Please excuse this resend. Harry Stephen A. Lawrence wrote:
> > > 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, Questioning the current party in power does not necessarily mean one is siding with the main opposition party. ;-) > and absent magnetic monopoles, all B fields can be > treated _as_ _if_ they are due to charges in motion. "_As if_" means don't take the _logic_ of an explanation seriously, because "_As if_" explanations are by definition useful but not necessarily truthful. For example Newton's theory of gravity was an _as if_ explanation of gravity. Newton never intended the logic of "a = g" to be taken seriously as Einstein did. If one takes the _logic_ of relativity seriously and if the magnetic force produced by a body depends on the relative motion of a test particle then when the relative motion of the test particle is zero then the magnetic field for the test particle is also zero. > 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! > You appear to be have a better grasp of the mathematics of EM theory than me, so I can't comment the rest of your post. Harry
