Mathematical Tools for Physics

12—Tensors 377

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

)

=

(

δcr

)

=

(

1 0

0 1

)

(33)

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 of the tensorg.
The question arises as before: how do you compute the components of a tensor in one basis when the
components are given in another basis. The question is answered as before: It is a problem in geometry to express
the new basis vectors in terms of the old one.
~e′i=aji~ej (34)

The covariant components in the new system are (say for a second rank tensor)

Tij′ =T(~e′i, ~ej′) =T(aki~ek, a`j~e`) =akia`jTk` (35)

Analogous equations will hold for the contravariant indices provided that you can find the reciprocal basis
vectors in the transformed system. That is, you need the vectors reciprocal to the~e′iin terms of the vectors
reciprocal to~ei. Assume a relation:
~e′i=bji~ej

You must find theb’s in terms of thea’s. The relationship must come from reciprocity relationships. Required:

~e′i.~e′j=δji=

(

bki~ek

)

.

(

~e`

)

=bkia`j~ek.~e`=bkia`jδk`=bkiakj (36)