2
$\begingroup$

I'm trying to learn the connection between special relativity and magnetism. I know that if I place a positive charge, at rest, next to wire with current, I should not observe any force on it because there is no electric field and there is no magnetic force as my charge is at rest.

But here is what confuses me - the wire contains moving electrons and according to what I learned, the stationary charge should observe a length contraction of those electrons and so the density of them will increase and a negative electric field should be observed.

This is definitely not the case and I wonder if someone can explain to me what is wrong in my analysts.

Thanks!

$\endgroup$
4
  • $\begingroup$ Nothing is wrong with your analysis. Here is a video from Veritasium. Here is a post from StackExchange. $\endgroup$ – mmesser314 Nov 26 '20 at 17:23
  • $\begingroup$ There is not electric field on my charge at rest from the wire. But because of length contraction of the electrons I expect one. This is what I want to resolve. $\endgroup$ – Gil Shafriri Nov 26 '20 at 19:04
  • $\begingroup$ @mmesser314 From the video, 1:38 wire-frame electrons are same density as lattice protons (Derek). 2:07 Henry explains they are thus dilated in the co-moving current frame, so OP's assumption of length contraction is wrong. $\endgroup$ – JEB Nov 27 '20 at 17:58
  • $\begingroup$ @JEB - True. The answers have explained it. The OP had some right ideas, but that was wrong. $\endgroup$ – mmesser314 Nov 27 '20 at 18:13
0
$\begingroup$

Consider a wire with no current flowing through it, and equal densities of negative charges (electrons) and positive charges (protons). Give the electrons some velocity (such that, in their rest frame, they have constant separation): then you will get a current due to the moving charge, and you will get a negative charge density, due to length contraction. This reasoning is correct. The wire now carries a current and has a negative charge.

Consider a wire with no current flowing through it, and a lower density of electrons than protons. This wire has an overall positive charge density. If you give the electrons a correctly chosen velocity (again, such that, they see no change in separation), then you will get a current due to the moving charge, and the negative charge density will rise (due to length contraction) to cancel out the positive charge. This reasoning is also correct, and this time you have a wire with a current but no overall charge.

You see that the current through a wire doesn't determine the charge on the wire: you also need to know the charge on the wire when it has no current. By adding or removing electrons from the wire, you can get any combination of current and charge (in principle—real matter will probably disintegrate at some point). In your scenario, the wire is constructed as in my second example, so that it carries a current and has no charge in the frame of the test charge. The test charge thus feels no force since there is no electric field.

Note that real wires are more complicated than this. Consider a loop of wire with no current and no charge. Apply some force (visiting bar magnet?) to get all the electrons circulating through it, like the first construction. In this scenario, the electrons can't see a constant separation in their rest frames, since that would imply contraction of the negative charge in the rest frame of the wire and the separation of charge into positive and negative zones. The density of electrons in the wire's rest frame must remain constant, so the electrons in their own frames see the other electrons pulled away from them. The end result, once the force is removed and we have circulating electrons at equilibrium, is that each small segment of the wire "microscopically" looks like it was made by the second construction, as the overall charge must remain zero. If this sounds weird, it probably should. Consider the Ehrenfest paradox and Bell's spaceship paradox to further understand how this works.

TL;DR: The charge on a wire is given by the relationship between electron density and proton density. The velocity of the electrons gives us the current and the relationship between electron density relative to the test charge's frame versus relative to the electrons' frame. The current does not give us the relationship between electron density and proton density: that remains a free parameter which we can adjust to get whatever overall charge we want. Specifically, the charge is controlled by whatever the wire is connected to at its ends, since that is what will be supplying/removing electrons.

$\endgroup$
2
  • $\begingroup$ Here is my understanding from the responses - the density of the electrons in the wire with current is derived from equilibrium and I can't simply transform the density of the electrons in the wire with no current to the density of the electrons of the wire with current because those are 2 different systems. You can only do that transformation when you observe, the same system from different reference frames. But this is not the case. Did I get it right ? $\endgroup$ – Gil Shafriri Nov 26 '20 at 19:53
  • $\begingroup$ @GilShafriri Yes, that works. $\endgroup$ – HTNW Nov 26 '20 at 19:55
1
$\begingroup$

The electrons themselves are length-contracted (not that such an effect is measurable). The distance between the electrons is not contracted. The density is constant in this frame.

This makes sense. Regardless of the speed of the electrons, the circuit path is the same length in this frame and the total charge (number of electrons) is fixed in this frame, so the density of charge remains fixed.

$\endgroup$
1
$\begingroup$

It's best to imagine this problem a simply as possible: a positive lattice with uniformly spaced ($a$) positive charge, each paired with an electron, also spaced at $a$...in the "wire frame".

So now the electrons are moving at speed $v$. They are still spaced at $a$, and the wire remains neutral. There is no reason to think that they would Lorentz contract in the wire frame, as the wire is, by definition, neutral, in the wire frame. Each electron is free to accelerate from zero to $v$ in it's own way so that the spacing remains $a$ (see: Bell's Spaceship Paradox for the implications of that).

The implication is that in a frame moving with the electrons, "the current frame", the electron spacing is dilated to $\gamma a$; moreover, the positive lattice is Lorentz contracted to a spacing of $a/\gamma$.

A moving charge (moving with the current), thus see and electric field cause by both an increase in positive charge density and a decrease in negative charge density, with respect to proper density.

$\endgroup$
0
$\begingroup$

There is no force because your charge is at rest; in this case, as you said, the magnetic force is zero as the charge does not move relative to the field, and the electric field is zero as the wire is overall neutral.

I believe your confusion would be because you're mixing up inertial frames. So now, let's instead consider an observer moving with the same velocity as the electrons in the wire. Then now this observer would see the charge moving with respect to them, but what they would also see is the positive charges in the wire move in the same direction with the same speed. So what we now see is an increase in the charge density of the positive charges which will repel the charge, but also a magnetic field produced by this motion that will attract the charge. So these forces will cancel and special relativity still works.

$\endgroup$
2
  • 1
    $\begingroup$ The test charge is positive: should the magnetic force be attractive and the electric one repulsive? $\endgroup$ – HTNW Nov 26 '20 at 19:25
  • $\begingroup$ Sorry, you're right; I'll edit this answer. $\endgroup$ – EigenFunction Nov 26 '20 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.