12—Tensors 320~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
(12.56)
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?)
As an example applying all this apparatus, do the transformation of the components of a second
rank tensor, the electromagnetic field tensor. This tensor is the function that acts on the current density
(four dimensional vector) and gives the force density (also a four-vector). Its covariant components are
(
Fij
)
=F
(
~ei,~ej
)
=
0 −Ex −Ey −Ez
Ex 0 Bz −By
Ey −Bz 0 Bx
Ez By −Bx 0
(12.57)
where theE’s andB’s are the conventional electric and magnetic field components. Compute a sample
component of this tensor in the primed coordinate system.
F 20 ′ =F(~e′ 2 ,~e′ 0 ) =F
(
~e 2 ,~e 0
1
√
1 −v^2 /c^2
+~e 1
v/c
√
1 −v^2 /c^2
)
=
1
√
1 −v^2 /c^2
F 20 +
v/c
√
1 −v^2 /c^2
F 21
or in terms of theEandBnotation,
Ey′ =
1
√
1 −v^2 /c^2
[
Ey−
v
c
Bz
]
Sincevis a velocity in the+xdirection this in turn is
E′y=
1
√
1 −v^2 /c^2
[
Ey+
1
c
(
~v×B~
)
y]
(12.58)
Except possibly for the factor in front of the brackets. this is a familiar, physically correct equation of
elementary electromagnetic theory. A charge is always at restin its own reference system. In its own
system, the only force it feels is the electric force because its velocity with respect to itself is zero. The
electric field that it experiences isE~′, not theE~ of the outside world. This calculation tells you that
this forceqE~′is the same thing that I would expect if I knew the Lorentz force law,F~=q
[~
E+~v×B~
]
.
The factor of
√
1 −v^2 /c^2 appears because force itself has some transformation laws that are not as
simple as you would expect.