Mathematical Tools for Physics

12—Tensors 391

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







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

Ey′=

1

√

1 −v^2 /c^2

[

Ey+

1

c

(

~v×B~

)

y

]

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 is
E~′, 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.