Harry Veeder wrote:
[EMAIL PROTECTED] wrote:
-----Original Message-----
From: Stephen A. Lawrence
The predicted magnetic field of a current can be obtained simply by
Lorentz transforming the electric field from the rest frame of the
charges making up the current to the frame of the observer moving
relative to them. Remarkably, the result is a first-order effect --
first order in the relative velocities -- unlike just about everything
else predicted by relativity.
<><><><><>
There are no charges (q) involved in permanent magnets. I hereby
extract myself from this discussion.
Charges may be involved. However, the _reality_ of a permanent magnetic body
is not recognised by a relativistic charged based model of magnetism. The
relativistic model implies that the permanence of a permanent magnetic body
is a matter of opinion since one could execute some motion relative to the
body and decide it is non-magnetic.
Actually, this isn't true. Given a pure magnetic field (with zero
electric field) there is no inertial frame in which there isn't any B
field. A typical permanent magnet has no associated electric field, and
so its field can't be transformed away. (Classically, as long as the
surface of the magnet is a conductor and the net charge contained in it
is balanced, there won't be an E field exterior to the magnet.)
You can't transform away a pure B field. Most other frames have a
nonzero E field as well, but they all also have a nonzero B field. A
simple argument shows this:
Consider a pure B field (no E field) in inertial frame S. Consider two
identical particles, particle P1, at rest in S, and particle P2, moving
in S. P1 feels no force, and is not accelerating. P2 feels a force,
and _is_ accelerating. The (Boolean-valued) existence of an
acceleration is absolute (at least as long as we stick with inertial
frames) -- a particle which is accelerating, is accelerating in all
frames; a particle which is "inertial" is inertial in all frames. So,
in all inertial frames, P1 will feel no net force, while P2 will feel a
net force. Since the only difference between the particles is their
velocity, yet they feel difference forces, they are clearly subject to a
velocity-dependent force. The E field isn't velocity dependent, so it
can't account for the difference. Ergo, there's a B field in every frame.
There's a fairly simple mathematical test that'll tell you right away
whether a B field (or E field) can be transformed away or not but off
hand I don't recall what it is off the top of my head. One can, of
course, also just write out the transform and look at it to check this
particular case:
Here's the transform for the B field (from MTW p.78 -- you can also get
it just by transforming the Faraday tensor):
B'(parallel) = B(parallel)
B'(perpendicular) = gamma*(B(perpendicular) - VxE(perpendicular))
B(parallel) obviously can't be transformed away since it doesn't change
under the Lorentz transform, so to get rid of the B field you need to be
moving perpendicular to it. But if there's no E field, the
perpendicular B field component transforms as:
B'(perp) = B(perp) / sqrt(1 - v^2)
and if B(perp) is nonzero, that will be nonzero too.
Harry
