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Current in wire + moving charge next to wire creates magnetic force in the stationary reference frame OR electric force in the moving reference frame from special relativity due to change in charge density etc.... I think I understand this and I think it's super cool. Now here's the rub...

Current in wire + stationary charge next to wire creates no net charge. This is how nature behaves. I get it. My question is why can't I use the same special relativity logic as above, that is, current in wire cause electrons in wire to contract in accordance with special relativity so there has to be a net charge on the wire, which then acts on stationary charge next to wire.

I have been reading and reading and reading and I have come up with the following:

(1) When electrons in wire get accelerated to create current the distance between them actually expands in accordance with special relativity - something to do with bells spaceship paradox - which I am not going to pretend to understand

(2) This expansion from (1) above is exactly opposite and equal in magnitude to the contraction special relativity then causes and the expansion and contraction cancel out to keep the charge density in the wire constant and therefore no net charge on the wire

Here are my questions:

Is the explanation above correct? If so, please elaborate because i dont understand

If not correct, what is going on?

This is driving me absolutely nuts.

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  • $\begingroup$ I'm not sure I can follow your question, but I hope you understand that for normal electromagnets the velocity of free electrons in the wire is quite slow, so I would not expect SR to have noticeable effect. Also there is no change in charge density, any more than pumping water in a pipe changes the amount of water in the pipe. $\endgroup$ Commented Oct 30, 2013 at 12:54
  • $\begingroup$ thanks mike - watch this verisatium video youtube.com/… $\endgroup$
    – Amit
    Commented Oct 30, 2013 at 13:36
  • $\begingroup$ I not sure if I completly understand your question but when there is a stationary charge next to the wire, you say that it is the magnetic force created by the current in the wire that repells the positive charge next to the wire. And when there is a charge that is moving along the wire with the same velocity and direction as the electron flow, then you say that because of SR the charge is repelled due to length contraction. I don't think I have much more knowledge of SR than you do, but I hope this helps. $\endgroup$
    – Reds
    Commented Oct 30, 2013 at 13:59
  • $\begingroup$ Related if not Duplicate : Why don't stationary charge feel force from a current carrying wire?. $\endgroup$
    – Voulkos
    Commented Mar 29, 2022 at 14:50

3 Answers 3

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Suppose you start with a linear charge density $\lambda^+$ of positive charges and $-\lambda^-$ of negative charges in the wire, everything at rest.

Case 1: No current, test charge stationary

You assume you have a neutral wire with no current. Therefore $\lambda^- = \lambda^+$. There's no other frame worth considering, since nothing is in motion anyway.

Even if you did go into another frame, any change in charge density will affect electrons and nuclei equally. Thus the wire is neutral in all frames, and test charges are entirely unaffected by it.

Case 2: Nonzero current, test charge moving with electrons

Now suppose you have a wire with a current. Again, the wire is neutral in the lab frame $S$, where the bulk of it is not moving. In this frame, we still must have $\lambda_S^- = \lambda_S^+$, even though the electrons are moving and the nuclei aren't.

If we slip into the rest frame $S'$ of the bulk electron motion, then the spacing between electrons must be different, and in fact it must be larger. Since charge doesn't change when changing frames, we know $\lambda_{S'}^- < \lambda_S^-$. Similarly, the nuclei spacing will be length-contracted, so $\lambda_{S'}^+ > \lambda_S^+$. In this frame, then, $\lambda_{S'}^+ > \lambda_{S'}^-$, so the wire looks positively charged, and any (positive) test charge at rest in this frame $S'$ will be repelled.

As you can check, this is exactly what the Lorentz force law tells you. If the electron bulk motion is in the $-z$-direction, then the current is in the $+z$-direction, and the magnetic field along the $+x$-axis (assuming the wire coincides with the $z$-axis) is in the $+y$-direction. A positive charge with velocity in the $-z$-direction in a magnetic field in the $+y$-direction will experience a force in the direction of $(-\hat{z}) \times (+\hat{y}) = +\hat{x}$, away from the wire.

Case 3: Nonzero current, test charge stationary

Now consider the setup as follows. In $S$, the nuclei and test charge are stationary, but the electrons are moving in the $-z$-direction. Just as before, we can transform into the electrons' rest frame, where we will find that the wire is positively charged. However, we also have that the test charge is moving in the $+z$-direction in $S'$, and that there is a current of positive charges in the $+z$-direction (which we could neglect earlier). Here the full Lorentz force law tells us there is a $qE$ repulsion, and also a $q \vec{v} \times \vec{B}$ attraction, and in fact they perfectly balance in this frame, so there is still no net force.

Summary

The space between electrons expands only if you keep yourself in their rest frame as you accelerate them. The spacing measured by an observer who doesn't accelerate is unchanged, in keeping with the assumption that the wire stays neutral in the lab frame. You can only use the electrostatic Coulomb's law if you are in the frame where the test charge of interest is stationary. If you are in a frame where the charge is still moving, you need the full Lorentz law, using whatever electric and magnetic fields are present in that frame.

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  • $\begingroup$ ok - are you saying the balancing attraction in case 3 is due to the magnetic fields generated by nuclei and test charge moving in the +z direction whilst we are in the electrons' rest frame? If so, very cool! $\endgroup$
    – Amit
    Commented Oct 30, 2013 at 20:40
  • $\begingroup$ final question - you say that one can only use coulomb's law in the test charge stationary frame. Let do that for case 3. Now the electrons get contracted and generate a net negative charge. This should affect test charge but it doesn't so something must be causing electrons to get expanded to compensate. What is doing this? Is it the magnetic field generated by the moving electrons themselves? Can moving charges interact with their own magnetic field? Am I making sense or am I losing the plot? $\endgroup$
    – Amit
    Commented Oct 30, 2013 at 20:49
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    $\begingroup$ First question: Yes. Second question: You have to be consistent which frame you are in, and you have to specify which frame has the neutral wire. In the test-charge-stationary frame in Case 3, I assume there is a current but the wire is neutral, as is often the case in laboratory setups. You can't assume the wire is neutral in some other frame - that would be a contradiction. If you want to change to another frame, you have to do 3 things: find the new speed of the test charge, find the charge of the wire, and find the current in the wire. Then apply Lorentz. $\endgroup$
    – user10851
    Commented Oct 30, 2013 at 22:26
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    $\begingroup$ Your argument is tautological because the physical reason our wire is neutral in the lab frame is because despite the length contraction of the electron density, physically the electrons of course must rearrange themselves until the electrostatic repulsion between them is balanced with the surrounding nuclei, ie until the wire is again neutral. The problem is that the same logic should hold in any other frame. If we consider a co-moving frame where the protons are moving, by symmetry the same physical process should again neutralize the wire. So I think your description is incomplete. $\endgroup$
    – user1247
    Commented Aug 16, 2017 at 20:33
  • $\begingroup$ @Amit User10851 answered your questions, but here is my understanding. It does not necessarily contradict to his. First, "one can only use coulomb's law in the test charge stationary frame" should be understood as "one can use coulomb's law (qE) instead of full Lorentz law in the test charge stationary frame". Second, "This should affect test charge but it doesn't" this is because of the magnetic force qv x B compensating the coulomb force, not because "something must be causing electrons to get expanded to compensate". $\endgroup$
    – verdelite
    Commented Mar 11, 2019 at 16:05
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Short answer: this type of exposition, and the video in particular, gloss over some of the details that complicate the situation.

I checked with Section 12.2 of Jackson ("On the Question of Obtaining the Magnetic Field, Magnetic Force, and the Maxwell Equations from Coulomb's Law and Special Relativity"). I think that a key statement is: "One key assumption or experimental fact is that in a frame K where all the source charges producing and electric field $\vec{E}$ are at rest, the force on a [test] charge $q$ is given by $\vec{F}=q \vec{E}$ independent of the velocity $\vec{u}$ of the charge in that frame."

This section refers to: D. H. Firsh and L. Willets, Am. J. Phys. 24 574, (1956) and Chapter 3 of M. Schwartz Principles of Electrodynamics McGraw-Hill, New York (1972) as more complete expositions of this way of obtaining Maxwell's equations in the context of special relativity.

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I am struggling with the same issue, can't figure out an answer. But there are a few things i realised, maybe these will help someone else in figuring out the answer.

  1. The relative motion of electrons and nuclei is present in a current carrying wire. That alone should give different charge densities in different frames.

2.Acceleration should not be necessary to answer this question.

  1. Moving the charge in one direction leads to attraction but moving it in the opposite direction causes repulsion, and not moving leads to no force

  2. The charge moving slower/same speed/faster than the current carrying electrons just increases the force, it doesn't lead to change in the direction of force. If you go by veritasium, this point is unanswered.

Veritasium's explanation is missing something for sure

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