Mathematical Tools for Physics - Department of Physics - University

12—Tensors 317

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 (12.50)

Compute the components of g in plane polar coordinates. The contravariant components ofgare from

Eq. (12.41)

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

)

12.8 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, and 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 in terms of the different coordinate functions. Call the two sets of coordinatesxiand

yi. Each of them defines a set of basis vectors such that a given velocity is expressed as

~v=~ei

dxi

dt

=~e′j

dyj

dt

(12.51)

What you need is an expression for~e′jin terms of~eiat each point. To do this, take a particular path

for the particle — along they^1 -direction (y^2 &y^3 constant). The right hand side is then

~e′ 1

dy^1

dt

Divide bydy^1 /dtto obtain

~e′ 1 =~ei

dxi

dt

/

dy^1

dt

But this quotient is just

~e′ 1 =~ei

∂xi

∂y^1

∣

∣∣

∣

y^2 ,y^3

And in general

~e′j=~ei

∂xi

∂yj

(12.52)

Do a similar calculation for the reciprocal vectors

gradf=~ei

∂f

∂xi

=~e′j

∂f

∂yj