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. > 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. Harry
