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Re: "Faraday's Disk" which started it all



----- Original Message -----
From: William Beaty <billb@ESKIMO.COM>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Thursday, July 01, 1999 7:07 PM
Subject: Re: "Faraday's Disk" which started it all


On Thu, 1 Jul 1999, Bob Sciamanda wrote:

For details of how a current loop (a magnetic dipole) is also an
electric
dipole when viewed in motion, see pg 340 of Panofsky & Philips (1rst
ed).
Essentially they show that an upright, rectangular current loop carrying
a
ccw current, when viewed moving to the right develops a positive charge
in
the upper leg and a negative charge in the bottom leg - an electric
dipole -
because of the different carrier velocities in the two legs.

This agrees with my own understanding. Because of Lorentz contraction,
the two parallel legs develop opposite charges, and an e-field appears
between them. (Even so, I can't clearly visualize what happens at the
corners of such a loop, where the altered charge-density of the
parallel-moving wire connects to the UN-altered charge density of the
perpendicularly-moving wire.)

As was true at the start of "simple magnet question", my main troubles
arise when the motion is rotational. If the two parallel legs of the
square current-loop do not move at the same speed, does one leg become
less charged than the other? Or a similar question: if we begin to rotate
a circular current-loop about its axis, does the charge density in the
loop appear to change because of Lorentz contraction? Does an uncharged
current loop appear to become charged if we force it to rotate? Such a
thing is impossible. Yet I cannot see any obvious way to avoid it. If
there is a solution to this, perhaps it is outside of SR.

From the hip: If a circular loop of current rotates about its own axis,
there is no definable up-down direction in which it can develop a charge
separation and an electrical dipole moment, as there is when it is
translated in a linear direction (as I indicated above). It would seem that
there is thus reason to speculate that the rotating magnet does NOT develop
an electrical polarization (and its E' field) as does a translating magnet.
This may be another instance of the pitfalls inherent in assuming that one
can apply SR to a rotating frame by considering an infinity of inertial
frames, each co-moving with a particular point fixed in the rotating frame.

In other words, a cyclotron can deflect an electron because, in the
inertial frame where the electron is (momentarily) stationary, all the
little dipoles in the cyclotron magnet are moving, and together they
are
creating an e-field which the electron can experience. Is this an
incorrect model?

The cyclotron is best understood in the lab frame (the rest frame of the
cyclotron, its magnet, the operating personnel, etc - the frame which
inspired its invention).

Then we have a disagreement in philosophy. I want to obey SR and insist
that the "lab frame" can vary: move a magnet past a test-charge, or move
a test-charge past a magnet. In either case a force will arise. I keep
bringing up the cyclotron because it is a clear example of a system where
a test-charge is deflected as a result of the *relative* motion between
a test-charge and a magnet, yet the b-field is uniform. If the force
depends only on the *relative* motion between the two, then we must not
shy away from holding the test-charge stationary and moving the magnet
instead.

I do not object to looking at phenomena from various frames, this is often
very helpful; but you somehow seem to ALWAYS prefer the rest frame of q as
preferred for purposes of understanding. F=q(E+VxB) is valid in any frame.
The frame in which V=0 is not to be preferred. Which frame is most useful
for understanding is dependent on the particular problem and the particular
tastes of the particular observer. But, that Maxwell's equations will
transform intact into any inertial frame is verifiable in principle, and
need not be tested in each individual case. (The actual constitution of
material reality, and the existence of accelerating frames are, of course,
confounding considerations.)

In this frame the electrons (while inside the dees) are subject only to
a vertical magnetic field and so are forced into a horizontal, circular
orbit by the QVxB magnetic force.

If I (now) understand things correctly, the qVxB force depends on the
frame of the observer: the force q(E + VxB) is "real" and unchanging, but
if the observer jumps to a different inertial reference frame, then E and
VxB will vary (although their sum will not.)

You fell off the wagon again, Bill! In F=Q(E+VxB) , only Q is numerically
the same in all inertial frames; V, E, B AND F transform into different
values V', E', B', AND F'. This is a substantive departure from Newtonian
mechanics. Even INERTIAL frames will differ in their evaluation of the
force experienced by an observed particle (this is a second order effect, so
that for non-relativistic velocities, you are approximately correct!).

I was confused earlier because E does vary as the relative motion of the
magnet varies, but this is not specifically expressed in the equation
above.

If you need to see the complete transformation equations for the general
situation, ask and I'll try to produce them.

Things are much more complicated when viewed from the non-inertial rest
frame of the electron and no more insightful understanding of the
apparatus
is gained for your trouble.

True. When discussing an electron which moves between the cyclotron's
poles, we must focus on the forces felt by an electron *at a particular
instant.* Either that, or assume that we could restrain the electron so
it moves uniformly, and then we could measure the perpendicular force.

F=q(VxB) is valid in the (inertial) LAB frame, EVEN THOUGH THE ELECTRON IS
ACCELERATING IN THAT FRAME. There is no need to stop it, or jump onto its
frame.

Finally, what happens when we dangle a test charge above the flat face
of
a *rotating* disk-magnet, where the plane of the disk is horizontal,
the
b-field points upwards, and the test-charge is not aligned with the
axis
of the disk? Won't this test-charge experience an e-field and
therefor a
force? After all, that test-charge is close to billions of tiny moving
dipoles within the magnet, and in the frame of that stationary
test-charge, each of those dipoles creates an e-field. If this is
wrong,
where does my error lie?

What you say is true. The rotational case is not easily analyzed, eg.
by a
simple Lorentz transformation, and I confess that I do not have a
detailed
understanding here.

Since this apparantly is a controversial topic, then there is a small
chance that *nobody* has a detailed understanding. Maybe not a "hole in
physics", but a foggy region which needs some illumination.


It seems that the external field will be very much
weaker than in the translational case, as I proposed in an earlier post.

I probably misunderstood. I thought you had that the e-fields of moving
magnets in general would be limited to the physical substance of the
magnet. Yet this contradicts q(E + VxB) if the test charge is stationary
and the magnet is moving. If the e-field outside of a linearly-moving
magnet is strong, then MAYBE the e-field outside of a rotating magnet is
also strong. Or maybe it is weak as a consequence of the geometry of
rotating systems.

A thought... If we magnetize a spherical shell such that the direction of
magnetization is radial, we do not obtain a monopole! Instead the
external fields sum to zero. Perhaps something similar applies to the
e-field of a rotating disk-magnet. (If so, then the concept-network by
which I understand Homopolar Generators is totally wrecked.)


A
lucid, detailed analysis of this is hard to come by, and what is
available
is often questionable. You have provoked me to make the time to
seriously
search for a better understanding here.

I hope those references in my other message end up shedding some light
rather than just deepening the controversy.


Anyway, experimentally a stationary
copper disc and a rotating disc magnet will NOT work as a generator.
The
homopolar generator only works when the QVxB force is used by rotating a
conductor in the magnetic field (as noted before, this conductor can be
the
magnet itself.) So, experimentally this external E field is zero or very
weak in the rotational case.

I must disagree. True, if we rotate only the magnet, then there can be no
"generator effect" and no current. This doesn't prove that there is no
e-field around the rotating magnet. If a cylindrical/radial e-field does
exist, then if we take the line-integral of the force it applies to each
charge in a nearby closed circuit, the force should sum to zero, and
therefor the e-field cannot create a direct current. Current can only
flow in the generator circuit when there is relative motion between the
conductive sections: motion as measured between and the wires of the
external circuit the conductive disk. If this is true regardless of the
strength of the e-field, then a Homopolar generator says nothing about the
existence of that e-field.


I do recall someone (a very early author) calling the unipolar generator
(simply brushes on a rotating magnet) the "electromagnetic version of
the
Foucault pendulum" since it shows that "absolute" rotation is
measureable.

I think this person is wrong. To generate a current, relative rotation
between the "conductor disk" and the "external circuit" is necessary. If
we keep the "external circuit" stationary and ignore that we have done so,
then it will *appear* we can detect the absolute rotation of the
conductive disk. If we hold the conductive disk stationary and rotate the
brushes, the meter, and the external circuit, we still obtain a current.
This shows that the Homopolar Generator does not rely on absolute
rotation. The rotation is always a relative motion between a "rotor" and
a "stator" section. The meter and its leads are the "stator". Hold the
"rotor" still and spin the "stator" around it, and the generator still
functions.

The rotating conducting magnet, all by its lonesome, will develop a charge
separation, from the Q(VxB) motional magnetic force on its moving carriers
if brushes are added, a current will be available). Again, the rotating
frame is more complicated than an infinity of inertial frames.

Thanks for the references; the first two are new to me - I will inquire (and
hopefully learn) from them.

William J. Beaty SCIENCE HOBBYIST website

-Bob

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor