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In the special relativity it is well established that, in the vacuum no one can ever travel faster than light, due to the relativistic velocity addition formula.

Recently I saw some silly statement claiming, that this conclusion holds for the translational speed addition doesn't imply it holds for the rotational speed. I think it is silly since the statement didn't give any justification reasoning.

However, it is legitimated to ask whether the no-faster-than-light derived from the translational speed addition could be extended, or be equivalent to, the no-faster-than-light for the rotational speed as well. I had the hunch that, there should be a proof to show the equivalency between the two. How to prove, or disprove (unlikely though), this equivalency?

Qmechanic
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user36125
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There is no such thing as "rotational speed". There is angular momentum, which has units of $\frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}}$ (and so is not a speed), and angular velocity which has units of $\frac{1}{\mathrm s}$ (and so is not a speed either). There is no maximum angular velocity in special relativity. So long as $\omega r\lt c$ (the velocities of the involved particles stay slower than light), you're fine.

Finally, unfortunately, a rigid body cannot rigidly spin up or spin down in special relativity! The key word here is "Born rigidity and angular acceleration". The quick explanation is that a disk whose edge spins near the speed of light will be length contracted. So a tape measure spinning with the edge would read more than $2\pi r$. Therefore, if you want to apply an angular acceleration to the disk and not cause any compressive/stretching forces, you have to add material to the disk!

Basically, you have to be sure to define what you want explicitly, because there are a lot of things you can do in Newtonian mechanics that you can't do in special relativity.