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Relativity and time (twin paradox)

  1. Dec 28, 2009 #1
    I have thought about it and I have found to be confused about something, regarding the twin paradox. As I know it, the twin paradox (set up to emphasize my confusion) is as follows: Two spaceships floating in space pointing away from each other, lets say A and B. When B accelerates and approaches the speed of light (as observed by A), B's time is slowed down, and when B returns to A, B is younger than A.
    My confusion is, if we make the "observer" to be B, then it is A that is approaching the speed of light, and when the two meet again, A should be younger than B. Why when analyzing the same scenario and just changing which ship is the observer changes which ship has gone further into time?
     
  2. jcsd
  3. Dec 28, 2009 #2
    Observer B feels himself pressed against the back of his pilot's seat when his ship accelerates.

    Observer A doesn't feel himself pressed against the back of his pilot's seat. Because his ship doesn't move at all.

    In other words A and B are not interchangeable. Simple as that really.
    B knows that his own ship is moving, A knows that his own ship isn't moving.
     
  4. Dec 28, 2009 #3
    yeah I thought of that too, but I thought that those relativistic affects are just based on the velocities, not accelerations. Is there more to this paradox than I know? That it is actually acceleration that dilates time and not simply going a high velocity will slow down your time? The lorentz factor does not include an acceleration term in it, only the speed of light constant and the velocity.
     
  5. Dec 28, 2009 #4
  6. Dec 28, 2009 #5
    Special Relativity effects ARE a result of the relative velocity of the reference frames, but don't forget that, in SR, only constant relative velocities between equivalent frames are allowed and, in the "twin" situation, there is always at least a point in B's trajectory where there is acceleration, which means that, at this point, B's spacecraft changes frames and these are not equivalent (because their relative velocity isn't constant), so you cannot use Lorentz's transformation to pass from one to the other; that's the source of the apparent asymmetry.

    By the way, you may have accelerated bodies in SR, and there are Lorentz formulas for the acceleration between equivalent frames provided that it doesn't arise from gravitational forces or these are weak enough, so that the spacetime curvature isn't significative.
     
  7. Dec 28, 2009 #6
    What do you mean Jsuarez when you say B changes frames, I thought we can chose where the reference frame is, can't we choose it to be B throughout it's acceleration. Also, there is another way to think of this, with the acceleration that is. The only time where there IS a difference in A and B's motion (apparently) is when B is accelerating. When this acceleration is over, NOW A and B have exactly the same motion as viewed from each to the other. Since at the end of the whole trip, B is younger than A, doesn't this mean that time dilation must have occurred during the acceleration of B, otherwise they could not have had any time differences, right?
     
    Last edited: Dec 28, 2009
  8. Dec 28, 2009 #7
    Yes, but you can't use the standard SR time dilation formula for that since it's only valid in inertial (unaccelerated) reference frames. And ship B isn't at rest in any single inertial reference frame for the entire trip.

    That leaves two choices:

    1. Assume an instantaneous turnaround and two completely different inertial frames for ship B that accelerates and a single inertial frame for ship A. From B's perspective, clock A "jumps ahead" during the turnaround due to a shift in simultaneity between the two inertial frames of ship B. That's the easiest method.

    2. Analyze the scenario for the accelerated reference frame of B during the turnaround. In this frame of ship B during the acceleration, clock A is running fast relative to the accelerating clock B due to gravitational time dilation. So in B's total reference frames, clock A runs slow during the outbound inertial motion, clock A runs very fast during the acceleration enough to go from being behind to being way ahead of clock B, then after the acceleration clock A runs slow again during the inbound inertial motion.

    Either choice will give the same answer: clock A will show less total elapsed time than clock B at the reunion.

    Interestingly, Einstein's own resolution of the Twins Paradox used the second method, and you can read it here: http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativity

    Both methods are actually equivalent, since method one uses a single "jump" in simultaneity between inertial frames, and gravitational time dilation in accelerated frames is nothing more than a series of "jumps" in simultaneity between successive inertial co-moving frames, where the interval between jumps is infinitesimal. This is how gravitational time dilation was originally derived.
     
    Last edited by a moderator: Dec 29, 2009
  9. Dec 29, 2009 #8
    It would be interesting to see an equation for this solution. It seems such a solution would have to have a term for the distance between A and B during the turnaround acceleration. Just in case it is not clear, I am not suggesting that there is no such solution and it would seem to be the logical outcome of the equivalence principle.

    It might also be worth mentioning in the context of this thread that the asymmetry of the situation becomes very clear when analysed in terms of the frequencies of signals sent between the twins at regular intervals. On the outward trip both see the signals of the other twin arriving at a slower rate than the rate they send signals out, but when B turns around he immediately sees an increase in the signal rate from A, while A does not see an increase in the signal frequency from B until B is nearly home. Both A and B agree that B sent less signals than A during the total trip.
     
    Last edited: Dec 29, 2009
  10. Dec 29, 2009 #9
    Einstein's resolution paper doesn't actually show the math, but I think there's a complete solution on the net somewhere, I don't remember where. It just uses the standard gravitational time dilation equation t(e) = T(1 +gh/c^2) to calculate the rate of the earth clock in the accelerated frame. It's a little complicated by the fact that h varies instead of being constant, since the earth clock isn't stationary in this accelerated frame. (like calculating the varying rate of a clock thrown up in the air relative to a clock at rest at earth's surface).

    I wouldn't say it's a result of the equivalence principle, I'd say it's the other way around. Gravitational time dilation was originally derived from considering an example similar to this one, and in turn predicted to apply in a gravitational field as a result of the equivalence principle.
     
    Last edited by a moderator: Dec 29, 2009
  11. Dec 29, 2009 #10
    wow thats great, so it IS the acceleration that causes the weird paradox, not the observed slowing of time each ship has on the other. that is so great thank you al, I am so glad I asked.

    (oh, and Al when you said "clock A will show less total elapsed time than clock B at the reunion" did you mean it the other way around?)
     
  12. Dec 29, 2009 #11
    Ooops, sorry. Yes, it's the other way around.
     
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