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Suppose you have two rings of equal radius exactly overlaid on each other. One of them has uniform charge density $+\lambda$ and other uniform charge density $-\lambda$. Clearly the charges will simply cancel and there will be no electric or magnetic fields anywhere.

Now suppose you start the positive ring rotating in place at relativistic speed. There will now be current flow and thus a magnetic field, but for simplicity I'm going to ignore that and just consider the Lorentz force on a charged particle $q$ at rest with respect to the non-spinning ring. Naively, I might expect the positive ring to get Lorentz-contracted and therefore appear to increase its linear charge density, thus creating a net outward electric field at $q$ and repelling it. But this can't be right, because the total charge across any fixed constant-time slice is both conserved and Lorentz-invariant, so it must stay zero. Why doesn't the positive ring Lorentz-contract and appear to gain charge and repel the charge $q$?

tparker
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1 Answers1

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Let $\theta$ parameterize locations on the fixed ring, and let $a(\theta,t)$ be the acceleration vector of the point on the moving ring at location $\theta$ and time $t$. If the acceleration "looks the same" at every point in the lab frame --- that is, if, for every $t$ and every $\theta$, $a(\theta,t)$ is just $a(0,t)$ rotated by $\theta$ --- then clearly the ring cannont contract in the lab frame. (It suffices to think about the limiting case where every point on the ring jumps from speed $0$ to speed $v$ all at once.)

So what happened to the Lorentz contraction? Answer: Imagine a traveler on the ring, holding a short circular-arc-shaped meter stick that lies on the ring. According to that traveler, the front of his stick will have started accelerating before the back of his stick started accelerating. Therefore the stick has stretched. The observer in the lab frame sees no stretching, hence sees a shorter stick than the observer on the ring.

Bottom line: In the lab frame, the ring is Lorentz contracted, in the sense that any given small arc looks smaller than it looks to a traveler on the ring, but at the same time, that arc (and the entire ring) looks no smaller than it did before it started moving. Hence no change in the charge density.

WillO
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  • What do you mean by "the front of his stick will have started accelerating before the back of his stick started accelerating"? Do you mean while the ring is speeding up in the lab frame, or long afterward, when everyone agrees it's moving uniformly?

    Also, what's special about circular motion? For a straight line of charge, the density does go up by a factor of $gamma$ after it starts moving.

     – tparker Feb 03 '17 at 22:16
  • I mean this: At time $0$, in the lab frame, the entire disk starts rotating counterclockwise at speed $v$, and continues to rotate at this speed. Let $E$ be the event that a lab observer says is at time $0$ and location $\theta$ and let $F$ be the event that a lab observer says is at time $0$ and location $\theta+\epsilon$. Let $\epsilon$ be small enough, and look over a small enough time interval, that we can interpret an observer at $\theta$ as moving in a straight line, so we can just use special relativity. Then Bob, (CONTINUED) – WillO Feb 03 '17 at 22:26
  • (CONTINUED), who rides on the disk and is present at event $E$, will say that $F$ took place earlier than $E$. He will still say the analogous thing at every time in the future as long as the disk is moving. It's complicated to show this if you insist on looking at Bob as a non-inertial observer, but easy if you look at Bob, over any short time interval, as an effectively inertial observer. – WillO Feb 03 '17 at 22:28
  • As for what's special about a circular charge, the answer is nothing. If every point of a straight-line stick accelerates by the amount $a(t)$ at time $t$, then the length of the stick cannot change in the lab frame and therefore the charge density does not change. Of course the charge density can change if different parts of the stick follow different acceleration paths according to a lab-based observer. In general, you can't just imagine that the "stick starts moving" without being more precise about which parts start moving when and according to whom. – WillO Feb 03 '17 at 22:34
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    Again, take the simplest case: At time $0$ (in the lab frame) all parts of the stick jump from velocity $0$ to velocity $v$. If the left and right ends of the stick are at points $0$ and $1$ to begin with, then they'll be at points $vt$ and $1+vt$ at time $t$ --- so the stick will still have length $1$ (again, all as measured in the lab frame). A rider on the stick will say the stick has stretched. – WillO Feb 03 '17 at 22:37
  • That last sentence cleared everything up for me! I was under the impression that the proper charge density (i.e. the local charge density as measured from a comoving frame) would stay the same during the acceleration process. But in fact, it decreases by a factor of $\gamma$, which is exactly canceled by the Lorentz contraction you get from boosting back into the lab frame. So the density observed in the lab frame is the same before and after the acceleration process. – tparker Feb 04 '17 at 00:56
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    I'm glad this helped. I remember asking myself exactly the same question some years ago, and being very happy when I got it figured out. – WillO Feb 04 '17 at 00:58