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B Time dilation of an orbiting ship

  1. May 10, 2017 #1

    Buckethead

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    I'm sure this has become a tedious question, but wasn't able to ween out an exact answer through searching.

    A ship orbits an asteroid using retro rockets to maintain an orbit. The gravity is insignificant and can be ignored. A clock on the ship will tick more slowly than a clock on the asteroid. Even though the ship feels acceleration due to the retro rockets, time dilation in this case is not due to acceleration by virtue of the Clock Hypothesis so I'm guessing we can ignore the forces felt on the ship as having anything to do with the time dilation. The ship is experiencing time dilation strictly because of its velocity relative to the asteroid. Now if we switch our observation from the asteroid to the ship then an observer on the ship would instead see the asteroid orbiting the ship at the same relative velocity. This appears to be a symmetric situation with the exception of the retro rockets. So is this situation the same as just two ships passing each other (each seeing the others clock move more slowly) or is this an asymmetric situation (similar to the twin paradox or a curved timeline in a world line diagram) where the ship has to be considered accelerating while the asteroid is not.
     
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  3. May 10, 2017 #2

    russ_watters

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    That's a big exception.
     
  4. May 10, 2017 #3

    Nugatory

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    If an accelerometer on the ship records acceleration - and it will in this case, because of the rockets - then the ship is accelerating, by definition.

    The general technique for solving these problems is to calculate the proper time along the worldlines of the two objects. In this case the gravitational effects are negligible so we have flat Minkowski spacetime, which makes the problem quite a bit easier. We only need two spatial dimensions because the asteroid and the ship lie in the same plane. If we put the origin of our coordinate system at the center of the asteroid, then the worldline of the asteroid is just the time axis; the worldine of the spaceship is a helix coiling around the time axis.
     
  5. May 10, 2017 #4

    Buckethead

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    OK, but can't we put the ship at the origin instead or does the fact that it feels acceleration prohibit that. If we can do that, then doesn't that make it a symmetrical situation like two passing ships?
     
  6. May 11, 2017 #5

    Orodruin

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    You could find a coordinate system where this would be true. However, that would not be an inertial frame and so you could not apply the expressions derived for inertial frames without thinking. You would need to redo the entire computation and in the end you would get the same result as that in the asteroid frame.
     
  7. May 11, 2017 #6

    Janus

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    Nothing ever "experiences" time dilation. Time dilation is something you measure as happening to clocks either with a relative motion with respect to you or at a different gravitational potential, or in a different position to you relative to the acceleration if you are in an accelerated frame. The clock postulate simply means that an accelerating clock won't exhibit any additional time dilation due to its acceleration as measured from any other frame. It does not apply to what time dilation an observer in the accelerated frame would measure in other clocks. The fact that the circling spaceship is in an accelerating frame does effect how it measures what is happening to the asteroid clock. Since it is constantly accelerating towards that clock, it will measure it as running fast compared to its own. You can't swap an accelerated frame's observations for an inertial frame's observations.
     
  8. May 11, 2017 #7

    Buckethead

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    I apologize for my sloppiness. I did mean to say that the ships clock moves slower relative to the asteroid. Accuracy is everything!
     
  9. May 11, 2017 #8

    PeroK

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    If you search for a thread titled "two twins moving around each other in a circular orbit", then in post #14 there is an analysis of this situation. This approximates the motion of the rocket as a sequence of instantaneous changes of direction.

    I'm on my phone and can't find a way to link to that thread.
     
  10. May 11, 2017 #9

    Nugatory

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  11. May 12, 2017 #10
    By the equivalence principle, this situation is effectively identical to a uniform gravity pulling everything in the direction the ship's rockets are firing, while the ship resists it perfectly. In this scenario, the ship is at a lower gravitational potential than the asteroid, so its time is slower as a result.
     
  12. May 13, 2017 #11

    PeroK

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    That is not correct. Acceleration does not cause time dilation. The equivalence principle concerns experiments carried out locally within an accelerating reference frame, but would not apply to observations made from outside the rocket.

    Moreover, if a rocket accelerates away from you, its time dilation is precisely related to its instantaneous velocity in your frame. There is not an additional component associated with its acceleration/"gravitational equivalence".
     
  13. May 15, 2017 #12
    I never said it did. Consider the twins paradox scenario. When the twin in the ship turns around, his acceleration to return home is equivalent to him resisting a uniform gravity causing everything to move in the direction of his rockets. Since Earth is at a higher potential in this field, its time is faster during the acceleration, thus explaining how the Earth twin ends up older upon the traveling twin's return.

    If you distill the scenario in the OP down to the basics, where you have two ships at the same location in an inertial reference frame, one ship moves away from the other, and then continually fires its engines so that it maintains what is essentially a circular orbit around the other ship for some time, and then returns. Both ships observe each other in relative motion, but the one doing the orbits will have passed less proper time during this maneuver. The equivalence principle and resulting gravitational time dilation can be used to explain this, though there are other ways of explaining it as well.
     
  14. May 15, 2017 #13

    PeroK

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    The time dilation entirely and precisely depends on the relative velocity. There is no reason I can see in complicating things by involving some sort of pseudo gravity. Even if that is a valid application of the equivalence principle, which I doubt.
     
  15. May 15, 2017 #14

    A.T.

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    The proper acceleration of the ideal clock doesn't affect the clock rate. But the proper acceleration of your reference frame does imply different clock rates between clocks resting at different positions in that non-inertial frame reference frame.

    Only in inertial frames. But the OP asks about a non-inertial frame.
     
  16. May 15, 2017 #15
    The weak equivalence principle states, among other things, that the local effects of motion in a curved spacetime (gravitation) are indistinguishable from those of an accelerated observer in flat spacetime, without exception. That means our accelerated ship in flat spacetime is the same for all intents and purposes as the same ship resisting a uniform gravity field (that keeps changing to reflect his constant shift in direction of acceleration), with all of the attendant effects such as gravitational time dilation.
     
  17. May 15, 2017 #16

    Nugatory

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    Yes, but note that word "local". An observer inside the ship and studying only what's going on inside the ship (light deflected towards the floor, dropped objects fall towards the floor, clocks near the nose run fast relative to clocks near the tail, ...) will find that these observations are consistent with either acceleration or gravity; that's the WEP at work. However, when he turns his attention to non-local phenomena such as a comparison of his clock and the clock on the asteroid, he will find that the WEP no longer works.

    You would be able to apply the WEP and understand the time dilation between ship and asteroid as equivalent to gravitational time dilation, if an accelerometer on the asteroid also read 1G in thevsame direction while the two remained at rest relative to one another. That's not the case here.
     
  18. May 15, 2017 #17

    A.T.

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    That would be the equivalent of gravitational time dilation in an uniform gravitational filed.

    But If we would use a rotating common rest frame of ship and asteroid, with a non-uniform proper acceleration along the radial line. Would that be equivalent to a non-uniform gravitational field (same proper accelerations) in terms of gravitational time dilation?
     
  19. May 15, 2017 #18

    PeroK

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    Let's assume that time dilation were related to proper acceleration. We know, however, that it is definitely related to relative velocity. But, relative velocity and proper acceleration are independent for a circular orbit. In the sense that by changing the radius you can have a different acceleration for the same speed.
     
  20. May 15, 2017 #19

    A.T.

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    Velocity is frame dependent. In the co-rotating frame both clocks are at rest, but still tick at different rates. In this frame that is explained by their different positions in the centrifugal potential, similar to how different positions in the gravitational potential imply gravitational time dilation.
     
  21. May 15, 2017 #20

    Nugatory

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    I'm not seeing how that works. Consider an exchange of light signals between the two clocks, with the the ship emitting its signal when it receives the signal from the asteroid. Is that different than an exchange of light signals between the asteroid and an observer moving inertially and on a path tangent to the circular path described by the accelerated observer, hence momentarily at rest in the corotating frame? In this case the time dilation will show up as a redshift, and unlike gravitational time dilation the redshift will be symmetrical because it's caused by the relative velocity.

    (We can take the duration of the signal to be arbitrarily small so that the difference between the straight-line inertial path and the circular path can also be made arbitrarily small and this is an ordinary transverse-Doppler problem).
     
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