The Faraday disk in the context of the theory of relativity

[Ivan Nikiforov]

I do not find such a discussion in my second edition. Maybe I am a careless reader. I should appreciate it if someone who has the book check it also.

The following is literally written in this fragment.

"Consider the so-called "Faraday disk" (Figure 22.4). Its action is based on the phenomenon discussed in paragraph 9.5. Many properties of the Faraday disk are consistent with the conclusions of this paragraph. In particular, all the results summarized in Table 9.1 regarding the possible relative movements of the field source and the external circuit remain correct. The important conclusion also remains valid that the rotation of the magnetic field source does not affect physical phenomena if the field remains constant. As before, the electromotive force can be found from Faraday's law (9.2), determining the change in the magnetic flux through the circuit along which the current carriers move. The correct answer is also provided by a consideration based on the "effective force acting on an electron." However, this is where the commonality of inertial and non-inertial reference systems ends."

Ниже я привел фрагменты из книги, на которые указаны ссылки в тексте выше.

 

[Ivan Nikiforov]

As I answered in #11, no.

Thanks for the comment. I will take it into consideration and try to understand the error of my conclusions.

 

[anuttarasammyak]

The second bottom line of the table 9.1 means movement of galvanometer circuit only. Horizontal metal part of circuit moving u in magnetic field B. You see it in the figure.

 

[Ivan Nikiforov]

The second bottom line of the table means movement of galvanometer circuit only. Horizontal metal part of circuit moving u in magnetic field B. You see this part is in B in the figure.

Regarding the understanding of Table 9.1, I completely agree with you. But I set myself the task of finding out whether an electric current will be observed if the horizontal metal part of the circuit is not in the magnetic field B. That's the whole point of the question.
The fact is that upon further consideration of the electromagnetic processes occurring in a magnetic bar, nowhere is the requirement that both frames of reference must be in a magnetic field stated or mentioned. Moreover, the process of the appearance of an electric field on the sides of a magnet, which is detected by a moving observer, has nothing to do with whether this moving observer is in a magnetic field.

 

[anuttarasammyak]

What kind of configuration for an example to make B=0 at the circuit ?

 

[Ivan Nikiforov]

What kind of configuration for an example to make B=0 at the circuit ?

A possible version of this configuration is indicated in comment No. 5. The figure shows that the magnetic field is concentrated in a magnetic core, while the external circuit is not in a magnetic field.
For the figure in comment No. 5, the logic of my conclusions is as follows: 1) if we rotate the disk in this model, then electric current will flow in the system; 2) rotation of the disk and rotation of the external circuit are equivalent; 3) therefore, if in this model we rotate the external circuit, then electric current will flow in the system as well.

 

[Ivan Nikiforov]

I understand the physics of the process as follows. When the disk rotates relative to the magnetic field, an observer located in the disk detects an electric field on the sides of the magnetic field. This electric field passes through the disk and closes to an external circuit, as a result of which an electric current flows in the system. When an external circuit rotates relative to the magnetic field, an observer located in the external circuit detects an electric field on the sides of the magnetic field. This electric field passes through an external circuit and closes onto a disk, as a result of which an electric current flows in the system. Thus, the process of detecting an electric field is the same in both cases, with the only difference being that in the first case, the observer is in a magnetic field, and in the second case, the observer is not in a magnetic field.
This understanding of the process fully corresponds to the fundamental statement: "An observer moving at a speed u relative to a medium with magnetization M detects an electric moment P equal to:

This fundamental statement does not specify that the observer must be in a magnetic field.

 

[renormalize]

This understanding of the process fully corresponds to the fundamental statement: "An observer moving at a speed u relative to a medium with magnetization M detects an electric moment P equal to:
View attachment 357079
This fundamental statement does not specify that the observer must be in a magnetic field.

Perhaps I missed it earlier, but could you please cite a specific source for this statement, preferably written in English? Thanks.

 

[anuttarasammyak]

A possible version of this configuration is indicated in comment No. 5. The figure shows that the magnetic field is concentrated in a magnetic core, while the external circuit is not in a magnetic field.

I am afraid metallic surface (blue lines) is kept equipotential during the process so no emf observed. If I misunderstand the configuration please correct me.

 

[TSny]

Regarding the figure in post #5, Maxwell's equations do not allow the B field to have the pattern shown. In particular, boundary conditions for $\mathbf B$ are violated at the interface between the core and the vacuum.

For example, in region $a$ shown above, the normal component of B is discontinuous at the interface. This violates $\nabla \cdot \mathbf B = 0$.

At the interface in region $b$, the tangential component of B is discontinuous. This violates $\nabla \times \mathbf B = \mu_0 \mathbf J + \varepsilon_0 \mu_0 \dfrac{\partial \mathbf E}{\partial t}$ assuming that there are no surface currents on the interface.

 

[anuttarasammyak]

Re:#31
I found a correspong page in my book "18-6 Transformation properties of the partial fields". I observe in "Fig.18-6 Faraday disk illustrating unipolar induction in rotation" that B perpendiculat to paper exist for both disk and galvanometer circuit. So I am not surprized that it behaves same as Table 9-1 of linear case.

 

[Ivan Nikiforov]

Regarding the figure in post #5, Maxwell's equations do not allow the B field to have the pattern shown. In particular, boundary conditions for $\mathbf B$ are violated at the interface between the core and the vacuum.

View attachment 357086

For example, in region $a$ shown above, the normal component of B is discontinuous at the interface. This violates $\nabla \cdot \mathbf B = 0$.

At the interface in region $b$, the tangential component of B is discontinuous. This violates $\nabla \times \mathbf B = \mu_0 \mathbf J + \varepsilon_0 \mu_0 \dfrac{\partial \mathbf E}{\partial t}$ assuming that there are no surface currents on the interface.

Thanks for the comment. Below, I have provided several drawings showing various generator options. In general, the design of a practical model can vary in a wide range of options. But this has nothing to do with the original question. For now, we may not consider the specific design at all, but simply set the initial condition: the external circuit is not in a magnetic field.

 

[Ivan Nikiforov]

Re:#31
I found a correspong page in my book "18-6 Transformation properties of the partial fields". I observe in "Fig.18-6 Faraday disk illustrating unipolar induction in rotation" that B perpendiculat to paper exist for both disk and galvanometer circuit. So I am not surprized that it behaves same as Table 9-1 of linear case.

I made a link to this part of the book in order to argue that, as applied to the problem, rotation and rectilinear motion are equivalent. This is exactly the doubt you originally had. I did not claim that this picture shows a variant in which the external circuit is not in a magnetic field. Moreover, you were able to make sure that the disk and the magnetic bar, as applied to the task, are the same thing.

 

[Ivan Nikiforov]

It seems to me that considering this process in the framework of the theory of relativity well explains why a charged particle moves along a circular trajectory under the influence of the Lorentz force. Usually, the reasons for the circular motion of a particle are not considered in detail in the literature. I'll make the following assumption. At the initial moment of time, an observer located in a charged particle detects an electric field on the sides of the magnetic field. This electric field acts on the particle and deflects its velocity vector. As a result of changing the direction of the velocity vector, the vector of the electric field that the particle detects changes. This in turn leads to a further deviation of the velocity vector. This process occurs continuously, as a result of which the charged particle moves along a circular trajectory. This assumption is in good agreement with the conclusions mentioned in comment No. 37 and may be their confirmation.

 

[anuttarasammyak]

I made a link to this part of the book in order to argue that, as applied to the problem, rotation and rectilinear motion are equivalent. This is exactly the doubt you originally had.

I believe I am not included in "you". I just said "no B no emf". I refer rotation only in [EDIT] of #18 for another context.

Moreover, you were able to make sure that the disk and the magnetic bar, as applied to the task, are the same thing.

It should be mentioned that Panofsky Philips said in the end as the difference with SR,
" "absolute" rotational motion of the disk (i.e. the motion relative to an inertial frame )
can in principle be dermined ".
[EDIT]
Rotation does not have same relativity as translational motion in SR has.

 

[renormalize]

I did not claim that this picture shows a variant in which the external circuit is not in a magnetic field.

I find the back and forth discussion here confusing. Could you please clarify what you're asking by describing your configuration in very general terms? In particular, can you answer the following:

1. Does your configuration of interest consist of at least two subsystems, each made of uncharged but electrically-conductive materials (e.g., metal wires, plates, disks, etc.)?
2. Are these subsystems in relative motion with respect to each other?
3. Are these subsystems immersed in an external magnetic field and/or electric field, or are they in no field whatsoever?

Thanks!

 

[Ivan Nikiforov]

I find the back and forth discussion here confusing. Could you please clarify what you're asking by describing your configuration in very general terms? In particular, can you answer the following:

1. Does your configuration of interest consist of at least two subsystems, each made of uncharged but electrically-conductive materials (e.g., metal wires, plates, disks, etc.)?
2. Are these subsystems in relative motion with respect to each other?
3. Are these subsystems immersed in an external magnetic field and/or electric field, or are they in no field whatsoever?

Thanks!

I will provide answers to your questions.: 1) Yes; 2) Yes; 3) The disk (or magnetic bar) is in a magnetic field, and the external circuit is not in a magnetic field. In order to get an answer to the first question of my topic, I suggest not delving into the specific construction yet, as this unproductively complicates the consideration. I suggest using the classic Faraday disk as a basis and simply making the scientific assumption that the external circuit is not in a magnetic field. Will current flow under such conditions if only the external circuit is rotating?

 

[Ivan Nikiforov]

I believe I am not included in "you". I just said "no B no emf". I refer rotation only in [EDIT].

It should be mentioned that Panofsky Philips said in the end as the difference with SR,
" "absolute" rotational motion of the disk (i.e. the motion relative to an inertial frame )
can in principle be dermined "

Please excuse me if I inadvertently made a tactless remark and chastised you in some way. In fact, I am very grateful to you and everyone who participates in this discussion. You can't even imagine how long I've been waiting for the moment to talk about this topic with the scientific community. I can only say that I first started researching this topic back in 2004. You can say that this is my life's work. However, of course, this does not mean that I am right about everything. As Descartes said, "Question everything."

 

[renormalize]

Will current flow under such conditions if only the external circuit is rotating?

Is the rotating external circuit in electrical contact with the disk through something like carbon motor brushes that touch the stationary disk at 2 different points outside of the magnetic field? Or is the circuit like a floating piece of disconnected wire traversing a circular path completely external to the disk?

 

[anuttarasammyak]

https://www.feynmanlectures.caltech.edu/II_17.html
It seems that we can make B at the circuit weak enough by magnet spot in this configuration. Does it help you @Ivan Nikiforov ?

 

[Ivan Nikiforov]

Is the rotating external circuit in electrical contact with the disk through something like carbon motor brushes that touch the stationary disk at 2 different points outside of the magnetic field? Or is the circuit like a floating piece of disconnected wire traversing a circular path completely external to the disk?

Yes, that's right. The rotating external electrical circuit has sliding contacts (carbon brushes), which are installed at the boundary of the magnetic field.
Brushes cannot be installed outside the boundaries of the magnetic field. At least one brush must be installed directly at the edge of the magnetic field of the disc.
This may seem paradoxical, but my experiments confirm it. I believe that this phenomenon can be explained as follows.
An observer located in an external circuit detects an electric field directly at the boundaries of the magnetic field. At the same time, the observer located in the disk does not detect any electric field. That is, this electric field exists only for a moving external circuit. In order for the electric field to enter the external circuit, it must be in direct contact with at least one brush. If the brushes are located at some distance from the observed electric field, despite the fact that they are attached to the metal of the disc, the electric field does not reach them, since this field does not exist for the disc and it does not conduct it. This is a very important point to consider when designing.

 

[Ivan Nikiforov]

View attachment 357117
https://www.feynmanlectures.caltech.edu/II_17.html
It seems that we can make B at the circuit weak enough by magnet spot in this configuration. Does it help you @Ivan Nikiforov ?

In fact, this model is not much different from the classic Faraday disc. The fact is that the magnetic field lines will leave the N pole, pass through the disk and turn towards the S pole. On their way, they will cross the wire of the external circuit. In fact, despite the apparent simplicity of such a generator, it is very difficult to choose a design in which the magnetic field will have the desired geometry and will not cross the external circuit. I have reviewed many designs and so far the best result is shown in the picture in comment No. 5.

 

[TSny]

I need to correct a statement that I made in post #40 regrading region $b$ in the figure below:

I blundered in stating the following

At the interface in region $b$, the tangential component of B is discontinuous. This violates $\nabla \times \mathbf B = \mu_0 \mathbf J + \varepsilon_0 \mu_0 \dfrac{\partial \mathbf E}{\partial t}$ assuming that there are no surface currents on the interface.

A discontinuous jump in the tangential component of the magnetic field at region $b$ does not violate the Maxwell equation. The magnetization of the material produces an effective "bound" current at the surface of the material. Therefore, there will be a discontinuous jump in tangential component of $B$ across the boundary at $b$. The B field in the region of the external circuit will be small compared to the field inside the magnetized material.

I believe this means that the apparatus will not be very effective in generating current in the external circuit. And this will be true whether you (1) rotate the material while keeping the external circuit at rest or (2) keep the material at rest and rotate the external circuit.

 

[anuttarasammyak]

As an amateur of electric engineering, I find shielding static magnetic field is a challenging task in real world ,e.g. in medical NMR.
Meisner effect may be useful by making the external line with superconductor. But a fundamental question here. Does Lorentz force, which is zero for zero B, apply electrons in moving superconductors or not ? Does Faraday disk made of superconductor generate emf ?

 

[pervect]

View attachment 356917
Good afternoon! I would like to understand how the Faraday disk works and get answers to two questions. The working conditions are as follows: 1) we rotate the external circuit with a voltmeter relative to a stationary disk; 2) the external circuit is not in a magnetic field. Questions: 1) will an electric current flow in such a system? 2) what forces will act on the external circuit with the voltmeter? I would like to consider these processes precisely from the point of view of the theory of relativity, since the Faraday disk is based on the principle of relativity of simultaneity. Thank you in advance!

I am a bit confused - isn't the whole point that the rotating part of the disk is in a magnetic field? I'll assume that this was a typo.

One way of describing the relativistic point of view is that the electric and magnetic fields are part of the Faraday tensor, which is a rank 2 anti-symmetric tensor. This tensor has six non-zero components (there are 16 compoonents in a rank 2 tensor, 4 of them are zero, and because it is anti-symmetric 6 of the 12 components are unique. We identify 3 of the components as the electric field, the other three as the magnetic field.

The wiki article on the Faraday tensor is here and describes the arrangement of the components, i.e. how you pack the 6 different components into a 4x4 matrix. There are some factors of c, the speed of light, when conventional units are used, but it is easier and common practice to use units where the speed of light is equal to 1 to avoid having to worry about this factor. The wiki articles will also illustrate the mathematical simplicity of the Lorentz force law and Maxwell's equations when the formulation using the Faraday tensor is viewed.

One can also describe the relativistic point of view without tensors, using the traditional E and B fields, though some of the transformation laws appear complicated. A key point of the relativistic formulation is that knowing E and B in one frame of reference, one can compute them in any frame of reference. This actually makes problems simpler to understand than the typical undergraduate formulation in many respects, as in the undergraduate approach, one does all the work to derive the solution in each difrent frame of reference, while the relativistic formulation one only needs to solve the problem in one frame of reference, and the solution can be transformed to any frame of reference. To do so successfully requires that one also transform the source fields relativistically, which is why the undergraduate solutions don't take advantage of the fact that knowing the solution in one frame of reference allows one to transform the solution to any frame of reference.

From a physical point of view, it is important that charges and curend densities be transformed in a relativistic manner. Charge density (also number density) depends on the frame of reference due to length contraction, for instance. There are other effects as well. The appropriate relativistic transformation laws for number density and charge density are the number-flux four vector and the charge-current 4-vector. This is discussed in Wikipedia at https://en.wikipedia.org/wiki/Four-vector.

Introductory treatments of electromagnetism can use the more familiar 3-vector formulations, but that's not the way I think of it. The 3-vector form of the transformation laws appears more complicated, see for instance https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#E_and_B_fields.

I believe Griffiths texbtook on electromagnetism does some motivation for why these rather complex appearing laws are the correct way to transform electric and magnetic field without using 4-vectors and tensors.

Basically, the relativistic explanation of the homopolar genrator is that in the non-rotating frame, there is only a magnetic field, and the relativistic transformation laws give rise to an electric field in the rotating frame of the disk.

You may prefer the non-tensor exposition of relativity - unfortunately, I've forgotten much of that. But do let us know which formulation works best for you, it's pointless to spend a lot of time discussing the 4-vector formalism if you're not interested in it and want to see a 3-vector treatment.

 

[Ivan Nikiforov]

Мне нужно исправить утверждение, которое я сделал в посте № 40, относительно области $b$ на рисунке ниже:

View attachment 357125

Я допустил ошибку, заявив следующее

Скачкообразный скачок тангенциальной составляющей магнитного поля в области $b$ не нарушает уравнение Максвелла. Намагничивание материала создает эффективный "связанный" ток на поверхности материала. Следовательно, произойдет прерывистый скачок тангенциальной составляющей $B$ через границу в точке $b$. Поле В в области внешней цепи будет небольшим по сравнению с полем внутри намагниченного материала.

Я полагаю, это означает, что устройство не будет очень эффективно генерировать ток во внешней цепи. И это будет справедливо независимо от того, (1) вращаете ли вы материал, удерживая внешнюю цепь в неподвижном состоянии, или (2) удерживаете материал в неподвижном состоянии, а внешнюю цепь вращаете.

Thanks for the comment. I agree with your conclusions. However, the fact is that the model you are talking about is conditional. It was given only in order to convey the basic meaning of how to exclude the influence of a magnetic field on an external circuit. The figure below shows the circuit of the generator, which is currently being manufactured, to test my questions. In this scheme, the electromotive force is removed between the axis and the edge of the disk. The voltmeter (electrical load) is connected in series to the axis circuit and is located in the reference frame of the disk. Previously, I made such a model on my own and it works, but in it part of the external circuit A-B was in a magnetic field. I expect that the model will work if I completely exclude the influence of the magnetic field on the external A-B circuit.

 

[Ivan Nikiforov]

As an amateur of electric engineering, I find shielding static magnetic field is a challenging task in real world ,e.g. in medical NMR.
Meisner effect may be useful by making the external line with superconductor. But a fundamental question here. Does Lorentz force, which is zero for zero B, apply electrons in moving superconductors or not ? Does Faraday disk made of superconductor generate emf ?

It seems to me that your question about the Faraday disk from a superconductor is still waiting for its persistent researcher. To solve the problem of excluding the external circuit from the magnetic field, I hypothetically considered this option if the external circuit were a superconductor. I have come to the conclusion that this may be a formal solution. That is, the volume of the conductor itself is not in a magnetic field. The magnetic field wraps around it. But in the context of my task, not only that is important. It is important that the current flowing through the external circuit does not interact with the external magnetic field with its magnetic field.

 

[Ivan Nikiforov]

I am a bit confused - isn't the whole point that the rotating part of the disk is in a magnetic field? I'll assume that this was a typo.

Good afternoon! Thanks for the comment. There is no mistake and my question is formulated correctly. I would like to consider just such a model in which the disk and the magnetic field are stationary, and the external electrical circuit rotates. However, it is not in a magnetic field. I appreciate your detailed explanation and will read it carefully.

 

[Ivan Nikiforov]

You may prefer the non-tensor exposition of relativity - unfortunately, I've forgotten much of that. But do let us know which formulation works best for you, it's pointless to spend a lot of time discussing the 4-vector formalism if you're not interested in it and want to see a 3-vector treatment.

Thank you again for the detailed comment. I have reviewed the materials on your links with interest. Indeed, the three-vector presentation is more understandable to me because it is simpler. I have seen three-vector equations for field transformations in the literature. I wanted to make sure that I understood them correctly and was drawing the right conclusions. One day I was reading a book about these equations and there was a neodymium magnet on the table. I measured its diameter, applied the average inductance and the orbital velocity of our planet. I came to the conclusion that an observer in the Sun should detect an electrical voltage of about 800 volts on the sides of the magnet. It was a revelation to me. It took me a while to get this in my head.
In the link text:
https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#E_and_B_fields
it is said that "...depending on the orientation of the magnetic field, the comma system can see the electric field, even if it is not present in the system without commas." Thus, this confirms the conclusion that a moving external circuit that is not in a magnetic field will detect an electric field on the sides of the magnetic field. Since the electrical circuit is closed, an electric current must flow in it. What forces will act on the moving external circuit in this case? An electric current flows through the external circuit, which creates a closed magnetic field around the conductor. In this case, the external circuit moves in a space in which there are no external magnetic and electric fields. Perhaps there are some forces unknown to me that are the result of relativistic phenomena?

 

[renormalize]

Thus, this confirms the conclusion that a moving external circuit that is not in a magnetic field will detect an electric field on the sides of the magnetic field. Since the electrical circuit is closed, an electric current must flow in it. What forces will act on the moving external circuit in this case? An electric current flows through the external circuit, which creates a closed magnetic field around the conductor. In this case, the external circuit moves in a space in which there are no external magnetic and electric fields. Perhaps there are some forces unknown to me that are the result of relativistic phenomena?

This statement contradicts Maxwell's equations. Per your post #56, your external circuit forms a closed path I label as $\partial\Sigma$, which is the boundary of an enclosed surface that I denote as $\Sigma$:

You declare that your magnetic shield is designed to entirely contain the magnetic field and prevent it from anywhere entering the area $\Sigma$ of the circuit. But the Maxwell-Faraday equation states that the electromotive force $\mathcal{E}$ around the path $\partial\Sigma\,$, which is what your voltmeter will register, is given by:$$\mathcal{E}=\oint_{\partial\Sigma}\vec{E}\cdot d\vec{l}=-\frac{d}{dt}\int\!\!\!\int_{\Sigma}\vec{B}\cdot d\vec{S}$$Since there is no normal-component of $\vec{B}$, nor indeed any magnetic-field component at all, on the surface $\Sigma\,$, the EMF $\mathcal{E}=0$ and your moving circuit will show zero voltage and generate no current whatsoever.