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Extrema of proper time

  1. Dec 13, 2009 #1
    Quick question:
    Suppose I hold two initially synchronized clocks on earth and throw one up and catch it when it comes back down. Now my (small amount of) knowledge of GR tells me that the proper time on the thrown clock should be maximized since it was on a geodesic.

    However, this seems like the twin paradox, and in the twin paradox, the thrown clock shows less time than the stationary clock.

    What am I wrong about?

    Thanks in advance!
     
  2. jcsd
  3. Dec 13, 2009 #2

    In the twin paradox, just as in your thought experiment, the clock on geodesic (the twin who stays at home) records a longer time interval than the clock on the distorted path (the twin who flies to a far away star and comes back).
     
  4. Dec 13, 2009 #3

    Dale

    Staff: Mentor

    You are correct. The thrown clock is on a geodesic (provided we measure from just after it leaves the thrower's hand until just before it returns) so the proper time is maximized. This is exactly analogous to the twin paradox since the thrown clock is the one that goes in a "straight" line. It travels inertially and does not measure any proper acceleration.
     
  5. Dec 13, 2009 #4

    A.T.

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    I think that only holds in SR. In GR geodesic world lines maximize proper-time only locally. You can also have two geodesics meeting twice with different proper-times intervals in between.

    See this thread:
    https://www.physicsforums.com/showthread.php?t=249722
     
    Last edited: Dec 13, 2009
  6. Dec 14, 2009 #5

    George Jones

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    Either tonight or tomorrow, I'll post (some of) the fairly simple details of a GR example for which the usual SR result doesn't hold, i.e., the elapsed proper time between meetings for an accelerated clock is greater than for a non-accelerated (geodesic) clock.

    The example consists of two clocks that have same [itex]r[/itex], with one clock in geodesic circular orbit (freely falling with no acceleration) and one clock hovering (accelerated).
     
  7. Dec 14, 2009 #6
    You're misidentifying which clock in the Twin Paradox is the "thrown" clock. In that example, it is the twin that remains on Earth that is analogous to the clock you throw in your own example. In the twin paradox, we are ignoring effects of gravity (which are minimal in any case), so we're really talking about a freely-floating twin (in place of the Earth) and a twin that goes to Vega and accelerates when he gets there. The accelerated twin knows that it was he who accelerated because he feels the acceleration (the acceleration is recorded by an accelerometer, i.e. it causes a mass-on-a-spring to stretch).

    In your example, it is the earthbound clock whose accelerometer stretches, while the freely falling clock experiences zero acceleration. Hence the freely falling clock will have the extreme aging.

    Note, however, that while in SR you are guaranteed that the inertial observer will have aged by the maximum amount, in GR you are only guaranteed that inertial observers (with no acceleration on their accelerometers) age by extreme amounts. That amount could be maximum, minimum, or even a saddle point. In this example, however, it will indeed be maximum.
     
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