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There are many questions on the Ehrenfest paradox but I couldn't find a duplicate (which may stil exist).

The rim of a rotating disc would be Lorentz-contracted as seen from a non-rotated observer but the radius would not. I visualize this using a number of equidistant points on the rim of the disc making up a regular polygon. The requirement that the sides are (Lorentz-) contracted implies that the radius must be contracted as well.

I assume that the solutions in general relativity has to do with the gravitational-equivalent centrifugal force. Light from the rim would appear gravitationally red-shifted as seen from the center, and it perhaps contracts the radius as well. I am aware that for example Wikipedia’s explanations, or suggested solutions, seem much more complicated.

Based on this reasoning, my question is now: Surely, this effect would apply to a rotating ring with no material present except for that which makes up the ring which might be taken to be (almost) arbitrarily thin - and the central observer.

So why does material science and Born rigidity make its way into the paradox?

Qmechanic
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Mikael Jensen
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  • I think of the materials science approach as a cop out. We can get rid of the disc entirely.

    Objects in the sky rotate at 360 degrees per 24 hours. Their speed is how far away they are from the earth's axis times that rotational speed. The speed of light is then exceeded by all objects over 3/4s of the way from here to Alpha-Centauri.

    Reconcile that, I think you have some ansers to the Ehrenfest paradox.

     – R. Romero Oct 23 '19 at 18:28
  • Related https://physics.stackexchange.com/q/705857/226902 – Quillo Apr 30 '22 at 09:32

2 Answers2

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This is a Special Relativity question. Your question is otherwise correct!

The disk does not feel its perimeter shrink. As well as time dilation and Lorenz contraction you have to remember the loss of simultaneity. Any small section of the disk rim will appear foreshortened to an outside observer, but the two end points of that section are seen at different times according to the disk. The leading point is seen at an earlier disk time than the trailing point, so the disk would expect the observer to get the 'wrong' answer when measuring the distance between them.

Reducing the disk to a ring makes no difference.

There is a problem with the concept of 'disk time' if you try to define it around the disk, but that is not a rigidity or a material science problem. Disk time can only be defined over a track that does not completely encircle the centre.

DrC
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If we arrange a mirror clock system in a lab , one mirror on the ceiling, another on the floor and count as a tick a rise from the bottom, a reflection from the ceiling, and then a return trip to the bottom mirror, we have in the lab frame each tick taking twice the height divided by the speed of light.

Now suppose we have some spinning centrifuge so that it's axis of rotation is parallel to the normal of the two mirrors. In this rotating frame we have both lateral motion of the tick beam as well as vertical motion. The distance risen remains constant between the two frames, but the horizontal distance increases in the rotating frame, dependent on distance to the axis of rotation. Since the distance increases, but the speed is constant, the time must increase. So we have a time dilation effect.

This alone might explain why alpha-centauri doesn't move across the sky faster than light.

Now suppose we have mirrors on straight line so that the tick beam runs parallel to the ground. Suppose they are as far from each other in this scenario as they were in the ceiling/floor scenario. How long does a return trip from the closest mirror to the axis of rotation to the farthest and back take vs. in the floor/ceiling scenario?

I don't know, but I think the analysis sheds light on the relativity of rotating frames without the need to consider the speed of sound in materials.

R. Romero
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