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Homework Help: Rotation in special relativity

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data
    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'.

    2. Relevant equations
    [tex]t'=\gamma (t-vx/c^2)[/tex]
    [tex]\theta '=\omega t'[/tex]

    3. 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: May 20, 2009
  2. jcsd
  3. May 20, 2009 #2


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    Hi ballzac! :smile:

    (have a theta: θ and an omega: ω and a gamma: γ :wink:)
    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.
  4. May 20, 2009 #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: May 20, 2009
  5. May 20, 2009 #4


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    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:
  6. May 20, 2009 #5
    Ah yes. Gotcha. It never occured 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. :)
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