What Happens to the Circumference of a Rotating Disk in the Lab Frame?

[jimmd]

I read the entry about the Ehrenfest rotating disk paradox on Wikipedia, but the entry does not actually answer my question about what happens to the circumference of the disk in the "resolution of the paradox" section.

All I want to know is what happens to the circumference of the rotating disk when observed from the non-rotating (lab frame) inertial reference frame? Is the circumference smaller than 2 pi r due to the Lorentz contraction? I have seen some relativity textbooks stating that the Lorentz contraction makes the circumference smaller by the factor 1/gamma, so that C = 1/gamma x 2 pi r.

Thanks in advance for your answers.

 

[Fredrik]

The short answer is that the circumference doesn't change. It is being Lorentz contracted by the usual gamma factor, but it's also being forcefully streched and the two effects cancel each other.

The easiest way to see that something like this must happen is to consider an instantaneous acceleration (a boost) of a rod, from velocity 0 to velocity v (almost c). Do the different parts of the rod accelerate at the same time? If they do, then at the same time in what frame? If they accelerate at the same time in the frame where the rod was at rest before the boost, then the front accelerates before the rear in the frame where it's at rest after the boost. If they accelerate at the same time in the other frame, then the rear accelerates before the front in the first frame.

It simply isn't possible to accelerate something without also forcefully stretching or compressing it. That's why people say that there are no rigid objects in SR. If you want to know more about this, look up "Born rigid".

 

[pervect]

I read the entry about the Ehrenfest rotating disk paradox on Wikipedia, but the entry does not actually answer my question about what happens to the circumference of the disk in the "resolution of the paradox" section.

All I want to know is what happens to the circumference of the rotating disk when observed from the non-rotating (lab frame) inertial reference frame? Is the circumference smaller than 2 pi r due to the Lorentz contraction? I have seen some relativity textbooks stating that the Lorentz contraction makes the circumference smaller by the factor 1/gamma, so that C = 1/gamma x 2 pi r.

Thanks in advance for your answers.

There's no problem with the circumference in the lab frame - it's just 2 pi r.

The wordlines of the disk all wrap around a cylinder. When you slice this cylinder with the appropriate plane of simultaneity in the lab frame, you get a well-defined closed circle whose circumference is just 2 pi r, see the space-time diagram in the wikipedia article, i.e.

http://en.wikipedia.org/wiki/Image:Langevin_Frame_Cyl_Desynchronization.png

As the wikipedia article points out, where people get confused is by thinking about the "rotating frame", which exists locally as a frame-field but not globally (i.e. you can't synchronize all the clocks on a rotating disk using Einstein's method, if you consider pairs of points that are picewise Einstein synchronized, this set of points does not form a closed curve).

However, if I understand your question, you are NOT asking about the confusing and ill-defined "rotating frame" but rather asking about the lab frame, and there is no problem there - the circumference in the lab frame is well defined and equal to 2 pi r.