# Rotation in special relativity

## Homework Statement

A cylinder rotates uniformly about the x' axis in S'. Show that in S, where it not only rotates but also travels forward, the twist per unit length is $$\gamma\omega v/ c^2$$, where $$\omega$$ is the angular speed of the cylinder in S'.

## Homework Equations

$$t'=\gamma (t-vx/c^2)$$
$$\theta '=\omega t'$$

## The Attempt at a Solution

Well, I actually got the required result, but I'm a little concerned that my derivation might be incorrect because it seems to indicate something that seems wrong.

I used the equations above to get
$$\theta =\omega \gamma (t-vx/c^2)$$
now, I have made the assumption that $$\theta=\theta'$$which makes sense to me because the cylinder is rotating IN S', and S' itself is not actually rotating with respect to S.

differentiating the with respect to x gives the twist per unit length $$\gamma\omega v/ c^2$$

All seems fine so far. Then I thought it would be interesting to differentiate with respect to time. This gives the angular frequency according to S as $$\gamma \omega$$

My problem with this is that it seems to contradict time-dilation. It seems to indicate that S will see the cylinder rotating FASTER than S' will. This can't be right though, because time dilation should cause it to rotate slower in S than in S'. I'm sure the resolution of this is quite simple, but I haven't been able to spot where I'm going wrong. Any help appreciated :)

Last edited:

tiny-tim
Homework Helper
Hi ballzac! (have a theta: θ and an omega: ω and a gamma: γ )
$$\theta'=\omega t'$$

now, I have made the assumption that $$\theta=\theta'$$which makes sense to me because the cylinder is rotating IN S', and S' itself is not actually rotating with respect to S.

θ = θ', and θ' = ω't', but ω ≠ ω' …

Imagine that the cylinder is a clock, which S' regards as rotating once a minute.

Then S regards the clock as going slow, and so regards the cylinder as taking longer than a minute to rotate once.

Yes, that's what I thought, that S regards the cylinder as taking longer to go around, hence ω' (thanks, lol) should be larger than ω (note that I've switched notation to match yours. The original question had ω as the angular speed according to S'). But ω (the angular speed according to S) is dθ/dt=γω', meaning that ω is larger than ω'. I think that you thought that I thought ( ) that ω=ω', but I was just using ω as the angular speed in S' because that was the notation used in the question. I didn't explicitly mention ω according to S because I didn't want to confuse things by priming it or calling it something else, but I did provide the formula that I derived for it, which is clearly wrong. Sorry for causing confusion, I should have just given them appropriate notation to begin with.

So anyway, I still have the original problem that my final expression relating ω to ω' had the Lorentz factor on the wrong side of the equation, yet my expression for the 'twist per unit length' is correct. So I'm quite confused.

Thanks for the input :)

Last edited:
tiny-tim
ah … got it! you need to choose x' = constant, ie x - vt = constant, so θ = θ' = ω't' = ω'(γt - γvx/c2) = ω'(γt - γv/c2(vt + constant)) = ω'γt(1 - v2/c2) + constant = ω't/γ + constant 