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In my previous question, entitled a problem with a current-carrying wire's magnetic field as viewed by an observer moving perpendicular to it, @Dale answered that:

so indeed, as you mentioned, in this frame the wire is electrically neutral. However, because the field is changing over time, as in your previous question, there is an electric field that is induced due to the changing magnetic field.

However, on this page, I want to redesign the problem by replacing the wire with a sheet of current so that there is no magnetic field changing over time, and thus there is no induced electric field along the $z$-axis. In this new version, I would like to emphasize that the problem persists. (The axes are relabeled the same as those in Dale's answer to the previous problem.)

Assume an infinitely large current-carrying sheet with a red positively charged test particle located somewhere above it and at rest with respect to the sheet's protons. One can either assume that the electrons have a velocity $u$ along the $z$-axis or a surface current of $k$ in the opposite direction, and the sheet is electrically neutral as viewed by the blue and red protons. [See Figure 1.] It is evident that there is a uniform magnetic field along the $x$-axis everywhere above the sheet from the viewpoint of the sheet's protons ($xyz$). Since the red test proton is at rest in this field, no Lorentz force acts on it, and there is no electric field either in space, as the sheet is neutral.

Figure 1 Figure 1

However, when we consider the perspective of the observer ($x'y'z'$) moving at $v$ relative to the sheet and along the $x$-axis, the circumstances become complicated. He indeed sees that the sheet of protons has a velocity $v$, while the sheet of electrons has a velocity $w=\sqrt{v^2+{\alpha_v}^2u^2}$, where $\alpha_v$ is the reciprocal of the Lorentz factor. These two separate sheets of current produce two different magnetic fields of, respectively, $B'_p$ and $B'_e$ so that $B'_p\perp v$ and $B'_e\perp w$. [See Figure 2.]

Figure 2 Figure 2

The resultant magnetic field is considered to be $B'_T$ uniformly distributed above the sheet. However, since the red test charge has a velocity $v$ relative to this observer ($x'y'z'$), a Lorentz force is expected to be exerted on it unless one proves that for any arbitrarily chosen values of $u$ and $v$, we have $B'_T\parallel v$ so that the Lorentz force is zero. [See Figure 3.]

Figure 3 Figure 3

For this configuration, note that there is no magnetic field changing over time from the viewpoint of both observers, or the changing magnetic field happens at infinity from the perspective of the latter observer ($x'y'z'$), which loses its strength at the location of the test particle. There is no electric field that can cancel out the Lorentz force, and the sheet is also electrically neutral from the perspective of the primed system, which may culminate in a paradox unless, as mentioned earlier, one can prove that for any arbitrarily chosen values of $u$ and $v$, we have $B'_T\parallel v$.

Edited and added according to @MichaelSeifert answer:

My calculations also confirm that the observer moving at $v$ observes no magnetic field along the $z'$-axis. Indeed, if the surface density of protons and electrons is equal from the viewpoint of the test particle ($\sigma_p=\sigma_e=\sigma$), the sheet is electrically neutral and the magnetic field is as follows:

$$B_T=\mu_0 \frac{k}{2}=\mu_0 \frac{\sigma u}{2}$$

From the viewpoint of the observer moving at $v$, two different sheets of current have the surface densities of $\sigma'_p=\sigma'_e=\alpha_v\sigma$, and the following magnetic fields:

$$B'_p=\mu_0 \frac{k'_p}{2}=\mu_0 \frac{\sigma'_p v}{2}=\mu_0 \frac{v\alpha_v\sigma}{2}$$

$$B'_e=\mu_0 \frac{k'_e}{2}=\mu_0 \frac{\sigma'_e w}{2}=\mu_0 \frac{w\alpha_v\sigma}{2}$$

Since the angle $\alpha'$ is the angle between the velocities, we have the following equations:

$$\sin \alpha'=\frac{u\alpha_v}{w}$$

$$\cos \alpha'=\frac{v}{w}$$

As shown in Figure 4, we expect that $B'_e\cos \alpha'=B'_p$, which can be easily proved using the above equations. Nonetheless, there is still a problem regarding the magnitude of $B'_T$ calculated as follows:

Figure 4 Figure 4

$$B'_T=B'_e\sin \alpha'=\mu_0 \frac{u{\alpha_v}^2\sigma}{2}$$

When comparing this equation with the first one, we get $B_T\neq B'_T$. This result contradicts the Lorentz transformation for the electromagnetic fields. Indeed, these transformations predict that when there is a uniform magnetic field in the unprimed system, the primed system, with a relative velocity along the magnetic field, calculates the magnetic field unchanged ($B_T= B'_T$). Where did I go wrong?

$\color{red}{\textbf{Updated:}}$ As I mentioned in a comment to @Dale's answer here, I made a mistake in calculating the surface densities for electrons and protons. Indeed, I assumed $\sigma'_p=\sigma'_e=\alpha_v\sigma$, whereas the correct form is $\sigma'_p=\sigma'_e=\gamma_v\sigma$. If one fixes it in my post and replaces it in my equations, they will see that the issue is resolved and there is no paradox even without applying four-vectors.

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  • $\begingroup$ It is not that four-vectors are the only possible method to get a right answer. They are just the easiest and most consistent. As you have demonstrated many times, your approach frequently leads to elementary mistakes when comparing frames. $\endgroup$ Commented Sep 28 at 13:26
  • $\begingroup$ @Dale My approach did not lead to an elementary mistake. I already knew that the densities increase by the Lorentz factor, and it was indeed a simple math typo in my calculations, which everyone can make, even if they use four-vectors. $\endgroup$ Commented Sep 28 at 13:37
  • $\begingroup$ That would be a more convincing claim if you hadn’t been posting such questions and mistakes (ones that would be avoided by using 4-vectors) here since 2017. $\endgroup$ Commented Sep 28 at 13:43

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The four-current density for protons in the sheet in the rest frame of the protons are given by $$ J_p^\mu = (\sigma, 0, 0, 0) \delta(y). $$ and that for the electrons is given by $$ J_e^\mu = (-\sigma, 0, 0, \sigma u) \delta (y) $$ Note that since the sheet is electrically neutral in this reference frame, the two charge densities cancel out.

If we transform these four-currents into a frame moving in the $x$-direction, they become $$ {J'}_p^\mu = (\gamma \sigma, -\gamma \beta \sigma, 0, 0) \delta(y'). $$ and $$ {J'}_e^\mu = (-\gamma \sigma, +\gamma \beta \sigma, 0, \sigma u) \delta (y') $$ where $\beta = v/c$ and $\gamma = 1/\sqrt{1- \beta^2}$. Thus, we see that the sheet remains electrically neutral, as predicted. However, the net current density is still in the $z$-direction in the new frame: $$ {J'}^\mu = (0, 0, 0, \sigma u) \delta(y'). $$
This implies that the magnetic field remains in the $x$-direction in this frame regardless of $u$ and $v$, and there is no paradox.

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  • $\begingroup$ I may have missed a common factor in the transformed four-currents, arising from the transformation of the delta-function in $y$; any comments on how this should be done corrently are welcome. However, this factor will be the same for the two four-currents, and the total will still point in the $z$-direction as described here. $\endgroup$ Commented Sep 24 at 11:54
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    $\begingroup$ excellent answer, +1. I think your delta function is fine. Since there is nothing of interest (to the delta function) happening in the y direction it is just a straight substitution $y=y’$ $\endgroup$ Commented Sep 24 at 21:01
  • $\begingroup$ I have updated my question. Figures are slightly modified. $\endgroup$ Commented Sep 25 at 11:28
  • $\begingroup$ @Dale I have updated my question. $\endgroup$ Commented Sep 25 at 11:28
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Since @Michael Seifert already answered the original question, I will skip that part and focus on the addition.

The reason that you are getting an incorrect calculation is, as always, because you are using a haphazard approach. Every single time that you ever get any electromagnetic "contradiction" or "paradox" between reference frames, you are 100% guaranteed that it is because you made a math error.

The relationship between the EM field and the sources is $$\partial_\alpha F^{\alpha\beta}=\mu_0 J^\beta$$ This relationship holds in all frames.

The EM field is constrained to follow $$\partial^\sigma F^{\mu\nu}+\partial^\mu F^{\nu\sigma}+\partial^{\nu} F^{\sigma \mu}=0$$ This relationship holds in all frames.

The EM force on a charge/current is $$f_\alpha=F_{\alpha\beta}J^\beta$$ This relationship holds in all frames.

So if you ever get any of them not holding in any frame it is because you made a mistake. It is literally impossible from first principles for there to ever be a conflict between frames in classical EM. This is precisely the reason that we use tensors in the math, to ensure that we do it correctly and we make it impossible to have a physical contradiction.

For this scenario, let us consider an infinite sheet of charge in the $x,z$ plane with a charge density of $\sigma$ and a current density of $j=\sigma u$ in the $z$ direction. For the protons $j=0$ and for the electrons $\sigma = -\sigma$, but we will solve for the general case.

Now, in the unprimed frame we have the four-current is $$J^\mu = \left(\sigma \ 0 \ 0 \ j\right)$$ This leads to an EM field $$F^{\mu\nu} = \left( \begin{array}{cccc} 0 & 0 & -\sigma & 0 \\ 0 & 0 & 0 & 0 \\ \sigma & 0 & 0 & -j \\ 0 & 0 & j & 0 \\ \end{array} \right)$$

Now, when we transform to the unprimed frame we get $$J'^\mu = \left(\gamma \sigma \ \gamma v \sigma \ 0 \ j\right)$$ so we get a charge density "dilation" as well as the introduction of a current density in the $x'$ direction. The current density in the $z'$ direction is the same as the current density in the $z$ direction. Now, this leads to an EM field of $$F'^{\mu\nu}=\left( \begin{array}{cccc} 0 & 0 & -\gamma \sigma & 0 \\ 0 & 0 & -v \gamma \sigma & 0 \\ \gamma \sigma & v \gamma \sigma & 0 & -j \\ 0 & 0 & j & 0 \\ \end{array} \right)$$ Notice that we have a E-field "dilation" corresponding to the charge density "dilation", and the introduction of a new component of the B-field in the $z'$ direction corresponding to the new component of the current density. The B-field in the $x'$ direction is the same as the B-field in the $x$ direction.

Now for the protons $\sigma=\sigma$ but for the electrons $\sigma=-\sigma$, so the $x'$ component of the B field of the protons exactly cancels out the $x'$ component of the B field of the electrons. So the total EM field in the primed frame is $$F'^{\mu\nu}=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -j \\ 0 & 0 & j & 0 \\ \end{array} \right)$$ which, as @Michael Seifert showed is identical to the EM field in the unprimed frame.

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  • $\begingroup$ If you think that four-vectors and matrices resolve the issue, I believe that finding a physical justification in accordance with my simple figures and calculations is a breeze for you! The only mistake in my post is that I wrote $\sigma'_p=\sigma'_e=\alpha_v\sigma$, whereas the correct form is $\sigma'_p=\sigma'_e=\gamma_v\sigma$. If one fixes it in my post and replaces it in my equations, they will see that the issue is resolved and there is no paradox even without applying four-vectors. Therefore, my approach has no fundamental deficit, and it works correctly, similar to your matrices. $\endgroup$ Commented Sep 25 at 19:24
  • $\begingroup$ I still think, though, that my method evokes more intuition and has greater visualization. $\endgroup$ Commented Sep 25 at 19:25
  • $\begingroup$ The point is that 1) tensors will consistently do it correctly and 2) it proves that there is never any possibility of a contradiction. You have shown repeatedly, over the course of dozens of questions over several years that your approach often gets wrong answers. Here it is $\alpha$ vs $\gamma$, but that is your fourth posted mistaken analysis in just a week. You may “evoke more intuition and visualization”, but the cost is consistently making mistakes. Is that really a good trade-off in your opinion? $\endgroup$ Commented Sep 25 at 21:12
  • $\begingroup$ @MohammadJavanshiry Also, regarding “evokes more intuition” Intuition is something that is a result of experience. I have used tensors a lot over the years so they are intuitive to me now. To the point that I now find it easier to understand an EM paper written with tensors than vectors. Any method that you use frequently will become intuitive. That is just how intuition works. The problem is that repeated use of your current method has built up some wrong intuition in you. Why not learn a new method and start to build some right intuition instead? $\endgroup$ Commented Sep 26 at 12:00

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