Mathematical Methods for Physics and Engineering : A Comprehensive Guide

TENSORS

Using (26.40), show thatAij=ijkpk.

By contracting both sides of (26.40) withijk, we find

ijkpk=^12 ijkklmAlm.

Using the identity (26.30) then gives

ijkpk=^12 (δilδjm−δimδjl)Alm

=^12 (Aij−Aji)=^12 (Aij+Aij)=Aij,

whereinthelastlineweusethefactthatAij=−Aji.

By a simple extension, we may associate a dual pseudoscalarswith every

totally antisymmetric third-rank tensorAijk, i.e. one that is antisymmetric with

respect to the interchange of every possible pair of subscripts;sis given by

s=

1

3!

ijkAijk. (26.41)

SinceAijkis a totally antisymmetric three-subscript quantity, we expect it to

equal some multiple ofijk(since this is the only such quantity). In factAijk=sijk,

as can be proved by substituting this expression into (26.41) and using (26.36).

26.12 Physical applications of tensors

In this section some physical applications of tensors will be given. First-order

tensors are familiar as vectors and so we will concentrate on second-order tensors,

starting with an example taken from mechanics.

Consider a collection of rigidly connected point particles of which theαth,

which has massm(α)and is positioned atr(α)with respect to an originO,is

typical. Suppose that the rigid assembly is rotating about an axis throughOwith

angular velocityω.

The angular momentumJaboutOof the assembly is given by

J=

∑

α

(

r(α)×p(α)

)

.

Butp(α)=m(α) ̇r(α)and ̇r(α)=ω×r(α), for anyα, and so in subscript form the

components ofJare given by

Ji=

∑

α

m(α)ijkx(jα) ̇x(kα)

=

∑

α

m(α)ijkx

(α)

j klmωlx

(α)

m

=

∑

α

m(α)(δilδjm−δimδjl)x(jα)x(mα)ωl

=

∑

α

m(α)

[(

r(α)

) 2

δil−x(iα)x(lα)

]

ωl≡Iilωl, (26.42)

whereIilis a symmetric second-order Cartesian tensor (by the quotient rule, see