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I have a question about the paricular part of the accepted answer in this post Does special relativity explains working of an electromagnet?. I will quote the whole answer here.

  1. Imagine you have a wire full of equally many positive and negative charges. The wire is neutral. You have a positive test charge outside of the wire. Since the wire is neutral, it experiences no electric force. Since it is not moving, it does not experience a magnetic force. Charge does not experience any force. enter image description here
  1. You start a current, making the negative charges flow to the right. Because of special relativity and length contraction, the negative charges are "squished together" slightly in the direction they travel, compared to their length in their rest frame (the frame where they are not moving).The wire will be negatively charged, since there are more negative charges per length unit than positive ones. Compensate for this by giving the wire a positive charge so that it is still neutral in the lab frame. Your test charge experiences no electric force, since the wire is neutral. It experiences no magnetic force since it is not moving. Charge still experiencing no force. enter image description here
  1. Get your test charge moving, in the same direction as the current, with the same velocity v for convenience. Since there is a current going through a wire, there is a magnetic field, and we can find its direction by the right hand rule. Since our test charge is moving in a magnetic field, it will experience a force. We can find this force using F=qvB, and using the right hand rule we find it is pointing away from the wire. There is still no electric force, since the wire is neutral. When the test charge moves, it is affected by a magnetic force enter image description here
  1. Now move to the reference frame that is moving along with the test particle. Here, the negative charges are still, and the positive charges are moving to the left. So there is still a current, but since the charge is not moving in this frame, there is no magnetic force. However, the positive charges are moving in this frame, so they will squish due to length contraction. At the same time, the negative charges will "un-squish", as we are now in their rest frame. The net result is that the wire is positively charged. The test charge will be repelled by an electric force. In the particle's rest frame, there is an electric force instead. enter image description here

Point 2 says that when the current starts in the lab frame, the electrons are "squished together". Thus, since the wire is nutral, there must be some extra positive charges come into it. But as far as I understand, positive charges are fixed in the nodes of the crystal structure and can not move.

So the questions are:

  1. Where do these extra charges come from?

  2. Why can not we apply the same logic to point 4 of the same answer, where the positive charges are squished now? Why there are no extra negative charges to compencate it in this case?

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Jan 4, 2023 at 17:10
  • $\begingroup$ The point 2 of the answer seems wrong to me also. Or perhaps with a very generous reading not wrong but super confusing $\endgroup$
    – Dale
    Commented Jan 4, 2023 at 18:07
  • $\begingroup$ Related : Why don't stationary charge feel force from a current carrying wire?. $\endgroup$
    – Frobenius
    Commented Jan 4, 2023 at 21:15

1 Answer 1

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This problem generates lots of confusion. As with most of special relativity problems, the root cause is mistreatment of the relativity of simultaneity. Another problem is forgetting that it is a thought experiment, hence we don't really need to worry about the fact that electron drift velocity in real conductors is absolutely non-relativistic$^1$.

The first point to get wrong is that once the current starts, and the (thought experiment) electron are moving to the right relativistically, that their spacing must Lorentz contract. This is false, as it it contradicted by the fact that the wire remains neutral. This is a stipulation of the problem, so there is no need to ask "why doesn't their spacing Lorentz contract?"--see Bell's Space Ship Paradox.

The moving electron spacing must match the stationary ion density, in the wire's rest frame for this to be true.

From there: just draw it in Minkowski space:

enter image description here

In the above figure, we see a line of stationary (blue) singly ionized atoms (heretofore: protons..it's a thought experiment after all), and matching red electrons moving along +x with $\gamma=2$. The past light cone is green, and the magenta axis shows the $x'$ axis of the moving electrons, along with their coordinate dilation:

$$ U' = U \times \sqrt{\frac{1+\beta^2}{1-\beta^2} }$$

In $S$, the wire rest frame, we have $E\propto \rho+ - |\rho_-|=0$ and $B\propto I = \rho_-\beta \equiv -\rho\beta$

Note that the graph has 1 charge per length unit, so $\rho=1$.

Now do what must be done to resolve confusion: The Lorentz Transformation:

enter image description here

Now the electrons are at rest (vertical red lines), but their density is not 1:

$$ \rho_-' = \rho/\gamma = \frac 1 2 $$

Again: See Bell's Spaceship Paradox.

The proton density has increased to

$$ \rho'_+ = \gamma\rho = 2 $$

Hence, in the electron frame there is now an electric field:

$$ E' \propto \rho'_+ - |\rho'_-| = \gamma - \frac 1 {\gamma} = \gamma\beta^2$$

so that:

$$ E'_{\perp} = \gamma(E_{\perp} + \beta \times B) = \gamma(0+\beta(\beta\rho)) = \gamma\beta^2I = \gamma\beta B$$

The protons now constitute a current $I' = \gamma(-\beta) = -\gamma\beta=\gamma I$ so that:

$$ B'_{\perp} = \gamma(B_{\perp} + \beta\times E) = \gamma B$$

Thereby satisfying the standard Lorentz transformation of electromagnetic field.

1 For realistic electron velocities, $\gamma\approx 1$, which seems like it should break the analysis (intuitively). The change in density is dominated by the relativity of simultaneity. If you imagine a time in the wire rest frame when each proton has an electron next to it (call that a series of coincident events $E_n(t_0, x_n)$ where $x_n = n \times [\rm lattice spacing]$ and then consider the times $t'_{0, n}$ in the moving frame, those events occur in the future (present) and ((past)) for $n<0$ ($n=0$) (($n>0$)), showing that the electron density is less than the proton density in the moving frame...and $\gamma$ his little influence.

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  • $\begingroup$ Thank you for your answer! But there are still few unclear spots. 1. You say: "the moving electron spacing must match the stationary ion density, in the wire's rest frame for this to be true". Ok, let's take as an experimental fact that Lorentz contraction does not work here. But what about the electron density in lab frame when there is no current? How does it relate to the stationary ion density and moving electron density? For now let's consider these three quantities only in lab frame. $\endgroup$ Commented Jan 4, 2023 at 18:35
  • $\begingroup$ when there is no current $\rho_+ = -\rho_-$, so it's neutral. In the transition to current, each electron takes its own path such that they have equal acceleration at all times in the lab frame. This mean in their instantaneous rest frames, they are spreading out. This is why I keep saying: See Bell's Spaceship Paradox, in which this apparent contradiction is resolved. $\endgroup$
    – JEB
    Commented Jan 4, 2023 at 18:39
  • $\begingroup$ It will take me a long time to figure out this paradox. So if you don't mind, just one last question, and then I will go and study:) You are saying that $ρ_- = ρ_-'$, where $ρ_-$ is the electron density in the wire without current and $ρ_-'$ - is the electron density in the wire with current both measured in lab frame, is it correct? And the Bell's Spaceship Paradox gives the explanation for this? $\endgroup$ Commented Jan 4, 2023 at 18:59
  • $\begingroup$ no, $\rho_-' = \rho_-/\gamma$. You can't measure $\rho'$ in the lab frame because "prime" always means the moving frame (not on PSE b/c ppl always don't think clearly, but in my answers, always....it's tradition). Bell's S/C paradox show how things that remain equidistant during acceleration in their initial rest frame must spread out in their local restrame...which is what happens to the electrons. Many YouTube commenters refuse to accept this fact, and the reason they don't get it is b/c they don't do the Lorentz transform. Do the LT for current in a neutral wire and all will be clear. $\endgroup$
    – JEB
    Commented Jan 5, 2023 at 11:17
  • $\begingroup$ I'm sorry for bad notation. Sure, I understand why $ρ′_−=ρ_−/γ$ when go from one frame to another. But please, note, that for now I am speaking only about lab frame. Let's denote electron density in the wire without current as $\rho_0$ and in the same wire with current $\rho$. Both in the same lab frame. My question is: are they equal, or not? If not, how are they related? $\endgroup$ Commented Jan 5, 2023 at 16:34

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