12—Tensors 386
~e^1 is perpendicular to thex^2 -axis, the linex 1 =constant, (as it should be). Its magnitude is the magnitude of
~e^1 , which is one.
To verify that this magnitude is correct, calculate it directly from the definition. The magnitude of the
gradient is the magnitude ofdφ/dswheresis measured in the direction of the gradient, that is, in the direction
~e^1.
dφ
ds
=
∂φ
∂x^1
dx^1
ds
= 1.1 = 1
Why 1 fordx^1 /ds? The coordinate lines in the picture arex^1 = 0, 1 , 2 ,.... When you move on the straight line
perpendicular tox^2 =constant (~e^1 ), and go fromx^1 = 1tox^2 = 2, then both∆x^1 and∆sare one.
Metric Tensor
The simplest tensor field beyond the gradient vector above would be the metric tensor, which I’ve been implicitly
using all along whenever a scalar product arose. It is defined at each point by
g(~a,~b) =~a.~b (44)
Compute the components of g in plane polar coordinates. The contravariant components ofgare from Eq. ( 38 )
and ( 40 )
gij=~ei.~ej=
(
1 0
0 1/r^2
)
Covariant:
gij=~ei.~ej=
(
1 0
0 r^2
)
Mixed:
gij=~ei.~ej=
(
1 0
0 1
)
12.7 Basis Change
If you have two different sets of basis vectors you can compute the transformation on the components in going
from one basis to the other, Eq. ( 34 ). In dealing with fields, a different set of basis vectors necessarily arises from
a different set of coordinates on the manifold. It is convenient to compute the transformation matrices directly