Mathematical Methods for Physics and Engineering : A Comprehensive Guide

TENSORS

26.4 First- and zero-order Cartesian tensors

Using the above example as a guide, we may consider any set of three quantities

vi, which are directly or indirectly functions of the coordinatesxiand possibly

involve some constants, and ask how their values are changed by any rotation of

the Cartesian axes. The specific question to be answered is whether the specific

formsv′iin the new variables can be obtained from the old onesviusing (26.4),

vi′=Lijvj. (26.9)

If so, theviare said to form the components of avectororfirst-order Cartesian

tensorv. By definition, the position coordinates are themselves the components

of such a tensor.The first-order tensorvdoes not change under rotation of the

coordinate axes; nevertheless, since the basis set does change, frome 1 ,e 2 ,e 3 to

e′ 1 ,e′ 2 ,v′ 3 , the components ofvmust also change. The changes must be such that

v=viei=v′ie′i (26.10)

is unchanged.

Since the transformation (26.9) is orthogonal, the components of any such

first-order Cartesian tensor also obey a relation that is the inverse of (26.9),

vi=Ljivj′. (26.11)

We now consider explicit examples. In order to keep the equations to reasonable

proportions, the examples will be restricted to thex 1 x 2 -plane, i.e. there are

no components in thex 3 -direction. Three-dimensional cases are no different in

principle – but much longer to write out.

Which of the following pairs(v 1 ,v 2 )form the components of a first-order Cartesian tensor

in two dimensions?:

(i) (x 2 ,−x 1 ), (ii) (x 2 ,x 1 ), (iii) (x^21 ,x^22 ).

We shall consider the rotation discussed in the previous example, and to save space we
denote cosθbycand sinθbys.
(i) Herev 1 =x 2 andv 2 =−x 1 , referred to the old axes. In terms of the new coordinates
they will bev 1 ′=x′ 2 andv′ 2 =−x′ 1 ,i.e.

v′ 1 =x′ 2 =−sx 1 +cx 2

v′ 2 =−x′ 1 =−cx 1 −sx 2.

(26.12)

Now if we start again and evaluatev 1 ′andv 2 ′as given by (26.9) we find that

v 1 ′=L 11 v 1 +L 12 v 2 =cx 2 +s(−x 1 )

v 2 ′=L 21 v 1 +L 22 v 2 =−s(x 2 )+c(−x 1 ).

(26.13)

The expressions forv′ 1 andv′ 2 in (26.12) and (26.13) are the same whatever the values
ofθ(i.e. forallrotations) and thus by definition (26.9) the pair (x 2 ,−x 1 )isa first-order
Cartesian tensor.