Re: simple magnets question

[Leigh Palmer <palmer@SFU.CA>]

Bill Beatty writes:

> ................ Unless I'm misunderstaing, people here keep talking
> as if fields can be stationary, yet in the same breath that the motion of
> a field is meaningless.  But then the stationary-ness of the field is
> meaningless too, and it cannot be used as part of the explanation.

If analysis can be simplified to the point that temporal change in
the fields can be made to go away then the analysis becomes easier.
That is the case for the present question, and that is the reason I
suggested changing the question to one that is more easily answered.
All that is really sought here is an explanation of the interaction
between a moving charged particle and magnetic field as it should
be described in the particle's instantaneous frame of rest.

Even though no one has invited me to do so, let me outline my
explanation in terms of sources. This won't be a complete
calculation, but it will get the interested student pointed in the
correct direction. He must then perform the calculation himself to
internalize the result, a process we call "learning".

An infinite (x-y) plane current sheet carries a constant, uniform
superficial current density A (which would have SI units of amperes
per meter) in the x direction. There are no other sources of field
in this system. Ampere's law tells us that the result is that the
space above the current carrying plane (z>0) is permeated with a
constant, uniform magnetic field of magnitude B pointing in the
negative y direction. (Check that with your right-hand rule to be
sure you're following my argument.) The magnitude of the field is

                              1
                         B = --- muzero A
                              2

Consider a particle of mass m carrying electric charge q which
moves with speed v in the positive x direction. The force which
acts on this particle is (in SI) found by the Lorentz force
equation to be

                                      q v
                         F = q v B = ----- muzero A
                                       2

The direction of this force is downward, toward negative z.

I claim that the question Bill has asked can be simplified to the
form: "How will an observer moving in the inertial frame in which
the charged particle is instantaneously at rest describe this
physical system?" I think by answering this simpler question I can
relieve some of the discomfort felt by some in this discussion.

Consider the source of the electromagnetic field in a little more
detail in the laboratory frame. The planar current sheet carries no
net charge, but it must contain moving charges. These can be
modelled in several ways, and please trust me that any model that
produces the uniform, constant superficial current density A will
yield the same result we are going to get in the moving frame for
the interaction.

Consider a model having a uniform sheet of positive electric charge
density sigma moving in the x direction with speed (drift velocity)
u. To restore charge neutrality we must add to this a static sheet
of charge density -sigma. The superficial current density in this
model system is given by

                          A = u sigma

and the force (in the laboratory frame) can be written in terms of
the model parameters as

                               q u v
                         F =  ------- muzero sigma
                                 2

I will not carry the calculation farther than this, but I will now
ask you to recognize that relativity tells us that the field source
looks different to an observer in the moving frame. Consider first
the static negative charge component. With respect to the moving
frame that static component is now moving with speed v, and it will
be seen as suffering a Lorentz contraction by an appropriate factor.
The result is that this negatively charged component will appear to
be *denser* by that same factor. Consider next the positive charge
component. that component is moving more slowly in the moving frame
than it was moving in the laboratory frame. The consequence of this
is that the positively charged component will be *less* contracted
in the moving frame than it was in the laboratory frame. Thus the
positive component will have a smaller surface charge density in
the moving frame than it has in the laboratory frame.

Both of these Lorentz contraction effects have the tendency to make
the plane more electronegative in the moving frame than it is in
the laboratory frame. The consequence is that the plane is seen to
have a net negative charge in the moving frame. Thus the force
which acts on the charge in this frame is of a purely electrical
nature.

The force will be seen to have the correct direction (downward)
here, but in order to make the calculation one must be careful to
cross all t's and dot i's on the way. For example, the speed of the
positive charge component in the moving frame is not u-v. One must
use the proper relativistic velocity transformation. More subtly,
force is not a direct observable; force is to be inferred from the
acceleration of the particle, and that will be different in the two
frames (due to time dilation as well). The force turns out to be
different in the two frames also.

Well, that shoots a couple of hours for me. I've got to get back to
work. Please let me know if this helps anyone to understand. It
works well when I do it with one or two students at a time; the
lightbulbs go on over their heads. I've never tried to do it in
ASCII with my visuals forbidden, however.

Leigh