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Time dilation question. . . .

  1. Sep 8, 2006 #1
    My understanding is that if X moves relative to a 'stationary' observer Y (say in direction +x), at close to the speed of light, time will pass 'more slowly' for X. That is to say, when X returns to Y, Y is older than X.

    But, isn't it just as accurate to say that Y is moving relative to X (in direction -x)? That being the case, Y would experience time dilation, and when X & Y meet up again, X should be older than Y.

    What gives?
     
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  3. Sep 8, 2006 #2

    radou

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  4. Sep 8, 2006 #3
    This is known as the twins paradox and a lot of smart people broke their heads over it in Einstein's day. The key is that there is a change in the inertial reference frame when X turns around, which is a no-no in relativity.
     
  5. Sep 9, 2006 #4
    Okey, let me put it another way:

    My understanding is that, when Y moves relative to a 'stationary' observer X, time passes more slowly on Y's watch than on X's.

    But isn't it just as accurate to say that X is moving relative to Y, and therefore Y's watch would tick "more slowly" than X's? Why would there be any net change?
     
  6. Sep 9, 2006 #5

    JesseM

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    The usual rules for time dilation in SR only apply to inertial observers, ie observers moving at a constant speed and direction (if you accelerate, you'll know you're not moving inertially because you'll feel G-forces). For a non-inertial observer, it's not true that a clock moving relative to her must always be ticking slower in a coordinate system where she's at rest.
     
    Last edited: Sep 9, 2006
  7. Sep 9, 2006 #6
    Right. . . .

    But the point is that, regardless of the accelerations involved, an entity Y will have its clock run IN SOME FASHION slower than X, if Y has a velocity relative to X.

    But, isn't it just as accurate to say that X has the velocity relative to Y, and consequently X would undergo the time dilation? Certainly, they can't both undergo time dilation because then there would be no net difference. . . .

    What gives?
     
  8. Sep 10, 2006 #7

    JesseM

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    I don't think that's right. For example, if from the perspective of some inertial frame Y is at rest while X is traveling in a circle around it at constant speed, then I think in some reasonably-constructed non-inertial rotating coordinate system where X is at rest while Y is the one moving, Y would be ticking at a constant rate faster than X.
     
  9. Sep 10, 2006 #8
    Well, let's look at it this way:

    Whatever the specifics, Y moves relative to X in direction +x, and therefore undergoes time dilation; time passes more slowly for Y.

    But isn't it just as accurate to say that X moved relative to Y, in an identical fashion except in direction -x? And consequently, X would have time pass more slowly?
     
  10. Sep 10, 2006 #9

    JesseM

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    But that's exactly what I just said is wrong. If the specifics are that the x-axis is part of a non-inertial coordinate system, then movement along this axis could cause Y to speed up rather than slow down in this coordinate system, and the rate of time dilation wouldn't necessarily just be a function of Y's velocity in this coordinate system either, even if Y moves at a constant rate it might be running faster at some times and slower at others.
     
  11. Sep 10, 2006 #10
    Well, certainly you concede that we can have a situation where Y is moving relative to X and Y undergoes time dilation.

    But why can we not say that it was X that moved relative to Y and X underwent the time dilation; where does the asymmetry come from?
     
  12. Sep 10, 2006 #11

    JesseM

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    Huh? That's what I keep saying is incorrect! Are my posts unclear?

    edit: sorry, I missed that you said "we can have a situation..." So yeah, I agree that you could have a non-inertial frame where Y was moving and its clock was slowed down. But this wouldn't be true at all times in a non-inertial frame of the travelling twin, for example--if you set up the non-inertial frame correctly, you'll still end up predicting that the earth-twin's clock shows more elapsed time, meaning at some point in the trip it must have been ticking faster rather than slower.
     
    Last edited: Sep 10, 2006
  13. Sep 10, 2006 #12
    If you are interested in time interval measurement by accelerating observers have a look please at

    arXiv.org > physics > physics/0608010
     
  14. Sep 10, 2006 #13
    Let's look at it this way: Y moves with a constant velocity, 0.5c, on a great circle path on a sphere. No accelerations involved in this example at all. The circumference of the sphere is 1 light year. X sits on the great circle, 'stationary'.

    In Y's reference frame, every 2yr it passes X. Yet in X's reference frame, Y passes it every 3yr or so due to time dilation. So after ten passes, X is ten years older than Y.

    But isn't it just as accurate to say that X is moving and Y is the stationary observer? In that case, every time X and Y meet, time dilation has caused Y to be a year older; accordingly, after ten passes, Y is ten years older than X. But then every time they meet, it would seem there is no net change in their ages: depending on the reference frame, one aged faster than the other.
     
  15. Sep 10, 2006 #14

    JesseM

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    Inertial motion refers exclusively to movement at constant speed and in a straight line--since velocity is a vector, any change in direction represents a change in velocity, even if speed is constant. And you can tell you're not moving inertially when you move in a circle at constant speed, because you'll feel G-forces (known in Newtonian physics as 'fictitious forces' because they are not genuine forces in inertial frames...see here and here)--in this case, the "centrifugal force". An observer moving inertially will always feel weightless.
     
    Last edited: Sep 10, 2006
  16. Sep 10, 2006 #15
    moving in a circular path = acceleration. Acceleration is not just a change in speed, it's a change in direction also. While constant motion is relative, acceleration is not.
     
  17. Sep 11, 2006 #16
    Is a body moving along a geodesic on the 2-sphere with a constant velocity not a inertial observer? I'm not talking about a higher-dimensional embedding: I'm saying that X (or Y) has constant velocity: no change in speed, no change in direction. He just happens to reside on S^2 instead of in Euclidean space.
     
  18. Sep 11, 2006 #17

    JesseM

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    A geodesic in curved spacetime is indeed "locally inertial", but once you introduce curved spacetime you can't assume the SR rules of time dilation will still apply outside of a local (infinitesimally small) region of spacetime. Also, it'd probably be better to talk about positively-curved 3-space, since my understanding is that general relativity would be seriously modified in 2-space (no gravitational waves, for example).
     
    Last edited: Sep 11, 2006
  19. Sep 11, 2006 #18
    A down to earth example (to make a play on words) is a GPS satellite clock - it is in an inertial frame - it experiences no acceleration since it is in free fall - and if you placed a clock on a tower at the North pole at the same elevation - the satellite clock will ACCUMULATE LESS TIME DURING EACH ORBIT than the north pole clock as it passes by - but the satellite clock really doesn't run slower - that is where the confusion arises - it simply has a space increment and a time increment that must be combined pathagorean like to total the time increment accumulated by the North Pole clock during each pass of the satellite - both clocks run at the same rate in their own frame.
     
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