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Time dilation and uniform circular motion

  1. Dec 10, 2013 #1
    Hi, I've taken a course in SR and studied GR on my own, but I do not know how to solve problems of this type. This is just purely for fun, not homework related at all.

    A particle of mass m is moving on a circle of radius R at constant linear velocity v = .8c. If the particle makes N revolutions around the circle, what is the relativistic time dilation?
     
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  3. Dec 10, 2013 #2

    PeterDonis

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    What solution techniques have you tried? There are a number of different ways of approaching a problem of this type.
     
  4. Dec 10, 2013 #3
    To be honest, my efforts are quite embarassing.

    Initially, I thought that I could just "unravel" the circle and use the regular time dilation formula for motion in a straight line Δt' = Δt/Sqrt[1 - v^2/c^2].

    But obviously this is incorrect since the particle is not moving in a straight line, nor is it in an inertial reference frame.

    The particle is experiencing the centripetal acceleration v^2/R directed towards the center.

    I then consulted Wikipedia, specifically http://en.wikipedia.org/wiki/Inertial_reference_frame
    Scrolled to the end, and saw

    "General relativity is based upon the principle of equivalence:

    There is no experiment observers can perform to distinguish whether an acceleration arises because of a gravitational force or because their reference frame is accelerating.
    —Douglas C. Giancoli, Physics for Scientists and Engineers with Modern Physics, p. 155."

    So then I thought I would translate the centripetal acceleration into some kind of gravitational interaction using the formalism of GR, but the textbook I had used had no examples of this type.

    I have been stuck going through many GR books, looking for examples and I would be very gracious for some help.
     
  5. Dec 10, 2013 #4

    WannabeNewton

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    Woah slow down there mate :) there's no need to bring GR into this!

    Let's say we start in an inertial frame with coordinates ##(t,\vec{x})## centered on the circular trajectory of the particle. At a single arbitrary instant ##t## in this frame, we boost to an inertial frame comoving with the particle at that instant. What is ##\gamma(t)##? Does ##\gamma(t)## even depend on ##t## in this scenario? The key thing to keep in mind is that ##\gamma## doesn't depend on the direction of ##\vec{v}##, only on it's magnitude (here ##\vec{v}## is the 3-velocity of the particle in this frame).
     
  6. Dec 10, 2013 #5

    pervect

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    It turns out that that's basically OK, as "time dilation" does not depend on acceleration, it only depends on velocity, as long as you view things from the perspective of an inertial frame.

    The particle is accelerating, but the drawing itself (and the coordinates it uses, if you use numerical coordinates) is a drawing from the perspective of an inertial observer. So it doesn't matter that the particle is accelerating, acceleration does not affect time dilation - because you are using an inertial frame to describe things, which is all that SR needs.

    Figuring out the "perspective" of an accelerating observer is a more advanced question, and maybe what you're interested in (hard to say) but you don't need to answer that question to solve for what you actually asked for, which is the proper time elapsed in the circular orbit.
     
  7. Dec 10, 2013 #6
    Looks like you're holding the wrong end of the lever there. The equivalence principle is supposed to be used to help understand hard problems in GR in terms of the simpler concept of an accelerating object in Minkowsky space-time. You semm to be using it the other way around. That won't help you. Turns out your "embarrassing" effort was correct all along.

    (Added) The rule of thumb is: If there is no gravity in the problem, don't use GR to solve it. (Not surprising since GR is a theory of gravity)
     
    Last edited: Dec 10, 2013
  8. Dec 11, 2013 #7

    Mentz114

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  9. Dec 11, 2013 #8
    Gravity and acceleration do not come into it, so your initial instincts were correct.
     
    Last edited: Dec 11, 2013
  10. Dec 11, 2013 #9

    WannabeNewton

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    There's no need for them for this problem. Read post #4 and you'll see it's a one line solution.
     
  11. Dec 11, 2013 #10

    Jonathan Scott

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    As has already been mentioned, the SR velocity time dilation works fine even though the velocity is in a circle.

    In this sort of case, it is also possible to work out the time dilation due to the effective "gravitational potential" by integrating the "acceleration field" from the centre "downhill" out to the radius of the motion assuming a constant angular velocity, although this is more complicated for relativistic speeds.

    Both calculations should give exactly the same result, of course.
     
  12. Dec 11, 2013 #11

    tom.stoer

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    Time dilation for non-constant velocity in SR can be calculated integrating

    ##d\tau = dt \, \sqrt{1 - \vec{v}^2(t)}##

    The result is

    ##\tau = \int_0^T dt \, \sqrt{1 - \vec{v}^2(t)}##

    Here t is the coordinate time of an inertial frame, and tau is the proper time.
     
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