I The Faraday disk in the context of the theory of relativity

[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.

 

[Ivan Nikiforov]

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 $\Sigma$:
View attachment 357169
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.

Please excuse me for the poor-quality model. Actually, it's not a shielding disk. This part is called a current plate. It is electrically isolated from the generator. I have drawn the path of the electric current with an intermittent line. In this model, the moving external circuit is a short section A-B. In the practical design of the generator, I made this circuit in the form of carbon brushes that pass through the pulley and rotate with it. I assume that instead of brushes, you can use a liquid conductor, such as mercury, which will rotate inside an annular channel with a rectangular cross-section.

 

[renormalize]

Thanks for the clarification, but that still begs the question: is there any component of the magnetic-field that is both nonzero within, and normal to, the area enclosed by the dashed orange line that is the closed path of the circuit? If not, Faraday's law of induction states that there can be no EMF and no current.

 

[Ivan Nikiforov]

Thanks for the clarification, but that still begs the question: is there any component of the magnetic-field that is both nonzero within, and normal to, the area enclosed by the dashed orange line that is the closed path of the circuit? If not, Faraday's law of induction states that there can be no EMF and no current.

I imagine the distribution of the magnetic field in the core of the generator to be something like this. I made an air gap in the generator core and non-gap surfaces (highlighted in red). It turns out that the external circuit in any case moves perpendicular to the lines of force of the magnetic field. In essence, the core of the generator is a disk, the edges of which are folded to the axis like an umbrella. I have indicated the orange line more precisely, as I imagine.

 

[renormalize]

It turns out that the external circuit in any case moves perpendicular to the lines of force of the magnetic field.

Then there is no magnetic-field component normal to the surface $\Sigma$ bounded by the orange path and the magnetic flux $\Phi_{\Sigma}=\int\!\!\!\int_{\Sigma}\vec{B}\cdot d\vec{S}$ through that surface is zero. Hence there is no EMF and no current. By Faraday's law of induction, you must have a nonzero, time-dependent flux through $\Sigma$ (i.e., $\dot{\Phi}_{\Sigma}\left(t\right)\neq0$) in order to develop an EMF around the orange path.

 

[Ivan Nikiforov]

Then there is no magnetic-field component normal to the surface $\Sigma$ bounded by the orange path and the magnetic flux $\Phi_{\Sigma}=\int\!\!\!\int_{\Sigma}\vec{B}\cdot d\vec{S}$ through that surface is zero. Hence there is no EMF and no current. By Faraday's law of induction, you must have a nonzero, time-dependent flux through $\Sigma$ (i.e., $\dot{\Phi}_{\Sigma}\left(t\right)\neq0$) in order to develop an EMF around the orange path.

It is possible that your question can be answered as follows. In Tamm's book "Fundamentals of the Theory of Electricity", the rule for calculating the electromotive force for contours, which opens during movement, is considered. The picture shows a model for such a case. The book claims that when calculating the flow change, the flow through the closed contour C'OAVBCC' is applied. This contour is obtained when contour C'OAVBC is deformed using a segment of the trajectory CC' along which the contour break point C' moves.

As I understand it, contours are breaking in my generator and the circular sector COC' must be taken into account to calculate the electromotive force. In addition, it is not yet clear to me what is the fundamental difference between the geometry of the magnetic field of my generator and the Faraday disk? In the drawing, I have depicted a part of the generator and a part of the Faraday disk. In both cases, the external chain moves inside the sheet. It seems to me that the configuration of the magnetic field and the velocity vector is the same in both cases.

 

[Ivan Nikiforov]

As I imagine, if the external circuit moves inside the sheet, it should detect something like this electric field pattern - the entire inner surface of the magnetic circuit is positively charged, and the entire outer surface of the magnetic circuit and the axis are negatively charged. It seems to me that the same process is obtained if an observer (a blue ball) flies through a toroid with a closed magnetic field. Therefore, if the entire inner surface of the magnetic circuit is positively charged, then the outer circuit will be in contact with an electric field.

 

[Ivan Nikiforov]

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.

I would like to consider this issue in more detail. I have analyzed the results of experiments over several years and have come to the following conclusion. In models where the brushes are located at some distance from the boundary of the magnetic field (and the observed E-field), the electromotive force is not observed. I even introduced a special term: "inert conductor" to denote the part of the circuit between the boundary of the M-field and the brush.
In fact, this phenomenon is explained as follows. It is important to understand that the electric field that is detected by a moving observer is by its nature an electrostatic field. That is, the electric fields in the Faraday disk and in most modern electric machines are inherently different.

The figure shows a metal conductor located in an electrostatic field. Under the influence of this field, the charges of the conductor are distributed in such a way that the electric field inside the conductor completely compensates for the external field. Thus, the electric field detected by a moving observer within the boundaries of the magnetic field is completely compensated by the internal E-field of the conductor (in our case, the disk). As a result, the observed electric field does not extend beyond the part of the conductor that is in the magnetic field. This is how the absence of an electromotive force can be explained if the brushes are installed outside the boundaries of the magnetic field.

 

[TSny]

It seems to me that the same process is obtained if an observer (a blue ball) flies through a toroid with a closed magnetic field. Therefore, if the entire inner surface of the magnetic circuit is positively charged, then the outer circuit will be in contact with an electric field.

View attachment 357195

When the toroid is at rest, there is no electric field anywhere and the magnetic field is confined to its interior.

Now, suppose we move toward the toroid. In our frame of reference, the toroid is moving out of the page towards us. In our frame, the toroid has surface charge as you indicated. There will exist both electric and magnetic fields inside the toroid. But there will still not be any B or E field outside the toroid. In our frame, a free charge carrier inside the toroid experiences a magnetic force and an electric force. The two forces cancel so that the charge carrier is in equilibrium.

Is this how you see it also?

 

[renormalize]

As I understand it, contours are breaking in my generator and the circular sector COC' must be taken into account to calculate the electromotive force. In addition, it is not yet clear to me what is the fundamental difference between the geometry of the magnetic field of my generator and the Faraday disk?

Tamm says:

So the cylinder in Fig. 90 is a permanent magnet similar to the one shown at right and thus it contains a nearly-uniform, vertical magnetic-field $\vec{B}$ inside. Note therefore that the moving closed current-path has segments $C,C'$ inside of, and orthogonal to, the field $\vec{B}$. Consequently, an EMF develops in the circuit and registers on the voltmeter. That's the fundamental difference: Fig. 90 has a current path partly inside the strong magnetic-field region, whereas you declare that your moving current path is entirely outside of the field. If that's truely the case, you will see no EMF nor current. Think about the situation this way: unless there is at least some portion of your moving current-path immersed inside the field $\vec{B}$, there can be no Lorentz force $q\,\vec{v}\times\vec{B}$ to drive electrons around the circuit as a current.

 

[Ivan Nikiforov]

Tamm says:
View attachment 357199
View attachment 357200View attachment 357201
So the cylinder in Fig. 90 is a permanent magnet similar to the one shown at right and thus it contains a nearly-uniform, vertical magnetic-field $\vec{B}$ inside. Note therefore that the moving closed current-path has segments $C,C'$ inside of, and orthogonal to, the field $\vec{B}$. Consequently, an EMF develops in the circuit and registers on the voltmeter. That's the fundamental difference: Fig. 90 has a current path partly inside the strong magnetic-field region, whereas you declare that your moving current path is entirely outside of the field. If that's truely the case, you will see no EMF nor current. Think about the situation this way: unless there is at least some portion of your moving current-path immersed inside the field $\vec{B}$, there can be no Lorentz force $q\,\vec{v}\times\vec{B}$ to drive electrons around the circuit as a current.

Thanks for the detailed explanation. As I understand it, my generator also has sections of an electrical circuit similar to OC and OC', that is, part of the electrical circuit passes inside a magnetic core, that is, in a magnetic field. In this case, only the external circuit physically rotates, followed by the OC section, which is essentially a current path. Thus, taking into account the small differences in geometry, my generator is an analog of the model in Figure 90.

 

[Ivan Nikiforov]

Когда тороид находится в состоянии покоя, электрическое поле отсутствует, а магнитное поле ограничено его внутренней частью.
View attachment 357197

Теперь предположим, что мы движемся к тороиду. В нашей системе отсчёта тороид движется по направлению к нам. В нашей системе отсчёта тороид имеет поверхностный заряд, как вы и указали. Внутри тороида будут существовать как электрическое, так и магнитное поля. Но за пределами тороида не будет ни магнитного, ни электрического поля. В нашей системе отсчёта свободный носитель заряда внутри тороида испытывает воздействие магнитной и электрической сил. Эти две силы компенсируют друг друга, и носитель заряда находится в равновесии.

View attachment 357198

Вы тоже так это видите?

Спасибо за интересное объяснение и иллюстрации. Да, вы правы, я понимаю процесс именно так, как вы описали. Наблюдаемое электрическое поле всегда находится в границах магнитного поля, то есть в данном случае электрическое поле будет находиться внутри тора.
The generator can be equivalently represented as a torus. The external circuit moves inside it in a circle, like a particle in a particle accelerator. In the figure below, a ring with a magnetic field is rolled into a ring in the vertical plane (the section of the torus is shown). The upper carbon brush (in black) is moving away from us, the lower carbon brush is moving towards us.

 

[renormalize]

As I understand it, my generator also has sections of an electrical circuit similar to OC and OC', that is, part of the electrical circuit passes inside a magnetic core, that is, in a magnetic field.

OK, I think we're now on the same page: we agree that there is a portion of your complete circuit that is inside the B-field region, as there must be to generate an EMF in that circuit. Then my only remaining question is: can you explain what is the practical advantage of your configuration, with its magnetic-field confining chambers and moving external pickup? How does that improve on the simplicity and performance of a simple Faraday-disk type homopolar/unipolar generator?

 

[pervect]

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?

Assuming that by "a moving circuit" you mean the same thing I mean, the Wikipedia example explicitly requires a magnetic field. Also, I assume you mean "unprimed" by "comma"?

If one of the fields is zero in one frame of reference, that doesn't necessarily mean it is zero in all other frames of reference. This can be seen by, for instance, making the unprimed electric field zero in the transformation to the primed electric field. In this case, depending on the orientation of the magnetic field, the primed system could see an electric field, even though there is none in the unprimed system.

If both the electric and magnetic field are zero in one frame, they will be zero in all frames. Again, I'm not sure if I'm understanding you correctly, but if I am, your remark appears to be incorrect.

The rules stated by wiki state that a component of the electric field in the direction of motion remains unchanged between the primed and unprimed frames. That's the meaning of

$$E^{\prime}_{\parallel} = E_{\parallel}$$

More interesting is the transformation of the component of the electric field perpendicular to the direction of motion, given by the other equation for the perpendicular component, $E_{\perp}$ which includes a term generated by motion, v $\times$ B, as well as the Lorentz factor gamma.

Potentially less confusing (I found it enough so that I went on to the second version of Wiki's statement) is the following section, which doesn't use the parallel and perpendicular components. Instead , it considers the case where motion is along the "x" axis. Then the parallel component of E is E_x, and both E_y and E_z are "perpendicular" components. THis gives the following (I've omitted E_z).

$$E^{\prime}_x = E_x \quad E^{\prime}_y = \gamma \left( E_y - v B_z\right)$$

From the complete version of the above (I've only written down a few of the many transformation equations) you can see that if E_x = E_y = E_z = B_x = B_y = B_z = 0, there are no electric or magnetic fields generated by motion. I'm unclear about which case you are interested in, but if you can write down the electric and magnetic field components E_x, E_y, E_z, B_x, B_y, B_z in some frame of reference, the equations will tell you how they transform. From your verbal description, I cannot write down these components to do it for you, you'll need to specify them clearly for me.

You also might want to consult a textbook to double check. I am not sure what the best source would be for you, I am sensing a high probability of some language differences between us, if that's the case it'd be best to find one in your native language (though I only speak English, myself so I can't help with that).

 

[Ivan Nikiforov]

Assuming that by "a moving circuit" you mean the same thing I mean, the Wikipedia example explicitly requires a magnetic field. Also, I assume you mean "unprimed" by "comma"?



If both the electric and magnetic field are zero in one frame, they will be zero in all frames. Again, I'm not sure if I'm understanding you correctly, but if I am, your remark appears to be incorrect.

The rules stated by wiki state that a component of the electric field in the direction of motion remains unchanged between the primed and unprimed frames. That's the meaning of

$$E^{\prime}_{\parallel} = E_{\parallel}$$

More interesting is the transformation of the component of the electric field perpendicular to the direction of motion, given by the other equation for the perpendicular component, $E_{\perp}$ which includes a term generated by motion, v $\times$ B, as well as the Lorentz factor gamma.

Potentially less confusing (I found it enough so that I went on to the second version of Wiki's statement) is the following section, which doesn't use the parallel and perpendicular components. Instead , it considers the case where motion is along the "x" axis. Then the parallel component of E is E_x, and both E_y and E_z are "perpendicular" components. THis gives the following (I've omitted E_z).

$$E^{\prime}_x = E_x \quad E^{\prime}_y = \gamma \left( E_y - v B_z\right)$$

From the complete version of the above (I've only written down a few of the many transformation equations) you can see that if E_x = E_y = E_z = B_x = B_y = B_z = 0, there are no electric or magnetic fields generated by motion. I'm unclear about which case you are interested in, but if you can write down the electric and magnetic field components E_x, E_y, E_z, B_x, B_y, B_z in some frame of reference, the equations will tell you how they transform. From your verbal description, I cannot write down these components to do it for you, you'll need to specify them clearly for me.

You also might want to consult a textbook to double check. I am not sure what the best source would be for you, I am sensing a high probability of some language differences between us, if that's the case it'd be best to find one in your native language (though I only speak English, myself so I can't help with that).

Thank you for the detailed comment. In my text, by the comma, I meant the word [ ' ], which the translator translated incorrectly. I quoted the same Wikipedia quote that you wrote. In general, I completely agree with you regarding the conversion of fields. The model I am investigating consists of two frames of reference. In the first frame of reference there is a permanent magnetic field and a conductor (disk) placed in this field. In the second frame of reference, there is an external electrical circuit, which is a conductor and sliding contacts connected to the conductor of the first frame of reference. In the second frame of reference, there are no electric and magnetic fields, that is, the external electrical circuit is not in a magnetic field. An option is considered in which the second reference frame, that is, an external electrical circuit, moves relative to the first reference frame, while the velocity vector is directed perpendicular to the magnetic induction vector. As I understand it, in this case, the second reference frame (a moving external electrical circuit) will detect an electric field within the boundaries of the magnetic field of the first reference frame. Since the electrical circuit is closed, an electric current must flow through both frames of reference under the influence of the observed electric field. This is the first question of my topic. The second question is a consequence of the first. If an electric current is flowing in such a model, then what forces will act in the second frame of reference on an external electrical circuit that moves in a space free from external electric and magnetic fields.

 

[renormalize]

In the second frame of reference, there are no electric and magnetic fields, that is, the external electrical circuit is not in a magnetic field.

Since there are no electric or magnetic fields in the second frame, there exists no frame at all in which those fields are nonzero because the electromagnetic field transforms as an antisymmetric tensor under Lorentz transformations: a zero-tensor in one frame is a zero-tensor in all frames. Period.

If an electric current is flowing in such a model, then what forces will act in the second frame of reference on an external electrical circuit that moves in a space free from external electric and magnetic fields.

If both frames are inertial then there will be no forces whatsoever acting on the "external electrical circuit that moves in a space free from external electric and magnetic fields". Period.

 

[Ivan Nikiforov]

Dear participants of the discussion! I am glad to inform you that the experimental sample is ready. It also needs to make excitation coils and purchase some equipment. I will certainly inform you about the results of the experiment. I may be away for a while, so I apologize in advance if I don't respond to comments.

 

[Ivan Nikiforov]

Good afternoon! I checked the operation of the prototype and got the following results. For measurements, I used a mirror galvanometer with a division price of 2.5 x 10.9 A. 1) When the external circuit rotates relative to the core at a speed of about 2 revolutions per second, the galvanometer deviates by 2 divisions, always in one direction, regardless of the direction of rotation. Thus, no EMF is detected when the external circuit is rotated. 2) When the core is rotated relative to a stationary external circuit, at a speed of about 30 degrees per second, the galvanometer confidently shows 20...30 divisions. It reacts correctly to the direction of rotation. Thus, when the core rotates, the EMF is detected. Previously, the following conclusions can be drawn from these results. 1) The external circuit is indeed not in a magnetic field. Otherwise, when it rotates, the result would be the same as when the core rotates. 2) The geometry of the magnetic circuit is executed correctly, since an EMF is observed when the core rotates.
That is, when the core rotates, the sample confidently operates like a unipolar generator. I'm trying to analyze the results, but I still can't figure out why there's no EMF when the external circuit is rotating. The external circuit moves relative to the magnetic field of the core. Therefore, it must detect the electric field on the sides of the magnetic field. Therefore, this electric field creates an EMF. But there is no EMF. It turns out that an observer moving relative to a medium with magnetization, in order to detect an electric field, must himself be in a magnetic field? I would appreciate it if you could comment on my results and explain why there is no EMF when the external circuit is rotating.

 

[weirdoguy]

and explain why there is no EMF when the external circuit is rotating.

Haven't you been given the explanation multiple times? Maybe you should re-read this thread.

 

[Ibix]

would appreciate it if you could comment on my results and explain why there is no EMF when the external circuit is rotating.

Remember that the magnetic field and electron velocities transform too. You'll find that if you transform to a local frame where the disc is moving that $\vec E=-\vec{v}\times\vec{B}$, whatever frame you use, and hence there's no force on the electrons in the disc and hence no current in a moving conductor. It's only when the electrons in the magnetic field (which does not include your conductor, by hypothesis) are moving in the maget frame that you get a net force.

 

[Ivan Nikiforov]

Haven't you been given the explanation multiple times? Maybe you should re-read this thread.

Yes, thanks, I've already reread it many times. I haven't found an answer yet.

 

[Ivan Nikiforov]

Remember that the magnetic field and electron velocities transform too. You'll find that if you transform to a local frame where the disc is moving that $\vec E=-\vec{v}\times\vec{B}$, whatever frame you use, and hence there's no force on the electrons in the disc and hence no current in a moving conductor. It's only when the electrons in the magnetic field (which does not include your conductor, by hypothesis) are moving in the maget frame that you get a net force.

I would like to understand the process precisely in terms of the transformation of fields in different frames of reference. I agree with your explanation. This explanation is confirmed by my experiment: in order to observe the EMF, the conductor must necessarily move in a magnetic field. But this statement contradicts, for example, the Wikipedia article: https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#E_and_B_fields
As I understand it, this article states the following: an observer moving relative to a magnetic field detects an electric field. However, it does not have to be in a magnetic field. This is the whole question - in order to detect an electric field (and EMF), a moving observer must be in a magnetic field or must not be in a magnetic field?

 

[PeterDonis]

this statement contradicts, for example, the Wikipedia article

No, it doesn't.

it does not have to be in a magnetic field

The EM equations are local; the fields that appear in them are the fields at the point where the observer is. If there is no magnetic field at the point where the observer is, then $\vec{B} = 0$ in the equations.

The Wikipedia article does not say anything different from that.

 

[Ibix]

I would like to understand the process precisely in terms of the transformation of fields in different frames of reference

Write the electromagnetic field using the Faraday tensor and the Lorentz transforms as a matrix. In matrix notation, the transformed Faraday tensor is then $\mathbf{\Lambda F\Lambda}^T$, where $\mathbf{F}$ is the original Faraday tensor (only some of the magnetic components and none of the electric ones are non-zero) and $\mathbf{\Lambda}$ is the Lorentz transform. Note that this is a matrix multiplication, so clearly at any event where all the field components are zero in one frame they are zero in all. Hence the electric field in the transformed frame only exists in the region where the magnetic field also exists.

You can read the transformed components off the transformed matrix and you will find that the $E$ components are a factor of $-v$ larger than the $B$ components.

This procedure only works locally for a rotating frame, since rotating frames' spatial planes are either not orthogonal to the time direction or do not close. However, using the corrected maths (you have a different matrix instead of the Lorentz transform, and you need to contract with an orthogonal basis instead of just reading off components) leads to different details but the same result.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After moderator review the thread will remain closed as the OP question has been more than answered.