Mathematical Tools for Physics

12—Tensors 364

take the same tensor and compute its components in a different basis. Note: The method I’ll describe here is
rather clumsy and inefficient. It is however conceptually simple. For an efficient way to do this you can skip to
the development following equation ( 22 ).

e e’

e

e 1

1

2

2

^

^

^

’ ^ ˆe′ 1 =ˆx√+ˆy

2

=

ˆe 1 +ˆe 2

√

2

ˆe′ 2 =

ˆy−ˆx

√

2

=

ˆe 2 −eˆ 1

√

2

(18)

In order to illustrate what is happening, I’ll carry out this computation by two methods, first: the hard way, and
second: the easy way.
You already have the componentsTijin one basis. They come from the defining equations ( 13 ),

1

1 T(ˆei) =Tjiˆej.

The analogous equation usingˆe′jdefines the components ofTin the other basis.

1

1 T

(

ˆe′i

)

=Tji′ˆe′j.

To relate one set of components to the other, the key is, as usual, linearity. The left side of this last equation is
(i= 1)

1

1 T

(

ˆe′ 1

)

=^11 T

(

eˆ 1 +ˆe 2

√

2

)

=

1

√

2

[ 1

1 T(ˆe^1 ) +

1

1 T(ˆe^2 )

]

=

1

√

2

[

T 11 ˆe 1 +T 21 ˆe 2 +T 12 ˆe 1 +T 22 ˆe 2

]

. (19)

The left side is (stilli= 1)

T 11 ′ˆe′ 1 +T 21 ′ˆe′ 2 =

1

√

2

[

T 11 ′(ˆe 1 +ˆe 2 ) +T 21 ′(ˆe 2 −eˆ 1 )

]

. (20)