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

Notes = Dowling

∧ = × British notation for Cross- 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 expressions in 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 – order your 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 unit dyad.

(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 be written in the dyadic form

(17) T = Tij eiej Pi,j In cartesian coordinates the of a tensor is a vector with components

(18) ( T )i = (∂Tji/∂xj ) ∇· Pj [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 vector of r, from the to the 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 enclosed by a S and dS = ndS, where n is the unit outward from V,

(27) dV f = dSf Z ∇ Z V S

(28) dV A = dS A Z ∇ · Z · V S

(29) dV T = dS T Z ∇· Z · V S

(30) dV A = dS A Z ∇ × Z × V S

(31) dV (f 2g g 2f) = dS (f g g f) Z ∇ − ∇ Z · ∇ − ∇ V S

(32) dV (A B B A) Z · ∇ × ∇ × − · ∇ × ∇ × V

= dS (B A A B) Z · × ∇ × − × ∇ × S If S is an open surface bounded by the contour C, of which the element is dl,

(33) dS f = dlf Z × ∇ I S C

5 (34) dS A = dl A Z · ∇ × I · S C

(35) (dS ) A = dl A Z × ∇ × I × S C

(36) dS ( f g) = fdg = gdf Z · ∇ × ∇ I − I S C C

6

EXAMPLES CLASS IN MATHEMATICAL

V Intro duction to

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

VI I I Tensors have many applications in theoretical physics not only in p olarization prob

lems like the ones describ ed below but in dynamics uid quantum

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

be familiar so this sheet contains quite a high prop ortion of formal material

Section A intro duces the concept of a tensor in this case a linear op erator that

transforms one vector into another in the physical contextofstudying the p olarisation of

an aspherical molecule

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

physics these include sux notation and concepts suchastheouter product of twovectors

and pro jection op erators These are mathematical ob jects which take the projection or

comp onent of a vector along a particular direction

Section C returns to a physical problem also ab out p olarization and leads to a dis

cussion of principal axes eigenvectors and eigenvalues of tensor quantities

Section D intro duces the prop erties of the antisymmetric third tensor which

ij k

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

metho ds This ground will be covered again in sheet VI I I so it do esnt matter if you run

out of time

A Intro duction

Weoften nd vectors that are linearly related to other vectors for instance the p olar

induced in a medium by applying a eld E isation P

P E

and E are parallel Their cartesian comp onents are related to each other by Here P

the same constant of prop ortionality for each comp onent Here we have denoted

vectors by underlinings as one would do if writing this out by hand Usually b o oks denote

vectors by b old typ e in the rest of this sheet we treat the two notations as completely

interchangeable

It is p ossible to imagine situations where the medium is anisotropic and the resp onse

factor in is dierent for comp onents in dierent directions The twovectors P and

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

use simple mo of molecular p olarisability to show the underlying physics and intro duce

the ideas of tensors

A An Electrostatics Problem The diacetylene molecule HCCCCH is

highly p olarisable along its long axis p E being the induced dip ole for a eld E

k k k k

applied along the molecule The resp onse to p erp endicular elds is p E where

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

k

E x the x direction E

x

i Prove

p E

x x

k

p E

y x

k

p

z

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

E p

where is the p olarisability tensor Tensors are written with a double underline as

matricies often are or in b o oks sometimes as doubly bold b old sans serif typ eface

characters In comp onent form is

p E E E

x xx x xy y xz z

p E E E

y yx x yy y yz z

p E E E

z zx x zy y zz z

ii What are the co ecients explicitly for the ab ove example Write as a of its

ij

comp onents Note that although the tensor is the same physical entity in all co ordinate

systems it relates p with E in whatever these are expressed its comp onents

ij

dep end on the co ordinate system used

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

are molecules per unit volume A hyp othetical diacetylene solid might consist of such

ro ds arranged in an arrayall directed along Then the macroscopic to replace

in would neglecting internal eld corrections be

B Miscellaneous Ideas and Notations

B Sux notation Commonly suces such as i and j are dummy symb ols and

when rep eated in an expression they are to b e summed over this is the Einstein Convention

For instance to can be succinctly written as

p E

i ij j

which is Eq rewritten in sux notation Note the order of the indices on the right

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

i

can be denoted vector is intended and not simply one of its comp onents Likewise

ij

In sux notation the dot pro duct also called pro duct or inner productoftwo

vectors a b is written as a b The op eration of a tensor on a vector also involves an inner

i i

pro duct sum over rep eated indices as describ ed ab ove The unit tensor by denition

ij

transforms any vector into itself v v

ij j i

B The outer pro duct Consider the ob ject Q dened as

ij

Q a b

ij i j

where a and b are vectors

i Show using sux notation that if p isavector Q p is also a vector nd its magnitude

and direction Note that Qp Q p We see that Q is a linear op erator that transforms

ij j ij

one vector p into another it is therefore a tensor This denes the or dyadic

product of two vectors

In nonsux notation sometimes called dyadic notation Eq is written as

Q ab

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

ab c ab c prove this using sux notation if you did not do so already ab ove

This notation is used in various areas of mathematics and physics but in general sux

notation is more versatile and can always be used instead

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

x x x y x z other terms

xx xy xz

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

x the xdirection E E

x

p E x E E

x xx yx zx x

iii Pro jection op erators In dyadic notation the identity tensor is often written

ij

Consider the action on a vector v of the op erators uu and as the unit op erator

uu where u is a Show that these p erform the job of resolving the vector

v into vectors v and v resp ectively parallel and p erp endicular to u Prove that these

k

pro jection op erators are idempotent The denition of an idemp otent op erator G is that

GG G Show that the p olarizability tensor in problem A can be written

uu uu

k

p

where u

C Use of Principal Axes

C A xed ob ject is placed in a uniform electric eld of kV m When the eld



is in the x direction it acquires a dip ole momentwith comp onents and C

m in the x y z directions For the same eld strengths in the y z directions the dip ole

 

moments are and C m resp ectively

i Write down the comp onents of the tensor that describ es the p olarisability of the ob ject

using the x y z co ordinate

p

ii What is the T p E on the ob ject when it is placed in an electric eld of



kVm in the direction Ans Nm

iii Supp ose there are eld directions uv w for which there is no torque on the ob ject

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 co ordinate system the unit vectors uv w there are no

odiagonal elements of the tensor when it is written as a matrix of comp onents The

three diagonal elements and are called the principal p olarisabilities of the ob ject

  



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

other two What are the vectors uv 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 p olarizability of any ob ject

the uv w co ordinate system can hence always b e found and is called the principal frame

often a very convenient frame to think ab out the physics in The same applies to any

problem involving a

into principal directions uv vi In a general co ordinate system the decomp osition of

w can be accomplished using pro jection op erators

u u v v w w

  

This is a relation between tensors so if it holds in one coordinate system it also holds in

any other Prove it for a convenient choice of coordinates system and hence generally

vii Eq expresses in a very direct way the op eration of a symmetric tensor on a

vector b To nd b the recip e is resolve b into comp onents along the principal

axes multiply each comp onent by the corresp onding eigenvalue add to form the

on b do es resultant Conrm from Eq that this is precisely what op erating with

C on a Dip ole

i By using sux notation and the ordinary for prove the vector

relation

r A rA A r

Derive a similar result for rA B Its best to leave the answer in sux notation

ii In a uniform electric eld E a dip ole p exp eriences a couple p E but no net force

Show that in a nonuniform eld a dip ole exp eriences a net force

E

i

F p

i j

x

j

E

i

is a tensor This is very general the gradientofavector is a second and explain why

x

j

rank tensor

iii A more general expression for the force on a dip ole derived from the

p E F

i i j

x

j

Show that this is equivalent to Eq only in the absence of a timevarying magnetic eld

iv A sp eck of matter has no p ermanent dip ole moment but the eld E induces a moment

p in it Under what conditions can we write a p E b p E Write down

i ij j i i



the force in each case Under what conditions can this be written F gradE

D The antisymmetric threetensor

ij k

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

ij

op eration on a vector gives another vector Avector is technically a tensor of rank one a

scalar of rank zero Tensors of higher rank can also be dened For example a tensor of

rank four Q acts on a rank two tensor to give another rank two tensor B Q A

ij

ij k l ij k l kl

with over rep eated indices For tensors higher than rank two dyadic notation

b ecomes extremely cumb ersome and sux notation is essential

D An imp ortant and useful tensor of rank three is the ful ly antisymmetric three

tensor denoted This op erates on either a second rank tensor or two vectors to givea

ij k

vector For i j k it is dened as if ij k is a cyclic of x y z eg

ij k

yzx and if the order is anticyclic eg xz y If any two of ij k are the same then

eg The tensor is very useful for writing down a vector pro duct

xxy

ij k ij k

in sux notation Show that

A B A B

i j

ij k k

The following relation

jm im

ij k klm il jl

is well worth rememb ering you should come across this in maths lectures It saves a

great deal of time when manipulating vector identities such as

B C A B C A

m

j

ij k klm l

A B C

m

j jm im

l il jl

B A C C A B A C B A B C

i j j i j j

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

The result Eq is esp ecially helpful when dealing with vector identities involving

the r op erator Because is a constant tensor it commutes with r and expressions

ij k

can be reordered accordingly For example

H r E H rE

i j

ij k k

r E H E r H H r E

i j j i i j

ij k k ij k k k

E r H H r E

j i i j

jik k k kij

E r H H r E

which is a result used in in with Poyntings vector in

i Make sure you understand each step of this derivation

ii Show using similar metho ds that



r rA rrA r A

r A rA A r

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

and I I I courses

EXAMPLES CLASS IN

VI I I 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 is reprinted here as an app endix

ij k

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

last time

As on sheet we use underlining and b old notation interchangeably Sux notation

and the summation convention are also used Carets denote unit vectors

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

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

formal development given here is go o d practice and will help deep en your understanding

of moments of inertia etc as well as of tensors

Section C intro duces the concepts of and strain tensors in a solid The linear

relation between these is governed by a fourth rank tensor whose prop erties are investi

gated

Section D contains problems on the transformation prop erties of tensors in mathemat

ical courses you mayhave seen tensors dened in terms of these transformation prop erties

A Rigid Bo dy Rotations

A rigid b o dy rotates with angular velo city ab out an axis through its centre of mass

points along the axis of and determines the velo city at the origin The vector

of rotation at r according to v r Make sure you can derivethisifyou are not sure

where it comes from

i Consider the body as a collection of point masses m with p osition vectors r relative

r

to the centre of mass

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

X

I r rr m

r

r

Show using Eq that the angular L and the T of the body

are

L I

I I T

i ij j

You will need to expand the vector triple pro duct f g h see App endix

Thus I sp ecies the linear dep endence of L on the angular velo city

i

via L I Eq and replaces the single the that

j i ij j

would be enough for a very symmetrical b o dy such as a sphere

y z ii Find the comp onents of I in a Cartesian system that is the usual basis x

ij

P P

xy m etc y z m I Answer I

r r xy xx

r r

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

ij

from the denition Deduce it also from the matrix of comp onents of I found in ii

ij

iv Take the continuum limit in Eq to obtain formallyasanintegral the inertia tensor

of a body of variable density r

v Write down the inertia tensor of a b o dy consisting of eight masses M at the corners of

a light cubic frame of side a Cho ose the co ordinate system most convenienttoyou What

is the momentof inertia of this cub e Why can we talk of the moment of inertia when we

strictly sp eaking have a tensor

In fact even for an asymmetric b o dy one sometimes sp eaks of the moment of inertia ab out

some sp ecic axis nsay This is written as nIn and gives the ratio of the n comp onent

n Since in general of angular momentum Ln to the n comp onent of angular velo city

L has comp onents which are not parallel to n this do es not oer a complete description

vi Angular velo city A b o dy rotates ab out an axis through its centre of mass with an

The velo city of a point in the body is v r Show that one can angular velo city

equally well express it as v r where is an with comp onents

z y

z x

y x

vii Show that any vector W constant with resp ect to axes xed in the body changes

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

dW dt W

and hence that if W is not b o dyxed but has a rate of change dW dt the vector

body

W in the spacexed frame ob eys

W dW dt dW dt

space

body

B Principal Axes and

Wenow use the ab ove results to solve simply a dicult problem the rotational motion

of an asymmetric b o dy

i What is the inertia tensor of four p oint masses M xed at the corners of a massless

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 cub e where the principal

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

whose angular momentum is conserved the result is free precession of the angular velo city

see iii b elow

ii It is clear from that in general I for a body is constant in b o dyxed axes and so

sp

in spacexed axes is not constant unless for an arbitrary motion the inertia tensor I

ij

trivially it is prop ortional to Show that in general the motion of a nontrivial body

ij

sp

whose angular momentum L is conserved in a spacexed frame involves an angular

sp sp

which changes in time Hint consider L I velo city

j

i ij

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

principal axis is b est solved by adopting a b o dy frame where the inertia tensor is constant

Show from that in the body frame L ob eys

dLdt L

iv Deduce that

I I

ij j j

ij k kl l

and show that by taking as the b o dyxed frame the principal frame of I this can be

ij

simplied 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 cost A sint

 

I I 



is not a constant vector unless it happ ens to and so conrm that where



I



p oint along one of the principal axes

C Stress and strain tensors

C Consider a small area element dS emb edded at a point r in an isotropic liquid

If the material is sub ject to P there is a normal force on eachside of the surface

element PdS These p oint in opp osite 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

sub ject to innite acceleration

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

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

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

dF dS

is the stress tensor at the p oint r For example the element of determines where

xz

the xforce per unit area on an element of surface with its normal pointing in the z

direction dF dS For a surface element of general orientation dF dS

x xz z x xx x

dS dS

xy y xz z

i Show that a small b oxofsides dx dy dz exp eriences a couple dx dy dz ab out

xz zx

the y axis The corresp onding 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 p osition

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

Z

ij

F dV

i

r

j

r where in a conventional notation

j

r

j

ii Strain Supp ose a solid material is sub jected to a small r r ur So

long as the deformation dep ends continuously on p osition we maywrite

u

i

dr du

j i

r

j

u

i

into its symmetric and antisymmetric parts Let us now decomp ose the tensor U

ij

r

j

U e R

ij ij ij

e U U

ij ij ji

R U U

ij ij ji

is symmetric and R antisymmetric Show also that R corresp onds to a pure Show that e

rotation ie without distortion of the bit of matter near r Hint compare Eq in

section A What typ e of deformation do es e represent Why may we assume that the

stress tensor dep ends on the strain tensor e but has no dep endence on R

iii Elasticity For small deformations we exp ect the stress and strain to be linearly

related but since b oth are describ ed by secondrank tensors the relation between them

generally involves a tensor of the fourth rank

E e

ij

ij k l kl

which is the generalization of Ho okes law to an arbitrary threedimensional material

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

ij k l

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

when these are taken into account E has indep endent comp onents

ij k l

iv Isotropic elastic solids are much simpler however and have only two indep endent

constants These are the Youngs mo dulus Y and Poissons ratio The denition of these

is that in the principal axes of the following relation applies

ij

Ye

xx xx yy zz

are zero in this co ordinate with similar relations for e e all odiagonal elements of e

yy zz

system By rearranging Eq obtain the result

Ye

xx xx

ii

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

ii

and zz comp onents Note that under the summation convention Show

ii

that by adding Eq to its analogues

Ye

ii

ii

Combine the ab ove two results to give expressions for etc in terms of the elements of

xx

and hence derive the following result in the principal frame e

Y

e Tre

ij ij ij

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

ij

indep endent form for the elasticity tensor in an isotropic solid

Y

ik jl il jk

E

ij

kl ij k l

C i Apply E to suitable deformations to recover expressions for the bulk

ij k l

mo dulus and shear mo dulus of an isotropic solid in terms of Y and

ii Show that the u inside a homogeneous isotropic solid sub ject to

only at its surface ob eys

D Transformation Laws

D Here we consider transformations of vectors and tensors b etween two sets of

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

 

  

sothatwe can use sux notation eg the fact that the e basis is orthonormal e e e

 

can be written

e e

i j ij



are linear combinations of the e The unit vectors e

i

i



e T e

ij j

i

 

is any vector and y A e Supp ose that x x e x x Prove that

i i

i i



i e e T

k ik

i

 



ii The inverse transformation e T e has T T

i

ki

k

ik ik



iii x T x

ik k

i



iv x T x

i

ki

k

 

v y T T A x

ij

lk jk

i

l

in the rotated co ordinate This means you have found the comp onents of the tensor A

system



A T T A

ij

lk jk

il

Indeed this prop erty under orthogonal coordinate transformations T is mathematically

often taken as the dening prop erty of a tensor A For tensors of higher rank than

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

detT are usually called proper rotations and those with detT improp er rota

tions Show that these are the only p ossible values of the for transformations

between orthonormal bases as dened ab ove

App endix The antisymmetric threetensor

ij k

Reprise of Sheet V section D

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

ij

op eration on a vector gives another vector Avector is technically a tensor of rank one a

scalar of rank zero Tensors of higher rank can also be dened For example a tensor of

rank four Q acts on a rank two tensor to give another rank two tensor B Q A

ij

ij k l ij k l kl

with summation over rep eated indices For tensors higher than rank two dyadic notation

b ecomes extremely cumb ersome and sux notation is essential

An imp ortant and useful tensor of rank three is the ful ly antisymmetric three tensor

denoted This op erates on either a second rank tensor or two vectors to givea vector

ij k

For i j k it is dened as if ij k is a cyclic permutation of x y z eg yzx

ij k

and if the order is anticyclic eg xz y If anytwoofij k are the same then

ij k

eg The tensor is very useful for writing down a vector pro duct in sux

xxy

ij k

notation Show that

A B A B

i j

ij k k

The following relation

jm im

ij k klm il jl

is tedious to prove using comp onents you should come across this in maths lectures but

is worth rememb ering It saves a great deal of time when manipulating vector identities

such as

A B C A B C

m

j

ij k klm l

A B C

m

j jm im

l il jl

B A C C A B A C B A B C

i j j i j j

which is a familiar result normally derived by more tedious means

The result Eq is esp ecially helpful when dealing with vector identities involving

the r op erator Because is a constant tensor it commutes with r and expressions

ij k

can be reordered accordingly For example

H r E H rE

i j

ij k k

r E H E r H H r E

i j j i i j

ij k k ij k k k

E r H H r E

j i i j

jik k k kij

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 metho ds that

r rA rrA r A

r A rA A r

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

III courses