Relativistic rotating disc: the Ehrenfest paradox

Question

This is a pedagogical effort in aid of understanding the Ehrenfest paradox. The Ehrenfest paradox concerns a rigidly rotating relativistic disc. The term paradox does not mean the physics is in any doubt; it means simply that an untrained intuition may lead one astray.

As a way in, consider the following problem:

Q1. $N$ electrons are equispaced around a circular ring of radius $1$ metre as observed in the laboratory (the laboratory is an inertial frame of reference). The ring rotates such that the speed of each electron, relative to the laboratory, is $0.99499\,c$. The proper distance between each electron and its neighbours is $62.8\,$centimetres. How many electrons are there in the ring?

Q2. A solid disc is initially at rest relative to the laboratory. It is then set in motion by forces which act in such a way that each part of the disc acquires its new momentum impulsively and simultaneously in the laboratory, with negligible change of position. Thereafter, centripetal forces (including, but not necessarily limited to, stresses in the disc) act in such a way that all parts follow circles in the laboratory frame, centred on the disc's centre, with the same angular velocity $\omega$. (This kind of motion is called rigid motion because once it is set up, distances between parts of the moving body are constant in time). What is the ratio between the circumference and the radius of this rotating disc,
(i) as observed by using standard rods fixed in the laboratory frame?
and
(ii) as observed by attaching standard rods to the disc, and counting the number across a radius and around the circumference?

[As always in relativity, a standard rod is a solid rod with unit proper length, i.e. unit length in its own inertial rest frame, and for an accelerating rod one considers an inertially moving rod whose worldline is tangential to that of some part of the accelerating rod (i.e. they momentarily have the same velocity at some given event), and one takes a limit where the unit of length is small compared to $c^2/a_0$ where $a_0$ is the proper acceleration.]

Q3. The following arguments are commonly applied to the above problem of the rotating disc:

Argument A: When set in motion, the disc suffers Lorentz contraction around the circumference but not along a radius. Consequently the ratio of circumference to radius, as observed in the laboratory, is $2\pi/\gamma$ where $\gamma$ is the Lorentz factor associated with the speed of the rim of disc.

Argument B: When set in motion, the disc acquires its new motion instantaneously in the laboratory, without significant change in position, and therefore the ratio of circumference to radius, observed in the laboratory, does not change: it is $2\pi$ and remains $2\pi$ thereafter.

Which of A or B (or neither) is correct? Your answer should also clarify where any failed argument(s) went wrong.

Answer 1

Answer

Q1. The Lorentz factor associated with the motion of each electron relative to the laboratory is $$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = 10.003 \simeq 10. $$ The proper distance between adjacent electrons is $62.8\,$cm, and this is the distance as observed in a frame moving relative to the laboratory at the speed $v = 0.99499\,c$. That distance will be Lorentz-contracted, such that the separation of adjacent electrons in the laboratory is $$ \frac{62.8\, \mbox{ centimetres}}{\gamma} \simeq 6.28\, \mbox{cm} $$ The ring's circumference (in the laboratory) is $2\pi \times 100\,$cm. so the number of electrons is $$ N = \frac{628.3}{6.26} = 100 $$ where we picked the nearest integer (on the understanding that the precision of the data is sufficient to find this integer exactly).

Some discussion will follow at the end.

Q2. (i) $2\pi$
(ii) $\gamma 2 \pi$

For explanation, see after Q3.

Q3. Argument B is correct; Argument A is not. Argument A is wrong because the change in physical dimensions of a body, when forces act upon it, is the response of that body to the forces; a response which can be considered in any one inertial frame (such as the laboratory frame here) and which therefore need not involve any consideration of Lorentz contraction whatsoever. Lorentz contraction concerns a comparison between a distance observed in one inertial frame and a distance observed in another. The reason these may be confused, for someone learning special relativity, is that we often consider a comparison between like bodies for which the forces have acted so as to preserve the proper length. But in the disc problem the forces do not preserve the proper length: in fact they stretch the circumference of the disc (so that it is thereafter in tension); two electrons on the disc's circumference which were separated by 6 cm before the disc was spun, for example, may now be separated by 60 cm in their joint instantaneous rest frame! The Lorentz contraction effect (which arises largely from the relativity of simultaneity) enables us to predict the observed separation in the laboratory for those two electrons: it will be 6 cm if $\gamma = 10$ (this is entirely analogous to Question 1.)

Questions 1 and 3 were introduced so as to help the reader understand the answers to Question 2. I hope they have succeeded. The answers to Question 2 are not in any doubt. And, note well, they do not require one to invoke the methods and concepts of General Relativity.

By asking about number of electrons, Question 1 should make it clear that the answer to Q2(ii) is unambiguous. The number of electrons in Q1 is a matter of counting: it has one answer and everyone must agree the answer, no matter what reference frame may have been adopted for the purposes of calculation or discussion. The number of rods in Q2 is, similarly, a matter of counting. The detail, and the conceptual issue, is to be clear on what is meant by 'proper distance' for a body in accelerated motion; the explanatory remark included in the question should have clarified that. Note that there is no need to consider any accelerated reference frame. (It is possible to define accelerated reference frames and this can be done for parts of the disc, but not for the entire disc, as a further study makes clear, but we don't need that to answer the questions here.)