functions usually have di�erent domains� namely U� � � � � � Up � V� � � � � � Vq and

V� � � � � � Vq� U� � � � �� Up� Even if their domains are the same� it is unusual to have

PQ � QP � For example� suppose p � q � � and U� � V� � V � Then P and Q are

vectors in V � say P � �u� Q � �v� For any �x and �y � V we have P ��x� � �u � �x� Q��y� � �v � �yso �PQ� ��x� �y� � P ��x�Q ��y� � ��u � �x� ��v � �y�� Similarly

�QP � ��x� �y� � Q ��x�P ��y� � ��v � �x� ��u � �y� �

If PQ � QP � then �PQ���x� �y� � �QP ���x� �y� for all �x� �y � V � so

��u � �x� ��v � �y� � ��v � �x� ��u � �y� �

Then �x � ��u ��v � �y�� � �x � ��v ��u � �y�� so

�x � ��u ��v � �y�� �v ��u � �y�� � �� ������

If we �x �y to be any particular vector in V � then ������ holds for all �x � V � Therefore

�u ��v � �y�� �v ��u � �y� � ��� �����

This holds for every �y � V � so it holds for �y � �u� If �u �� �� then �u��u �� � so �u��v��u����u��u��v �

�� shows that �u and �v are linearly dependent� And if �u � ��� then of course �u and �v are

also linearly dependent� Thus PQ � QP implies �u and �v are linearly dependent� The

converse is obvious�

Remark ������ There are no divisors of zero� That is if P � U� � � � � � Up and Q �V� � � � � � Vp then

PQ � � �� P � � or Q � �� �����

To prove this� suppose PQ � � and Q �� �� We will show P � �� Choose ��v� � � � � �vq� �V� � � � � � Vq so that Q��v�� � � � � �vq� �� � �by hypothesis Q �� � so this is possible�� Then

for any ��u�� � � � � �up� � U� � � � � � Up� PQ � � implies P ��u�� � � � � �up�Q��v�� � � � � �vq� � ��

Cancelling Q��v�� � � � � �vq�� which is �� �� gives P ��u�� � � � � �up� � �� Since �u�� � � � � �up were

arbitrary� P � ��

���� PROPERTIES OF TENSOR PRODUCTS �

Corollary ������ P� � � �Ps � � �� one of P�� � � � � Ps � �� �Induction on s gives proof��

Remark ������ Associative law� If P � U� � � � � � Up� Q � V� � � � � � Vq� and

R � W� � � � � �Wr then

�PQ�R � P �QR� � ������

Hence� we can write PQR without ambiguity�

Proof�

For any u � ��u�� � � � � �up� � U�� � � ��Up� any v � ��v�� � � � � �vq� � V�� � � ��Vq�and any w � ��w�� � � � � �wr� � W� � � � � �Wr� we have

��PQ�R� � u� v� w� �� ��PQ�� u� v��R� w�

�� P � u� ��Q v�R� w�� �rule of arithmetic in R�

�� P � u� ��QR�� v� w��

�� �P �QR�� � u� v� w��

Remark ����� Right and left distributive laws� Suppose a�� � � � � am � R� P�� � � � � Pm �U� � � � � � Up� and Q � V� � � � � � Vq� Then

�aiPi�Q � ai�PiQ� and Q�aiPi� � ai�QPi�� ������

Proof�

For any u � ��u�� � � � � �up� � U��� � ��Up and any v � ��v�� � � � � �vq� � V��� � ��Vqwe have

��aiPi�Q� � u� v� � ��aiPi�� u��Q� v�

� fai �Pi� u��gQ� v�

� ai �Pi� u�Q� v�� �real arithmetic�

� ai ��PiQ�� u� v�� � �ai�PiQ�� � u� v��

The second of equations ������ is proved in the same way�

�� CHAPTER �� TENSOR PRODUCTS

Remark ����� Multilinearity� For i � f�� � � � � sg� suppose Pi � Ui�� � � � � � U i�

pi�

Then P�P� � � �Ps is separately linear in each of P�� P�� � � � � Ps�

Proof�

To prove linearity in Pr� write P � P� � � �Pr��� Q � Pr� R � Pr�� � � �Ps� �If

r � �� P � �� if r � s� R � ��� Then we want to show that PQR depends

linearly on Q� We take Q�� � � � � Qm � Ur�� � � � � � U r�

pr and a�� � � � � am �R� Then P �aiQi�R � �by ���� P ��aiQi�R� �by ����� � P �ai�QiR�� �

�by ����� � ai �P �QiR�� � �by remark ���� � ai�PQiR��

�� Polyads

De�nition ����� If ��v�� � � � � �vq� � V�� � � �� Vq� where V�� � � � � Vq are Euclidean spaces�

then the tensor product �v��v� � � ��vq is called a polyad of order q� It is a member of

V� � � � � � Vq� It is a �dyad� if q � � a �triad� if q � � a �tetrad� if q � �� a �pentad�

if q � �� By the de�nition ������ of a tensor product� for any ��x�� � � � � �xq� � V� � � � � � Vq

we have

��v� � � ��vq� ��x�� � � � � �xq� � ��v� � �x�� � � � ��vq � �xq� � ������

By remark ������� �v��v� � � ��vq depends linearly on each of �v�� � � � � �vq when the others are

�xed� We de�ne a mapping

a� � � V��� � � � ��Vq V� � � � � � Vq

by requiring that for any ��v�� � � � � �vq� � V� � � � � � Vq we have

b� � ��v�� � � � � �vq� � �v��v� � � ��vq�Then � is multilinear� That is

c� � �M �V� � � � � � Vq� � V� � � � � � Vq� � �����

The tensor product PQ is sometimes written P �Q� and �v��v� � � ��vq is written �v� � �v� �� � � � �vq� This notation makes ����b� look thus� ���v�� � � � � �vq� � �v� � � � � � �vq� We will

usually avoid these extra ��s�

���� POLYADS ��

Remark ����� �u��u� � � ��uq � � i� at least one of �u�� �u�� � � � � �uq is ��

Proof�

�� follows from remark ����� and the multilinearity of � in ������ ��follows by induction on q from remark �������

Remark ������ Suppose for each p � f�� � � � � qg that �up� �vp � Vp and ap � R� and

�vp � ap��up�� Suppose also a� � � �aq � �� Then �u� � � ��uq � �v� � � ��vq�

Proof�

Since � is multilinear� �v��v� � � ��vq � �a��u���a��u�� � � � �aq��uq�� �ha�a� � � �aq�

i�u��u� � � ��uq � �u�� �u� � � ��uq�

Remark ������ Suppose that for each p � f�� � � � � qg� we have �up� �vp � Vp� Suppose also

that �u� � � ��uq � �v� � � ��vq �� �� Then �a�� � � � � aq � R such that a�a� � � �aq � � and for each

p � f�� � � � � qg we have �vp � ap��up��

Proof�

By remark ����� no �up or �vp is ��� By hypothesis� for any ��x�� � � � � �xq� �V� � � � � � Vq we have

��u� � �x�� � � ���uq� � �xq�

�� ��v� � �x�� � � �

��vq� � �xq�

�� �����

Choose any p � f�� � � � qg� and take �xr � �vr if r �� p� Then for any �xp � Vp

����� implies

�vp� � �xp� � ��u� � �v�� � � � ��up��� � �vp������up� � �xp����up��� � �vp���� � � � ��uq� � �vq����v� � �v�� � � � ��vp��� � �vp������vp��� � �vp���� � � � ��vq� � �vq��

or

�vp� � �xp� � ap��up� � �xp� ������

where

ap� ���u� � �v�� � � � ��up��� � �vp������up��� � �vp���� � � � ��uq� � �vq����v� � �v�� � � � ��vp��� � �vp������vp��� � �vp���� � � � ��vq� � �vq�� � ������

�� CHAPTER �� TENSOR PRODUCTS

Since ������ is true for all �xp � Vp� it follows that

�vp � ap��up��

Then �v� � � ��vq � �a� � � �aq���u� � � ��uq�� But �v� � � ��vq � �u� � � ��uq by hypothesis�

so �a� � � �aq � ����u� � � ��uq� � �� Since �u� � � ��uq �� �� a� � � �aq � � � �� or

a� � � �aq � � �see facts about ��� p�s��

Chapter �

Polyad Bases and Tensor Components

�� Polyad bases

It is true �see exercise �� that some tensors of order � are not polyads� However� every

tensor is a sum of polyads� We have

Theorem ������ For p � f�� � � � � qg� suppose Vp is a Euclidean space andBp � ��bp�� � � � � ��bp�np �

is an ordered basis for Vp� Then the n�n� � � �nq polyads �b��i��b��i� � � ��bq�iq are a basis for

V� � � � � � Vq�

Proof�

First these polyads are linearly independent� Suppose that Si����iq is an n� �n� � � � � nq dimensional array of scalars such that

Si����iq�b��i� � � ��bq�iq � �� �������

We want to prove Si����iq � �� Equation ������� is equivalent to the assertion

that for any ��x�� � � � � �xq� � V� � � � � � Vq�

Si����iq��b��i� � �x�

�� � �

��bq�iq � �xq

�� �� ������

Let BDp � ��b�p�� � � � ��bnpp�� be the dual basis for Bp� Choose a particular q�tuple

of integers �j�� � � � � jq� and set �x� � �bj���� � � � � �xq � �bjqq� in ������� The result is

Si����iq��b��i� ��bj���

����bq�iq ��bjqq�

�� ��

�� CHAPTER �� POLYAD BASES AND TENSOR COMPONENTS

But for dual bases� ��bi ��bj� � �ij� so

Si����iq�i�j� � �i

jq � ��

Hence� Sj����iq � �� as we hoped�

Second� the n�n� � � �nq polyads �b��i� � � ��bq�iq span V� � � � � � Vq� For let T �

V� � � � � � Vq� De�ne

T i����iq �� T��bi���� � � � ��biqq�

�������

and

% �� T i����iq�b��i� � � ��bq�iq � �������

Clearly % is a linear combination of the polyads �bi�i� � � ��bq�iq � We will show that

T � %� Since �BD� � � � � � BD

q � is a basis sequence for V� � � � � � Vq� by remark

����� it su!ces to show that

%��bj���� � � � ��bjqq�

�� T

��bj���� � � � ��bjqq�

��

But

%��bj���� � � � ��bjqq�

�� T i����iq

h��b��i� � � ��bq�iq

� ��bj���� � � � � ��bq�iq

�i� T i����iq

��b��i� ��bj���

�� � �

��bq�iq ��bjqq�

�� T i����iq�i�

j� � � � �iq jq

� T j����iq � T��bj���� � � � ��bjqq�

��

QED�

Corollary ������ If T � V��� � ��Vq and p � f�� � � � � qg then T is a sum of n� � � �nq�nppolyads� where n� � dimV�� � � � � nq � dimVq�

Proof�

From T � % and ������� we can write

T � �b��i� � � ��b p���ip�� �v i����ip��ip�����iq�b

p���ip��

� � ��b q�iq

where

�vi����ip��ip�����iqp� � T i����iq�b

p�ip � Vp�

���� COMPONENTS OF A TENSOR RELATIVE TO A BASIS SEQUENCE� DEFINITION ��

Corollary ������ Suppose V�� � � � � Vq are Euclidean spaces and W is a real vector space�

not necessarily Euclidean� Suppose L � V� � � � � � Vq W and M � V� � � � � � Vq W

are linear� and L�T � � M�T � for every polyad T � Then L � M �

Proof�

For an arbitrary T � V� � � � � � Vq� write T �PN

j � Tj where the Tj are

polyads� Then L�T � �PN

j � L�Tj� �PN

j �M�Tj� � M�T � because L and M

are linear�

�� Components of a tensor relative to a basis

sequence� De�nition

De�nition ����� For each p � f�� � � � � qg� suppose Vp is a Euclidean space and Bp �

��bp�� � � � � ��bp�np � is an ordered basis for Vp� with dual basis BD

p � ��b�p�� � � � ��bnpp��� Suppose

T � V� � � � � � Vq� Then T has q arrays of components relative to the basis sequence

�B�� � � � � Bq� for V� � � � � � Vq� Each array has dimension n� � � � � � nq� The arrays are

as follows� �������������������������������������������������������

T i����iq � T��bi���� � � � ��biqq�

T i����iq��iq � T

��b��

i�� � � � ��bq��� iq�� ��bq�iq

���

Ti����iq��iq � T

��b��i� � � � � ��biq�� q����bq�

iq�

Ti����iq � T��b��i� � � � � ��b q�iq

������

An array is said to be �covariant� in its subscripts and �contravariant� in its super�

scripts� The �rst array in ������ is the array of �contravariant components of T re�

lative to �B�� � � � � Bq��� The last array is the array of �covariant components relative

�� CHAPTER �� POLYAD BASES AND TENSOR COMPONENTS

to �B�� � � � � Bq��� The other q � arrays are arrays of �mixed components relative

to �B�� � � � � Bq��� Note that each array in the list ������ is the array of contravariant

components of T relative to a suitable basis sequence� For example� T i����iq��iq is the

contravariant array relative to the basis sequence �B�� � � � � Bq��� BDq �� and Ti����iq is the

contravariant array relative to �BD� � B

D� � � � � � BD

q �� Similarly� each array is the covariant

array relative to some basis sequence�

From ������ and ������� and the fact that T � % in those equations� we conclude

T � T i����iq�b��i��b��i� � � ��bq�iq � �����

That is� if T � V� � � � � � Vq� then the coe!cients required to express T as a linear

combination of the basis polyads from the basis sequence �B�� � � � � Bq� are precisely the

contravariant components of T relative to that basis sequence� By considering all q basis

sequences� �B�� � � � � Bq�� �B�� � � � � Bq��� BDq �� � � � � �BD

� � � � � � BDq � we obtain from ����� the

q equations �������������������������������������������������������

T � T i����iq�b ��i� � � ��b q�iq

T � T i����iq��iq�b��i� � � ��bq��iq��

�biqq�

���

T � Ti����iq��iq � �bi��� � � ��b iq��q����

�bq�iq

T � Ti����iq�bi��� � � ��b iqq��

�����

Note ����� Suppose T � T i����iq�b��i� � � ��bq�iq for some array T i����iq � Since the polyads

�b��i� � � ��b q�iq are a basis for V� � � � � � Vq� therefore T i����iq � T i����iq � By using others of the

q basis sequences one can reach the same conclusion for any of the q equations ������

Note ����� If S and T � V� � � � � � Vq and there is one basis sequence for V� � � � � � Vq

relative to which one of the q component arrays of S is the same as the corresponding

component array of T �e�g� Ti�i�i����iq � Si

i� i� � � � iq� then S � T �

���� CHANGING BASES ��

Note ����� The notation has been chosen to permit the restricted index conventions� A

double index is summed only when it is a superscript at one appearance and a subscript

at the other� The equation ai � bi holds for all possible values of i� but ai � bi should

not occur unless we introduce orthonormal bases�

Note ����� If each of the bases B�� � � � � Bq is orthonormal� then Bp � BDp � and all q

component arrays ����� are identical� and all q equations ����� are the same� One

uses the non�restricted index convention�

�� Changing bases

Suppose T � V��� � ��Vq� If we know the array of contravariant components of T relative

to one basis sequence �B�� � � � � Bq�� we know T � Therefore� we ought to be able to calculate

the contravariant components of T relative to any other basis sequence � B�� � � � � Bq�� The

procedure is this� Let

Bp ���bp�� � � � � ��b p�np

�� Bp �

� �bp�

� � � � � � �bp�

np

BDp

�� �b�p�� � � � ��bnpp�

�� BD

p �

� �b�

p�� � � � � �bnp

p�

��

The array of contravariant components of T relative to � B�� � � � � Bq� is

T j����jq � T

� �bj�

��� � � � � �bjq

q�

��T i����iq�b

��i� � � ��b q�iq

�� �b j���� � � � � �b jqq��

� T i����iq

���b��i� � � ��b q�iq

�� �b j���� � � � � �bjqq���

� T i����iq��b��i� ��b j���

�� � �

� �bq�

iq � �bjq

q�

��

Thus

T j����jq � T i����iq

��b��i� � �b

j�

��

�� � �

��bq�iq � �b

jq

q�

�� ������

�� CHAPTER �� POLYAD BASES AND TENSOR COMPONENTS

If we replace some of the bases in �B�� � � � � Bq� and � B�� � � � � Bq� by their duals� we

can immediately obtain ������ with some of the j�s as subscripts and some of the i�s as

subscripts on the components and superscript on the �b�s�

For example�

Tj�j����jq � T i����iq��

iq

��b��i� � �b

��

j�

���b��i� � �b

j�

��

�� � �

� � ���bq���iq�� �

�bjq��

q���

���biqq� � �b

jq

q�

������

As an interesting special case of ������� we can take for � B�� � � � � Bq� the basis sequence

obtained from �B�� � � � � Bq� by replacing some of the Bp by BDp � For example� if B� � BD

and Bp � Bp for p � � ������ becomes

Tj�j����jq � T i�j����jqg

��i�j� �����

where g��ij � �b ��i ��b ��j is the covariant metric matrix of B�� Similarly� if � B�� � � � � Bq� �

�BD� � � � � � BD

q �� ������ becomes

Tj����jq � T i����iqg��i�j� � � �gq�iqjq � ������

Formulas like ����� and ������ �there are �q such� are said to raise or lower the indices

of the component array of T �

� Properties of component arrays

Remark ������ If S� T � V��� � ��Vq and a� b � R� then relative to any basis sequence

for V� � � � � � Vq�

�aS � bT �i����iq � aSi����iq � bT i����iq �

Proof�

Immediate from de�nition of components� ������

���� SYMMETRIES OF COMPONENT ARRAYS �

Remark ������ Suppose P � U�� � � ��Up and Q � V�� � � ��Vq� Then relative to any

basis sequence for U� � � � � � Up � V� � � � � � Vq

�PQ�i����ipj����jq � P i����ipQj����jq �

Proof�

Immediate from de�nition of components� ������

Remark ������ Suppose T � V� � � � � � Vq and �vp� � Vp� p � �� � � � � q� Taking compon�

ents relative to any basis sequence gives

T��v��� � � � � �vq�

�� T i����iqv

��i� � � � vq�iq �

Proof�

�v p� � vp�ip�bipp� so T ��v

��� � � � � �v q�� � T�v��i�

�b i���� � � � � vq�iq�biqq�

�� v

��i� � � � vq�iq T

��bi���� � � � ��b iqq�

�� v

��i� � � � vq�iq T

i����iq �

� Symmetries of component arrays

If V� � � � � � Vq � V � it is usual to consider only basis sequences �B�� � � � � Bq� in which

B� � � � � � Bq � B� This is not necessary� but it does help if one wants to study the

symmetries of T � V� � � � � � Vq � �qV by looking at its arrays of components� If

B� � � � � � Bq � B� then one speaks of the arrays of components of T relative to the

basis B�

If we do have B� � � � � � Bq � B � ��b�� � � � ��bn�� and if T � �qV � then for any

permutation � � Sq we have

��T �i����iq � ��T ���b i� � � � � ��b iq

�� T

�b i���� � � � � ��b i��q�

�� T i�������i��q� � �������

�� CHAPTER �� POLYAD BASES AND TENSOR COMPONENTS

Here BD � ��b�� � � � ��bn� is the basis dual to B� For mixed arrays of components� things

are not quite so simple� For example� if �rs� is a transposition with r s then

��rs�T �i����ir���is�� is is�����iq � ��rs�T ���b i� � � � � ��b is�� ��bis ��b is��� � � ��b iq

�� T

��b i� � � � � ��b ir����bis��b ir��� � � � ��b is����bir ��bis�� � � � ��biq

�� T i����ir��

isir�����is��iris�����iq �

Whether T � �qV is symmetric or antisymmetric under �rs� can be tested by looking

at one array of components of T relative to one basis� as long as the r and s indices in that

array are both covariant or both contravariant� If one index is covariant and the other is

contravariant� two di�erent component arrays must be compared to test for symmetry�

For example� if T � V � V antisymmetric� then relative to any basis for V �

T ij � �T ji� T ij � �Tj i� Tij � �Tji� ������

Proof�

T ��bi��bj� � �T ��bj��bi� � T ��bi��bj� � �T ��bj��bi� � T ��bj��bi� � �T ��bj ��bi� �

Moreover� if any one of the three equations ������ is true relative to one basis B for

V � then T is antisymmetric� For example� if T ij � �Tj i then T and ����T take the

same values on the basis sequence �BD� B� for V � V � Therefore� by remark ������

T � ����T �

�� Examples of component arrays

Example ����� Let I be the identity tensor on Euclidean vector space V � By de�nition�

I is that member of V � V such that for any �u��v � V we have

I��u��v� � �u � �v�

Let B � ��b�� � � � ��bn� be any ordered basis for V � with dual basis BD � ��b�� � � � ��bn�� Let

gij � �bi ��bj and gij � �bi ��bj be the covariant and contravariant metric matrices relative to

���� EXAMPLES OF COMPONENT ARRAYS ��

B� Then

Iij � I��bi��bj

�� �bi ��bj � gij

Ii j � I��bi��bj

�� �bi ��bj � �i j

Ii j � I��bi��b

j�� �bi ��bj � �i

j

Iij � I��bi��bj

�gij�

�The reason for calling I the �identity tensor� will appear later�� Relative to an orthonor�

mal basis� any component array of I is �ij�

From ����� it follows that

I � gij�bi�bj � �i j�bi�bj � �i

j�bi�bj � gij�bi�bj�

Thus�

I � gij�bi�bj � �bi�bi � �bi�bi � gij�b

i�bj� �������

In particular� if � x�� � � � � xn� is an orthonormal basis for V �

I � xi xi ������

Example ����� Let V � B� BD be as in example ����� Let A be any alternating tensor

over V �i�e� any member of �nV �� Then from �����

Ai����in � A��b i� � � � � ��b in

�� �i����inA

��b �� � � � ��bn

�� ������

Ai����in � A��bi� � � � ��bin

�� �i����inA

��b�� � � ��bn

�� �������

It is not true that

Ai�i����in � �i�

i����inA��b���b

�� � � � ��bn��

If �V�A� is an oriented Euclidean vector space� then A is unimodular� If B �

� x�� � � � � xn� is an orthonormal basis for V then from �������

Ai����in � �i����in if � x�� � � � � xn� is positively oriented�

Ai����in � ��i����in if � x�� � � � � xn� is negatively oriented�

�� CHAPTER �� POLYAD BASES AND TENSOR COMPONENTS

Chapter

The Lifting Theorem

We must learn to perform a number of simple but useful operations on tensors of order q�

Most of these operations will be easy to perform on polyads� so we express T � V��� � ��Vqas a sum of polyads and perform the operations on the individual polyads� adding the

results� The procedure works quite well once we have overcome a di!culty which is best

understood by considering an example�

Suppose � � r s � q and V�� � � � � Vq are Euclidean vector spaces and Vr � Vs� For

any T � V� � � � � � Vq� we want to de�ne the �trace on indices r and s�� written trrsT �

First� suppose P is a polyad in V� � � � � � Vq� that is

P � �u� � � ��ur � � ��us � � ��uq� �������

Then we de�ne

trrsP �� ��ur � �us� �u� � � � ��ur � � � ��us � � ��uq�� ��ur � �us� �u� � � ��ur���ur�� � � ��us���us�� � � ��uq� ������

If T � V� � � � � � Vq� we write

T �mXi �

Pi� Pi a polyad ������

and we de�ne

trrsT �mXi �

trrsPi� �������

There are two serious objections to this apparently reasonable procedure�

��

�� CHAPTER �� THE LIFTING THEOREM

�i� Suppose P � �u� � � ��ur � � ��us � � ��uq � �v� � � ��vr � � ��vs � � ��vq � Is it true that ��ur ��us��u� � � � � �ur � � � � �us � � ��uq � ��vr � �vs��v� � � � � �vr � � � � �vs � � ��vq $ If not� trrsP is not

uniquely de�ned by �������

�ii� Suppose T �Pm

i � Pi �Pn

j �Qj where Pi and Qj are polyads� Is it true thatPmi � trrsPi �

Pnj � trrsQj$ If not� trrsT is not uniquely de�ned by �������

To attack this problem� we return to the very �rst sloppiness� in ������� We hope that

trrsP will be a unique polyad in V�� � � �� �Vr� � � �� �Vs� � � ��Vq� so that trrs maps the

set of polyads in V��� � ��Vq into the set of polyads in V��� � �� �Vr�� � �� �Vs�� � ��Vq�But di!culty i� above leads us to look carefully at ������� and ������� and to recognize

that until we have provided some theory all we have really done in ������ is to show

how to take an ordered q�tuple ��u�� � � � � �uq� and assign to it a polyad of order q � in

V� � � � �� �Vr � � � �� �Vs � � � � � Vq� The polyad assigned to ��u�� � � � � �uq� is

M ��u�� � � � � �uq� � ��ur � �us� �u� � � � ��ur � � � ��us � � ��uq� �������

This equation does clearly and unambiguously de�ne a function M � F�V� � � � � � Vq V� � � � �� �Vr � � � �� �Vs � � � � � Vq�� In fact� from de�nition �������� property d� on

page D��� for dot products� and remark ���� for tensor products� it is clear that M is

multilinear� Hence�

M � M�V� � � � � � Vq V� � � � �� �Vr � � � �� �Vs � � � � � Vq�� �������

This M is all we really have� The process of constructing a linear mapping trrs � V��� � ��Vq V��� � �� �Vr�� � �� �Vs�� � ��Vq from the M of ������� is a very general one� That

such a construction is possible and unique is the intent of the �lifting theorem�� This

theorem holds for all tensors� not just those over Euclidean spaces� and it is sometimes

�as in Marcus� Multilinear Algebra� taken as the de�nition of V� � � � � � Vq�

A formal statement of the lifting theorem for Euclidean vector spaces is as follows�

Theorem ������ �Lifting theorem�� Let V�� � � � � Vq be Euclidean vector spaces and let

W be any real vector space� Let M � M�V�� � � � � Vq W �� Then there is exactly one

��

x

xx x. . .

. . . x x

V VM

W

MVV q

q1

1

Figure ����

M � L�V� � � � � � Vq W � such that

M � M � � �������

where � �M�V� � � � � � Vq V� � � � � � Vq� is de�ned by ���v�� � � � � �vq� � �v� � � ��vq�

Before proving the theorem� we discuss it and try to clarify its meaning� To make

������� less abstract we note that it is true ��

M ��v�� � � � � �vq� � M �� ��v�� � � � � �vq�� � � ��v�� � � � � �vq� � V� � � � � � Vq�

This is equivalent to

M ��v�� � � � � �vq� � M ��v� � � ��vq� � �������

In other words� the lifting theorem asserts the existence of exactly one linear mapping

M � V� � � � � � Vq W such that when M is applied to any polyad the result is the

same as applying M to any q�tuple of vectors whose tensor product is that polyad�

A diagram of the three functions may clarify matters� M maps V� � � � � � Vq into W �

and � maps V� � � � � � Vq into V� � � � � � Vq� Both mappings are multilinear� The lifting

theorem �lifts�M from V��� � ��Vq to V��� � ��Vq by producing a unique linear mapping

M � V� � � � � � Vq W which satis�es ������� or �������� These equations mean that if

we start at any ��v�� � � � � �vq� in ��V� � � � � � �Vq�� and go to W either via M or via � and

M� we will reach the same vector in W � Usually� when people draw the diagram as in

Figure �� they mean f � U W � g � U V � h � V W � and f � h � g�

Proof of the lifting theorem�

The proof is actually rather simple� Choose a basis sequence �B�� � � � � Bq� for V��� � ��Vq� with Bp � ��bp�� � � � � ��bp�np �� Let B

Dp � ��b�p�� � � � ��bnpp�� be the basis for Vp dual to Bp� First

�� CHAPTER �� THE LIFTING THEOREM

U

WVh

fg

Figure ���

we will prove thatM is unique� i�e� there can be no more than oneM � L�V��� � ��Vq W � which satis�es ������� or �������� SupposeM is such a mapping� Let T � V��� � ��Vq�By ������ T � T i����iq�b

��i� � � ��bq�iq so M�T � � T i����iqM��b

��i� � � ��bq�iq � because M is linear�

Then M�T � � T i����iqM��b��i� � � � � ��b q�iq � from ������� with �v� � �b

��i� � � � � � �vq � �b

q�iq � Thus

if such an M exists� then for any T � V� � � � � � Vq we must have

M�T � � T��bi���� � � � ��biqq�

�M��b��i� � � � � ��bq�iq

�� �������

In other words� M is determined because M�T � is known for all T � V� � � � � � Vq�

It still remains to prove that there is an M � L�V� � � � � � Vq W � which satis�es

������� and �������� Our uniqueness proof has given us an obvious candidate� namely

the M � V� � � � � � Vq W de�ned by �������� This M is certainly a well de�ned

function mapping V� � � � � � Vq into W � By the de�nitions of ad and sc in V� � � � � � Vq�

M�T � depends linearly on T � i�e� M � L�V� � � � � � Vq W �� �For a formal proof�

let T�� � � � � Tn � V� � � � � � Vq and a�� � � � � an � R� Then we want to show M�ajTj� �

ajM�Tj�� From �������� and the de�nition of ajTj� and the rules of vector arithmetic in

W � given on page D���� we compute

M

�ajTj

��

�ajTj

� ��b i���� � � � ��b iqq�

�M��b��i� � � � � ��b q�iq

��

najhTj��b i���� � � � ��b iqq�

�ioM��b��i� � � � � ��b q�iq

�� aj

hTj��b i���� � � � ��b iqq�

�M��b��i� � � � � ��b q�iq

�i� ajM�Tj��

QED��

��

It remains only to prove that M satis�es �������� Let ��v�� � � � � �vq� � V� � � � � � Vq�

Then if T � �v� � � ��vq we have T��b i���� � � � ��b iqq�

�� ��v� ��b i���� � � � ��vq ��b iqq�� � v i�� � � � v iqq�� so

M��v� � � ��vq� � vi�� � � � viqq�M��b��i� � � � � ��bq�iq �� By hypothesis� M is multilinear� so

M��v� � � ��vq� � M�vi�� �b��i� � � � � � v iqq �b q�iq �� But this is just �������� QED�

As an application of the lifting theorem� let us return to trrsT � The M de�ned in

������� will be the M of the lifting theorem� The vector space W in that theorem will be

W � V� � � � �� �Vr � � � �� �Vs � � � � � Vq� Then� M � L�V� � � � � � Vq V� � � � ��� Vr � � � �� � Vs � � � � � Vq� is given to us by the lifting theorem� and we de�ne for any

T � V� � � � � � Vq

trrsT �� M�T �� ��������

If a T is a polyad� T � �v� � � ��vq� then

trrs ��v� � � ��vq� � M ��v� � � ��vq� � M ��v�� � � � � �vq�� ��vr � �vs��v� � � � ��vr � � � ��vs � � ��vq� ��������

Thus the lifting theorem answers both the objections on page ���

We agree to de�ne

trsrT �� trrsT� �������

As an application of this �machinery�� let us �nd the component arrays of tr��T

relative to the basis sequence �B�� � � � � Bq� for V� � � � � � Vq� We have

T � Ti�i�i����iq�bi���

�b��i��b��i� � � ��bq�iq �

We must suppose V� � V� if tr��T is to be de�ned� Then we take B� � B�� so �bi�� �

�bi���

Since tr�� is a linear mapping�

tr��T � Ti�i����iq tr��

h�b i���

�b��i��b��i� � � ��b q�iq

i� �������

By ��������

tr��h�bi���

�b��i��b��i� � � ��bq�iq

i���bi��� ��b��i�

��b��i� � � ��bq�iq �

We have agreed to take B� � B�� so �bi��� ��b��i� � �i� i� � Thus�

tr��h�bi���

�b��i��b��i� � � ��bq�iq

i���i� i��b

��i� � � ��bq�iq

�� ��������

�� CHAPTER �� THE LIFTING THEOREM

Then� substituting this in ��������

tr��T � Tjji����iq�b��i� � � ��bq�iq �

so

�tr��T �i����iq � Tj

ji����iq � ��������

Similarly

�tr��T �i����iq � T j

ji����iq ��������

is proved from

T � T i�i�i����iq�b ��i�

�b i����b��i� � � ��b q�iq �

Recall that for orthonormal bases we need not distinguish between superscripts and

subscripts� Thus� if all bases are orthonormal and B� � B��

�tr��T �i����iq � Tjji����iq � ��������

Chapter

Generalized Dot Products

�� Motivation and de�nition

Suppose U � V � and W are Euclidean vector spaces� and �u � U � �x � V � �y � V � �w � W �

Then �u�x � U�V and �y �w � V �W � What could be more natural than to de�ne �following

Willard Gibbs�

��u�x� � ��y �w� �� �u ��x � �y� �w� �������

which is the same as ��u�x� � ��y �w� �� ��x � �y� �u�w$ Then the de�nition can be extended so

that if P � U � V and R � V �W we de�ne P �R by writing

P �mXi �

Pi� R �nXj �

Rj ������

where Pi and Rj are dyads� then we de�ne

P �R ��mXi �

nXj �

Pi �Rj ������

where Pi �Rj is de�ned by �������� It is in U �W � as is P �R�Does all this work$ If �u�x � �u ��x � and �y �w � �y � �w �� is it true that ��x � �y� �u�w �

��x � � �y �� �u � �w �$ And if P �Pm

i � Pi �Pm �

i � P�i and R �

Pnj �Rj �

Pn �

j �R�j� is it true

thatPm

i �

Pnj � Pi �Rj �

Pm�

i �

Pn�

j � P�i �R�

j$

The lifting theorem will show that all of this works� and that it can even be generalized

as follows� Suppose U�� � � � � Up� V�� � � � � Vq� W�� � � � �Wr are Euclidean spaces� and �ui � Ui�

��

�� CHAPTER � GENERALIZED DOT PRODUCTS

�xj and �yj � Vj� �wr � Wr� We would like to de�ne a generalized dot product of order q as

��u� � � ��up�x� � � ��xq� hqi ��y� � � ��yq �w� � � � �wr�

�� ��u� � � ��up� ��x� � �y�� � � � ��xq � �yq� ��w� � � � �wr�

� ��x� � �y�� � � � ��xq � �yq� ��u� � � ��up �w� � � � �wr� � �������

Then for any P � U��� � ��Up�V��� � ��Vq and any R � V��� � ��Vq�W��� � ��Wr

we would like to de�ne P hqiR � U� � � � � � Up �W� � � � � �Wr by writing ������ and

P hqiR ��mXi �

nXj �

PihqiRj� �������

�Incidentally� note the mnemonic device that

��u� � � ��up�x� � � ��xq� hqi ��y� � � ��yq �w� � � � �wr�

�� �u� � � ��up�BBBBB��x�

��xq

CCCCCA� � �

� � �

�BBBBB��y�

��yq

CCCCCA�w� � � � �wr

� �������

Some authors de�ne

��u� � � ��up�x� � � ��xq� hqi ��yq � � ��y� � � � �w� � � � �wr�

��

��xq � �yq����

���

��x� � �y���u� � � ��up ��x� � �y� � �w� � � � �wr

� ��xq � �yq� � � � ��x� � �y�� �u� � � ��up �w� � � � �wr�

De gustabus non disputandum est� but I think the �rst de�nition has more convenient

properties��

To prove that these de�nitions are unambiguous� we de�ne

M � M�U� � � � � � Up � V� � � � � � Vq � V� � � � � � Vq �W� � � � � �Wr U� � � � � � Up �W� � � � � �Wr�

��� MOTIVATION AND DEFINITION ��

by requiring

M ��u�� � � � � �up� �x�� � � � � �xq� �y�� � � ��yq� �w�� � � � � �wr� ��

� ��x� � �y�� � � � ��xq � �yq� �u� � � ��up �w� � � � �wr� �������

Since M is multilinear� the lifting theorem provides a unique

M � L�U� � � � � � Up � V� � � � � � Vq � V� � � � � � Vq �W� � � � � �W U� � � � � � Up �W� � � � � �Wr�

satisfying

M ��u� � � � � � �up � �x� � � � � � �xq � �y� � � � � � �yq � �w� � � � � � �wr�

� M ��u�� � � � � �up� �x�� � � � � �xq� �y�� � � � � �yq� �w�� � � � � �wr�

� ��x� � �y�� � � � ��xq � �yq� �u� � � ��up �w� � � � �wr �������

for all

��u�� � � � � �up� �x�� � � � � �xq� �y�� � � � � �yq� �w�� � � � � �wr� �

� U� � � � � � Up � V� � � � � � Vq � V� � � � � � Vq � V� � � � � � Vq �W� � � � � �Wr�

For any P � U�� � � ��Up� V�� � � ��Vq and any R � V�� � � �� Vq�W�� � � ��Wr�

we de�ne

P hqiR �� M�PR�� �������

Then if P and R are polyads� ������� says we do have ��������

We can calculate P hqiR for any tensors and P and R by writing them as sums of

polyads and doing the arithmetic justi�ed by

Remark ����� Suppose P�� � � � � Pm � U� � � � � � Up � V� � � � � � Vq and R�� � � � � Rn �V� � � � � � Vq �W� � � � � �Wr and a�� � � � � am� b�� � � � � bn � R� Then

�aiPi� hqi �bjRj� � aibj �Pi �Rj� ��������

Proof�

�� CHAPTER � GENERALIZED DOT PRODUCTS

�aiPi� hqi �bjRj� � M ��aiPi� �bjRj�� by de�nition of hqi� M �aibj�PiRj�� by multilinearity of the tensor product

� aibjM�PiRj� because M is linear

� aibj�PihqiRj� by de�nition of hqi�

Equation �������� is true for any Pi and Rj� but when Pi and Rj are polyads�

�������� and ������� give a convenient way to calculate P hqiR for any tensors

P and R�

�� Components of P hqiR

Suppose P � U��� � ��Up�V��� � ��Vq and R � V��� � ��Vq�W��� � ��Wr� Suppose

��b��� � � � � ��bi�mi

� is an ordered basis for Ui� and ���j�� � � � � ��j�nj

� is an ordered basis for Vj� and

���R�� � � � ���R�Pk

� is an ordered basis for Wk� We would like to compute the components of

P hqiR relative to the basis sequence for U� � � � � � Up �W� � � � � �Wr� We have

P � P i����ipj����jq�b��i� � � ��b p�ip

����j� � � � �� q�

jq

R � Rk����kql����lr �� k�

�� � � � ��kqq��� ��l�� � ��� r�lr

���kpp� are the dual basis vectors to ��

p�kp �� Then ������� gives

P hqiR � P i����ipj����jqRk����kql����lr

�����j� � �� k�

��

�� � �

���q�jq � ��kqq�

��b��i� � � ��b p�ip

����l�� � ��r�lr

� P i����ipj����jqRk����kql����lr� k�j� � � � � kqjq

����i� � � ��b p�ip �

��l�� � ��� r�lr

But Rk����kql����lr�j�

k� � � � �jq kq � Rj����jql����lr so

P hqiR ��P i����ipj����jqRj����jq

l����lr��b��i� � � ��b p�ip

����l�� � ��� r�lr �

Therefore� by note �������

�P hqiR�i����ipl����lr � P i����ipj����jqRj����jql����lr � ������

��� PROPERTIES OF THE GENERALIZED DOT PRODUCT ��

It is important to remember that for ������ to hold� the same bases for V�� � � � � Vq �and

their dual bases� must be used to calculate the components of P and R�

Covariant or contravariant metric matrices can be used to raise or lower any index i or

l or pair j in ������� The same collection of p�q�r formulas can be obtained immediately

from ������ by choosing to regard certain of the original bases for the Ui� Vj or Wk as

dual bases� the duals becoming the original bases� For example�

�P hqiR�i� i����ipl����lr��lr �

� Pi�i����ipj�

j����jgRj�j����jql����lr��

lr

All the p�q�r variants of ������ collapse into a single formula when all the bases are

orthonormal�

�� Properties of the generalized dot product

Property ���� Bilinearity� For i � f�� � � � � mg and j � f�� � � � � ng� suppose Pi �U� � � � � � Up � V� � � � � � Vq and Rj � V� � � � � � Vq �W� � � � � �Wr and ai� bj � R�

Then �aiPi�hqi�bjRj� � aibj�PihqiRj��

This has already been proved as remark ��������

Property ���� Suppose P � U� � � � � � Up � V� � � � � � Vq and R � V� � � � � � Vq �W� � � � � �Wq� Then if P � � or R � �� P hqiR � ��

Proof�

Suppose P � �� Then �P � P so P hqiR � ��P �hqiR � ��P �R� � ��

Note that the converse is false� We can have P hqiR � � even though P �� � and R �� ��

For example �uh�i�v � �u � �v and this can vanish even though �u �� �� and �v �� ���

De�nition ����� Since �uh�i�v � �u � �v for vectors� it is usual to write P h�iR as P � R�P hiR as P � R� P hiR as P

�� � R and P h�iR as P � � R� The notation P�� � R for P h�iR

seems not to be used�

�� CHAPTER � GENERALIZED DOT PRODUCTS

Property ���� P h�iR � PR�

Proof�

If P and R are polyads� this is obvious from ������� with q � �� Both P h�iRand PR depend linearly on R� so if R �

PRj where the Rj are polyads� and

if P is a polyad� we have P h�iR � P h�i�Pj Rj� �P

j�P h�iRj� �P

j�PRj� �

P �P

j Rj� � PR� Therefore P h�iR � PR if P is a polyad and R is any

tensor� If P �P

i Pi where the Pi are polyads� then for any tensor R� P�R �

�P

i Pi�h�iR �P

i�Pih�iR� �P

i�PiR� � �P

i Pi�R � PR� Thus P h�iR � PR

for any tensors P and R�

Property ���� P hqiR and RhqiP need not be the same even if both are de�ned� This is

obvious from property ��� and the corresponding result for tensor products� An example

with q � � is this� Let �u��v� �w� �x � V � Let P � �u�v� R � �w�x� Then P �R � ��v � �w��u�x and

R � P � ��x � �u��w�v� If these two dyads are equal and not �� remark ������ shows that �u

must be a multiple of �w and �x must be a multiple of �v�

Property ���� �Associativity�� Suppose

P � U� � � � � � Up � V� � � � � � Vq

R � V� � � � � � Vq �W� � � � � �Wr �X� � � � � �Xs

T � X� � � � � �Xs � Y� � � � � � Yt�

Then

P hqi �RhsiT � � �P hqiR� hsiT� ������

Proof�

First we prove ������ when P� R� T are polyads� In that case ������ follows

from ������� by simple computation� which we leave to the reader� That

������ holds for polyads can also be seen immediately from the mnemonic

diagram �������� The generalized dot products hqi and hsi are formed by

��� PROPERTIES OF THE GENERALIZED DOT PRODUCT ��

{ {

r

s

t

{

p { {q

P

R

T

Figure ����

dotting vertical pairs of vectors in the regions of overlap of P � R� T � It is

clearly irrelevant to the outcome� whether we dot �rst the q vectors in P with

those below them in R or the s vectors in R with those below them in T �

Next� suppose P � R and T�� � � � � Tn are polyads and T �P

i Ti� Then P hqi�RhsiT � �P hqi �Rhsi�Pi Ti�� � P hqi �Pi�RhsiTi�� �

Pi �P hqi�RhsiTi�� �

Pi �P hqi�RhsiTi�� �

�P hqiR� hsi �PTi� � �P hqiR� hsiT� This proves ������ when P and R are

polyads and T is any tensor�

Next suppose P � R�� � � � � Rn are polyads and T is any tensor and R �P

iRi�

Then P hqi�RhsiT � � P hqi ��PiRi�hsiT � � P hqi �Pi�RihsiT �� � Pi �P hqi�RihsiT �� �P

i ��P hqiRi�hsiT � � �P

i�P hqiRi�� hsiT � �P hqi�EiRi�� hsiT � �P hqiR�hsiT �Thus ������ holds when P is a polyad and R and T are any tensors�

Finally� suppose P�� � � � � Pn are polyads and R and T are any tensors� and

P �P

i Pi� Then P hqi�RhsiT � � �P

i Pi�hqi�RhsiT � �P

i �Pihqi�RhsiT �� �

�� CHAPTER � GENERALIZED DOT PRODUCTS

P

R

T

Figure ���

Pi ��PihqiR�hsiT � � �

Pi�PihqiR�� hsiT � ��

Pi Pi�hqiR� hsiT � �P hqiR�hsiT �

This proves ������ for any tensors P � R� T �

Figure ����� also shows how associativity can fail� If P and T overlap as

in �gure ����� there is trouble� The tensors P hqi�RhsiT � and �P hqiR� hsiTmay be di�erent even if both are de�ned �they may not be�� As an example�

suppose all tensors are over a single space V � Let �u��v� �w� �x� �y�� V � Let

P � �u�v� R � �w� T � �x�y� Then we have the diagram

P � ��

R � �

T � ��

which exhibits the fatal P � T overlap shown in Figure ����� We have P ��R � T � � �u�v � ���w � �x��y� � ��w � �x� ���u�v� � �y� � ��w � �x���v � �y��u and �P � R� � T �

��u��v � �w�� � ��x�y� � ��v � �w� ��u � ��x�y�� � ��v � �w���u � �x��y� If we choose �x and �y to

be mutually perpendicular unit vectors� and take �v � �y� �u � �w � �x� we have

P � �R � T � � �u � �x and �P �R� � T � ���

Property ���� Suppose P � R � V� � � � � � Vq� Then

P hqiR � RhqiP� �����

��� PROPERTIES OF THE GENERALIZED DOT PRODUCT ��

v = y

u = w = x

Figure ���

Proof�

If P � �x��x� � � ��xq and R � �y��y� � � ��yq then P hqiR � ��x� � �y�� � � � ��xq � �yq� �

��y� � �x�� � � � ��yq � �xq� � RhqiP � so ����� is true of P and R are polyads� But

both RhqiP and P hqiR are bilinear in P and R� so the proof of ����� for

general tensors P and R in V� � � � � � Vq is like the proof of �������

Property ��� Suppose P � V� � � � � � Vq and P �� �� Then

P hqiP � �� �����

Proof�

For i � f�� � � � � qg� let � xi�� � � � � � xi�ni � be an orthonormal basis for Vi� By

�������

P hqiP � Pi����iqPi����iq � ������

This is a sum of squares of real numbers� It is � � unless every term is �� i�e� Pi����iq � ��

In that case P � ��

Theorem ���� On the vector space V� � � � � � Vq� de�ne the dot product dp�P�R� �

P hqiR for all P � R � V� � � � � � Vq� With this dot product� V� � � � � � Vq is a Euclidean

vector space�

Proof�

�� CHAPTER � GENERALIZED DOT PRODUCTS

Properties ������� ������ ����� show that dp satis�es a� b�� c�� d� on page D�

���

Corollary ����� Let � xi�� � � � � � xi�ni � be an orthonormal basis for Vi� i � �� � � � � q� Then

the n�n� � � �nq polyads x��i� x

��i� � � � x��iq are an orthonormal basis for V� � � � � � Vq�

Proof�

By theorem ������ they are a basis� To prove them orthonormal� we note

� x��i� � � � xq�iq �hqi� x��j� � � � xq�jq � � � x��i� � x��j� � � � � � xq�iq � xq�jq � � �i�j��i�j� � � � �iqjq �

De�nition ����� If P � V� � � � � � Vq� de�ne the �length� of P � kPk� in the way one

usually does for vectors in a Euclidean space� That is� kPk is the non�negative square

root of P hqiP �

Remark ����� Suppose P � U� � � � � � Up � V� � � � � � Vq and R � V� � � � � � Vq �W� � � � � �Wr� Then

kP hqiRk � kPkkRk� ������

Proof�

Introduce an orthonormal basis for each space Ui � Vj and Wk� Take compon�

ents relative to these bases� Abbreviate the p�tuple of integers �i�� � � � ip� as I�the q�tuple �j�� � � � � jq� as J � the r�tuple �l�� � � � lr� as L� Then ������ can be

abbreviated

�P hqiR�IL �XJ

PIJRJL�

By the ordinary Schwarz inequality for real numbers��XJ

PIJRJL

��

��X

J

P �IJ

��XK

R�KL

where K stands for the q�tuple �k�� � � � � kq�� �We have simply changed the

summation index from J to K�� Summing over I and L gives

XI�L

�XJ

PIJRJL

�����X

I�J

P �IJ

A��XK�L

R�KL

A �

��� APPLICATIONS OF THE GENERALIZED DOT PRODUCT �

. . .

. . .

u u1 p

T

v1 vq

Figure ����

or XI�L

�P hqiR��IL ���X

I�J

P �IJ

A��XK�L

R�KL

A �

By equation ������� this last inequality is

kP hqiRk� � kPk�kRk��

Taking square roots gives �������

� Applications of the generalized dot product

Application �� The value of a tensor�

Suppose T � U� � � � � � Up � V� � � � � � Vq and ��u�� � � � � �up� �v�� � � � � �vq� � U� � � � � �Up � V� � � � � � Vq� Then

T ��u�� � � � � �up� �v�� � � � � �vq� � � ��u��u� � � ��up� hpiT hqi ��v��v� � � ��vq� � �������

Note that ������ applies� so no parentheses are needed to say whether hpi or hqi iscalculated �rst� There is none of the overlap shown in �gure ��� The diagram for �������

is shown in Figure ���� Therefore� at least ������� is unambiguous� To prove it true�

suppose �rst that T is a polyad� T � �x� � � ��xp�y� � � ��yq� Then

��u� � � ��up� hpi T hqi ��v� � � ��vq� � ���u� � � ��up� hpiT � hqi ��v� � � ��vq�

�� CHAPTER � GENERALIZED DOT PRODUCTS

� ���u� � �x�� � � � ��up � �xp� �y� � � ��yq� hqi ��v� � � ��vq�� ��u� � �x�� � � � ��up � �xp� ���y� � � ��yq� hqi ��v� � � ��vq��� ��x� � �u�� � � � ��xp � �up� ��y� � �v�� � � � ��yq � �vq�� T ��u� � � � � �up� �v�� � � � � �vq� �

This proves ������� when T is a polyad� But both sides of ������� depend linearly on T �

so if ������� is true for polyads T � it is true when T is a sum of polyads� That is� �������

is true for any tensor T � U� � � � � � Up � V� � � � � � Vq�

Corollary ���� Suppose P is a �xed tensor in U�� � � � � Up� V�� � � � � Vq� with the

property that P hqiR � � for every R � V��� � ��Vq�W��� � ��Wr� Then P � �� �Also�

if R is a �xed tensor in V� � � � � � Vq �W� � � � � �Wr with the property that P hqiR � �

for every P � U� � � � � � Up � V� � � � �Vq then R � � ��

Proof�

Choose a �xed nonzero T � W��� � ��Wr� For any ��v�� � � � � �vq� � V��� � ��Vq�we know by hypothesis that P hqi ���v� � � ��vq�T � � �� Figure ���� works� so

�P hqi��v� � � ��vq��T � �� By remark ������� P hqi��v� � � ��vq� � �� Then for any

��u�� � � � � �up� � U� � � � � � Up we have ��u� � � ��up�hpi�P hqi��v� � � ��vq�� � �� By

equation �������� P ��u�� � � � � �up� �v�� � � � � �vq� � �� Since ��u�� � � � � �up� �v�� � � � � �vq� isan arbitrary member of U� � � � � � Up � V� � � � � � Vq� therefore P � �� �The

other half of the corollary� in parentheses� is proved similarly�

Corollary ���� Suppose P � P � � U� � � � � � Up � V� � � � � � Vq and for every R �V� � � � � � Vq �W� � � � � �Wr we have P hqiR � P �hqiR� Then P � P �� �Also� if R�

R� � V� � � � � � Vq �W� � � � � �Wr and for every P � U� � � � � � Up � V� � � � � � Vq we

have P hqiR � P hqiR�� then R � R���

Application �� Vector cross product�

Suppose �V�A� is an oriented three dimensional Euclidean space whose alternating

tensor is A� For any �u� �v � V � we de�ne their A�cross product as

�uA� �v � Ahi ��u�v� � ������

��� APPLICATIONS OF THE GENERALIZED DOT PRODUCT ��

Usually we omit the A and write simply �u� �v� but it should always be remembered that

there are two oriented �spaces� �V�A� and �V��A�� and so two ways to de�ne �u� �v�

We will now show that our cross product is the usual one� First� if ai� bj � R and �ui�

�vj � V for i � �� � � � � m and j � �� � � � � n

�ai�ui�� �bj�vj� � aibj ��ui � �vj� � ������

For �ai�ui�� �bj�vj� � Ahi ��ai�ui��bj�vj�� � Ahi ��aibj�ui�vj�� � aibj �Ahi��ui�vj�� � aibj��ui��vj�� Second� �w � ��u� �v� � �w � �Ahi��u�v�� � A��w� �u��v� � A��u��v� �w� so

��u� �v� � �w � A ��u��v� �w� � �������

Therefore� since A is totally antisymmetric��������u� �v� � �w � ��v � �w� � �u � ��w � �u� � �v� � ��v � �u� � �w � � ��u� �w� � �v � � ��w � �v� � �u�

�������

In particular� ��u� �v� � �w � ���v � �u� � �w for all �w� so

�v � �u � ��u� �v� �������

Therefore� �u � �u � ��� Setting �w � �u in ��u� �v� � �w � ��w � �u� � �v gives ��u � �v� � �u �

��u� �u� � �v � �� Thus� using ��������

��u� �v� � �u � ��u� �v� � �v � �� �������

Relative to any ordered basis ��b���b���b�� for V we have

��u� �v�i � Aijkujvk � A

��b���b���b�

��ijku

jvk�

If the ordered basis is positively oriented and orthonormal� A��b���b���b�� � �� so

��u� �v�i � �ijkujvk� �������

If the ordered basis is negatively oriented and orthonormal� A��b���b���b�� � �� so

��u� �v�i � ��ijkujvk� �������

�� CHAPTER � GENERALIZED DOT PRODUCTS

The notorious formula

�u� ��v � �w� � ��u � �w��v � ��u � �v� �w ��������

can be derived from ������� as follows� relative to any positively oriented orthonormal

basis

��u� ��v � �w��i � �ijkuj��v � �w�k

� �ijkuj�klmvlwm

� �ijk�klmujvlwm

� ��il�jm � �im�jl� ujvlwm see page D��

� ujwjvi � ujvjwi

� ��u � �w� vi � ��u � �v�wi � ���u � �w��v � ��u � �v� �w�i �

Since the vector on the left of ������� has the same components as the vector on the right�

those two vectors are equal�

From ������� and �������� we have ��u��v����u��v� � �u���v � ��u� �v�� � �u��k�vk��u� ��v � �u��v� �k�uk�k�vk� � ��u � �v��� so

k�u� �vk� � k�uk�k�vk� � ��u� �v��� ��������

We have de�ned the angle � between �u and �v as the � in � � � � � such that �u � �v �

k�ukk�vk cos �� With this de�nition of �� �������� implies

k�u� �vk � k�ukk�vk sin �� �������

Equations ������� and ������� determine that �u � �v is ��u and �v and has length

k�ukk�vk sin �� This leaves two possibilities for a nonzero �u��v� The correct one is determ�

ined by the fact that

A ��u��v� �u� �v� � � if �u� �v �� �� �������

To prove ������� we note that A��u��v� �u � �v� � A��u � �v� �u��v� � ��u � �v� � �Ahi��u�v�� ���u� �v� � ��u� �v�� Inequality ������ is usually paraphrased by saying that ��u��v� �u� �v is

a positively oriented ordered triple�� If we orient the space we live in by choosing A so

��� APPLICATIONS OF THE GENERALIZED DOT PRODUCT ��

Figure ����

that our right thumb� index �nger and middle �nger are positively oriented ordered triple

of vectors when extended as in Figure ������ we will obtain the usual de�nition of �u� �v

in terms of the right hand screw rule�

Application �� Let U�� � � � � Up� V�� � � � � Vq be Euclidean spaces� Then

U� � � � � � Up � V� � � � � � Vq � �U� � � � � � Up�� �V� � � � � � Vq� � ��������

Proof�

What is meant is that there is a natural isomorphism between the spaces

on the two sides of ��������� Before proving this� we note that the tensor

product of U� � � � � � Up and V� � � � � � Vq on the right in �������� is de�ned�

because U� � � � � � Up and V� � � � � � Vq are Euclidean spaces� Their dot

products are hpi and hqi� To construct the isomorphism whose existence

is asserted by ��������� let T � U� � � � � � Up � V� � � � � � Vq and de�neeT � �U��� � ��Up���V��� � ��Vq� R by requiring for any P � U��� � ��Up

and any Q � V� � � � �Vq that

eT �P�Q� � P hpiT hqiQ� ��������

�� CHAPTER � GENERALIZED DOT PRODUCTS

Then eT is bilinear� so eT � �U� � � � � � Up� � �V� � � � � � Vq�� From ���������g�ajTj� � aj eTj� so T � eT is a linear mapping of U� � � � � � Up � V� � � � � � Vq

into �U��� � ��Up���V��� � ��Vq�� We claim this mapping is bijective� To see

that it is injective� suppose T� � T� � U��� � ��Up�V��� � ��Vq and eT� � eT��Then for every P � U��� � ��Up and Q � V��� � ��Vq we have P hpiT�hqiQ �

P hpiT�hqiQ� so P hpi�T�hqiQ�T�hqiQ � � so P hpi��T��T��hqiQ� � �� Hence�

�T� � T��hqiQ � � for all Q � V� � � � � � Vq� Hence� T� � T� � �� To see

that T � eT is a surjection� suppose % � �U� � � � � � Up� � �V� � � � � � Vq��

De�ne T � U��� � ��Up�V��� � ��Vq by requiring T ��u�� � � � � �up� �v�� � � ��vq� �%��u� � � ��up� �v�� � � ��vq�� Then P hpiT hqiQ � %�P�Q� if P and Q are polyads�

Hence P hpiT hqiQ � %�P�Q� for all P � U�� � � ��Up and Q � V�� � � �� Vq�

By ��������� % � eT � Thus T � %� and the mapping T � eT is a surjection of

U� � � � � � Up � V� � � � � � Vq onto �U� � � � � � Up�� �V� � � � � � Vq��

Application �� Linear mappings as second order tensors�

De�nition ����� Suppose U and V are Euclidean vector spaces and L � L�U V ��

De�ne�

L� F�U � V R� by�

L ��u��v� � L��u� � �v� ��������

Then�

L ��� �v� and �

L ��u� �� are linear� so�

L� M�U � V R��

That is��

L� U � V� ��������

Thus� �������� can be written in the equivalent form

�u� �L ��v � L��u� � �v� ��������

For a �xed �u � U � �������� is true for all �v � V � Hence� since �u � �L and L��u� are both in

V � we have

�u� �L� L��u�� ��������

Let � � L�U V � U � V be de�ned by ��L� ��

L� If we know�

L� ��L�� we can

recover L from ��������� so � is an injection� We claim � is also a surjection� That is�

��� APPLICATIONS OF THE GENERALIZED DOT PRODUCT ��

every T � U�V is ��L� for at least one L � L�U V �� Suppose T is given and de�ne L

by L��u� � �u �T � Then �

L ��u��v� � �u� �L ��v � L��u� ��v � �u �T ��v � T ��u��v�� so T ��

L� ��L��

We claim that � � L�U V � U � V is also linear� If L�� � � � � Ln � L�U V � and

a�� � � � � an � R� we claim�

�aiLi�� ai�

Li � �������

The proof is this� for any �u � U and �v � V �

�u��

�aiLi� ��v � ��aiLi���u�� � �v � fai �Li��u��g � �v� ai �Li��u� � �v� � ai��u�

Li ��v�� �u �

�ai

Li

�� �v�

Thus ��

aiLi���u��v� � �ai�

L���u��v� for all �u � U � �v � V � Hence ��������

We have now shown that � is a linear bijection between the two vector spaces L�U V � and U � V � We can use � to regard any linear mapping L � U V as a tensor�

L� U � V and vice�versa� The process of taking linear combinations can be done either

to the tensors or to the linear mappings� so confusing tensors with linear mappings does

no harm to linear combinations�

Linear mappings can also be �multiplied� by composition� If K � L�U V � and

L � L�V W � then L � K � L�U W �� Our identi�cation of tensors with linear

mappings almost preserves this multiplication� We have

L �K��

K � �L � �������

The proof is simple� For any �u � U �

�u��

�L �K� � �L �K���u� � L �K��u��

� K��u�� �L� ��u � �K�� �L� �u � � �K� �L� by �������

Since this is true for all �u � U � corollary �������� gives ��������

If IU � U U is the identity operator on U �that is� IU��u� � �u for all �u � U� then for

any �u�� �u� � U ��

I U ��u�� �u�� � IU ��u�� � �u� � �u� � �u�� ������

�� CHAPTER � GENERALIZED DOT PRODUCTS

That is��

I U is what we have already called the identity tensor on U � see example �����

The origin of that name is now clear� And ������ can also be written

�u���

I U ��u� � �u� � �u�� ������

This is true for all �u� and �u� � U � so we have for any �u � U

�u� �I U��

I U ��u � �u� �������

If L � U V then L � IU � IV � L � L� It follows then from ������� that

I U ��

L��

L � �I V��

L � �������

Actually� a more general result is true� If U � V � W�� � � � � Wq are Euclidean spaces and

T � U �W� � � � � �Wq � V then

I U �T � T � �I V� T� �������

The proof follows the usual lines� Since all three expressions in ������� are linear in T � it

su!ces to prove ������� for polyads T � But for T a polyad� ������� follows immediately

from associativity and ��������

Thus��

I U acts like a multiplicative identity� Then some second order tensors can have

inverses�

De�nition ����� If L � L�U V � has an inverse L�� � L�V U�� then we de�ne��

L

���to be

L���

Thus� using �������� we conclude

L�� � �L��

I V ��

L ��

L����

I U � �������

Since the tensors in U � V and the linear mappings in L�U V � can be thought of

as essentially the same objects� any concept de�ned for one can be de�ned immediately

for the other� De�nition ���� is an example of this process� Other examples are below�

De�nition ����� If�

L� V � V then det�

L�� detL�

��� APPLICATIONS OF THE GENERALIZED DOT PRODUCT ��

From de�nition ���� and ����� it is apparent that for any�

L� V � V ��

L�� exists i�

det�

L �� ��

De�nition ����� If�

L� U � V then��

L

�T��

LT � Thus�

LT� V � U �

Here LT is the transpose of L � U V � as de�ned on page D��� From the de�nition

of LT we have��

L

�T��v� �u� �

LT ��v� �u� � LT ��v� � �u � �v � L��u� � L��u� � �v ��

L ��u��v�� Thus

LT� ����

L � �������

It is also true that for any ��u��v� � U � V � �v��

LT ��u ��

LT ��v � �u� ��

L ��u��v� � �u� �L ��v�Thus �v �

� �LT ��u� �u� �L

�� �� This is true for that for any �v � V so for all �u � U

LT ��u � �u� �L � �������

Similarly���v�

LT � �

L ��v�� �u � � for all �u � U so if �v � V

�v� �LT��

L ��v� �������

Application �� Linear mappings as higher order tensors�

Suppose U�� � � � � Up� V�� � � �Vq are Euclidean vector spaces and L � L�U�� � � ��Up V� � � � �Vq�� Both U� � � � � � Up and V� � � � � � Vq are Euclidean vectors spaces� with

dot products hpi and hqi� Therefore� L corresponds to a second order tensor�

L in �U� �� � � � Up� � �V� � � � � � Vq�� exactly as in ��������� That is if P � U� � � � � � Up and

Q � V� � � � � � Vq then�

L �P�Q� � L�P �hqiQ

or

P hqi �L hqiQ � L�P �hqiQ� �������

Here hpi and hqi are to be interpreted as the dot products on U��� � ��Up and V��� � ��Vq�and

L is a second order tensor in the space �U�� � � ��Up�� �V�� � � �� Vq�� which is the

tensor product of two Euclidean spaces�

�� CHAPTER � GENERALIZED DOT PRODUCTS

However� as we saw on page ����

L can also be interpreted as a tensor of order p � q

in U� � � � � � Up � V� � � � � � Vq� Equation ������� continues to hold� but now hpi andhqi are interpreted as generalized dot products� and

L� U� � � � � � Up � V� � � � � � Vq�

Therefore� the linear mappings L � L�U� � � � � � Up V� � � � � � Vq� can be viewed as

tensors�

L� U��� � ��Up�V��� � ��Vq� If we know L� we �nd�

L from ������� by taking

P and Q to be arbitrary polyads� If we know�

L� U� � � � � � Up � V� � � � � � Vq� then we

know it in �U� � � � � � Up�� �V� � � � � � Vq� so we �nd L by appealing to ��������� That

is� for any P � U� � � � � � Up�

L�P � � P hpi �L � ������

The generalization from application � to application � is particularly useful in con�

tinuum mechanics�

Chapter �

How to Rotate Tensors �and why

��� Tensor products of linear mappings

For j � f�� � � � � qg� suppose Vj andWj are Euclidean vector spaces and Lj � L�Vj Wj��

We would like to de�ne a linear mapping

L� � � � � � Lq � V� � � � � � Vq W� � � � � �Wq

by requiring that for any polyad �v� � � ��vq � V� � � � � � Vq we have

�L� � � � � � Lq� ��v� � � ��vq� � L� ��v�� � � �Lq ��vq� � �������

The lifting theorem will show that there is exactly one such L� � � � � � Lq� To �nd it� we

invoke that theorem� We de�ne

M � V� � � � � � Vq W� � � � � �Wq

by requiring for any ��v�� � � � � �vq� � V� � � � � � Vq that

M ��v�� � � � � �vq� � L���v�� � � �Lq��vq��

Since L�� � � � � Lq are linear� M is multilinear� so we can use the lifting theorem with M as

above and W � W��� � ��Wq� The picture is shown in ���� Then we de�ne L��� � ��Lq

to be the M provided by the lifting theorem� With this choice� L� � � � � � Lq is linear

and satis�es ��������

��

�� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

x

x

x x x x. . .

. . . x x

. . . V VM

W W1 q 1 q

V VM

1 q

Figure ����

Remark ���� Suppose��bj�� � � � � ��bj�nj

�is an ordered basis for Vj andQ � Qi����iq�b��i� � � ��bq�iq

� V� � � � � � Vq� Then

�L� � � � � � Lq� �Q� � Qi����iqhL�

��b��i�

�� � �Lq

��bq�iq

�i� ������

Proof�

L��� � ��Lq is linear� so �L��� � ��Lq��Q� � Qi����iqh�L� � � � � � Lq�

��b��i� � � ��bq�iq

�i�

Now use ��������

Remark ���� Suppose for j � f�� � � � � qg that Uj� Vj� Wj are Euclidean and Kj �L�Vj Wj� and Lj � L�Uj Vj�� Then

�K� � � � � �Kq� � �L� � � � � � Lq� � �K� � L��� � � � � �Kq � Lq� � ������

Proof�

We want to show for every P � U� � � � � � Uq that

��K� � � � � �Kq� � �L� � � � � � Lq�� �P � � ��K� � L��� � � � � �Kq � Lq�� �P ��

By linearity it su!ces to prove this when P is any polyad �u� � � ��uq� But

��K� � � � � �Kq� � �L� � � � � � Lq�� ��u� � � ��uq�� �K� � � � � �Kq� ��L� � � � � � Lq���u� � � ��uq��� �K� � � � � �Kq� �L���u�� � � �Lq��uq��

� K� �L��u���K� �L��u��� � � �Kq �Lq�uq�� ��K� � L����u�� ��K� � L����u���

� � � ��Kq � Lq���uq��

� ��K� � L��� � � � � �Kq � Lq�� ��u� � � ��uq��

���� TENSOR PRODUCTS OF LINEAR MAPPINGS ��

Another way to try to understand L� � � � � � Lq is to exhibit explicitly for any P �V� � � � � � Vq the tensor �L� � � � � � Lq��P � � W� � � � � � Wq by evaluating it at any

��w�� � � � � �wq� � W� � � � � �Wq� We have

Remark ���� Suppose that for j � f�� � � � � qg we have Lj � L�Vj Wj� for Euclidean

spaces Vj and Wj� Suppose P � V� � � � � � Vq and ��w�� � � � � �wq� � W� � � � � �Wq� Then

��L� � � � � � Lq��P �� ��w�� � � � � �wq� � PhLT� ��w��� � � � � LT

q ��wq�i� �������

Proof�

Both sides of this equation are linear in P � so it su!ces to prove the equation

when P is a polyad� P � �v� � � ��vq� In that case�

��L� � � � � � Lq��P �� ��w�� � � � � �wq� � �L���v�� � � �Lq��vq�� ��w�� � � � � �wq�

� �L���v�� � �w�� � � � �Lq��vq� � �wq�

�h�v� � LT

� ��w��i� � �

h�vq � LT

q ��wq�i

� PhLT� ��w��� � � � � LT

q ��wq�i�

Finally� it will be useful to understand how permutations a�ect tensor products of

mappings� To do this we need

Lemma ����� Suppose V�� � � � � Vq are Euclidean spaces� �vj � Vj for j � f�� � � � � qg� and� � Sq� Then

� ��v� � � ��vq� � �v������� � ��v���

�q�� �������

Proof�

From the de�nition ���������

�� ��v� � � ��vq����w������ � � � � �w���q�

�� ��v� � � ��vq� ��w�� � � � � �wq�

� ��v� � �w�� � � � ��vq � �wq�

���v����� � �w�����

�� � �

��v����q� � �w���q�

��

��v����� � � ��v���q�

� ��w������ � � � � �w���q�

�for any ��w�� � � � � �wq� � V� � � � � � Vq� QED�

� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

Now we can prove

Remark ����� For j � f�� � � � � qg� suppose Lj � L�Vj Wj�� and suppose � � Sq�

Then

� �L� � � � � � Lq� ��L����� � � � � � L���q�

��� �������

Proof�

The domain of L� � � � � � Lq is V� � � � � � Vq� We want to prove that if

P � V�� � � ��Vq then ���L�� � � ��Lq���P � � ��L������ � � ��L���q�����P ��

i�e�� ���L� � � � � � Lq���P � � �L����� � � � � � L���q����P �� Both sides of the

last equation are linear in P � so it su!ces to prove that equation when P is a

polyad� say P � �v� � � ��vq� In this case

� ��L� � � � � � Lq� ��v� � � ��vq�� � � �L���v�� � � �Lq��vq��

� L�����

��v�����

�� � �L���q�

��v���q�

�by �����

��L����� � � � � � L���q�

� ��v����� � � ��v���q�

��

�L����� � � � � � L���q�

��� ��v� � � ��vq�� by �����

QED�

Special Case�

Suppose V� � � � � � Vq � V and W� � � � � � Wq � W and L� � � � � � Lq � L� Then

we write L� � � � � � Lq as �qL� Then

�qL � L ��qV �qW � � �������

and �qL is de�ned by requiring for all ��v�� � � � � �vq� � xqV that

�qL��v� � � ��vq� � L��v�� � � �L��vq�� �������

If P � �qV � ������� leads to the �gure of speech that ��qL��P � is obtained by applying

L to P � �Really �qL is applied to P ��

���� APPLYING ROTATIONS AND REFLECTIONS TO TENSORS �

If K � L�V W � and L � L�U V � then from ������

��qK� � ��qL� � �q �K � L� � �������

If L � L�V W � and P � �qV and ��w�� � � � � �wq� � xqW then from �������

���qL��P �� ��w�� � � � � �wq� � PhLT ��w��� � � � � LT ��wq�

i� ��������

If L � L�V W � and � � Sq then from �������

���qL� � ��qL��� ��������

��� Applying rotations and re�ections to tensors

Recall that L � L�V V � is �orthogonal� if LT � L��� and �proper orthogonal� if also

detL � �� rather than ��� Recall the de�nitions on page D���

"�V � � set of all orthogonal operators on V

"��V � � set of all proper orthogonal operators on V�

Recall that according to corollary PP��� every proper orthogonal operator is a rotation�

As remarked on page PP��� if L is improper orthogonal �detL � ��� then L is the

product of a rotation and a re�ection�

If L is an orthogonal operator on V andQ � �qV � then two more facts about ��qL��Q�

make it even more reasonable to think of that tensor as the result of applying L to Q�

These facts are remark ����� and its corollary�

Remark ����� If L � "�V �� applying L to tensors over V preserves their generalized

dot products� More precisely suppose p� q� r are any non�negative integers� Suppose

P � �p�qV � R � �q�rV � and L � "�V �� Then

��p�rL��P hqiR� �h��p�qL��P �

ihqi

h��q�rL��R�

i� ������

Proof�

� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

Both sides of this equation are linear in P and R� so it su!ces to prove

the equation when P and R are polyads� say P � �u� � � ��up�x� � � ��xq and R �

�y� � � ��yq�v� � � ��vr � Then ��p�rL��P hqiR� � ��p�rL� ���x� � �y�� � � � ��xq � �yq��u� � � ��up�v� � � ��vr�� ��x� � �y�� � � � ��xq � �yq���p�rL���u� � � ��up�v� � � ��vr�� ��x� � �y�� � � � ��xq � �xq� �L��u�� � � �L��up�L��v�� � � �L��vr���

So far we have not used L � "�V �� Now we use it to set �x � �y � L��x� � L��y��Therefore the above polyad is ��p�rL��P hqiR� �

� �L��x�� � L��y��� � � � �L��xq� � L��yq�� �L��u�� � � �L��up�L��v�� � � �L��vr��� �L��u�� � � �L��up�L��x�� � � �L��xq�� hqi �L��y�� � � �L��yq�L��v�� � � �L��vr���

h��p�qL���u� � � ��up�x� � � ��xq�

ihqi

h��q�rL���y� � � ��yq�v� � � ��vr�

i�

h��p�qL��P �

ihqi

h��q�rL��R�

i�

QED

Corollary ���� The value of a rotated q�tensor at a rotated q�tuple of vectors is the

value of the original tensor at the original q�tuple of vectors� The same is true for re�ec�

tions� in general� if L � "�V � and Q � �qV and ��v� � � � � �vq� � xqV then

���qL� �Q�� �L ��v�� � � � � � L ��vq�� � Q ��v�� � � � � �vq� �

Proof�

���qL� �Q�� �L ��v�� � � � � � L ��vq��

� ���qL� �Q�� hqi �L ��v�� � L ��v�� � � �L ��vq��

� ���qL� �Q�� hqi ���qL� ��v��v� � � ��vq��� Qhqi ��v� � � ��vq� by remark �����

� Q ��v�� � � � � �vq� �

���� PHYSICAL APPLICATIONS �

��� Physical applications

Some physical properties of physical systems are tensors� For example� Ohm�s law in a

crystal says that if an electric �eld �E is applied� the resulting current density �J depends

linearly on �E� This means that there is a second�order tensor�

K such that �J � �E� �K�

This�

K is the �conductivity tensor� of the crystal� In a liquid� �J � �EK where K is a

scalar� so�

K� K�

I � but in a crystal all that can be said in general is that if there is

no magnetic �eld��

K is symmetric ��

KT��

K�� This is Onsager�s theorem for dissipative

thermodynamics�

A second example is an elastic crystal� As we will see later� its local state of deforma�

tion is described by a symmetric second order tensor�� � the strain tensor� Its local state

of stress is described by another symmetric second order tensor�

S � the stress tensor� For

small�� �i�e� k �

� k ������

S is a linear function of�� � so there is a tensor of fourth

order over ordinary �space� the elasticity tensor E� such that�

S��� hiE� Thermody�

namics assures that ���E � ���E � ������E � E� but E has no other symmetries

in general so the dimenson of the space of all possible elasticity tensors is �� �We will

discuss all this in detail later��

A third example is the second moment tensor�

M of a collection of mass points� If

there is a mass mv at position �rv for v � �� � � � � N � then

M ��NXv �

mv�rv�rv� ������

This is the underlying tensor from which the inertia tensor and gravitational second degree

harmonic tensor of a solid body are calculated�

If V is real �space and L � L�V V �� then �subjecting a physical system to the

mapping L� means that we move all the mass points �atoms� which make up the system�

so that a mass point which was originally at �rv is moved to the new location L��rv� � �rv��

L�

If L � "��V �� L subjects the body to a rigid rotation� If L � "�V � but detL � ��� Lsubjects the body to a rigid rotation and also a re�ection in some plane� If L � "�V �

but detL � � then L is the product of a rigid rotation and a symmetric linear operator�

The latter stretches the body by di�erent amounts in di�erent directions� If detL �� a

� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

H

F Br

Cl

Right Handed

H

F Cl

Br

Left Handed

Figure ���

re�ection also occurs�

It is actually possible to imagine subjecting a real body to a re�ection if we allow

ourselves to disassemble and reassemble the body atom�by�atom� For example� iceland

spar �calcite� crystals come in right�handed or left�handed version� Liquids can also be

re�ected� For most� re�ection has no e�ect because the distribution of molecules in the li�

quid has no preferred directions and the molecules either have no right� or left�handedness�

or the right� and left�handed versions are present in equal amounts� A liquid which is

a�ected by re�ection is bromochloro�uoromethane� CHFClBr� At one atmosphere of

pressure� it is liquid between �����C and ��C� the molecule is a tetrahedron with H� F�

Cl� Br at the four vertices and C at the center� It can come in two forms� left handed or

right handed� To re�ect a pot of left�handed CHFClBr means to replace it by a pot of

right�handed CHFClBr�

If we subject a system of mass points mv at positions �rv to the mapping L � L�V V �� where V is real �space� then the second moment tensor

M of that system �see �������

changes to ��L��

M�� To see this� note that the second moment tensor of the deformed

���� PHYSICAL APPLICATIONS �

system is

ML �NXv �

mvL ��rv�L ��rv� �NXv �

mv �� L ��rv�rv�

����L

�� NXv �

mv�rv�rv

�����L

�� �

M

��

If we subject a homogeneously stressed crystal to the mapping L� the stress tensor�

S in

that crystal will not change to ��L��

S�� Rather it will be determined by the elasticity

tensor E of the crystal� These examples show that it is not always obvious how to calculate

the e�ect of subjecting a physical system to a linear mapping L� Some tensor properties

of that system simply have L applied to them� and others do not�

The situation is much simpler if L � "�V �� To see this� suppose �b�� �b�� �b� is any basis

for real �space V � Suppose Q � �qV is a measurable tensor property of a particular

physical system� Then

Q � Qi����iq�bi��bi� � � ��biq � �����

�Measuring� Q means measuring the real numbers Qi����iq � Suppose we apply to the

physical system a mapping L � "�V �� The new physical system will have instead of

Q the tensor QL to describe that particular physical property� Suppose we apply L to

the apparatus we used to measure Qi����iq � and we use this mapped apparatus to measure

the contravariant components of QL relative to the basis L��b��� L��b��� L��b��� We assume

either that no gravity or electromagnetic �eld is present or� if they are� that they are

also mapped by L� Then the experiment on QL is identical to that on Q� except that it

has been rotated and possibly re�ected relative to the universe� As far as we know� the

univese does not care� the laws of nature are invariant under re�ection and rigid rotation�

The universe seems to be isotropic and without handedness� �As a matter of fact� a small

preference for one orientation or handedness exists in the weak nuclear force governing

��decay� Yang� Lee and Wu won Nobel prizes for this discovery� it has no measurable

e�ects at the macroscopic level of continuum mechanics in an old universe�� Therefore�

the numbers we read on our dials �or digital voltmeters� will be the same in the original

and mapped systems� That is� the contravariant components of QL relative to L��b���

� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

L��b��� L��b�� will be the same as the contravariant components of Q relative to �b�� �b�� �b��

Thus

QL � Qi����iqL��bi��� L��bi�� � � �L��biq �

� Qi����iq��qL���bi��bi� � � ��biq�� ��qL�

hQi����iq�bi�

�bi� � � ��biqi�

Therefore� if L � "�V � and the physical system is mapped by L� the new tensor property

QL is obtained by applying L to the original Q�

QL � ��qL��Q�� �����

A tensor property of a physical system rotates with that system� and is re�ected if the

system if re�ected�

�� Invariance groups

Sometimes� subjecting a physical system to certain rotations or re�ections will not a�ect

any of its measurable physical properties� For example� suppose I work all morning on

a cubic crystal of NaCl� While I am at lunch� you rotate it in such a way as simply to

permute its three crystal axes� When I return from lunch� I will be unable to detect that

anything has happened to the crystal� The same is true if you manange to work so fast

as to disassemble and reassemble the crystal so as to subject it to a re�ection along one

of its crystal axes while I am at lunch� If I was working not on NaCl but on a pot of

left�handed CHFCLBr before lunch� you can subject it to any rotation whatever� I will

not be able to detect the rotation when I return� However� if you re�ect it� I can detect

that now I have a pot of right�handed CHFClBr� For example� it will rotate the plane

of polarized light oppositely before and after lunch�

Let S be any physical system in real �space V � For any L � "�V �� let LS be the

physical system obtained by subjecting S to L� Let G�S� be the set of all members of

"�V � which� when applied to S� change none of the measurable properties of S� That is�L � ��V � is a member of G�S� if no measurement can distinguish between S and LS�

���� INVARIANCE GROUPS

These de�nitions have some simple but useful implications� Obviously

I � G�S� �������

where I is the identity operator on V � And if L�� L� � G�S�� then L�S is indistinguishable

from S� so L��L�S� is indistinguishable from S� Therefore

if L�� L� � G�S� then L� � L� � G�S�� ������

Finally� we claim

if L � G�S� then L�� � G�S�� ������

For LS and S are indistinguishable� Hence so are L���LS� and L��S� But L���LS�is S� so S and L��S are indistinguishable� Properties �������� ������� ������ show that

G�S�� is a subgroup of "�V �� It is called the invariance group of the physical system S�Examples of invariance groups are these�

Example ���� If S is water� G�S� � "�V ��

Example ���� If S is liquid� left�handed CHFClBr � G�S� � "��V ��

Example ��� If S is polycrystalline calcite� with the microcrystals randomly oriented�

and with equal numbers of right and left�handed microcrystals� and if we make only

measurements involving lengths � � microcrystal diameter �e�g� study only seismic

waves very long compared to the size of the microcrystals�� G�S� � "�V ��

Example ��� If in example ������ �& of the microcrystals are left handed and ��&

are right handed� G�S� � "��V ��

Example ��� If S is an NaCl crystal� G�S� is the �cubic group�� It is the group

of orthogonal operators which send into itself a cube centered on ��� If x�� x�� x� are

unit vectors parallel to the edges of the cube� G�S� consists of those L � "�V � such that

L� xi� � �i� x�i� for i � �� � � where � � S� and ��� ��� �� can be �� or �� independently�

� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

If Q � �qV describes some measurable property of physical system S� then clearly

��qL��Q� � Q for all L � G�S�� This fact limits the possible forms which Q can take�

and is therefore of physical interest� It leads us to introduce

De�nition ����� Let V be a Euclidean vector space� Let G be a subgroup of "�V ��

Then Iq�G� is the set of all Q � �qV such that

��qL� �Q� � Q for all L � G� �������

The set Iq�G� is called the space of q�th order tensors invariant under G� The tensors in

Iq�"�V �� are unchanged by any orthogonal operator� They are called �isotropic tensors��

The tensors in Iq�"��V �� are unchanged by any rotation� but may be changed by re�ec�

tions� They are called �skew isotropic tensors�� If G is the cubic group� the tensors in

Iq�G� are unchanged by any rotation or re�ection which is physically undetectable in an

NaCl crystal�

The following are simple consequences of de�nition ������

Corollary ����� Iq�G� is a subspace of �qV �

Proof�

Clearly Iq�G� � �qV � so all we need prove is that any linear combination of

members of Iq�G� is a member� If Q�� � � � � QN � Iq�G� and a�� � � � � aN � Rand L � G� then ��qL��Qi� � Qi� so ��qL��aiQi� � ai��qL��Qi� � aiQi

because �qL is linear�

Corollary ����� If G� � G� then Iq�G�� � Iq�G���

Proof�

If Q � Iq�G��� Q is unchanged by any L � G�� and then certainly by any

L � G�� Hence Q � Iq�G��� �The bigger the group G� the harder it is to be

unchanged by all its members��

���� INVARIANCE GROUPS �

Corollary ����� If Q � Iq�G�� and � � Sq then �Q � Iq�G��

Proof�

For any L � L�V V �� equation ������ gives ��qL� � � � � � ��qL�� Then

��qL���Q� � � ���qL��Q��� If Q � Iq�G� then for any L � G� ��qL��Q� � Q�

so ��qL���Q� � �Q� Hence �Q � Iq�G��

Corollary ����� If P � Ip�q�G� and R � Iq�r�G� then P hqiR � Ip�r�G��

Proof�

For any L � G� L � "�V �� so ��p�rL��P hqiR� � ���p�qL��P �� hqi ���q�rL��Q�� �

P hqiR �

Corollary ����� If Q � Iq�G� then the component array of Q is the same relative to

any two ordered orthonormal bases � x�� � � � � xn� and � x��� � � � � x�n� such that x�iL� xi� for

some L � G�

Proof�

For any Q � �qV �

Q � Qi����iq xi� � � � xiq � Q�i����iq

x�i� � � � x�iq � �������

For any L � L�V V ��

��qL� �Q� � Qi����iqL � xi�� � � �L� xiq�� �������

If Q � Iq�G� and x�i � L� xi� for some L � G� then ��qL��Q� � Q and

comparing ������� with ������� gives Qi����iq � Q�i����iq

Corollary ����� Suppose Q � �qV � Suppose V has one ordered orthonormal basis

� x�� � � � � xn� such that whenever L � G and x�i � L� xi�� then Q has the same component

array relative to � x�� � � � � xn� and � x��� � � � � x�n�� Then Q � Iq�G��

Proof�

Suppose L � G� Let x�i � L� xi�� By hypothesis Q�i����iq � Qi����iq � Comparing

�������� and ������� gives ��qL��Q� � Q� Since this is true for every L � G�Q � Iq�G��

�� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

�� Isotropic and skew isotropic tensors

De�nition ���� Let V be a Euclidean vector space� The tensors in Iq�"�V �� are

called �isotropic�� The tensors in Iq�"��V �� are called �skew isotropic��

Any tensor describing a measurable property of water� or of polycrystalline calcite with

randomly oriented microcrystals and equal numbers of left� and right�handed microcrys�

tals� is isotropic� Any tensor describing a property of left�handed liquid CHFClBr � or of

polycrystalline calcite with randomly oriented microcrystals and more of one handedness

than the other� is skew isotropic� Since "��V � � "�V �� corollary ������� implies

Iq�"�V �� � Iq�"��V ��� �������

That is� every isotropic tensor is skew isotropic�

By corollaries ����� and ������ a tensor is isotropic i� it has the same array of

components relative to every orthonormal basis for V �

Suppose �V�A� is an oriented Euclidean space� By corollary ������ a tensor is skew

isotropic if it has the same array of components relative to every positively oriented

orthonormal basis for V � Also� it is skew isotropic if it has the same array of components

relative to every negatively oriented orthonormal basis for V � If a tensor is skew isotropic�

by corollary ����� it has the same array of components relative to every positively oriented

orthonormal basis� and the same array relative to every negatively oriented orthonormal

basis� The two arrays need not be the same �indeed� if they are� the tensor is isotropic��

Example ����� The identity tensor�

I is isotropic� For� relative to any orthonormal

basis � x�� � � � � xn� for V � Iij ��

I � xi� xj� � xi � xj � �ij�

Example ����� If A is a unimodular alternating tensor� A is skew isotropic but not

isotropic� For as we saw on page �� A has the same component array relative to every

positively oriented orthonormal basis� but its component array changes sign for negatively

oriented bases�

���� ISOTROPIC AND SKEW ISOTROPIC TENSORS ��

Example ����� By corollary ������

I�

I is isotropic� So is�

I�

I�

I ��

I�

I�

I�

I � etc� By

corollary ���� � every permutation of these tensors is isotropic� By corollary ������

every linear combination of such permuted tensors is isotropic�

Example ����� All the tensors in example ����� are skew isotropic� By corollary

����� so are�

IA��

I�

IA��

I�

I�

IA� etc� By corollary ����� so are all permutations of these

tensors� By corollary ������ so is every linear combination of such permuted tensors� We

gain nothing new by using more than one factor A� The polar identity ����� shows that

if n � dimV then AA is a linear combination of n� permutations of�

I�

I � � � �I �with n

factors�

I ��

On p� �� of his book �The Classical Groups� �Princeton� ������ Hermann Weyl shows

that there are no isotropic tensors except those listed in example ������ and no skew

isotropic tensors except those listed in example ������ Note the corollary that all nonzero

isotropic tensors are of even order� We cannot consider Weyl�s general proof� but we will

discuss in detail the tensors of orders q � �� �� � � �� which are of particular interest in

continuum mechanics� We assume dimV � �

q � � Tensors of order � are scalars� They are una�ected by any orthogonal transform�

ation� so are isotropic and therefore also skew isotropic�

q � � If dimV � � the only skew isotropic tensor of order � is ��� Therefore the only

isotropic tensor of order � is ���

Proof�

Tensors of order � are vectors� We are claiming that if dimV � then

the only vector which is una�ected by all rotations is the zero vector�

A formal proof of this intuitively obvious fact is as follows� Suppose

�v � I��"��V ��� Let x and y be any two unit vectors� Each is the �rst

vector of a positively oriented ordered orthonormal basis �pooob�� so by

corollary ������ �v� x� � �v� y�� That is �v � x � �v � y for any two unit vectors

�� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

x and y� If dimV � � we can always �nd a y such that �v � y � �� Then

�v � x � � for all x� so �v � ���

q �

I��"�V �� � spf�I g ������

I��"��V �� � spf�I g if dimV � ������

I��"��V �� � spf�I � Ag if dimV � and

A is any nonzero member of ��V� �������

Proof�

The containments � have all been established in examples ����� and

������ so we need prove only � in ������ ����� ������� In all these cases�

suppose T � I��"��V ��� If x and y are any unit vectors� each is the

�rst vector of an ordered orthonormal basis �oob�� and by changing the

sign of one later basis vector� if necessary� we can assume that these oob�s

have the same orientation� Therefore T � x� x� � T � y� y� because T has

the same component array relative to all oobs with the same orientation�

Thus there is a scalar aT such that

T � x� x� � aT for every unit vector x� �������

If f x� yg is orthonormal� � x� y� is the �rst pair of vectors in an oob� Re�

placing � x� y� by � y�� x� in that oob gives a second oob with the same

orientation� Hence� T � x� y� � T � y�� x� � �T � y� x��

T � x� y� � �T � y� x� if f x� yg is orthonormal � �������

When dimV � � choose an oob � x�� x�� and let a � aT � b � T � x�� x��

�A� x�� x��� Then relative to this oob� Tij � a�ij � bA� x�� x���ij � �a�

I

�bA�ij� so T � a�

I �bA� and ������� is proved� To treat ������ ������ let

���� ISOTROPIC AND SKEW ISOTROPIC TENSORS ��

� x� y� be orthonormal� Both � x� y� and � y� x� are the �rst pair of vectors

in an oob� so if T � I��"�V ���

T � x� y� � T � y� x� if f x� yg is orthonormal� �������

If dimV � � both � x� y� and � y� x� are �rst pairs in oobs with the same

orientation� namely � x� y� x�� � � � � xn� and � y� x�� x�� x�� � � � � xn�� If T �I�"��V ��� we again have ������ In either case we have ������� and �������

so

T � x� y� � � if f x� yg is orthonormal� �������

Now let � x�� � � � � xn� be an oob� By ������� and �������� relative to this

oob we have Tij � aT �ij� so T � aT�

I � QED

q �

I��"�V �� � f�g �������

I��" � �V �� � f�g if dimV �� ��������

I��"��V �� � spfAg if dimV � and A �� �� A � ��V� ��������

Proof�

If dimV is even� let � x�� � � � � xn� be an oob for V � Then �� x�� � � � �� xn�is another oob with the same orientation� Then if T � I��"��V ���

T � xi� xj� xk� � T �� xi�� xj�� xk� � �T � xi� xj� xk� � � so Tijk � � �

If dimV is odd and dimV � �� let � x� y� z� be any ordered orthonor�

mal sequence� It is the �rst triple of an oob � x� y� z� x�� � � � � xn� and

�� x� y� z�� x�� x�� � � � xn� is an oob with the same orientation� so

T � x� y� z� � T �� x� y� z� � �T � x� y� z� � �� Similarly� T � x� y� y� �

�T � x� y� y� � � and T � x� x� x� �� �T � x� x� x� � �� Thus if x�� x�� x�

are orthonormal� T � xi� xj� xk� � � for fi� j� kg � f�� � � g� This com�

ment applies to any three vectors from an arbitrary oob for V � so Tijk � �

�� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

for fi� j� kg � f�� � � � � ng relative to any oob� and T � �� Finally� sup�

pose dimV � � If f x� yg is orthonormal� � x� y� is the �rst pair of an

oob � x� y� z�� and the oob �� x� y�� z� has the same orientation� Thus

T � x� y� y� � T �� x� y� y� � �T � x� y� y� � �� Similarly T � y� x� y� � � and

T � y� y� x� � �� Also T � x� x� x� � T �� x�� x�� x� � �T � x� x� x� � ��

Thus if T � I��"��V ��

T � x� x� x� � � for every unit vector x �������

T � x� y� y� � T � y� x� y� � T � y� y� x� if f x� yg is orthonormal� �������

Finally� if � x� y� z� is an oob� then the following are oobs with the same

orientation� � y� z� x� � � z� x� y� � � y� x�� z� � � x�� z� y� � �� z� y� x� � Then

if T � I��"��V ��

T � x� y� z� � T � y� z� x� � T � z� x� y� � �T � y� x� z� � �T � x� z� y� � �T � z� y� x���������

if f x� y� zg is orthonormal� Let � x�� x�� x�� be an oob for V � Let c �

T � x�� x�� x���A� x�� x�� x��� Then ������ � ����� � ������� imply Tijk �

cA� x�� x�� x���ijk � cAijk so T � cA� QED�

Note that in all the foregoing arguments� only mutually � vectors were used� These

arguments remain valid if G is the cubic group and T � Iq�G� with q � �� � and

we work only with unit vectors and oobs in the crystal axis directions� Therefore

the conclusions apply to tensors of orders �� � known only to be invariant under

the cubic group of Na Cl� For example� it follows that the conductivity tensor�

K

of Na Cl must have the form K�

I � In Na Cl� Ohm�s law �J � �E� �K reduces to the

isotropic form �J � �EK� even though an NaCl crystal is not isotropic�

q � �

I��"�V �� � sp��

I�

I � ���

I�

I � ����

I�

I

���������

���� ISOTROPIC AND SKEW ISOTROPIC TENSORS ��

I��"��V �� � ��if dimV �� f� �g ��������

I��"��V �� � sp��

I�

I � ���

I�

I � ����

I�

I � A�

if dimV � � ��������

I��"��V �� � spf�I�I � �� �I�I � ��� �I�I ��

IA� ���

IA� ����

IA�

A�

I � ��A�

I � ���A�

I g if dimV � � ��������

Here A is any nonzero alternating tensor� A � �nV with n � dimV �

Proof�

The containments � have all been established in examples ����� and ������

so we need prove only �� We consider only �������� �������� and leave ��������

������� to the reader with Sitz�eisch� Therefore we assume T � I��"��V ���

and if dimV � f� �g we assume T � I��"�V ��� If dimV � �� then any four

orthonormal vectors � w� x� y� z� are the �rst four vectors of and oob for V � say

� w� x� y� z� x�� � � � � xn�� The vectors �� w� x� y� z� are the �rst four vectors of

an oob with the same orientation� namely �� w� x� y� z�� x�� x�� � � � � xn�� SinceT � I��"��V ��� T has the same component array relative to both oob�s� so

T � w� x� y� z� � T �� w� x� y� z� � �T � w� x� y� z�� Thus for dimV � ��

T � w� x� y� z� � � if f w� x� y� z�g is orthonormal� ��������

This result is also vacuously true for dimV � � because then we cannot �nd

four orthonormal vectors in V �

Next� suppose dimV � and f x� y� zg is orthonormal� Then f x� y� zg and

�� x� y�� z� are both the �rst triple in oobs with the same orientation� so

T � x� y� z� z� � T �� x� y�� z�� z� � �T � x� y� z� z� � �� Similarly� if f x� y� zg isorthonormal

T � x� y� z� z� � T � x� z� y� z� � T � x� z� z� y�

� T � z� x� y� z� � T � z� x� z� y� � T � z� z� x� y� � �� �������

�� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

Equations ������� are vacuously true if dimV � because then we cannot

�nd three orthonormal vectors in V �

Next� suppose f x� yg is orthonormal� Then � x� y� and �� x� y� are the �rst pairsin oobs� and if dimV � we may assume these oobs to have the same orient�

ation� say � x� y� x�� � � � xn� and �� x� y�� x�� x�� � � � xn�� Then T � x� y� y� y� �

T �� x� y� y� y� � �T � x� y� y� y� � �� Similarly

T � x� y� y� y� � T � y� x� y� y�

� T � y� y� x� y� � T � y� y� y� x� � � if f x� yg orthonormal��������

Next� suppose that f x� yg and f x�� y�g are both orthonormal� Each pair � x� y�

and f x�� y�g is the �rst pair in an oob� so if T � I��"�V �� then T � x� x� y� y� �

T � x�� x�� y�� y��� If dimV � � then � x� y� and � x�� y�� are �rst pairs in oobs with

the same orientation� say � x� y� x� � � � � xn� and � x�� y��� x�� x�� � � � � xn�� Then in

that case also� T � x� x� y� y� � T � x�� x�� y�� y��� In either case� there is a scalar

T depending only on T such that

T � x� x� y� y� � T if f x� yg is orthonormal� ������

Similarly� there are scalars uT and vT such that

T � x� y� x� y� � uT if f x� yg is orthonormal� ������

T � x� y� y� x� � vT if f x� yg is orthonormal� �������

Finally� if x and y are unit vectors� each is the �rst member of an oob� and

the two oobs can have the same orientation� so T � x� x� x� x� � T � y� y� y� y��

Therefore� there is a scalar �T depending only on T such that

T � x� x� x� x� � �T for every unit vector x� �������

Then tensor�

I�

I satis�es ���������������� with � � �� � �� u � �� v � ��

The tensor ���

I�

I satis�es ���������������� with � � � � � � � u � ��

���� ISOTROPIC AND SKEW ISOTROPIC TENSORS �

v � �� The tensor ����

I�

I satis�es ���������������� with � � � � � � �

u � �� v � �� Therefore� the tensor

S � T � T�

I�

I �uT ���

I�

I �vT ����

I�

I �������

satis�es ���������������� with �S � �T �T �uT �vT and S � uS � vS � ��

Now let � x� y� be orthonormal in V and let � x� y� x�� � � � � xn� be an oob for V �

Then � x�� y�� x�� � � � � xn� is an oob with the same orientation if we take

x� � x cos � � y sin �� y� � � x sin � � y cos ��

To see this� let A � �nV � A �� �� Then

A� x�� y�� x�� � � � xn� � A� x cos � � y sin ��� x sin � � y cos �� x�� � � � � xn�� � cos � sin �A� x� x� x�� � � � � xn� � cos� �A� x� y� x�� � � � � xn�� sin�A� y� x� x�� � � � � xn� � sin � cos �A� y� y� x�� � � � � xn�

��cos� � � sin� �

�A� x� y� x�� � � � � xn� � A� x� y� x�� � � � � xn��

But if � x� y� x�� � � � � xn� and � x�� y�� x�� � � � � xn� have the same orientation and

S � I��"��V ��� the S� x� x� x� x� � S� x�� x�� x�� x��� The S of ������� is a linear

combination of members of I��"��V ��� so it is in I��"��V ��� In addition� S

satis�es ���������������� with S � uS � vS � �� Therefore

S� x� x� x� x� � S� x�� x�� x�� x��

� S� x cos � � y sin ��

x cos � � y sin �� x cos � � y sin �� x cos � � y sin ��

� cos� �S� x� x� x� x� � sin� �S� y� y� y� y� �������

since all other �� terms in the multilinear expansion of S vanish� Thus �S �

�cos� � � sin� ���S� It is possible to choose � so that cos� � � sin� � �� ��

Therefore �S � �� Therefore S � �� Therefore

T � T�

I�

I �uT ���

I�

I �vT ����

I�

I � �������

�� CHAPTER � HOW TO ROTATE TENSORS �AND WHY�

This completes the proofs of �������� and ���������

Notice that the proof of ������� works equally well if all the unit vectors are

parallel to the crystal axes of an Na Cl crystal� Only ������� fails� But if

x� y� z are crystal axes� then S � �S� x x x x� y y y y� z z z z�� Therefore� we have

proved that if T is invariant under the cubic group we can write

T � �S � x x x x� y y y y � z z z z�

� T�

I�

I �uT ���

I�

I �vT ����

I�

I � �������

where �S � �T � T � uT � vT �

Chapter �

Di�erential Calculus of Tensors

��� Limits in Euclidean vector spaces

De�nition ���� Let U be a Euclidean vector space� Let �u � U � Let t � R� t � �� We

de�ne three sets�

i� B��u� t� �� f�u� � �u� � U and k�u� � �uk tg

ii� B��u� t� �� f�u� � �u� � U and k�u� � �uk � tg

iii� 'B��u� t� �� f�u� � �u� � U and k�u� � �uk � tg �

B��u� t� � B��u� t� and 'B��u� t� are called� respectively� the open ball� the sphere� and the

closed ball centered at �u with radius t� See �gure ���

De�nition ���� A subset D of Euclidean space U is �open� if every point in D is the

center of a ball lying entirely in D� That is� if �u � D then �� � � � B��u� �� � D� See

�gure ���

De�nition ����� Suppose U and V are Euclidean vector spaces� D � U � �f � D V �

�u� � D and �v� � V � We say that lim�u��u��f��u� � �v�� or� equivalently �f��u� �v�� as �u �u�

if for every � � � there is a ���� � � such that whenever �u � D and � k�u� �u�k ����

then kf��u�� �v�k ��

��

�� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

u

t

B(u, t)

B(u, t) ∂

Figure ����

���� LIMITS IN EUCLIDEAN VECTOR SPACES ��

D

.

.

.

Figure ���

�� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

De�nition ����� If �f��u�� � �v� in de�nition ����� then �f is �continuous at �u���

De�nition ����� If x and y are real numbers� denote the algebraically smaller of them

by x � y �read �x meet y�� and the algebraically larger by x � y �read �x join y��� Thus

� � � � �� � � � � �� ���� � � � ��� ���� � � � ��

Corollary ����� Suppose U and V are Euclidean vector spaces� D � U and f � D V �

Suppose �u� � D� �v� � V and �f��u� �v� as �u �u�� Then if �u � D and k�u� �u�k �����

we have

kf��u�k � � k�v�k�

Proof�

If �u � D and k�u��u�k ����� then kf��u���v�k � so k�f��u�k � k�f��u���v���v�k � kf��u�� �v�k� k�v�k � � k�v�k�

Corollary ���� Suppose U and V are Euclidean vector spaces� D � U � �f � D V �

�f � � D V � �u� � D� �v� and �v�� � V � Suppose that �f��u� �v� and �f ���u� �v �� as �u �u��

Then ��f � �f ����u� �v� � �v �� as �u �u��

Proof�

For any given � � � we must �nd a ������ � � such that �u � D and k�u��u�k ������ implies k�f � f ����u���v���v ��k �� By hypothesis� there are ����� � �

and ������ � � such that if �u � D then

k�f��u�� �v�k �� if k�u� �u�k ����� �������

k�f ���u�� �v ��k �� if k�u� �u�k � ������ ������

Let ������ � ����� � ������� If k�u � �u�k ������� then both ������� and

������ are true� so k��f � �f ����u� � �v� � �v ��k � k�f��u� � �v� � �f ���u� � �v ��k �k�f��u�� �v�k� k�f ���u�� �v ��k � �� � �� � �� QED�

���� LIMITS IN EUCLIDEAN VECTOR SPACES ��

Corollary ���� Suppose U � V � W � X are Euclidean vector spaces and D � U and

R � D V � W and T � D W � X� De�ne R � T � D V � X by requiring

�R � T ���u� � R��u� � T ��u� for each �u � D� Suppose �u� � D and as �u �u� we have

R��u� R� and T ��u� T�� Then �R � T ���u� R� � T��

Proof�

Choose � � �� We must �nd ���� � � such that if �u � D and k�u� �u�k ����

then k�R � T ���u�� R� � T�k �� We note

�R � T ���u�� R� � T� � R��u� � T ��u�� R� � T�� R��u� � T ��u�� R��u� � T� �R��u� � T� �R� � T�� R��u� � �T ��u�� T�� � �R��u�� R�� � T��

Thus

k�R � T ���u�� R� � T�k � kR��u� � �T ��u�� T�� k� k �R��u��R�� � T�k�

The generalized Schwarz inequality� remark ����� gives

kR��u� � �T ��u�� T�� k � kR��u�kkT ��u�� T�kk�R��u�� R�� � T�k � kR��u��R�kkT�k�

Thus if �u � D

k�R � T ���u�� R� � T�k � kR��u�kkT ��u�� T�k� kR��u��R�kkT�k� ������

Now if �u � D and�����������������

if k�u� �u�k �R��� then kR��u�k � � kR�k �see corollary ������

if k�u� �u�k �T�

����kR�k�

�then kT ��u�� T�k �

���kR�k�

if k�u� �u�k �R�

����kT�k�

�then kR��u��R�k �

���kT�k��

�������

�� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Take ���� � �R��� � �T�

����kR�k�

�� �R

��

���kT�k�

�� Then if k�u� �u�k �����

we have all three of �������� so by �������

k�R � T ���u�� R� � T�k �� � kR�k� �

�� � kR�k� �kT�k�

�� � kT�k� ��

QED�

Corollary ���� Suppose U and V are Euclidean vector spaces� and D � U � and

�u� � D and �v� � V � Suppose �f � D V is de�ned by �f��u� � �v� for all �u � D� Then

�f��u� �v� as �u �u��

Proof�

Obvious from de�nition ������

Corollary ����� Suppose U and V are Euclidean vector spaces� and D � U � and

�u� � D and �v� � V � Suppose �f � D V �

i� If �f��u� �v� as �u �u�� then �bi � �f��u� �bi � �v� as �u �u�� for any basis ��b�� � � � ��bn� of

V � with dual basis ��b�� � � � ��bn��

ii� If there is one basis for V � ��b�� � � � ��bn�� such that �bi � �f��u� �bi � �v� as �u �u�� then

�f��u� �v� as �u �u��

Proof�

Regard �bi as a function whose domain is D and whose value is �bi for each

�u � D� Then i� follows from corollaries ����� and ������ And ii� follows

from corollaries ����� � ������ and ������

Notice that the ranges of our functions can lie in tensor spaces since these are Euclidean

vector spaces� The corollaries all apply verbatim� The dot products are� of course�

the generalized dot products used to make tensor spaces Euclidean� The only corollary

which perhaps deserves comment is ������ It now reads thus� Suppose U � V�� � � � � VW�� � � � �Ws� X�� � � � � Xt are Euclidean vector spaces� and D � U and R � D V� �

���� GRADIENTS� DEFINITION AND SIMPLE PROPERTIES ��

� � � � Vr �W� � � � � �Ws and T � D W� � � � � �Ws �X� � � � � �Xt� De�ne RhsiT �

D V� � � � � � Vr � X� � � � � � Xt by requiring �RhsiT ���u� � R��u�hsiT ��u� for each

�u � D� Suppose �u� � D and as �u �u� we have R��u� R� and T ��u� T�� Then

�RhsiT ���u� R�hsiT��

��� Gradients� De�nition and simple properties

De�nition ����� We de�ne the �dot product� of two real numbers to be their ordinary

arithmetic product� With this de�nition� R is a one�dimensional Euclidean vector space�

and it has two orthonormal bases� f�g and f��g� which we write as � and � � when

thinking of R as a Euclidean space� When we think of a real number as a vector� we

write it with an arrow� like �� Thus � � �� and � � � � � � ���

De�nition ����� Suppose U and V are Euclidean vector spaces� D is an open subset

of U � �f � D V � and �u � D� A linear mapping L � U V is a �gradient mapping for �f

at �u � if there is a function �R � U V �which may depend on �f and �u and L � such that

a� lim�h���

�R��h� � ��� and

b� �f��u� �h� � �f��u� � L��h� � k�hk�R��h� for any�h � U such that �u� �h � D� ������

Remark ����� �f has at most one gradient mapping at �u�

Proof�

If L� and L� are both gradient mappings for �f at �u then whenever �u��h � D

we have

�f��u� �h� � �f��u� � L���h� � k�hk�R���h�

� �f��u� � L���h� � k�hk�R���h��

Therefore if �u� �h � D�

L���h� � k�hk�R���h� � L���h� � k�hk�R���h�� �����

�� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Now D is open� so there is an � � � such that if k�hk � then �u � �h � D�

Let �k be any vector in U � and let t be any nonnegative real number� Then

kt�kk � as long as � � t ��k�kk� Therefore� if t is in this range� ��� is

true for �h � t�k� That is

L��t�k� � kt�kk�R��t�k� � L��t�k� � ktkkR��t�k��

But Li�t�k� � tLi�k� and kt�kk � tk�kk� so if t � �

L���k� � k�kk�R��t�k� � L���k� � k�kk�R��t�k�� �����

This is true for all t in � t ��k�kk� Let t �� and both �R��t�k� �� and

�R��t�k� ��� Therefore L���k� � L���k�� Since this is true for all k � U � we

have L� � L�� QED

De�nition ����� In de�nition ���� if �f � D V has a gradient mapping L at �u�

then �f is �di�erentiable at �u �� and �R is the �remainder function for �f at �u �� The tensor�

L� U � V is called the �gradient tensor of f at �u � or the �gradient of �f at �u �� It is

written �r�f��u�� Since L��h� � �h� �L� we can rewrite ��� as

a� lim�h���

�R��h� � �� and

b� �f��u� �h� � �f��u� � �h �h�r�f��u�

i� k�hk�R��h� if �u� �h � D� ������

Remark ����� Suppose U and V are Euclidean vector spaces� D is an open subset of

U � �u � D� and �f � D V � Suppose there is a tensor�

L� U�V and a function �R � D V

with these properties�

a� lim�h���

�R��h� � �� and

b� �f��u� �h� � �f��u� � �h � L� k�hk�R��h� if �u� �h � D� ������

Then �f is di�erentiable at �u� its gradient tensor at �u is�

L� and its remainder function at

�u is �R�

Proof�

���� GRADIENTS� DEFINITION AND SIMPLE PROPERTIES �

�h � �h� �L is a linear mapping L� so ������ � ������� Therefore �f is di�er�

entiable� By Remark ����� L is the gradient mapping for �f at �u� so�

L is the

gradient tensor� Then the �R�s in �����b� and �����b� must be the same�

Example ����� Suppose �f � D V is constant� Then �f is di�erentiable at every

�u � D and �r�f��u� ����

Proof�

The hypotheses of remark ���� are satis�ed with�

L����

R ��h� � ���

Example ����� Suppose �f � D U is the identity mapping on D� i�e� �f��u� � �u for

all �u � D� Then �f is di�erentiable everywhere in D� and for each �u � D� �r�f��u� ��

I U �

the identity tensor on U � �This result is often abbreviated �r�u ��

I U ��

Proof�

The hypotheses of remark ���� are satis�ed with�

L��

I U � �R��h� � ���

Example ����� Suppose U and V are Euclidean spaces� D is an open subset of U and�

T is a �xed tensor in U � V � Suppose �f � D V is de�ned by �f��u� � �u� �T for every

�u � D� Then at every �u � D� �f is di�erentiable and �r�f��u� ��

T � �Often abbreviated

�r��u� �T � ��

T ��

Proof�

The hypotheses of remark ���� are satis�ed with�

L��

T � �R��h� � ���

Example ���� Suppose that for i � f�� � � � � ng we have ai � R and �fi � D V �

Suppose all the �fi are di�erentiable at �u � D� Then so is ai �fi � D V � and �r�ai �fi���u� �ai��r�fi���u��

Proof�

Each �fi satis�es equation ������� so the hypotheses of remark ���� are sat�

is�ed for ai �fi with�

L� ai��r�fi��u�� and �R��h� � ai �Ri��h��

�� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Example ���� Suppose thatD is an open subset of Euclidean space U and f � D R�

In elementary calculus classes� f is de�ned to be di�erentiable at �u � D if there is a vector

�L � U and a function R � U R such that

a� lim�h���

R��h� � � and

b� f��u� �h� � f��u� � �h � �L� k�hkR��h� if �u� �h � D� ������

The vector �L is called the gradient of f at �u� written �rf��u�� If we think of R as a one�

dimensional Euclidean space� then we can think of f as a vector�valued function �f � f �

�the vector is one dimensional�� The scalar valued function f is di�erentiable in the sense

of equations ������ i� �f � f � is di�erentiable in the sense of equation ������� and then

�r�f��u� � �rf��u� �� �rf��u� � �r�f��u� � �� ������

Proof�

To obtain ������ from ������� multiply all terms in ������ on the right by ��

To obtain ������ from ������ dot � on the right in all terms in ������� Here

clearly�

L� �L � � �rf��u� ��

Example ���� Suppose D is an open subset of R� regarded as a one�dimensional

Euclidean space� Suppose �f � D V has an ordinary derivative at u � D �we will write

this derivative as u �f�u� rather than d�f�du� for reasons made clear later�� Then �f is

di�erentiable at �u and

�r�f�u� � �u �f�u�� ������

Proof�

By hypothesis� limh��

��fu�h���fu�

h

�� u �f�u�� De�ne �R�h� � � if h � � and

for h �� � then

�R�h� �h

jhj

��� �f�u� h�� �f�u�

h� u �f�u�

� �

���� GRADIENTS� DEFINITION AND SIMPLE PROPERTIES ��

Then limh���R�h� � ��� and

�f�u� h� � �f�u� � hu �f�u� � jhj�R�h�� ������

But h � �h � � so hu �f��u� � �h � �u �f�u�� Also jhj � k�hk� Therefore ������ is

������ with�

L� �u �f�u��

Example ����� Suppose f � D R and �g � D V are di�erentiable at �u � D� Then

so is the product function f�g � D V � and

�r�f�g���u� �h�rf��u�

i�g��u� � f��u�

h�r�g��u�

i� �������

Proof�

By hypothesis� if �u� �h � D then

f��u� �h� � f��u� � �h � �rf��u� � k�hkRf��h

�g��u� �h� � �g��u� � �h � �r�g��u� � k�hkRg��h��

Multiplying these two equations gives

f��u� �h��g��u� �h� � f��u��g��u� � �h �hf��r�g� � ��rf��g

i� k�hk�Rfg��h� �������

where� by de�nition�

�Rfg��h� � Rf ��h�h�g��u� � �h � �r�g��u�

i��Rg��h�

hf��u� � �h � �rf��u�

i� k�hkRf ��h��Rg��h�

�h�h � �rf��u�

i h�h � �r�g��u�

i�k�hk�

As �h ��� the three terms in �Rfg� which involve Rf or �Rg� obviously �� So does the last term� because by Schwarz�s inequality its length is �k�hkk�rf��u�kk�r�g��u�k� Therefore remark ���� can be applied to ��������

QED�

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Example ����� The Chain Rule� Suppose U � V � W are Euclidean vector spaces Df

is an open subset of U � Dg is an open subset of V � and f � Df Dg and g � Dg W �

Suppose �u � Df and �v � �f��u�� Suppose �f is di�erentiable at �u and �g is di�erentiable at

�v� Then their composition� �g � �f � Df W � is di�erentiable at �u and

�r��g � �f���u� � �r�f��u� � �r�g��v�� ������

Proof�

Let�

Lf� �r�f��u� and�

Lg� �r�g��v�� Let �Rf and �Rg be the remainder functions

for �f at �u and �g at �v� Then �Rf ��h� �� as �h ��� and �Rg��k� �� as �k ���

and

�f��u� �h� � �f��u� � �h� �Lf �k�hk�Rf ��h� if �u� �h � Df � ������

�g��v � �k� � �g��v� � �k� �Lg �k�kk�Rg��k� if �v � �k � Dg� �������

We hope to �nd �Rg�f � U W such that �Rg�f ��h� �� as �h �� and

��g � �f ���u��h� � ��g � �f ���u���h ���

Lf ��

Lg

�� k�hk�Rg�f ��h� whenever �u��h � Df �

This equation is the same as

�gh�f��u� � �h�

i� �g

h�f��u�

i� ��h� �Lf ��

Lg �k�hk�Rg�f ��h�� �������

To obtain �������� choose any �h � U such that �u� �h � Df and de�ne

�k � �f��u� �h�� �f��u� � �f��u� �h�� �v� �������

Then �k � V and �v��k � �f��u��h� � Dg� so ������� holds for the �k of ��������

Moreover� from ������

�k � �h� �Lf �k�hk�Rf��h�� �������

Substituting ������� in ������� gives

�gh�f��u� �h�

i� �g

h�f��u�

i���h� �Lf �k�hk�Rf ��h�

�� �Lg �k�kk�Rg��k��

���� GRADIENTS� DEFINITION AND SIMPLE PROPERTIES ���

This is exactly ������� if we de�ne �Rg�f���� � � and� for �h �� ��� de�ne

�Rg�f ��h� � �Rf ��h���

Lg �k�kkk�hk

�Rg��k�� �������

Applying to ������� the triangle and Schwarz inequalities gives

k�kk � k�hk�k �Lf k� k�Rf��h�k

�� �������

Therefore� as �h ��� we have �k ��� and k�kk�k�hk remains bounded� Hence

�Rf�g��h� �� as h �� QED�

Example ����� Suppose Df is an open subset of Euclidean space U � and Dg is an open

subset of R� and f � Df Dg and g � Dg R� Suppose f is di�erentiable at �u � Df

and that g has a derivative� vg�v�� at v � f��u�� Then g � f � Df R is di�erentiable at

�u and

�r�g � f���u� � �vg�v�� �rf��u�� v � f��u�� ������

Proof�

This fact is presumably well known to the reader� It is mentioned here only

to show that it follows from the chain rule� example ������� If we regard Ras a one�dimensional Euclidean space and consider �f � f � and �g � g �� then

�f is di�erentiable at �u� �g is di�erentiable at �v � v �� so �g � �f is di�erentiable

at �u and satis�es ����� But �r�f � �rf� �� and �r��g � �f� � �r�g � f� �� and�r�g � �v�g � � �vg� Thus ���� is �r�g � f� � � ��r f� � � ��vg � �vg��rf ��Dot � on the right to obtain �����

Example ����� Suppose Df is an open subset of R� Dg is an open subset of V � and

�f � Df Dg� �g � Dg W � Suppose �f has a derivative� u �f�u�� at u � Df � and suppose

�g is di�erentiable at �v � �f�u�� Then �g � �f has a derivative at u and

u��g � �f��u� � u �f�u� � �r�g��v�� �v � �f�u�� ������

Proof�

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

�f is di�erentiable at �u � u �� with �r�f��u� � �u �f�u�� Therefore� by the chain

rule� �g� �f is di�erentiable� Therefore u��g� �f� exists and �r��g� �f� � �u��g� �f��Substitute these expressions in ������ and dot � on the left to obtain �������

Example ����� Suppose �f and �g are inverse to one another� and �f is di�erentiable at

�u and �g is di�erentiable at �v � �f��u�� Then �r�f��u� and �r�g��v� are inverse to one another�

Proof�

LetDf be the domain of �f � an open subset of Euclidean space U � LetDg be the

domain of �g� an open subset of Euclidean space V � By hypothesis� �f ��g � IDg

and �g � �f � IDf� By the chain rule� �r�f��u� � �r�g��v� � �rIDf

��u� ��

I U and

�r�g��v� � �r�f��u� � �rIDg��v� ��

I V � QED�

��� Components of gradients

Throughout this chapter� U and V are Euclidean vector spaces� ��b�� � � � ��bm� is a �xed

basis for U with dual basis ��b�� � � � ��bm�� and ����� � � � � ��n� is a �xed basis for V � with dual

basis ����� � � � � ��n�� Df is an open subset of U � and �f � Df V �

De�nition ����� The i�th covariant component function of �f relative to ���� � � � � ��n is

fi � Df R where� for every �u � Df �

fi��u� �h�f��u�

ii� �f��u� � ��i� ������

The ith contravariant component function of �f relative to ���� � � � � ��n is the i�th covariant

component function of �f relative to ���� � � � ��n� That is� it is f i � Df R where� for any

�u � Df �

f i��u� �h�f��u�

ii� �f��u� � ��i�

Evidently

�f � f i��i � fi��i� �����

���� COMPONENTS OF GRADIENTS ���

Theorem ���� Let ���� � � � � ��n be any basis for V � Let Df be an open subset of U �

Then �f � Df V is di�erentiable at �u � Df i� all n covariant component functions

fi � Df R are di�erentiable at �u� If they are� then

�rfi��u� �h�r�f��u�

ii� �r�f��u� � ��i �����

�r�f��u� �h�r�f��u�

ii��i � �r�fi��u���

i� ������

Proof�

� Suppose �f is di�erentiable at �u� Then it satis�es ������� Dotting ��i on the

right in each term of ������ produces

lim�h���

Ri��h� � �

fi��u� �h� � fi��u� � �h �h�r�f��u� � ��i

i� k�hkRi��h�

where Ri � �R � ��i� By remark ����� fi is di�erentiable at �u� and its gradient

is ������ Then ������ follows from �D���� and ������

� Suppose all the fi are di�erentiable at �u� We can regard ��i as a constant

function onDf � so it is di�erentiable at �u� Then by example ���� the product

fi���i� is di�erentiable at �u� By example ����� so is the sum fi��

i � �f � QED�

De�nition ���� Let �b�� � � � ��bm be any basis for U � Let Df be an open subset of U

and suppose �f � Df V � Then �f�uj�bj� is a V �valued function of the real variables

u�� � � � � un� If the partial derivative �f�uj�bj��ui exists at �u � uj�bj� we abbreviate it as

ui �f��u� or i �f��u� � Let �b�� � � � ��bm be the basis dual to �b�� � � � ��bm� If the partial derivative

�f �uj�bj��ui exists at �u � uj�b

j� abbreviate it as ui�f��u� or if��u� � In summary

i �f��u� �� ui �f��u� �� �f �uj�bj�

ui������

i �f��u� �� ui�f��u� ��

�f �uj�bj�

ui� ������

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Theorem ���� Suppose �b�� � � ��bm is any basis for U � and Df is an open subset of U �

and �f � Df V � If �f is di�erentiable at �u � Df then the partial derivatives i �f��u� and

i �f��u� all exist� and

i �f��u� � �bi � �r�f��u� ������

i �f��u� � �bi � �r�f��u� ������

�r�f��u� � �bii �f��u� � �bii �f��u�� ������

Proof�

By example ���� if �f is di�erentiable at �u � uj�bj then the partial derivative

of �f�uj�bj� with respect to ui exists �hold all uj �xed except for ui� and is

equal tohi�u

j�bj�i� �r�f��u�� But obviously� by the de�nition of i� iu

j � �ij

so

i�uj�bj� � �bi�

Thus we have ������� ���� is obtained similarly� To get ������ use the fact

that�

I U� �bi�bi � �bi�bi� Then �r�f��u� �

I U ��r�f��u� � �bi�bi � �r�f��u� � �bii �f��u� and

�r�f��u� ��

LU ��r�f��u� � �bi�bi � �r�f��u� � �bi

i �f��u��

Note� If dimU � �� the converse of theorem ����� is true� See example ������

However� if dimU � � that converse is false� A famous example is this� Let dimU � �

V � R� Let x� y be an orthonormal basis for U and de�ne f � U R by f���� � ��

f�x x � y y� � xy�px� � y� if x� � y� �� �� Then xf���� � �� yf���� � �� so if f is

di�erentiable at �� then �rf���� � � x � � y � ��� Then by example ���� tf�t x � t y� �

�t�t x � t y�� � �rf���� � � at t � �� But f�t x� t y� � t� so tf�t x� t y� � � at t � �� This

contradiction shows that f is not di�erentiable at ���

Theorem ����� Suppose ��b�� � � � ��bm� and ����� � � � � ��n� are bases for U and V respect�

ively� with dual bases ��b�� � � � ��bm� and ����� � � � � ��n�� Suppose Df is an open subset of

U and �f � Df V and �u � Df � Suppose f is di�erentiable at �u� Then the partial

���� GRADIENTS OF DOT PRODUCTS ���

derivatives i �f��u� and ifj��u� exist� and so do the gradients �rfj��u�� Moreover

hi �f��u�

ij

� ifj��u� �������h�rfj��u�

ii

� ifj��u� �������h�r�f��u�

iij

� ifj��u�� ������

Proof�

By theorem ������ fj is di�erentiable at �u so �rfj��u� exists� By theorem

������ i �f��u� and ifj��u� exist because �f and fj are di�erentiable at �u� By

�����

�r�f��u� �h�rfj��u�

i��j� ������

By ������� �bi � �r�f��u� � i �f��u� and �bi � �rfj��u� � ifj��u� � Therefore dotting

�bi on the left throughout ������ gives

i �f��u� � ifj��u���j� �������

This is equivalent to �������� Equation ������ applied to fj gives

�rfj��u� � �biifj��u�� �������

This is equivalent to �������� Substituting it in ������ gives

�r�f��u� � �biifj��u���j� �������

This is equivalent to �������

�� Gradients of dot products

In this chapter we suppose that U � V � W � X are Euclidean spaces� that D is an open

subset of U � and that�

f � D V �W and�g � D W � X� We suppose that

f and�g

are di�erentiable at �u � D� We want to show that�

f � �g � D V �X is di�erentiable at

�u� and we want to calculate �r��f � �g ���u��

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Method �� No Bases�

By hypothesis�

f ��u� �h� ��

f ��u� � �h � �r �

f ��u� � k�hk �Rf ��h�

�g ��u� �h� �

�g ��u� � �h � �r �

g ��u� � k�hk �Rg ��h��

Dotting the �rst equation into the second gives

��

f � �g ���u� �h� � ��

f � �g ���u� ���h � �r �

f ��u��� �g ��u�

��

f ��u� �h�h � �r �

g ��u�i� k�hk �Rf �g ��h� �������

where

Rf �g ��h� ��

f ��u�� �Rg ��h� ���h � �r �

f ��u���h�h � �r �

g ��u�i�k�hk

���h � �r �

f ��u��� �Rg ��h��

Rf ��h�� �g ��u�

��

Rf ��h� �h�h � �r �

g ��u�i� k�hk �Rf ��h�� �Rg ��h��

An application of Schwarz�s inequality proves that�

Rf �g ��h� �� as �h ��� In ������� the

expression ��h � �r �

f ��u��� �g ��u��

f ��u� �h�h � �r �

g ��u�i

������

depends linearly on �h� so ������� shows that�

f � �g is di�erentiable at �u and that �r��f � �g���u� is that tensor in U �V �X such that �h � �r��f � �g ���u� is equal to ������ for all �h � U �

But what is �r��f � �g ���u�$ We need

Lemma ����� If P � V �W � �h � U � and Q � U �W �X then

P �h�h �Q

i� �h �

����

hP � ���Q

i��

Proof�

Since both sides of this equation are linear in P and Q� it su!ces to prove

the equation when P and Q are polyads� say P � �v �w� Q � �u�w��x� Then

���� GRADIENTS OF DOT PRODUCTS ��

�P �h�h �Q

i� ��h � �u���w � �w���v�x � and

�h � f��� �P � ���Q�g � �h � f��� �P � �w��u�x�g� �h f��� ��v��w � �w���u�x�g� ��w � �w���h � ��� ��v�u�x�� ��w � �w���h � ��u�v�x� � ��w � �w����h � �u��v�x�

From lemma ������ it follows that we can write ������ as

�h ���r �

f ��u�� �g ��u� � �����

f ��u� � ��r �g ��u�

���

Therefore

�r��f � �g ���u� � �r �

f ��u�� �g ��u� � �����

f ��u� �h����r �

g ��u�i�

� ������

Method ��

Introduce bases in U � V � W � X and take components� This is the procedure likely to be

used in practical calculations� and the results are generally easier to use than �������

Since�

f and�g are di�erentiable at �u� so are their component functions fjk and gk l�

By example ������� so are the products of fjk�gk�

l� By example ������� so are the

sums fjkgkl� and from examples ���� and �����

�r�fjkgk l� � ��rfjk�gk l � fjk��rgk l�� �������

Then ������� and ������� imply

i�fjkgkl� � �ifjk�g

kl � fjk�ig

kl�� �������

Now fjkgkl � �

f � �g �jl and so ������� and ������� imply�i�

f � �g ��jl��i

f�jkgk l � fjk

hi

�gik

l�

therefore �i�

f � �g ��jl���i

f�� �g

�jl���

f ��i

�g��

jl�

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

and

i��

f � �g ���u� ��i

f ��u��� �g ��u��

f ��u� �hi

�g ��u�

i� �������

If we multiply ������� on the right by �bi� where �b�� � � ��bn is the dual to the basis we have

introduced in U � we obtain

�r��f � �g � � ��r �

f �� �g ��bi��

f ��i�g ��� �������

The last terms in ������ and ������� can be shown to be equal by lemma ����� In fact�

neither ������� nor ������ is very useful� One usually works with ������� or ��������

The hoped for formula� �r��f � �g � � �r��f �� �g ���

f � � ��r �g �� is generally false� It is

true if one of �r or�

f � is one�dimensional� If U is one dimensional� ������� reduces to

u��

f � �g ��u� � u�

f �u�� �g �u���

f �u� � u�g �u�� �������

If�

f is one dimensional� ������� is

�r�f �g ��u� � �rf��u� �g �u� � f�u��r �

g ��u�� �������

�� Curvilinear coordinates

The set Rn of real n�tuples �u � �u�� � � � � un� is a vector space if� for �u and �v �

�v�� � � � � vn� � R we de�ne �u � �v � �u� � v�� � � � � un � vn� and� for �u � Rn and

a � R we de�ne a�u � �au�� � � � � aun�� Rn becomes a Euclidean vector space if we de�ne

�u��v � uivi� An orthonormal basis forRn is e� � ��� �� � � � � ��� e� � ��� �� �� � � � � ��� � � � en �

��� � � � � �� ���

De�nition ���� Suppopse DV is an open subset of Euclidean space V � A �coordinate

system� on DV is an open subset DU of Rn� together with an everywhere di�erentiable

bijection �V � DU DV whose inverse mapping �C � DV DU is also everywhere

di�erentiable� The real number �i � ci��v� � ei � �C��v� is the �value of the i�th coordinate

at the point �v� and �v is uniquely determined by the values of its n coordinates u�� � � � � un�

���� CURVILINEAR COORDINATES ���

We have

�v�u�� � � � � un� � �V�ui ei� � �V��u�ui � ci��v� � ei � �C��v��

De�nition ���� Suppose W is a Euclidean space and �g � DU W is di�erentiable�

Then we will write i�g for the partial derivative with respect to ui� That is

i�g��u� ��g�uj ej�

ui� �������

If �f � DV W is di�erentiable� then so is �f � �V � DU W �by the chain rule�� We will

abbreviate i��f � V� as i �f � Thus

i �f��v� � �f ��V�u�� � � � � un��

ui������

if �v � �V�u�� � � � � un��

Note that by theorem ������ the coordinate functions ci � DV R are di�erentiable

at every �v � DV � And by theorem ����� the partial derivatives i�V exist at every �u � DU �

The key to understanding vector and tensor �elds in curvilinear coordinates is

Theorem ����� Suppose �V � DU DV is a curvilinear coordinate system on open

subset DV of Euclidean vector space V � Suppose n � dimU is the number of coordinates�

Then dimV � n� For any �v � DV � let �u � C��v�� Then the two n�tuples of vectors in V �

���V���u�� � � � � n�V��u� and ��rc���v�� � � � � �rcn��v��� are dual bases for V �

Proof�

By the de�nition of a coordinate system� �C � �V � IU jDU � By the chain rule�

�r��C � �V���u� � �r�V��u� � �r�C��v�� By example ������ �r�IU jDU� ��

I U � Hence�

�r�V��u� � �r�C��v� ��

I U � ������

Similarly� �V � �C � IV jDV � so

�r�C��v� � �r�V��u� ��

I V � �������

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Now �r�V��u� � U � V so �r�V��u� ��

L for some L � L�U V �� and �r�C��v� �V � U � so �r�C��v� �

M for some M � L�V U�� According to ������ and

���������

L � �M��

I U so

M � L � IU � �������

According to ������� and ���������

M � �L��

I V so

L �M � IV � �������

According to �D���� ������� and ������� together imply that L and M are

bijections and M � L��� Therefore U and V are isomorphic� Hence they

have the same dimension�

Lemma ����� If U and V are �nite dimensional vector spaces and L � U V is a

linear bijection� and �u�� � � � � �un is a basis for U � then L��u��� � � � � L��un� is a basis for V �

Proof of lemma�

Suppose aiL��ui� � �� for a�� � � � � an � R� Then L�ai�ui� � ��� Since L is a bijec�

tion� ai�ui � ��� Since �u�� � � � � �un are linearly independent� ai � � � � � an � �� Hence

L��u��� � � � � L��un� are linearly independent� Next� let �v � V � There is a �u � U such that

�v � L��u�� There are a�� � � � � an � R such that �u � ai�ui� Then �v � L��u� � L�ai�ui� �

aiL��ui�� Hence L��u��� � � � � L��un� span V �

Now by ������� ei � �r�V��u� � i�V��u�� By ����� � �r�C��v� � ej � �rcj��v�� And ei��

I U

� ej � ei � ej � �ij� Therefore if we dot ei on the left and ej on the right on each side of

������ we obtain

i�V��u� � �rcj��v� � �ij� �������

Thus� the two n�tuples of vectors ���V��u�� � � � � n�V��u�� and ��rc���v�� � � � � �rcn��v�� are dualto one�another� Hence they are linearly independent� Since dimV � n� they span V �

Hence they are bases for V � and each is the basis dual to the other��

�A more intuitive way to put the proof of ������� is as follows� uj � cj��V�u�� � � � � un�� so �ij � �iu

j �

�i�cj��V�u�� � � � � un�� � �i�V � �rc

j �

���� CURVILINEAR COORDINATES ���

Corollary ����� If �f � DV W is di�erentiable �W being a Euclidean space� then

�r�f��v� � �rci��v�i �f��v�� �������

Proof�

������� is ������ with �bi � �rci��v��

A function �f � DV W is often called a vector �eld on DV � If W is a tensor

product of other vector spaces� �f is called a tensor �eld� It is often convenient

to express �f in terms of a basis forW which varies from point to point in DV �

Suppose that for i � f�� � � � � Ng� �i � DV W is di�erentiable everywhere

in DV � Suppose also that at each �v � DV � �����v�� � � � � ��N��v� is a basis for W �

with dual basis �����v�� � � � � ��N��v�� Then ��i � DV W is also di�erentiable in

DV � At each �v � DV we can write

�f��v� � f j��v���j��v� � fj��v���j��v��

Then� by ��������

�r�f��v� � ��rci�ihf j��j

i� ��rci��if j���j � f j��rci��i��j��

Now i��j��v� is a vector in W � so there are coe!cients (ijk��v� � R such that

i��k��v� � (ijk��v����j��v�� �������

Then

�r�f��v� �hif

j � (ijkf

ki��rci���j��v�� ��������

Equation ������� is called the �connection formula� for the basis �����v�� � � � � ��N��v�in the coordinate system �V � DU V � The scalars (i

jk��v� are the �Chris�

to�el symbols� at �v� The expression ifj �(i

jkf

k is often abbreviated Difj

and� in the older literature� is called the covariant derivative of the vector �eld

�f � DV W �

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

If W � �qV � then ����� � � � � ��N� is often taken to be a q�th order polyad

basis constructed from ��rc�� � � � � �rcn� and ���V� � � �n�V�� For example� if

W � V � V and �f � DV V � V � we might write

f� f iji�Vj�V� ��������

The Christo�el symbols (ijk for i�V� � � � � n�V are de�ned by

i�k�V� � (ijk�j�V�� �������

If we use �k�V��l�V� as the basis ���� � � � � ��N for V �V � we have from �������

ih�k�V��l�V�

i� (i

jk�j�V��l�V� � (i

jl�k�V��j�V�� �������

This is the connection formula ������� for �k�V��l�V�� It leads to

�r �

f� �Difjk���rci��j�V��k�V� ��������

where

Difjk �� if

jk � (ijlf

lk � (iklf

jl� ��������

Note that ������� implies

(ijk � (k

ji� ��������

In general this will not be true for (ijk in ��������

Instead of ��������� we might prefer to write

f� fij��rci���rcj�� ��������

Then we need the connection formula for ��rck���rcl� analogous to ��������

Therefore we need the analogue of ������� for �rck� The Christo�el symbols

introduced by that analogue will not be new� From �������� �rcj � k�V � �j k�

Then i��rcj� � k�V � �rcj � i�k�V� � �� From ��������

���� MULTIPLE GRADIENTS AND TAYLOR�S FORMULA ���

i��rcj� � k�V � �(i l k �rcj � l�V � �(i l k�j l � �(i j k� Thus �(i j k �rck �

i��rcj� � �k�V���rck� � i��rcj���

I V� i��rcj�� Hence�

i��rcj� � �(i j k��rck�� ��������

Therefore�

�r �

f ��u� � �Difjk���rci���rcj���rck� ��������

where

Difjk � ifjk � (iljflk � (i

lkfjl� �������

��� Multiple gradients and Taylor�s formula

Let U and V be Euclidean spaces with �xed bases� ��b�� � � � ��bm� and ����� � � � � ��n� respect�ively� Take all components with respect to these bases� Let D be an open subset of U

and suppose �f � D V is di�erentiable at each �u � D� Then �r�f��u� � U � V for each

�u � U � so we can de�ne a function

�r�f � D U � V� �������

This function may itself be di�erentiable everywhere in V � in which case we have another

function�

�r�r�f � D U � U � V� ������

The process can continue� If �rp �f � D ��pU�� V exists� we say �f is p times di�eren�

tiable in D�

Now suppose �f � D V is P times di�erentiable in D� For any �u � D write �u � ui�bi�

Let i � �i� Then

�r�f � �bii �f

�r�r�f � �rh�bii �f

i� �bjj

h�bii �f

i� �bj�biji �f

���

�rp �f � �bi� � � ��bipi� � � �ip �f�

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Thus� for any �h � U � if we write

�hp � �h�h � � ��h

�p factors � we have

�hphpi�rp �f � hi� � � �hipi� � � �ip �f�

Fix �h � U so that �u� �h � D� Let

�g�t� � �f��u� t h� ������

where

h � �h�k�hk�

By the ordinary Taylor formula in one independent variable

�g�t� �PXp �

tp

p�pg ��� � jtjP �RP �T �

where �RP �t� � as t �� But

t�g�t� � h � r�f��u� t h� � hii �f��u� t h�

�t �g�t� � � h � �r�h hii �f��u� t h�

i� hjj

h hii �f��u� t h�

i� hj hiji �f��u� t h�

� � h��hi�r� �f��u� t h��

Similarly pt �g�t� � � h�phpi�rp �f��u� t h� so

pt �g��� � � h�phpi�rp �f��u��

and

tppt �g��� � �t h�phpi�rp �f��u��

Thus�

�g�t� �PXp �

p��t h�phpi�rp �f��u� � jtjp �Rp�t��

��� DIFFERENTIAL IDENTITIES ���

Setting t � k�hk in this formula gives

�f��u� �h� �PXp �

p���h�phpi��r�p �f��u� � k�hkp �Rp�t�

�PXp �

p�hi� � � �hipi� � � �ip �f��u� � k�hkp �Rp��h� �������

with �Rp��h� � as �h ���

�� Di�erential identities

De�nition ���� Suppose U and V are Euclidean spaces� D is an open subset of U �

and�

T � D U � V is di�erentiable at �u � D� Then �r �

T ��u� � U � U � V � We de�ne

the divergence of�

T at �u to be

�r� �T ��u� ��

I U hi�r�

T ��u�� �������

Remark ���� Suppose �b�� � � � ��bm is a �xed basis for U and ���� � � � � ��n is a �xed basis

for V � Write �u � ui�bi� i ���ui

��

T� T ik�bi��k� Then

�r� �T ��u� � iTik��k or� equivalently� ������

��r� �T ��u�

�k� iT

ik� ������

Proof�

In theorem ����� replace V by U � V and �f � D V by�

T � D U � V �

For U � V use the basis �bj��k� Then by ������� �r �

T ��u�iTjk�bi�bj��k� Then

�r� �T ��u� ��

I U hi�r�

T ��u� ��

I U hihiT

jk�bi�bj��ni� iT

jk

��

I U hi�bi�bj���k �

iTjk��bi ��bj���k�� iT

jk�i j��k � iTik��k�

Corollary ���� Suppose f � D U and �g � D V � Then

�r � ��f�g� � ��r � �f��g � �f � ��r�g�� �������

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

^z

x

y

Figure ���

Proof�

Use components� Then� because of ������� ������� is equivalent to i�figk� �

�ifi�gk � f i�ig

k�� In this form it is obvious�

De�nition ���� Suppose �U�A� is an oriented three�dimensional Euclidean space and

V is another Euclidean space� Suppose D is an open subset of U and�

T � D U � V

is di�erentiable at �u � D� Then the A�curl of�

T at �u� written �rA� �

T ��u�� or simply

�r� �

T ��u�� is de�ned to be

�r� �

T ��u� � Ahi�r �

T ��u�� �������

Note ��� There are two curls for�

T � D U � V � one for each of the two unimodular

alternating tensors over U � We will always choose the right handed A de�ned on page ��

so Figure �� is positively oriented� The other curl is the negative of the one we use�

because the two A�s di�er only in sign�

Remark ���� Suppose ��b�� � � � ��bm� is a basis for U and ����� � � � ��n� is a basis for V

and�

T� Tkl�bk��l� Then

�r� �

T�hAijkjTkl

i�bi��

l� �������

or equivalently� ��r� �

T

�il � AijkjTkl� �������

��� DIFFERENTIAL IDENTITIES ��

Proof�

��r �

T �jkl � jTkl and �Ahi�r �

T �i l � Aijk��r �

T �jkl�

Corollary ���� If �f � D V is twice continuously di�erentiable on D �i�e� �r�r�f �

D U � U � V exists and is continuous at every �u � D� then

�r� �r�f ��� � �������

Proof�

��r� �r�f�i l � Aijkj��r�f�kl � Aijkjkfl� Because of the continuity assump�

tion� jk � kj so Aijkjkfl � Aijkkjfl � Aikjjkfl � �Aijkjkfl � ��

Corollary ���� If�

T � D U � V is twice continuously di�erentiable on D then

�r � ��r� �

T � � �� �������

Proof���r � �r� �

T

�l� i��r�

T �i l � iAijkjTkl � AijkijTkl�

Again this � � because ij � ji and Aijk � �Aikj�

Corollary ���� If�

T � D U�V is twice continuously di�erentiable then �r���r� �

T �

� �r��r� �T �� �r � ��r �

T �� �N�B� �r � ��r �

T � is often written r��

T ��

��� CHAPTER � DIFFERENTIAL CALCULUS OF TENSORS

Proof�

Let ��b���b���b�� be orthonormal and positively oriented� so ������� reads

��r� �

T �il � �ijkjTkl� Then��r� ��r� �

T ��il

� �ijkj��r��

T �kl

� �ijkj ��kmnmTnl�

� �ijk�kmnjmTnl � ��im�jn � �in�jm� jmTnl

� jiTjl � jjTil � i�jTjl�� j�jTil�

� i��r��

T �l � j��r�

T �jil

���r��r� �T �

�il���r � ��r �

T ��il�

Thus���r� ��r� �

T ��il���r��r� �T �� �r � ��r �

T ��il� QED�

Chapter ��

Integral Calculus of Tensors

���� De�nition of the Integral

������ Mass distributions or measures

Let D be any subset of Euclidean space U � A function m which assigns real numbers

to certain subsets of D is called a mass distribution� or measure� on D if it has these

properties� �if S is a subset of D to which m does assign a mass� that mass is written

m�S�� and S is called �measurable�m��

i� If S is measurable�m� then m�S� � ��

ii� D and the empty set are measurable�m�

iii� If S�� S�� � � � is any �nite or in�nite sequence of sets which are measurable�m � then

S�S�� S�� S� and S��S��� � � are measurable�m � and m�S��S��� � �� � P m�S�

if no Si and Sj have common points �i�e�� if S�� S�� � � � are mutually disjoint��

The general theory of mass distributions is discussed in Halmos� �Measure Theory�� Van

Nostrand� ����� An example of a �mass distribution� is �dimU�� dimensional volume�

Some subsets of D are so irregular that a volume cannot be de�ned for them� All bounded

open sets D do have volumes� As with volume� any mass distribution can be extended

���

��� CHAPTER ��� INTEGRAL CALCULUS OF TENSORS

to permit in�nite masses� The reader will be presumed to be familiar with the following

mass distributions �but not perhaps with the general theory��

Example ������� D is an open subset of U � ���u� is the density of mass per unit of

�dimU��dimensional volume at �u� and the mass in any measurable set S is m�S� �RS dV ��u����u� where dV ��u� is the in�nitesimal element of �dimU��dimensional volume in

U �

Example ������� D is a �dimU�� � dimensional �surface� in U � ���u� is the density of

mass per unit of �dimU�� � dimensional �area� at �u� dA��u� is the in�nitesimal element

of such area at �u� and the mass in any measurable set S is m�S� �RS dA��u����u��

Example ������ D is a curve in U � and ���u� is the density of mass per unit of length

at �u� while dl��u� is the in�nitesimal element of such length at �u� Then the mass in any

measurable subset S of D is m�S� �RS dl��u����u��

Example ������ D is a �nite set of points� �u�� � � � � �uN � and these points have masses

m�� � � � � mN � The mass of any subset S of D is the sum of the masses M for which

�u � S�

������ Integrals

Let F � D R be a real�valued function on D� The reader is presumed to know how to

calculate the integral of f on D with respect to a mass distribution m� We will denote

this integral byRD dm��u�f��u�� In the foregoing examples� this integral is as follows�

Example �������

ZDdm��u�f��u� �

ZDdV ��u����u�f��u��

Example �������

ZDdm��u�f��u� �

ZSdA��u����u�f��u��

����� DEFINITION OF THE INTEGRAL ���

Example ������ ZDdm��u�f��u� �

ZCdl��u����u�f��u��

Example ������

ZDdm��u�f��u� �

NXv �

mvf��uv��

It is possible to invent functions f � D R so discontinuous thatRD dm��u�f��u�

cannot be de�ned in examples ������ � ������� ������� �ClearlyRD dm��u�f��u� is always

de�ned in example �������� If the integral can be de�ned� f is said to be integrable with

respect to the mass distribution m� or integrable�dm� We will not need the general theory

of mass distributions� All we need are the following properties of integrals�

Remark ������� SupposeD�� � � � � DN are subsets of U � with mass distributionsm�� � � � � mN �

OnD � D��� � ��DN � de�ne a mass distributionm bym�S� �PN

v �mv�SDv�� Suppose

f � D R and for each v� f jDv is integrable�dmv� Then f is integrable�dm and

ZDdm��u�f��u� �

NXv �

ZDv

dmv��u��f jDv���u�� ��������

Remark ������ Suppose m is a mass distribution on D and for each v � f�� � � � � Ng�cv � R and fv � D R is measurable�m� Then cvfv � D R is integrable�dm andZ

Ddm��u��cvfv���u� � cv

ZDdm��u�fv��u�� �������

Remark ������ Suppose m is a mass distribution on D� and c � R� and S is an m�

measurable subset of D� Suppose f � D R is de�ned by f��u� � c if �u � S� f��u� � � if

�u �� S� Then f is integrable�dm andZDdm��u�f��u� � cm�S�� �������

Remark ������ Suppose m is a mass distribution onD and f � D R and g � D Rare both integrable�dm and f��u� � g��u� for every �u � U � ThenZ

Ddm��u�f��u�� �

ZDdm��u�g��u�� ��������

��� CHAPTER ��� INTEGRAL CALCULUS OF TENSORS

Remark ������� Suppose m is a mass distribution on D and f � D R is integrable�

dm� Then so is jf j � D R� and �by �����������ZDdm��u�f��u�

���� � ZDdm��u� jf��u�j � ��������

Amended version of remark ��������

Remark ������� Suppose f � D R is measurable�m� Then so is jf j� and f is

integrable�dm i� jf j is integrable�dm� If f is integrable�dm then����ZDdm��u�f��u�

���� � ZDdm��u� jf��u�j �

Remark ������� Suppose f � D R and g � D R are measurable�m� and g is

integrable�dm� and jf j � g� Then f is integrable�dm�

Our goal is simply to extend the idea of integrating a function f with respect to

a mass distribution so that we can integrate vector�valued and tensor�valued function�

Since V�� � � ��Vq is a Euclidean vector space when V�� � � � � Vq are� it su!ces to consider

vector�valued functions�

De�nition ������� Let U and V be Euclidean vector spaces� Let D be a subset of U �

and let m be a mass distribution on D� Let �f � D V � We say that �f is integrable�dm

if for every �xed �v � V the function ��v � �f� � D R is integrable�dm� �Obviously we

de�ne �v � �f by requiring ��v � �f���u� � �v � �f��u� for all �u � D�� We de�ne Im��f � V R by

requiring for each �v � V

Im��f��v� �ZDdm��u���v � �f���u�� ��������

Corollary ������� If �f � D V is integrable�dm� then Im��f � L�V R��

Proof�

Suppose �v�� � � � � �vN � V and c�� � � � � cN � R� Then cv�vv � V � so �cv�vv� � �f is

integrable�dm� Moreover� �cv�vv� � �f � cv��vv � �f�� so by remark ���������

Im��f�cv�vv� �ZDdm��u�

h�cv�vv� � �f

i��u�

�ZDdm��u�

hcv��vv � �f�

i��u�

� cv

ZDdm��u���vv � �f ���u� � cvIm��f��vv��

����� INTEGRALS IN TERMS OF COMPONENTS ���

Corollary ������ If �f � D V is integrable�dm� there is a unique vector �V�m� �f � � V

such that for all �v � V � Im��f��v� � �v � �V�m� �f ��

Proof�

Immediate from theorem �� and corollary �������

De�nition ������� If �f � D V is integrable�dm� we denote the vector �V�m� �f � byRD dm��u��f��u� and call it the integral of �f with respect to m� From this de�nition and

��������� it follows that for every �v � V

�ZDdm��u��f��u�

�� �v �

ZDdm��u���v � �f���u� ��������

and that the vectorRD dm��u��f��u� is uniquely determined by the demand that ��������

hold for all �v � V �

Note ������ If V � V� � � � � � Vq� then the dot product on V is the generalized dot

product hqi� Then T � D V� � � � � � Vq is integrable�dm if for every Q � V� � � � � � Vq�

the function QhqiT � D R is integrable�dm� If it is� thenRD dm��u�T ��u� is the unique

tensor in V� � � � � � Vq such that for every Q � V� � � � � � Vq�ZDdm��u�T ��u�

�hqiQ �

ZDdm��u��QhqiT ���u�� ��������

Here �QhqiT ���u� �� Qhqi�T ��u��� If Q � �v��v� � � ��vq then �QhqiT ���u� � �T ��u��hqiQ �

�T ��u����v�� � � � � �vq� so �������� implies the special case

�ZDdm��u�T ��u�

���v�� � � � � �vq� �

ZDdm��u�f �T ��u�� ��v�� � � � � �vq�g ��������

for any ��v�� � � � � �vq� � V� � � � � � Vq�

���� Integrals in terms of components

Remark ������� Suppose U and V are Euclidean spaces� D � U � and m is a mass

distribution on D� Suppose �f � D V is integrable�dm� Suppose ����� � � � � ��n� is a basis

��� CHAPTER ��� INTEGRAL CALCULUS OF TENSORS

for V � with dual basis ����� � � � � ��n�� De�ne f i �� ��i � �f � Then f i � D R is integrable�dm

and �ZDdm��u��f��u�

�i�ZDdm��u��f

i��u�� �������

Proof�

This is a special case of de�nitions �������� ������� with �v � ��i in ���������

Lemma ������� Suppose V�� � � � � Vq are Euclidean spaces and� for p � f�� � � � � qg� ���p�� � � � � � ��p�np �

is a basis for Vp� with dual basis ����p�� � � � � ��npp��� Then the polyad bases f����i� � � � ��q�iq g andf��i��� � � � ��iqq�g are dual to one another�

Proof�

�����i� � � � ��q�iq

�hqi

���j��� � � � ��jqq�

�� �i�

j� � � � �iq jq �

Corollary ������ Suppose U � V�� � � � � Vq are Euclidean vector spaces� D� m is a mass

distribution on D� and T � D V� � � � � � Vq is integrable�dm� For p � f�� � � � � qg�suppose ���

p�� � � � � � ��p�np � is a basis for Vp� with dual basis ����p�� � � � � ��npp��� Suppose T i����iq �

���i��� � � � ��iqq��hqiT � Then T i����iq � D R is integrable�dm and

ZDdm��u�T i����iq��u� �

�ZDdm��u�T ��u�

�i����iq� ������

Proof�

Immediate from lemma ����� and remark ����� by setting� in remark

����� � f���� � � � � ��ng � f����i� � � � ����iq g and f���� � � � � ��ng � f��i��� � � � ��iqq�g�

Theorem ������� Suppose U and V are Euclidean vector spaces� D is a subset of U� m

is a mass distribution on D� and �f � D V � Suppose ����� � � � � ��n� is a basis for V and

f i � ��i � �f � Suppose that for each i � f�� � � � � ng� f i � D R is integrable�dm� Then �f

is integrable�dm�

����� INTEGRALS IN TERMS OF COMPONENTS ���

Proof�

Let �v � V � and vi � ��i ��v� Then �v � �f � vifi� Each f i is integrable�dm� so vif

i

is integrable�dm by remark �������� Thus� �v � �f is integrable�dm� Since this is

true for any �v � V � de�nition ������ is satis�ed� and �f is integrable�dm�

Corollary ������ Suppose U � V�� � � � � Vq are Euclidean spaces� D � U � m is a mass

distribution on D� and T � D V� � � � � � Vq� Suppose ���p�� � � � � � ��qnp� is a basis for

Vp with dual basis ���p�� � � � � ��npp��� Suppose T i����iq �� ��i��� � � � ��iqq��hqiT and that each

T i����iq � D R is integrable�dm� Then T is integrable�dm�

Proof�

Immediate from theorem ���� and lemma ������

Now it becomes obvious that remarks �������� �������� ������� are true if �f � D V

instead of f � D R� and if in remark �������� �c � V � We simply apply those remarks to

the component functions of �f relative to a basis for V � Remark ������� has no analogue

for vector�valued functions� because �f �g is not de�ned for vectors� However� ��������

does have a vector analogue� Suppose �f � D V � and suppose �f and also k�fk � D Rare integrable�dm� Then

kZDdm��u��f��u�k �

ZDdm��u�k�f��u�k� ������

Proof�

For any �xed �v � V � �v � �f is integrable�dm and so is k�vk k�fk� Also� j�v � �f j �k�vk k�fk� Therefore� using ��������������v � Z

Ddm��u��f��u�

���� �����ZDdm��u���v � �f���u�

���� � ZDdm��u�k�vkk�f��u�k

� k�vkZDdm��u�k�f��u�k�

If we set �v �RD dm��u��f��u� in this inequality� we obtain

k�vk� � k�vkZDdm��u�

����f��u���� �Cancelling one factor k�vk gives �������

��� CHAPTER ��� INTEGRAL CALCULUS OF TENSORS

We also have from remark ������ and theorem �����

Remark ������� Suppose U � V � W � X� Y are Euclidean spaces� D � U � and m is a

mass distribution on D� Suppose P � V � V �W � Q � X � Y � and T � D W �X is

integrable�dm� Then so are P � T � D V �X and T �Q � D W � Y � Moreover

ZDdm��u��P � T ���u� � P �

ZDdm��u�T ��u� �������

�ZDdm��u��T �Q���u�

���Z

Ddm��u�T ��u�

��Q� �������

Proof�

We consider P �T � The proof for T �Q is similar� Choose bases for V �W � X� Y

and take components with respect to these bases� Then �P �T ���u� � P � �T ��u��so

�P � T �ik��u� � ��P � T ���u��ik � �P � T ��u��ik � P ijT

jk��u��

By hypothesis� T is integrable�dm� Hence so is T jk� Hence so is P ijT

jk�

Hence so is �P � T �ik� Hence so is P � T � And�Z

Ddm��u��P � T ���u�

�ik�ZDdm��u��P � T ik���u�

�ZDdm��u�P i

jTjk��u� � P i

j

ZDdm��u�T jk��u�

� P ij

�ZDdm��u�T ��u�

�jk��P �

ZDdm��u�T ��u�

�ik�

Corollary ������� In remark ������� take W � W��� � ��Wp and X � X��� � ��Xq�

Then ZDdm��u��P hpiT ���u� � P hpi

ZDdm��u�T ��u� �������Z

Ddm��u��T hqiQ���u� �

�ZDdm��u�T ��u�

�hqiQ� �������

Chapter ��

Integral Identities

���� Linear mapping of integrals

Remark ������� Suppose U � V � W are Euclidean spaces� D � U � m is a mass distri�

bution on D� L � L�V W �� and �f � D V is integrable�dm� Then L � �f � D W is

integrable�dm and ZDdm��u��)L � �f���u� � L

�ZDdm��u��f��u�

�� ��������

In other words� L can be applied toRD dm��u��f��u� by applying it to the integrand�

Proof�

Let�

L� V �W be the tensor corresponding to L� Then �L� �f���u� � L��f��u�� �

�f��u�� �L� ��f � �L���u� so L� �f � �f � �L� and remark ������ says this is integrable

and that

ZDdm��u��L � �f���u� �

ZDdm��u���f � �L���u�

��Z

Ddm��u��f��u�

�� �L

� L�Z

Ddm��u��f��u�

��

QED�

��

��� CHAPTER ��� INTEGRAL IDENTITIES

.

..

τ( )u u

u

u

D

v

1

2C

Figure �����

Corollary ������� Suppose U � V � W � D� L and m as in remark �������� Suppose

T � D �qV is integrable�dm� Then ��qL� � T � D �qW is integral�dm and

ZDdm��u� ���qL� � T � ��u� � ��qL�

�ZDdm��u�T ��u�

�� �������

Proof�

A special case of remark �������� with V � W � L replaced by �qV � �qW and

�qL�

���� Line integral of a gradient

Theorem ������� Suppose U and V are Euclidean spaces� D is an open subset of U �

and �f � D V is continuously di�erentiable at each �u � D� Suppose C is a smooth

curve in D whose two endpoints are �u� and �u�� Suppose dl��u� is the in�nitesimal element

����� LINE INTEGRAL OF A GRADIENT ���

of arc length on C� and ���u� is the unit vector tangent to C at �u and pointing along C in

the direction from �u� to �u�� Then

ZCdl��u�� � � �r�f���u� � �f��u��� �f��u��� �������

Proof�

We assume that the reader knows this theorem when V � R� and �f is a real�

valued function� Let ����� � � � � ��n� be a basis for V � with dual basis ����� � � � � ��n��and de�ne f i �� ��i � �f � Since �f is di�erentiable at each �u� so is f i� Thus �rf i��u�exists� Moreover� �r��i �

��� so by equation �������� �rf i��u� � �r��f � ��i���u� �

�r�f��u� � ��i� Both �r�f and ��i are continuous� and then so is �r�f � ��i� Thus

f i � D R is continously di�erentiable� By the known scalar version of

�������� ZCdl��u� � � �rf i��u� � f i��u��� f i��u���

If we multiply this equation by ��i and sum over i we obtain

�f��u��� �f��u�� ��Z

Cdl��u� � � �rf i��u�

���i

�ZCdl��u�

h� � � �rf i���i

i��u� by ����� with q � �

�ZCdl��u�� � � �r�f���u� because

� � � �rf i���i � � �h��rf i ���i

i� � � �r�f i��i� � � � �r�f�

QED�

Corollary ������� Suppose D is open and arcwise connected �any two points in D can

be joined by a smooth curve lying wholly in D�� Suppose �f � D V is continuously

di�erentiable and �r�f��u� ��� for all �u � D� Then �f is a constant function�

Proof�

For any �u�� �u� � D� �f��u�� � �f��u�� by theorem �����

��� CHAPTER ��� INTEGRAL IDENTITIES

���� Gauss�s theorem

Theorem ������� Suppose U and V are Euclidean spaces� D is an open subset of U � D

is the boundary of D� and n��u� is the unit outward normal at �u � D� Let 'D � D�D �

closure ofD� Suppose�

f � 'D U�V is continuous and�

f jD is continuously di�erentiable

at each �u � D� Let dV ��u� and dA��u� be the in�nitesimal volume element in D and the

in�nitesimal surface area element on D respectively � dV ��u� has dimension dimU � and

dA��u� has dimension one less�� ThenZ�D

dA��u�� n� �f ���u� �ZDdV ��u���� �f ���u�� �������

Proof�

We assume that the reader knows the corresponding theorem for vector�valued

functions� �f � 'D U � Let ����� � � � � ��n� be a basis for V � with dual basis

����� � � � � ��n�� De�ne �fi� 'D U by �f

i��

f ���i� so �f i��u� ��

f ��u����i� Note that�fi��u� is not a component but a vector in U � By corollary ������ �f

i� 'D U

is continuous� and by section ���� �fiis di�erentiable� By corollary ������

��fiis continuous� Thus� �f

i� 'D U is continuous and �f

ijD is continuously

di�erentiable� Therefore� �fiobeys Gauss�s theorem�Z

�DdA��u�� n � �f i

���u� �ZDdV ��u���r � �f i

���u��

If we multiply on the right by ��i and sum over i� ������� with q � � givesZ�D

dA��u��� n � �f i

���i

���u� �

ZDdV ��u�

���r � �f i

���i

���u�� ������

But �fi��i �

f ����i��i� ��

f ����i��i� � ��

f � �I V � ��

f so � n��f i���i � n���f i��i� � n� �f �

and by section ���

��r � �f i���i �

��

I U hi�r�f i���i �

I U hi���r�f i��i�

��

I U hih�r��f i��i�

i��

I U hi�r�

f� �r� �f �

Substituting these expressions in ������ gives ������

����� STOKES�S THEOREM ���

Figure ����

��� Stokes�s theorem

Theorem ������� Suppose �U�A� is an oriented Euclidean �space� and A is used to

de�ne cross products and curls� Suppose D is an open subset of U � and S is a smooth

two�sided surface in D� with boundary curve S� Suppose n � S U is one of the two

continuous functions such that n��u� is a unit vector in U perpendicular to S at �u � S� If

�u � S� let � ��u� be the unit vector tangent to S at �u and pointing so that n��u�� � ��u�

points into S� Suppose that�

f � D U � V is continuously di�erentiable everywhere in

D� Suppose that dA��u� is the element of area on S and dl��u� is the element of arc length

on S� Then Z�Sdl��u�� � � �f ���u� �

ZSdA��u�

� n � ��r� �

f ����u�� ��������

Proof�

Again� we assume the reader knows this theorem when V � R� i�e� for

vector �elds �f � D U � Let ����� � � � � ��n� be a basis for V � with dual basis

����� � � � � ��n�� and de�ne �f i � �f � ��i� Then we can use Stoke�s theorem for �fi�

i�e�� replace�

f by �fiin ��������� If we multiply both sides of the result on the

��� CHAPTER ��� INTEGRAL IDENTITIES

right by ��i� sum over i� and use ������� with q � �� we getZ�Sdl��u�

�� � � �f i

���i

���u� �

ZSdA��u�

�� n � ��r� �f

i����i

���u��

As in the proof of Gauss�s theorem� �fi��i �

f � so � � � �f i���i � � � ��f i��i� � � � �f �

And � n � ��r� �f i����i � n � ���r� �fi���i�� Finally�

�r� �fi���i �

�Ahi�r�f i

���i

� Ahi���r�f i

���i

�� Ahi

��r��f i��i�

�� Ahi

��r �

f�

� �r� �

f �

This �nishes the proof�

��� Vanishing integral theorem

Lemma ������� Suppose D is an open subset of Euclidean space U � f � D R is

continuous� andRD� dV ��u�f��u� � � for every open subset D� of D� Here dV ��u� is the

volume element in U � Then f��u� � � for all �u � D�

Proof�

Suppose �u� � D and f��u�� �� �� If f satis�es the hypotheses so does �f � soif f��u��� � we may replace f by �f � Therefore� we may assume f��u�� � ��

Then there is a � � � such that if j�u��u�j � then �u � D and jf��u��f��u��j f��u���� But then �f��u��� f��u�� f��u�� f��u���� so f��u� � f��u��� if

�u � B��u�� ��� the open ball of radius � centered at �u�� ThenZB�u���

dV ��u�f��u� �ZB�u���

dV ��u�f��u��

� jB� �u�� ��jf��u��

where jB��u�� ��j � volume of B��u�� ��� But jB��u�� ��jf��u��� � �� and

B��u�� �� is an open subset of D� so we must haveRB�u���

dV ��u�f��u� � ��

This contradiction shows f��u�� � ��

����� TIME DERIVATIVE OF AN INTEGRAL OVER A MOVING VOLUME ���

n

V

K(t+ t)

{

IIIIII

.

δW t

δK(t)

xx

xx

x

x

x

x

x

u

∂ ∂

Figure ����

Theorem ������� Suppose U and V are Euclidean spaces� D is an open subset of U �

and �f � D V is continuous� Suppose that for every open subset D� of D� we haveRD� dV ��u��f��u� � ��V � Then �f��u� � ��V for all �u � D�

Proof�

Let ����� � � � � ��n� be a basis for V � with dual basis ����� � � � � ��n�� Let f i �

��i � �f � Then f i � D R is continuous� and for every open subset D� of D�RD� dV ��u�f i��u� � �� Hence� f i � � by lemma ������� Hence �f � f i��i � ��V �

��� CHAPTER ��� INTEGRAL IDENTITIES

���� Time derivative of an integral over a moving

volume

Suppose that for each instant t we are given an open set K�t� in Euclidean vector space

U � with boundary surface K�t� and outward unit normal n� Suppose that K�t� changes

with time su!ciently smoothly that its surface K�t� has a well de�ned speed normal to

itself� At �u � K�t�� this speed is W ��u� t�� and it is � � if K�t� is moving outward from

K�t�� � if inward�

In �gure ���� W � � on the part of K�t� marked with x�s and W � on the part

of K�t� marked with o�s�

Suppose that all the sets K�t� are subsets �K�t� � K�t��K�t� � closure of K�t�� of

an open set D� and �f � D �R V � where V is another Euclidean space� Thus for each

�u � D and t � R� �f��u� t� is a vector in V � We suppose this vector depends continuously

on �u and is continuously di�erentiable in t �i�e� �f��� t� is continuous for each t � R� and

�f��u� �� is continuously di�erentiable for each �u � D �� Then

d

dt

ZKt�

dV ��u��f��u� t� �ZKt�

dV ��u�t �f��u� t�

�Z�Kt�

dA��u��W �f���u� t�� ��������

Here t �f��u� t� stands for ��f ��u� ����t�� or �f ��u� t��t�Proof�

Let K�t�� stand for the part of K�t� marked with x�s in �gure ���� while

K�t�� stands for the part marked with o�s� Let II be the set K�t�K�t��t��

while I is K�t�nK�t� �t�� and III is K�t� �t�nK�t�� See �gure ���� Let

�F �t� �ZKt�

dV ��u��f��u� t��

Then

�F �t� �ZIdV ��u��f��u� t� �

ZIIdV ��u��f��u� t��

�F �t� �t� �ZIIdV ��u��f��u� t� �t� �

ZIIIdV ��u��f��u� t� �t��

���� CHANGE OF VARIABLES OF INTEGRATION ���

Correct to �rst order in �t�

�f��u� t� �t� � �f��u� t� � ��t� t �f��u� t� in D

dV ��u� � ��t�W ��u� t�dA��u� in III� where W � �

dV ��u� � ���t�W ��u� t�dA��u� in I� where W ��

Thus� correct to �rst order in �t�ZIdV ��u��f��u� t� � ���t�

Z�Kt��

dA��u�W �f��u� t�ZIIIdV ��u��f��u� t� � ��t�

Z�Kt��

dA��u�W �f��u� t��

Therefore�ZIIIdV ��u��f��u� t��

ZIdV ��u��f��u� t� � ��t�

Z�Kt�

dA��u�W �f��u� t�� �������

Also� ZIIdV ��u�

h�f��u� t� �t�� �f��u� t�

i� ��t�

ZIIdV ��u�t �f��u� t�

� ��t�ZKt�

dV ��u�t �f��u� t�� ��t�ZIdV ��u�t �f��u� t��

ButRI dV ��u�t �f � ���t� R�Kt�� dA��u�Wtf � correct to �rst order in �t� There�

fore� correct to �rst order in �t�ZIIdV ��u�

h�f��u� t� �t�� �f��u� t�

i� ��t�

ZKt�

dV ��u�t �f��u� t�� �������

Combining ������� ������� we have� correct to �rst order in �t�

�F �t� �t�� �F �t� � �t

�Z�Kt�

dA��u�W �f��u� t� �ZKt�

dV ��u�t �f��u� t�

��

Dividing by �t and letting �t � gives ���������

��� Change of variables of integration

Suppose �U�AU� and �V�AV � are two oriented n�dimensional Euclidean spaces� Suppose

L � L�U V �� For any �u�� � � � � �un � U � de�ne A��u�� � � � � �un� � AV �L��u��� � � � � L��un���

��� CHAPTER ��� INTEGRAL IDENTITIES

Clearly A � �nU � so there is a scalar kL such that A � kLAU � We denote this scalar by

detL and call it the determinant of L� Then

AV �L��u��� � � � � L��un�� � �detL�AU ��u�� � � � � �un� ��������

for all ��u�� � � � � �un� � XnU � Note that detL changes sign if we replace either AU or AV by

its negative� Thus detL depends not only on L but on how U and V are oriented� For a

given L� detL can have two values� one the negative of the other� and which value it has

depends on which of the two unimodular alternating tensors �AU and �AV are used to

orient U and V �

If � x�� � � � � xn� is a pooob in U and � y�� � � � � yn� is a pooob in V and Lij � xi��

L � yj�then� L� xi� � Lij yj and so

detL � �detL�AU� x�� � � � � xn� � AV �L� x��� � � � � L� xn��� AV �Lij� yj�� � � � � Lnjn yjn� � L�j� � � �LnjnAV � yj�� � � � yjn��

Therefore detL � L�j� � Lnjn�i����in � det�Lij�� �������

Now suppose H and K are open subsets of U and V respectively and �r � H K is

a continuously di�erentiable bijection of H onto K� For �x � H write �x � xi xi and de�ne

rj � yj � �r � At any �x � H we have irj � ��r�r �ij � xi � ��r�r � � yj so the determinent of

the matrix irj� x� is det �r�r ��x�� as we see from ��������

Finally� suppose f � K R is integrable with respect to volume in V � Then f � r �

H R is integrable with respect to volume in U and

ZKf�rj yj�dr� � � �drn �

ZHf ��r �xk xk��

�����det r jxi

����� dx� � � �dxn� �������

The reader is presumed to know this formula for changing variables in a multiple integral�

The determinant det�rj�xi� is the Jacobian of the coordinate transformation� We note

that rj�xi � irj� so by �������� at �x � H

det

�rjxi

�� det �r�r ��x�� ��������

�Proof Lij �yj � xi��

L ��yj �yj � �xi��

L

���� CHANGE OF VARIABLES OF INTEGRATION ��

The local volume elements in U and V are dVU��x� � dx� � � �dxn and dVV ��r� � dr� � � �drn�so ������� can be written

ZKdVV ��r�f��r� �

ZHdVU��x�

���det �r�r ��x���� ��f � �r ���x�� ��������

Now suppose W is another Euclidean space� with basis ����� � � � � ��p� and dual basis

����� � � � � ��p�� Suppose �f � K W is integrable� Then so is fi � ��i � �f � and �������� holds

for each fi� If we multiply the resulting equations by ��i and sum over i we obtain

ZKdVV ��r��f��r� �

ZHdVU��x�

���det �r�r ��x���� ��f � �r ���x�� ��������

��� CHAPTER ��� INTEGRAL IDENTITIES

Part II

Elementary Continuum Mechanics

��

Chapter ��

Eulerian and Lagrangian Descriptions

of a Continuum

���� Discrete systems

For some purposes� some physical systems �for example� the solar system� can be regarded

as consisting of a �nite number N of mass points� A complete description of the motion

of such a system consists simply of N vector�valued functions of time� the value of the

��th function at time t is the position of the ��th particle at time t� If the values of the

positions �r��t�� � � � � �rN�t� of all particles are known for all times� nothing more can be said

about the system�

To predict the motion of such a �discrete� system� one must usually add other prop�

erties to the model� In the case of the solar system� one ascribes to each particle a mass�

m being the mass of the ��th particle� Then� if all the velocities and positions are known

at some time t�� all the functions �r�t� can be calculated from Newton�s laws of motion

and gravitation�

If the system of particles is a classical �pre�quantum� atom� one needs not only their

masses but their charges in order to calculate their motion�

And� of course� for some purposes the discrete model of the solar system is too crude�

It cannot model or predict the angular acceleration of the earth�s moon due to the tidal

���

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

bulges on the earth� A point mass has no tidal bulges� Nevertheless� the discrete model

of the solar system is very useful� No one would suggest abandoning it because the solar

system has properties not described by such a model�

���� Continua

A parcel of air� water or rock consists of such a large number of particles that a discrete

model for it would be hopelessly complicated� A di�erent sort of model has been developed

over the last three centuries to describe such physical systems� This model� called a

�continuum�� exploits the fact that in air� water and rock nearby particles behave similarly�

The continuum model for a lump of material regards it as consisting of in�nitely many

points� in fact too many to count by labeling them even with all the integers� �There are

too many real numbers between � and � to label them as x�� x�� � � � � Any in�nite sequence

of real numbers between � and � will omit most of the numbers in that interval� This

theorem was proved by Georg Cantor in about ������ Nearby points are given nearby

labels� One way to label the points in the continuum is to give the Cartesian coordinates of

their positions at some particular instant t�� relative to some Cartesian coordinate system�

Then each particle is a point to which is attached a number triple �x�� x�� x�� giving the

Cartesian coordinates of that particle at the �labelling time� t�� This number triple is a

label which moves with the particle and remains attached to it� Nearby particles have

nearly equal labels� and particles with nearly equal labels are close to one another� The

collection of all particles required to describe the motion will usually �ll up an open set

in ordinary real �space�

Of course there are many ways to label the particles in a continuum� The label attached

to a particle could be the number triple �r� �� � giving the values of its radius� colatitude

and longitude in a system of polar spherical coordinates at the labelling time t��

Another label would be simply the position vector �x at time t� of the particle relative

to some �xed origin and axes which are unaccelerated and non�rotating �so that Newton�s

laws can be used�� In fact� labelling by number triples can also be thought of as labelling

����� CONTINUA ���

by vectors� since a number triple is a vector in the Euclidean vector space R�� �Rn is the

set of real n�tuples� with addition and multiplication by scalars de�ned coordinate�wise

and with the dot product of �u�� � � � � un� and �v�� � � � � vn� de�ned to be uivi��

In order to leave ourselves freedom to choose di�erent ways of labelling� we will not

specify one particular scheme� We will simply assume that there is a three�dimensional

oriented real Euclidean space� �L�AL�� from which the labels are chosen� The labels

describing a continuum will be the vectors �x in a certain open and subset H of label

space L� We will denote real physical space by P � and we will orient it in the usual way�

denoting the �right handed� unimodular alternating tensor by AP � or simply A� We use

this notation�

�r L��x� t� �� position at time t of the particle labelled �x� ������

The label �x is in the open subset H of label space L� and the particle position �r L��x� t�

is in P � One special case which is easy to visualize is to take L � P � and to choose

some �xed time t�� and to label the particles by their positions at t�� With this labelling

scheme� called �t��position labelling�� we have

r L��x� t�� � �x� for t��position labelling� �����

The motion of a continuum is completely described by giving the position of every

particle at every instant� i�e� by giving �r L��x� t� for all �x � H and all t � R� This amounts

to knowing the function

r L � H �R P� �����

This function is called the Lagrangian description of the motion of the continuum�

We denote by K�t� the set in position space P occupied by the particles of the con�

tinuum at time t� We assume that particles neither �ssion nor coalesce� each particle

retains its identity for all time� Therefore we need never give two particles the same label

�x� and we do not do so� But then if �r � K�t� there is exactly one particle at position �r at

time t� We denote its label by �x E��r� t�� Thus

x E��r� t� �� label of particle which is at position �r at time t� ������

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

From the de�nitions� it is clear that if �r � K�t� and �x � H then

�r � �r L��x� t�� �x � �x E��r� t�� ������

But ������ says that �r L��� t� � H K�t� is a bijection� whose inverse is �x E��� t� �

K�t� H� If we know the function �x E��� t�� we can �nd the function �r L��� t�� For

any �x � H� �r L��x� t� is the unique solution �r of the equation �x � �x E��r� t�� Therefore�

knowing the label function �x E is equivalent to knowing �r L� Either function is a complete

description of the motion of the continuum� and either can be found from the other� For

any �xed t�

�r L��� t��� � �x E��� t�� ������

In order to enforce that nearby particles have nearby labels we will assume not only

that �r L and �x E are continuous� but that they are continuously di�erentiable as many

times as needed to make our arguments work� �Usually� twice continuously di�erentiable

will su!ce�� Since ������ implies �r L��� t� � �x E��� t� � IP jK�t�� the identity function

on P � restricted to K�t�� and also �x E��� t� � �r L��� t� � ILjH� the identity function on

L� restricted to H� therefore the chain rule implies that if �r � �r L��x� t� or� equivalently�

�x � �x E��r� t�� then ��������������r�r L��x� t� � �r�x E��r� t� �

I L

�r �x E��r� t� � �r�r L��x� t� ��

I P �

������

���� Physical quantities

If the motion of a continuum is to be predicted from physical laws� more properties of

the continuum must be considered than simply the positions of its particles at various

times� Most of the useful properties are local� i�e� they have a value at each particle at

each instant� Examples are temperature� mass density� charge density� electric current

density� entropy density� magnetic polarization� the electrical conductivity tensor and� in

a �uid� the pressure�

����� PHYSICAL QUANTITIES ���

A local physical property f can be described in two ways� at any instant t� we can

say what is the value of f at the particle labelled �x� or we can say what is the value of f at

the particle whose position is �r at time t� The two values are the same� but they are given

by di�erent functions� which we denote by fL and fE� We call these the Lagrangian and

the Eulerian descriptions of f � They are de�ned thus�

fL��x� t� � value of physical quantity f at time t at the particle labelled �x� ������

fE��r� t� � value of physical quantity

f at time t at the particle whose position is �r at time t� �����

The physical quantity f can be a scalar� vector or tensor� From the de�nitions� obviously�������������if �r � �r L��x� t� or �x � �x E��r� t� then

fL��x� t� � fE��r� t��

�����

This is equivalent to asserting�������������fL��x� t� � fE

h�r L��x� t�� t

iand

fE��r� t� � fLh�x E��r� t�� t

i�

������

These two assertions are equivalent in turn to

�������������fL��� t� � fE��� t� � �r L��� t�

fE��� t� � fL��� t� � �x E��� t��������

If we have the Lagrangian description of the motion� the function �r E� then we can �nd

the labelling function �x E� Then using ������ or ������� we can �nd both fE and fL

if we know either one of them�

So far� our discussion of �physical quantities� has been somewhat intuitive� We have

found some mathematical rules which are satis�ed by certain entities that we feel com�

fortable to call physical quantities� In the spirit of mathematical model building� it is

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

useful to give a precise de�nition to what will qualify as a �physical quantity�� Because

of what we have learned above� we introduce

De�nition ������� Given a continuum whose motion is described by �r L � H�R P �

and whose labelling function is �x E� with �x E��� t� � K�t� H� a physical quantity f is

an ordered pair of functions �fL� fE� with these properties�

i� The domain of fL is H �R

ii� The domain of fE is a subset of P �R� and the domain of fE��� t� is K�t�

iii� If �r L��x� t� � �r � or �x E��r� t� � �x� then fL��x� t� � fE��r� t� �

The function fL is the Lagrangian description of the physical quantity f � and fE is the

Eulerian description of f � Property iii� is equivalent to�������������fL��x� t� � fE

h�r L��x� t�� t

ior to

fE��r� t� � fLh�x E��r� t�� t

ifor all t � R�

������

These in turn are equivalent to�������������fL��� t� � fE��� t� � r L��� t� or to

fE��� t� � fL��� t� � �x ��� t� for all t � R�������

Temperature� mass density� and the other items enumerated beginning on page ���

are physical quantities in the sense of de�nition ������ The advantage of having such

a formal de�nition is that it enables us to introduce new physical quantities to suit our

convenience� We can de�ne a new physical quantity f by giving either its Eulerian

description fE or its Lagrangian description fL� The missing description can be obtained

from ������ or ������� assuming that we have �r L� the Lagrangian description of the

motion of the continuum� A somewhat trivial example is the f such that fL��x� t� � � for

all �x� t� Clearly fE��r� t� � �� and we can regard � �or any other constant� as a physical

quantity�

����� DERIVATIVES OF PHYSICAL QUANTITIES ��

A more interesting example is a physical quantity which we call �particle position��

and denote by �r � It is de�ned by requiring that its Eulerian description be

�r E��r� t� � �r for all t � R and �r � K�r�� ������

From ������� its Lagrangian description is the �r L of ������� the Lagrangian de�

scription of the motion�

Another interesting example is a physical quantity we call �particle label�� and denote

by �x � It is de�ned by requiring that its Lagrangian description be

�x L��x� t� � �x for all ��x� t� � H �R� ������

From ������� its Eulerian description is �x E� the label function� The fact that the physical

quantity �x is conserved at each particle turns out to be very useful in some variational

formulations of the equations of motion of a �uid� although at �rst glance the fact seems

rather trivial�

��� Derivatives of physical quantities

It will be useful to introduce the following notation for the derivatives of a physical

quantity f � �fL� fE��

DtfL��x� t� � lim

���

fL��x� t� ��� fL��x� t�

� time derivative of fL��x� t� at the particle label �x� �������

�DfL��x� t� � �rhfL��� t�

i��x� � gradient tensor of the function fL��� t�

at the particle label �x� ������

tfE��r� t� � lim

���

fE��r� t� ��� fE��r� t�

� time derivative of fE��r� t� at spatial position �r� ������

�fE��r� t� � �rhfE��� t�

i��r� � gradient tensor of the function fE��� t�

at the position �r in P� �������

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

From these de�nitions� ������� and the chain rule� it is clear that if �r � �r L��x� t� �or

�x � �x E��r� t�� then

�DfL��x� t� � �D�r L��x� t� � �fE��r� t� �������

�fE��r� t� � � �x E��r� t� � �DfL��x� t�� �������

It is also useful to relate DtfL and tf

E� We will calculate the former from the latter�

We assume that �r and �x are chosen so

�r � �r L��x� t��

Then we de�ne �h��� for any real � by

�h��� � �r L��x� t� ��� �r L��x� t��

We have ��������������h��� � �Dt�r

L��x� t� � � �R���� where

�R���� � as � ��

�������

We also have

fL��x� t� �� � fEh�r L��x� t� ��� t� �

i� fE

h�r � �h���� t � �

i

� fEh�r � �h���� t

i� �tf

Eh�r � �h���� t

i� �R����

where R���� � as � �� De�ne

R���� � R���� � tfEh�r � �h���� t

i� tf

E ��r� t� �

Since �h��� �� as � � therefore�������������R���� � as � � and

fL��x� t� �� � fEh�r � �h���� t

i� �tf

E��r� t� � �R�����

�������

����� DERIVATIVES OF PHYSICAL QUANTITIES ���

Also �������������fE��r � �h� t� � fE��r� t� � �h � �fE��r� t� � k�hkR���h�

where R���h� � as �h ���

�������

Substituting ������� in ������� gives

fE��r � �h� t� � fE��r� t� � �Dt�rL��x� t� � �fE��r� t�

��R���� ��������

where

R���� � �R���� � �fE��r� t�����Dt�r

L��x� t� � �R�������R�

h�h���

i�

Thus R���� � as � �� Moreover� fE��r� t� � fL��x� t�� so substituting �������� in

������� gives

fL��x� t� �� � fL��x� t� � �htf

E��r� t� �Dt�rL��x� t� � �fE��r� t�

i�� �R���� �R����� ��������

where R���� �R���� � as � �� From ������ it follows immediately that�������������Dtf

L��x� t� � tfE��r� t� �Dt�r

L��x� t� � �fE��r� t�

where �r � �r L��x� t� � or �x � �x E��r� t���

�������

The formal mathematical relationship between labels and positions is symmetrical�

That is� we can interpret �r as label and �x as position� Therefore� we can infer immediately

from ������ that

tfE��r� t� � Dtf

L��x� t� � t �xE��r� t� � �DfL��x� t�

where r � �r L��x� t� �or �x � �x E��r� t��� �������

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

Alternatively� ������� can be computed in the same way as was �������� Note the

consequence of ������� ������ that for any physical quantity f �

Dt�rL��x� t� � �fE��r� t� � t �x

E��r� t� � �DfL��x� t� � �� ��������

The notation has become somewhat cumbersome� and can be greatly simpli�ed by

introducing some new physical quantities�

De�nition ������� Let �r be the physical quantity �particle position� �see bottom of

page ���� and let f be any physical quantity� The physical quantities �v� �a � �Df � �f �

Dtf � tf are de�ned as follows�

�vL��x� t� � Dt�rL��x� t� ��������

�aL��x� t� � Dt�vL��x� t� ��������

� �Df�L��x� t� � �DfL��x� t� ��������

��f�E��r� t� � �fE��r� t� ��������

�Dtf�L��x� t� � Dtf

L��x� t� ��������

�tf�E��r� t� � tf

E��r� t�� �������

The names of these physical quantities are as follows�

�v � particle velocity

�a � particle acceleration

�Df � label gradient of f

�f � spatial gradient of f

Dtf � substantial time derivative of f � material time derivative of f � time derivative

moving with the material�

tf � partial time derivative of f �

����� DERIVATIVES OF PHYSICAL QUANTITIES ���

With these de�nitions� the relations among the derivatives of a physical quantity become

�v � Dt�r �������

�a � Dt�v ������

�Df � �D�r � �f ������

Dtf � tf � �v � �f �������

�f � � �x � �Df �������

tf � Dtf � t �x � �Df �������

The function �vE is called the Eulerian description of the motion of the continuum��

If the Lagrangian description is known� the Eulerian description can be obtained from

�������� The converse is true� in the following sense�

Remark ������� Suppose the Eulerian description �vE of a continuum is given� Suppose

that at one instant t�� the function �r L��� t�� is known� that is� for each �x� the position

at time t� of the particle labelled �x is known� Then the Lagrangian description of the

motion can be calculated� That is� the position �r L��x� t� of the particle labelled �x can be

calculated for every time t�

Proof�

By de�nition of �v� we have for �r � �r L��x� t� the relation Dt�rL��x� t� � �vE��r� t�

or

Dt�rL��x� t� � �vE

h�r L��x� t�� t

i� �������

But ������� is an ordinary di�erential equation in t if �x is �xed� Therefore�

for any �xed particle label �x� we can solve ������� for �r L��x� t� if we know

�r L��x� t�� for one t� � QED

Corollary ������� If �vE is known� then the Lagrangian description of the motion can

be calculated without further information if the particles are �t��position labelled�� i�e� if

their labels are their positions at some tme t��

�If �t�vE � ��� the motion is called steady��

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

Proof�

With t��position labelling �r L��x� t�� � �x� so �r L��x� t�� is known for all particle

labels �x�

��� Rigid body motion

For this discussion we need a lemma about linearity�

Lemma ������ Suppose V and W are real vector spaces and f � V W � Suppose f

has the following properties� for any nonzero �x� �y� �z � V �

i� f���� � ��

ii� f���x� � �f��x�

iii� f�a�x� � af��x� for any positive real a �i�e�� a � R� a � ���

iv� If �x� �y � �z � �� then f��x� � f��y� � f��z� � ���

In that case� f is linear�

Proof�

a� For any c � R and �v � V � f�c�v� � cf��v�

a�� If c�v � ��� either c � � or �v � ��� so f�c�v� � �� � cf��v��

a� If c�v �� �� and c � �� use iii� above�

a� If c�v �� �� and c �� then f���c��v� � ��c�f��v� by iii� above� But

f���c��v� � f��c�v� � �f�c�v� by ii� above�

Hence �f�c�v� � �cf��v��

b� For any �u��v � V � f��u� �v� � f��u� � f��v��

b�� If �u or �v is ��� one of f��u� and f��v� is �� by i� above� E�g�� if �u � ���

f��u� �v� � f��v� � f��u� � f��v��

����� RIGID BODY MOTION ���

b� If �u and �v are nonzero but �u � �v � ��� then �v � ��u so f��u � �v� �

f���� � �� � f��u�� f��u� � f��u� � f���u� � f��u� � f��v��

b� If �u� �v and �u��v are nonzero� let �w � �u��v� Then �u��v����w� � �� so

by iv�� f��u��f��v��f���w� � ��� By ii� this is f��u��f��v��f��w� � ���

or f��w� � f��u� � f��v��

QED

The motion of a continuum is called �rigid body motion� if the distance separating

every pair of particles in the continuum is independent of time� We want to study rigid

body motions�

Step �� Choose one particular particle in the continuum� and call it the pivot particle�

Step �� Choose t� � R� Choose a reference frame for real physical space P so that its

origin is at the position of the pivot particle at time t��

Step �� Introduce t��position labelling to give the Lagrangian description of the motion�

�r L� Let H denote the open subset of L � P consisting of the particle labels� The

pivot particle is labelled ��� so �� � H� and

�r L��x� t�� � �x for all �x � H�

Step �� For each t � R� de�ne a mapping �rt � H P by

�rt��x� � �r L��x� t�� �r L���� t�

and de�ne �R � R P as �R�t� � �r L���� t�� Thus

�r L��x� t� � �R�t� � �rt��x� �������

for all �x � H and t � R� Moreover� for any t � R and any particle labels �x� �y � H�

�r L��x� t�� �r L��y� t� � �rt��x�� �rt��y�� so� by the de�nition of a rigid body�

k�rt��x�� �rt��y�k � k�x� �yk for all t � R and �x� �y � H� ������

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

Also� from the de�nition of �R�t� and ��������

�rt���� � �� for all t � R� ������

The main result implied by the above is contained in

Theorem ������ Suppose f � H P where H is an open subset of real Euclidean

vector space P � Suppose f���� � �� and for any �x� �y � H� kf��x�� f��y�k � k�x� �yk� Thenthere is a unique L � "�P � �i�e�� L is linear and orthogonal� such that for all �x � H

f��x� � �x� �L� L��x�� �������

Proof�

Lemma �� kf��x�k � k�xk for any �x � H�

Proof� kf��x�k � kf��x����k � kf��x�� f����k � k�x���k � k�xk �

Lemma � f��x� � f��y� � �x � �y for all �x� �y � H�

Proof� kf��x�� f��y�k� � k�x� �yk� so

kf��x�k� � f��x� � f��y� � kf��y�k� � k�xk� � �x � �y � k�yk��

But by lemma �� kf��x�k� � k�xk� and kf��y�k� � k�yk��

Lemma � Suppose a�� � � � � aN � R� �x�� � � � � �xN � H� and ai�xi � H� Then

f�ai�xi� � aif��xi��

Proof� kf�ai�xi��aif��xi�k� � kf�ai�xi�k��ajf��xj��f�ai�xi� �kaif��xi�k� �kai�xik� � aj �f��xj� � f�ai�xi�� �aiajf��xi� � f��xj�� By lemma � this

is kai�xik�� aj �xj � �ai�xi�� � aiaj��xi � �xj� � kai�xik�� aiaj�xi � �xj � ��

Lemma �� Choose � � � so small that if k�xk � then �x � H� De�ne

L � P P as follows�

i� L���� � ��

ii� L��x� � �k�xk���f���x�k�xk� if �x �� ���

����� RIGID BODY MOTION ���

Then L��x� � f��x� for all �x � H� and L satis�es conditions i��iv� of

lemma ������

Proof� First� if �x � H then in lemma take N � �� a� � ��k�xk� �x� � �x�

Lemma then implies f���x�k�xk� � ��k�xkf��x�� and from ii� above�

obviously L��x� � f��x�� Now for i��iv� of lemma ������

i� is obvious

ii� L���x� � �k��xk��� f����x�k��xk� � �k�xk���f����x�k�xk�� In lemma

above� take N � �� a� � ��� �x� � ��x�k�xk� Then f����x�k�xk� ��f���x�k�xk�� Thus L���x� � �L��x��

iii� L�a�x� � �ka�xk���f��a�x�ka�xk� � a�k�xk���f��a�x�ak�xk�� a�k�xk���f���x�k�xk� � aL��x� if a � � and �x �� ���

iv� Suppose �x � �y � �z � �� and �x� �y� �z nonzero� Choose � so small

�� � R� � � �� that ��x� ��y� ��z � H� Then ��x� ��y� ��z � ��� so�

using lemma � �� � f���� � f���x���y���z� � f���x��f���y��f���z��

Since ��x � H� f���x� � L���x�� and similarly for �y� �z� Thus �� �

L���x� � L���y� � L���z�� By iii� above� L���x� � �L��x�� etc�� so

�� � ��L��x� � L��y� � L��z��� Thus L��x� � L��y� � L��z� � ���

It follows from lemma � above and lemma ����� that L � P P is

linear� For any �x �� �� in P � choose � so small �� � R� � � �� that

��x � H� Then ��kL��x�k� k�xk� � k�L��x�k� k��xk � kL���x�k� k��xk �kf���x�k � k��xk � k��xk � k��xk � ��

Thus kL��x�k � k�xk for all �x � P � Therefore L � "�P �� QED

By theorem �������� for each t � R there is an orthogonal tensor�

L �t� � P �P such

that in equation ������� we can write

�r L��x� t� � �R�t� � �x� �L �t� for all �x � H� �������

Then clearly

�D�r L��x� t� ��

L �t� for all �x � H� �������

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

We are assuming that �D�r L��x� t� is continuous in �x and t� so�

L �t� is a continuous function

of t� At the labelling time we have �R�t�� � �r L���� t�� � �� and so �x� �L �t�� � �r L��x� t�� � �x�

Thus�

L �t�� ��

I P�R�t�� � �� �������

and det�

L �t�� � ��� But detL�t� depends continuously on t since L�t� does� and detL�t�

is always �� or �� since L�t� is orthogonal� Therefore

detL�t� � ��� L�t� � "��P � for all t � R� �������

Every rigid motion can be written as in ������� with ������� and ��������

Now let us �nd the Eulerian description of the motion whose Lagrangian description

is �������� We have

�vL��x� t� � Dt�rL��x� t� � �t �R�t� � �x � �t

L �t�� �������

�Note� Since �R and�

L are functions of t alone� we write their time derivatives as �t �R and

�t�

L� so as not to confuse them with t or Dt��

To obtain the Eulerian description from ������� we must express �x in terms of �r�

where as usual

�r � �r L��x� t�� or

�r � �R�t� � �x� �L �t�� ��������

Since�

L� "��P ���

LT��

L��� Therefore the solution of �������� is

�x �h�r � �R�t�

i� �L

T� ��������

�Note� It follows that x E��r� t� � ��r� �R�t��� �L �t�T �� Substituting this ������� and using

�vL��x� t� � �vE��r� t� gives

�vE��r� t� � �t �R�t� �h�r � �R�t�

i� �L �t�T � �t

L �t��

We introduce these abbreviations�

�V �t� � �t �R�t� �������

����� RIGID BODY MOTION ��

* �t� ��

L �t�T � �t�

L �t�� �������

Then the Eulerian description of the motion is

�vE��r� t� � �V �t� �h�r � �R�t�

i� �* �t�� ��������

Remark ������ ��t�

L �t��T � �tLT �t��

Proof�

For any �x � P � �x� �L �t� ��

LT �t� � �x� so �x � ��t�

L �t�� � ��t�

LT �t�� � �x� But

�x � ��t�

L� � ��t�

L�T � �x� so ��t�

L �t��T � �x � ��t�

LT �t�� � �x for all �x � P �

Hence� we have the remark

Remark �������

* �t� � � �

* �t�T � I�e���

* �t� is antisymmetric�

Proof�

Since�

L �t� is orthogonal��

LT �t�� �L �t� ��

I P for all t� Hence� ��t�

LT � � �L ��

LT

���t�

L� � �t�

I P��� for all t� But

*��

LT � �t

L and�

*T� ��t�

L�T � ��

LT �T �

��t�

LT � � ��L�� Thus�

*T ��

*���� QED

Now de�ne�

* �t� ��

Ahi �* �t� ��������

where A is the unimodular alternating tensor we have chosen to orient real physical space

P � By exercise ���

* �t� � A � �*�t�� ��������

We claim that for any vector �x � P we have

�x� �* �t� � �*�t�� �x� ��������

Proof�

Take components relative to a pooob� Then��x� �*

�j

���x � A � �*

�j� xi�ijk*k

� �kij*kxi ��Ahi�*�x

�j���*� �x

�j�

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

Thus �������� can be written

�vE��r� t� � �V �t� � �*�t��h�r � �R�t�

i ��V � �t �R

�� ��������

The vector �*�t� is called the angular velocity of the motion relative to the pivot particle�

Obviously �V �t� is the velocity of the pivot particle� and �R�t� is its position�

An obvious question not answered by the foregoing discussion is this� if we are given

two arbitrary di�erentiable functions �R � R P � �* � R P � and de�ne �V � �t �R� then

�������� gives the Eulerian description of a continuum motion� Is this always a rigid

body motion� or do rigid body motions produce particular kinds of functions �R � R P

and �* � R P $

In fact� any choice of �R and �* will make �������� the Eulerian description of a

rigid body motion� To see this� use the given�

* �t� as in ����������

* �t� is� of course�

antisymmetric� Now de�ne�

L �t� as follows�

�t�

L��

L � �* � ��������

�This is obtained from ����� as if�

LT��

L��� but we don�t assume that�� We obtain�

L �t�

by solving �������� with the initial condition

L �t�� ��

I P � �������

Remark ������ �

L �t� obtained from �������������� is proper orthogonal�

Proof�

�t

��

L � �LT�

� �t�

L � �LT�

L ���t

L

�T�

L � �* � �LT�

L ���

L � �*�T

��

L � �* � �LT�

L � �*T � �L

T

��

L ���

* ��

*T�� �L

T��

L � �O � �LT��

O �

Thus��

L �t�� �L �t�T is independent of t� At t � t�� it is�

I P � so it is always�

I P � This proves�

L �t� � "�P �� And det�

L �t� is continuous in t and always

�� and �� at t�� so it is always ��� Thus�

L �t� � "��P � for all t� QED

����� RELATING CONTINUUM MODELS TO REAL MATERIALS ���

Now we can write �������� in the form ��������� As we have seen in deriving ���������

the motion whose Lagrangian description is ������� then has Eulerian description ��������

or� equivalently ��������� But as long as L�t� � "��P � for all t� ������� is a rigid�body

motion� Indeed� for any �x and �y � P � k�r L��x� t�� �r L��y� t�k � k��x� �y�� �L �t�k � k�x� �ykif L�t� � "�P ��

���� Relating continuum models to real materials

In a rock there is some jumping of atoms out of lattice positions� i�e�� �self�di�usion��

but crystal structures stay mostly intact� In water and air� however� individual molecules

move at random� After t seconds� two water molecules which were initially neighbors

are separated� on average� by about �����t���� cm and two air molecules at sea level

are separated by ���t���� cm� What is the physical meaning of a �particle� in water or

air$ We cannot mean a molecule� because then �r L��x� t� would quickly become a very

discontinuous function of �x�

In reality� continuummodels are �two scale� approximations� To see what we mean� let

us examine how to make a continuum model of a gas in which the individual molecules

are moderately well modelled as randomly colliding very small hard spheres� with all

their mass at the center� Let B��r� � be the open ball of radius with center at �r� Let

Mt�B��r� �� be the sum of the masses of all the molecules whose centers are in B��r� � at

time t� Let jB��r� �j � ���� � volume of B��r� �� The average density in B��r� � at

time t is

h�i���r� t� � �

jB��r� �jMt �B��r� �� � �������

To use a continuum model� we must be able to choose so large that the jumps in

h�i���r� t� as a function of �r and t� which occur when individual molecules enter or leave

B��r� � as �r and t vary� are very small fractions of h�i���r� t�� But at the same time�

must be so small that ������� does not sample very di�erent physical conditions inside

B��r� �� That is� must be so small that k��rk can be many times and yet h�i���r���r� t�

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

λ

r

Figure ����

will di�er from h�i���r� t� by only a very small fraction� Therefore� the average physical

conditions must not change appreciably over a length �

For example� if we want to use a continuum approximation to study sound waves

of wavelength � in air� we must use � in �������� and yet we must have ��

average distance between nearest neighbors� Actually� to do all of continuum mechanics

accurately �including momentum and heat� we must have �� mean free path of air

molecules� We cannot �nd such a to use in ������� unless � �� mean free path of air

molecules� This mfp varies from ���� cm at sea level to �� cm at ��� km altitude� and

determines the shortest sound wave treatable by continuum mechanics�

In the spirit of ������ we would de�ne the average momentum in B��r� � at time t as

h�pi���r� t� � �

jB��r� �j�Pt �B��r� �� ������

where �Pt�B��r� �� is the sum of the momenta m�v for all the molecules in B��r� � at time

t� Then we de�ne the average velocity at �r� t to be

h�vi���r� t� � h�pi���r� t�h�i���r� t� ��Pt �B��r� ��

Mt �B��r� ��� ������

If � is the shortest length scale in our gas over which occur appreciable fractional changes

in average physical properties� and if � �� mean free path� we can choose any in

mfp �� and model our gas as a continuum whose Eulerian velocity function is

�vE��r� t� � h�vi���r� t�� �������

����� RELATING CONTINUUM MODELS TO REAL MATERIALS ���

The Lagrangian description of this continuum is obtained by choosing a labelling scheme

and then integrating ������� with �vE given by �������� Usually t��position labelling is

convenient�

Evidently in a gas� the �particles� of the continuum model are �ctitious� In a solid�

they can be thought of as individual molecules� at least for times short enough that self

di�usion is unimportant� An example of the breakdown of the continuum model is a

shock wave in air� They are considerably less thick than one mean free path� In fact�

continuum models work better inside shocks than is at present understood�

��� CHAPTER ��� EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM

Chapter ��

Conservation Laws in a Continuum

In this whole chapter we will use the following notation� �L�AL� is oriented label space and

�P�AP � is oriented real physical space� H is the open subset of L consisting of the labels �x

of all the particles making up the continuum andH � is any open subset ofH with piecewise

smooth boundary H � �i�e� H � consists of a �nite number of smooth pieces�� The open

set in P occupied by the particles at time t will be written K�t�� and the open subset of

K�t� consisting of the particles with labels in H � will be written K ��t�� As t varies� K�t�

and K ��t� will change position in P � but will always contain the same particles� The sets

K�t� and K ��t� are said to move with the material� The outward normal velocity W ��r� t�

of K ��t� at �r � K ��t�� to be used in ��������� is W ��r� t� � nP ��r� t� � �vE��r� t�� where �v is

the particle velocity and nP ��r� t� is the unit outward normal to K ��t� at �r � K ��t��

We will use dVL��x� to denote a very small open subset of H � containing the particle

label �x� We will denote the positive numerical volume in L of this small set by the same

symbol� dVL��x�� We use dVP ��r� to denote a very small open subset of K ��t� containing the

particle position �r� we also use dVP ��r� for the positive numerical volume of this small set

in P � We will always assume when using this notation for the sets that �r � �r L��x� t� and

dVP ��r� � ��r L��� t���dVL��x��� so dVP ��r� consists of the positions at time t of the particles

whose labels are in dVL��x�� It follows that� for the numerical volumes�

dVP ��r� ����det �D�r L��x� t�

��� dVL��x�� �������

��

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

We assume that the reader knows this Jacobian formula from calculus� �We will acci�

dentally prove it later�� Note that L means Lagrangian description and L means label

space�

���� Mass conservation

������ Lagrangian form

The mass of the collection of particles with labels in dVL��x� is

dm � �E��r� t�dVP ��r� �������

where �E��r� t� is the mass per unit of volume in physical space� This is our mathematical

model of how mass is distributed in a continuum�

Because of �������� we can also write

dm � �L��x� t�dVL��x� ������

if we set �L��x� t� � �E��r� t�j det �D�r L��x� t��� But �E��r� t� � �L��x� t� since we are assuming

�r � �r L��x� t�� Therefore �L��x� t� � �L��x� t�j det �D�r L��x� t�j � ��j det �D�r j�L��x� t�� so

� � �j det �D�r j� ������

The physical quantity � is the mass per unit of volume in label space�

The material with labels in H � always consists of the same particles� so it cannot

change its mass� Therefore� dm in ������ must be independent of time� Since dVL��x� is

de�ned in a way independent of time� it follows that �L��x� t� must be independent of t�

Therefore� for all �x � H and all t� t� � R�

�L��x� t� � �L��x� t��� �������

An equivalent equation is Dt �L��x� t� � �� or �Dt��

L��x� t� � �� so

Dt � � �� �������

����� MASS CONSERVATION ���

Either equation ������� or ������� expresses the content of a physical law� the law of

conservation of mass� The mathematical identity ������ makes this law useful� From

������ and ������� we deduce

�L��x� t����det �D�r L��x� t�

��� � �L��x� t�����det �D�r L��x� t��

��� �������

for all �x � H and all t� t� � R� If we use t��position labelling� then �r L��x� t�� � �x�

so �D�r L��x� t�� ��

I P and det �D�r L��x� t�� � det�

Ip� det IP � �� Thus� with t��position

labelling�

�L��x� t����det �D�r L��x� t�

��� � �L��x� t��� �������

������ Eulerian form of mass conservation

An informal physical argument like ������� ����� is possible here� but rather confusing�

The bookkeeping becomes clearer if we account for real� �nite masses and volumes� The

total mass in the set K ��t� in physical space P at time t is

ME �K ��t�� �ZK�t�

dVP ��r��E��r� t�� �������

The physical law of mass conservation is

d

dtME �K ��t�� � �� �������

because K ��t� always consists of the same material�

Using �������� with w � nP�vE��r� t� we have the mathematical identity

d

dtME �K ��t�� �

ZK�t�

dVP ��r�t�E��r� t� �

Z�K�t�

dAP ��r� nP � �vEP ��r� t��

where nP is the unit outward normal to K ��t�� An application of Gauss�s theorem to the

surface integral gives

d

dtME �K ��t�� �

ZK�t�

dVP ��r�ht� � � � ���v�

iE��r� t�� ��������

Comparing ������� and ��������� we see that if K � � K ��t� thenZK�

dVP ��r�ht� � � � ���v�

iE��r� t� � �� ��������

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Now let K � be any open subset of K�t�� with piecewise smooth boundary K �� Let

H � be the set of labels belonging to particles which are in K � at time t� Then for this

set of particles� K ��t� � K �� so �������� holds� In other words� �������� holds for every

open subset K � of K�t�� as long as K � is piecewise smooth� Therefore� by the vanishing

integral theorem� �t�� � � ���v��E��r� t� � � for all t � R and all �r � K�t�� Therefore

t�� � � ���v� � �� �������

Equation ������� is the Eulerian form of the law of mass conservation� It is called the

�continuity equation��

Since � � ���v� � ���� � �v � ��� � �v� and Dt� � t� � �v � ����� the continuity equation

can also be written

Dt�� ��� � �v� � �� �������

By the chain rule for ordinary di�erentiation� Dt ln � � Dt���� so

Dt�ln �� � � � �v � �� ��������

A motion of a continuum is called incompressible if it makes �L��x� t� independent

of t at each particle �x� The density is constant at each particle� but can vary from

particle to particle� A material is incompressible if it is incapable of any motion except

incompressible motion� A motion is incompressible i� Dt�L��x� t� � � for all �x � H and

t � R� i�e� i� Dt� � �� Comparing �������� the motion is incompressible i�

� � �v � �� ��������

An extremely useful consequence of the continuity equation follows from

Lemma ������ Suppose U and V are Euclidean spaces� D is an open subset of U � and

�f � D U and �g � D V are di�erentiable� Then

� � ��f�g� � �� � �f��g � �f � ���g�� ��������

����� MASS CONSERVATION ��

Proof�

Take components relative to orthonormal bases in U and V � Then ��������

is equivalent to i�figj� � �ifi�gj � fi�igj�� But this is the elementary rule

for the partial derivative of a product�

Remark ������� Suppose �f is any physical quantity taking values in a Euclidean space

V � Suppose K ��t� is any open set moving with the material in a continuum �i�e�� always

consisting of the same particles�� Then

d

dt

ZK�t�

dVP ��r����f�E��r� t� �

ZK�t�

dVP ��r���Dt�f�E��r� t�� ��������

Proof�

By ��������� we haved

dt

ZK�t�

dVP ��r����f�E

��r� t� �ZK�t�

dVp��r�t���f�E��r� t�

�Z�K�t�

dAP ��r� nP � ��v��f�E��r� t�

where nP is the unit outward normal to K ��t� and dAP is an element of area�

Applying Gauss�s theorem to the surface integral gives

d

dt

ZK�t�

dVP ��r����f�E��r� t�

�ZK�t�

dVP ��r�ht���f� � � �

���v �f

�iE��r� t��

The integrand is �t���f � ��t �f� �h� � ���v�

i�f � ��v � �� �f� because of lemma

�������� This integrand can therefore be written �t�� � � ���v���f ���t �f ��v ��� �f��� From ������������� and ������� this is ��f � �Dt

�f � which proves

���������

An alternative proof of �������� may be physically more enlightening� We

use ������� to change the variable of integration in �������� to �x instead of

�r� Then ZK�t�

dVP ��r��E��r� t��fE��r� t�

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

�ZH�

dVL��x����det �D�r L��x� t�

��� �L��x� t��fL��x� t��use �����

�ZH�

dVL��x� �L��x��fL��x� t��

Since H � and �L��x� do not vary with t� �dVL��x� �L��x� � dm�

d

dt

ZK�

�t�dVP ��r����f�E��r� t� �

ZH�

dVL��x� �L��x�Dt

�fL��x� t�

�ZH�

dVL��x����det �D�r L��x� t�

��� �L��x� t��Dt�f�L��x� t�

�ZKt�

dVP ��r��E��r� t��Dt

�f�E��r� t��

QED

���� Conservation of momentum

������ Eulerian form

We use the notation introduced on page ��� We denote by �PE�K ��t�� the total momentum

at time t of the material in subset K ��t� of P at time t� This material consists of the

particles whose labels are in H � and whose positions at time t are in K ��t�� The law of

conservation of momentum is

d

dt�PE �K ��t�� � �FE �K ��t�� � ������

where �FE�K ��t�� is the total force at time t on the material in K ��t�� As it stands� ������

is essentially a de�nition of �FE�K ��t��� and our task is to calculate this force and then to

extract from ������ a local form of the law of conservation of momentum in the same

way that ������� was extracted from �������� To do this� we also need an expression

for �PE�K ��t��� but this is simple� The mass in dVP ��r� is dVP ��r��E��r� t� at time t� so its

momentum is dVP ��r��E��r� t��vE��r� t�� Summing over all the volume elements dVP ��r� in

K ��t� gives

�PE �K ��t�� �ZK�t�

dVP ��r����v�E��r� t� �����

����� CONSERVATION OF MOMENTUM ���

If we apply to this equation both �������� and ������������� we obtain the identity

d

dt�PE �K ��t�� �

ZK�t�

dVP ��r����a�E��r� t� �����

where �a � Dt�v � t�v � �v � ��v � particle acceleration�

Finding �FE�K ��t�� is less simple� The expression for it is one of the major eighteenth

century advances in mechanics� If + is the gravitation potential� so that �g � ��+ is the

local acceleration of gravity� then the gravitational force on the matter in dVP ��r� at time

t is dVP ��r��E��r� t��gE��r� t�� If �E and �B are the electric and magnetic �elds� and �Q and �J

are the electric charge density and electric current density then the electromagnetic force

on the matter in dVP ��r� at time t is dVP ��r���Q �E � �J � �B�E��r� t�� The total gravitational

plus electromagnetic force on the matter in dVP ��r� is

d �FB��r� � dVP ��r��fE��r� t� ������

where

�f � ��g � �Q �E � �J � �B� ������

The total gravitational plus electromagnetic force on the matter in K ��t� is

�FEB �K ��t�� �

ZK�t�

dVP ��r��fE��r� t�� ������

This is called the �body force� on K ��t�� and �f is the body force density per unit of

physical volume� the physical density of body force�

If we accept �FEB as a good model for �FE� the only force acting on a cubic centimeter

of ocean or rock is ��g �assuming �E � �B � ���� and yet neither is observed to fall at

���cm��sec� Something important is still missing in our model of �FE�

The physical origin of our di!culty is clear� The forces in ������ are calculated from

the average distribution of molecules and charge carriers� as in Figure ���� In addition

to these long range average forces� we expect that the molecules just outside K ��t�� will

exert forces on the molecules just inside K ��t�� and these contribute to �FE�K ��t��� Also�

in gases and� to some extent� in liquids� individual molecules will cross K ��t�� and the

entering molecules may have� on average� di�erent momenta from the exiting molecules�

�� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

r

dA ( r )P

Back

Front

^

S

n ( r )P

Figure ����

This will result in a net contribution to the rate of change of the total momentum in

K ��t�� Therefore� it is part of d�P �K ��t���dt and hence� by ������� part of �FE�K ��t���

Both the intermolecular force and the momentum transfer by molecular motion can be

modelled as follows� Fix time t and choose a �xed �r � K�t�� Choose a very small nearby

plane surface S in K�t� such that S passes through �r� Choose an even smaller surface

dAP ��r� which passes through �r and lies in S� Arbitrarily designate one side of S as its

�front�� and let nP ��r� be the unit normal to S extending in front of S� Assume that S

is so small that the molecular statistics do not change appreciably across S� but that S is

considerably larger in diameter than the intermolecular distance or the mean free path�

Then the total force exerted by the molecules just in front of dAP ��r� on the molecules

just behind dAP ��r� will be proportional� to the area of dAP ��r�� We write this area as

dAP ��r�� The proportionality constant is a vector� which we write �Sforce� It depends on �r

and t� and it may also depend on the orientation of the surface S� i�e� on the unit normal

nP ��r�� If the material is a gas or liquid� there will also be a net transfer of momentum

from front to back across dAP ��r� because molecules cross dAP ��r� and collide just after

crossing� and the population of molecules one mean free path in front of dAP ��r� may be

�if the linear dimension of dA is �� distance between molecules

����� CONSERVATION OF MOMENTUM ��

nP

P

r

dA ( )r

K’(t)

Figure ���

statistically di�erent from the population one mean free path behind dAP ��r�� The net

rate of momentum transfer from just in front of dA��r� to just behind dAP ��r� will produce

a time rate of change of momentum of the material just behind dAP ��r�� that is� it will

exert a net force on that material� This force will also be proportional to dAP ��r�� and

the proportionality constant is another vector� which we write �Smfp� This vector also

depends on �r�t and nP ��r�� The sum �S��r� t� nP � � �Sforce��r� t� nP � � �Smfp��r� t� nP � is called

the stress on the surface �S� nP �� The total force exerted by the material just in front of

dAP ��r� on the material just behind dAP ��r� is

d �FS��r� � dAP ��r��S��r� t� nP �� ������

This is called the surface force on dAP ��r�� Summing ������ over all the elements of area

dAP ��r� on K ��t� gives

�FES �K ��t�� �

Z�K�t�

dAP ��r��S ��r� t� nP ��r� t�� ������

where nP ��r� t� is the unit outward normal to K ��t� at �r � K ��t�� The expression

������ is called the surface force on K ��t�� It is the total force exerted on the material

just inside K ��t� by the material just outside K ��t�� The total force on K ��t� is the sum

�� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

of the body and surface forces� �FE�K ��t�� � �FEB �K

��t�� � �FES �K

��t��� so

�FE �K ��t�� �ZK�t�

dVP ��r��fE��r� t� �

Z�K�t�

dAP ��r��S ��r� t� nP ��r� t�� � ������

Combining the physical law ������ with the mathematical expressions ����� and

������ gives

ZK�

dVP ��r����a� �f�E��r� t� �Z�K�

dAP ��r��S ��r� t� nP ��r�� � �������

where K � � K ��t� and nP ��r� is the unit outward normal to K � at �r � K ��

As with ��������� ������� must hold for every open subset K � of K�t�� as long as

K � is piecewise smooth� This conclusion is forced on us if we accept the mathematical

models ������ for �FE�K ��t�� and ����� for PE�K ��t���

To convert ������� to a local equation� valid for all �r � K�t� at all times t� �i�e�� to

�remove the integral signs�� we would like to invoke the vanishing integral theorem� as we

did in going from �������� to �������� The surface integral in ������� prevents this�

Even worse� ������� makes our model look mathematically self�contradictory� or

internally inconsistent� Suppose that K � shrinks to a point while preserving its shape�

Let be a typical linear dimension of K �� Then the left side of ������� seems to go to

zero like �� while the right side goes to zero like �� How can they be equal for all � �$

Cauchy resolved the apparent contradiction in ���� He argued that the right side

of ������� can be expanded in a power series in � and the validity of ������� for all

shows that the �rst term in this power series� the � term� must vanish� In modern

language� Cauchy showed that this can happened i� at every instant t� at every �r � K�t��

there is a unique tensor�

SE��r� t� � P � P such that for each unit vector n � P �

�S��r� t� n� � n� �SE��r� t�� �������

If equation ������� is true� we can substitute it in the surface integral in ������� and

use Gauss�s theorem to write

Z�K�

dAP ��r� nP ��r���

SE��r� t� �

ZK�

dVP ��r����

SE��r� t��

����� CONSERVATION OF MOMENTUM ��

Then ������� becomes

ZK�

dV ��r����a� �f � �� �S

�E��r� t� � ��

Since this is true for all open subsets K � of K�t� with piecewise smooth boundaries K ��

the vanishing integral theorem implies that the integrand vanishes for all �r � K�t� if� as

we shall assume� it is continuous� Therefore

��a � �� �S ��f� ������

This is the Eulerian form of the momentum equation� The tensor�

SE

��r� t� called the

Cauchy stress tensor at ��r� t�� The physical quantity�

S is also called the Cauchy stress

tensor�

The argument which led Cauchy from ������� to ������� is fundamental to con�

tinuum mechanics� so we examine it in detail� In �������� t appears only as a parameter�

so we will ignore it� Then ������� implies ������� because of

Theorem ������ �Cauchy�s Theorem� Suppose U and V are Euclidean spaces� K

is an open subset of U � NU � set of all unit vectors n � U � �f � K V is continuous�

and �S � K � NU V is continuous� Suppose that for any open subset K � of K whose

boundary K � is piecewise smooth� we have

ZK�

dVU��r��f��r� �Z�K�

dAU��r��S ��r� nU��r�� � ������

where nU��r� is the unit outward normal to K � at �r � K �� Then for each �r � K there

is a unique�

S ��r� � U � V such that for all n � NU �

�S��r� n� � n� �S ��r�� �������

�This theorem is true for any value of dimU � � We give the proof only for dimU � �

the case of interest to us� For other values of dimU � a proof can be given in exactly the

same way except that �u� � �u� must be replaced by the vector in Exercise ���

Proof�

�� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

K

Figure ���

Two lemmas are required�

Lemma ������ Suppose �f and �S satisfy the hypotheses of theorem ����� Let �r�

be any point in K and let K � be any open bounded �i�e�� there is a real M such that

�r � K � � k�rk � M� subset of U � with piecewise smooth boundary K �� We don�t need

K � � K� Then Z�K�

dAU��r��S ��r�� nU��r�� � ��V �������

if nU��r� is the unit outward normal to K � at �r � K � and dAU��r� is the element of area

on K ��

Proof of Lemma ������ � For any real in � �� de�ne �r � � U U by

�r ���r� � �r� � ��r � �r�� for all �r � U � Since �r���r� � �r� � ��r � �r��� �r � shrinks U

uniformly toward �r� by the factor � The diagram above is for � ��� De�ne

K �� � �r ��K

�� so K �� � �r ��K

��� Choose �r � K � and let �r� � �r ���r�� Let dA��r�

denote a small nearly plane patch of surface in K �� with �r � dA��r�� and use dA��r�

both as the name of this set and as the numerical value of its area� Let the set

dA���r�� be de�ned as �r ��dA��r��� and denote its area also by dA���r��� Then by

����� CONSERVATION OF MOMENTUM ��

geometric similarity

dA���r�� � �dA��r�� �������

Let n��r� be the unit outward normal to K � at �r� and let n���r�� be the unit outward

normal to K �� at �r�� By similarity� n��r� and n���r�� point in the same direction�

Being unit vectors� they are equal�

n���r�� � n��r�� �������

Since �r� is �xed� it follows thatZ�K�

dA���r���S ��r�� n���r��� � �Z�K�

dA��r��S ��r�� n��r�� � �������

If is small enough� K �� � K� Then� by hypothesis� we have ������ with K � and

K � replaced by K �� and K �

�� ThereforeZK�

dV ��r��f��r� �ZK�

dA���r��n�S ��r�� n���r���� �S ��r�� n���r���

o�Z�K�

dA���r���S ��r�� n���r��� �

From ������� it follows thatZ�K�

dA��r��S ��r�� n��r�� ��

ZK�

dV ��r��f��r�

��

Z�K�

dA���r����S ��r�� n���r���� �S ��r�� n���r���

�� �������

Let m�S�� � maximum value of k�S��r�� n�� �S��r� n�k for all �r � K �� and all n � NU �

Let m�f�� � maximum value of j�f��r�j for all �r � K ���

Let jK ��j � area of K �

�� jK �j � area of K ��

Let jK ��j � volume of K �

�� jK �j � volume of K ��

Then jK ��j � �jK �j and jK �

�j � �jK �j� so ������ and ������� imply����Z�K�

dA��r��S ��r�� n��r������ � jK �jm�f�� � jK �jm�S��� ������

�� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

As �� m�f�� remains bounded �in fact kf��r��k� and m�S�� � because

�S � K � NU V is continuous� Therefore� as �� the right side of ������

�� Inequality ������ is true for all su!ciently small � �� and the left side is

non�negative and independent of � Therefore the left side must be �� This proves

������� and hence proves lemma �����

We also need

Lemma ������� Suppose �S � NU V � Suppose that for any open set K � with piecewise

smooth boundary K �� �S satis�es

Z�K�

dA��r��S � n��r�� � ��V ������

where n��r� is the unit outward normal to K � as �r � K �� Suppose F � U V is de�ned

as follows�

F ���U� � ��V and if �u �� ��U � F ��u� � k�uk�S�

�u

k�uk�� �����

Then F is linear�

Proof of Lemma �������� a� F ���u� � �F ��u� for all �u � U � To prove this� it su!ces

to prove

�S�� n� � ��S� n� for all n � NU � �����

Let K � be the �at rectangular box shown at upper right� For this box� ������

gives

L��S� n� � L��S�� n� � �Lh�S� n�� � �S�� n�� � �S� n�� � �S�� n��

i� ��V �

Hold L �xed and let � �� Then divide by L� and ����� is the result�

b� If c � R and �u � U � F �c�u� � cF ��u��

i� If c � � or �u � ��U � this is obvious from F ���U� � ��V �

ii� If c � � and �u �� ��U � F �c�u� � kc�ukk�S�c�u�kc�uk� � ck�uk�S�c�u�ck�uk� �

ck�uk�S��u�k�uk� � cF ��u� �

����� CONSERVATION OF MOMENTUM �

- n^

^n

^n

L

ε

L

- n^

1

2

^n

1

- n^

2

Figure ����

�� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

iii� If c �� F �c�u� � �F ��c�u� by a� above� But �c � � so F ��c�u� �

��c�F ��u� by ii�� Then F �c�u� � ���c�F ��u� � cF ��u��

c� F ��u� � �u�� � F ��u�� � F ��u�� for all �u�� �u� � U �

i� If �u� � ��U � F ��u� � �u�� � F ��u�� � ��V � F ��u�� � F ��u�� � F ��u���

ii� If �u� �� ��U and �u� � c�u� then F ��u� � �u�� � F ��� � c��u�� � �� � c�F ��u�� �

F ��u�� � cF ��u�� � F ��u�� � F �c�u�� � F ��u�� � F ��u���

iii� If f�u�� �u�g is linearly independent� let �u� � ��u� � �u�� We want to prove

F ���u�� � F ��u�� � F ��u��� or �F ��u�� � F ��u�� � F ��u��� or

F ��u�� � F ��u�� � F ��u�� � ��V � ������

To prove ������ note that since �u� � �u� are linearly independent� we can de�ne the unit

vector � � ��u� � �u���k�u� � �u�k� We place the plane of this paper so that it contains �u�

and �u�� and � points out of the paper� The vectors �u�� �u�� �u� form the three sides of a

nondegenerate triangle in the plane of the paper� � � �ui is obtained by rotating �ui ���

counterclockwise� If we rotate the triangle with sides �u� � �u�� �u� ��� counterclockwise�

we obtain a triangle with sides � � �u�� � � �u�� � � �u�� The length of side � � �ui is

k � � �uik � k�uik� and �ui is perpendicular to that side and points out of the triangle� Let

K � be the right cylinder whose base is the triangle with sides ���ui and whose generators

perpendicular to the base have length L� The base and top of the cylinder have area

A � k�u� � �u�k� and their unit outward normals are �� and ���� The three rectangular

faces of K � have areas Lk�uik and unit outward normals �ui��k�ui�k� Applying ������ to

this K � gives

A�S� �� � A�S�� �� ��Xi �

Lk�uik�S��ui�k�uik� � ��V �

But �S� �� � ��S�� �� so dividing by L and using ����� gives ������

Corollary ������� �to Lemma ������ Under the hypotheses of lemma ����� there

is a unique�

S� U � V such that for all n � NU

�S� n� � n� �S � ������

����� CONSERVATION OF MOMENTUM ��

.

L

ν

x

ν

ν

−ν

xν u

u

u

u

u

u

u

u

2

u1

3

1

2

2

3

1

3

Figure ����

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Proof�

Existence� Take�

S��

F � the tensor in U�V corresponding to the F � L�U V � de�ned by ���� Then n � �S� n� �F� F � n� � �S� n��

Uniqueness� If n� �S�� n� �S� for all n � NU then�c n�� �S�� �c n�� �S� for all

n � NU and c � R� But every �� � U is c n for some c � R and n � NU � so

�u� �S� for all �u � U � Hence�

S���

S� �

Now we return to the proof of Cauchy�s theorem �theorem ������ If �r� is

any �xed point in K� then by lemma ����� the function �S��r�� �� � NU V

satis�es the hypothesis of lemma ����� Therefore by corollary ����� there

is a unique�

S ��r�� � U�V such that for any unit vector n � U � ��S��r�� ���� n� � n� �S ��r��� But ��S��r�� ���� n� � �S��r�� n� so we have ��������

Having completed the proof of Cauchy�s theorem� we have proved that if

������� holds for all K � then Cauchy�s stress tensor�

SE

��r� t� exists and

satis�es �������� And as we have seen� ������� leads automatically to the

Eulerian momentum equation ������� It is important to have a clear phys�

ical picture of what the existence of a Cauchy stress tensor�

SE��r� t� means�

If dAP is any small nearly plane surface in the material in physical space P

at time t� and if �r � dAP � choose one side of dAP to be its front� and let nP

be the unit normal extending in front of dAP � Then the force d �FS exerted by

the material just in front of dA on the material just behind dA is� at time t�

d �FS � dAP nP ��

SE��r� t�� ������

The stress� or force per unit area� exerted on the material just behind dA by

the material just in front is

�S��r� t� nP � � nP ��

SE��r� t�� ������

If K � is any open subset of K�t�� with piecewise smooth boundary K �� then

at time t the total surface force on the matter in K � is� according to �������

����� CONSERVATION OF MOMENTUM ���

.back

front

^

r

dA

nP

P

Figure ����

and �������

�FES �K �� �

Z�K�

dAP ��r � nP ��r���

SE��r� t�� ������

Using Gauss�s theorem� we can write this as

�FES �K �� �

ZK�

dVP ��r����

SE��r� t�� ������

The total force on the matter in K � is� according to ������ and �������

�FE �K �� �ZK�

dVP ��r���f � �� �S

�E��r� t�� ������

If the particles are all motionless� �r L��x� t� is independent of time� so

�vL��x� t� � Dt�rL��x� t� � ��� and �aL��x� t� � Dt�v

L��x� t� � ��� and the momentum

equation ������ reduces to

�� � �� �S ��f� ������

the �static equilibrium equation�� In this case� we see from ������ that the

surface force on the matter in K � exactly balances the body force� for every

open subset K � of K�t��

If we have both �a � �� and �f � ��� then

�� �S� ��� �����

The physical quantity�

S is called a �static stress �eld� if it satis�es ������

An example is an iron doughnut in orbit around the sun �to cancel gravity��

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Figure ����

Cut out a slice� close the gap by pressure� weld� and remove the pressure�

as in �gure ��� The resulting doughnut will contain a nonzero static stress

�eld� The net force on every lump of iron in the doughnut vanishes because

the body force vanishes and the surface force sums to ���

������ Shear stress� pressure and the stress deviator

We want to discuss some of the elementary properties of stress and the Cauchy stress

tensor� In �gure ������ choosing which side of dA is the �front� is called �orienting dA��

and the pair �dA� n� is an oriented surface� Two oriented surfaces can be produced from

dA� namely �dA� n�� and �dA�� n� depending on which side of dA we call the front�

We want to study the state of stress at one instant t at one position �r � K�t�� so we

will omit �r� t and also the superscript E� We will work only with Eulerian descriptions

at the single point �r and time t� Thus the stress �S��r� t� n� on small nearly plane oriented

area �dA� n� will be written �S� n�� and�

SE��r� t� will be written

S�

De�nition �������

Sn� n� �� n � �S� n� � n� �S � n �����

�SS� n� �� �S� n�� nSn� n� � n� �S � n� n� �S � n

�������

pn� n� �� �Sn� n� � � n��

S � n� ������

The vector nSn� n� is called the normal stress acting on the oriented surface �dA� n�� the

vector �SS� n� is called the tangential or shear stress acting on �dA� n�� and pn� n� is the

����� CONSERVATION OF MOMENTUM ���

pressure acting on �dA� n��

Corollary �������

n � �SS� n� � � ������

�S� n� � nSn� n� � �SS� n� � � npn� n� � �SS� n�� ������

The shear stress is always perpendicular to n� parallel to dA� The normal stress acts

oppositely to n if pn� n� � ��

To visualize �S as a function of n� note that the set NP of unit vectors n � P is precisely

B���� ��� the spherical surface of radius � centered on �� in P � Thus �S attaches to each n

on the unit sphere NP a vector �S� n� whose radial part is nSN� n� and whose tangential

part is �SS� n�� This suggests

De�nition ������ The average value of pn� n� for all n � NP is called the average or

mean pressure at �r� t� It is written hpni��r� t�� We do not write the ��r� t� in this chapter�

Thus

hpni � �

��

ZNP

dA� n�pn� n� ������

where dA� n� is the element of area on NP �

Remark ������� hpni � ��� tr�

S

Proof�

Let y�� y�� y� be an orthonormal basis for P � and take components relative to

this basis� Then

pn� n� � � n��

S � n � �niSijnj � �Sijninj�

By the symmetry of NP �ZNP

dA� n�ninj � � if i �� j and

ZNP

dA� n�n�� �ZNP

dA� n�n�� �ZNP

dA� n�n�� �ZNP

dA� n��n�� � n�� � n���

��

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Therefore� ZNP

dA� n�ninj ���

�ij and

hpni � � �

� �Sij

ZNP

dA� n�ninj � ��

�ijSij

� ��

Sii � ��

tr

S �

Only part of�

S produces the shear stress �SS� n�� To see this we need a brief discussion

of tensors in U � U � where U is any Euclidean space�

Lemma ������� tr�

I U� dimU �

Proof�

Let y�� � � � � yn be an orthonormal basis for U � Then�

I U� yi yi and tr�

I U�

yi � yi � n � dimU �

Lemma ������� If�

S� U � U � tr�

S��

I U hi�

S�

Proof�

With yi as above� tr�

S� Sii � �ijSij � ��

I U�ijSij ��

I U hi�

S�

De�nition �������

S� U � U is �traceless�� or a �deviator� if tr�

S� ��

Corollary ��������

S is a deviator� �

S is orthogronal to all isotropic tensors in U �U �

Proof�

The isotropic tensors are the scalar multiples of�

I U � But�

S is orthogonal to�

I U��

S hi �I U� �� Now use lemma ����

Remark ������� If�

S� U � U � there is exactly one isotropic tensor ��

I U and one

deviator�

SD such that�

S� ��

I U ��

SD � ������

����� CONSERVATION OF MOMENTUM ���

Proof�

Uniqueness� Since tr�

SD� �� ������ implies tr�

S� � tr�

I U� ��dimU��

Thus � is uniquely determined by�

S � Then� from ������� the same is true

of�

SD�

Existence� Choose � � �tr�

S�� dimU and de�ne�

SD by ������� Then

tr�

S� ��tr�

I U� � tr�

SD� tr�

S �tr�

SD so tr�

SD� ��

De�nition ������ In ������� ��

I U is the isotropic part of�

S� and�

SD is the deviatoric

part of�

S� If U � P and�

S is a Cauchy stress tensor��

SD is the stress deviator�

Remark ������� Suppose ��

I P and�

SD are the isotropic and deviatoric parts of Cauchy

stress tensor�

S� Then

i� hpni � �� �

ii� If�

S� ��

I P �i�e���

SD���� then pn� n� � �� for all n and

SS � n� � �� for all n�

iii��

S and�

SD produce the same shear stress for all n �i�e�� the shear stress is due entirely

to the deviatoric part of�

S��

Proof�

i� By ������ tr�

S� �� By remark �������� tr�

S� �hpni�

ii� Substitute ��

I P for�

S in ������ and �������

iii� From ������� if n is �xed� �SS� n� depends linearly on�

S� By �������

�SS� n� is the sum of a contribution from ��

I P and one from�

SD� By ii�

above� the contribution from ��

I P vanishes�

Remark ������� If a Cauchy stress tensor produces no shear stress for any n� it is

isotropic�

Proof�

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

The hypothesis is� from ������� that n� �S� nSn� n� for all unit vectors n�

Let y�� y�� y� be an orthonormal basis for P � Then setting n � yi gives

yi��

S� yi�Sn� yi��� But�

S��

I P ��

S� � yi yi���

S� yi� yi��

S� �P

yi yiSn� yi��

That is�

S��Xi �

yi yiSn� yi��

But then�

ST��

S� so n� �S � yi � yi��

S � n� and thus nSn� n� � yi � yiSn� yi� � n� orSn� n�� n � yi� � Sn� yi�� n � yi�� This is true for any n� so we may choose an n

with n � yi �� � for all of i � �� � � Then Sn� n� � Sn� y�� � Sn� y�� � Sn� y���

so�

S� Sn� n� yi yi � Sn� n��

I P � QED

Fluid Motion�

As an application of Cauchy stress tensors and the momentum equation� we consider

certain elementary �uid problems�

De�nition ������� A �uid is a material which cannot support shear stresses when it

is in static equilibrium� A non�viscous �uid is a material which can never support shear

stresses�

Corollary ������ In a non�viscous �uid� or in a �uid in static equilibrium� the Cauchy

stress tensor is isotropic� and�

SE��r� t� � �hpni��r� t�

I P �

Proof�

From remark ���������

SE

��r� t� � ���r� t��

I P � From remark ����� i��

� � �hpni�

De�nition ������� In a non�viscous �uid� or in a �uid in static equilibrium� hpni iscalled simply �the� pressure� written p� Thus

S� �p �

I P � �������

Remark ������� � � �p �

I � � �p�

����� CONSERVATION OF MOMENTUM ��

Proof�

By exercise ��a� � � �p �

I P � � �p� �I P �p�� �I P � But �� �I P��

I P hi��

I P� �

because�

I P is constant and ��

I P� �� Also �p� �I P� �p� QED�

Therefore� the Eulerian momentum equation in a non�viscous �uid is

��a � ��p� �f� �������

Since �a � Dt�u � t�u� �u � ��u� this is often written

��t�u� �u � ��u

�� ��p� �f� ������

In a �uid in static equilibrium� ��a � ��� so

�p � �f� ������

Therefore� � � �f � ��� We have

Remark ������� Static equilibrium in a �uid subjected to body force �f is impossible

unless � � �f � ���

Remark ������ If a �uid is in static equilibrium in a gravitational �eld� then the pres�

sure and density of the �uid are constant on gravitational equipotential surfaces which

are arcwise connected�

Proof�

Let + be the gravitational potential� Then �f � ���+� so ������ shows

that �p is a scalar multiple of �+� And � � �f � ���� �+� Since � � �f � ���

�� is also a scalar multiple of �+� Therefore� if C is any curve lying in a

gravitational equipotential surface� and if � ��r� is the unit tangent vector to C

at �r� then �� ��r� � ����r� � ���r� � �p��r� � �� Therefore� by theorem ����� �

and p are the same at both ends of C�

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

���� Lagrangian form of conservation of momentum

We use the notation introduced on page ��� If �x is a point in label space L� we use

dVL��x� to stand for a small set containing �x and also for the volume of that set� The

mass per unit volume in label space is �L��x�� independent of time� so the total mass of the

particles whose labels are in dVL��x� is �L��x�dVL��x�� Their velocity at time t is �vL��x� t��

so their momentum is dVL��x� �L��x��vL��x� t�� The total momentum of the particles whose

labels are in open subset H � of H is� at time t�

�PLt �H �� �

ZH�

dVL��x� �L��x��vL��x� t� �

ZH�

dVL��x�� ��v�L��x� t��

Let �FLt �H

�� denote the total force exerted on those particles at time t� Then the law of

conservation of momentum requires

d

dt�PLt �H �� � �FL

t �H �� � ������

Since the set H � does not vary with time� we also have the mathematical identity

d

dt�PLt �H �� �

ZH�

dVL��x� �L��x�Dt�v

L��x� t��

ord

dt�PLt �H �� �

ZH�

dVL ��x� � ��a�L ��x� t� � �����

The particles whose labels are in H � occupy the set K ��t� in physical space P at time

t� Therefore �PLt �H

�� � �PE�K ��t��� and ����� should agree with ������ That it does

can be seen by changing the variables of integration from �r to �x in ����� by means of

�������� and then using �������

Let �FLB�t�H

�� denote the total body force on the particle with labels in H �� Then

�FLB�t�H

�� � �FEB �K

��t��� By changing the variables of integration in ������ from �r to �x

via ��������� we �nd

�FLB�t �H

�� �ZH�

dVL��x����det �D�r L��x� t�

��� �fL��x� t��or

�FLB�t �H

�� �ZH�

dVL��x�� fL

��x� t� �����

����� LAGRANGIAN FORM OF CONSERVATION OF MOMENTUM ���

.back

front

^

( )

x

dAL x

nL

Figure ����

where

� f ����det �D�r

��� �f� ������

The vector � f is the body force density per unit of label�space volume� Formula ������

can be seen physically as follows� The body force on the particles in the small set dVP ��r�

is dVP ��r��fE��r� t� at time t� But these are the particles whose labels are in dVL��x�� so by

������� this force is dVL��x�� fL

��x� t� with � f given by �������

The surface force at time t exerted by the particles whose labels are just outside H �

on the particles whose labels are just inside H � we will denote by �FLS�t�H

��� The particles

with labels just outside �inside� H � are those whose positions at time t are just outside

�inside� K ��t�� so

�FLS�t �H

�� � �FES �K ��t�� � ������

Then �FLt �H �� � �FL

B�t �H�� � �FL

S�t �H��� It remains to study the surface force �FL

S�t�H���

Let dAL��x� be any small nearly plane patch of surface in label space L containing the

label �x � H� Choose one side of dAL to be the front� and let nL be the unit normal to

dAL��x� on its front side� Figure ��� is the same as �gure ����� except for the labelling�

Now� however� the picture is of a small patch of surface in label space� not physical space�

The force d �FS exerted by the particles with labels just in front of dAL on the particles

with labels just behind dAL is proportional to dAL� as long as the orientation of that small

surface does not change� i�e�� as long as nL is �xed� We denote the vector �constant� of

proportionality by � S� Then d �FLS � � SdAL� The vector � S will change if nL changes� and it

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

^ ( )

x

L xn

H H

.

∂ ′ ′

Figure ����

also will depend on �x and t� so at time t at label �x

d �FLS � � S��x� t� nL�dAL��x�� ������

�Compare ������� The vector � S we call the �label stress�� The total surface force on

the particles in H � is then

�FLS�t �H

�� �Z�H�

dAL��x�� S ��x� t� nL��x�� ������

where nL��x� is the unit outward normal to H � at �x � H �� Now we can combine �����

and ������ to obtain

�FLt �H �� �

ZH�

dVL��x� fL��x� t� �

Z�H�

dAL��x�� S ��x� t� nL��x�� � ������

Inserting this and ����� in ������ givesZH�

dVL��x�� ��a� � f

�L��x� t� �

Z�H�

dAL��x�� S ��x� t� nL��x�� � ������

Equation ������ is true for every open subset H � ofH for which H � is piecewise smooth�

It is mathematically of the same form as ������� and the same application of Cauchy�s

theorem ������ shows that there is a unique tensor� SL

��x� t� � L�P such that for every

unit vector nL � L�

� S��x� t� nL� � nL��

SL��x� t�� �������

The tensor� SL

��x� t� is the Piola�Kircho� or body stress tensor� We would prefer to call

it the label stress tensor� but history prevents us� Proceeding in the way which led from

����� LAGRANGIAN FORM OF CONSERVATION OF MOMENTUM ���

������� to ������ we note that by Gauss�s theoremZ�H�

dAL��x� nL��

SL��x� t� �

ZH�

dVL��x� �D��

SL��x� t�

so ZH�

dVL��x�

� ��a� � f � �D �

� S

�L

��x� t� � ��

Since this is true for all open H � � H with H � piecewise smooth� the vanishing integral

theorem gives

��a � � f � �D �� S� �������

This is the Lagrangian form of the momentum equation� Equation ������ can now be

written

d �FLS � dAL��x� nL �

� SL

��x� t�� ������

where d �FLS � dAL��x� and nL refer to �gure ����

Equations ������� and ������ refer to the same motion� The relation between �

and � is ������� The relation between � f and �f is ������� which can also be written

� f

��

�f

�� ������

both sides of this equation being the force per unit mass on the material� But what is the

relation between� S and

S$

Suppose that dAL��x� is a small nearly plane surface in label space L� and that we

have chosen one of its sides as the front� Let nL be the unit normal to dAL��x� on its

front side� At time t� let �r � �r L��x� t�� Let dAP ��r� be the small nearly plane surface

in physical space P consisting of all positions at time t of particles whose labels are in

dAL��x�� Let dAL��x� and dAP ��r� denote both the sets of points which make up the small

surfaces and also the areas of those sets� Choose the front side of dAP ��r� so that particles

are just in front of dAP ��r� if their labels are just in front of dAL��x�� Let nP be the unit

normal to dAP ��r� on its front side� Then the force exerted on the particles just behind

dAP ��r� by the particles just in front can be written either as dAP ��r� nP ��

SE��r� t� or as

dAL��x� nL �� SL

��x� t�� Therefore� these two expressions must be equal�

dAP ��r� nP ��

SE��r� t� � dAL��x� nL �

� SL

��x� t� �������

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

.front

backback

front

^

( )

^

( )

n

dA

L

L x

nP

rdA rP

.x

Figure �����

if

�r � �r L��x� t� �������

and

dAP ��r� �h�r L��� t�

i�dAL��x�� � �������

Notice that in �������� dAP and dAL are positive real numbers� areas� while in �������

they stand for sets� the small surfaces whose areas appear in ��������

Our program is to express dAP ��r� nP in terms of dAL� x� nL in �������� to cancel

dAL� x� nL� and thus to obtain the relation between�

S and� S� Equation ����� holds�

whatever the shape of dAL��x�� It is convenient to take it to be a small parallelogram

whose vertices are �x� �x � ���� �x � ���� �x � ��� � ���� Then dAP ��r� will be a very slightly

distorted parallelogram with vertices

�r � �r L��x� t�

�r � ��� � �r L��x� ���� t�

�r � ��� � �r L��x� ���� t�

�r � ��� � ��� � �r L��x� ��� � ���� t��

Correct to �rst order in the small length k��ik� we have

�r � ��i � �r L��x� t� � ��i � �D�r L��x� t��

so

��i � ��i � �D�r L��x� t�� �������

����� LAGRANGIAN FORM OF CONSERVATION OF MOMENTUM ���

nP

.front

back

ξ

+ ξ+ ξ + ξ

( )

ξ + ξ+ ρ

ρ

+ ρ

( )

+ ρ + ρ

2

x2

x

1 x1

dA xL

x1 2

1

nL

r r2

dA rP

r11 2r

. .

.

Figure �����

We naturally take ��� and ��� in di�erent directions� so k�������k �� � �recall that L and P are

oriented �spaces� so cross products are de�ned�� We number ��� and ��� in the order which

ensures that ��� � ��� points into the region in front of dAL��x�� so ��� � ��� � k��� � ���k nL�But k��� � ���k � dAL��x�� so

dAL��x� nL � ��� � ���� �������

Also� clearly� k�������k � dAP ��r�� and ������� is perpendicular to dAP ��r�� so ������� ��k��� � ���k nP � Thus

dAP ��r� nP � ���� � ���� �������

But which sign is correct$ We have to work out which is the front side of dAP ��r�� because

that is where we put nP � If ��� � L is small� �x � ��� is in front of dAL��x� � ��� � nL � ��

and hence ���� �

���� � ���

�� �� ������

The particle with label �x� ��� has position �r � ���� with ��� given by �������� Then���������������������������

��� � ���� � ���� � AP ����� ���� ���� � AP ����� ���� ����

� AP

���� � �D�r L� ��� � �D�r L� ��� � �D�r L

�hdet �D�r L��x� t�

iAL

����� ���� ���

��hdet �D�r L��x� t�

i��� �

���� � ���

��

������

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Thus the position �r � ��� is in front of dAP ��r� � ��� � ���� � ���� has the same sign as

det �D�r L��x� t�� Let sgn c stand for the sign of the real number c� That is� sgn c � �� if

c � �� � �� if c �� and � � if c � �� Then position �r � ��� is in front of dAP ��r��

sgn ��� � ���� � ���� � sgn det �D�r L��x� t�� �����

We have chosen the front of dAP ��r�� and the direction of nP � so that �r � � nP is in front

of dAP ��r� when � � �� Therefore sgn nP � ���� � ���� � sgn det �D�r L��x� t�� Thus

��� � ��� � k��� � ���k nP sgn det �D�r L��x� t�

and

dAP ��r� nP � ���� ���� sgn det �D�r L��x� t�� �����

Using ������� and ������ we hope to relate to dAP nP and dAL nL� so we try to

relate ��� � ��� and ��� � ���� The two ends of ������ give

��� � ���� � ���� � ��� ����� � ���

�det �D�r L��x� t� ������

for any ���� ���� ��� � L� if ���� ���� ��� are given by ��������

Substituting ������� and ����� in ������ and using c sgn c � jcj� we �nd

��� � �dAP ��r� nP � � ��� � �dAL��x� nL����det �D�r L��x� t�

��� �But ��� � ��� � �D�r L��x� t�� so

��� � �D�r L��x� t� � �dAP ��r� nP � � ��� � �dAL��x� nL����det �D�r L��x� t�

��� �Since this is true for any ��� � L� we must have

�D�r ���x� t� � �dAP ��r� nP � � dAL��x� nL���det �D�r L��x� t�

���or� letting T stand for transpose�

dAP ��r� nP �h�D�r L��x� t�

iT� dAL��x� nL

���det �D�r L��x� t���� � ������

����� LAGRANGIAN FORM OF CONSERVATION OF MOMENTUM ���

If we dot� SL

��x� t� on the right of each side of ������� we obtain

dAP ��r� nP � �D�r L��x� t�T �� SL

��x� t�

� dAL��x� nL �� SL

��x� t����det �D�r L��x� t�

��� � ������

But �see ������

d �FS � dAL��x� nL �� SL

��x� t� � dAP ��r� nP ��

SE��r� t�

where �r � �r L��x� t�� Thus

dAL��x� nL �� SL

��x� t� � dAP ��r� nP ��

SL��x� t��

Substituting this on the right side of ������ gives

�dAP ��r� nP � ���D�r L��x� t�T �

� SL

��x� t�

� �dAP ��r� nP � ���

SL��x� t�

���det �D�r L��x� t����� � ������

We have proved ������ when dAP is a small parallelogram� as in Figure �������� But

��� and ��� can be arbitrary as long as they are small� Hence so can dAP ��r� nP � ��� � ����

Thus ������ is true when dAP ��r� nP is any small vector in P � Since ������ is linear

in dAP ��r� nP � it is true whatever the size of that vector� Therefore ������ implies

�D�r L��x� t�T �� SL

��x� t� ��

SL��x� t�

���det �D�r L��x� t���� �

or ��D�r

�T � � S ��

S���det �D�r

��� � ������

This gives�

S in terms of� S� By ������ ������ �D�r L��x� t� has an inverse� � �x E��r� t��

where �r � �r L��x� t�� so �D�r L��x� t�T has the inverse � �x E��r� t�T � Dotting this on the left

in ������ gives� S �

�� �x

�T � �S ���det � �x ����� ������

where we use det� �D�r ��� � �det �D�r����

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

By appealing to ������� we can write ������ and ������ as���

S

A ���D�r

�T ��B�� S

CA ������

�B�� S

CA ��� �x

�T ����

S

A � ������

��� Conservation of angular momentum

����� Eulerian version

Refer again to the notation on page ��� All torques and angular momenta are calculated

relative to the origin �OP in physical space P � Let �LE�K ��t�� denote the total angular

momentum of the particles in K ��t�� while �ME�K ��t�� denotes the total torque on them�

The law of conservation of angular momentum is

d

dt�LE �K ��t�� � �ME �K ��t�� � �������

Our problem is to express this law locally�

The momentum of the particles in dVP ��r� is dVP ��r��E��r� t��vE��r� t� and this produces�

relative to �OP � the angular momentum

�r �hdVP ��r����v�

E��r� t�i� dVP ��r���r � ��v�E��r� t��

It is possible that the individual atoms in dVP ��r� might have angular momentum� this

might be from the orbits of electrons not in closed shells� from the spins of unpaired elec�

trons� or from nuclear spin� Orbital angular momentum is quantized in integer multiples

of 'h� while spin angular momentum is in units of 'h�� Here 'h � h�� and h is Planck�s

constant� so 'h � ������ � ���� joule sec� If the atoms are not randomly aligned� there

may be an intrinsic atomic angular momentum of �lE��r� t� per kilogram at position �r at

time t� Then the intrinsic angular momentum in dVP ��r� is dVP ��r����l�E��r� t�� and the total

angular momentum in dVP ��r� is

d �LE � dVP ��r�h���r � �v ��l

�iE��r� t��

����� CONSERVATION OF ANGULAR MOMENTUM ��

Figure ����

The total angular momentum in K ��t� is

LE �K ��t�� �ZK�t�

dVP ��r�h���r � �v ��l

�iE��r� t��

From ��������

d

dtLE �K ��t�� �

ZK�t�

dVP ��r�h�Dt

��r � �v ��l

�iE��r� t��

Now

Dt��r � �v� � �Dt�r�� �v � �r �Dt�v � �v � �v � �r � �a � �r � �a�

sod

dtLE �K ��t�� �

ZK�t�

dVP ��r�h���r � �a �Dt

�l�iE

��r� t�� ������

To calculate the torque on the particles in K ��t� we note that the torque exerted by the

body force dVP ��r��fE��r� t� is �r�dVP ��r��fE��r� t�� or dVP ��r���r� �f �E��r� t�� The torque exerted

by the surface force dAP ��r��S��r� t� nP � is �r�dAP ��r��S � �dAP ��r��S��r � �dAP ��r�� n��

SE

��r� t�� � �r� If�

Q� P � P is a polyad� � n� �Q� � �r � n � ��Q ��r�� so this is true for all�

Q�

�Exercise ��b gives a de�nition of �r� �

Q� and�

Q ��r is de�ned similarly�� Thus

�r � dAP ��r��S��r� t� nP � � �dAP ��r� nP ���

S ��r�E

��r� t��

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

.back

front^

( )

r

n

dA rP

P

Figure ����

Therefore� the torque about �OP which is exerted on the particles in K ��t� by the body

and surface force acting on them is

ZK�t�

dVP ��r��vecr � �f

�E��r� t��

Z�K�t�

dAP ��r� nP ���

S ��r�E

��r� t��

By applying Gauss�s theorem to the surface integral� we can write this torque as

ZK�t�

dVP ��r���r � �f � � �

��

S ��r��E

��r� t�� ������

In addition to this torque� there may be a torque acting on each atom in dVP ��r��

This would be true� for example� if the material were a solid bar of magnetized iron

placed in a magnetic �eld �B� If the magnetization density was �M � there would be a

torque dVP ��r� �M � �B acting on dVP ��r� and not included in ������� In general� we might

want to allow for an intrinsic body torque of �m joules�meter�� so that ������ must be

supplemented by a term ZK�t�

dVP ��r��mE��r� t��

Finally� in a magnetized iron bar� atoms just outside K ��t� exert a torque on atoms

just inside K ��t�� so there is a �torque stress� acting on K ��t�� The torque exerted on

the material just behind dAP ��r� by the material just in front is proportional to dAP as

long as the orientation of that small patch of surface� i�e�� its unit normal� stays �xed�

We write the proportionality constant as �M��r� t� nP �� The torque exerted on the material

just behind dAP by the material just in front� other than that due to nP ��

SE

��r� t�� is

����� CONSERVATION OF ANGULAR MOMENTUM ���

�M��r� t� nP �dAP ��r�� so we must also supplement ������ by a term

Z�K�t�

dAP ��r� �M ��r� t� nP ��r� t�� �

where nP is the outward unit normal to K ��t�� Thus we �nally have

�ME �K ��t�� �ZK�t�

dVP ��r���r � �f � � �

��

S ��r�� �m

�E��r� t�

�Z�K�t�

dAP ��r� �M ��r� t� nP ��r� t�� � �������

If we substitute ������ and ������� in �������� we obtain

RK�t� dVP ��r�

����r � �a�Dt

�l�� �r � �f � � �

��

S ��r�� �m

�E��r� t�

�R�K�t� dAP ��r� �M ��r� t� nP ��r� t�� � �������

This equation has the same mathematical form as �������� and leads via Cauchy�s the�

orem to the same conclusion� At �r� t there is a unique tensor�

ME ��r� t� � P � P � which

we call the �Cauchy� torque�stress tensor� such that for every unit vector nP � P we have

�M ��r� t� nP � � nP ��

ME ��r� t� �� �������

Substituting this on the right in �������� applying Gauss�s theorem to the surface integral�

and the vanishing integral theorem to the volume integral� gives

���r � �a �Dt

�l�� �r � �f � � �

��

S ��r�� �m � � � �M� �������

Before we accept this as the Eulerian angular momentum equation� we note that it can be

greatly simpli�ed by using the momentum equation ������� We have ���r � �a� � �r � �f

� �r ����a� �f

�� �r � ��� �S�� so ������� is

�Dt�l � �r �

��� �S

�� � �

��

S ��r�� �m� �� �M � �������

This can be still further simpli�ed� Let y� � y�� y� be a poob in P � � x�� x�� x�� a pooob in

L� and take components relative to these bases� Then

��r � ��� �S�

�i� �ijkrj���

S�k � �ijkrjlSlk�

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Also� for �u � �v� �w � P � ��u�v�� �w � �u��v� �w� �this is how we use the lifting theorem to

de�ne�

Q ��w for�

Q� P � P and �w � P � so

���u�v�� �w�ij � ��u ��v � �w��ij � ui ��v � �w�j

� ui�jkl�kwl

� ��u�v�ik �jklw�

Therefore for any�

Q� P � P �

��

Q ��w�ij� Qik�jklwl�

Therefore

��

S ��r�li

� Slk�ikjrj�� �

��

S ��r��

i� l

��

S ��r�li� l �Slk�ikjrj�

� �ikjl �Slkrj�

� �ikj ��lSlk� rj � Slklrj� �

Referring to page ��� and using �ijk � �ijk � �� we get

��r �

�� � �S

�� � �

��

S ��r��

i� �ikjSlklrj�

But lrj � �lj� so this is �ikjSjk � ��ijkSjk � ��AP hi

S

�i� Thus

�r ���� �S

�� � �

��

S ��r�� �AP hi

S � �������

Thus ������� is

�Dt�l � AP hi

S ��m� �� �M � ��������

This is the Eulerian form of the angular momentum equation� In non�magnetic materials

it is believed that �l� �m are�

M and all negligible� and that the essential content of ��������

is

AP hi�

S� ��� ��������

����� CONSERVATION OF ANGULAR MOMENTUM ���

That is� �ijkSjk � �� Multiply by �ilm and sum on i� and this becomes �see page D���

��jl�km � �jm�kl�Sjk � ��

or

Slm � Sml � ��

or�

ST��

S � �������

The conservation of angular momentum requires that the Cauchy stress tensor be sym�

metric� if intrinsic angular momentum� body torque and torque stress are all negligible�

����� Consequences of�

ST�

S

In the absence of intrinsic angular momentum� torque and torque stress� the Cauchy stress

tensor is symmetric� This greatly simpli�es visualizing the stress at a point �r at time t�

We will abbreviate�

SE��r� t� as

S in this chapter� because we are looking at one time t and

one position �r in physical space� We denote by �S� n� the stress on the oriented small area

�dA� n� at �r� t �i�e� we write �S��r� t� nP � as �S� n�� and nP as n� dAP as dA�� We denote by

S the linear operator on P corresponding to the tensor�

S� Then for any unit vector n�

�S� n� � n� �S� S� n�� �������

Since S � P P is a symmetric operator� P has an orthonormal basis x� y� z consisting

of eigenvectors of S� That is� there are scalars a� b� c such that

S� x� � a x� S� y� � b y� S� z� � c z� ��������

Then x� �S� a x� y� �S� b y� z� �S� c z so

S��

I P ��

S� � x x� y y � z z� � �S� x �a x� � y �b y� � z �c z�

so�

S� a x x � b y y � c z z� ��������

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

The unit vectors x� y� z are called �principal axes of stress� and a� b� c are �principal

stresses�� By relabelling� if necessary� we can always assume that a � b � c� Also if we

change the sign of any of x� y� z both �������� and �������� remain unchanged� Therefore

we may assume that � x� y� z� is positively oriented�

For any vector �n � P we write

�n � x x � y y � z z� ��������

The unit sphere NP consists of those �n � P for which

x� � y� � z� � �� ��������

The pressure pn� n� acting on �dA� n� is pn� n� � � n��

S � n � ��ax�� by�� cz��� It can be

thought of as a function of position n � x x � y y � z z on the spherical surface ���������

In the earth� pn � �� so it is convenient to introduce the principal pressures A� B� C�

de�ned by

A � �a� B � �b� C � �c� ��������

Then A � B � C� The cases where equality occurs are special and are left to the reader�

We will examine the usual case�

A B C� ��������

We want to try to visualize pn� n� and the shear stress �SS� n� acting on �dA� n� as func�

tioning of n� i�e�� as functions on the spherical surface ���������

We have

pn� n� � Ax� �By� � Cz� �������

�S� n� � �SS� n�� pn� n� n �������

and� since �S� n� � n� �S��S� n� � �Ax x� By y � Cz z� ������

First consider pn� Its value is unchanged if we replace any of x� y� z by its negative

in ��������� Therefore pn is symmetric under re�ections in the coordinate planes x � ��

����� CONSERVATION OF ANGULAR MOMENTUM ���

y � �� z � �� Therefore it su!ces to study pn� n� when n is in the �rst octant of ���������

i�e�� when

x � �� y � �� z � �� ������

If p is any constant� then the level line pn� n� � p on Np satis�es

Ax� �By� � Cz� � p �������

x� � y� � z� � �� �������

Hence� it satis�es

�A� p� x� � �B � p� y� � �C � p� z� � �� �������

This equation and �������� have no real solutions �x� y� z� unless

A � p � C� �������

so the values of pn� n� all lie in the interval �������� The value p � A requires y � z � ��

x � ��� so pn� n� � A only for n � � x� The value p � C requires x � y � �� so pn� n� � C

only for n � � z� The value of p � B makes ������� into xpB � A � �zpC � B�

and these two planes intersect the sphere �������� in great circles� one of which lies in

������� For any value of p in A p C� ������� and ������� imply all three of the

following� ��������������C � A�x� � �C �B� y� � C � p

�B � A� x� � �C � B� z� � B � p

�B � A� y� � �C � A� z� � p� A�

�������

Therefore the projection of the level curve �������� �������� onto the xy or yz plane is

part or all of an ellipse� and its projection on the xz plane is part of a hyperbola with

asymptotes xpB � A � �zpC � B� The level lines pn� n� � B on Np� with p � B�

are the great circles where Np intersects the planes xpB � A � �zpC � B� The level

lines of pn on NP are the dashed curves in �gure ����� Next� consider the shear stress

�SS� n�� At any n � NP � this is a vector tangent to NP � It is de�ned by �������� and

from ������ �S� n� � ��� �pn� n� where pn is to be considered as a function de�ned

for all �n � P by �������� even when k�nk �� �� At a point n � NP � let �� n� be the unit

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

np = C

p = Bn

p = An

p = Bn

^

^

^

z

x

y

Figure �����

����� CONSERVATION OF ANGULAR MOMENTUM ���

vector tangent to the level curve of pn passing through n� Then n � � � n� � �� and since

�� n� is tangent to a curve lying in the level surface pn��n� � constant in P � it follows that

�� n� � �pn� n� � �� Therefore� from �������� we have

� � n� � �S� n� � �

and

�� n� � n � ��

Then from �������� � � n� � �SS� n� � �� Thus the shear stress �SS� n� at n � NP is

perpendicular to the level line of pn which passes through n� The �lines of force� of the

shear stress on NP are the curves which are everywhere tangent to �SS� n�� They are the

solid curves in �gure ����� and are everywhere perpendicular to the level lines of pn�

These �lines of force� give the direction of �SS� n� everywhere on NP �

Finally� we would like to �nd k�SS� n�k on NP � which we abbreviate as

�� n� � k�SS� n�k� �������

Again we consider only the �rst octant� x � �� y � �� z � �� The other octants are

obtained by re�ection in the three coordinate planes x � �� y � �� z � �� First consider

the edges of the �rst octant� where x � � or y � � or z � �� On the quarter circle

n � x cos � � y sin �� � � � � ��� subjecting ������� and �������� ������ to a little

algebra yields

�SS� n� � sin � cos � � x cos � � y sin �� on n � x cos � � y sin �� �������

Therefore� from ��������

�� n� ��

�B � A� sin � if n � x cos � � y sin �� �������

This reaches a maximum value of �B � A�� at � � ���� half way between x and y on

the quarter circle� And �� x� � �� y� � �� Similarly

�� n� � ���C � A� sin � if n � x cos � � z sin � ������

�� n� � ���C � B� sin � if n � y cos � � z sin �� ������

To see what happens to �� n� when x � �� y � �� z � �� we appeal to

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Lemma ������� On any level line of pn in the �rst octant on NP � �� n� decreases mono�

tonically as y increases�

Proof�

From �������� k�S� n�k� � �� n�� � pn� n�� so

�� n�� � k�S� n�k� � pn� n��� �������

We can parametrize the level curve pn� n� � p on NP by the value of y at n�

namely y � n � y� We want to show that

d

dy�� n� � on the curve pn� n� � p on NP

�s �rst octant� �������

On that level curve� of course pn� n� is constant� so ������� shows that

������� is equivalent to

d

dyk�S� n�k� � on the curve pn� n� � p on Np

�s �rst octant� �������

The vector d n�dy on pn� n� � p is tangent to that level curve� Hence n �d n�dy � � and �p� n� � d n�dy � �� so �S� n� � d n�dy � �� Hence� for some

d n

dy� n� �S� n��

Now d n�dy � x�dx�dy� � y � z�dz�dy� on the level curve� Hence

y � d ndy

� �� so � � y � n� �S� n��

But n � xx � yy � zz� ��S� n�Ax x �By y � Cz z� so

� y � n� �S� n� �

������������ � �

x y z

Ax By Cz

������������ �xz�C � A��

Thus � ��xz�C � A�� and

d n

dy�

xz�C � A� n� �S� n�� �������

����� CONSERVATION OF ANGULAR MOMENTUM ��

By the chain rule� ddyk�S� n�k� � d�n

dy� �k�S� n�k�� From �������

k�S��n�k� � A�x� �B�y� � C�z�� so

�k�S��n�k� � A�x x � B�y y � C�z z�

Therefore

d

dyk�S� n�k� �

xz�C � A�

h n� �S� n�

i��A�x x �B�y y � C�z z

xz�C � A�

�����������x y z

�Ax �By �CzA�x B�y C�z

������������ � y

�C � A�

������������ � �

A B C

A� B� C�

������������ �y�C � B��B � A� ��

This proves lemma ����� �

QED�

From ������� ����� ����� and lemma ����� we infer that the largest value of

�� n�� the magnitude of the shear stress� is �C�A��� and this occurs at n � �� x� z��p�

and for no other n�

It is of some interest to see which pairs �pn� n�� �� n�� are possible on NP � On the three

quarter�circles in ������� ����� ����� we have�B� pn� n�

�n� n�

CA �

�B� A�B�� B�A

�cos �

B�A�

sin �

CA if n � x cos � � y sin � �z � �� �������

�B� pn� n�

�n� n�

CA �

�B� A�C�� C�A

�cos �

C�A�

sin �

CA if n � x cos � � z sin � �y � �� �������

�B� pn� n�

�n� n�

CA �

�B� B�C�� C�B

�cos �

C�B�

sin �

CA if n � y cos � � z sin �� �x � �� ��������

For z � �� the possible pairs �pn� n�� �� n�� are given by ������� and trace out the

seimicircle marked z � � in Figure ����� For y � �� the possible pairs are given by

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

increases in the direction of the arrows.θ

MOHR

MOHR

.

.start

stop

A B C^

σ( )^

y = 0

z = 0

x = 0

n

n np ( )p

Figure �����

����� CONSERVATION OF ANGULAR MOMENTUM ���

������� and trace out the semicircle marked y � � in Figure ����� For x � �� use the

third semicircle in Figure ������

In �gure ����� suppose we start at a point with y � � and move along the level line

of pn till we hit either x � � or z � �� Then in the ��� pn� plane we will start at the point

marked �start� in Figure ����� and we will remain on the vertical dashed line pn � p�

According to lemma ����� as we increase y on the level curve pn� p in Figure �������

we will decrease �� n�� so we move down the dashed line in Figure ���� until we strike

the point marked �stop�� Therefore� the possible pairs �pn� n�� �� n�� which can occur on

NP are precisely the points in the shaded region in Figure �����

Coulomb suggested that a rock will break when the maximum value of �� n� exceeds

a critical value characteristic of that rock� Navier pointed out that when two plates are

pressed together� the greater this pressure the harder it is to slide one over the other�

Navier suggested that the fracture criterion is that there be an orientation n for which

�� n� � �c��pn� n�� where �c and � are constants characteristic of the rock �when p� n� � ��

Navier�s fracture criterion reduces to Coulomb�s�� The constant � is a sort of internal

coe!cient of friction� Mohr suggested that Coulomb and Navier had over�simpli�ed the

problem� and that the true fracture criterion is ��n� � f �p� n��� where f is a function

characteristic of the rock� which must be measured empirically� and about which one can

say in advance only that

f�p� � �� df�dp � �� ��������

Even with such a general criterion� much can be said� The curve � � f�p� is marked

�MOHR� in Figure ����� If A� B� C are moved so as to produce an overlap of the

shaded region with the Mohr curve� the rock will break at �rst contact of the curve and

the shaded region� This �rst contact will always occur on the plane y � �� at a value of

� in ���� which is between � and ���� so the normal to the plane of fracture will lie in

the xz plane� closer to x than to z� The circles in Figure ���� are called Mohr circles�

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

��� Lagrangian Form of Angular Momentum

Conservation

Refer again to the notation introduced on page ��� If �LLt �H �� is the angular momentum

of the collection of particles with labels in H �� and �MLt �H

�� is the total torque on them�

the law of conservation of angular momentum is

d

dt�LLt �H �� �ML

t �H��� �������

Using the notation introduced on page ��� and page ����

�LLt �H �� �ZH�

dVL��x�h ��l � ��r � �v

iL��x� t�

sod

dt�LLt �H �� �

ZH�

dVL��x�h ��Dt�l � �r � �a

�iL��x� t� � ������

Moreover�

MLt �H

�� �ZH�

dVL ��x���r � � f � �m

�L��x� t�

�ZH�

dAL ��x�

��r �

� nL �

� SL�

� � M ��x� t� nL ��x��

where nL��x� is the unit outward normal to H � at �x � H �� � m is body torque per unit

of volume in label space� and � M��x� t� nL��x��dAL is the force exerted by the material with

labels just in front of �dAL� nL� on the material with labels just behind �dAL� nL�� Again

we have �r � � nL �� S� � �� nL �

� S�� �r � � nL � �

� S � �r�� so using Gauss�s theorem�

MLt �H

�� �ZH�

dVL ��x�

��r � � f � � m� �D �

�� S � �r

��L��x� t� ������

�ZH�

dAL ��x�� M ��x� t� nL��x�� � �������

Therefore

RH� dVL ��x�

� ��Dt�l � �r � �a

�� �r � � f � � m � �D �

�� S � �r

��

�R�H� dAL ��x�

� M ��x� t� nL��x�� � �������

����� CONSERVATION OF ENERGY ���

Applying Cauchy�s theorem to ������� shows that for each ��x� t� there is a unique� ML��x� t� � L� P such that for any unit vector n � L�

� M ��x� t� n� � n �

� ML

��x� t� � �������

If we substitute this on the right in ������� and use Gauss�s theorem and the vanishing

integral theorem� we obtain

�Dt�l � ��r � �a� �r � � f � � m � �D �

�� S � �r

�� �D �

� M�

Using the Lagrangian momentum equation �������� we can write this as

�Dt�l � �r �

��D �

� S

�� �D �

�� S � �r

�� �D �

� M � m�

Finally� by the sort of argument which led to �������� we can show

�r ���D �

� S

�� �D �

�� S � �r

�� �AP hi

���D�r�T � � S� �

so the Lagrangian form of the angular momentum equation is

�Dt�l � �D �

� M �

� m � AP hi

���D�r�T � � S� � �������

The relation between� M and

� M is the same as that between

� S and

S � If the intrinsic

angular momentum �l� body torque m� and torque stress� M are all negligible� �������

reduces to

AP hi���D�r�T � � S� � �� �������

A glance at ������ shows that ������� is the same as ��������� so we learn nothing

new from ������� in this case�

���� Conservation of Energy

����� Eulerian form

The velocity �vE��r� t� imparts to the matter in dVP ��r� a kinetic energy dVP ��r��E��r� t��

�k

�vE��r� t�k�� In addition� that material has kinetic energy of molecular motion� and the

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

potential energy of the intermolecular forces� The sum of these latter is called the internal

energy� We denote by UE��r� t� the internal energy per kilogram in the matter at �r� t� Then

the total energy in dVP is

dVP ��r�����

v� � U

��E��r� t� �

The total energy in K ��t� is

EE �K ��t�� �ZK�t�

dVP ��r�����

v� � U

��E��r� t� � �������

The law of conservation of energy is

d

dtEE �K ��t�� �WE �K ��t�� ������

where WE�K ��t�� is the rate at which energy is being added to the matter in K ��t�� i�e��

the rate at which work is being done on that matter�

The body force �f does work on dVP ��r� at the rate of dVP ��r���v � �f�E��r� t� watts� If

�r � K ��t� then the surface stress nP ��

SE

��r� t� does work on the particles just inside

K ��t� at the rate of dAP ��r� nP ��

SE

��r� t� � �vE��r� t� watts� Thus the purely mechanical

component of WE�K ��t�� is

ZK�t�

dVP ��r���v � �f

�E��r� t� �

Z�K�t�

dAP ��r� nP ���

S ��v�E

��r� t� �

Using Gauss�s theorem� this can be written

ZK�t�

dVP ��r���v � �f � � �

��

S ��v��E

��r� t� � ������

In addition to this mechanical method of adding energy to K ��t� there are non�

mechanical methods� Examples are

i� ohmic �heating� at a rate of dVP ��r�k �Jk��� watts� where �J � electric current density

and � � electrical conductivity�

ii� �heating� by absorption of light �e�g�� sunlight in muddy water��

iii� radioactive �heating��

����� CONSERVATION OF ENERGY ���

.back

front

( )

^

dA r

nP

P

r

Figure �����

All these �heating� mechanisms together will add energy to dVP ��r� at a total rate of

dVP ��r�hE��r� t�� where hE is the sum of the heating rates per unit volume� The resulting

contribution to WE�K ��t�� is

ZK�t�

dVP ��r�hE ��r� t� watts� �������

Finally� energy can leak into K ��t� through the boundary� K ��t�� The physical mech�

anisms for this are molecular collision and di�usion� but that does not concern us� We are

interested only in the possibility that if dAP ��r� is a small nearly plane surface containing

�r� with a designated front to which a unit normal nP is attached� the energy can leak

across dAP ��r� from front to back at a rate proportional to dAP as long as the orientation

of dAP is not varied� i�e�� as long as nP is �xed� We write the constant of proportionality

as �H ��r� t� nP �� so that energy �ows from front to back across dA��r� at a net rate of

�H��r� t� nP �dAP ��r� watts� This �heat �ow� contributes to WE �K ��t�� the term

�Z�K�t�

dAP ��r�H ��r� t� nP � watts�

The minus sign in the de�nition of the proportionality constant is a convention established

for two centuries� and we must accept it� Indeed� there is aesthetic reason to put a minus

sign in the de�nition of �S� to avoid the �p �

I term� Our choice of signs for �S and�

S was

also dictated by history� and it is unfortunate that the same choice was not made in both

cases�

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

The sum of all the foregoing rates at which energy is added to the matter in K ��t�

gives

WE �K ��t�� �ZK�t�

dVP ��r���v � �f � � �

��

S ��v�� h

�E��r� t� �������

�Z�K�t�

dAP ��r�H ��r� t� nP � � �������

From ������� and ��������

dEE�K ��t��

dt�ZK�t�

dVP ��r�

��Dt

��

v� � U

�E���r� t� � �������

Substituting ������� and ������� in ������ gives

ZK�t�

dVP ��r���Dt

��

v� � U

�� �v � �f � � �

��

S � �v�� h

�E��r� t� �������

� �Z�K�t�

dA ��r�H ��r� t� n� � �������

Applying Cauchy�s theorem ���� to ������� shows that at each ��r� t� there is a

unique vector �HE��r� t� � P such that for all unit vectors n � P

H ��r� t� n� � n � �HE ��r� t� � ��������

The vector �HE��r� t� is called the heat��ow vector� Heat �ows from front to back across

�dAP � nP � at the rate of �dAP ��r� nP � �HE��r� t� watts� Inserting �������� in ������� and

applying Gauss�s theorem converts ������� to

� �ZK�t�

dVP ��r���Dt

��

v� � U

�� �v � �f � � �

��

S ��v�� h � � � �H

�E��r� t� � ��������

SinceK ��t� can be any open subset ofK�t� with piecewise smooth boundary� the vanishing

integral theorem gives

�Dt

��

v� � U

�� � � �H � h � �v � �f � � �

��

S ��v�� �������

This equation can be simpli�ed� We have

Dt

�v��� Dt ��v � �v� � �v �Dt�v � �v � �a�

����� CONSERVATION OF ENERGY ���

Also� taking components relative to an orthonormal basis�

� ���

S ��v�

� i �Sijvj� � �iSij� vj � Sijivj �������

���� �S

�� �v� �

S hi��v� ��������

Therefore ������� is

��a � �v � �DtU � � � �H � h� �f � �v ���� �S

�� �v� �

S hi��v�

But the momentum equation says ��a � �f � �� �S� so those terms cancel� leaving only

�DtU � � � �H � h��

S hi��v� ��������

This is the Eulerian form of the internal energy equation� or heat equation�

����� Lagrangian form of energy conservation

The argument should be familiar by now� The total energy of the material with labels in

H � is� at time t�

ELt �H �� �ZH�

dVL ��x�� ���

v� � U

��L��x� t� � ��������

The rate at which energy is added is

WLt �H �� �

ZH�

dVL ��x�

�� f � �v � �D �

�� S � �v

�� 'h

�L��x� t� ��������

�Z�H�

dAL ��x� � H ��x� t� nL��x�� � ��������

The law of conservation of energy is

d

dtELt �H �� �WL

t �H �� ��������

and from �������� we have the identity

d

dtELt �H �� �

ZH�

dVL��x�� �Dt

��

v� � U

��L��x� t�� �������

Substituting �������� and ������� in �������� givesZH�

dVL ��x�

� �Dt

��

v� � U

�� � f � �v � �D �

�� S � �v

�� 'h

�L��x� t�

� �Z�H�

dAL��x� �H ��x� t� nL��x�� � �������

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

At any ��x� t�� Cauchy�s theorem ���� assures us of the existence of a unique vector

� �HL��x� t� � L such that for any unit vector n � L� �H��x� t� n� � n� � HL

��x� t�� Substituting

this on the right in �������� and applying Gauss�s theorem and the vanishing integral

theorem gives

�Dt

��

v� � U

�� � f � �v � �D �

��

S ��v�

'h � � �D� � �H�

But Dtv� � �v �Dt�v � �v � �a� and

�D ��� S � �v

��

��D �

� S

�� �v �

� Shi �D�v

so

��a � �v � �DtU � � f � �v ���D �

� S

�� �v� �

S hi �D�v � 'h � � �D� � �H�

The Lagrangian momentum equation� ��a� � f � �D �� S � ��� permits some cancellation and

leaves us with

�DtU � �D� � �H � 'h �� Shi �D�v� ������

This is the Lagrangian form of the internal energy equation�

Using ������ and the identity

dAP n � �H � dAL nL� � �H�

one proves�H

����D�r�T � � �H

�������

in the same manner as ������ was proved�

��� Stu�

Mass� momentum� angular momentum and energy are all examples of a general math�

ematical object which we call a �stu��� The idea has produced some confusion� so we

discuss it here�

���� STUFF ��

De�nition ������ A �stu�� is an ordered pair of functions � ��E��

FE

� with these prop�

erties� V is a Euclidean space� and for each t � R� K�t� is an open subset of physical

space P � and

��E��� t� � K�t� V �������

FE��� t� � K�t� P � V� ������

The function ��Eis called the spatial density of the stu��

FE

is the spatial current density

or spatial �ux density of the stu�� And ��E� the creation rate of the stu�� is de�ned to be

��E � t ��E� �� �F

E

� ������

De�nition ������ Suppose t � R� �r � K�t�� and dV ��r� is a small open subset of K�t�

containing �r� dV ��r� will also denote the volume of that subset� Suppose dA��r� t� is a small

nearly plane oriented surface containing �r� and n is the unit normal on the front side of

dA��r� t�� We also use dA��r� t� to denote the area of the small surface� Suppose dA��r� t�

moves with speed W normal to itself� with W � � when the motion is in the direction of

n� Then

i� dV ��r� ��E��r� t� is called the amount of stu� in dV ��r� at time t�

ii� dA��r� t�� n� �FE

��r� t� �W ��E��r� t�� is called the rate at which the stu� �ows across

�dA��r� t�� n� from back to front�

iii� dV ��r���E��r� t� is called the rate at which stu� is created in dV ��r� at time t�

De�nition ������ Let K ��t� be any open subset of K�t�� with piecewise continuous

boundary K ��t� and unit outward normal n��r� at �r � K ��t�� Suppose K ��t� moves so

that the outward velocity of K ��t� normal to itself at �r � K ��t� is W ��r� t�� Then

i�RK�t� dV ��r� ��

E��r� t� is called the amount of stu� in K ��t� at time t� Denote if by

�,�K ��t��

ii�R�K�t� dA��r�

� n��r�� �F

E

��r� t��W ��r� t� ��E��r� t�

�is called the rate at which stu� �ows

out of K ��t� across K ��t� at time t� Denote it by �F �K ��t���

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

iii�RK�t� dV ��r���E��r� t� is called the rate at which the stu� is created in K ��t� at time t�

Denote it by �(�K ��t���

Remark ����� d�dt �,�K ��t�� � �(�K ��t��� �F �K ��t�� for any open subset K ��t� of K�t��

moving in any way whatever�

Proof�

d

dt�, �K ��t�� �

ZK�t�

dV ��r�t ��E��r� t� �

Z�K�t�

dA��r�W ��E��r� t�

�F �K ��t�� �ZK�t�

dV ��r��� �FE

��r� t��Z�K�t�

dA��r�W ��E��r� t��

Therefore�

d

dt�, �K ��t�� � �F �K ��t�� �

ZK�t�

dV ��r�

�t ��

E� �� �F

E�

�ZK�t�

dV ��r���E��r� t� � �( �K ��t�� �

De�nition ������ Suppose the K�t� in de�nition ����� is the set of positions occu�

pied by the particles of a certain continuum at time t� Suppose � and �v are the density

of mass and the particle velocity in the contiuum� Then de�ne

i� ��E��r� t� � ��E��r� t���E��r� t�

ii��

FE��r� t� �

FE

��r� t�� �vE��r� t� ��E��r� t��

Let ��� �� ��

F ��

F be� respectively the physical quantities whose Eulerian descriptions

are ��E� ��E��

FE��

FE

� Then the de�nitions of ��E and�

FEimply

�� � �� �� or �� � ��� �������

�� ��

F ��v �� or�

F��

F ���v��� �������

The physical quantity �� is called the density of stu� per unit mass of continuum material

��material density of the stu��� for short�� while �� is the density of stu� per unit volume

of physical space� The physical quantity�

F is the �ux density of the stu� relative to the

material in the continuum� or the material �ux density� while�

F is the �ux density of the

stu� in space�

���� STUFF ���

Remark ����� Suppose the two pairs of functions � �� ��

F � and ��� �F � are related by

�������� Then

t �� � �� �F� �Dt��� �� �F �������

so the creation rate of the stu� � �� ��

F � is

�� � �Dt��� �� �F � �������

Proof�

t �� � �t�� ��� ��t��

�� � �F � �� �F �

h� � ���v�

i��� ���v� � ���

t � � � � �F �ht�� � � ���v�

i��� �

ht��� �v �

�����i

� �� �F� ���� �Dt

��� �� �F �

QED�

Stu�s can be added if their densities and �uxes have the same domain and same range�

We have

De�nition ������ Suppose � �� ��

F � for each v � �� � � � � N is a stu�� and that all

these stu�s have the same K�t� and V in de�nition �������� Suppose c�� � � � � cN � R�

ThenPN

� c��� �

F � stands for the stu� �PN

� c�� �� �

PN � c

F ��

Corollary ����� The stu�PN

� c��� �

F � has creation ratePN

� c��

Proof�

Obvious from �������

Corollary ����� In a continuum with particle velocity �v and density �� the material

density and material �ux density of the stu�PN

� c�� � �F � are� respectivelyPN

� c�

andPN

� c �Fv�

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Proof�

Obvious from ������� and ��������

Now we consider a number of stu�s which arise out of the conservation laws� The

name of each stu� is capitalized and underlined� and then its spatial density �� � spatial

�ux�

F � material density ��� material �ux�

F � and creation rate �� are given� The stu� is

not de�ned by �� or �� alone� Both �� and�

F � or both �� and�

F must be given� Then �

is calculated from ������� The choice of�

F is dictated by convenience in applications�

Mass� � � �� �F � ��v� � � �� �F � ��� � � t � � � � �F � t� � � � ���v� � �� Mass is

not created� and its material �ux is zero�

Momentum� �� � ��v��

F� ��v�v� �

S� �� � �v��

F� � �

S� �� � �Dt��� � �F � �Dt�v����

S� �f

Intrinsic Angular Momentum� �� � ��l ��

F� ��v�l� �

M � �� � �l��

F� � �

M � so � �

�Dt�l � �� �M� �m � AP hi

S �

Kinetic Angular Momentum�

�� � ���r � �v���

F� ��v��r � �v���

S ��r ��� � �r � �v�

F��

S ��r�� � �Dt ��r � �v� � � �

��

S ��r�� ��r � �a� �r �

��� �S

�� AP hi

S

�� � �r � �f � AP hi�

S �

Total Angular Momentum�

�� � ���l � �r � �v

��

F� ��v��l � �r � �v

�� �

M ��

S ��r� � �l � �r � �v�

F� � �

M ��

S ��r �

Since total angular momentum is the sum of intrinsic and kinetic angular momentum

in the sense of de�nition �������� its creation rate is the sum of their creation rates�

by corollary ��������� Thus

�� � �m� �r � �f�

���� STUFF ���

Internal Energy�

� � �U� �F � ��vU � �H

� � U� �F � �H

� � �DtU � � � �H

so

� � h��

S hi��v�

An alternative de�nition of internal energy suggests itself if h is entirely hnuc� the

radioactive heating rate� Let �Lnuc��x� t� be the nuclear energy in joules per kilo�

gram of mass near particle �x� As the radioactive nuclei near �x decay� they emit

��rays and massive particles� We will assume that these lose most of their energy

by collision with molecules so close to �x that the macroscopic properties of the

continuum are nearly constant over the region where the collisions occur� Then

kinetic energy of molecular motion is added to the material near �x at the rate

�Dt�Lnuc��x� t� watts�kg� To convert to watts�meter �� we must multiply by the

density� so hLnuc��x� t� � ��L��x� t�Dt�Lnuc��x� t� or

hnuc � ��Dt�nuc� �������

We can regard U � �nuc as total internal energy density� and de�ne a stu� called

TOTAL INTERNAL ENERGY with

� � � �U � �nuc� � �F � ��v �U � �nuc� � �H

� � U � �nuc� �F � �H�

The creation rate of this energy is � � �Dt� � � � �F � �DtU � � � �H � �Dt�nuc

� h��

S hi��v � hnuc� If h � hnuc� then � ��

S hi��v�

This �total internal energy� is not very useful when the value of �nuc is una�ected

by what happens to the material� That is the situation at the low temperatures and

pressures inside the earth �with one possible small exception� see Stacey� Physics of

the Earth� p� ��� In such situations� hnuc�� is a property of the material which is

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

known at any particle �x at any time t� independent of the motion of the continuum�

By contrast� the molecular U is not known at �x� t until we have solved the problem

of �nding how the continuum moves� In the deep interiors of stars� pressures and

temperatures are high enough to a�ect nuclear reaction rates� so it is useful to

include �nuc with U �

Kinetic Energy�

� ��

�v�� �F � ��v

��

v��� �

S ��v

� ��

v�� �F � � �

S ��v

� � �Dt�� � � �F ��

�Dtv

� � � ���

S ��v�

� ��v �Dt�v � i �Sijvj� relative to an orthonormal basis for P

� ��v � �a� �iSij� vj � Sijivj

����a� �� �S

�� �v� �

S hi��v

� � �f � �v� �

S hi��v�

IK Energy� �Internal plus Kinetic Energy��

This is the sum of internal energy and kinetic energy in the sense of de�nition

������� so it has

� � ����v� � U

�� �F � ��v

���v� � U

�� �H� �

S ��v� � �

�v� � U� �F � �H� �

S ��v� � h� �f � �v�

Potential Energy� Suppose the body force density �f is given by �f � ���� where

the potential function �E��r� t� is independent of time� i�e�� t� � �� Then Dt� �

t� � �v � �� � �v � �� so �Dt� � ��f � �v�

Then the stu� �potential energy� is de�ned by

� � �� �F � ��

� � ��� �F � ��v�

���� STUFF ���

so its creation rate is � � �Dt� � ��f � �v � ��

IKP Energy� �Internal plus kinetic plus potential energy��

If �f � ���� with t� � �� then we de�ne the stu� �IKP energy� as the sum of

internal� kinetic and potential energies in the sense of de�nition ��������� Then

� ��

v� � U � �� �F � �H� �

S ��v

� � ���

v� � U � �

�� �F � ��v

��

v� � U � �

�� �H� �

S ��v�

By corollary ������� � is the sum of the three ��s for kinetic� internal and potential

energy� so � � h� If we include nuclear potential energy in U and neglect other

sources of h �e�g� solar heating�� then � � �� Usually� this is not done� and IKP

energy is called �total energy��

������ Boundary conditions

So far we have dealt only with physical quantities which were continuously di�erentiable

to the extent needed to justify using Gauss�s theorem and the vanishing integral the�

orem� Now we analyze what happens when there is a surface S in the continuum� across

which physical quantities may have jump discontinuities� That is� lim�E��r� t� exists as �r

approaches �r� � S from either side of S� but the two limits are di�erent�

Examples of such surfaces are the interface between the ocean and atmosphere� the

ocean bottom� the core�mantle boundary� and a shock wave in air� water or rock� �Shock

waves in rock are excited only by nuclear blasts and meteor impacts� Earthquakes are

too weak to excite them��

We will assume that it is possible to choose a front side of S �M-obius strips are

excluded� and that we have done so� We will denote the unit normal to S on its front� or

positive� side by �� and we will call its back side the negative side� S is a large surface�

not nearly plane� but piecewise smooth� so � can vary with position �r � S� Moreover�

we will permit S to move� so that S�t� may vary with time t� Then so will ���r� t�� the

normal to S�t� at �r � S�t�� We will write W ��r� t� for the velocity with which S�t� moves

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

. front+

back

^

r

r

S

ν( )

Figure �����

normal to itself at point �r � S�t�� and we take W � � if the motion is in the direction of

�� W � if opposite�

If the surface is given in physical space P � we write it as SE�t�� If it is given in label

space L� we write it as SL�t�� If � is any physical quantity with a jump discontinuity

across S�t�� we write �E��r�� t�� for the limiting value of �E��r� t� as �r �ro � SE�t� from

in front of SE�t�� We write �E��r�� t�� for the limit of �E��r� t� as �r �r� from behind

SE�t�� And we de�ne

h�E ��r�� t�

i��� �E��r�� t�

� � �E��r�� t��� �������

This quantity is called the jump in �E across SE�t�� We introduce a similar notation for

label space� with h�L ��x�� t�

i��� �L ��x�� t�

� � �L ��x�� t�� �

We will need the surface version of the vanishing integral theorem� This says that if

V is a Euclidean vector space� S is a surface in P or L� and �f � S V is continuous� andRS� dA��r�

�f��r� � �OV for every open patch S � on S� then �f��r� � �OV for all �r � S� Here an

�open patch� on S is any set S � � K � S� where K � is an open subset of P or L� The

���� STUFF ���

proof of the surface version of the vanishing integral theorem is essentially the same as

the proof of the original theorem ������� so we omit it�

We will consider only the Eulerian forms of the conservation laws� so we will omit the

superscript E� The reader can easily work out the Lagrangian forms of the law�

We use the notation introduced on page ��� but now we assume that the surface S�t�

passes through K ��t�� Therefore� we must augment the notation� We use K ��t�� for the

part �excluding S�t�� of K ��t� which is in front of S�t�� �K ��t� for the part of K ��t� in

front of S�t�� and K ��t�� for the whole boundary of K ��t��� including K ��t� S�t�� If �is a function de�ned and continuous on the closed set K ��t� � K ��t��K ��t�� except for a

jump discontinuity on S�t�� we write �� for the continuous function on K ��t�� de�ned as

� except on K ��t� S�t�� where its value is the �� of equation �������� We label objects

behind S�t� in the same way� except that � is replaced by �� The unit outward normal to

K ��t�� will be written n�� The unit outward normal to K ��t�� will be written n�� The

unit outward normal to K ��t� will be written n� Then n� � n on �K ��t� and n� � � �on K ��t� S�t�� And n� � n on �K ��t�� while n� � � on K ��t�S�t�� As on page ���

we choose K ��t� so that it always consists of the same particles� Then the outward normal

velocity of K ��t�� relative to itself is n ��� on �K ��t� and �W on K ��t�S�t�� Here �v is

the particle velocity in the continuum� The outward normal velocity of K ��t�� relative

to itself is n � �v on �K ��t� and W on K ��t� S�t��

������ Mass conservation

Let M �K ��t�� denote the mass of the material in K ��t�� Since K ��t� always consists of the

same particles� the physical law of mass conservation is

d

dtM �K ��t�� � �� ��������

We need a mathematical identity like �������� We begin with

M �K ��t�� �ZK�t�

dV ��r�� ��r� t� � ��������

This looks obvious until we recognize that there are physical situations which we might

want to model by placing a mass per unit area on S�t�� which would have to be added

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Figure �����

���� STUFF ��

to ��������� Examples are S�t� � the soap �lm in a soap bubble in air� or S�t� � the ice

pack on a polar sea� We ignore these possibilities and assume

mass on S�t� � �� �������

Next� we try to di�erentiate �������� using ��������� The jump discontinuity on

S�t� K ��t� prevents this� so we break �������� into

M �K ��t�� � MhK ��t��

i�M

hK ��t��

iwith �������

MhK ��t��

i�ZK�t��

dV ��r�����r� t� ��������

MhK ��t��

i�ZK�t��

dV ��r�����r� t�� ��������

The two integrands in �������� and �������� are continuous� so we can apply �������� to

obtain

d

dtMhK ��t��

i�ZK�t��

dV ��r� t���

Z��K�t�

dA n ��v����ZK�t��St�

dAW��� ��������

We would like to use Gauss�s theorem on the right in ��������� but �K ��t� is not a

closed surface�

To close it we must add K ��t� S�t�� ThenZ��K�t�

dA n � ���v�� �Z�K�t��

dA n� � ���v�� �ZK�t��St�

dA n� � ���v��

�ZK�t��

dV � � ���v�� �ZK�t��St�

dA � � ���v��

Using this in �������� gives

d

dtMhK ��t��

i�ZK�t��

dVht�� � � ���v�

i��ZK�t��St�

dA �� � � � v �W ��� �

But �t�� � � ���v��� in K ��t�� is just the value of t�� � � ���v� there� so it vanishes� and

d

dtMhK ��t��

i�ZK�t��St�

dA ��� � � �v �W ��� � ��������

By an exactly similar argument�

d

dtMhK ��t��

i�ZK�t��St�

dA ��� � � �v �W ��� � ��������

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

Therefore�d

dtM �K ��t�� �

ZK�t��St�

dA ��� � � �v �W ���� � ��������

Combining �������� and �������� gives for K � � K ��t�

ZK�t��St�

dA ��� � � �v �W ���� � �� �������

Since K � can be any open subset of K�t�� K � S�t� can be any open patch on S�t��

Therefore� by the surface version of the vanishing integral theorem

�� � � � �v �W ���� � � on S�t�� �������

By analogy with de�nition ����� ii�� we call either ��� � ��v�W ��� or ��� � ��v�W ���

the �ux of mass across S�t�� and write it m��r� t� since it depends on �r and t� Thus

m � ��� ��v �W ��� � ��� ��v �W ��� � ������

If m � �� S�t� is a boundary between materials� It is then called a contact surface� It is

not crossed by particles� Then

� � �v � � � � �� � W� ������

If m �� �� S�t� is a shock wave or shock front�

������ Momentum conservation

Again refer to �gure ����� Let �P �K ��t�� be the momentum in K ��t�� while �F �K ��t�� is

the force on that material� Then mometum conservation requires

d

dt�P �K ��t�� � �F �K ��t�� � �������

We must �nd mathematical identities for the two sides of �������� Because of ��������

we have

�P �K ��t�� �ZK�t�

dV ��v� �������

���� STUFF ���

Again we cannot appeal directly to �������� to di�erentiate �������� because of the

jump discontinuity on S�t�� Therefore we imitate the procedure beginning on page ��

We have

�P �K ��t�� � �PhK ��t��

i� �P

hK ��t��

i�������

where

�PhK ��t��

i�ZK�t��

dV ���v�� �������

�PhK ��t��

i�ZK�t��

dV ���v��� �������

Then� appealing to ��������� we have

d

dt�PhK ��t��

i�

ZK�t��

dV t���v��

�Z��K�t�

dA n � ��v��v��

�ZK�t��St�

dAW ���v��

and

Z��K�t�

dA n � ��v��v�� �Z�K�t��

dA n� � ���v�v��

�ZK�t��St�

dA n� � ���v�v��

�ZK�t��

dV � � ���v�v��

�ZK�T ��St�

dA � � ���v�v�� �

Thus

d

dt�PhK ��t��

i�

ZK�t��

dVht ���v� � � � ���v�v�

i��ZK�t��St�

dA �� � � �v �W � ��v �� �

Now

t���v� � � � ���v�v� �ht�� � � ���v�

i�v � �

ht�v � �v � ��v

i� �Dt�v � ��a � �� �S ��f

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

so� using Gauss�s theorem

d

dt�PhK ��t��

i�

Z�K�t��

dA n�� �S�

�ZK�t��

dV �f �ZK�t��St�

dA �� � � �v �W � ��v��

�ZK�t��

dV �f �Z��K�t�

dA n� �S

�ZK�t��St�

dA�� �� �S ��� � � �v��W � ��v

���

Using ����� we �nally get

d

dt�PhK ��t��

i�

ZK�t��

dV �f ��Z��K�t�

dA n� �S

�ZK�t��St�

dA�� �� �S �m�v

��� �������

An exactly similar argument gives

d

dt�PhK ��t��

i�

ZK�t��

dV �f �Z��K�t�

dA n� �S

�ZK�t��St�

dA�m�v � �� �S

��� �������

Therefore� adding ������� and ��������

d

dt�P �K ��t�� �

ZK�t�

dV �f �Z�K�t�

dA n� �S

�ZK�t��St�

dA�m�v � �� �S

���� �������

We still need a mathematical identity for �F �K ��t�� in �������� This seems easy�

�F �K ��t�� �ZK�t�

dV �f �Z�K�t�

dA n� �S � ������

But ������ does involve assumptions� If S�t� is an air�water interface holding an electric

charge per unit area� there will be an electrostatic force per unit area on S�t� which must

be added to ������� In addition� the surface integral on the right in ������ assumes

that the stress on the material just inside K ��t� by the material just outside K ��t�

is n� �S�

behind S�t� and n� �S�

in front of S�t�� all the way to S�t�� In fact� if S�t�

is an air�water interface� for example� the water molecules just behind S�t� and the air

���� STUFF ���

molecules just in front will be arranged di�erently from those in the bulk of the air and

water� Therefore� along the curve K ��t� S�t� there will be an additional force per unit

length exerted by the molecules just outside K ��t� on the molecules just inside K ��t��

This force per unit length is called the surface stress in S�t�� For an air�water interface it

is tangent to S�t� and normal to K ��t� S�t�� and its magnitude is the surface tension�

We assume

there is no surface stress or surface force per unit area on S�t�� ������

Then ������ is correct� and substituting ������� and ������ in ������� givesZK��St�

dA�m�v � �� �S

���� � �������

for any open subset K � of K�t�� By the surface version of the vanishing integral theorem�

it follows that �m�v � �� �S

���� � on S�t�� �������

If S�t� is a boundary between two materials rather than a shock� then m � � so

������� reduces to � �� �S

���� � on S�t�� �������

At a boundary between two materials �a contact surface�� the stress on the boundary �the

�normal stress�� is continuous across that boundary�

Equation ������� has a simple physical interpretation� Consider a small patch dA

on S�t�� Mass m�t dA crosses dA from back to front in time �t� and gains momentum

m�t ��v��� dA � �m�v��� �tdA� Therefore� the patch requires momentum to be supplied to

it at the rate �m�v���dA per second� Where does this come from$ In the material the

�ux density of momentum is � �

S� so momentum arrives at the back of dA at the rate

� �� �S�dA and leaves the front at the rate � �� �S

�dA� The net momentum accumulation

rate in dA� available to supply the required �m�v���dA� is � � ��

S� dA � � � �

S�

�dA

� � � ��S���dA� Thus ������� simply says that the di�erence in momentum �ux into the

two sides of dA supplies the momentum needed to accelerate the material crossing dA� If

m � �� this reduces to action � reaction� i�e� ��������

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

����� Energy conservation

The physical law of conservation of energy is

d

dtE �K ��t�� �W �K ��t�� �������

where E �K ��t�� is the total kinetic plus internal energy in K ��t� and W�K ��t�� is the rate

at which energy is being added� We refer again to �gure ���� and follow the by now

well�trodden path to �nd mathematical identities for both sides of �������� We have

E �K ��t�� � EhK ��t��

i� E

hK ��t��

iwhere

EhK ��t��

i�ZK�t��

dV����

v� � U

����������

with a similar formula for E �K ��t���� Then

d

dtEhK ��t��

i�

ZK�t��

dV t

����

v� � U

����Z��K�t�

dA n ���v���

v� � U

����ZK�t��St�

dAW����

v� � U

����

Also

R��K�t� dA n �

h��v���v� � U

�i��R��K�t�� dA n� �

h��v���v� � U

�i�� RK�t��St� dA n� �

h��v���v� � U

�i��RK�t�� dV

� �h��v���v� � U

�i��RK�t��St� dA � �

h��v���v� � U

�i��

Thus�

d

dtEhK ��t��

i�

ZK�t��

dV�t

����

v� � U

��� � �

���v��

v� � U

�����ZK�t��St�

dA�� � � �v �W � �

��

v� � U

����

Now on page � we see that IK energy is a stu� with

, � ���

v� � U

�� �F � ��v

��

v� � U

�� �H� �

S ��v�

���� STUFF ���

Therefore t, � � � �F � � � h � �f � �v for this stu�� Thus th���

�v� � U�

i� � �h

��v���v� � U

�i� � � �H � � �

��

S ��v�� h� �f � �v� and

d

dtEhK ��t��

i�

ZK�t��

dV��� � �H � � �

��

S ��v�� h � �f � �v

���ZK�t��St�

dA m��

v� � U

���

ZK�t��

dV�h � �f � �v

���Z�K�t��

dA n� ��� �H�

S ��v��

�ZK�t��St�

dA m��

v� � U

��

d

dtEhK ��t��

i�

ZK�t��

dV�h� �f � �v

���Z��K�t�

dA n ��� �H�

S ��v�

�ZK�t��St�

dA�m��

v� � U

�� � �

��H� �

S ��v���

� �������

Similarly�

d

dtEhK ��t��

i�

ZK�t��

dV�h� �f � �v

���Z��K�t�

dA n ��� �H�

S ��v�

�ZK�t��St�

dA�m��

v� � U

�� � �

��H� �

S ��v���

� ��������

Therefore

d

dtE �K ��t�� �

ZK�t�

dV�h � �f � �v

��Z��K�t�

dA n ��� �H�

S ��v�

�ZK�t��St�

dA�m��

v� � U

�� � �

��H� �

S ��v���

�� ��������

Also�

W �K ��t�� �ZK�t�

dV�h� �f � �v

��Z�K�t�

dA n ��� �H � �S � �v

��

Therefore� ������� implies

ZK��St�

dA�m��

v� � U

�� � �

��H� �

S ��v���

�� � �������

for every open subset K � ofK�t�� By the surface version of the vanishing integral theorem��m��

v� � U

�� � � �H � �� �S ��v

���� � on S�t�� �������

��� CHAPTER ��� CONSERVATION LAWS IN A CONTINUUM

If the material is stationary� m � � and �v � �� so ������� reduces to the thermal

boundary condition h � � �H

i��� �� ��������

At a material boundary which is not a shock� m � �� so � �� �S �� � � �� �S��� and

� �� �S ��v��� � � �� �S�� � �v� �� �� �S�� � �v� � � �� �S� � ��v��� where �� �S� � �� �S�� � � �� �S���Also� from ������� if �v � v � � �vS where vv � � � �v� then v� � v� � so ��v��� � ��vS�

��

Therefore ������� becomes h � � �H

i���� �� �S

�� ��vS��� � ��������

The quantity � � � �H��� is the net heat �ow per unit area per second from the boundary into

the surrounding medium� Equation �������� says that heat is generated mechanically

when the two materials slip past one another at the relative velocity ��vS��� and when the

surface stress � �� �S� on the boundary has a component in the direction of the tangential

relative slip velocity� Brune� Henyey and Roy �J� Geophys� Res� ��� ��� used the

absence of an observed heat �ow anomaly along the San Andreas fault� together with an

estimate of ��vS��� on the fault� to put an upper bound on the component of �� �S in the

direction of slip� They used ���������

Even when m �� �� ������� can be made to resemble ������� We have��

mv�

���

��

m��v � � �v �

����v � � �v �

��

� ���

S� � �

S�����v � � �v �

�from �����

so ��

mv� � �� �S ��v

���

��

� ��� �

S� ��v �� �

S� ��v ��

S� � �v�� �

S� � �v�

�� � � � �

��

S��

S��� ��v��� �

De�ne

h�Si�� ��

��

S��

S��� average

S on S� ��������

Then ������� can be written in general ash � � �H �m U

i��� � � h�Si�� � ��v��� � ��������

Chapter ��

Strain and Deformation

��� Finite deformation

Consider a continuum which at time t occupies open subset K�t� of physical space P �

Suppose K ��t�� is a very small open subset of K�t��� We would like to be able to visualize

the particles in K ��t��� and to see their spatial positions at a particular later time t� The

time di�erence t � t� is not small� but K ��t�� is a very small set� in a sense to be made

precise shortly�

We use t��position labelling� Then label space L is the same as physical space P �

and �r L��x� t� � position at time t of the particle which was at position �x at time t�� The

particles in which we are interested have their labels in the set H � � K ��t��� First choose

a particular particle �xA � K ��t��� Let �rA be its position at time t� Then �rA � �r L��xA� t��

Consider any other particle �x � K ��t��� and let �r be its position at time t� Then �r �

�r L��x� t�� De�ne �� � �r � �rA� �� � �x� �xA� Then

�� � �r L��x� t�� �r L��xA� t� � �r L��xA � ��� t�� �r L��xA� t�

so

�� � �� � �D�r L ��xA� t� � k��k�R����

��������

where �R���� is the remainder function for �r L��� t� at �xA�

��� CHAPTER ��� STRAIN AND DEFORMATION

..

..

ξ

0

ρ

AAx

r

r

x

r

.( , t)

H =K (t ) K (t)

L

′ ′ ′

Figure �����

Now we assume that K ��t�� has been chosen so small that for every �x � K ��t���

k�R����k k �D�r L ��xA� t� k� �������

This is the sense in which K ��t�� must be a small set� If ������� is satis�ed� then we can

neglect the remainder term in �������� and write

�� � �� � �D�r L ��xA� t� � �������

If we measure initial positions in K ��t�� as displacement vectors from an origin at �xA�

and �nal positions in K ��t� as displacement vectors from an origin �rA� the mapping from

initial to �nal positions �� �� �� is linear� and is �������� Thus the e�ect of the motion

on the particles in K ��t�� is clear� The particle at �xA is displaced to �rA� The relative

positions of nearby particles� relative to particle �xA�s position� su�er the linear mapping

��������

As we have noted �see ������ j det �D�r L��xA� t��j is never � for any t�� But �r L�x� t�� �

�x� so �D�r L��x� t�� ��

I P and det �D�r L��xA� t�� � �� Then� since det �D�r L��xA� t�� depends

continuously on t� and is never �� it cannot change sign� so it is always positive� Therefore

det �D�r L ��xA� t� � �� ��������

It follows from the polar decomposition theorem �page D��� that there are unique tensors�

C and�

R� P � P with these properties�

����� FINITE DEFORMATION ��

i��

C is symmetric and positive de�nite�

ii��

R is proper orthogonal �i�e�� a rotation through some angle � between � and � about

some axis w��

iii�

�D�r L ��xA� t� ��

C � �R � ��������

Thus ������� is

�� � ��� �C � �R � ��������

Thus the e�ect of the motion on K ��t�� is to displace �xA to �rA� then to subject the relative

position vectors �relative to particle �xA� to the symmetric operator C� and then to rotate

everything through angle � about an axis passing through �rA in the direction of w�

The displacement �xA �� �rA and the rotation�

R are easy to visualize� The mapping�

C may be worth discussing� Since�

CT��

C there is an orthonormal basis y�� y�� y� for P

which consists of eigenvectors of C� That is� there are scalars c�� c�� c� such that

C � yi� � ci� yi�� i � �� � � ��������

Then yi��

C� ci� yi� so�

C��

I P ��

C� � yi yi� ��

C � yi

� yi�

C

��P�

i � ci� yi� yi�� Thus

C� c� y� y� � c� y� y� � c� y� y�� ��������

Since�

C is positive de�nite� c� and c� and c� must all be positive� Now �� � �� y���� y���� y�

so ��� �C� c��� y� � c��� y� � c��� y�� If we imagine a cube with edges of unit length parallel

to the axes y�� y�� y�� the e�ect of�

C is to multiply the i�th edge of the cube by ci�

transforming that cube to the rectangular parallelipiped in �gure ����

Suppose that in K ��t�� we imagine a small cube� with edges of length � parallel to the

axes y�� y�� y� which are eigenvectors for�

C� We can use the cube on the left in ��� by

shrinking its edges to � and placing the hidden vertex at �xA� Then the e�ect of �r L��� t�on this cube is

i� to translate it so the hidden vertex is at �rA�

��� CHAPTER ��� STRAIN AND DEFORMATION

^

^

^

y1

1

y3

1

1

y2

^

^

^

y

c

y

y

c

1

3

2

3

1

2

c

Figure ����

����� FINITE DEFORMATION ���

ii� to stretch or compress it by the factor ci in the direction yi� as shown in Figure ����

iii� to rotate it through angle �� with � � � � �� about an axis passing through �rA in the

direction w�

Now �������� implies

�D�r L ��xA� t� ��

R ��

C � ��������

where�

C ���

R�� � �C � �R��Xi �

ci

� yi�

R

�� yi�

R

�� ���������

Therefore we can perform the rotation�

R as step ii�� and the stretching or compression

as step iii�� But then� from ���������� the orthogonal axes along which stretching or

compression occurs are the rotated axes y���

R� y���

R� y���

R�

All this motivates the following de�nition�

De�nition ������ Suppose a continuum is labelled with t��positions� and �r L � K�t���R P is the resulting Lagrangian description of the motion� Suppose t � R� Then

�r L��� t� � K�t�� K�t� is called the �deformation� of the continuum from t� to t� For

any �xA � K�t��� �D�r L��xA� t� is the �deformation gradient� at particle �xA� The rotation�

R in �������� and �������� is called the �local rotation at particle �xA�� and�

C in ��������

is the �prerotation stretch tensor at �xA�� while�

C � in ���������������� is the �postrotation

stretch tensor at �xA�� We will write �D�r L��xA� t� as�

Gt�L��xA� t�� For any t��

Gt� � is a

physical quantity�

Various tensors obtained from�

C or�

C � are called �nite strain tensors� The commonest

is probably�

C � �

I P � and the most useful is probably

�Xi �

�ln ci� yi yi�

All these strain tensors reduce to the in�nitesimal strain tensor when the deformation is

nearly the identity mapping �see next section�� and to�� when �r L��� t� ��

I P �

In the foregoing discussion� the displacement of particle �xA to position �rA was so easy

to visualize that we said very little about it� It is worth considering� If we �x t�� the

��� CHAPTER ��� STRAIN AND DEFORMATION

displacement vector of any particle �x at time t is

�sL ��x� t� � �r L ��x� t�� �r L ��x� t�� � ���������

As the notation indicates� we will consider the physical quantity �s� called displacement�

whose Lagrangian description is the �sL de�ned by ���������� The de�nition requires that

we �x t�� If we change it� we change the de�nition of �s�

For t��position labelling� �r L��x� t�� � �x � �x L��x� t� so �sL��x� t� ���r L � �x L

���x� t��

Therefore

�s � �r � �x for t��position labelling� ��������

Equation �������� makes it convenient to use t��position labelling when studying dis�

placements� In particular� for such labelling� � �D�x �L��x� t� � �D�x ��

I P so �D�x ��

I P �

Thus� from ��������

�D�r ��

I P � �D�s for t��position labelling� ��������

Also� ���r �E��r� t� � ��r ��

I P so ��r ��

I P � Then from ��������

� �x ��

I P ���s for t��position labelling� ���������

But �� �x � � � �D�r � ��

I P � so from �������� and ���������� ��

I P ��

�s�� ��I P � �D�s� ��

I P � or�

I P � �D�s� ��s� ��s � �D�s ��

I P � or

�D�s � ��s� ��s � �D�s � ��s ���

I P � �D�s�� ���������

��� In�nitesimal displacements

We use t��position labelling throughout this section� and we discuss the displacement �s

de�ned by ��������� or ���������

De�nition ������ A tensor�

T� P � P is �small� if k �

T k �� That is k �

T hi �

T

k �� �� A tensor�

Q� P � P is �close to the identity� if�

Q � �

I P is small�

����� INFINITESIMAL DISPLACEMENTS ���

Remark ������ Suppose�

T� P � P and k �

T k �� De�ne ��

T �n ��

T n� so�

T ���

I P ��

T ���

T ��

T ���

T � �T ��

T ���

T � �T � �T � etc� Then ��

I P ��

T ��� and limN� PN

n �����n�

T n

both exist� and they are equal�

Proof�

kNX

n M

����n�

T n k �NX

n M

k �

T kn � k �

T kM�� k �

T k�

As M � N �� this �� so the Cauchy convergence criterion assures the

existence of limN� PN

n �����n�

T n� Then

��

I P ��

T

��

NXn �

����n �

T n �NXn �

����n �

T n �NXn �

����n�

T n��

�NXn �

����n�

T n �N��Xn �

����n�

T n��

I P �����N �

TN��

But k �

TN�� k � k �

T kN��� so letting N � on both sides gives

��

I P ��

T

�� Xn �

����n �

T n��

I P �

QED

Corollary ������� If k �D�sk �� then k��sk �� and to �rst order in �D�s�

�D�s � ��s� �������

Proof�

By remark �������� ��

I P � �D�s��� exists and equalsP

n �����n��D�s�n� Then

k��I P � �D�s���k � P n � k �D�skn � �����kD�sk�� By ��������� � ��s � �D�s � ��I P

� �D�s ���� so

k�sk � k �D�skk�IP � �D�s

��� k � k �D�sk� ��� k �D�sk� �

��� CHAPTER ��� STRAIN AND DEFORMATION

Remark ������ If�

T� P � P and�

T is symmetric and k �

T k �� then�

I P ��

T is

positive de�nite and its positive de�nite square root is

��

I P ��

T

�����

Xn �

�B� ��

n

CA �

T n ������

where �B� N

n

CA � N�N � �� � � � �N � n � ���n� �� � if n � ���

Proof�

If k �

T k �� all T �s eigenvalues lie between �� and �� Therefore�

I P ��

T has

all its eigenvalues between � and � Therefore� it is positive de�nite� ThatP n �

����n�

T n exists� is positive de�nite� and when squared gives�

I P ��

T

can be proved in the same way as remark ��������� Alternatively� the e�ect

ofPN

n �

����n�

T n can be studied on each of the vectors in an orthonormal

basis for P consisting of eigenvectors of�

T � Then one simply uses the known

validity of ������ when�

I P is replaced by � and�

T by any of its eigenvalues�

Corollary ������� Suppose�

T T��

T and k �

T k �� Then correct to �rst order in k �

T k���

I P ��

T

������

I P ��

T �

Theorem ������ Use the notation of de�nition ��������� and de�ne the displacement �s

by ��������� Suppose that k �D�sk �� Then to �rst order in �D�s we have

��s � �D�s ������

C��

C���

I P ��

h�D�s ��xA� t� � �D�s ��xA� t�

Ti

�������

R��

I P ��

h�D�s ��xA� t�� �D�s ��xA� t�

Ti� �������

Proof�

����� INFINITESIMAL DISPLACEMENTS ���

We have already proved ������� For the rest� abbreviate �D�s��xA� t� as�

T �

Since �D�r ��

C � �R we have � �D�r � � � �D�r �T ��

C � �R ��

RT ��

CT ��

C � �R ��

R��

� �C��

C � �I P ��

C��

C�� so

C ����D�r

����D�r

�T �����

���

I P ��

T

����

I P ��

T T

�����

��

I P ��

T ��

T T ��

T ��

T T

����

��

I P ��

��

T ��

T T ��

T ��

T T

��O

�k �

T ��

T T ��

T ��

T T k��

��

I P ��

��

T ��

T T

��O

�k �

T k���

Also� �D�r ��

R ��

C �� so��D�r

�T � ��D�r��

C �T ��

RT � �R ��

C � ��

C � ��

R�� � �R ��

C �

��

C � � �I P ��

C ����

C �

��so

�C � ����D�r

�T � � �D�r�����

���

I P ��

T T

����

I P ��

T

�����

��

I P ��

T T ��

T ��

T T � �T�

��

I P ��

��

T T ��

T ��

T T � �T��O

�k

T T ��

T ��

T T �� �T k��

��

I P ��

��

T ��

TT��O

�k �

T k���

Finally��

R��

C�� � �D�r ��

C�� ���

I P ��

T

�and

C����

I P ���

��

T ��

T T

��O

nk�Tk�

oso

R �

��

I P ��

��

T ��

T T

��O

nk�Tk�

o����

I P ��

T

��

I P ��

��

T ��

T T

��O

nk�Tk�

o�

QED�

De�nition ������ Use t��position labelling so the displacement �s is de�ned by ���������

Suppose k �D�sL��x� t�k �� Then the tensor

��L��x� t� �

h�D�sL��x� t� � �D�sL��x� t�T

i�������

��� CHAPTER ��� STRAIN AND DEFORMATION

is called the in�nitesimal strain tensor at particle �x at time t� and

wL ��x� t� ��

h�D�sL��x� t�� �D�sL��x� t�T

i�������

is the in�nitesimal rotation tensor� The vector

�wL ��x� t� ��

AP hi �

wL��x� t� �������

is called the in�nitesimal rotation vector at particle �x at time t�

Corollary ������� �wL��x� t� can be written �� �D��sL��x� t� or ���� � �s

�L��x� t� because

�wL ��x� t� ��

AP hi �D�sL ��x� t� �������

and� to �rst order in �D�s�

�wL ��x� t� ��

AP hi

���s�L

��x� t� � ��������

Proof�

�������� follows from ������� and ������� To prove ������� note that for

any tensor�

T we have

AP hi�

T T� �AP hi�

T � ��������

To see this� look at components relative to a pooob in P � Then ��������

is equivalent to �ijkTkj � ��ijkTjk� But by changing index names we get

�ijkTkj � �ijkTjk � ��ijkTjk� Applying �������� to ������� ������ with�

T� �D�sL��x� t� gives �������� QED

Corollary ������� Let y�� y�� y� be an orthonormal basis for P consisting of eigenvectors

of�� ��xA� t�� where �xA is as in de�nition ������� Then to �rst order in �D�sL��xA� t��

C��

C���

I P ��� ��xA� t� �

�Xi �

�� � �i� yi yi �������

where ��� ��� �� are the eigenvalues of�� ��xA� t� belonging to the eigenvectors y� � y�� y��

And�

R is a rotation in the right�handed�screw sense through angle k�wL��xA� t�k about theaxis through �OP in the direction �wL��xA� t��

����� INFINITESIMAL DISPLACEMENTS ���

δφ

ω

ξ sin θ

ξ

ω ξx

OP

θ

Figure ����

Proof�

We show�� ��xA� t� �

P�i � �i yi yi in the same way we showed ��������� The

rest of ������� follows from �������� To prove the results about�

R� we note

from exercise �c that

�wL��xA� t� � AP � �wL ��xA� t� �

Then from �������� to �rst order in �D�s

R��

I P �AP � �wL��xA� t��

If we apply�

R to a vector ��� the result is

��� �R� �� � �� � AP � �wL ��xA� t� �

We claim �� � AP � �v � �v � ��� Looking at components relative to a pooob�

��� � Ap � �v�i � �j�jikvk � �ikjvk�j��v � ��

�i� Therefore

��� �R� �� � �wL ��xA� t�� ��� �������

Abbreviate �wL ��xA� t� by �w� and examine �gure ��� where �� � �R � ��� �w� ���

Since k�w � ��k � k�wkk��k sin �� clearly �� � k�wk� QED�

��� CHAPTER ��� STRAIN AND DEFORMATION

Now suppose

t � t� � �t ��������

where �t is small enough that it is a good approximation to write

�r L ��x� t� � �r L ��x� t�� � �tDt�rL ��x� t�� �O ��t�� � ��������

Then� since Dt�r � �v and �sL��x� t� � �r L��x� t�� �r L��x� t��� we have

�sL ��x� t� � �t�vL ��x� t�� �O ��t�� ��������

Then� to �rst order in �t�

��L��x� t� �

��L��x� t� � �t� � �t

h�D�vL ��x� t�� � �D�vL ��x� t��

Ti

��������

�wL��x� t� �

�wL��x� t� � �t� � �t

h�D�vL ��x� t��� �D�vL ��x� t��

Ti

��������

�wL��x� t� �

�wL��x� t� � �t� � �t

Aphi

h�D�vL ��x� t��

i� �t

�D � �vL ��x� t�� � ��������

Consider the material near particle �xA at various times t�� �t� with small �t� Relative to

particle �xA� the material stretches as in �������� the fractional stretching �i being propor�

tional to �t� �i � #�i�t� where #�i is one of the eigenvalues of ��h�D�vL ��xA� t�� � �D�vL ��xA� t��

Ti�

and yi in ������� is the corresponding eigenvector� And relative to particle �xA� the mater�

ial also rotates through angle �tk���D��vL��xA� t�k about an axis through �xA and in direction

���D��vL��x�� t�� In other words� it rotates with angular velocity �� �D��vL��xA� t�� To �rst

order in �t�

C ���

C� so the rotation can occur before or after the stretching� We prefer to

think of them as occurring simultaneously� The foregoing analysis began by choosing a

time t� at which to introduce t��position labels� We can choose t� to be any time we like�

the foregoing results are true for all t� � R� Therefore� we can de�ne physical quantities��� �

��w� #�w by requiring for any �x � P and any t� � r that

���E

��x� t�� ��

h�D�vL ��x� t�� � �D�vL ��x� t��

Ti

�������

����� INFINITESIMAL DISPLACEMENTS ��

��wE

��x� t�� ��

h�D�vL ��x� t��� �D�vL ��x� t��

Ti

�������

��wE

��x� t�� ��

�D � �vL ��x� t�� � ������

Here it is understood that to evaluate the left side of ������� ������ ����� at any

particular t�� we use that t� to establish t��position labels in establishing the Lagrangian

descriptions needed on the right� Each t� on the left calls for a new Lagrangian description

on the right� the one using t��position labels�

But with t��position labels� �D�r L��x� t�� ��

I P � so �EfE��x� t�� � �DLfL��x� t�� for any

physical quantity f � Therefore �D�vL can be replaced by ��vE on the right in ������ Since

the Eulerian description of physical quantities are independent of the way particles are

labelled� this replacement leads to the conclusions

����

���v �

���v�T �

������

��w�

���v �

���v�T �

�������

#�w ��

� � �v� �������

The quantity #�w is called the vorticity� and��� is the strain�rate tensor�

��� CHAPTER ��� STRAIN AND DEFORMATION

Chapter ��

Constitutive Relations

��� Introduction

Let us examine the conservation laws as a system� They are

Eulerian Form Lagrangian Form

Mass Dt�� �� � �v � � Dt � � � ��������

Momentum �Dt�v � �� �S ��f �Dt�v � �D�� S �� f �������

Internal Energy �DtU � � � �H � h��

S hi�v �DtU � �D � � H � h�� S hi �D�v�

�������

If we knew the initial values of �� �v and U � or �� �v and U � at some instant t�� we might

hope to integrate ������� forward in time to �nd later values of these quantities� To do

this� clearly we must know �f and h� or � f and h� and we will assume henceforth that these

are given to us� But there still remains the question of �nding �H and�

S� or� H and

� S� The

equations which give their values when the state of the material is known are called the

�constitutive equations� of the material� Ideally� we would like to be able to deduce these

from the molecular pictures of matter� but theory and computational techniques are not

yet adequate to this task for most materials� We must rely on experimental measurements�

The outcomes of these experiments suggest idealized mathematical models for various

classes of real materials�

��

��� CHAPTER ��� CONSTITUTIVE RELATIONS

K

n

K

Figure �����

In modelling a real material� one usually begins by considering its homogeneous ther�

mostatic equilibrium states �HTES�� Then one studies small deviations from thermostatic

equilibrium� by linearizing in those deviations� Large deviations from thermodynamic

equilibrium are not fully understood� and we will not consider them� They make plasma

physics di!cult� for example�

A homogeneous thermostatic equilibrium state �HTES� is said to exist in a material

when no macroscopically measurable property of that material changes with time� and

when the macroscopic environments of any two particles are the same� The second law of

thermodynamics then implies that there can be no heat �ow in the material� The material

can have a constant velocity� but if it does we study it from a laboratory moving with

the same velocity� so we may assume �v � �� in an HTES� Then DtU � tU � But tU � �

so� from �������� h � �� The Cauchy stress tensor�

S must be the same everywhere at all

times in an HTES� so �� �S� ��� and ������ �momentum�� implies �f � ��� Therefore� in an

HTES we have

� � constant� �H � �� �v � ���

U � constant� h � ��

S� constant� �f � ��

If the material occupies a volume K with boundary K and unit outward normal n�

then in order to maintain the material in an HTES we must apply to the surface K a

����� INTRODUCTION ���

stress n� �S� and we must prevent surface heat �ow and volume heating� That is we need�������������surface stress applied to K at �r is n��r�� �S n��r� � �H � � at every �r � K

h � � at every �r � K�

��������

Two HTES�s are di�erent if they di�er in any macroscopically measurable quantity�

including chemical composition� Most pure substances require only the values of a few of

their physical properties in order to specify their HTES�s� For example� two isotropically

pure samples of H�O �no deuterium� only O��� liquid which have the same density and

the same internal energy per unit mass will have the same values of pressure� temperature�

entropy per gram� electrical resistivity� index of refraction for yellow light� neutron mean

absorption length at �� kev� etc� For water� the set of all possible HTES�s is a two�

parameter family� �We have ignored� and will ignore� electrical properties� It is possible

to alter U for water by polarizing it with an electric �eld��

Changes from one HTES to another are produced by violating some of the conditions

��������� Changes produced by altering the surface stress applied to K are said to �do

work�� while changes produced by letting h �� � are said to involve �internal heating or

cooling�� and n � �H �� � on K involves �heat �ow at K�� For the materials in which

we are interested� these changes can be described as follows� We pick one HTES as a

reference state� and call it HTES� We use t��position labelling� where t� is any time when

the material is in HTES�� We carry out a change by violating ��������� and wait until time

t� when the material has settled into a steady state� HTES�� The Lagrangian description

of the motion is �r L��x� t�� and it is independent of t for t � t� and for t � t�� We write

�r �� ��x� � �r L ��x� t�� � ��������

G�� ��x� � �D�r �� ��x� � ��������

Since the environments of all particles must be alike in HTES�� they must have been

subjected to the same stretch and the same rotation so

G�� is constant� independent of �x� ��������

��� CHAPTER ��� CONSTITUTIVE RELATIONS

��

G�� is called the deformation gradient of HTES� relative to HTES��� Then from ���������

�r ����x� � �x� �G�� ��r ������� If �r ������ �� ��� we may move the material with a uniform

displacement so �r ������ � ��� Thus we may assume

�r ����x� � �x� �G�� � ��������

Here �x is particle position in HTES� and �x� �G�� is particle position in HTES�� If we

want to study another HTES� say HTES�� we can adopt either HTES� or HTES� as the

reference state� If we use HTES� for refrence� then particle position in HTES� is �x���

G��

where �x� is particle position in HTES� and�

G�� is calculated from �������� but with t�

position labelling� If a particle has position �x in HTES�� then �x� � �x� �G�� and so �x� �

position in HTES� � �x���

G�� � �x� �G�� ��

G��� But also �x� � �x� �G��� so

G����

G�� ��

G�� ��������

for any HTES�� HTES�� HTES��

��� Elastic materials

The observed HTES�s of many material are approximated well by the following idealiza�

tion�

De�nition ������� An elastic material is one whose HTES�s have these properties�

i� If HTES� is any HTES� there is an M� � � such that for any�

G� P � P with k �

G

� �

I P k M�� there is an HTES� whose deformation gradient relative to HTES� is�

G����

G �

ii� Once a reference state HTES� is chosen� any other equilibrium state HTES� is com�

pletely determined by its deformation gradient�

G�� relative to HTES�� and by its

entropy per unit mass N��

iii� U � the internal energy per unit mass� is unchanged by rigid rotations� That is� if�

G� P � P and�

R� "��P � then

U��

G�N�� U

��

G � �R�N�� �������

����� ELASTIC MATERIALS ���

Now consider two di�erent HTES�s of an elastic material� say HTES� and HTES��

neither of them the reference state HTES�� Suppose �rL � K��R P is the Lagrangian

description of the motion which carries the material from state � to state � Here K� is

the region occupied by the material in HTES�� so the total mass of the material is

M � jK�j � ������

where jK�j � volume of K�� The total internal energy of the material is

U i � MUi � jK�j �Ui in HTESi� ������

Let ��W be the work done on the material and ��Q the heat added to it in going from

HTES� to HTES�� Energy conservation requires

��W � ��Q � U � � U �� �������

Now suppose HTES� and HTES� are not very di�erent� Then we will write df � f��f�for any property f of the two states� Thus ������� is dU � ��W � ��Q� According

to the second law of thermodynamics� correct to �rst order in small changes�

��Q � �dN � ��N � � N �� �������

where N i � jK�j �Ni � total entropy of state i� and � is the smaller of �� and ��� It

is possible to �nd processes �r L in which equality is approximated arbitrarily closely in

�������� These are called �reversible processes�� In such processes

��Q � �jK�j �dN� �������

In any transition from state � to state � reversible or not�

�DtU � � �D � � H � h�� S hi �D�v

so� integrating over K� and using Gauss�s theorem�

�ZK�

dVL��x�DtU �ZK�

dVL h�Z�K�

dAL nL � �H �ZK�

dVL� S hi �D�v

��� CHAPTER ��� CONSTITUTIVE RELATIONS

ord

dtU � #Q�

ZK�

dVL �Shi �D�v

where #Q is the rate at which heat is added to the material� Thus� integrating from t� to

t�

U � � U � � ��Q�Z t�

t�dtZK�

dVL� S hi �D�v�

Comparing with ������� we see that

��W �Z t�

t�dtZK�

dVL� S hi �D�v�

Now if the transformation is done slowly�� S will be nearly constant� about equal to its

value in HTES� or HTES� so

��W �� S hi

ZK�

dVL

Z t�

t�dt �D�v�

But �v � dt�r so �D�v � �DDt�r � � �D�v�L��x� t� � �DDt�r

L��x� t� � Dt�D�r L��x� t�

� �Dt

G�L��x� t�� Thus� �D�v � Dt

G� and

��W �� S hi

ZK�

dVL��x���

G ��x� t����

G ��x� t����

But�

G ��x� t�� and�

G ��x� t�� refer to HTES�s� They are independent of �x� Thus

��W � jK�j� S hid �

G � �������

From �������� ������� and �������� dU � jK�j� S hid �

G �jK�j ��dN � Since

dU � jK�j �dU � dividing by jK�j � gives

dU � d�

G hi� S

�� �dN �������

for small reversible changes between HTES�s� Here � and� S refer to either the initial or

the �nal state� since they are close�

But U is a function of�

G and N � Equation ������� says this function is di�erentiable�

and that

���

G�N��

NU��

G�N�

�������

����� ELASTIC MATERIALS ���

� S �

G�N�

�� ��

GU��

G�N�� ��������

where ��GU�

G�N� is an abbreviation forh�U��� N�

i� �G�� Since P�P is a Euclidean space�

the theory of such gradients is already worked out� In particular� if y�� y�� y� is a pooob

for P � then the dyads yi yj are an orthonormal basis for P � P � and

G� Gij yi yj� ��������

Then

�GU��

G�N�� yi yj

U��

G�N�

Gij

so� from �������� Sij�

�U�

G�N�

Gij� �������

To obtain the Cauchy stress tensor we use ������� namely

���

S

A ��

GT ��B�� S

CA � �������

The equation

U � U��

G�N�

��������

is called the equation of state of the elastic material� It can be solved for N as a function

of�

G and U �

N � N��

G�U�� ��������

This serves as the basis for

De�nition ������� A perfectly elastic material is one in which the small region near

any particle �x at any time t behaves as if it were in an HTES with�

G� �D �r L��x� t� and

U � UL��x� t��

In a perfectly elastic material� we will have �see page ��� h � �� �H � ��� and� S will be

given by ��������� so the missing information needed to integrate �������� ������ ������

is supplied by �������� which is known if the equation of state is known�

��� CHAPTER ��� CONSTITUTIVE RELATIONS

��� Small disturbances of an HTES

In seismology� the earth is at rest before an event� This resting state is used to label

particles by their positions in the resting state� Then particle displacements are given by

�sL ��x� t� � �r L ��x� t�� �x� �������

Usually it is a very good assumption in seismology that

k �D�sL ��x� t� k �� ������

Expressed in terms of deformation gradients� we have

G� �D�r ��

I P � �D�s

so�

G��

I P ���

G ������

where

��

G� �D�s

and

k� �

G k �� �������

We want to study HTES�s which are close to the zeroth or initial state in the sense

of �������� We use subscripts � to refer to the initial state� Then� since particles are

labelled by their positions in HTES��

G���

I P �������

�� � �� �������� S�

S� � �������

Furthermore� if we write �f � f� � f� for any physical quantity� f� being its value in

HTES�� �since f is constant throughout the material� we needn�t distinguish between fL

and fE� then from �������� to �rst order in �N and ��

G we have

�U � ��N ��

��

S� hi��

G �

����� SMALL DISTURBANCES OF AN HTES ��

If ��

GT� �� �

G� then�

G��

I P ���

G is a rotation to �rst order in ��

G� and therefore �N � �

implies �U � �� Consequently�

S� hi��

G� � if ��

GT� �� �

G �

But then for any ��

G��

S� hi���

G ���

GT

�� �� so

��

S� ��

S�T�hi� �

G� �� Thus

S���

ST� � �������

The Cauchy stress tensor is symmetric in every HTES of an elastic material�

We can go further� Since HTES� is close to HTES�� all the changes �f are small� so

we can linearize ������� and �������� The changes �� � �� � ��� �� S�

� S� �

� S�

are

given in terms of �N � N� �N� and ��

G��

G� ��

G� by

�� � �N���

G�� N�

�N

� ��

G hi��G���

G�� N�

� �S � �N

�S

��

G�� N�

�N

� ��

G hi�G

�S

��

G�� N�

��

We de�ne the following tensors�

Q�� ��P ��

W �� ��P � B� � R�

Q�� ��G

�S

��

G�� N�

�� ����

G��GU��

G�� N�

��������

W � �

N

�S

��

G�� N�

�� �

N��GU��

G�� N�

�� ���

G

NU��

G�� N�

�� ����

G����

G�� N�

���������

B� �

N���

G�� N�

��

N�U��

G�� N�

�� ��������

Then

�� � �NB� � ��

G hi�� �

W �

��

A �������

�Note� � � ��� Also

�� S� �N

W � ���

G hi��

Q� � �������

��� CHAPTER ��� CONSTITUTIVE RELATIONS

Furthermore� if y�� y�� y� is any pooob for P � and we write�

G as in �������� and write

��

Q�� Qijkl yi yj yk yl�

then

Qijkl � �U�GijGkl � �U�GklGij

so

Qijkl � Qklij or ������

��

Q��

��

Q� � ��������

We would also like to calculate ��

S��

S� ��

S�� the change in the Cauchy stress tensor�

From ������� and �������

S������det �G��

������ �

GT�� �

S��

Now det�

G �t� depends continuously on t in going from HTES� to HTES�� and can never

vanish� It is � in HTES�� so is always � �� Hence j det �G j � det�

G and

S���det

G��

��� �

GT�� �

S�� ��������

Now�

G����

G��

I P ���

G and� taking components relative to y�� y�� y�

det�

G �

������������ � �G�� �G�� �G��

�G�� � � �G�� �G��

�G�� �G�� � � �G��

������������ � � �Gii �O

�k� �

G k��� � � tr �

G �Ok �� Gk��

Thus �det

G

���� �� tr �

G �O�k� �

G k��� ��������

Then�

S����� tr �

G

���

I P ���

GT

����

S� �� �S

��

where we have used �������� and have linearized in ��

G� Keeping only terms up to �rst

order in ��

G� we have

S���

S� � �tr ��G���

S� ���

GT � �S� �� �S�

����� SMALL DISTURBANCES OF AN HTES ���

or

��

S� � �S � �

GT � �S� ��

S�

�tr �

G

�� ��������

If we take components relative to the pooob� y�� y�� y� from P � and write

S�� Sij yi yj ��������

then �������� becomes

�Skl � � Skl � �GikSil � �GjjSkl

� ��Gij�Qijkl � �NWkl � �Gij ��jkSil � �ijSkl�

� ��Gij� �Qijkl � Sil�kj � �ijSkl� � �NWkl�

We de�ne a tensor��

F� ��P by

��

F �� �Qijkl � Sil�kj � �ijSkl� yi yj yk yl�

or���

F ��

��

Q� ������

S�

I P

�� �

I P�

S� � ��������

Then

Fijkl � Qijkl � Sil�kj � �ijSk �������

and

��

S� ��

G hi��

F � ��N�

W � � �������

This equation is of exactly the same form as �������� but

��

Q� is replaced by��

F ��

Now ��

S is the di�erence between the Cauchy stress tensors in two HTES�s� so ��

ST�

��

S for any ��

G whatever and any �N � First� take ��

G� �� Then we must have

WT

���

W � � ������

Next� take �N � �� Then

�Skl � �GijFijkl � �Slk � �GijFijlk�

��� CHAPTER ��� CONSTITUTIVE RELATIONS

so

�Gij �Fijkl � Fijlk� � �

for every ��

G� Hence

Fijkl � Fijlk ������

or

���

��

F ��

��

F � � �������

Now in general��

G is a joint property of two HTES�s� the zeroth or initial state used

to label particles� and the �nal state� However��

S�� B� ��

W ��

��

Q� and��

F � are properties

of the initial state alone� HTES�� As we have seen�

S� and�

W � are symmetric� and

��

Q�

satis�es ��������� while��

F � satis�es �������� If the initial HTES is isotropic� all these

tensors must be isotropic� From this follows

Remark ������ In an isotropic HTES of an elastic material� there are constants p� ��

� � such that �� � are the Lam.e parameters of the material����������������������

S� �p �

I P�

W� �� �

I P��

F� �

I P�

I P �� ��� � �����

I P�

I P or

Fijkl � �ij�kl � � ��ik�jl � �il�kj� relative to any pooob for P�

�������

Proof�

In our discussion of I�"��P �� we proved all of this except that for��

F we proved

only the existence of three scalars � �� � such that

Fijkl � �ij�kl � ��ik�jl � ��il�kj�

But by ������� ��ik�jl���il�kj � ��il�jk���ik�lj� or �������ik�jl��il�jk� ��� Setting i � k � �� j � l � gives � � � � �� so � � �� and we have the

expression for��

F given in the remark ������

����� SMALL DISTURBANCES OF AN HTES ���

Corollary ������� For HTES�s close to an isotropic HTES��������� � ��N�B �

�tr

��������

��

S� �tr

��� �

I P �u��

�������

where��� ����

G ���

GT � � ����D�s�

��D�s�T�

� in�nitesimal strain tensor�

In most cases� the HTES� of the elastic material is not isotropic� either because the

material itself is anisotropic or because�

S� is anisotropic� Then the sti�ness tensor��

F �

gives the response of�

S to small changes ��

G in the deformation gradient�

G away from�

G���

I P � The sti�ness tensor has two sets of symmetries �see ������

Fijkl � Fijlk� �������

The second symmetry is obtained �������� and �������� Let

Rijkl �� �jkSil � �ijSkl� �������

Then from ������� Qijkl � Fijkl � Rijkl� so from ��������

Fijkl � Fklij �Rijkl � Rklij� �������

Since Fijkl have to be measured experimentally� and there are �� such components� we

would like to use ������� and ������� to see how many of these components are ind�

pendent� Only those need be measured and recorded�

To answer this question we need one more symmetry� deducible from ������� and

�������� We have from those two equations

Fijkl � Fklji �Rijkl � Rklij�

Interchange i and j in �������� so

Fkljk � Fjikl � Rjikl �Rklji�

Then

Fijkl � Fjikl �Rijkl � Rklij �Rklji �Rjikl� �������

��� CHAPTER ��� CONSTITUTIVE RELATIONS

Now de�ne

��

E�� ��P by

Eijkl ��

��Fijkl � Fjikl � Fklij � Flkij� � �������

Then it is a trivial algebraic identity that

Fijkl � Eijkl ��

��Fijkl � Fijkl� �

��Fijkl � Fjikl�

��

��Fijkl � Fklij� �

��Fijkl � Flkij�

� Eijkl ��

��Fijkl � Fjikl� �

��Fijkl � Fklij�

��

��Fijkl � Fklij � Fklij � Flkij�

� Eijkl ��

��Fijkl � Fjikl� �

�Fijkl � Fklij�

��

��Fklij � Flkij� �

Using ������� once as written and once with the interchanges i ! k and j ! l� and

using ������ once as written� we get

Fijkl � Eijkl ��

��Rijkl �Rklij �Rklji � Rjikl�

��

��Rklij � Rijkl �Rijlk �Rlkij�

��

�Rijkl � Rklij�

� Eijkl ��

�Rijkl �Rklij� �

��Rijlk � Rlkij�

��

��Rjikl �Rklji�

� Eijkl ��

��jkSil � �ijSkl � �liSkj � �klSij�

��

���jlSik � �ijSlk � �kiSlj � �lkSij�

��

���ikSjl � �jiSkl � �ljSki � �klSji� �

Therefore

Fijkl � Eijkl ��

��klSij � �ijSkl � �jlSik � �ikSjl � �jkSil � �ilSjk� ������

����� SMALL DISTURBANCES OF AN HTES ���

where we have used �ij � �ji and Sij � Sji� Now we must measure the six independ�

ent components of Sij and the components of Eijkl in order to know��

F �� How many

independent components does

��

E have$ From �������

Eijkl � Ejikl � Eijlk � Eklij� ������

Therefore we can think of Eijkl as a �� � matrix whose rows are counted by �ij� � �����

���� ���� ��� ��� �� and whose columns are counted by �kl� � ����� ���� ����

��� ��� ��� We need not consider �ij� � ���� ���� or �� because of the �rst of

equations ������� The fact that Eijkl � Eklij means that this ��� matrix is symmetric�

Therefore it contains only � independent components� These can be chosen arbitrarily

but not all choices describe stable materials�

A tensor in ��P with the symmetries ������ is called an �elastic tensor�� The set

of all such tensors in a subspace of ��P � which we call E��P �� We have just proved

that dimE��P � � �� We have also shown that if

��

F� ��P has the symmetries �������

������ ������ then it can be written in the form ������ with

��

E� E��P �� Conversely�

if

��

E� E��P � and�

ST��

S and

��

F is given by ������� then

��

F has the symmetries �������

������ ������� a fact which is easily veri�ed and is left to the reader� Therefore it is not

possible to reduce the problem further� In principle� to measure

��

F �� it may be necessary

to measure the six independent components of�

S� and the � independent components of��E �� Actually� only �ve of the components of

S� need be measured� Suppose� � is the

stress deviator for�

S�� so

S�� �p��

I P �� �� with � p� � tr

S� � �������

Then substituting ������� in ������� shows that the isotropic part of�

S� makes no

contribution to

��

F �� and we can rewrite ������ as

Fijkl � Eijkl ��

��kl ij � �ij kl � �jl ik � �ik jl � �jk il � �il jk� � �������

Therefore

��

F � has � independent components� � in

��

E� and � in

�� ��

��� CHAPTER ��� CONSTITUTIVE RELATIONS

Suppose a material is originally isotropic in an HTES with

��

S������ Suppose the mater�

ial is strongly compressed so that��

S�� �p��

I P where p� is very large �say of the order of

mantle pressure� ��� to ��� megabar�� The new HTES will still be isotropic� and��

F ����

E�

will have the form �������� which involves only two constants� the Lam.e parameters

and �� Suppose then that a small �a few kilobars� stress deviator is added� say� �� Then

��

E� will change by a small amount �

��

E which depends linearly on

�� �� and

��

F � will change

by �

��

F� �

��

E � terms involving

�� � in �������� Now �

��

E� J�hi� � where J� � ��P �

This tensor is a property of the original highly compressed isotropic HTES� Therefore it is

a member of I�����V ��� i�e� it is isotropic� By Weyl�s theorem� it is a linear combination

of all possible permutations of�

I P�

I P�

I P � These are as follows�

�ij�kl�mn �ij�km�ln �ij�kn�lm

�ik�jl�mn �ik�jm�ln �ik�jn�lm

�il�jk�mn �il�jm�kn �il�jn�km

�im�jk�ln �im�jl�kn �im�jn�kl

�in�jk�lm �in�jl�km �in�jm�kl�

Since mn � nm� we can combine two terms which di�er only in interchangingm and n�

Since mm � �� we can discard terms with �mn� Thus there are scalars �� �� �� �� � such

that Jijklmn � ���ij�km�ln��ij�kn�lm�����ik�jm�ln��ik�jn�lm� ����il�jm�kn��il�jn�km�

����im�jk�ln � �in�jk�lm� ����im�jl�kn � �in�jl�km� ����im�jn�kl � �in�jm�kl�� Moreover�

Jijklmn must satisfy ������ for any �xed m� n� so � � � � � � � and � � �� Therefore

J� involves only two independent constants� and

Jijklmn � ��ij ��km�ln � �kl�lm� � ��kl ��im�jn � �in�jm�

��

��ik ��jm�ln � �jn�lm� � �il ��jm�kn � �jn�km�

��jk ��im�ln � �in�lm� � �jl ��im�kn � �in�km�

��

Then

�Eijkl � Jijklmn mn � � ��ij kl � �kl ij�

�� ��ik jl � �il jk � �jk il � �jl ik� �

����� SMALL DISTURBANCES IN A PERFECTLY ELASTIC EARTH ���

Thus� �nally� correct to �rst order in ij� ������� gives

Fijkl � �ij�kl � u ��ik�jl � �il�jk� �������

���� �

��ij kl �

���

��kl ij �������

��� � �

���ik jl � �il jk� �

�� �

���jk il � �jl ik� � �������

If the anisotropy in

��

F � arises solely because an isotropic material has been subjected to

a small �anisotropic� deviatoric stress then��

F � involves only � independent constants� the

�ve independent components of

�� � and the four constants � u� �� �� If the anisotropy

was already present in the crystal structure of the material� either because it was a single

crystal� or because its micro�crystals were not randomly oriented� then�

E� can involve �

and��

F � � independent parameters�

�� Small disturbances in a perfectly elastic earth

Neglecting dissipation� we will treat the earth as a perfectly elastic body in studying its

response to earthquakes� Before an earthquake� we assume the earth is at rest in static

equilibrium� We label the particles by their equilibrium positions� so we use t��position

labelling where t� is any time before the earthquake� When the earth is at rest� we have������r L ��x� t� � �x�

GL ��x� t� � �D �r �L ��x� t� ��

I P��������

so the Eulerian and Lagrangian descriptions of any physical quantity are the same� and

the two kinds of density and two kinds of stress tensor are the same� Using subscript �

to denote the equilibrium state�

�� ��x� � �� ��x�

S� ��x� �� S� ��x� �

The body force is �f ��x� � �f ���x� � ����x��g���x� where �g is the acceleration of gravity� and

�g� is its equilibrium value� The conservation equations reduce to

�� � �� �S� ����g� � �D � �S� � ���g�� �������

��� CHAPTER ��� CONSTITUTIVE RELATIONS

Now suppose an earthquake occurs� after which

�r L ��x� t� � �x � �sL ��x� t� � �������

with

k �D�sL ��x� t� k � for all �x� t� ��������

Then�

GL ��x� t� � �D �r L ��x� t� ��

I P � �D �sL ��x� t� � ��������

For earthquakes� �s as well as �D�s is quite small� and it is almost always possible to

replace the exact conservation laws and constitutive relations by linearized versions in

which all powers of �s or �D�s higher than the �rst are omitted� In these problems it is

useful to be able to convert easily between Eulerian and Lagrangian representations� For

example� gravity is easy to calculate in the Eulerian representation� and the constitutive

relations are simple in the Lagrangian representation� We discuss this conversion now�

For any physical quantity f � we have

fL ��x� t� � fE ��r� t� where �r � �r L ��x� t� �

If �r L��x� t� � �x � �s L��x� t� then to �rst order in �s we have

fL ��x� t� � fE��x� �sL ��x� t� � t

�� fE ��x� t�

��s L ��x� t� � �fE ��x� t� � ��������

If f is small of �rst order in �s� we have

fL ��x� t� � fE ��x� t� or fL � fE if f � Ok�s k� ��������

If f is small of order �s� there is no need to distinguish between its Eulerian and Lagrangian

representations� so we can write both fE ��x� t� and fL ��x� t� simply as f ��x� t�� That is

f ��x� t� �� fE ��x� t� � fL ��x� t� if f � Ok�s k� ��������

This is true in particular for f � �s� If f is of order �s� we can use f not merely for the pair

of functions �fL� fE� but for either function alone� since to order �s they are equal� Even

����� SMALL DISTURBANCES IN A PERFECTLY ELASTIC EARTH ��

when f is not small of order �s� fE��x� t��f���x� is� so to �rst order in �s� �sL��x� t� � �fE��x� t�� �s��x� t� � �f���x� and �������� for any f is

fL��x� t� � fE��x� t� � �s��x� t� � �f���x� ��������

or� as a relationship between functions�

fL � fE � �s � �f�� ���������

for any physical quantity f �

Relations between derivatives are also needed� For any physical quantity f � Dtf �

tf � �v � �f � But �v � Dt�s � Ok�sk� and f � f� is of order �s� so correct to �rst order in �s�

�v � �f � �v � �f�� andDtf � t � �v � �f�� ���������

If f is Ok�sk then �v � �f� is Ok�sk�� so correct to Ok�s k

Dtf � tf if f � Ok�sk� ��������

In particular� correct to Ok�sk

�v � Dt�s � t�s� �a � Dt�s � �t �s� ��������

If f is any physical quantity� �Df � � �D�r � � �f � ��

I P � �D�s� � �f � �f � �D�s � �f � If

f � Ok�s k then correct to Ok�s k

�Df � �f if f � Ok�sk� ���������

In particular� correct to Ok�sk� �D�s � ��s� Therefore� correct to Ok�sk� �D�s � �f � ��s � �f� ��s � �f�� Thus for any f

�Df � �f � ��s � �f� correct to Ok�sk� ���������

For any physical quantity f we de�ne �Lf and �Ef � the Lagrangian and Eulerian

derivatives of f from its equilibrium value� Both are physical quantities� and they are

de�ned by ��Lf

�L��x� t� � fL ��x� t�� f� ��x� ���������

��� CHAPTER ��� CONSTITUTIVE RELATIONS

��Ef

�E��r� t� � fE ��r� t�� f� ��r� � ���������

Both �Lf and �Ef are small of order k�sk� so we can apply to them the convention ��������

even when f is not small of order �s� Thus� for the functions�

�Lf � fL � f� ���������

�Ef � fE � f�� ���������

Then �Lf � �Ef � fL � fE � �s � �f� from ���������� so

�Lf � �Ef � �s � �f� to �rst order in �s� ��������

From ����������

G���

I P so �s � � �

G����� Therefore

�L�

G� �E�

G� ��s to �rst order in �s� ��������

With the help of this machinery� the Eulerian and Lagrangian conservation equa�

tions and the constitutive relations can be linearized� They produce either Eulerian or

Lagrangian equations of motion� and it is conventional to work with the former� even

in seismology� We will give the derivation of these equations� based on the Lagrangian

constitutive relations and Eulerian conservation equations�

The Eulerian mass conservation equation is

t�E ��r� t� � � � �E ��r� t��vE ��r� t� � ��

We have �E � �� � �E� and t�� � �� so� correct to �rst order in �s�

t��E�

�� � � ����v� � ��

But �v � Dt�s � t�s so� correct to �rst order in �s�

th�E� � � � ����s�

i� ��

For t � t� �before the earthquake� �E� � � and �s � �� Therefore

�E� � �� � ����s� � �������

����� SMALL DISTURBANCES IN A PERFECTLY ELASTIC EARTH ���

The Eulerian momentum conservation equation is

��D�

t�s�E

��r� t� � �� �SE��r� t�� �E ��r� t��gE ��r� t� �

Here�

SE��

S� ��E�

S� �E � �� � �E�� �gE � �g� � �E�g� so correct to �rst order in �s�

���t �s �

�� �S� �� ���E

S

�� ���g� � ��

��E�g

����E�

��g��

But then we can apply ������� to obtain

���t �s �

� ���E

S

�� ��

��E�g

����E�

��g�� �������

We �nd �E� from �������� We �nd �E�g as �E����� � ���E � ��� � ����E ����� so

�E�g � ����E�

���������

where � is the gravitational potential� It is obtained from �� � ���( where ( �

Newton�s gravitational constant� Subtracting ��� � ����( gives

��E� � ��(�E�� ��������

In order to solve �������� it remains to express �E�

S in terms of �s� We approximate

the neighborhood of each particle in the earth as a HTES� and we assume that�

S can be

calculated everywhere at all times as if the local material were in such a state� Thus at

each particle �x we can use ���������

�L�

S� �L�

G hi��

F � ��x� ���LN

� �

W � ��x� � ��������

We must use the Lagrangian representation here� because ������� applies to a particular

lump of matter� the matter close to particle �x� The initial HTES� at particle �x is the HTES

appropriate for the region near that particle before the earthquake� Thus this HTES� will

vary from particle to particle� and the tensors��

F � and�

W � really will depend on �x� The

heat conductivity of the earth is so low that during the passage of earthquake waves almost

no heat has time to �ow� so there is no change in entropy per gram at each particular

particle� That is

�LN � � everywhere at all times� ��������

�� CHAPTER ��� CONSTITUTIVE RELATIONS

Thus equation �������� gives �L�

S in terms of �s� since �L�

G� ��s� And then ��������

gives �E�

S as

�E�

S � �L�

S ��s � � �

S�

�E�

S �

���shi

��

F �

�� �s � � �

S� �

Thus ������� �nally becomes� to �rst order in �s�

���t �s � � �

����s�hi

��

F � ��x�

�� � �

��s � � �

S�

����

��E�g

��h� � ����s�

i�g� ��������

where �E�g is obtained from �s via �������� �������� and ���������

For waves short compared to the radius of the earth� it is safe to neglect �g and ��

S��

If we assume that the undisturbed state is the same everywhere and is isotropic� then

Fijkl � �ij�kl � � ��ik�jl � �il�jk�

so ���s�hi

��

F �

�kl

� �� � �s

��kl � � �ksl � lsk�

���s�hi

��

F � � �� � �s

� �

I P ��h� �s� ��s�T

i�

If we assume that there is no �x dependence in the equilibrium state �homogeneous

earth� then and � are independent of �x� and �������� becomes

���t sl � k

����s�hi

��

F �

�kl

� l�� � �s

�� �

�kksl � klsk

���t �s � �� �� �

�� � �s

�� ���s

� �� �� ��� � �s

�� �

h��s� �

�� � �s

�i� �� �� �

�� � �s

�� � �

�� � �s

��

Divide this equation by �� and de�ne

c�p �� �

��� c�s �

��� ��������

����� SMALL DISTURBANCES IN A PERFECTLY ELASTIC EARTH ��

Then

�t �s � c�p��� � �s

�� c�s� �

�� � �s

�� ��������

Since we are assuming ��� � � independent of �x� the same is true of c�p and c�s� Taking

the divergence of �������� gives

�t�� � �s

�� c�p

��� � �s

�� ��������

The scalar � ��s propagates in waves with velocity cp� These are called P �waves �primary

waves� or compression waves� Taking the curl of �������� and using �����v � ��� � �v

��

��v gives

�t�� � �s

�� c�S

��� � �s

�� �������

The vector ���s propagates in wave with velocity cs� These are called S�waves �secondary

waves� or shear waves�

There is an interesting inequality involving cs and cp which we can derive by studying

������� when �N � � and��

F � is isotropic� In that case

��

S� �

I P

�tr �

G

�� �

���

G ����

G

�T��

For a pure shear� �s��r� � ��r � y�� y�� we have ��

G� y� y� so ��

S� �� y� y� � y� y�� and

y� � ��

S� � y�� If � � �� the material either does not resist shear or aids it once it has

begun� so we must have

� � �� �������

For a pure dilatation �s��r� � ��r� we have ��

G� ��

I P � so ��

S� �� � ���

I P � If � � ��

the material expands slightly� If this does not produce a drop in pressure� the material

will explode� Therefore

� � � �� ��������

Combining these two relations gives �� �� � �� � �� so

� c�sc�p

�� �

cscp

������ ��������

Therefore the P wave always travels faster than the S wave� and arrives �rst� Hence it is

the �primary� wave�

�� CHAPTER ��� CONSTITUTIVE RELATIONS

�� Viscous �uids

A �uid is a material whose HTES�s support no shear� so the Cauchy stress tensor is

always isotropic� We need now a more precise de�nition� A �uid supports no shear in an

HTES� and each HTES is uniquely determined by the density and entropy per unit mass�

or by volume per unit mass� � � and entropy per unit mass� N � Clearly � � ���� and by

our assumption

U � U ��� N� � ��������

From elementary thermodynamics� it is well known that

dU � �pd� � �dN so �������

p ��� N� � �U��� N�

�� �������

� ��� N� �U��� N�

N� ��������

In an HTES� the Cauchy stress tensor is

S� �p �

I P � ��������

Now suppose the �uid deviates slightly from an HTES� At each point in space� �r� and

each instant t� there will be a well de�ned density �E��r� t� and a well de�ned energy per

unit mass� UE��r� t�� We set � � ���E��r� t�� U � UE��r� t� and use �������� to calculate

NE��r� t� as if the �uid were in an HTES at ��r� t�� Then with N � NE��r� t� and � �

���E��r� t� we use ������� to calculate pE��r� t� as if the �uid were in an HTES at ��r� t��

We call this pressure the thermostatic equilibrium pressure at �r� t and write it pEHTES��r� t��

The �uid at ��r� t� is not in fact in the HTES de�ned by UE��r� t� and �E��r� t�� so the actual

Cauchy stress tensor�

SE ��r� t� is not given by �������� but by

SE ��r� t� � �pEHTES ��r� t��

I P ��

V E ��r� t� ��������

where �������� de�nes�

V E� The tensor�

V E ��r� t� is called the viscous stress tensor at ��r� t��

It vanishes in HTES�s�

����� VISCOUS FLUIDS ��

The deviation from an HTES can be measured by the failure of � to be constant and

by the failure of �v to vanish� Thus�

V E ��r� t� can be expected to depend on the functions

�E and �vE in the whole �uid� We expect�

V E to vanish when �E is constant for all �r� t� In

fact� �vE � constant for all �r� t is simply an HTES moving with constant velocity� so we

expect

V E ��r� t� ��� if �E and �vE are constant

for all �r and t�

For ordinary �uids� we would also expect�

V E ��r� t� to be a�ected only by the behavior of

the functions �E and �vE near �r� t� This behavior is determined by the derivatives� so we

would expect�

V E ��r� t� to be determined by the values at ��r� t� of�������������E� t�

E� ���E� �t��E� �t �

E� � � �

� �vE� t �vE� �� �vE� �t� �vE� �t�vE� � � �

��������

We will assume that the deviation from an HTES is small� In that case� �V E��r� t� can be

expanded in Taylor series in the quantities ��������� and we can drop all but the linear

terms� We will also assume that the in�uence of distant �uid on ��r� t� drops o� very rapidly

with distance in either space or time� so that only �rst derivatives need be considered in

��������� In fact� physical arguments suggest that a term �mnt �E or �mnt

�vE in ��������

contributes to�

V ��r� t� in the ratio ��L�m�t�T �n where and t are mean free path and

mean time between collisions for a molecule and L and T are the macroscopic wavelength

and time period of the disturbance being studied�

On the basis of the foregoing approximations� we expect that�

V E ��r� t� will depend

linearly on ��E��r� t�� t�E��r� t�� ��vE��r� t� and t�v

E��r� t��

We are thus led to a model in which at any point ��r� t� there are tensors

��

FE ��r� t� ���P �

��

JE ��r� t� and

��

KE ��r� t� � ��P � and�

HE ��r� t� � ��P such that

V����v�hi

��

F ��������

J ��t�v� ���

K ��t���

H � ��������

�� CHAPTER ��� CONSTITUTIVE RELATIONS

It seems reasonable to assume that the �uid is skew isotropic� Then the same must be

true of

��

F �

��

J �

��

K� and�

H� Therefore at each ��r� t� there are scalars � �� �� �� �� � such

that relative to any pooob for P

Fijkl � �ij�kl � ��ik�jl � ��il�jk

Jijk � ��ijk

Kijk � ��ijk

Hij � ��ij�

It is observed experimentally that ��v� ��� t�v and t� can be adjusted independently�

Therefore we can take all but �� to be �� Then Vij � ��ijkk�� This�

V is antisymmetric�

If we assume that the intrinsic angular momentum� body torque and torque stress in the

�uid are negligible� then Sij � Sji� so Vij � Vji� Therefore we must have � � �� By the

same argument� � � � and � � �� Thus we conclude that

Vij � �ij�� � �v

�� � �ivj � jvi� � ��ijt�� ��������

If �vE � ���r� the �uid is being compressed at a uniform rate� If t� � � then ��������

gives Vij � ��ij� � ���� An extra pressure� over and above pHTES� is required to

conpress the �uid at a �nite rate� The quantity � � � �� is therefore called the bulk

viscosity�

� � � ��� ���������

Then

Vij � ��ij�� � �v

�� �

�ivj � jvi �

�ij�� � �v

��� ��ijt�� ���������

The quantity � is called the shear viscosity because in a pure shear �ow� �vE��r� � w�r � y� y���

V� w�� y� y� � y� y�� and y���

V� � y�w� The tangential stress y���

V is � times the rate of

shear� w� In the ocean� ��� � ���� However� except in sound waves� � � �v � �� so the

large value of � in the ocean is not noticed except in the damping of short sound waves�

The second law of thermodynamics gives more information about ���������� namely

� � �� � � �� � � �� ��������

����� VISCOUS FLUIDS ��

The argument is as follows� At a particle �x at time t� U and � are measurable physical

quantities even though the material is not in an HTES� If the local values of U and � are

used in ��������� a local value of N can be found for particle �x at time t� Then �������

and �������� give local values of p and � at particle �x at time t� These local values of N �

p� � are not measured� They are calculated from the measured � and U as if the material

near �x at time t were in an HTES� Therefore we denote them by NLHTES��x� t�� p

LHTES��x� t��

�LHTES��x� t�� Note that pHTES is not ��� tr �S� the latter is a measurable quantity� If we

follow particle �x� ������� implies

DtU � �pHTESDt� � �HTESDtNHTES� ��������

There are no other candidates for � and N � So we drop the subscripts HTES on � and N �

We do not do so on pHTES� because ��� tr�

S is another candidate for the pressure and

we do not want to confuse the two� If we multiply �������� by � and use �������� we

obtain

�� � �H � h��

S hi��v � �pHTES�Dt� � ��DtN� ���������

But �Dt� � ���Dt� � Dtln � � �Dtln� � � � �v� Therefore ��������� is

��DtN � �H � h� Shi�v � pHTES�� � �v

�� ���������

But pHTES�� ��v� � pHTES�

I P hi��v so�

S hi��v�pHTES�� ��v� � ��

S �pHTES�

I P �hi���v���

V hi��v from ��������� If we divide ��������� by � and use the identity � � � �H��� �

�� � �H��� �H � ������� ��������� becomes

�DtN � � ��� �H

A �h

��

V hi��v � �H � ���

�� ���������

If we de�ne the stu� �entropy� as the stu� whose density per unit mass is N and

whose material �ux density is �H��� then ��������� shows that its creation rate � is

� �h

�� �H � �

��

��

V hi��v� ���������

Now h can be positive �e�g� ohmic heating� or negative �e�g� heat radiation by a trans�

parent liquid into empty space if the liquid is a blob in interstellar space�� However�

�� CHAPTER ��� CONSTITUTIVE RELATIONS

� � h�� is the creation rate of entropy by local events in the �uid� The second law of

thermodynamics is generalized to motions near HTES by requiring that everywhere at all

times

� � h

�� �� ���������

To use ���������� we must calculate�

V hi��v in ���������� Since Vij � Vij� we have

V hi��v � Vijivj ��

�Vij � Vji� ivj �

Vijivj �

Vjivj

��

Vijivj �

Vijjvi � Vij

�ivj � jvi�

��

V hi�

���v �

���v�T �

Now suppose we write ��������� as

Vij � �ij��� � �v � �t�

�� �

�ivj � jvi �

�ij� � �v

����������

and write

�ivj � jvi� � �ij

��

� � �v

��

�ivj � jvi �

�ij� � �v

�� ��������

Then we have expressed�

V and �����v����v�T � as the sum of their isotropic and deviatoric

parts� Since an isotropic and a deviatoric tensor are orthogonal� there are no cross terms

when we calculate�

V hi�� ���v � ���v�T � from ��������� and ��������� Therefore

V hi��v ��� � �v

� ��� � �v � �t�

���

k��v �

���v�T �

I P � � �vk��

Thus for a viscous �uid ��������� requires

� �H � ���

���

�� � �v

����

������v � ���v�T �

I P�� � �v

������ � ��� � �v

�t� � �� ��������

We can arrange experiments in which �� � ��� and t� � �� and �vE��r� t� � ��r � ���r � y�� y�� Then ��v � �Ip � � y� y�� � � �v � �� and �������� becomes ����� ��� � �� Since

this inequality must hold for � � �� � �� � and for � �� �� � � �� we must have � � ��

� � �� We can also arrange experiments with �vE��r� t� � ��r� �� � ��� and t� of any value�

In such an experiment� �������� becomes

����� ��t� � �� �������

����� VISCOUS FLUIDS �

If � �� �� take � � � and �t� ���� Then ������� is violated� Therefore we must

have � � �� The fact that �������� are observed experimentally is one of the arguments

for ��������� as an extension of the second law of thermodynamics�

�� CHAPTER ��� CONSTITUTIVE RELATIONS

Part III

Exercises

��

Set One

Exercise �

Let ��b�� � � � ��bn� be an ordered basis for Euclidean vector space V � Let ��b�� � � � ��bn� be its

dual basis� Let gij and gij be its covariant and contravariant metric matrices� Prove

a� �bi � gij�bj

b� �bi � gij�bj

c� gijgjk � �i

k

d� gijgjk � �i k

Note that either c� or d� says that the two n� n matrices gij and gij are inverse to each

other� Thus we have a way to construct ��b�� � � � ��bn� when ��b�� � � � ��bn� are given� This is

the procedure�

Step �� Compute gij � �bi ��bj�

Step �� Compute the inverse matrix to gij� the matrix gij such that gijgjk � �i

k or

gijgjk � �i k �either equation implies the other��

Step �� �bi � gij�bj �

Exercise �

V is ordinary Euclidean �space� with the usual dot and cross products� The function M

is de�ned by one of the following six equations� where �v�� �v� �v�� �v� are arbitrary vectors

��

���

in V � In each case� state whether M is multilinear� whether M is a tensor� and whether

M is totally symmetric or totally antisymmetric� State why you think so�

i� M��v�� �v�� � �v� � �v��

ii� M��v�� �v�� � �v� � �v��

iii� M��v�� �v�� � ��v� � �v���v��

iv� M��v�� �v�� � ��v� � �v�� � �v��

v� M��v�� �v�� �v�� � ��v� � �v�� � �v��

vi� M��v�� �v�� �v�� �v�� � ��v� � �v����v� � �v���

Solutions

��a� For any vector �v � V � we know that

�v ���v ��bj

��bj

See equations �D���� and �D�����

Let i � f�� � � � � ng� and apply this result to the vector �v � �bi� Thus we obtain

�bi ���bi ��bj

��bj� or �bi � gij�b

j�

��b� For any �v � V � �v � ��v ��bj��bj �see equations �D��� and �D���� Apply this result to

�v � �bi� Thus �bi � ��bi ��bj��bj� or �bi � gij�bj� Note that part b� can also be obtained

immediately by regarding ��b�� � � � ��bn� as the original basis and ��b�� � � � ��bn� as the

dual basis�

��c� Dot �bk into a� and use �bi ��bk � �ik�

�d� Dot �bk into b� and use �bi ��bk � �i k�

���

�i� M is multilinear because a dot product is linear in each factor�

M is a tensor because its values are real numbers �scalars�� �It is of order ��

M is totally symmetric because �v� � �v� � �v� � �v��

�ii� M is multilinear because a cross product is linear in each factor�

M is not a tensor because its values are vectors in V � not scalars�

M is totally antisymmetric because �v� � �v� � ��v� � �v��

�iii� M is not multilinear because M��v�� �v�� � �M��v�� �v���

M is not a tensor and is neither totally symmetric nor totally antisymmetric because

M��v�� �v�� is generally �� M��v�� �v�� and �� �M��v�� �v���

�iv� M��v�� �v�� � �� so M is multilinear�

M is a tensor because its values are real numbers� �It is of order ��

M is both totally symmetric and totally antisymmetric because

M ��v�� �v�� � M ��v�� �v�� � �M ��v�� �v��

for all �v�� �v� � V �

�v� M is multilinear because it is linear in each of �v�� �v�� �v� when the other two are

�xed�

M is a tensor because its values are scalars� �It is of order ��

M is totally antisymmetric� To show this� we must show that

���M � ���M � ��M � �M�

We have

M ��v�� �v�� �v�� � ��v� � �v�� � �v� � � ��v� � �v�� � �v� � �M ��v�� �v�� �v��

so ���M � �M � And we have

M ��v�� �v�� �v�� � ��v� � �v�� � �v� � ��v� � �v�� � �v� � �M ��v�� �v�� �v��

���

so ���M � �M � Finally

M ��v�� �v�� �v�� � ��v� � �v�� � �v� � ��v� � �v�� � �v� � �M ��v�� �v�� �v��

so ��M � �M �

�vi� M is multilinear because it is linear in each of �v�� �v�� �v�� �v� when the other three

are �xed�

M is a tensor because its values are scalars� �It is of order ���

M is neither totally symmetric nor totally antisymmetric� It is true that ���M � M

and ���M � M � However� ��M is neither M nor �M � To see this� we must �nd

�u�� �u�� �u�� �u� � V such that M��u�� �u�� �u�� �u�� �� M��u�� �u�� �u�� �u��� and we must �nd

�v�� �v�� �v�� �v� � V such thatM � ��v�� �v�� �v�� �v�� �� �M � ��v�� �v�� �v�� �v��� Let x� y� z be

an orthonormal basis for V and let �u� � �u� � �v� � �v� � x� �u� � �u� � �v� � �v� � y�

Then M��u�� �u�� �u�� �u�� � � and M��u�� �u�� �u�� �u�� � �� �Note� One might be tempted

to say that ��v� � �v�� ��v� � �v�� is �obviously� di�erent from ���v� � �v����v� � �v��� Yet if

dimV � �� we do have ��v� ��v����v� ��v�� � ��v� ��v����v� ��v��� andM is totally symmetric�

Extra Problems

In what follows� V is an n�dimensional Euclidean vector space and L � L�V V ��

�E� Recall that l � R is an �eigenvalue� of L i� there is a nonzero �v � V such that

L��v� � l�v�

a� Show that if l is an eigenvalue of L and p is any positive integer� then lp is an

eigenvalue of Lp�

b� If � � Sq �i�e�� � � f�� � � � � qg f�� � � � � qg is a bijection �� show that there is a

positive integer p � q� such that �p � e� the identity permutation�

c� Regard � � Sq as a linear operator on �qV � Suppose c � R is an eigenvalue of

� � �qV �qV � Show that c � ���

���

E�

a� For any c � R� show that det�cL� � cn detL�

b� Recall that L�� exists i� the only solution �u of L��u� � �� is �u � ��� Use this

fact to give a pedantically complete proof that l � R is an eigenvalue of L i�

det�L� lIV � � �� where IV is the identity operator on V �

E�

a� Show that det�L� lIV � is a polynomial in l with degree � n� That is� det�L �lIV � � c�l

n � c�ln�� � � � �� cn� �Hint� Choose a basis B � ��b�� � � � ��bn� for V �

Let Lij be the matrix of L relative to B� i�e�� L��bi� � Li

j�bj� Show that relative

to B the matrix of L� lIV is Lij� l�i j� Finally� use detL � L�

i� � � �Lnin�i����in

with L� lIV replacing L��

b� Show that c� � ����n and cn � detL� �Thus the degree of det�L� lIV � is n��

c� ����n��c� is called the trace of L� written tr L� Calculate tr L in terms of the

matrix Lij of L relative to B�

Solutions to Extra Problems

�E�

a� By hypothesis� �v �� �� and L��v� � l�v� Then L���v� � L�L��v�� � L�l�v� � lL��v� �

l��v� And L���v� � L�L���v�� � L�l��v� � l�L��v� � l��v� Etc� �or use mathematical

induction on p��

b� Sq has q� members� Therefore� among the q��� permutations e� �� ��� ��� � � � � �q��two must be the same� say �a � �b� a b� Then ��a�a � e � ��a�b � �b�a�

Take p � b� a�

c� If c is an eigenvalue of �� and p is as in b� above� then cp is an eigenvalue of

�p� so there is a nonzero M � �qV such that eM � cpM � But eM � M � so

cp � �� Since c � R� c � ���

���

E�

a� Let A � �nV � A �� �� Let ��b�� � � � ��bn� � B be a basis for V � Then A��cL���b��� � � � � �cL���bn��� det�cL�A��b�� � � � ��bn�� But A��cL���b��� � � � � �cL���bn�� � A�cL��b��� � � � � cL��bn��� cnA�L��b��� � � � � L��bn�� � cn�detL�A��b�� � � � ��bn�� Since A��b�� � � � ��bn� �� ��

det�cL� � cn�detL��

b� l is an eigenvalue of L � ��v � V � �v �� �� and L��v� � l�v� �De�nition of eigen�

value��

� L��v�� lIV ��v� � �� � def� of IV �

� �L� lIV ���v� � �� � def� of L� lIV ��

But ��v � V � �v �� �� and �L� lIV ���v� � ��� det�L� lIV � � ��

E�

a� �L � lIV ���bi� � L��bi� � lIV ��bi� � Lij�bj � l�bi � �Li

j � l�ij��bj� so the matrix

of L � lIV relative to B is Lij � l�i

j� / Therefore det�L � lIV � � �Lli� �

l�li�� � � � �Ln

in � l�nin��i����in � This is a sum of n� terms� Each term is a

product of n monomials in l� whose leading coe!cients are � or ��� Thereforeeach term is a polynomial in l of degree � n� Hence so is the sum of all n�

terms�

b� cn is the value of the polynomial when l � �� Hence it is the value of det�L� lIV �when l � �� i�e�� it is detL�

b and c� To �nd c� and c�� note that �/� in the solution E�a� is a sum of nn terms�

of which n� have a nonzero factor �i����in � Among these nonzero terms� one has

i� � �� i� � � � � � � in � n� It is �y��L�

� � l�ll�� � �

�Ln�

n� � l�n�n�������n

��L�

� � l� �

L�� � l

�� � �

�Ln�

n� � l��

In all other nonzero terms� there must be at least two failures among i� � ��

i� � � � � � � in � n� because fi�� � � � � ing � f�� � � � � ng� Therefore� in each

��

nonzero term in �/� except �y� above� at least two factors have no l� so the

total degree as a polynomial in l is � n� � Therefore� all the terms in ln and

ln�� in det�L� lIV � come from �y�� But �y� is

����nln � ����n�lhL�

� � L�� � � � �� Ln�

n�iln�� � � � � �

Thus c� � ����n and c� � ����n��Lii� Hence trL � Li

i�

���

Set Two

Exercise �

Let � x�� x�� � � � � xn� � B be an ordered orthonormal basis for Euclidean space V � Let

B� � � x�� x�� x�� � � � � xn� �i�e�� just interchange x� and x� �� Show that B and B� are

oppositely oriented�

Exercise

If V is Euclidean vector space� the �identity tensor� on V is de�ned to be that I � V �Vsuch that for any �x� �y � V

I ��x� �y� � �x � �y�

Suppose that �u�� � � ��un� �v�� � � ��vn are �xed vectors in V such that

I �nXi �

�ui�vi�

a� Prove that n � dimV � �Hence� if dimV � � I is not a dyad ��

b� Prove that if n � dimV then ��u�� � � � � �un� is the basis of V dual to ��v�� � � ��vn��

Solutions

� Let A be one of the two unimodular alternating tensors over V � Let � � Sn� Then

A� x���� � � � � x�n�� � � sgn ��A� x�� � � � � xn�� Hence � x�� � � � � xn� and � x���� � � � � x�n��have the same or opposite orientation according as sgn � � �� or ���

��

���

�� Change notation so as to use our restricted summation convention� Write I � �ui�vi�

Then for any �x� �y � V we have

�x � �y � I��x� �y� � ��ui � �x���vi � �y� � �x �h�ui��vi � �y

�i� ��

Thus

�x �h�y �

��y � �vi

��uii� �� ���

For any �xed �y � V � ��� is true for all �x � V � Thus �y � ��y � �vi��ui � ��� or

�y ���y � �vi

��ui� for any �y � V� ���

a� ��� shows that f�u�� � � � � �ung spans V � Hence� n � dimV �

b� If n � dimV � then ��u�� � � � � �un� is an ordered basis for V � Let ��u�� � � � � �un� be

its dual basis� Choose �xed j� k � f�� � � � � ng and set �x � �uj� �y � �uk in ���

Then

�uj � �uk ���ui � �uj

� ��vi � �uk

�or

�j k � �ij��vi � �uk

�� �vj � �uk�

Thus ��v�� � � � � �vn� is a sequence dual to ��u�� � � � � �un�� Since ��u�� � � � � �un� is a

basis for V � there is exactly one such dual sequence� namely ��u�� � � � � �un�� Hence�vi � �ui�

Exercise

Let V be an n�dimensional Euclidean space� Let A be any nonzero alternating tensor

over V �that is� A � �nV and A �� ��� Let ��b�� � � � ��bn� be an ordered basis for V � with

dual basis ��b�� � � � ��bn�� Let � A��b�� � � � ��bn�� For each i � f�� � � � � ng de�ne

�vi ��

����i��Ahn� �i��b� � � � �� bi � � ��bn

��

Show that �vi � �bi� When n � � express ��b���b���b�� explicitly in terms of ��b���b���b��� using

only the dot and cross products� not A�

���

Exercise �

Suppose L � L�V V ��

a� Show that there are unique mappings S� K � L�V V � such that S is symmetric

�ST � S�K is antisymmetric �KT � �K� and L � S�K� �Hint� then LT � S�K��

b� Suppose dimV � and A is one of the two unimodular alternating tensors over V �

Suppose K � L�V V � is antisymmetric� Show that there is a unique vector

�k � V such that�

K� A ��k� Show that �k � ��Ahi �K and that

K� �k �A� �Hint� Takecomponents relative to a positively�oriented ordered orthonormal basis� Use �����

and page D����

Exercise

Let ��b�� � � � ��bn� be a basis for Euclidean vector space V � its dual basis being ��b�� � � � ��bn��Let L � L�V V ��

a� Show that�

L� �biL��bi� � �biL��bi��

b If L is symmetric� show that V has an orthonormal basis x�� � � � � xn with the property

that there are real numbers l�� � � � � ln such that�

L�Pn

v � lv� xv xv� �Hint� Let lv� be

the eigenvalues of L� See page D����

Solutions

��a� We show that �bj � �v i � �ji� and then appeal to the fact that the dual sequence of an

ordered basis is unique� We have

�bj � �v i ��

����i���bj �Ahn� �i��b�� � � � �� bi � � ��bn

��

����

i��

A��bj��b�� � � � ��bi����bi��� � � � ��bn

��

���

It takes i� � interchanges to change ��bj��b�� � � � ��bi����bi��� � � � ��bn� into��b�� � � � ��bi����bj��bi��� � � � ��bn�� and each changes the sign of A� so

�bj � �v i ��

A��b�� � � � ��bi����bj��bi��� � � � ��bn

��

If j �� i� �bj appears twice� so A��b�� � � � ��bi����bj��bi��� � � � ��bn� � � and �bj � �v i � �� If

j � i� then A��b�� � � � ��bi����bi��bi��� � � � ��bn� is � so �bj � �v i � �� QED�

��b� When n � � � A��b���b���b�� � �b� � Ahi��b��b�� � �b� � ��b� ��b��� and�b�

��

Ahi��b ��b�� � �

�b� ��b�

�b�

� � �

Ahi��b��b�� � � �

�b� ��b� � �

�b� ��b�

�b� ��

Ahi��b��b�� � �

�b� ��b��

��a� Uniqueness� If L � S � K with S � ST and KT � �K then LT � S � K so

S � ���L� LT � and K � ���L� LT ��

Existence� de�ne S � ���L � LT � and K � ���L � LT �� Then ST � S�

KT � �K� and L � S �K�

��b� Uniqueness� Suppose�

K� A � �k� Relative to any pooob we have Kij � �ijkkk so

Kij�ijl � �ijl�ijkk

k � �l kkk � kl� Thus� �k �

K hiA� Also kl � �lijKij� so

�k � Ahi �K�

Existence� Given�

K antisymmetric� de�ne �k � ��Ahi �

K� We have relative to

any pooob �A � �k

�ij

� �ijkkk �

�klmKlm

��

��i

l�jm � �i

m�jl�Klm

��

Kij � �

Kji � Kij�

Hence��

K� A � �k� Also Kij � �ijkkk � kk�kij so

K� �k � A�

��a� �bi� �L� L��bi�� Multiply by �bi and sum over i to get �biL��bi� � �bi��b

i� �L� � ��bi�bi�� �L

��

I V ��

L��

L� By viewing �bi as the original and �bi as the dual basis� one then also

�nds �biL��bi� ��

L�

���

��b� From page D��� V has an orthonormal basis x�� � � � � xn consisting of eigenvectors

of L� That is� there are numbers l�� � � � � ln � R such that L� x�� � l� x�� � � � � L� xn� �ln xn� By �a�

L�nX

v �

xvL� xv� �nX

v �

xv �lv xv� �nX

v �

lv � xv xv� �

Extra problems

�E� Suppose f�b�� � � � ��bmg is a linearly independent subset of Euclidean space V �so m �dimV � m dimV is possible��� Let ��u�� � � � � �um� and ��v�� � � � � �vm� be any m�tuples

of vectors in V � Show that �ui�bi � �vi�bi implies �ui � �vi�

�E� Suppose V is a Euclidean vector space and a� b� c� d� e are non�negative integers�

Suppose A � �aV � C � �cV � E � �eV � We want to study whether it is true that

Ahbi �ChdiE� � �AhbiC� hdiE� ���

i� Using p �� of the notes� �nd conditions on a� b� c� d� e which imply ����

ii� If dimV � � show that whenever a� b� c� d� e violate the conditions found in i��

it is possible to have A � �aV � C � �cV and E � �lV which violate ����

�Hint� use polyads��

�E� In oriented Euclidean �space �V�A�� show that �u� �v� �w can be chosen so that �u ���v � �w� �� ��u� �v�� �w�

Solutions to Extra Problems

�E� Let U � spf�b�� � � � ��bmg� Then U is a subspace of V and hence a Euclidean vector

space under V �s dot product� Also� ��b�� � � � ��bm� is an ordered basis for U � so in

U it has a dual basis ��b�� � � � ��bm�� If �u i�bi � �v i�bi then ��u i�bi� � �bj � ��v i�bi� � �bj� so�u i��bi ��bj� � �v i��bi ��bj�� so �u i�i

j � �v i�ij� Then �u j � �vj� QED�

���

a

p q

c

q sC r

b

A

s

AE t

d

e

Figure Ex��� Exercise �E�i

�E�i� The picture that makes ��� work is in �gure Ex�� There must be p� q� r� s� t � �

such that a � p � q� b � q� c � q � r � s� d � s� e � s � t� solving for p� q� r� s� t

gives p � a � b� q � b� r � c � b � d� s � d� t � e � d� The conditions that

p� q� r� s� t� a� b� c� d� e � � reduce to

� � b � a� � � d � e� b� d � c�

�E�ii� The question is� is it possible to violate the inequalities in i� in such a way that

Ahbi�ChdiE� and �AhbiC�hdiE are both de�ned� Let hki�P � � order of tensor P �

Thus hki�A� � a� hki�AhbiC� � a � c � b� etc� A product P hqiR is de�ned only

if hki�P � � q and hki�R� � q� If Ahbi�ChdiE� is de�ned we must have a � b�

hki�ChdiE� � c � e� d � b and c � d� e � d� If �AhbiC�hdiE is de�ned we must

have d � e and hki�AhbiC� � d� or a � c � b � d� and also b � a� b � c� If both

triple products are de�ned we must have b � a� b � c� d � c� d � e� b � d � c � e

and b � d � a � c� The only way i� can fail if all these hold is to have b � d � c�

Choose ���� � � � � ��a� ���� � � ���c� ���� � � ���e � V and let A � ��� � � � ��a� C � ��� � � ���c�E � ��� � � ���e� Then we have �gure Ex��ii And AhbiC � ���a�b�� � ���� � � � ���a �

���

a

b

A

C c

E

d

e

Figure Ex�� Figure Ex���ii

��b���� � � � ��a�b��b�� � � ���c �AhbiC�hdiE � ���a�b�� � ���� � � � ���a � ��b����a�c��b�d�� � ����� � � ���a�b � ��d�b�c� ���b�� � ��d�b�c��� � � � ���c � ��d� ��� � � � ��a�c��b�d ��d�� � � ���e�

Similarly ChdiE � ���c�dn � ���� � � � ���c � ��d���� � � ���c�d��d�� � � ���e and Ahbi�ChdiE� �

���a�b�� ����� � � � ���a�b�c�d ���c�d����a�b�c�d�� ���d��� � � � ���a ���b��d�c� ���c�d�� ����� � � � ���c ���d���� � � � ��a�b��b��d�c�� � � ���e�

Choose all ��� ��� �� so the dot products are �� �� Now a � b � a � c � b � d� so

�AhbiC�hdiE � Ahbi�ChdiE� implies that ��a�b and ���d�b�c are linearly dependent�

By choosing them linearly independent we have �AhbiC�hdiE �� Ahbi�ChdiE��

�E�

�u� ��v � �w� � ��u � �w��v � ��u � �v��w��u� �v�� �w � �w � ��v � �u� � ��w � �u��v � ��w � �v��u�

These are equal i� ��u � �v��w � ��v � �w��u� Choose �u and �w mutually � and of unit

length� with �v � �u� �w� and this equation will be violated�

���

Exercise �

Let U be a Euclidean vector space� Let D be an open subset of U � Let�

T� U � U � For

each �u � U � let �f��u� ��

T ��u and g��u� � �u � �u� Show that at every �u � D� �f � D U

and g � D R are di�erentiable� Do so by �nding �r�f��u�� �rg��u�� and the remainder

functions� and by showing that R��h� � as �h ���

Exercise �

Suppose thatDf andDg are open subsets of Euclidean vector spaces U and V respectively�

Suppose that �f � Df Dg and �g � Dg Df are inverse to one another� Suppose there

is a �u � Df where �f is di�erentiable and that �g is di�erentiable at �v � �f��u�� Show that

dimU � dimV � �Hint� See page �����

Exercise ��

Suppose V is a Euclidean vector space� �U�A� is an oriented real three�dimensional Eu�

clidean space� D is an open subset of U � and �f � D U and �g � D V are both

di�erentiable at �u � D�

i� Which proposition in the notes shows that �f�g � D U � V is di�erentiable at �u$

ii� Show that at �u

�r���f�g����r� �f

��g � �f � �r�g�

�Note� In this equation� one of the expressions is unde�ned� De�ne it in the obvious way

and then prove the equation��

Solutions

�a�

�f��u� �h

��

T ���u� �h

���

T ��u� �

T ��h

��

� �f��u� � �h� �TT�

Therefore �r�f��u� ��

TTand �R��h� � �� for all �h�

�b�

g��u� �h

��

��u� �h

����u� �h

�� �u � �u� �h � �u� �h � �h� g ��u� � �h � ��u� �

����h��� ����h��� �Therefore �rg��u� � �u� and R��h� �

����h���� As �h �� R��h� � because �h ��

means����h��� ��

�� De�ne�

P� �r�f��u� and�

Q� �r�g��v�� By example ������

P ��

Q��

I U and�

Q � �P��

I V �

Therefore� by ������ the linear mappings P � U V and Q � V U satisfy

Q�P � IU and P �Q � IV � By iii� on page D��� P and Q are bijections� Since they

are linear� they are isomorphisms of U to V and V to U � By preliminary exercise

� dimU � dimV �

���i� Method � on page ��� or method on page ���� applied to �f � � D U �R� and

��g � D R� � V �

���ii� For �u � U and�

P� U � V we de�ne �u� �

P �� Ahi��u�

P

�� Then� relative to any

basis �b�� �b�� �b� for U and ���� � � � � ��n for V � we have ��u� �

P �i l � Aijk�jPkl� Then

h�r�

��f�g�ii

l � Aijkj��f�g�kl� Aijkj �fkgl�

� Aijk ��jfk� gl � fkjgl�

���r� �f

�igl � Aikjfkjgl

�h��r� �f

��gii

l � Aijkfj��r�g

�kl

�h��r� �f

��gii

l ���f � �r�g

�il

�h��r� �f

��g � �f � �r�g

iil� QED�

���

Extra Problems

�E� Let U be a Euclidean vector space� Let�

T� U � U � Let D be the set of all �u � U

such that �u� �T ��u � �� If �u � D� de�ne f��u� � ln��u� �T ��u��

i� Show that D is open�

ii� Show that f � D R is di�erentiable at every point �u � D� and �nd �rf��u��

�E� Let V be the set of all in�nite sequences of real numbers� �x � �x�� x�� � � ��� such thatP n � x

�n converges� If �y � �y�� y�� � � �� is also in V and a� b � R� de�ne a�x � b�y �

�ax� � by�� ax� � by�� � � ��� De�ne �x � �y �P

n � xnyn� �The sum converges by the

Cauchy test because� for anyM and N � Schwarz�s inequality implies jPNn m xnynj �

�PN

n m x�n�����

PNn m y�n�

���� Furthermore� j�x��yj� � j limN� PN

n � xnynj� � limN�

�PN

n � x�n��

Pnn � y

�n� � k�xk�k�yk�� so Schwarz�s inequality works in V �� V is a real

dot�product space but not a Euclidean space� because it is not �nite dimensional�

If �v�� �v�� �v�� � � � is a sequence of vectors in V � de�ne limn� �vn � �v� by the obvious

extension of de�nition ������ i�e� limn� k�v� � �vnk � ��

i� Suppose limn� �vn � �v�� Show that for any �x � V � limn� �x � �vn � �x � �v��

ii� Construct a sequence �v�� �v�� � � � in V such that limn� �x ��vn � � for every �x � V �

but limn� �vn does not exist�

�For di�erentiability� V has two di�erent de�nitions� leading to Frechet and Gateau

derivation respectively based on norms and on components��

Solutions

�E� If�

T���� D is empty and i� and ii� are trivial� Therefore assume k �

T k � �� De�ne

g��u� � �u� �T ��u�

���

i� g��u � �h� � g��u� � �h� �

T ��u � �u� �

T ��h� �

T ��h� Therefore� by the triangle and

generalized Schwarz inequalities�

���g ��u� �h�� g ��u�

��� � k�hkk �

T k�k�uk� k�hk

��

Assume �u � D so g��u� � �� Then �u �� ��� Let � be the smaller of k�uk and

g��u���k�ukk �

T k�� If k�hk � then jg��u � �h� � g��u�j �k �

T k�k�uk � k�hk� �k �

T kk�uk � g��u�� Hence g��u��h�� g��u� � �g��u�� so g��u��h� � �� That

is� if �u � D and � � smaller of �u and g��u���k�ukk �

T k� then the open ball

B��u� �� � D� Hence D is open�

ii� By i� above� g��u � �h� � g��u� � �h � ��T ��

TT� � �u � k�hkR��h� where R��h� �

��h� �T ��h��k�hk for �h �� �� and � �� for �h � ��� Thus g is di�erentiable at �u� with

�rg��u� � ��

T ��

TT���u� By the chain rule� f �g is di�erentiable� with f�v� � lnv�

and its gradient is vf�v��rg��u� where v � g��u�� This is v���rg��u�� so

��rln

��u� �T ��u

����u� �

��

T ��

TT�� �u

�u� �T ��u�

�E� i�

j��x � �vn � �x � �v��j � j�x � ��vn � �v��j � k�xkk�vn � �v�k �

if k�vn � �v�k ��

�E� ii� Let �vn � ��� � � � � �� �� �� � � ��� all � except for a � in the n�th place� If �x � V �

�x ��vn � xn � as n � becauseP

� x�n converges� But if k�vn��vk � for some

�v � V then there is an N so large that if n � N then k�vn � �vk ��� If m� n � N

then k�vm��vnk � k�vm��v��v��vnk � k�vm��vk�k�vn��vk � �� But k�vm��vnk �p�

Exercise ��

A mass distributionm occupies a subset D of ordinary Euclidean �space U � Let n be any

unit vector in U and denote by nR the straight line through U �s origin �� in the direction

���

of n� Let w ��r� be the perpendicular distance of the point �r from the axis nR� so that

the moment of inertia of the mass distribution about that axis is J� n� �RD dm��r� w��r���

De�ne�

T � D U�U by requiring for each �r � D that�

T ��r� � r��

I U ��r�r� Here r� � �r ��rand

I U is the identity tensor in U�U � The tensor�

J�RD dm��r�

T ��r� is called the inertia

tensor of the mass distribution� and is usually written�

J�RD dm��r��r�

I U ��r�r ��

a� Show that�

J is symmetric�

b� Show that J� n� � n� �J � n�

c� Show that to calculate�

J it is necessary to calculate only six particular integrals of

real�valued functions on D with respect to the mass distribution m� �Once these

six functions are calculated� the moment of inertia of the body about any axis is

found from ��b��

Exercise ��

In problem ��� suppose the mass distribution rotates rigidly about the axis nR with

angular velocity �* � * n� so that the mass at position �r has velocity �v��r� � �* � r� The

angular momentum of the mass distribution about �� is de�ned as �L �RD dm��r���r��v��r���

and the kinetic energy of the mass distribution is de�ned asK � ��RD dm��r�v��r��� where

v� means �v � �v� Show the following�

a� �L ��

J ��*

b� K � �� �*� �J ��*

c� �L need not be J� n��*�

Exercise ��

The second moment tensor of the mass distribution in problem �� is de�ned as�

M�RD dm��r��r �r�

���

a� Show that tr�

M�RD dm��r�r��

b� Express�

J in terms of�

M � tr�

M and�

I U �

c� Express�

M in terms of�

J � tr�

J and�

I U � ��

J can be observed by noting the reaction

of the mass distribution to torques� Then �c� gives�

M and �a� givesRD dm��r�r�

from such observations��

d� If the mass distribution is a cube of uniform density �� side a and center at ��� �nd

J� n� when nR is the axis through one of the corners of the cube�

Exercise �

Let y�� y�� y� be an orthonormal basis for physical space P � Let c be a �xed scalar� For

any �r � P � write �r � ri yi� Consider a steady �uid motion whose Eulerian description �vE

is given by

�vE��r� t� � r� y� � c r� y� � �r � � y� y� � c y� y��

for all �r � P and all t � R�

a� Find the Lagrangian description of this motion� using t��position labels�

b� Show that the paths of the individual particles lie in planes r� � constant� and are

hyperbolas if c � �� and straight lines if c � ���

c� Find the label function �x E � P �R P � �It will depend on which �xed t� is used to

establish t��position labels��

d� Find the Eulerian and Lagrangian descriptions of the particle acceleration �a�

Solutions

�� a� The permutation operator ���� P � P P � P is linear� and ����

T ��r� ��

T ��r��

so

���ZDdm��r �

T ��r � �ZDdm��r� ���

T ��r � �ZDdm��r �

T ��r ��

���

rn

.

.

.n r^

w( r )~

Figure Ex��

b�

ew ��r �� � r� � � n � �r ��

� r� n� �I P � n� n � r r � n� n �

�r�

I P ��r �r�� n�

Hence

J� n� �ZDdm��r� n �

hr�IP � �r �r

i� n ���

� n ��Z

Ddm ��r�

�r�

I P ��r �r��� n � n� �J � n� ���

c� Choose a pooob� x�� x�� x�� Then�

J� Jij xi xj and since�

JT��

J � Jij � Jji� Thus�

J is

known if we know J��� J��� J��� J��� J��� J�� relative to one pooob� But

Jij �ZDdm � r�

hr��ij � rirj

i�

� a�

�L �ZDdm ��r��r �

h�*� �r

i�ZDdm ��r��r

h�*� �r

��r � �*

�i�

ZDdm ��r �

��r�

I P ��r �r�� �*�

�ZDdm ��r �

�r�

I P ��r�r�� �* �

J ��*�

���

b�

K ��

ZDdm ��r �

��*� �r

����*� �r

��

ZDdm ��r� * �

h�r �

��*� �r

�i�

ZDdm ��r � �* �

hr��*� �r

��r � �*

�i�

ZDdm ��r�

��* �

�r�

I P ��r �r�� �*�

��

�* �

�ZDdm ��r�

�r�

I P ��r�r��� �*

��

�*� �J ��*�

c� Let the mass distribution consist of a single mass point m at position �r� Then

J � m�r�

I P ��r �r�

�L ��

J ��* � m*�r�

I P ��r �r�� n

� m*hr� n� �r ��r � n�

iJ � n� �* �

� n� �J � n

� n* � m*

hr� � � n � �r ��

i n�

If �L � J� n��* then �r�� � n ��r��� n � r� n��r��r � n�� so ��r � n�� n � ��r � n��r� To violate

this condition choose n so it is neither parallel nor perpendicular to �r�

� a� tr � P � P R is linear� so

tr�

M � trZDdm ��r��r �r �

ZDdm ��r� tr�r�r

�ZDdm ��r� r��

b��

J�Rdm��r�

�r�

I P ��r �r�� �

RD dm ��r� r��

I P � RD dm ��r��r �r

� ����J��

I P

�tr

M

�� �

M �

c� From the above equation� since tr � P � P R is linear�

tr�

J��tr

I P

��tr

M

�� tr

M� tr�

M �tr �

M� tr�

M �

Hence� tr�

M� �� tr�

J and�

M� ���

I P �tr�

J �� �

J �

���

d� Let x�� x�� x� be a pooob parallel to the edges of the cube� Then

Mij �Zcube

dx� dx� dx� xixj�

� �Z a

�adx�

Z a

�adx�

Z a

�adx� xi xj�

Thus Mij � � if i �� j and M�� � M�� � M�� � �a��R a�a x

� � �a���� Therefore�

M� �a����

I P and tr�

M� �a�� so from ��� above�

J���a��

I P �

Then for any n� n� �J � n � ��a����

�� a�

dr�

dt� r��

dr�

dt� cr��

dr�

dt� � so

r� � A�et�t��� r� � A�ect�t��� r� � A��

The label �x � xi yi belongs to the particle which was at that position at t � t�� For

this particle when t � t�� ri � xi� so xi � Ai� Thus

��� r� � x�et�t��� r� � x�ect�t��� r� � x��

�r � ri yi � et�t��x� y� � ect�t��x� y� � x� y�

� �x �h y� y�e

t�t�� � y� y�ect�t�� � y� y�

i�

�r L ��x� t� � �x �h y� y�e

t�t�� � y� y�ect�t�� � y� y�

i�

b� y� � �r is independent of time for each particle� so each particle lies in a plane y� � �r �

constant� And for a particular particle� r�r� � x�x�e��c�t�t�� � x�x� if c � ���which is a hyperbola� And r��r� � ec���t�t��x��x� � x��x�� if c � ��� Then this

is a straight line through the y� axis�

c� From ��� above� x� � r�e�t�t��� x� � r�e�ct�t��� x� � r�� so

�x � xi yi � r�e�t�t�� y� � r�e�ct�t�� � r� y�

� �r �h y� y�e

�t�t�� � y� y� e�ct�t�� � y� y�

i�

�x E ��r� t� � �r �h y� y� e

�t�t�� � y� y� e�ct�t�� � y� y�

i�

���

d�

�aL ��x� t� � D�t �r

L ��x� t� � �x �h y� y� e

t�t�� � y� y� ect�t��

i�aE ��r� t� � �x E ��r� t� �

h y� y� e

t�t�� � c� y� y� ect�t��

i

�aE ��r� t� � �r �h y� y� e

�t�t�� � y� y� e�ct�t�� � y� y�

i�h y� y� e

t�t�� � c� y� y� ect�t��

i� �r �

h y� y� � c� y� y�

i�

Another way to get the same result is to observe that

�vE ��r� t� � �r � � y� y� � c y� y��

has no explicit t dependence� so t�vE��r� t� � ��� Also ��vE��r� t� � y� y� � c y� y�� so

�vE � ��vE ��r� t� � �r � � y� y� � c� y� y��� But

�aE ��r� t� � Dt�vE ��r� t�

� t�vE ��r� t� � �vE � ��vE ��r� t�

� �� � �r �� y� y� � c� y� y�

��

Exercise �

a� Given a continuum and a function f � R W which depends on t alone �W is a

Euclidean vector space�� invent a physical quantity which can reasonably be called

f � and show that �Dtf�L�t� � �tf�

E�t� � �tf�t��

b� A continuum undergoes rigid body motion� At time t the position� velocity and

acceleration of the pivot particle are �R�t�� �V �t� � �t �R�t� and �A�t� � �t�V �t�� and the

angular velocity relative to the pivot particle is �*�t�� Find the Eulerian description

of particle acceleration in the material�

���

Exercise ��

Several decades ago� before the big bang was accepted� Fred Hoyle suggested that the

expanding universe be explained by the continuous creation of matter� Then mass is not

quite conserved� Suppose that a certain region of space is occupied by a continuum� and

that at position �r at time t new matter is being created at the rate of �E��r� t� kilograms

per cubic meter per second� Make the necessary changes in the derivation of the Eulerian

form of the law of conservation of mass and in equations ������� and ���������

Exercise �

At time t� in a certain material� the Cauchy stress tensor�

S is symmetric�

a� Let K � be an open set in the region occupied by the material� and suppose its boundary

K � is piecewise smooth� Show that the total force �F and the total torque �L about

�� exerted on K � by the stress on K � are the same as those exerted by a �ctitious

body force with density �� �S newtons � m� acting in K ��

b� Write the Taylor series expansion of�

S about �� in physical space P as�

S ��r� ��

S��

��r � S�� � ����r �r�hiS�� � � � � where �

S����

S����

S��� � � � are constant tensors in

P �P � ��P � ��P � � � � and ���S�� � S��� Neglect all terms except those involving�

S����

S��� and

S��� and calculate �F and �L of a� for K � � B���� c�� the solid ball of

radius c centered on ���

Hint �� Do all calculations relative to an orthonormal basis in P �

Hint � Use the symmetry of B���� c� to evaluateRB dV ��r�rirj�

Solutions

�� a� The physical quantity has as its Lagrangian description fL��x� t� � f�t�� Then its

Eulerian description is fE��r� t� � f�t�� Since � �f � �� Dtf � tf � �v � � �f � tf �

��

b�

�vE ��r� t� � �V �t� � �*�t��h�r � �R�t�

i� �V �t� �

h�r � �R�t�

i� �* �t��

��vE ��

* so

�vE � ��vE � �vE� �*� �*� �vE � �*� �V � �*�h�*�

��r � �R

�it�v

E � �A�t�� �V �t�� �* �h�r � �R

i� t

*

� �A� �*� �V � t�*���r � �R

��

�Dt�v�E � �aE �

�t�v � �v � ��v

�E� �A� t�*�

��r � �R

�� �*�

h�*�

��r � �R

�i�aE ��r� t� � �A�t� �

h�r � �R�t�

i��t

* ��

* � �*��

��� The amount of mass in K ��t� is

M �K ��t�� �ZK�t�

dVP ��r��E��r� t��

It is a mathematical identity �derivative over moving volume� that

d

dtM �K ��t�� �

ZK�t�

dVPt�E �

Z�K�t�

dAp np ���vE�E

��

ZK�t�

dVpht� � � � ���v�

iE��r� t� �

Hoyle�s new physical law is that

d

dtM �K ��t�� �

ZK�t�

dVp �E��r� t��

Therefore ZK�t�

dVpht�� � � ���v�� �

iE��r� t� � ��

Since this is true for every open set K ��t� with piecewise smooth boundary� the

vanishing integral theorem gives

t� � � � ���v� � ��

���

Then for any other physical quantity �f �

d

dt

ZK�t�

dVP���f

�E�

ZK�t�

dVPh�t�� �f � �

�t �f

�iE�ZK�t�

dAP n �h�v���f

�iE�

ZK�t�

dVPh�t�� �f � �

�t �f

�� � �

���v �f

�iE�

ZK�t�

dVPn�t�� �f � �

�t �f

��h� � ���v�

i�f � ��v � �f

oE�

ZK�t�

dVPn��t �f � �v � t �f

��ht�� � � ���v�

i�foE

�ZK�t�

dVPh�Dt

�f � � �fiE

�� a�

Z�K�

dA n� �S�ZK�

dV��� �S

��Z

�K�

dA�r �� n� �S

��i� �

�ZK�

dA�n� �

S ��r�i

� �ZK�

dV�� �

��

S ��r��

i� �

ZK�

dV j �Sjkrl�kli�

� ��iklZK�

dV �rljSjk � �jlSjk�

� ��ikjZK�

dV �rjlSlk � �jlSlk�

� �ijk

ZK�

dV �rjlSlk � Sjk�

�ZK�

dV��r � �� �S

�i�

b�

Sij � S��ij � rkS

��kij �

rkrlS

��klij

iSij � S��iij �

��k ir

l � �l irk�S��klij

� S��iij �

�rlS

��ilij � rkS

��kiij

�iSij � S

��iij � rkS

��kiij � vj � rkwkj

� de�nes vj and wkj��

���

�� �S � �v � �r� �w �

�r � �� �S � �r � �v � �r ���r� �w

���r � �� �S

�i

� ��r � �v�i � �ijkrjrlwlkZBdV �� �S � jBj�v �

�ZBdV �r

�� �w�

ZBdV �r �

�� so

�F ���

c��v �

��

c� tr�� S

��

�L �ZBdV �r � �� �S� Li �

ZBdV �ijk �rjvk � rjrlwlk� �

ZBdV rj � � and

RB dV rjrl � ����� c��jl so Li � ����� c��ijkwjk� �L � ����� c�

Ahitr��S�� �

Exercise ��

a� The angular momentum of a single dysprosium atom is ��h��� where h is Planck�s

constant� The density of dysprosium is ����gm�cm�� and its atomic weight is

������ Compute the angular momentum of a stationary ball of solid dysprosium

with radius r cm if all the dysprosium atoms are aligned in the same direction� If

the dysprosium atoms had no angular momentum� how rapidly would the ball have

to rotate as a rigid body in order to have the same angular momentum$

b� Let y�� y�� y� be a positively oriented ordered orthonormal basis for physical space P �

A permanent alnico magnet has magnetization density �M � M y� where M � ���

amp�meter �close to the upper limit for permanent magnets�� The magnet is held at

rest in a uniform magnetic �eld �B � B y� with B � � tesla �about the upper limit for

commercial electromagnets�� Then the volume torque on the magnet is �m � �M � �B

joules �meter�� Suppose that the magnet�s intrinsic angular momentum density �l

does not change with time� and that its torque stress tensor�

M vanishes� Let�

S be

the Cauchy stress tensor in the magnet� Find ����

S � �

ST�� the antisymmetric part

of�

S� in atmospheres of pressure �one atmosphere � ����� ��� newtons�meter� �

�������� dynes �cm� �� If the symmetric part of�

S� ����

S ��

ST�� vanishes� sketch

���

1 meter

1 meter

^

^^

^

^

^

n 5

S

n

5

S

4

4

n3

S3

x

y

z S1

n1

S2

n 2

Figure Ex���

the stresses acting on the surface of a small spherical ball of material in the magnet�

Exercise ��

a� Give the obvious de�nition of the stu� �y component of momentum�� Give its �,� �F ��

its ��� �F �� and its creation rate ��

b Show that in a material at rest with no body forces� the material �ux density �F of y

momentum satis�es � � �F � �� �If �F were the velocity of a �uid� the �uid would be

incompressible��

c� An aluminum casting �� cm thick and � meter on each side is shaped as in �gure Ex���

A steel spring is compressed and placed between the jaws of the casting as shown�

The spring and casting are at rest and gravity is to be neglected� Roughly sketch

���

the �eld lines of �F �the material �ux density of y momentum� in the casting� �If

�F were a force �eld� the �eld lines would be the lines of force� if �F were a velocity

�eld� they would be the streamlines��

Hint for c�� Estimate qualitatively the sign of the y component of the force exerted

by the aluminum just in front of �Si� ni� on the aluminum just behind it for

the surfaces i � �� � � �� � sketched in the �gure�

Note for c�� The x axis points out of the paper� the y axis is parallel to the axis of

the spring� and the z axis points up� as shown in the �gure�

Solutions

�� a� There are ����� ���������� � ��� ���� atoms in a gram of Dysprosium� Each

has angular momentum ������������ � ����� erg sec� If all are aligned� one

gram of Dy has angular momentum l � �� � ���� erg sec�gm� A Dy ball of

radius r cm has intrinsic angular momentum ������r�l� If this angular momentum

were not intrinsic� but due to an angular velocity � of rigid rotation� the angular

momentum would be ����r�������r��� Thus l � ���� r��� or � � ���� l�r� �

��� �����r� radians�sec� with r in cm�

b�

�� � �m� Ahi �S� so relative to the pooob y�� y�� y�

� � mi � �ijkSjk� Then

� � mi�ilm � �ilm�ijkSjk

� � mi�ilm � ��lj�mk � �lk�mj�Sjk � mi�ilm � Slm � Sml�

��� �

�Slm � Sml� � ��

�lmimi�

In our problem� �m � MB y� so

�S�� � S��� � ��

MB�

�S�� � S��� �

�S�� � S��� � �� and

���

....

m

Figure Ex���

��

S � �

ST�

��

��

S � �

ST�ij yi yj� so

��

S � �

ST�

��

MB � y� y� � y� y�� �

MB �

���

joules�meter� �

���

newtons�meter�

����

� ����� ���atm � ���� atmospheres�

For any�

S��

S� ����

S ��

ST� ����

S � �

ST�� If ���

S ��

ST� � �� then

S� ����

S

� �

ST�� From �/��

S� ���A � �m� The stress �S� n� is n� �S� ��� n � A � �m so

Sj� n� � ���ni�ijkmk � ���jiknimk � ��� n� �m�j� Thus

�S� n� ��

n� �m�

The stress is directed along lines of latitude about the north pole �m� points west�

and has magnitude equal to the sine of the colatitude of n relative to �m�

�� a�

, � ��v � �y � � �v � y�F �

���v�v� �

S

�� y �F � � �

S � y

� � �Dt�� � ��� �

S � y����Dt�v � �� �S

�� y � �f � y�

���

n4

n5^

^

^

^

^

x

z

y

^

^

n3

n1

n2

Figure Ex���

b� If �f � � and �v � �� then �� �S� �� so ��� �S� � y � � � ��S � y� � ��

c� �F � � �

S � y� The double arrows show ni��

S on the �ve surface sections� Then

on S�� n� � �F � y ��� �

S � y�� � y� �S � y � �

� n��

S

�� y � �

on S�� n� � �F � � z ��� �

S � y�� �

�� z� �S

�� y � �

� n��

S

�� y � �

on S�� n� � �F � � y ��� �

S � y�� �

�� n��

S

�� y � � above

� below

on S�� n� � �F � z ��� �

S � y�� �

� n��

S

�� y � �

on S�� n� � �F � y ��� �

S � y�� �

� n��

S

�� y � ��

Thus the �ow lines are as in �gure Ex���

���

Figure Ex���

Exercise ��

Geological evidence indicates that over the last � million years� the west side of the San

Andreas fault has moved north relative to the east side at an average rate of about �

cm�year� On a traverse across the fault� Brune� Henyey and Roy ������ �J�G�R� ���

��� found no detectable extra geothermal heat �ow due to the fault� They estimate that

they would have detected any anomalous heat �ow larger than ��� ���� watts�meter��

Use �������� to obtain an upper bound on the northward component of the average stress

exerted by the material just west of the fault on the material just east of the fault� Assume

that the fault extends vertically down from the surface to a depth D� and that the shape

of the heat �ow anomaly due to the fault is a tent function whose width is D on each side

of the fault�

���

D

D D

North Pacific Plate

San Andreas Fault

east

American Plate

up

east

Extra heat flow due to fault, watts /m2

. .Figure Ex���

���

Solutions

�� We assume that the heat �ow is steady� Since we are interested only in averages� we

may assume that ��

S��� is constant on the fault� as is � � � �H���� D is probably � km

�see Brune et al� ����� so we may treat the fault as in�nitely long� Then �H lies in

east�west vertical planes� Consider a length L along the fault� The amount of heat

produced by this section of the fault in LD� � � �H���� It all �ows out of the surface of

the earth in a rectangle of length L along the fault and width D across the fault� the

pro�le being the triangular one sketched as the bottom �gure in the exercise� The

area under that triangle is Dh where h is the triangle�s height� so the total heat �ow

out in the length L and width D is LDh� This must equal LD� � � H��� in the steady

state� Thus h � � � � H���� Since no anomaly was detected� h � �� � ����w�m��

Thus � � � H��� � � � ��S��� � ��v��� � ��� ����w�m�� Let x be a unit vector pointing

north� and let y � � point west� Then ��v��� � � x cm�yr � � � ���� x meters�yr

� � ���� x meters�sec� The heat production must be positive� so

� � � ���

S

���� x�� ����

�� ��� �����

or

� �� �� �S

���� x � ��� ��� pascals�

The material just west of the fault exerts on the material just east of the fault� a

stress � �� �S���� The northward component of this stress is � �� �S��� � x� and it lies

between � and �� bars ���� pascal ��

See Brune et al�� for a discussion of energy loss from �v � ��S��� � ��v��� by seismic

radiation� This exercise works only for the parts of the fault where most energy is

not radiated� i�e�� the parts of the fault which creep�

Exercise ��

Below are the Eulerian descriptions of three motions of continua� In these descriptions�

f x� y� zg is a right�handed orthonormal basis for physical space P � � is a constnat real

��

number� and �* is a constant vector� For each motion� �nd the local strain rate tensor��� �

its eigenvectors� and its eigenvalues� Also �nd the local rotation rate tensor���� and the

local angular velocity�

�w�

a� �vE��r� t� � ��r � � x x� y y� � ��x x� y y�

b� �vE��r� t� � ��r � � y x� � �y x

c� �vE��r� t� � �*� �r �

Exercise ��

Let �v�#�� �

#�� be the Eulerian descriptions of the velocity� local strain rate tensor and local

rotation rate tensor in a certain continuum� Let x�� x� x� be a right�handed orthonormal

basis in physical space P � Let position vectors be written �r � ri xi� Let i denote the

partial derivative with respect to ri� Take tensor components relative to x�� x�� x��

a� Show that i��jk� j

��ki �k �

�ij

b� Show that ij��kl �jk �

�li �kl��ij �li �

�jk� ��

c� Show that if��� � P P � P is any tensor�valued function of �r such that everywhere

��ij�

��ji and b� holds� then a function

���� P P � P can be found which satis�es

��ij� � �

�ji and also a� above� Show that this function is unique up to an additive

constant tensor

��

* which is antisymmetric�

d� Show that if��� is the function in c�� there is a velocity function �v � P P whose local

strain rate tensor is��� � Show that any two such functions �v� and �v� di�er by the

velocity �eld of a rigid motion� that is� there are �xed vectors �* and �V such that

for all �r � P � �v���r�� �v���r� � �V � �*� �r�

Hint for c� and d�� Suppose gi � P R for i � �� � � You know from elementary

vector calculus that there is a �potential function� f � P R such that gi � if i�

igj � jgi everywhere� and if this is so� then f��r� � f���� �R �r� drigi �line integral��

���

Exercise ��

Use the notation beginning on page �� for discussing small disturbances of an HTES of

an elastic material whose reference state is HTES��

a� Using only ����

S�� B���

W �� ���

��

Q�� explicitly express U � U� as a function of �N and

��

G� correct to second order �i�e�� including the terms ��N��� �N��

G and ��

G ��

G��

b� The tensor

��

F �� is called the isentropic or adiabatic sti�ness tensor because if �N � �

�no change in entropy� then

��

S� ��

G hi��

F �

Find the isothermal sti�ness tensor

��

F ��� the tensor such that if �� � � then

��

S� ��

G hi��

F �� �

Express

��

F �� in terms of

��

F ���

W � B�� ���

Exercise �

Using the notation on page ��� give the results of a�� b� when HTES� is isotropic�

Solutions

� a�

��vE � � � x x� y y� ����vE

�T�

so���� � � x x� y y� �

eigenvectors and eigenvalues being � x� ��� � y����� � z���������

��� so

��� ���

���

b�

��vE � � y x����vE

�T� � x y�

��� �

� x y � y x� �

����

� y x� x y�

eigenvectors and eigenvalues of��� are�

x� yp��

��

� x � yp

���

�� � z� �� �

�w��

Ahi

����

� y � x� x� y� � ��

� z

c� Let�

*� A � �*� Then

�vE � Ahi�*�r � ��* � A � �r� � �

* ��r � �r� �* �

Then ��vE ��

*����vE

�T� � �

*� so����

�� �so all eigenvalues are � and any vector is

an eigenvector�����

*��

��� �*�

� a�

#�jk ��

�jvk � kvj�

#�ij ��

�ivj � jvi�

k #�ij ��

�kivj � kjvi�

Replace �ijk� by �kij� and this becomes

j #�ki ��

�jkvi � jivk� �

i #�jk ��

�ijvk � ikvj� �

Comparing these three equations gives a��

���

b� il #�jk � li #�jk so� by a��

i �j #�kl � k #�lj� � l �j #�ki � k #�ij�

or� since #�kl � #�lk and ij � ji�

ji #�kl � ik #�lj � ki #�li � lj #�ik � ��

Now make the replacements j i� i j� and one proves b��

c� Fix i and j� a� can be written

��� k #�ij � i #�jk � j #�ki�

Then b� implies kl #�ij � lk #�ij� which are exactly the integrability conditions for

�/�� They insure that the line integralZ �r

��drk �i #�jk � j #�ki�

will be independent of path� Thus the solution of �/� exists� and has the form

#�ij��r� � � #�ij���� �Z �r

��drk �i #�jk � j #�ik� �

The integral is antisymetric in ij� If #�ij���� is chosen antisymmetric in ij� #�ij��r�

will be antisymmetric for all �r�

d� Let #�ij��r� be given as in c�� We want to try to solve

ivj � #�ij � #�ij�

This will be possible� and the solution will be

vj��r� � Vj �Z �r

��dri � #�ij � #�ij� �

as long as for each �xed j the integrability condition kivj � ikvj is satis�ed�

i�e��

�y� k � #�ij � #�ij� � i � #�kj � #�kj� �

But #�ij was obtained by solving �/� on the preceding page� Substituting �/� in �y�makes �y� an identity�

���

� a�

U � U� � �NNU � ��

G hi��GU

��

��N�� �NU � �N�

G hi��GNU

��

���

G ��

G

�h�i��

G��GU � third order terms

� �N�� � ��

G hi�

S�

���

��N��B�

��N��

G hi�

W�

���

���

G ��

G

�h�i

��

Q�

��� third order term�

b�

�� � �NB� � ��

G hi�

W �

��

��

S � �N�

W � ���

G hi��

F � �

If �� � �� then �N � �� �

G hi�

W�

��B�� so

��

S� ��

G hi�� ��

F � ��

W �

W �

��B�

A �

Therefore��

F ���

��

F� ��

W �

W �

��B��

�� In an isotropic material�

S� � �p��

I�

W � � �� �

I

Fijkl � �ij�kl � � ��ik�jl � �il�kj�

Qijkl � Fijkl � �ijS��kl � �jkS

��il

� Fijkl � p� ��jk�il � �ij�kl�

Qijkl � �� p�� �ij�kl � � �ik�jl � ��� p�� �jk�il

���

a�

U � U� � �N�� � p���

�tr �

G

��

��N��B�

� �

����N�

�tr �

G

��

��

n�� p��

�tr �

G

��� � �

G hi� �

G

���� p�� ��

G hi� �

GTo

� third order terms�

b�

��ij�kl � �� ��ik�jl � �il�jk�

� �ij�kl � � ��ik�jl � �il�jk�� ��

��B��ij�kl�

Thus �� � �� � � � �����B��