B Faraday's disk and "absolute" magnetic fields

[Dale]

So a wire accelerating though an inertial coil, or a coil accelerating past an inertial wire generate the same voltage? It was mentioned in post 61 that the two situations are not symmetrical,

Post 61 didn't mention anything about this scenario. Post 61 was about the circular "wiggling" acceleration.

now I see what was meant was only the asymmetry between the two objects but this does not affect the outcome. Either one accelerating will generate the same voltage.

No. In post 61 the coil wiggling will not immediately generate voltage while the wire wiggling will. No wiggling acceleration is equivalent to linear acceleration or uniform gravity. Please do not blindly copy-and-paste answers like this.

If this is the case, then this gets me back to an earlier post where moving a wire in a circle (without rotating it) in a magnetic field is the same as rotating the coil in the same plane and leaving the wire stationary. In both cases a voltage will be generated in the wire. So again, this must mean that moving the coil in a circle (accelerating it) must also move the field lines such that they are forced to cut across the wire. So again, it seems you can move a single field line in a circle, but you cannot rotate the field lines about themselves (which would happen if you rotated the magnet instead of the disk).

No! This is highly frustrating. Please go back and re read the earlier comments. Gravity is not equivalent to wiggling something in a circle! Do not confound the two. Nothing we discuss here alters the previous answer.

 

[Dale]

then it is back to my statement that a wire inside a coil and stationary to it, mounted on an accelerating ship would generate a voltage and thus would also generate a voltage sitting on the Earth

Yes, this is the correct application of the equivalence principle.

This clearly does not happen, so what is going on?

It is not at all clear to me that this does not happen. Have you actually worked it out?

in the case of the objects resting on Earth, there is no voltage

Are you sure? What are Maxwell's equations in gravity, and have you solved them for this case? I haven't, so I would not be surprised to learn there is a voltage, but perhaps you have more experience with Maxwell's equations in gravity than I do

This means that moving a wire in a circle or moving the coil in a circle (both acceleration) would also generate the equivalent voltages.

This is completely inequivalent. It is frankly starting to irritate me. Go back to everything that was previously said. I am uninterested in repeating it again.

 

[Benbenben]

That is substantially different fromThe "or the model" part in particular is objectionable.

The model (relativity in this case) is consistent with all of the current and historical empirical data within its domain of applicability. People do make mistakes, but those are strictly "implementation" mistakes rather than "implementation or the model".

"Particularly ojectionable": that is bordering on dogmatism. I was merely pointing out the possibilities. While I doubt relativity is inconsistent with the outcome of these experiments, it should never be beyond question. That is a basic tenet of sound science.

What is far more likely is that your interpretation of relativity has gotten off track somewhere. Are you not claiming that a spinning magnet will induce a current in a set up like the faraday disk without relatove motion between the conducting disk and the remainder of the circuit? If you are, that is inconsistent with observation.

If that is not your assertion and if you insist that you are only making a claim about a spherical conducting magnet being spun, realize that your emphasis on the spinning of the magnet is wrongheaded. It is the conductive disk or sphere spinning in relation to the remainder of the circuit in a magnetic field that developes current. There is no spinning field. That is the beauty of separating the magnet from the conductor from the remainder of the circuit.
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[vanhees71]

Instead of whining, you should finally give a concise description of the experiments which make you doubt in the very foundation of physics which is among the best tested fundamental models ever, i.e., (special) relativity. My example of a spherical conducting magnet has been chosen, because you can solve the problem analytically with simple functions. It's not in principle different from any other shape of the magnet. Of course the details of the em. field depend on the shape of the spinning magnet, but it's not qualitatively different.

 

[Dale]

"Particularly ojectionable": that is bordering on dogmatism. I was merely pointing out the possibilities. While I doubt relativity is inconsistent with the outcome of these experiments, it should never be beyond question. That is a basic tenet of sound science.

This isn't how sound science works. In science an extraordinary claim requires extraordinary evidence. Nothing is beyond question, but all claims require evidence. Any claim seeking to overturn more than a century of data without any evidence is "particularly objectionable" and scientifically unsound.

There is no spinning field

I know that, I have been telling that to the OP from the beginning.

 

[Benbenben]

This isn't how sound science works. In science an extraordinary claim requires extraordinary evidence. Nothing is beyond question, but all claims require evidence. Any claim seeking to overturn more than a century of data without any evidence is "particularly objectionable" and scientifically unsound.

I know that, I have been telling that to the OP from the beginning.

What we have here is a failure to communicate. I don't have any reason to doubt general or special relativity. What I was saying was that if an interpretation of a theory doesn't fit observation then either the theory or application thereof must be flawed...with the implication that it must be the application.
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It did seem pretty clear that you were stating that the spinning magnet was motivating the voltage/current. If I misunderstood that or replied to the wrong comment, then I made a mistake.
If you are not saying the spinning magnet motivates the current/voltage, then I am not disagreeing with you,

 

[Dale]

It did seem pretty clear that you were stating that the spinning magnet was motivating the voltage/current.

Where? I certainly did not intend that, but perhaps I wrote something particularly poorly.

 

[sweet springs]

Hi. There is a famous "paradox" about a charged particle at rest in magnetic field. No force is working on the particle but In another Lorentz coordinate the particle is moving by v thus Lorentz force vXB works. The answer is in new Lorentz coordinate electric field appears and cancels Lorentz force.
This is a case of translation not rotation, but we can apply this result in parts of rotating magnet where the part is moving $v=r\omega$ and get similar results.

 

[vanhees71]

I still don't know the exact setup you are discussing. That's why it's not clear what's going on here. In general, a rotating conducting magnet also implies an electric field as demonstrated in my notes on the homopolar generator:

http://th.physik.uni-frankfurt.de/~hees/pf-faq/homopolar.pdf

To discuss why for some setup you don't measure this electric field, one needs to know the precise setup!

 

[sweet springs]

The answer is in new Lorentz coordinate electric field appears and cancels Lorentz force.

The electric field comes from plus and minus charged parts of the moving coil. Velocities of free electrons in the coil are $V+v$ or $V-v$ or so according to positions in the coil where v is velocity of free electron and V is velocity of Lorenz transformation.
Current is constant in the circuit so slower velocity part contains more electrons so is charged minus and faster velocity part contains fewer electrons so is charged plus.

 

[sweet springs]

This is a case of translation not rotation, but we can apply this result in parts of rotating magnet where the part is moving $v=r\omega$ and get similar results.

Applying the case of translation to small circuits that consist permanent magnet, it seems
\mathbf{P}=-(\omega \times \mathbf{r})\times \mathbf{M}
Revolving magnet is dielectric whose polarization is proportional to revolving radius at least in first order approximation of v/c.

 

[sweet springs]

Electric field outside revolving thus polarized magnet is order of (v/c)^2. As of first order approximation of v/c, electromagnetic field by revolving magnet, that was called "rotating magnetic field" by OP, is not different from (electro)magnetic field by magnet at rest. Re: post #46 and its improvement #60. For more precise consideration of (v/c)^2 or higher, we should go deeper into theory of both special and general relativity.

 

[vanhees71]

special relativity is sufficient here :-).

 

[Benbenben]

Okay, I think a simple experiment can establish that relative motion between parts of the circuit is a necessary part to generate current and that rotation of the conductor around an axis parallel with magnetic field lines will not generate a current if there is not relatove motion between the rotating conductor and the return path, even if the return path could be corotating (no relative motion) AND avoid cancelling the potential developed in the other portion of the circuit, what I am calling the conductor.
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Please bare with my explanation of the setup:
Start from a typical Faraday disk setup:
a conductive disk free to rotate about the axis of revolution;
with sliding electrical contacts one around the circumference of the disk and another on an area on and around the center;
concentric and offset below is a circular magnet just larger than the conductive disk which in this case is solidly attached to the conductive disk so as to spin with it.

If a return circuit is provided between the inner and outer sliding contacts and the disk (with accompanying magnet) is rotated sliding in the contacts, we are in agreement current flows in the circuit.

I believe I will still have general agreement if instead of the complete disc, we use just a pie slice on the disk, we will still get a current (albeit reduced) in the circuit as the inner and outer portions of the rotating pie slice of the conductor (with a matching pie slice magnet carried below) slide on the inner and outer contacts.

Taking a second similar pie slice of conductor and magnet, but this time with the polarity flipped along the axis of rotation, if we swap it out between the sliding contacts for the first pie piece and spin it the same direction, the current is reversed.
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The return part of the circuit connecting the inner sliding contact to the outer sliding contact can be removed if we complete the circuit by using the original pie piece and the reversed pole pie piece at the the same time, attached to rotate together on opposite sides of the circular path but electically isolated except for the sliding contacts at the center and outer edge.
As long as the sliding contacts are not in isolated segments (as might be use for commutation), but instead provide a low resistance path between the center of the original pie piece to the center of the pole reversed pie piece and also between the outer edge of the two pie pieces, the combination should function as a dc motor or generator.
If anyone believes this will not operate as a motor if current is made to flow through, or as a generator with a prime mover, please let me know why.
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If we are okay up to that point, then I can show why rotation of the conductors in the field alone is insufficient to develop current. ... the explanation is undeniably simple and doesn't require even the most cursory dip into relativity.
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If the contacts at center as well as the contacts at the outer edge are altered so that instead of sliding, the connections are fixed and rotate the the two connected pie pieces, then the assembly could not function as a homopolar motor. If a battery were mounted and current were driven from the center out the rim of one pie piece, along the circumference then back in the pie piece with the opposite polarity magnet, no torque could be developed without breaking conservation of (angular) momentum, as there is nothing off which it might produce countertorque.
...and if you can't get any rotation by passing current through the nonsliding embodiment, then no current will be produced by rotating the nonsliding embodiment.
Relative motion between parts of the circuit appears more and more as a sine qua non. Rotation (parallel to field lines) of a conductor in a magnetic field does appear insufficient without relative motion in the circuit.

 

[Buckethead]

I came across this which demonstrates the capacitor experiment I mentioned earlier. Both videos are well done and worth watching.

 

[rrogers]

I came across this which demonstrates the capacitor experiment I mentioned earlier. Both videos are well done and worth watching.

Okay, I understand.
Now if somebody could show me the mathematical derivation in terms of Maxwell's equations and/or the electromagnetic two-form, I would appreciate it. It would save me some trouble and thinking. Preferably with very few words.

 

[Dale]

Now if somebody could show me the mathematical derivation in terms of Maxwell's equations and/or the electromagnetic two-form,

The mathematical derivation of what?

 

[rrogers]

The mathematical derivation of what?

The voltage measured with different experimental setups?

 

[Paul Colby]

I'm struggling with this one. It seems the $v\times B$ gives the correct answer. The problem is what is $v$? In the video $v$ is the velocity relative to the lab frame and $B$ is the field in the lab frame. What I only partially get is that spinning the magnet doesn't change the magnetic field or induce an electric field in the lab (this is what I think is shown by the experiment). The one case not done is spinning magnet+disk+stator. My guess is this would produce no EMF which I find, well, confusing.

 

[Dale]

The voltage measured with different experimental setups?

I would start here: http://www.doiserbia.nb.rs/img/doi/0353-3670/2011/0353-36701102221B.pdf

 

[PAllen]

I'm struggling with this one. It seems the $v\times B$ gives the correct answer. The problem is what is $v$? In the video $v$ is the velocity relative to the lab frame and $B$ is the field in the lab frame. What I only partially get is that spinning the magnet doesn't change the magnetic field or induce an electric field in the lab (this is what I think is shown by the experiment). The one case not done is spinning magnet+disk+stator. My guess is this would produce no EMF which I find, well, confusing.

No, see the second video, which corotates the stator as well and still shows electric polarization.

 

[Paul Colby]

No, see the second video,

Thanks, I gave up too early. The experiment shown removes one of my major issues with the previous experiment which uses a stationary electroscope. Having the co-rotating data logger really removes changes of the circuit/current path while rotating. I haven't worked out the math but I would expect an $E$ field in the rotating frame since from the lab frame the rotating observer is after all rotating. Why is the claim made that this violates relativity? It now seems consistent to me.

 

[PAllen]

Thanks, I gave up too early. The experiment shown removes one of my major issues with the previous experiment which uses a stationary electroscope. Having the co-rotating data logger really removes changes of the circuit/current path while rotating. I haven't worked out the math but I would expect an $E$ field in the rotating frame since from the lab frame the rotating observer is after all rotating. Why is the claim made that this violates relativity? It now seems consistent to me.

I think it clearly does not violate relativity. Whether, historically, some relativity experts including possibly Einstein had an incomplete or incorrect understanding of the problem is not something I can give informed comment on.

 

[Dale]

I came across this which demonstrates the capacitor experiment I mentioned earlier. Both videos are well done and worth]

I just watched the video. I am not sure why he thinks his results contradict Einstein in any way. I think that the author has some misunderstanding about relativity.

 

[rrogers]

I would start here: http://www.doiserbia.nb.rs/img/doi/0353-3670/2011/0353-36701102221B.pdf

Nice paper :) Thanks, I think I can do better at the mathematics but I think this is concrete enough to write the equations.

 

[vanhees71]

Okay, I understand.
Now if somebody could show me the mathematical derivation in terms of Maxwell's equations and/or the electromagnetic two-form, I would appreciate it. It would save me some trouble and thinking. Preferably with very few words.

I don't know, what the movies are about, and I've not the time to watch them. I gave a derivation for the most simple example of a homopolar generator, i.e., a homogeneously magnetized rotating sphere, here:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/

Something seems to be wrong with our web server, but it should get online again very soon (so I hope ;-)).

 

[rrogers]

I don't know, what the movies are about, and I've not the time to watch them. I gave a derivation for the most simple example of a homopolar generator, i.e., a homogeneously magnetized rotating sphere, here:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/

Something seems to be wrong with our web server, but it should get online again very soon (so I hope ;-)).

Worked fine for me. I do have one complaint about the comments beyond Eq: 32. I think it is misleading since a loop of wire in a time-varying magnetic field certainly does demonstrate a curl-type electric field; i.e. a transformer. It's true that the original source might be electric but the local field doesn't care. All it needs is the boundary conditions and controlling PDE; which can be made around a very small region. That's the reason PDE's rule FEA's

 

[vanhees71]

Maybe, the statement is a bit misleading in the way I expressed it, but for sure one should write the Maxwell equations in the following form, grouping them in terms of homogeneous
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$
and inhomogeneous equations (from the point of view of the electromagnetic field),
$$\vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
When solving the equations for the full time-dependent problem, you can easily derive inhomogeneous wave equations for the electromagnetic field components, and you solve these for the usual situation that you have given charge-current distributions at sources of outgoing electromagnetic waves, using the retarded Green's function, which leads to the socalled Jefimenko equations. These are of a form, to be expected from a relativistic local field theory: They lead to local expressions, taking into account the finite speed of propagation (in this case of a massless field it's the speed of light). This solution has also the great advantage of being manifestly Lorentz covariant (when written in the appropriate way as integrals over Minkowski-space tensor fields).

You can, of course, write solutions of the time-dependent Maxwell equations in a way, only using the 3D Helmholtz fundamental theorem of vector calculus at a fixed time. Then you get a kind of "instantaneous" form of integral solutions. However they look very ugly and unintuitive by just looking at them ;-)). They are also completely useless to solve the problem. You find a nice discussion with a lot of equations in

https://arxiv.org/abs/1609.08149

 

[rrogers]

Maybe, the statement is a bit misleading in the way I expressed it, but for sure one should write the Maxwell equations in the following form, grouping them in terms of homogeneous
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$
and inhomogeneous equations (from the point of view of the electromagnetic field),
$$\vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
When solving the equations for the full time-dependent problem, you can easily derive inhomogeneous wave equations for the electromagnetic field components, and you solve these for the usual situation that you have given charge-current distributions at sources of outgoing electromagnetic waves, using the retarded Green's function, which leads to the socalled Jefimenko equations. These are of a form, to be expected from a relativistic local field theory: They lead to local expressions, taking into account the finite speed of propagation (in this case of a massless field it's the speed of light). This solution has also the great advantage of being manifestly Lorentz covariant (when written in the appropriate way as integrals over Minkowski-space tensor fields).

You can, of course, write solutions of the time-dependent Maxwell equations in a way, only using the 3D Helmholtz fundamental theorem of vector calculus at a fixed time. Then you get a kind of "instantaneous" form of integral solutions. However they look very ugly and unintuitive by just looking at them ;-)). They are also completely useless to solve the problem. You find a nice discussion with a lot of equations in

https://arxiv.org/abs/1609.08149

Actually, the using the differential notation/antisymmetric tensor form reduces the symbolic clutter a lot. You can fit rotations and velocity changes on one line. Actually, the physics fit's on four lines of crisp geometric objects and then everything else is Minkowski geometry. Simple things for simple people like me. I always try to keep in mind that our mathematics is just a language describing an underlying reality (I am a die-hard Neo-Platonist) and there are probably other better languages possible. Maxwell's equations versus the geometric differential forms is an example.