On Thu, 28 Aug 2008 06:56:52 +1000 Timo A. Nieminen
<timo@[EMAIL PROTECTED]
> wrote:
| 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.
I say there is no Lorentz force on the disc in #3, and that the current is
the result of the effect of the rotating field on the external wiring that
connects to the disk.
| 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 in #3, that current is not produced in the disc itself.
|> 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.
I'm saying the magnetic field is NOT identical between #1 and #3. In #1
it is
not rotating. In #3 it is rotating. I _am_ saying the field rotates with
the
magnet.
|> 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 disagree. I say case #3 is effectively 1 battery, in the wiring alone.
I say no potential is induced in the disk itself.
|> |> 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.
I suggest it is functionall the same as rotating the disc AND magnet
together
with the wiring stationary (as in case #3). In case #4 the wiring cuts
through
the stationary magnetic field, inducing a potential that, being unopposed,
will
result in a current if the loop is complete.
|> |> | 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.
I agree that assuming the field does not rotate explains all the cases and
the observed effects. What I am saying is that assuming the field really
DOES rotate with the magnet ALSO explains all the cases and the observed
effects.
I have reasons outside of the scope of these 5 cases to believe that the
field rotates with the magnet. All I can say within the context of these
cases is that the observations do not rule out that the field rotates with
the magnet.
|> 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.
And why is this not equivalent to a moving field?
Consider case #0. It's on the surface of a moving planet. Is it moving
with
respect to what you describe?
| 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.
I do treat electric and magnetic fields as a single entity, with separate
ways to observe them.
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