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1 Introduction

This is a quick review of some index notation that will be useful in simplifyingexpressions to perform computations on various objects known as tensors, butwe will simply use them in the context of proving various identies involvingthe vector scalar and cross products. We can represent the components of avector a R3 by (a)i = ai R for i = 1, 2, 3 and adopt Einstein notation i.esummation is implied by repeated indices, which will be explained as we goalong.

2 Index Notation

2.1 Some tools

Kronecker delta:

ij =

0 for i = j1 for i = j

e.g 11 = 1, 12 = 0 etc.Properties:

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j=1ijaj = ai

3i=1

aiij = aj

3k=1

ikkj = ij

Levi-civita symbol:

ijk =

1 even permutations of (1, 2, 3)1 odd permutations of (1, 2, 3)

0 repeated indices

e.g 123 = 1, 321 = 1, 112 = 0 etc.

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We can represent this as:

ijk =(ij)(j k)(k i)

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for i,j,k = 1, 2, 3

Relation to Kronecker delta:

ijkkmn = imjn injm

We can write the scalar and cross products in a more compact form:

Scalar Product:

a b =3

i,j=1

ijaibj = ijaibj

Cross Product:

The ith component of the cross product can be represented as

(ab)i =3

j,k=1

ijkajbk = ijkajbk

Simularly, we can repesent the ith component of of the Gradient (Del) oper-ator by ()i =

xi

= i where x are the usual cartesian coordinates i.e(x)1 = x1 = x, x2 = y and x3 = z. With this machinery, it is now quitesimple to prove identieis that involve scalar and cross products of variousobject. You just have to keep track of the indices; it shouldnt be repeatedmore than twice as it could lead to ambigiuity in the expression, also dummyindices i.e those which are being summed can always be replaced by anotherletter and it is useful to do so when trying to compare different objects.

Let me illustrate the notation by applying it to prove an idenity on the

problem sheet. If we represent the ith component of a vector field by (v)i =vi(x,y,z ) = vi, then for every vector field v(x,y,z ): ( v) = 0.

( v) = iji( v)j = ijijmnmvn

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= ijjmn2

xixm

vn = imn2

xixm

vn = min2

xmxi

vn = imn2

xixm

vn = 0

Where in the last two lines we applied the anti-symmetry of the Levi-Civita symbol upon exchanging indices, symmtry of the partial derivativesand replacing dummy indices to put it in the form that we can compare withwhat we have before. We have just shown that the above is equal to minusitself i.e x = x and so x = 0 is the only solution.

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