Re: is the magnetic field really moving?

On Wed, 27 Aug 2008, phil-news-nospam@[EMAIL PROTECTED]
 wrote:

> On Wed, 27 Aug 2008 13:01:51 +1000 Timo Nieminen
<timo@[EMAIL PROTECTED]
> wrote:
> | On Wed, 27 Aug 2008 phil-news-nospam@[EMAIL PROTECTED]
 wrote:
> |
> |> On Wed, 27 Aug 2008 06:56:27 +1000 Timo A. Nieminen
<timo@[EMAIL PROTECTED]
> wrote:
> |>
> |> | In the Faraday Paradox, there are 3 components: magnet, disc, and
wire.
> |> | That the rotating magnet vs stationary magnet cases are the same,
for
> |> | either the rotating disc or stationary disc case is one where
relative
> |> | motion doesn't make much difference (but not _no_ difference). When
the
> |> | disc is stationary, it's equivalent to putting a loop of wire near
a
> |> | rotating or stationary magnet. No current will flow around the
loop. Yet,
> |> | there will be a physical effect, since when the magnet rotates,
there'll
> |> | be an electric field where the loop is, which will electrically
polarise
> |> | it. Small effect, so usually ignored.
> |>
> |> 0.  Disk is stationary, magnet is stationary.
> |> 1.  Disk is rotating, magnet is stationary.
> |> 2.  Disk is stationary, magnet is rotating.
> |> 3.  Disk is rotating, magnet is rotating.
> |>
> |> I believe you are referring to case #2.
> |
> | Yes, but you'll also get the same effect due to the electric field of
the
> | rotating magnetic in case #3. Neither gives a current (in both cases,
> | curl(E)=0, so you don't get any current around the loop. It's an
> | electrostatic field.
>
> I disagree.  In case #3 there will be a current.

Yes, but this current is due to the Lorentz force on the disc, not the 
rotation-produced electric field. It's a linear system; the effects just 
add together, and can be considered independently.

All I'm saying is that in #2 and #3, the effect of rotation of the magnet 
is the same, and in neither case does this rotation produce a current. 
Likewise, in #1 and #3, the effect of rotation of the disc is the same, 
and in both cases it produces a current.

> But this is different from
> case #1 (also has a current).  In case #1 the current is induced into
the
> disk itself (in the "spokes" of the disk).  In case #3 the current is
induced
> into the wiring that connects between the disk axle and the disk rim. 
This
> is due to the rotating magnet, relation to the stationary wiring.

Why do you say the mechanism producing the current is different between #1

and #3? In both cases, the magnetic field is identical.

> Case #3 does have current running through the disk.  The current will
have
> a magnetic field that will oppose the rotational relation****p between
the
> disk and the magnet.  I have not determined how this will act in the
case
> of the disk and magnet rigidly connected to rotate together.  But I
believe
> it will just present an amount of inductance to the circuit.
>
> | The loop (i.e., the wire + disc) can be treated as a single conductor.
Put
> | a conductor in an electrostatic field, and you'll get induced surface
> | charges: one end will become positive, the other negative; it will
become
> | electrically polarised.
>
> In case #2 there is a potential induced in the disk and _also_ a
potential
> induced in the wiring.  These potentials oppose each other.  The net
result
> is no current, but there will be a charge developed at 2 points in the
loop,
> one negative and one positive.  It would be equivalent to 2
batteries/cells
> wired in parallel.

Yes. Case #1 is effectively 1 battery in the disc ****tion of the circuit, 
driving a current. Case #2 is 1 battery in each of the wire and disc 
****tions, opposing, so no current. Case #3 is both of these added 
together, so effectively 3 batteries, with 2 in series in the disc 
****tion.

> |> I believe there is as much potential
> |> induced in the disk in #2 as there is in #1.
> |
> | My first guess is yes, you are right. The electric field due to
rotation
> | of a magnet depends on vxB, which is the same term we see in the
Lorentz
> | force.
>
> A new case:
>
> 4.  Disk is stationary, magnet is stationary, and the wiring moves
around
>    the disk (with no explanation of how this can be achieved,
mechanically).
>
> No potential is induced in the disk.  A potential is induced in the
wiring.
> If the wiring completes a loop, current would flow (including through
the
> disk).  The relation between the wiring and the magnet matters.  One has
to
> be moving relative to the other to induce a potential, just as in cases
#2
> and #3.

Functionally the same as rotating the disc. All you need to do is only 
move _part_ of the circuit relative to the magnet. Move the whole circuit,

and you get the effect of two opposed batteries you noted above. Move part

of the circuit, and you only have one effective battery. Same thing with 
the linear motion version as well.

> |> | You weren't asking about them. In the Faraday Paradox, electric
fields due
> |> | to the rotation of the magnet don't produce a current, just an
electric
> |> | polarisation of the disc+wire. So ignore it; all you need to
explain it
> |> | working is the Lorentz force.
> |>
> |> What do you mean by an electric polarisation?
> |>
> |> I take it we are not pursing the same thing.
> |
> | As I said, you initially just asked about a wire and a magnet, not the
> | Faraday Paradox. So I don't know what you're pursuing. But the physics
is
> | fundamentally the same for all 3 things that have been discussed:
rotating
> | magnets, moving magnets, and the Faraday Paradox.
>
> I'm seeking to solidify my understanding of the Faraday Paradox.  The
way I
> understand this is that the field _does_ rotate with the magnet, and all
the
> cases Faraday considered with the disk are explained when one includes
the
> effect of the magnet (particularly when rotating) on the attachment
wiring.
>
> I believe this is the correct explanation.  If the opposite is true,
such that
> the field does not rotate with a rotation of the magnet, then that opens
up a
> number of possibilities that seem to me to defy relativity.

I find that all of the cases #0 - #4 above are explained without any 
problems by the standard relativistic theory which assumes that the field 
does _not_ rotate.

> In particular,
> if a conductor and magnetic field moving together can induce a potential
in
> that conductor, this would have to happen as a result of motion due to
the
> rotation of the earth, the earth revolving around the sun, the solar
system
> revolving around the galaxy, etc.  And that's not even the end of it. 
If
> you can get a potential induced as a result of these motions, you can
then
> determine the vector of these motions (over some time to account for the
> changes since you're dealing with a vector sume of several motions). 
This
> would reveal what our motion is relative to a "universe stationary"
point,
> which relativity says cannot exist.

Which is why you need to use the _relativistic_ theory. In particular, the

relativistic transformations of the fields. The conductor and magnet 
moving together is instructive. The conductor is in a constant magnetic 
field, and moving, so you have the Lorentz force, pro****tional to vxB. How

can this be reconciled with the stationary magnet and conductor case? The 
relativistic transformation of the magnet field gives you an electric 
field pro****tional to vxB - this exactly cancels the Lorentz force.

It's common enough to see statements such that the magnetic field is a 
necessary relativistic consequence of the electric field (which predates 
special relativity, going back to Heaviside or perhaps earlier). It works 
the other way too, with an electric field being a necessary relativistic 
consequence of a magnetic field.

The fundamental relativistic EM field is the 4x4 combined electric and 
magnetic field tensor (or, if you prefer - and are prepared to choose a 
gauge - the 4-potential). You can't consider the electric and magnetic 
fields as separate entities.

-- 
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
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