Mathematical Tools for Physics - Department of Physics - University

12—Tensors 319

The basis vectors associated with this coordinate system point along the directions of the axes.
This manifold isnotEuclidean however so that these vectors are not unit vectors in the usual
sense. We have

~e 0 .~e 0 =− 1 ~e 1 .~e 1 = 1 ~e 2 .~e 2 = 1 ~e 3 .~e 3 = 1

and they are orthogonal pairwise. The reciprocal basis vectors are defined in the usual way,

~ei.~ej=δij

so that

~e^0 =−~e 0 ~e^1 =~e 1 ~e^2 =~e 2 ~e^3 =~e 3

The contravariant (also covariant) components of the metric tensor are

gij=







−1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1





=gij (12.54)

An observer moving in the+xdirection with speedvwill have his own coordinate system with

which to describe the events of space-time. The coordinate transformation is

x′^0 =ct′=

x^0 −vcx^1

√

1 −v^2 /c^2

=

ct−vcx

√

1 −v^2 /c^2

x′^1 =x′=

x^1 −vcx^0

√

1 −v^2 /c^2

=

x−vt

√

1 −v^2 /c^2

x′^2 =y′=x^2 =y x′^3 =z′=x^3 =z

(12.55)

You can check that these equations represent the transformation to an observer moving in the+x

direction by asking where the moving observer’s origin is as a function of time: It is atx′^1 = 0or

x−vt= 0, givingx=vtas the locus of the moving observer’s origin.

The graph of the coordinates is as usual defined by the equations (say for thex′^0 -axis) thatx′^1 ,

x′^2 ,x′^3 are constants such as zero. Similarly for the other axes.

x^0 =ct

x′^0 =ct′

x′^1 =x′

x^1 =x

~e 0

~e 0 ′

~e 1 ′

~e 1

Find the basis vectors in the transformed system by using equation (12.52).

~e′j=~ei

∂xi

∂yj

In the present case theyjarex′jand we need the inverse of the equations (12.55). They are found by

changingvto−vand interchanging primed and unprimed variables.

x^0 =

x′^0 +vcx′^1

√

1 −v^2 /c^2

x^1 =

x′^1 +vcx′^0

√

1 −v^2 /c^2