Here are some notes on vector and dyadic tensor notation similar towhat I will be using in class, with just a couple of changes innotation.

Notes = Dowling

∧ = ×   British notation for Cross-Product vs. American (Dowling’s)

A =A  Notation for Vector vs. Standard (Dowling’s).

B =B Notation for Dyadic Tensor vs. Standard (Dowling’s).

With these notes you should be able to make sense of the expressionsin the vector identities pages below that involve the dyadic tensor T.

(Notes courtesy of Peter Littlewood, University of Cambridge.)

The vector identity pages are from the NRL Plasma Formulary – orderyour very own online or download the entire thing at:

http://wwwppd.nrl.navy.mil/nrlformulary/ --JPD

VECTOR IDENTITIES4

Notation: f, g, are scalars; A, B, etc., are vectors; T is a tensor; I is the unitdyad.

(1) A ·B×C = A×B ·C = B ·C×A = B×C ·A = C ·A×B = C×A ·B

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

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

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

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

(6) ∇(fg) = ∇(gf) = f∇g + g∇f

(7) ∇ · (fA) = f∇ ·A + A · ∇f

(8) ∇× (fA) = f∇×A +∇f ×A

(9) ∇ · (A×B) = B · ∇ ×A−A · ∇ ×B

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

(11) A× (∇×B) = (∇B) ·A− (A · ∇)B

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

(13) ∇2f = ∇ · ∇f

(14) ∇2A = ∇(∇ ·A)−∇×∇×A

(15) ∇×∇f = 0

(16) ∇ · ∇ ×A = 0

If e1, e2, e3 are orthonormal unit vectors, a second-order tensor T can bewritten in the dyadic form

(17) T =∑

i,jTijeiej

In cartesian coordinates the divergence of a tensor is a vector with components

(18) (∇·T )i =∑

j(∂Tji/∂xj)

[This definition is required for consistency with Eq. (29)]. In general

(19) ∇ · (AB) = (∇ ·A)B + (A · ∇)B

(20) ∇ · (fT ) = ∇f ·T+f∇·T

4

Let r = ix + jy + kz be the radius vector of magnitude r, from the origin tothe point x, y, z. Then

(21) ∇ · r = 3

(22) ∇× r = 0

(23) ∇r = r/r

(24) ∇(1/r) = −r/r3

(25) ∇ · (r/r3) = 4πδ(r)

(26) ∇r = I

If V is a volume enclosed by a surface S and dS = ndS, where n is the unitnormal outward from V,

(27)

∫V

dV∇f =

∫S

dSf

(28)

∫V

dV∇ ·A =

∫S

dS ·A

(29)

∫V

dV∇·T =

∫S

dS ·T

(30)

∫V

dV∇×A =

∫S

dS×A

(31)

∫V

dV (f∇2g − g∇2

f) =

∫S

dS · (f∇g − g∇f)

(32)

∫V

dV (A · ∇ × ∇×B−B · ∇ × ∇×A)

=

∫S

dS · (B×∇×A−A×∇×B)

If S is an open surface bounded by the contour C, of which the line element isdl,

(33)

∫S

dS×∇f =

∮C

dlf

5

(34)

∫S

dS · ∇ ×A =

∮C

dl ·A

(35)

∫S

(dS×∇)×A =

∮C

dl×A

(36)

∫S

dS · (∇f ×∇g) =

∮C

fdg = −

∮C

gdf

6

EXAMPLES CLASS IN MATHEMATICAL PHYSICS

V� Introduction to Tensors ��������

Commentary� This is the �rst of two sheets on tensors �the second will come in class

VIII�� Tensors have many applications in theoretical physics� not only in polarization prob�

lems like the ones described below� but in dynamics� elasticity� uid mechanics� quantum

theory� etc�� The mathematical apparatus for dealing with tensor problems may not yet

be familiar� so this sheet contains quite a high proportion of formal material�

Section A introduces the concept of a tensor �in this case� a linear operator that

transforms one vector into another� in the physical context of studying the polarisation of

an aspherical molecule�

Section B summarizes various mathematical and notational ideas that are useful in

physics these include su�x notation� and concepts such as the outer product of two vectors�

and projection operators� �These are mathematical objects which take the projection or

component of a vector along a particular direction��

Section C returns to a physical problem �also about polarization� and leads to a dis�

cussion of principal axes� eigenvectors and eigenvalues of tensor quantities�

Section D introduces the properties of the antisymmetric third�rank tensor �ijk which

can be used to derive many results in vector calculus much more rapidly than by other

methods� This ground will be covered again in sheet VIII so it doesn�t matter if you run

out of time�

A� Introduction

We often �nd vectors that are linearly related to other vectors� for instance the polar�

isation P induced in a medium by applying a �eld E

P ���E ���

Here P and E are parallel� Their cartesian components are related to each other by

the same constant of proportionality ��� for each component� �Here we have denoted

vectors by underlinings as one would do if writing this out by hand� Usually books denote

vectors by bold type in the rest of this sheet we treat the two notations as completely

interchangeable��

It is possible to imagine situations where the medium is anisotropic and the response

factor ��� in ��� is di�erent for components in di�erent directions� The two vectors P and

E are no longer necessarily parallel and what connects them is called a tensor� We will

use simple model of molecular polarisability to show the underlying physics and introduce

the ideas of tensors�

A�� An Electrostatics Problem� The diacetylene molecule� H�C�C�C�C�H is

highly polarisable along its long axis� pk �kEk being the induced dipole for a �eld Ek

applied along the molecule� The response to perpendicular �elds is p� ��E� where

�� � �k� Let the cylindrical molecule point in the ��� �� �� direction and apply a �eld in

the x direction� E Ex�x

�i� Prove�

px ��k � ���Ex�� ���

py ��k � ���Ex�� ���

pz � ���

�

Hence one sees that p and E are not parallel and in general we must write

p � �E ���

where � is the polarisability tensor� Tensors are written with a double underline �as

matricies often are� or in books sometimes as �doubly bold� �bold sans serif typeface�

characters� In component form ��� is

px �xxEx � �xyEy � �xzEz ���

py �yxEx � �yyEy � �yzEz ���

pz �zxEx � �zyEy � �zzEz ���

�ii� What are the coe�cients �ij explicitly for the above example� Write � as a matrix of its

components� Note that� although the tensor � is the same physical entity in all coordinate

systems �it relates p with E in whatever basis these are expressed�� its components �ij

depend on the coordinate system used�

Remark� If many molecules are present with the same orientation� P is �p where there

are � molecules per unit volume� A �hypothetical� diacetylene solid might consist of such

rods arranged in an array� all directed along ��� �� ��� Then the macroscopic � �to replace

� in ���� would �neglecting internal �eld corrections� be ���

B� Miscellaneous Ideas and Notations

B�� Su�x notation� Commonly� su�ces such as i and j are dummy symbols and

when repeated in an expression they are to be summed over this is the Einstein Convention�

For instance ��� to ��� can be succinctly written as

pi �ijEj

�

which is Eq���� rewritten in su�x notation� �Note the order of the indices on the right��

Thus we have written the vector p simply as pi and it will be clear from context that a

vector is intended and not simply one of its components� Likewise � can be denoted �ij �

In su�x notation the dot product �also called scalar product� or inner product� of two

vectors a � b is written as aibi� The operation of a tensor on a vector also involves an inner

product �sum over repeated indices� as described above� The unit tensor �ij by de�nition

transforms any vector into itself� �ijvj vi�

B�� The outer product� Consider the object Qij de�ned as

Qij � ai bj ���

where a and b are vectors�

�i� Show using su�x notation that if p is a vector� Q � p is also a vector �nd its magnitude

and direction� �Note thatQ�p � Qijpj �� We see thatQij is a linear operator that transforms

one vector p into another it is therefore a tensor� This de�nes the outer product or dyadic

product of two vectors�

In non�su�x notation �sometimes called dyadic notation�� Eq���� is written as

Q ab ����

or Q a b� where there is no dot between the two vectors� In this notation we have

�ab� � c a�b � c�� prove this using su�x notation if you did not do so already above�

This notation is used in various areas of mathematics and physics� but in general su�x

notation is more versatile and can always be used instead�

�ii� We can write the tensor � in Q� A� in the following form�

� �xx �x �x� �xy �x �y � �xz �x �z � � other terms ����

Check that ���� indeed agrees with ��� to ��� by �nding the result of applying a �eld in

the x�direction� E Ex�x�

p � �E � � � �xEx ��xx� �yx� �zx�Ex �

�

�iii� Projection operators� In dyadic notation� the identity tensor �ij is often written

as the �unit operator� �� Consider the action on a vector v of the operators �u�u and

�� �u�u� where �u is a unit vector� Show that these perform the job of resolving the vector

v into vectors vk and v�� respectively parallel and perpendicular to �u� Prove that these

�projection operators� are idempotent� �The de�nition of an idempotent operator G is that

GG G�� Show that the polarizability tensor in problem A� can be written

� �k�u�u� ����� �u�u�

where �u ��� �� ���p��

C� Use of Principal Axes�

C�� A �xed object is placed in a uniform electric �eld of � kV� m� When the �eld

is in the x direction� it acquires a dipole moment with components �� �� and �� ����� C

m in the x� y� z directions� For the same �eld strengths in the y� z directions� the dipole

moments are ��� �� ��� ����� and ��� �� ��� ����� C m respectively�

�i� Write down the components of the tensor that describes the polarisability of the object

�using the x� y� z coordinate frame��

�ii� What is the torque T p �E on the object when it is placed in an electric �eld ofp�

kV�m in the ������� direction� �Ans� �������� ����� Nm��

�iii� Suppose there are �eld directions �u��v� �w for which there is no torque on the object

what relation do E and p then have� Show that �u� �v� and �w are the eigenvectors of ��

�iv� Show that� if we use as a coordinate system the unit vectors �u��v� �w there are no

o��diagonal elements of the tensor � when it is written as a matrix of components� The

three diagonal elements� �� � and � are called the principal polarisabilities of the object

and are the eigenvalues of �� For this problem� one eigenvalue is �� ������ �nd the

�

other two� What are the vectors �u��v� �w in terms of �x� �y� �z�

�v� In this example� � is symmetric and hence has real eigenvalues and orthogonal eigen�

vectors� An energy argument shows this to be true for the polarizability of any object

the �u��v� �w coordinate system can hence always be found and is called the principal frame

� often a very convenient frame to think about the physics in� The same applies to any

problem involving a symmetric tensor�

�vi� In a general coordinate system� the decomposition of � into principal directions �u��v�

�w can be accomplished using projection operators�

� ��u�u� ��v�v � ��w�w ����

This is a relation between tensors� so if it holds in one co�ordinate system it also holds in

any other� Prove it for a convenient choice of co�ordinates system �and hence generally��

�vii� Eq����� expresses in a very direct way the operation of a symmetric tensor � on a

vector b� To �nd � � b� the recipe is� ��� resolve b into components along the principal

axes ��� multiply each component by the corresponding eigenvalue ��� add to form the

resultant� Con�rm from Eq����� that this is precisely what operating with � on b does�

C�� Force on a Dipole�

�i� By using su�x notation� and the ordinary chain rule for derivatives� prove the vector

relation

r � �A� r �A�A � r

Derive a similar result for r�A �B�� �It�s best to leave the answer in su�x notation��

�ii� In a uniform electric �eld� E� a dipole p experiences a couple p� E but no net force�

Show that in a non�uniform �eld� a dipole experiences a net force

Fi �Ei

�xjpj ����

and explain why �Ei

�xjis a tensor� �This is very general the gradient of a vector is a second�

rank tensor��

�

�iii� A more general expression for the force on a permanent dipole� derived from the

energy� is F grad�p�E�� or�

Fj �

�xjpiEi �

Show that this is equivalent to Eq���� only in the absence of a time�varying magnetic �eld�

�iv� A speck of matter has no permanent dipole moment� but the �eld E induces a moment

p in it� Under what conditions can we write �a� pi �ijEj �b� pi �Ei� Write down

the force in each case� Under what conditions can this be written F � grad�E���� �

D� The antisymmetric three�tensor �ijk�

A tensor with two indices such as �ij is called a tensor of second rank� and its linear

operation on a vector gives another vector� A vector is technically a tensor of rank one a

scalar of rank zero� Tensors of higher rank can also be de�ned� For example a tensor of

rank four� Qijkl acts on a rank two tensor to give another rank two tensor� Bij QijklAkl

�with summation over repeated indices�� For tensors higher than rank two� dyadic notation

becomes extremely cumbersome and su�x notation is essential�

D�� An important and useful tensor of rank three is the fully antisymmetric three

tensor denoted �ijk� This operates on either a second rank tensor or two vectors to give a

vector� For i � j � k it is de�ned as �ijk � if ijk is a cyclic permutation of x� y� z �e�g��

yzx� and �� if the order is anticyclic �e�g�� xzy�� If any two of ijk are the same then

�ijk � �e�g�� �xxy ��� The tensor �ijk is very useful for writing down a vector product

in su�x notation� Show that

�A �B�i �ijk Aj Bk ����

The following relation

�ijk �klm �il �jm � �im �jl ����

�

is well worth remembering �you should come across this in maths lectures� � It saves a

great deal of time when manipulating vector identities such as

A � �B � C� �ijk Aj �klmBl Cm

Aj Bl Cm ��il �jm � �im �jl�

BiAj Cj � CiAj Bj �A � C�B � �A �B�C

which is a familiar result normally derived by far more tedious means�

The result Eq����� is especially helpful when dealing with vector identities involving

the r operator� Because �ijk is a constant tensor� it commutes with r� and expressions

can be re�ordered accordingly� For example

r � �E �H� � ri��ijk Ej Hk� ����

�ijkri�Ej Hk� �ijk�EjriHk �HkriEj � ����

�Ej �jikriHk �Hk �kijriEj ����

�E � r �H �H � r � E ����

which is a result used in in connection with Poynting�s vector in electromagnetism�

�i� Make sure you understand each step of this derivation�

�ii� Show� using similar methods� that

r� �r�A� r�r �A��r�A

r� �A� r�A�A � r

These relations are useful in electromagnetism and other topics in IB� and in the parts II

and III courses�

�

EXAMPLES CLASS IN MATHEMATICAL PHYSICS

VIII� Further Applications of

Tensor and Vector Calculus ���������

Commentary� This is the second sheet on tensors� It follows on from Sheet �� Section D

of that sheet �on uses of the antisymmetric ��tensor �ijk� is reprinted here as an appendix

and you should complete it before starting the main part of this sheet if you didnt do so

last time�

As on sheet � we use underlining and bold notation interchangeably� Su�x notation

and the summation convention are also used� Carets ��� denote unit vectors�

In Sections A and B tensor ideas are used to work through some results in rigid

body dynamics� You should have covered this material in IB dynamics courses and the

formal development given here is good practice and will help deepen your understanding

of moments of inertia etc� as well as of tensors�

Section C introduces the concepts of stress and strain tensors in a solid� The linear

relation between these is governed by a fourth rank tensor whose properties are investi�

gated�

Section D contains problems on the transformation properties of tensors in mathemat�

ical courses you may have seen tensors de�ned in terms of these transformation properties�

A� Rigid Body Rotations

A rigid body rotates with angular velocity � about an axis through its centre of mass

at the origin �� The vector � points along the axis of rotation and determines the velocity

of rotation at r according to v � � � r� �Make sure you can derive this if you are not sure

where it comes from��

�i� Consider the body as a collection of point masses �mr with position vectors r relative

to the centre of mass ��

The tensor I is called the inertia tensor of the body� It is given by

I �Xr

r���� �r�r

��mr ���

Show using Eq���� that the angular momentum L and the kinetic energy T of the body

are

L � I � � ���

T ��

�� � I � � �

�

��i Iij �j ���

�You will need to expand the vector triple product f � �g � h� � see Appendix��

Thus I speci�es the linear dependence of angular momentum Li on the angular velocity

�j �via Li � Iij �j Eq����� and replaces the single number the moment of inertia that

would be enough for a very symmetrical body such as a sphere�

�ii� Find the components of Iij in a Cartesian system that is the usual basis set �x �y �z�

�Answer Ixx �P

r�y� � z���mr Ixy � �

Pr�xy��mr etc���

�iii� Iij is a symmetric tensor that is I�v � v�I for any vector v which should be clear

from the de�nition ���� Deduce it also from the matrix of components of Iij found in �ii��

�iv� Take the continuum limit in Eq���� to obtain formally as an integral the inertia tensor

of a body of variable density ��r��

�v� Write down the inertia tensor of a body consisting of eight masses M at the corners of

�

a light cubic frame of side a� Choose the coordinate system most convenient to you� What

is the moment of inertia of this cube� Why can we talk of the moment of inertia when we

strictly speaking have a tensor�

�In fact even for an asymmetric body one sometimes speaks of the moment of inertia about

some speci�c axis �n say� This is written as �n�I��n and gives the ratio of the �n component

of angular momentum �L��n� to the �n component of angular velocity ����n�� Since in general

L has components which are not parallel to �n this does not o�er a complete description��

�vi� Angular velocity� A body rotates about an axis through its centre of mass with an

angular velocity �� The velocity of a point in the body is v � � � r� Show that one can

equally well express it as v � � � r where � is an antisymmetric tensor with components �� ��z �y�z � ��x��y �x �

����

�vii� Show that any vector W constant with respect to axes �xed in the body changes

with time in a space��xed frame by an amount dW in time dt

dW � ��dt� �W

and hence that if W is not body��xed but has a rate of change �dW�dt�body the vector

W in the space��xed frame obeys

�dW�dt�space � �dW�dt�body � � �W � ���

B� Principal Axes and Precession

We now use the above results to solve simply a di�cult problem the rotational motion

of an asymmetric body�

�

�i� What is the inertia tensor of four point masses M �xed at the corners of a massless

square frame of side a� What are the principal axes and principal moments of inertia�

Why are two of the axes indeterminate�

This is an example of a nontrivial body �unlike the sphere or cube� where the principal

moments of inertia are not all the same� Now L need not be parallel to �� for a body

whose angular momentum is conserved the result is free precession of the angular velocity

�see �iii� below��

�ii� It is clear from ��� that in general I for a body is constant in body��xed axes and so

for an arbitrary motion the inertia tensor I�sp�ij in space��xed axes is not constant �unless

trivially it is proportional to �ij�� Show that in general the motion of a nontrivial body

whose angular momentum L�sp� is conserved �in a space��xed frame� involves an angular

velocity � which changes in time� �Hint consider L�sp�i � I

�sp�ij �j ��

�iii� The problem of �nding � for a body of �xed angular momentum �not parallel to a

principal axis� is best solved by adopting a body frame where the inertia tensor is constant�

Show from ��� that in the body frame L obeys

dL�dt � �� � L ���

�iv� Deduce that

Iij ��j � ��ijk �j Ikl �l

and show that by taking as the body��xed frame the principal frame of Iij this can be

simpli�ed to

I� ��� � ��� ���I� � I��

I� ��� � ��� ���I� � I��

I� ��� � ��� ���I� � I��

where I� I� I� are the principal moments of inertia�

�v� For the inertia tensor of the square frame found in question �i� I� � I� �such a body

�

is known as a �symmetric top��� Prove for this case that � precesses in the body frame

according to

�� � const� � �� � A cos��t� � �� � A sin��t�

where � ��I��I��

I��� and so con�rm that � is not a constant vector unless it happens to

point along one of the principal axes�

C� Stress and strain tensors�

C �� Consider a small area element dS embedded at a point r in an isotropic liquid�

If the material is subject to pressure P there is a normal force on each side of the surface

element PdS� These point in opposite directions so the net force on the surface element

is zero � which is as it must be since the element has no mass and would otherwise be

subject to in�nite acceleration�

In the case of a solid there can also be a shear force exerted in equal and opposite

directions on the two sides of the element� such a force lies in the plane of the element

rather than normal to it� In the general case we can write

dF � � � dS ���

where � is the stress tensor at the point r� For example the element �xz of � determines

the x�force per unit area on an element of surface with its normal pointing in the z�

direction dFx � �xzdSz� For a surface element of general orientation dFx � �xxdSx �

�xydSy � �xzdSz�

�i� Show that a small box of sides dx� dy� dz experiences a couple ��xz� �zx�dx dy dz about

the y axis� The corresponding moment of inertia is of order �dx���dz��dy� Explain why

this means that � must be symmetric� Show also that if ��r� is a function of position

�

then the net force on a �nite piece of material is given by

Fi �

Z��ij�rj

dV � ���

where �in a conventional notation� ��rj

� rj �

�ii� Strain� Suppose a solid material is subjected to a small deformation r� r�u�r�� So

long as the deformation depends continuously on position we may write

dui ��ui�rj

drj � ���

Let us now decompose the tensor Uij ��ui

�rjinto its symmetric and anti�symmetric parts

Uij � eij � Rij

eij ��

�

�Uij � Uji

�Rij �

�

�

�Uij � Uji

�Show that e is symmetric and R antisymmetric� Show also that R corresponds to a pure

rotation �i�e� without distortion� of the bit of matter near r� �Hint compare Eq���� in

section A�� What type of deformation does e represent� Why may we assume that the

stress tensor � depends on the strain tensor e but has no dependence on R�

�iii� Elasticity� For small deformations we expect the stress and strain to be linearly

related but since both are described by second�rank tensors the relation between them

generally involves a tensor of the fourth rank

�ij � Eijkl ekl ����

which is the generalization of Hookes law to an arbitrary three�dimensional material�

Since � is symmetric Eijkl must be symmetric under exchange of i and j� various other

symmetries can be found from considering the form of the stored elastic energy but even

when these are taken into account Eijkl has �� independent components�

�

�iv� Isotropic elastic solids are much simpler however and have only two independent

constants� These are the Youngs modulus Y and Poissons ratio � The de�nition of these

is that in the principal axes of �ij the following relation applies

Y exx � �xx � ��yy � �zz� ����

with similar relations for eyy� ezz� all o��diagonal elements of e are zero in this coordinate

system� By rearranging Eq����� obtain the result

Y exx � �� � ��xx � �ii

�where x indicies are not summed over but �ii is summed� with similar relations for yy

and zz components� �Note that �ii � Trace��� under the summation convention�� Show

that by adding Eq����� to its analogues

�ii �Y eii�� �

Combine the above two results to give expressions for �xx etc in terms of the elements of

e and hence derive the following result in the principal frame

�ij �Y

� �

�eij �

�� ��ijTr�e�

��

�Bear in mind that eij � � for i �� j in this frame�� Hence derive the general �frame

independent� form for the elasticity tensor in an isotropic solid

Eijkl �Y

� �

��ik �jl � �il �jk

��

�

��� ��

��ij �kl

�

C �� �i� Apply Eijkl to suitable deformations to recover expressions for the bulk

modulus and shear modulus of an isotropic solid in terms of Y and �

�ii� Show that the displacement u inside a homogeneous isotropic solid subject to

external forces only at its surface obeys

grad divu� ��� ��r�u � ��

�

D� Transformation Laws�

D �� Here we consider transformations of vectors and tensors between two sets of base

vectors each of which is orthonormal� We call our two sets of base vectors e�� e�� e� and

e��� e�

�� e�

� so that we can use su�x notation e�g� the fact that the e basis is orthonormal

can be written

ei � ej � �ij ����

The unit vectors e�i are linear combinations of the ei

e�i � Tijej � ����

Suppose that x � xiei � x�ie�

i is any vector and y � Ax� Prove that

�i� e�i � ek � Tik

�ii� The inverse transformation ei � T��ike�k has T��

ik� Tki

�iii� x�i � Tikxk

�iv� xi � Tkix�

k

�v� y�i � TijTlkAjkx�

l

This means you have found the components of the tensor A in the rotated coordinate

system

A�

il � TijTlkAjk

Indeed this property under orthogonal coordinate transformations T is mathematically

often taken as the de�ning property of a tensor A� For tensors of higher rank than �

one factor of the transformation matrix T is needed for each index� Transformations with

det�T � � � are usually called proper rotations and those with det�T � � �� improper rota�

tions� Show that these are the only possible values of the determinant for transformations

between orthonormal bases as de�ned above�

�

Appendix� The antisymmetric three�tensor �ijk�

�Reprise of Sheet V section D��

A tensor with two indices such as �ij is called a tensor of second rank and its linear

operation on a vector gives another vector� A vector is technically a tensor of rank one� a

scalar of rank zero� Tensors of higher rank can also be de�ned� For example a tensor of

rank four Qijkl acts on a rank two tensor to give another rank two tensor Bij � QijklAkl

�with summation over repeated indices�� For tensors higher than rank two dyadic notation

becomes extremely cumbersome and su�x notation is essential�

�� An important and useful tensor of rank three is the fully antisymmetric three tensor

denoted �ijk� This operates on either a second rank tensor or two vectors to give a vector�

For i �� j �� k it is de�ned as �ijk � � if ijk is a cyclic permutation of x� y� z �e�g� yzx�

and � �� if the order is anticyclic �e�g� xzy�� If any two of ijk are the same then �ijk � �

�e�g� �xxy � ��� The tensor �ijk is very useful for writing down a vector product in su�x

notation� Show that

�A �B�i � �ijk Aj Bk ����

The following relation

�ijk �klm � �il �jm � �im �jl ����

is tedious to prove �using components � you should come across this in maths lectures� but

is worth remembering� It saves a great deal of time when manipulating vector identities

such as

A � �B � C� � �ijk Aj �klmBl Cm

� Aj Bl Cm ��il �jm � �im �jl�

� BiAj Cj � CiAj Bj � �A � C�B � �A �B�C

�

which is a familiar result normally derived by more tedious means�

The result Eq����� is especially helpful when dealing with vector identities involving

the r operator� Because �ijk is a constant tensor it commutes with r and expressions

can be re�ordered accordingly� For example

r � �E �H� � ri��ijk Ej Hk� ����

� �ijkri�Ej Hk� � �ijk�EjriHk �HkriEj � ����

� �Ej �jikriHk �Hk �kijriEj ����

� �E � r �H �H � r � E ����

which is a result used in in connection with Poyntings vector in electromagnetism�

�i� Make sure you understand each step of this derivation�

�ii� Show using similar methods that

r� �r�A� � r�r �A��r�A

r� �A� � r�A�A � r

These relations are useful in electromagnetism and other topics in the IB and parts II and

III courses�

��