Mathematical Tools for Physics

12—Tensors 390

You can check that these equations represent the transformation to an observer moving in the+xdirection by
asking where the moving observer’s origin is as a function of time: It is atx′^1 = 0orx−vt= 0, givingx=vt
as 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 = x

x = ct

1

1

(^00)
e
e^0
0
1
1
x = x e
e
’ ’
’
’
’
x = ct
Find the basis vectors in the transformed system by using equation ( 46 ).
~e′j=~ei
∂xi
∂yj
In the present case theyjarex′jand we need the inverse of the equations ( 49 ). They are found by changingv
to−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
~e′ 0 =~ei
∂xi
∂x′^0
=~e 0

1

√

1 −v^2 /c^2

+~e 1

v/c

√

1 −v^2 /c^2

~e′ 1 =~ei

∂xi

∂x′^1

=~e 0

v/c

√

1 −v^2 /c^2

+~e 1

1

√

1 −v^2 /c^2

(50)

It is easy to verify that these new vectors point along the primed axes as they should. They also have the
property that they are normalized to plus or minus one respectively as are the original untransformed vectors.
(How about the reciprocal vectors?)