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 [tex]\gamma\omega v/ c^2[/tex], where [tex]\omega[/tex] is the angular speed of the cylinder in S'.
[tex]\theta '=\omega t'[/tex]
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
[tex]\theta =\omega \gamma (t-vx/c^2)[/tex]
now, I have made the assumption that [tex]\theta=\theta'[/tex]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 [tex]\gamma\omega v/ c^2[/tex]
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 [tex]\gamma \omega[/tex]
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 :)