Dear Vo,
Which what is-are information a given magnet is
[1] "a conductor" ... OR
[2] not "a conductor"
[3] uh........ what IS the conductivity of ...uh..... some given magnet?
[4] OR... is-are there any measure of conductivity in some given
"""" permanent"""" magnet???
HMmmmmm ?
[5] Oh welll????
[6] Can we get-see- decide onr of these '''magnets....
[7]
can we see the - uh ......... this magnet PLEASE
???????????
On 6/18/06, Harry Veeder <[EMAIL PROTECTED]> 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.
> 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
111111111111 > 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
