5
$\begingroup$

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 $q$?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – tparker
    Commented Feb 3, 2017 at 22:16
  • $\begingroup$ 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) $\endgroup$
    – WillO
    Commented Feb 3, 2017 at 22:26
  • $\begingroup$ (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. $\endgroup$
    – WillO
    Commented Feb 3, 2017 at 22:28
  • 1
    $\begingroup$ 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. $\endgroup$
    – WillO
    Commented Feb 3, 2017 at 22:37
  • 1
    $\begingroup$ 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. $\endgroup$
    – WillO
    Commented Feb 4, 2017 at 0:58
0
$\begingroup$

First, lets examine the linear analogue without the complication of circular motion. Imagine we two very long rods with the charges evenly distributed such that the combined charge of the two rods is neutral. Lets say the positively charged rod is accelerated in Born rigid manner. The number of charges on the positive rod remain constant, but the rod length contracts and so the charge density of the positive rod increases. The net charge of two rods become positive and test charge at rest with the stationary rod (assuming it is positive) will be repelled. Veritasium does a nice analysis of a very similar set up in this video.

the total charge across any fixed constant-time slice is both conserved and Lorentz-invariant, so it must stay zero.

While the total number of charges is Lorentz invariant, the charge density is not. I have just demonstrated that in the linear case the relative charge density does change and the system does not remain neutral. Is there any particular reason you think things would be different in the circular case? The linear case is a reasonable approximation of a segment of a very large ring. Centrifugal force is proportional to $mv^2/r$, so for a given tangential velocity the centrifugal force can be made arbitrarily small by making r arbitrarily large. The change in direction per unit time ($\omega$) is also arbitrarily small for arbitrarily large r since $\omega = v/r$. In the limit as r goes to infinity, travelling on the perimeter of the ring becomes indistinguishable from travelling in a straight line. The same applies when we consider an infinitesimal portion of the ring.

Naively, I might expect the positive ring to get Lorentz-contracted and therefore appear to increase its linear charge density

Your first instinct was correct and there is nothing naïve about length contraction. It is a solid prediction of relativity and if the radius is allowed to shrink as would happen naturally without any external forces holding the radius constant, the charge density does increase.

Why doesn't the positive ring Lorentz-contract and appear to gain charge and repel the charge q?

Returning to the linear case, the positive rod length contracts. If we attached powerful rockets to either end of the rod whose job is maintain the coordinate length of the positive rod in the rest frame of the negative rod and in this particular contrived instance, there would be no increase in charge density. The rockets will have to exert enormous force to prevent the positive rod length contracting (this is essentially Bell's rocket paradox). Similarly a force would have to be applied to the rotating ring to prevent its radius shrinking. You cannot assume the radius remains constant without specifying any external force or mechanism to hold the radius of the rotating ring constant. While centrifugal force could provide such a balancing force, the amount of centrifugal force is not directly related to relativistic length contraction and it would have to be a very specific set up to get the centrifugal force to balance the contraction. As mentioned before, for a very large ring, the centrifugal force becomes negligible. If we ignore centrifugal force the ring would contract. Of course we cannot simply ignore centrifugal force so whether the ring contracts or expands, depends on the relationship between the tangential velocity and the radius.

While spacetime is non-Euclidean in a rotating reference frame, from the point of view of a non-rotating inertial observer at rest with the centre of the ring spacetime is Euclidean and the the relationship between circumference and radius remains $C = 2\pi r$ and if the circumference is shrinking so is the radius in the flat spacetime of the inertial observer.

In summary, the ring does length contract and the charge density does increase and the test particle is repelled unless external forces are artificially applied to keep the radius of the ring constant. Any answer claiming the charge density remains constant should specify exactly what external forces are applied to keep the ring radius constant.

Improve this answer
$\endgroup$
14
  • $\begingroup$ The radius wouldn’t naturally shrink, it would naturally expand as the material strains under the rotational acceleration. The material tends to expand so much that once it reaches about the speed of sound it will expand uncontrollably and rupture. $\endgroup$
    – Dale
    Commented Jul 19 at 11:20
  • $\begingroup$ @Dale You are talking about centrifugal force and I specifically addressed that issue in my answer where I pointed out that centrifugal force can be made arbitrarily small by making r arbitrarily large. whether the circumference increases or decreases depends on the ratio between the length contraction and the centrifugal force which depends on the initial radius. There is no single answer. It depends on the initial radius. $\endgroup$
    – KDP
    Commented Jul 19 at 11:30
  • $\begingroup$ The natural strain is outwards. By making the ring arbitrarily large it will rupture outwards at arbitrarily low $\omega$. I am talking about the strain in rotational acceleration. You have the “natural” direction of the strain backwards $\endgroup$
    – Dale
    Commented Jul 19 at 11:33
  • $\begingroup$ Are you talking about the strain induced while we are accelerating the system from rest? We could for example accelerate the system over a arbitrarily long time period making the angular acceleration arbitrarily small. $\endgroup$
    – KDP
    Commented Jul 19 at 11:37
  • 1
    $\begingroup$ I have edited my answer to remove the slightly contentious statement that the radius would naturally contract since I accept what is considered natural is not clearly defined. $\endgroup$
    – KDP
    Commented Jul 19 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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