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Time dilation without acceleration

  1. May 28, 2007 #1
    Clock #1 is on a rocket with constant velocity .8c moving in direction of earth where clock #2 is located

    When the rocket is 5 light hours away from earth, clocks #1 and #2 are photographed at t=0. At the moment the rocket passes directly overhead clock#2 on earth both clocks are again photographed.
    Clock #2 reads 6:15 (5 light hrs/.8c)
    Clock #1 reads 3:45 (.6 x 6.25 hrs.)

  2. jcsd
  3. May 28, 2007 #2
    Photographed by whom?
  4. May 28, 2007 #3


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    I assume the photographer for the rocket clock is on the rocket. When he photographs the clock "when the rocket is 5 light hours away from earth", do you mean that the rocket is 5 light hours away from the Earth according to his (the rocket photographer's) perspective?

    Similarly, the photographer for the Earth clock is presumably on the Earth. When he photographs the clock "when the rocket is 5 light hours away", is he using his own, different prespective?

    Note that the two events are not the same - draw a spacetime diagram. Because of Lorentz contraction, when the rocket is 5 light hours distant from Earth in the Earth frame, at the same event the rocket is much closer in the rocket frame.
  5. May 29, 2007 #4
    I would like to restate time dilation w/o acceleration by changing path of rocket:Now clock #1 is on a rocket in a very low earth orbit with constant velocity .8c.
    Clock # 2 is stationed on earth. t=0 for both clocks is when the rocket passes directly over earth clock #2 ( if it can be done on a train platform it can be done here ) The rocket orbits the earth for a distance of 5 light hours and both clocks are mechanically photographed as the rocked passes over earth clock #2 at the completion of that distance.
    The earth clock reads 6:15 (5 light hrs/.8c)
    The rocket clock reads 3:45 (.6 x 6.25)
  6. May 29, 2007 #5


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    You've just added acceleration to the problem. Clock #1 is following a circular path, and anything following a circular path is constantly accelerating. IOW, it can't have a constant velocity (as velocity is both speed and direction), even though it can have a constant speed.
  7. May 29, 2007 #6


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    Aside from what Janus mentinoned, I'm confused by the details. Is the object orbiting the Earth at .8c? If so, it can't be in a free-fall orbit. (Possibly it could be in a powered orbit, i.e. accelerating via a rocket very very hard). But this seems to be inconsistent with the intent to have "no acceleration. (Though it turns out this intent is impossible, as Janus has noted).

    Possibly, it was intended that the orbiting clock be orbiting a body with a strong enough gravity so that an orbit with an orbital velocity of .8c was possible, which would make it a black hole.

    There would be effects in this sort of scenario due to both GR and SR.

    It might be helpful to consider real-life examples of satellites in orbit around the Earth which have been measured. While the velocities for such orbits are certainly less than .8c, our clocks are sensitive enough nowadays to detect the relativistic effects of velocities much lower than .8c.

    See for instance http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

    The actual clocks were at a high enough altitude so that the GR effect was most important. For a hypothetical orbiting clock at sea level, though, the GR effect would be absent and the SR effect due to time dilation would be the most important.

    Of course the actual measurements I refer to were only done for orbiting clocks at very high altitudes, though the Hafle-Keating experiment is closely related (and consistent with SR).

    Note that From the GR perspective one would say that the orbiting clock is in free fall, but the Earth clock is accelerating. A newtonian perspective might (confusingly) take a different perspective, but the important thing to note is that the problem definitely does involve acceleration no matter how one tries to pose it, as Janus has noted.
    Last edited: May 29, 2007
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