Rotation in special relativity

In summary, the cylinder rotates uniformly about the x' axis in S'. The twist per unit length is gamma(omega t' - vx/c^2)
  • #1
ballzac
104
0

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 [tex]\gamma\omega v/ c^2[/tex], where [tex]\omega[/tex] is the angular speed of the cylinder in S'.

Homework Equations


[tex]t'=\gamma (t-vx/c^2)[/tex]
[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 :)
 
Last edited:
Physics news on Phys.org
  • #2
Hi ballzac! :smile:

(have a theta: θ and an omega: ω and a gamma: γ :wink:)
ballzac said:
[tex]\theta'=\omega t'[/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.

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.
 
  • #3
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 (:biggrin:) 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:
  • #4
ah … got it! :wink:

ok … you're saying that the cylinder itself is a clock, and therefore its speed should be slowed down …

but the time dilation of γ only applies to a moving clock (as seen by S), and by fixing x = constant to get θ = θ' = ω't' = ω'γt, you're choosing a clock moving in S'

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 :smile:
 
  • #5
Ah yes. Gotcha. It never occurred to me that x is a function of t and hence does not disappear when I differentiate wrt t. Thanks heaps for that. It means that my original answer is actually correct as far as calculating the twist per unit length (which was the task) and the error only arose when I went further than what was asked by trying to explain the rotation in terms of time-dilation. Thanks heaps Tim. :)
 

1. What is rotation in special relativity?

Rotation in special relativity is the phenomenon where an object appears to rotate when observed from different inertial reference frames. This is due to the relative motion between the observer and the object, and is a consequence of Einstein's theory of special relativity.

2. How does rotation affect time dilation in special relativity?

Rotation can affect time dilation in special relativity because it causes a difference in the relative motion between two objects. This difference in motion can result in a difference in the elapsed time between the two objects, leading to time dilation effects.

3. Is there a limit to how fast an object can rotate in special relativity?

According to special relativity, there is no limit to how fast an object can rotate. However, as an object approaches the speed of light, the effects of time dilation become more significant and eventually prevent the object from reaching the speed of light.

4. How does rotation affect the length contraction in special relativity?

Rotation can affect length contraction in special relativity because it changes the relative velocity between two objects. This change in velocity can lead to a difference in the measured length of the object, as observed from different inertial reference frames.

5. Can rotation in special relativity lead to a violation of causality?

No, rotation in special relativity does not violate causality. While it may appear that events are happening in a different order when viewed from different reference frames, causality is still preserved. This is because the order of events is always consistent within a given reference frame.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
803
  • Advanced Physics Homework Help
Replies
4
Views
379
  • Advanced Physics Homework Help
Replies
8
Views
683
Replies
4
Views
984
Replies
7
Views
1K
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
9
Views
846
Replies
1
Views
762
  • Introductory Physics Homework Help
Replies
32
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
3K
Back
Top