Harry Veeder wrote:
Stephen A. Lawrence wrote:
>> 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.)
The point is, it is "true" according to the theory (dogma?) that
_all_ magnetism is simply an effect of charges in motion.
No, it's not true according to "current dogma". It's true according to
a certain model, but that model is not intrinsic to relativity theory,
nor is it built into Maxwell's equations, nor is it accepted by all
mainstream scientists.
As far as anyone knows at the present time, there is no such thing as a
magnetic monopole, and absent magnetic monopoles, all B fields can be
treated _as_ _if_ they are due to charges in motion. Any magnetic
dipole can be treated as though it's caused by a current loop;
mathematically, it might just as well be, even if it's actually
intrinsic to a single particle which happens to have a nonzero magnetic
moment.
The model I've occasionally mentioned, which treats the A field as
fundamental and the Faraday tensor as being the exterior derivative of
A, with the E and B fields serving as components of the Faraday tensor,
does indeed require that there be no magnetic monopoles. However, as I
just said, that model is not a consequence of relativity, and not a
consequence of Maxwell's equations, though it incorporates the latter.
It's elegant but not necessary, and in fact work has been done on a
version which allows monopoles (since, supposedly, GUTs generally
predict their existence, this is a big issue in some circles). I
haven't yet managed to grok the version which allows monopoles, but in
any case it doesn't have much of anything to do with ordinary magnets,
which don't include monopoles in anybody's theory AFAIK.
> 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.
Maxwell's equations do not actually state that all magnetism is
simply effect of charges in motion. Such a theory is complementary
to Maxwell's equations, much like the kinetic theory of heat is
complementary to the laws of thermodynamics.
That's correct, as I just said.
However, it's also true that a static magnetic field due to a current
loop _cannot_ be transformed away via the Lorentz transforms. The
dipole field of a bar magnet is the same sort of field, and it can't be
transformed away either.
Electromagnetic radiation, which is also a phenomenon handled by
relativity (indeed, it's a large part of what relativity theory was
designed to "explain"), generally can't be transformed away, either,
please note!
Harry
