B Faraday's disk and "absolute" magnetic fields

[sweet springs]

So you say post #46 was wrong? I would appreciate your correction for my study.

 

[jartsa]

So, there is a case of interest to me in which I'd like opinions on. If the brushes are replaced by fixed connections, say a screw and a length of wire attached to the disk. If the disk is reciprocated sinusoidally in angle and the magnetic is fixed, is there a voltage generated? Also, very important to me, is there a corresponding tongue associated with a sinusoidal applied current?

My opinion is:

1: There is a voltage generated, between the center of the disk and the rim of the disk. Because there is a Lorentz force exerted on the electrons on the disk ... if the electrons are forced to follow the rotation of the disk.

2: There is a torque generated, the magnet applies a torque on the disk, the disk applies a reaction torque on the magnet. To be more specific the current carrying electrons of the disk feel a Lorentz force, and those electrons transmit that force to the rest of the disk somehow, I guess the force is transmitted by friction, in other words by resistance.

I mean when voltage is applied between the center of the disk and the rim of the disk, then there exists those torques.

 

[Buckethead]

That's not what I said. I said that moving the coil would not produce a voltage in the wire because the change in the magnetic flux is zero (Faraday's law). I did not say that moving the wire in the field would not produce a voltage (Lorentz's law).

Perhaps you're getting confused because there are two different laws here?

It is my understanding that moving the coil will produce a voltage. We are looking at a relative situation here, moving the wire relative to the coil or moving the coil relative to the wire. I understand the formulas used in each case is different but the result has to be the same, this is just relativity. I must be misunderstanding you.

 

[Buckethead]

No, I wasn't kidding.

But I think I must be misinterpreting what you mean by "absolute rotation of the disk". Does that mean rotation "relative to the fixed stars" as in Newton's rotating bucket experiment?

More accurately, an object is considered rotating if it experiences rotational forces (proper acceleration due to rotation). If the disk experiences these forces than I call this absolute rotation but it is probably more proper to say a disk is rotating if its angular velocity is > 0.

 

[TSny]

More accurately, an object is considered rotating if it experiences rotational forces (proper acceleration due to rotation).

But we know that a current can be generated when the disk is not experiencing rotational forces if we rotate the external wires of the circuit instead of rotating the disk.

But, if I understand some of your previous remarks, you are not claiming that the circular disk must have absolute rotation. You are claiming that some part of the apparatus must have absolute rotation. So, in the picture on the left, it would be the disk that has absolute rotation while in the picture on the right it is the wires that have absolute rotation. But how could you prove that it's absolute rotation that matters, rather than relative motion of the disk and wires?

 

[Buckethead]

But we know that a current can be generated when the disk is not experiencing rotational forces if we rotate the external wires of the circuit instead of rotating the disk.
View attachment 206432
But, if I understand some of your previous remarks, you are not claiming that the circular disk must have absolute rotation. You are claiming that some part of the apparatus must have absolute rotation. So, in the picture on the left, it would be the disk that has absolute rotation while in the picture on the right it is the wires that have absolute rotation. But how could you prove that it's absolute rotation that matters, rather than relative motion of the disk and wires?

Because the disk has no effect on the wires. Move two parallel wires past each other in a magnetic field and each wire experiences forces only because of each wire and the field, not because you have two wires going past each other.

 

[Drakkith]

It is my understanding that moving the coil will produce a voltage. We are looking at a relative situation here, moving the wire relative to the coil or moving the coil relative to the wire. I understand the formulas used in each case is different but the result has to be the same, this is just relativity.

I can't disagree with that and I think we're a bit beyond my knowledge level here. I think I'll step out of this conversation before I put my foot in my mouth.

 

[Buckethead]

I can't disagree with that and I think we're a bit beyond my knowledge level here. I think I'll step out of this conversation before I put my foot in my mouth.

Oh, that's too bad. I've enjoyed your input and insight. It's been a helpful conversation. Thanks for sharing your thoughts.

 

[Drakkith]

Oh, that's too bad. I've enjoyed your input and insight. It's been a helpful conversation. Thanks for sharing your thoughts.

Thanks so much! One last thing, I think part of the problem in our final example is that when you're moving the Helmholtz coil in small circles, you're probably generating EM waves in the field, which we didn't take into account. I'm not sure how to account for those, so I'll have to let someone else step in.

 

[sweet springs]

Thanks TSny for post #55 that suggests to improve my post #46 , with signature to mention the direction of emf, i.e.

(d,c,m)
----------
(s,s,s)=0
(r,s,s)=+1
(s,r,s)=-1
(r,r,s)=0
(s,s,r)=0
(r,s,r)=+1
(s,r,r)=-1
(r,r,r)=0
----------
Best.

 

[Dale]

The wire is perpendicular to the field and the circle scribed by the wiggle is in the plane that is also perpendicular to the field

I haven't worked it out in detail, but I think there will be no EMF induced on such a wire from the static field. I.e. This is a physically different scenario from when the wire is wiggled in the homogenous region of the coil.

We are looking at a relative situation here, moving the wire relative to the coil or moving the coil relative to the wire.

It is not relative. Accelerometers attached to the wires or the coils will be different in the two different situations. These two situations are not symmetric, they are two physically different scenarios.

 

[Benbenben]

A Faraday disk (even one that allows rotation of the magnet, the disk, and the return circuit independently) alone cannot demonstrate whether or not the magnetic field rotates with the magnet or is stationary or is somethi ng for which rotatio is not an apt descriptor.
A Faraday dish (with indipendently rotating components) shows that the requirement for current is relative rotation between the disk and the remainder of the circuit.
.
Perhaps with a setup where the return ircuit is significantly shielded and a magnetic flux return path is built away from the return circuit more certainty woukd be available.

 

[Buckethead]

It is not relative. Accelerometers attached to the wires or the coils will be different in the two different situations. These two situations are not symmetric, they are two physically different scenarios.

This is a really good point and one that slipped by me entirely. If the wire is moving at a constant velocity linearly or the coil is doing so instead then I believe the situation is symmetric, but circular motion introduces acceleration and I can see how that would be different. I suppose it follows that even linear acceleration of the coil or wire is not symmetric. Is this true?

 

[Buckethead]

A Faraday disk (even one that allows rotation of the magnet, the disk, and the return circuit independently) alone cannot demonstrate whether or not the magnetic field rotates with the magnet or is stationary or is somethi ng for which rotatio is not an apt descriptor.
A Faraday dish (with indipendently rotating components) shows that the requirement for current is relative rotation between the disk and the remainder of the circuit.
.
Perhaps with a setup where the return ircuit is significantly shielded and a magnetic flux return path is built away from the return circuit more certainty woukd be available.

The Faraday disk is a clumsy setup when trying to analyze the relationship between the disk and the field but this relationship is put to rest in an experiment done in the early 20th century (sorry I don't have a name for this or a reference). The setup consists of 2 concentric metal cylinders with a small gap between them and a single wire connecting the two cylinders located halfway down their length. A round magnet is located at one end of the cylinders around the axis of the cylinders.

The cylinders are rotated and the magnet is either held stationary or rotated relative to the cylinders (doesn't matter). After a while, remove the wire and stop the rotation and measure the voltage between the two cylinders (which is essentially a capacitor). In both cases there is a voltage. You can now discharge the cylinders and just rotate the magnet and no charge will show. This indicates the rotation of the cylinder relative to the field and not the magnet is what is causing the voltage.

 

[Dale]

If the wire is moving at a constant velocity linearly or the coil is doing so instead then I believe the situation is symmetric,

Yes, I agree.

I suppose it follows that even linear acceleration of the coil or wire is not symmetric. Is this true?

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.

 

[rrogers]

I may have missed it but: You can represent the EM field (in this case a magnet) as a 4by4 antisymmetric matrix, a good representation of a tensor. This tensor is cast into other reference frames by multiplying the tensor with a lorentz matrix/tensor. Both the rows and columns have to be transformed so multiplication has to be on both sides when viewing the situation as a matrix. Now the electrons have their own reference frame that determines the Lorentz force they feel.
Just because the magnetic field isn't changing in magnitude; it doesn't imply it isn't changing. The changing can always be interpreted as the generated electric field. Orthogonal to the pointwise magnetic field and relative velocity vectors.
If the electron were rotating with the field it shouldn't experience a Lorentz force.
Unfortunately, I don't recall if there are second order effects due to the acceleration the electron would be undergoing. One is certain Cherenkov radiation.
I pretty sure the above is a good representation of the mathematics/physics but am open to constructive criticism; I have done any research on this for ages.

 

[Benbenben]

The Faraday disk is a clumsy setup when trying to analyze the relationship between the disk and the field but this relationship is put to rest in an experiment done in the early 20th century (sorry I don't have a name for this or a reference). The setup consists of 2 concentric metal cylinders with a small gap between them and a single wire connecting the two cylinders located halfway down their length. A round magnet is located at one end of the cylinders around the axis of the cylinders.

The cylinders are rotated and the magnet is either held stationary or rotated relative to the cylinders (doesn't matter). After a while, remove the wire and stop the rotation and measure the voltage between the two cylinders (which is essentially a capacitor). In both cases there is a voltage. You can now discharge the cylinders and just rotate the magnet and no charge will show. This indicates the rotation of the cylinder relative to the field and not the magnet is what is causing the voltage.

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.

 

[rrogers]

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.

I really need to have a reference for this experiment!

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.

I really need to have a reference for this! I did have the Faraday disk in mind at one point and the cylinders seem to contradict what I remember (not hard but I would like explicit details).

 

[jartsa]

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.

 

[Paul Colby]

There is no meaning to absolute rotation of the disk. As Paul Colby points out in post #20, the return path must also be considered.
Here is a video that shows various cases of parts of the setup moving relative to other parts.

The video link in your post is quite helpful. Relative motion of the disk and stator are clearly shown to be important. I think I converged to the wrong answer to the question I posed in post #47.

 

[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).
.
"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.
.
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.