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Re: simple magnets question

Hi Folks --

As I said before, this question cuts to the core of what electromagnetism is.

Let me try to make the discussion a little more quantitative, so people can
see why and how a magnetic field in one frame is (and *must* be) an
electric field in another frame. It's not magic. It's just physics.

Suppose Joe is in the lab frame. He sets up an electromagnet consisting of
a wire with current flowing in it. According to Joe, at some typical point:
* The wire has stationary positive metal ions with a density rho_p = rho_0.
* The wire contains electrons moving with rapidity u.
* The wire is macroscopically electrically neutral.
Therefore the electron density rho_e must be equal to rho_0
(ignoring terms of order u**2).

Now suppose a test charge whizzes by at a rapidity v. Joe sees the test
charge deflected by the magnetic field of his magnet, and not by any
electric field. (Remember, Joe has no electric field.)

But now consider what this situation looks like to Moe, an observer
comoving with the test charge. According to Moe:
% There may be some magnetic field, but we won't bother to calculate its
magnitude, since the charge is not moving (according to Moe) and cannot
possibly be affected by any magnetic field.
% The wire contains positive metal ions moving with rapidity -v.
% The wire contains electrons moving with rapidity u-v.
% The wire is *not* electrically neutral, as we now calculate.

Since we know the densities rho_p and rho_e in Joe's frame, we can
calculate the densities in Moe's frame. They are Lorentz-contracted
versions of the rest-frame densities:
Ions: rho_0 cosh(v)
Electrons: rho_0 cosh(v-u)
Net: -rho_0 sinh(v) u using Taylor expansion for u << 1
Or just: -rho_0 v u using Taylor expansion for v << 1

We have ignored all sorts of higher-order terms.

This electric charge density gives rise to *electric* fields in Moe's frame
in proportion to the current density (rho_0 u) and proportional to some
velocity (v) which just happens to be what Joe thought was the velocity of
the test charge. Moe says this electric field deflects the charge, while
Joe says the magnetic field does the job. They agree on the magnitude of
the deflection, so (....drum roll, please....) physics is saved.

At 04:21 PM 6/25/99 -0700, William Beaty wrote:

Another puzzling or interesting feature of the
electric field given by (8) is that its divergence is NOT equal to zero
anywhere in space, which Sommerfeld refers to as "an interesting
mathematical difficulty".

Right. If you spin a uniformly-magnetized disk, the local velocity (v) is
nonuniform, so the charge density in Moe's frame is everywhere nonzero, and
these charges cause a divergence in the electric field, in the usual way in
accordance with Maxwell's equations. I have no idea why Sommerfeld thought
this posed mathematical difficulties. Mathematically it is analogous to
the heat conduction in a disk that is being uniformly internally heated
(e.g. by internal radioactive decay). And it certainly isn't any sort of
physics paradox.

Finally, I must retract part of what I said at 03:35 PM 6/25/99 -0400
alleging the equivalence of

QQQQQ
QQQQQQQQQ
QQQQQQQQQQQ (1)
QQQQQQQQQ
QQQQQ


and

___
/ \
/ \
| | (2)
\ /
\___/


If all you care about is magnetic fields, then you can apply Stokes-type
arguments: the current in the left edge of one Q cancels the current in
the right edge of the neighboring Q. So the things are magnetically
equivalent, to first order. But if you spin the thing, the two currents
Lorentz-contract differently, so there is a charge at that point and no
Stokes-type cancellation of the charge. So when you spin them, the things
are not equivalent with respect to their electric fields.