Faraday's disk and "absolute" magnetic fields

[vanhees71]

So, when a disk-shaped magnet spins, no electric field is detected. Why is that?

Let's say the disk is magnetic because there are microscopic current loops all over the disk. When the disk is spinning, we can say a microscopic current loop is approximately in linear motion. Now we know that a moving current loop is an electric dipole, there was a discussion about that some time ago here.

So, what kind of macroscopic electric field is caused by those microscopic dipoles? Well the rim of the disk becomes charged. Every point of such disk has the same potential. So there are no currents in static wires that are brushing the disk.

Read my FAQ article! It's just using a spherical magnet, because then everything can be calculated in terms relatively simple standard functions. The short message is: When a magnet spins in general you have an electric field, and that's where the voltage measured in the homopolar-generator setup comes from. It teaches us once more that electrodynamics is a relativistic phenomenon, and thinking in non-relativistic terms can be misleading. All socalled paradoxes are gone, as soon as you use fully relativistic reasoning, among other things the apparent Faraday-disk paradox, the phenomenon of socalled "hidden momentum", which is not hidden but just an incomplete balance equation when using the electromagnetic momentum and at the same the non-relativistic approximation for mechanical momentum of the moving charges etc. etc.

Here's the link to the FAQ article again:

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

 

[Benbenben]

... When a magnet spins in general you have an electric field, and that's where the voltage measured in the homopolar-generator setup comes from. It teaches us once more that electrodynamics is a relativistic phenomenon, and thinking in non-relativistic terms can be misleading. All socalled paradoxes are gone, as soon as you use fully relativistic reasoning, among other things the apparent Faraday-disk paradox...

.
A 'belief' in the absolute superiority of one particular model is also a path that can mislead. Regardless of how elegant and complete a model (relativity in this case) might seem, when the predictions don't align with empirical data, then the implementation or the model is misleading
The data show spinning the magnet has no effect on voltage in a faraday disc. Only the relative movement of the two parts of the circuit (disc and return path).
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"Essentially, all models are wrong. Some are useful" - George E. P. Box.

 

[rrogers]

.
A 'belief' in the absolute superiority of one particular model is also a path that can mislead. Regardless of how elegant and complete a model (relativity in this case) might seem, when the predictions don't align with empirical data, then the implementation or the model is misleading
The data show spinning the magnet has no effect on voltage in a faraday disc. Only the relative movement of the two parts of the circuit (disc and return path).
.
"Essentially, all models are wrong. Some are useful" - George E. P. Box.

"The data show spinning the magnet has no effect on voltage in a faraday disc"
Can I have a reference for the experiment?

Reference https://www.physicsforums.com/threa...te-magnetic-fields.918885/page-4#post-5798780

 

[Benbenben]

"The data show spinning the magnet has no effect on voltage in a faraday disc"
Can I have a reference for the experiment?

Reference https://www.physicsforums.com/threa...te-magnetic-fields.918885/page-4#post-5798780

References abound. Reply #46 details in table form the results of the various combinations of revolving or stationary for the magnet, disk and return circuit. These results occur every time. Note that spinning the magnet or holding it stationary has no effect and that current is only produced for conditions with relative motion between the disc and return circuit.

 

[Dale]

Regardless of how elegant and complete a model (relativity in this case) might seem, when the predictions don't align with empirical data, then the implementation or the model is misleading

Uhh, except that relativity does align with the data. That should be clear since Maxwell's equations are relativistic.

 

[jartsa]

Read my FAQ article! It's just using a spherical magnet, because then everything can be calculated in terms relatively simple standard functions. The short message is: When a magnet spins in general you have an electric field, and that's where the voltage measured in the homopolar-generator setup comes from. It teaches us once more that electrodynamics is a relativistic phenomenon, and thinking in non-relativistic terms can be misleading. All socalled paradoxes are gone, as soon as you use fully relativistic reasoning, among other things the apparent Faraday-disk paradox, the phenomenon of socalled "hidden momentum", which is not hidden but just an incomplete balance equation when using the electromagnetic momentum and at the same the non-relativistic approximation for mechanical momentum of the moving charges etc. etc.

Here's the link to the FAQ article again:

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

Well, I understood some of that.

My idea was that there is a disk-shaped magnet, and there is a charge floating nearby the magnet. If the magnet moves linearly, the charge feels a Lorentz-force. When the magnet spins, the charge feels different Loretz-forces from different parts of the magnet, as the different parts move almost linearly to different directions. The sum of those Lorentz-forces is zero.

 

[vanhees71]

Hm, but as with my example of the rotating spherical magnet, also a rotating disk-shaped magnet should result in an electric field, and thus there should be a force on the charge near the magnet.

Concerning the other discussed setup with the rotating cylinder, I guess it's more demonstrating the Hall effect, i.e., conduction-charge separation when the cylinders are rotating due to the Lorentz force. If you rotate the magnet instead, the electric field may be way too weak to have a measurable effect on the charges of the cylinders.

The idea that relativity is disproven by Faraday's findings is absurd, because to the contrary only with relativity you can correctly describe these effects, and the Maxwell equations are the paradigmatic example for a relativistic field theory. That's why relativity was discovered by analyzing the Maxwell equations.

 

[Benbenben]

Uhh, except that relativity does align with the data. That should be clear since Maxwell's equations are relativistic.

Relativity does provide a useful model, however if your interpretation predicts a meaningful effect on current produced the spinning of the magnet (in the classic Faraday disc set-up); your implementation is doing a disservice to relativity.
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Please take notice of the bias favoring theory over impiricism in the short response you did leave. Your argument is nonsequitur. The fact that Maxwells equations are relativistic has exactly squat to do with alignment with a real set of data.

 

[Paul Colby]

The idea that relativity is disproven by Faraday's findings is absurd,

Agreed. However, your rotating magnet example has no return path for a current. Topologically it's not a circuit so any electric field will move charge till there is no field because the magnet is also conducting. If I understand correctly, it's only when the stator-return-path part rotates relative to the conducting disk part that there is a current. The "reason" for the current in the relatively rotating parts is the flux enclosed by the circuit is changing and the charge is free to flow around a closed circuit. If the stator-return-path part is stationary with respect to the conducting disk path there is no net flux change and no current flow.

 

[Benbenben]

Hm, but as with my example of the rotating spherical magnet, also a rotating disk-shaped magnet should result in an electric field, and thus there should be a force on the charge near the magnet.

Concerning the other discussed setup with the rotating cylinder, I guess it's more demonstrating the Hall effect, i.e., conduction-charge separation when the cylinders are rotating due to the Lorentz force. If you rotate the magnet instead, the electric field may be way too weak to have a measurable effect on the charges of the cylinders.

The idea that relativity is disproven by Faraday's findings is absurd, because to the contrary only with relativity you can correctly describe these effects, and the Maxwell equations are the paradigmatic example for a relativistic field theory. That's why relativity was discovered by analyzing the Maxwell equations.

I don't see anyone claiming relativity is disproven. I would say that if you predict that spinning just the magnet in a classic Faraday disc set up will lead to current, then your implementation of relativity is flawed.
Also, your claim that 'only with relativity can you correctly describe there effects is unsupportable. Moreover it mistakes a modep of reality for the fundamental causes.

 

[Dale]

Your argument is nonsequitur. The fact that Maxwells equations are relativistic has exactly squat to do with alignment with a real set of data.

You are correct I did not explicitly make the complete argument, but I thought the rest of the argument would be understood by anyone with a basic understanding of the homopolar motor, Maxwell's equations, and relativity.

Maxwell's equations correctly predict the empirically observed behavior of the homopolar motor, Maxwell's equations are relativistic, therefore relativity also does "align with the data" for the homopolar motor.

 

[vanhees71]

No, I just calculated the electric field. To measure the corresponding voltages you should use a volt meter with high resistance. I guess, it's not a good generator in the sense that you can get high currents from it.

It's also very difficult to discuss only very vaguely defined setups. Is there somewhere a clear description of the experiments with rotating cylinders discussed? I have a vague idea what it might be about, and if so, it's very easy to understand that this is a rotating conductor in an external magnetic field. Then you have of course charge separation due to the Lorentz force on the electrons building up a counteracting electric field, leading to a charge of the cylindrical capacitor, which you can measure with help of a volt meter after taking away the connecting wire. This is very similar to the DC Hall effect, where the only difference is that in the usually discussed version of the Hall effect the current is a conduction current due to a voltage applied to the conductor, while here it is a convection current due to the rotation of the cylinder. The physics is the same: Maxwell's equations + the law of the Lorentz force $\vec{F}=q (\vec{E}+\vec{v}/c \times \vec{B})$, both together forming a set of equations that are of course fully compatible with relativity (but not with Galilei symmetry).

Another question is what happens, if you rotate the magnet. Since, if I'm guessing the setup right, the magnet is just under the cylinders, when rotating the electric field induced along the cylinder is very weak, so that its effect may be well negligible. Of course, without a clear description of the experiment, it's impossible to say anything concrete. One thing, however, I'm pretty sure about is that it won't contradict relativity and the Maxwell equations. Of so, we'd all know this sensation very well ;-)).

 

[vanhees71]

You are correct I did not explicitly make the complete argument, but I thought the rest of the argument would be understood by anyone with a basic understanding of the homopolar motor, Maxwell's equations, and relativity.

Maxwell's equations correctly predict the empirically observed behavior of the homopolar motor, Maxwell's equations are relativistic, therefore relativity also does "align with the data" for the homopolar motor.

More so for the homopolar motor/generator you even need relativity to explain it right!

 

[vanhees71]

Can you please give a concrete setup of your experiment? What I completely disagree with is the claim that the Faraday disk experiment of any kind contradicts relativity. If so, it would be a sensational finding leading with a very high probability to a trip to Stockholm in December ;-).

Taking the setup in

https://en.wikipedia.org/wiki/Homopolar_generator#Disk-type_generator

you an pretty easily describe what's going on. Let's make the simplifying assumption that the magnetic field $\vec{B}=\text{const}$ along the disk. Then in the stationary state
$$\vec{j}=\sigma (\vec{E}+\vec{v}/c \times \vec{B})=0.$$
We have $\vec{B}=(0,0,B)$ and $\vec{\omega}=(0,0,\omega)$ and thus $\vec{v}=\vec{\omega} \times \vec{r}$ and thus
$$\vec{E}=-\frac{\vec{v}}{c} \times \vec{B} = -\frac{\omega B}{c} (\vec{e}_z \times \vec{r}) \times \vec{e}_z.$$
Now we have
$$(\vec{e}_z \times \vec{r}) \times \vec{e}_z =\vec{r}-z \vec{e}_z =(x,y,0).$$
Thus we have (inside the disk)
$$\vec{E} = -\frac{\omega B}{c} (x,y,0)=-\frac{\omega B}{2c} \vec{\nabla} (x^2+y^2).$$
Thus the electric potential is
$$\phi=\frac{\omega B}{2c} (x^2+y^2).$$
The voltage measured from the center to the rim of the disk thus is
$$U=\frac{\omega B R^2}{2c},$$
where $R$ is the radius of the cylinder.

All this is completely relativistic (particularly I used the complete relativistic version of Ohm's Law in the very beginning!).

 

[Benbenben]

You are correct I did not explicitly make the complete argument, but I thought the rest of the argument would be understood by anyone with a basic understanding of the homopolar motor, Maxwell's equations, and relativity.

Maxwell's equations correctly predict the empirically observed behavior of the homopolar motor, Maxwell's equations are relativistic, therefore relativity also does "align with the data" for the homopolar motor.

Like I continue to reitterate, yes relativity does provide robust predictions of reality. However if your interpretwtion of relativity has you predicting the rotation of the magnet alone meaningfully affects current in a classic faraday disk setup, then you are misapplying relativity.

 

[Benbenben]

Can you please give a concrete setup of your experiment? What I completely disagree with is the claim that the Faraday disk experiment of any kind contradicts relativity. If so, it would be a sensational finding leading with a very high probability to a trip to Stockholm in December ;-).

Please point out the person claiming relativity has been contradicted. ..all I see are people claiming someone has state relativity has been contradicted.

 

[vanhees71]

You plainly contradict my calculation about the rotating magnetic, which is based on relativity. The homopolar generator, however, works. You find it in many good textbook on electrodynamics (e.g., Becker+Sauter, Sommerfeld vol. III).

 

[jartsa]

Hm, but as with my example of the rotating spherical magnet, also a rotating disk-shaped magnet should result in an electric field, and thus there should be a force on the charge near the magnet.

Ok I admit that there is a force on a charge near a spinning magnet.

So the Lorentz-forces from all the microscopic magnets do not cancel out, like I said before. I only said the forces cancel out because I though there should be no force on the charge.

 

[Dale]

Like I continue to reitterate, yes relativity does provide robust predictions of reality. However if your interpretwtion of relativity has you predicting the rotation of the magnet alone meaningfully affects current in a classic faraday disk setup, then you are misapplying relativity.

That is substantially different from

Regardless of how elegant and complete a model (relativity in this case) might seem, when the predictions don't align with empirical data, then the implementation or the model is misleading

The "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".

 

[Buckethead]

I'm not clear on the setup you are describing.

Are the cylinders locked to rotate together? If this is the case, I am surprised this is occurring without relative motion between the two.

...or are the cylinders rotating with respect one another? If that is the case, it is difficult to understand how you can distinguish between the net of relative motion between each of the cylinders and the magnet as opposed to relative motion just between the cylinders.

The cylinders are always rotating together. Regardles if the magnet rotates or not, if the cylinders spin, there is a charge buildup. This is a beautiful example to show it is the relationship between the spinning object and the magnetic field and not between any two rotating objects that is key in the Faraday disk.

 

[Buckethead]

Yes, I agree.

I think that is correct also. If you were to write Maxwell's equations in an accelerating reference frame then they would look different than the usual form.

So something isn't working out here. If a wire accelerating and within the Helmholtz field is not symmetrical to a Helmholtz coil being accelerated with a non-accelerating wire in the field then this means if you do one or the other you will get two different voltages right? This would imply that if a wire stationary to the coil were to be mounted in a ship and if the ship were to accelerate then there would be a voltage generated. This couldn't be the case or due to the equivalence principle, such a stationary setup in a gravitational field would also generate a voltage.

 

[Dale]

This couldn't be the case or due to the equivalence principle, such a stationary setup in a gravitational field would also generate a voltage

What would be wrong with that?

 

[PAllen]

So something isn't working out here. If a wire accelerating and within the Helmholtz field is not symmetrical to a Helmholtz coil being accelerated with a non-accelerating wire in the field then this means if you do one or the other you will get two different voltages right? This would imply that if a wire stationary to the coil were to be mounted in a ship and if the ship were to accelerate then there would be a voltage generated. This couldn't be the case or due to the equivalence principle, such a stationary setup in a gravitational field would also generate a voltage.

The two gravitational equivalents would be a wire at rest on a planet with the coil free falling around it ( equivalent to case of inertial coil in SR), and coil at rest on planet with wire falling through (equivalent to accelerating coil case in SR). I would indeed expect these equivalent cases to behave identically if they are not too large (the principle of equivalence is a local principle).

 

[Buckethead]

The two gravitational equivalents would be a wire at rest on a planet with the coil free falling around it ( equivalent to case of inertial coil in SR), and coil at rest on planet with wire falling through (equivalent to accelerating coil case in SR). I would indeed expect these equivalent cases to behave identically if they are not too large (the principle of equivalence is a local principle).

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, but 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.

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).

I find this peculiar.

 

[PAllen]

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, but 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.

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).

I find this peculiar.

I think you misunderstood me. A wire accelerating through an inertial coil is POE equivalent to a wire resting on Earth as a free fall coil passes. In no way did I claim the equivalence you suggest. Please reread my prior post more carefully. Please note that a free fall coil is not accelerating per GR.

 

[Buckethead]

The two gravitational equivalents would be a wire at rest on a planet with the coil free falling around it ( equivalent to case of inertial coil in SR), and coil at rest on planet with wire falling through (equivalent to accelerating coil case in SR). I would indeed expect these equivalent cases to behave identically if they are not too large (the principle of equivalence is a local principle).

You said "( equivalent to case of inertial coil in SR)" i.e. an inertial coil with a wire accelerating through it at 1G and "(equivalent to accelerating coil case in SR)" i.e. an inertial wire with a coil accelerating around it at 1G. What did I misunderstand?

Oh I get it, you meant the wire at rest on Earth and the accelerating wire in space, were equivalent, not the two experiments.

 

[Buckethead]

OK back to sorting this out. So if an accelerating wire through an inertial coil does not generate the same voltage as an accelerating coil around an inertial wire, 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. This clearly does not happen, so what is going on?

 

[PAllen]

OK back to sorting this out. So if an accelerating wire through an inertial coil does not generate the same voltage as an accelerating coil around an inertial wire, 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. This clearly does not happen, so what is going on?

This looks like a totally separate case. Coaccelerating wire and coil, equivalent to wire and coil resting in earth. It is different from other cases discussed in the past several posts.

 

[Buckethead]

This looks like a totally separate case. Coaccelerating wire and coil, equivalent to wire and coil resting in earth. It is different from other cases discussed in the past several posts.

Not so different. But first, in the case of the objects resting on Earth, there is no voltage which means accelerating the coil or accelerating the wire must be equivalent. So far so good? 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 means the field lines when moving the coil must also be moving in a circle.

 

[PAllen]

Not so different. But first, in the case of the objects resting on Earth, there is no voltage which means accelerating the coil or accelerating the wire must be equivalent. So far so good? 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 means the field lines when moving the coil must also be moving in a circle.

No, the resting on Earth case suggests NOTHING other than linearly coaccelerating wire and coil will produce no current as measured by a coaccelerating detector.

 

[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.
.

 

[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.