12—Tensors 310of the tensor are reflected in the symmetry of the covariant or the contravariant components (but not
usually in the mixed components).
Metric Tensor
Take as an example the metric tensor:
g(~u,~v) =~u.~v. (12.37)
The linear function found by pulling off the~ufrom this is the identity operator.
g(~v) =~v
This tensor is symmetric, so this must be reflected in its covariant and contravariant components. Take
as a basis the vectors
~e^1
~e 1
~e 2
~e^2
Let|~e 2 |= 1and|~e 1 |= 2; the angle between them being 45 ◦. A little geometry shows that
|~e^1 |=
1
√
2
and |~e^2 |=
√
2
Assume this problem is two dimensional in order to simplify things.
Compute the covariant components:
g 11 =g(~e 1 ,~e 1 ) = 4
g 12 =g(~e 1 ,~e 2 ) =
√
2
g 21 =g(~e 2 ,~e 1 ) =
√
2
g 22 =g(~e 2 ,~e 2 ) = 1
(
grc
)
=
(
4
√
√^2
2 1
)
Similarly
g^11 =g(~e^1 ,~e^1 ) = 1/ 2
g^12 =g(~e^1 ,~e^2 ) =− 1 /
√
2
g^21 =g(~e^2 ,~e^1 ) =− 1 /
√
2
g^22 =g(~e^2 ,~e^2 ) = 2
(
grc
)
=
(
1 / 2 − 1 /
√
2
− 1 /
√
2 2
)
The mixed components are
g^11 =g(~e^1 ,~e 1 ) = 1
g^12 =g(~e^1 ,~e 2 ) = 0
g^21 =g(~e^2 ,~e 1 ) = 0
g^22 =g(~e^2 ,~e 2 ) = 1
(
grc
)
=
(
δrc
)
=
(
1 0
0 1
)
(12.38)
I usedrandcfor the indices to remind you that these are the row and column variables. Multiply
the first two matrices together and you obtain the third one — the unit matrix. The matrix
(
gij
)
is
therefore the inverse of the matrix
(
gij
)
. This last result isnotgeneral, but is due to the special nature