Now that we have this in hand, we can easily see how to transform the electric and magnetic fields when we boost a frame. Of course, that does not guarantee that the result will be simple.

To convert
from
to
, we must contract its indices
with the transformation tensors,
F'^&alpha#alpha;&beta#beta; = x^&prime#prime;&alpha#alpha;x^&gamma#gamma;
x^&prime#prime;&beta#beta;x^&delta#delta; F^&gamma#gamma;&delta#delta; .
Note that since
is a *linear* transformation:
A^&alpha#alpha;_ &gamma#gamma; = x^&prime#prime;&alpha#alpha;x^&gamma#gamma;
(where I have deliberately inserted a space to differentiate the first
index from the second) we can write this in terms of the components of
as:
F'^&alpha#alpha;&beta#beta; & = & A^&alpha#alpha;_ &gamma#gamma; F^&gamma#gamma;&delta#delta; A^ &beta#beta;_&delta#delta;

& = & A^&alpha#alpha;_ &gamma#gamma; F^&gamma#gamma;&delta#delta;
Ã^&beta#beta;_ &delta#delta;
or (in a compressed notation):
F' = AFÃ
This is just a specific case of the general rule that
can be used in
general to transform any nth rank tensor by contracting it appropriately
with each index.

As we saw in our discussion of Thomas precession, we will have occasion
to use this result for the particular case of a pure boost in an
arbitrary direction that we can without loss of generality pick to be
the 1 direction. Let's see how this goes. Recall that
for a pure
boost in the one direction is the matrix formed with a lower right
quadrant identity and an upper left quadrant
with
on the diagonal and
on the corners). Thus:
so:
F'^01 & = & A^0_ 0 F^0 1 A^ 1_1 + A^0_ 1F^1 0A^ 1_1

-E'_1c & = & - &gamma#gamma;^2 E_1c - &gamma#gamma;^2&beta#beta;^2
E_1c

E'_1 & = & (&gamma#gamma;^2 + &gamma#gamma;^2&beta#beta;^2) E_1

E'_1 & = & E_1
Note that we have extracted the ordinary cartesian components of
and
from
after transforming it. I leave the rest of them to
work out yourself. You should be able to show that:
E_1' & = & E_1

E_2' & = & &gamma#gamma;(E_2 - &beta#beta;B_3)

E_3' & = & &gamma#gamma;(E_3 + &beta#beta;B_2)

B_1' & = & B_1

B_2' & = & &gamma#gamma;(B_2 + &beta#beta;E_3)

B_3' & = & &gamma#gamma;(B_3 - &beta#beta;E_2)

The component of the fields in the direction of the boost is unchanged,
the perpendicular components of the field are mixed (almost as if they
were space-time pieces) by the boost. If you use instead the general
form of
for a boost and express the components in terms of dot
products, you should *also* show that the general transformation is
given by:
**E**' & = & &gamma#gamma;(**E** +
×**B**) -
&gamma#gamma;^2&gamma#gamma;+ 1
(
·**E**)

**B**' & = & &gamma#gamma;(**B** -
×**E**) -
&gamma#gamma;^2&gamma#gamma;+ 1
(
·**B**) .

A purely electric or magnetic field in one frame will thus be a mixture of electric and magnetic fields in another. We see that truly, there is little reason to distinguish them. We have to be a little careful, of course. If there is a monopolar (static) electric field in any frame, we cannot transform it completely into a magnetostatic field in another, for example. Why? Because the equations above will lead to some mixture for all , and in nature as a constraint.

I encourage you to review the example given in Jackson and meditate upon
the remarks therein. We will not spend valuable class time on this,
however. Instead we will end this, after all, purely
mathematical/geometrical *kinematical* interlude (no Hamiltonians or
Lagrangians = no physics) and do some *physics*. Let us deduce the
covariant dynamics of relativistic particles in (assumed fixed)
electromagnetic fields.