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Twin Paradox with a twist

  1. Jan 11, 2015 #1
    Hi all,


    I know the twin paradox has been discussed many times, so I hope you'll bear with me. My version has a slight twist.

    Suppose twin's A and B start off in a "rest" frame in dead space, frame F, at x = 0. At t = 0, A moves in the direction x with velocity c/2, and B moves in direction -x with velocity c/2. The numbers are just to remove any ambiguity. At t1, both A and B reverse their directions and head back to x = 0. From A's point of view, B has taken a journey at a relativistic speed more than c/2 and less than c. Therefore, A should expect B to be younger than him after the journey. From B's point of view, A has gone on a "relativistic" journey, and expects that A is the younger.

    If tpN is the time passed for N, then

    According to A:

    tpA > tpF > tpB

    According to F:

    tpA = tpB < tpB

    According to B:

    tpB > tpF > tpA


    Intuitively, A and B should have aged similarly. So how is this paradox resolved?


    thanks,
    Aaron
     
    Last edited by a moderator: Jan 27, 2015
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  3. Jan 11, 2015 #2

    mfb

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    t1 in which frame?
    Directly before B reverses, he'll think of A moving away at this point in time. Then he changes his direction - and suddenly the "simultaneous" point happens to be when A is returning already. If you just calculate the time A has on its clock, in the frame of B, you'll suddenly see a jump forwards when B reverses.
    This is not a physical jump, of course, it is a consequence of a poor way of time-keeping if you accelerate.

    A similar effect: The andromeda galaxy is about 2.5 million light years away. You see events "now" as they happened 2.5 million years ago (in the frame of earth). If you walk towards the galaxy, relativity tells you its distance became one day less. So you see events as they happened (2.5 million years minus one day) ago. You still see the same events as others on earth, of course, but the definition of "now" at that large distance changed significantly.
     
  4. Jan 11, 2015 #3

    Dale

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    They have indeed aged the same. You can calculate that in any inertial frame using the standard time dilation formula and you will get that answer.

    A's frame and B's frame are non inertial so the standard time dilation formula doesn't apply.
     
  5. Jan 11, 2015 #4
    Thanks for the replies.

    Hi mfb, t1 refers to F. At t1 in F, both A and B reverse directions. I'm sorry, but I didn't quite get your point. It sounds like it has to do more with GR than SR, and unfortunately, I haven't reached that part in Einstein's book yet :D I'll come back to your explanation when I do.

    Hi DaleSpam, so you're saying that in my example, I can only use the rest frame, F, for my calculations. Can I not simply transform between reference frames based on the instantaneous velocity? i.e. integrate the changes in spacetime?

    Alternatively, suppose A and B only experience acceleration for a fraction of the journey. An example of a possible journey would look like this for A wrt F:

    1) accelerate in +x direction for 0.1t
    2) move at constant velocity, c/2 for 0.3t in +x direction.
    3) decelerate for 0.1t, change directions and accelerate in -x direction for 0.1t
    4) move at constant velocity c/2 for 0.3t in -x direction.
    5) decelerate for 0.1t and stop.

    For stages 2 and 4, both A and B are in inertial frames so I should be able to apply the standard time dilation formulas. Furthermore, as we already agreed, during stages 1, 3, and 5, both A and B cover similar distances and ages wrt F.


    thanks,
    Aaron
     
  6. Jan 11, 2015 #5

    Dale

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    No, I said:
    That means any inertial frame whatsoever. F was the only inertial frame that you identified, but there are an infinite number of other inertial frames, all of which would be valid.

    I do however mean that you cannot use A's frame nor B's frame, at least not without deriving the correct time dilation formula in their frame. The reason is that those frames are non-inertial so the simplified inertial-frame result does not apply.

    The usual approach of doing this runs quickly into some mathematical problems that render it an invalid way of defining a coordinate chart across all spacetime. Most discussions of the twin paradox on this forum contain a discussion of exactly this problem.
     
  7. Jan 11, 2015 #6

    ghwellsjr

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    As mfb asked, t1 in which frame? We could say it is the Coordinate Time in frame F and that is what I think you mean but A and B will not have any way of knowing when that time occurs. We could also say that they each individually turn around when their own clocks reach a certain Proper Time value which will be that same Coordinate Time for themselves in their individual rest frames up to the point of turn around but it won't be the same Coordinate Time for the other one. I'll show you some spacetime diagrams to illustrate these different times starting with frame F:

    TwinsWithTwist1.PNG

    A is shown in red, B in black and blue is an observer who remains at rest in F throughout the scenario. The dots represent one-year increments of time. So the Proper Time for A and B when they turn around is 7 in this example and they spend another 7 years getting back so their total age accumulation during their trips is 14 years and they get back to F when he has aged just over 16 years.

    This is only partially true. In A's rest frame up to the point of turn around, B is traveling away from him at 0.8c but B doesn't turn around until some later at which point he comes to rest in that frame:

    TwinsWithTwist2.PNG

    If we look at A's rest frame after he turns around, we see that B is approaching him at 0.8c but B has already turned around much earlier:

    TwinsWithTwist3.PNG

    No, not if he is applying SR correctly. He should expect them to be the same age as we can see from any of the above three Inertial Reference Frames.

    B has the same issues that A has and should come to the same conclusions. You can just interchange the last two diagrams to apply everything to B.

    As I have pointed out, this statement is only partially true, that is, it is only true up to the point of A's turn around.

    I think you meant:

    tpA = tpB < tpF

    And this is a true statement.

    Again, this is only partially true, that is, up to the point of B's turn around.

    Hopefully, by now, the paradox has been resolved. If not, ask.
     
  8. Jan 11, 2015 #7
    Hi DaleSpam,

    .
    thanks for the clarification, but I did understand what you meant, I was only referring to the reference frames pointed out in my example.

    I see. I assume that you are referring to the sudden jump when A turns around and changes its reference frame. Just out of curiosity, has the problem been mathematically resolved?


    Hi George,


    thank you for taking the time to draw those very informative space-time diagrams. I can see how you worked it out. Based on the diagrams, just before the turn-around point of A (from A's frame), B appears to have a ways to go, but just after A turns around, he sees that not only has B turned around, but B has completed more than half of the return journey. This sudden jump must be what mfb was talking about. What I understood from DaleSpam's post is that trying to treat the problem in the usual way of finding change in x wrt t and then integrating leads to difficulties. So has the problem been resolved mathematically?


    thanks,
    Aaron
     
  9. Jan 11, 2015 #8

    mfb

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    There is no problem, at least not with physics or mathematics - nothing to solve.
    It is just a problem of understanding the different reference frames and that things change if you change frames.
     
  10. Jan 11, 2015 #9
    thanks for the reply. What I mean is that if you were to take a gradual deceleration instead of an instantaneous reversal of velocity could you avoid this jump in the reference frames, and has anybody got a simple example of how this is solved mathematically?

    Aaron
     
  11. Jan 11, 2015 #10

    mfb

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    You could record a very fast "advance of the clock of A", but that is an artifact of your calculation that constantly changes inertial frames.
    For every point within the acceleration procedure, you can check the spacetime diagram and draw the line of constant time. You can also calculate it, of course. I don't think the result is so interesting.
     
  12. Jan 11, 2015 #11
    I understand your point, thanks. My curiosity comes from DaleSpam's comment,

    regards,
    Aaron
     
  13. Jan 11, 2015 #12

    Dale

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    There are many valid mathematical solutions for defining the reference frame of a non inertial object. The problems with the naive approach do not mean that there is no valid solution. Just that specific approach is flawed.

    The other problem is that there is no one standard convention. So saying "the reference frame of X" is ambiguous when X is non inertial. You could mean any one of an infinite number of possible coordinate systems.
     
  14. Jan 11, 2015 #13

    Nugatory

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    No. Either way, there is no inertial frame in which you are at rest during the turnaround.
     
  15. Jan 12, 2015 #14
    Sure, I agree with that. By the naive approach, I assume you mean the approach that most people would initially attempt, and that pointed out by mfb:

    Well, I don't really know what you mean here. I figure that once you define the deceleration in the "rest" frame, then what A sees (i.e. from his frame of reference) is defined, and has only one possibility. I will try the "naive" approach and maybe I'll understand what you mean.

    Did I understand you correctly and you're saying that, "no, there is no way to avoid the jump in reference frames". So if I were person A, then as I turn around, I would suddenly see B "jump", which implies that for that instant, he would be travelling faster than the speed of light (in my point of view, which I understand is not an inertial frame)? It sounds like there's a lot of information behind your statement which I'm not familiar with, so I'll have to do some more reading.


    thanks for the help,
    Aaron
     
  16. Jan 12, 2015 #15

    Fredrik

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    There's no obviously correct definition of "his frame of reference". If what you mean is a different inertial coordinate system (the momentarily comoving one) at each point on his world line, then what he "sees" as he turns around is that the other guy starts moving and aging much faster than before. This happens because these comoving inertial systems have very different simultaneity lines. In fact, if the turnaround is near instantaneous, the other guy's age will make a huge jump, because A just before the turnaround considers himself simultaneous with an event where the other guy recently left the starting point, and A just after the turnaround considers himself simultaneous with an event where the other guy is soon going to be back to the starting point.

    Note that the word "sees" is sometimes used very loosely in these discussions. What an observer "sees" may have nothing to do with the light that hits his eyes, and everything to do with the numbers assigned by the coordinate system(s) that we have chosen to think of as describing his "point of view". For example, when we say that he "sees" two events as simultaneous, it means that the coordinate system that we're associating with his motion at a certain event assigns the same time coordinate to those two events.

    Edit: A better option than using the word "see" in this counterintuitive way is to avoid it entirely in these discussions.

    The concept you need to understand is simultaneity lines in spacetime diagrams.

    My favorite introduction to SR is the one in chapters 1-2 in "A first course in general relativity", by Schutz.
     
    Last edited: Jan 12, 2015
  17. Jan 12, 2015 #16

    Nugatory

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    What A sees is always clearly defined and is the same no matter what reference frame you or he choose to use. He's seeing light hitting the retina of his eye, he's seeing the position of the hands of his clocks and the readings on dials of his lab equipment. All of these phenomena are local. The reference frame is just a convention for assigning times to non-local events; for example if light from an event three light-seconds away reaches my eyes at noon, it is natural to say that it happened three seconds ago at 11:59:57 AM. But I didn't actually see it happen at 11:59:57; I received a light signal at noon and assigned the label "11:59:57" to the emission event. An observer moving at a different speed might assign a different label to that event; or he and I could standardize on my labeling scheme (agree to use the frame in which I am at rest and he is not); or we could standardize on his labeling scheme (agree to use the frame on which he is at rest and I am not).

    No. An easy way to avoid the jump is to just use one inertial reference frame throughout the entire thought experiment. Of course the traveller won't be always be at rest in that frame, but that's to be expected because the traveller is changing speed.
    No, you do not see B jump. What you will see is explained here: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html and it is the same no matter what reference frame (convention for assigning time labels to distant events) you use.
     
  18. Jan 12, 2015 #17
    Hi all,


    thank you both for your patience. I understand the confusion I created by using the word "see". What I meant was what Nugatory explained about assigning a time to a non-local event. I will go ahead and use the word "measure" for this, and the measurement de-embeds the time it takes for light to reach the observers eyes.

    Yes, this is what I meant. So A would "measure" that B has essentially jumped through space-time. This is allowed because A is in a non-inertial frame, and is basically experiencing the shift in his reference frame. i.e. he is not an observer who could believe himself to be in a "rest" frame.

    thank you for the reference.

    I understand that, but I am more interested in observer A's "measurements". I guess the crux of the matter is that being in a non-inertial frame allows A to observe B jumping through space-time. Thanks for the link on the doppler effect.


    best regards,
    Aaron
     
  19. Jan 12, 2015 #18

    Nugatory

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    That's still not quite right. What A observes and measures is completely independent of any reference frame, and doesn't lead to B "jumping through space-time" - B is moving through space-time on whatever path he is following and that's unaffected by anything that A does or observes. All that A's acceleration does is make the times that he assigned to events back on earth before he accelerated not match up properly with the times that he assigned to events back on earth after he he accelerates.

    Perhaps the most intuitive way of seeing this is to imagine that instead of turning his spaceship around, the traveller grabs his notebook and jumps from his outbound ship onto a ship that conveniently just happens to be travelling in the opposite (inbound) direction at the appropriate speed. When he compares his notes with the logbooks of the second ship and discusses his and his twin's past history with the crew of the second ship, he will find a completely consistent story in which he was ageing less quickly than the stay-at-home twin the whole time. The fact that the traveller's diary doesn't line up well with the ship's logbook just tells us that the two logs were maintained using a different standard of time.
     
  20. Jan 12, 2015 #19



    Hi Nugatory,


    thank you for the clarification. I do understand of course that B from his own perspective has not leaped through time, however, I don't see exactly where my thinking was wrong. From A's perspective, he really does appear to measure B leap through space-time, however, having experienced acceleration/deceleration, he is aware that there will be an inconsistency in his result. Of course, as you say, when they meet-up back on Earth, they will be able to resolve the puzzle into a consistent story. Feel free to correct me if I have misunderstood the situation.


    regards,
    Aaron
     
  21. Jan 12, 2015 #20

    Dale

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    This is precisely the process that is not uniquely defined for a non inertial observer.

    One approach is to take the momentarily co moving inertial frame and use that to assign the time to a non local event. This has several problems, the least of which is the "jump forward" of the home twin's clock.

    But that is not the only possibility. Any smooth one to one mapping will work. Here is another common approach. http://arxiv.org/abs/gr-qc/0104077
     
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