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Simple question regarding the twin paradox

  1. May 27, 2012 #1
    Hi,

    I am a little confused with this paradox. I asked my professor about it and he didnt really give a convincing answer. So the scenario basically seems to be some twins on earth(or anywhere) at rest, and then one leaves at relativistic speed for some time then comes back to see that his/her twin is much older than them.

    My question is, how come you can tell which one would age more? Why couldn't it just as well be the one on the ship? From the twin on earths reference frame, they are at rest and then the rocket flies away from them, while in the rocket frame it is at rest and the earth flies away from them. From each of the twins perspective the other one moves and they are stationary in their own frames. How come the same thing wouldn't happen to the twin on earth and find the rocket twin older when the earth arrived back at the rocket?

    I just took this paradox for granted for a long time but now I seem to be confused.

    Thanks for reading.
     
  2. jcsd
  3. May 27, 2012 #2
    In order for twin on the rocket to return they must undergo acceleration, and therefore do not have an inertial frame of reference during the entire round trip.
     
  4. May 27, 2012 #3

    ghwellsjr

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    You need to pick one inertial frame and stick with it from start to finish. And you need to understand that the faster you go in that frame, the slower your clock ticks.

    Now can you see that from the earth's frame, only the rocket twin's clock will run slow?

    And can you see that if you use the rocket's frame during the first half of the trip, only the earth's clock will run slow but during the last half of the trip, the rocket has to travel much faster than the earth in order to catch up with it and so its clock has to run even slower such that it ends up with less time on it when it gets back to earth?
     
    Last edited: May 27, 2012
  5. May 27, 2012 #4
    If you are at rest in a given inertial reference frame then you do not feel any proper acceleration. When the travelling twin turns around to head back he experiences proper acceleration and he can now be certain that he is not at rest in inertial reference frame. The stay at home twin does not feel proper acceleration during the turn around event, so the situation is not symmetrical.
     
  6. May 27, 2012 #5

    mathman

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    There is a simple way to understand it based on the Lorentz transformation. Say the traveler goes to a place ten light years away from earth. He quickly speeds up to almost the speed of light. The distance in his frame to his destination is now foreshortened and the time to get there is is greatly reduced. Once he gets there, stops and turns around to go back to earth, the same foreshortening of distance and time takes place. So in his frames the total time would be a lot less than twenty years, while the earth bound twin would age more than twenty years.
     
  7. May 27, 2012 #6

    PeterDonis

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    Visceral, you might want to read the Usenet Physics FAQ entry on the Twin Paradox:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

    There are a number of different ways of understanding what's going on in this scenario (most of which have been mentioned in this thread), but the FAQ entry ties them all together.

    My personal preference is to look at everything using a spacetime diagram (the FAQ entry discusses this in some detail). Looking at it this way makes the solution obvious, in my view: you have two twins, each of which takes a different path through spacetime between the same pair of events (the event where they part, and the event where they come back together). The two paths they take have different lengths, so they experience different amounts of proper time passing between those two events (since the "length" of a worldline is just the proper time elapsed for someone following that worldline). Working out the actual math tells you that the stay-at-home twin's path is longer, so he ages more and is older when the two meet up again. It's no different in principle than the fact that two paths through Euclidean space that start and end at the same point can have different lengths; it's just geometry.
     
  8. May 28, 2012 #7
    I think that's why it is called the twin paradox
     
  9. May 28, 2012 #8

    HallsofIvy

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    Actually, no, it isn't. The term "paradox" is being used ironically- it is only a seeming paradox which can be resolved as others have said here.
     
  10. May 28, 2012 #9

    DrGreg

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    Well, actually, the word "paradox" has multiple meanings including:
    1. a seemingly absurd or self-contradictory statement that is or may be true
    2. a self-contradictory proposition
    (my emphasis). Most of the paradoxes in physics and maths (including the twins paradox) turn out to be of the first type.

    Source: Collins Concise Dictionary, 4th Ed 1999
     
  11. May 28, 2012 #10
    Well, the OP said he suddenly thought to himself, why isn't this story symmetrical -- why is one twin different from the other?? But that's exactly the point of the story, isn't it? That's why it is the 'twin paradox' not the 'twin effect' or some such. As DrG points out, it seems contradictory, but it isn't.
     
  12. May 28, 2012 #11
    If the traveler twin, on both his outgoing trip and on his returning trip, says that his home twin is aging slower than he is, then how can he find the home twin to be older when he finally gets back?

    I think the answer is that the traveler twin says his home twin ages a lot during his turnaround.
     
  13. May 28, 2012 #12

    PeterDonis

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    This is one way of looking at it, yes. Check out the Usenet Physics FAQ entry I linked to in post #6.
     
  14. May 28, 2012 #13
    Thanks. I read it, but it says that the gap is an accounting error. Seems like if the gap is an error, then it would also have to be an error for the traveler twin to say that the stay-home twin ages slower when the traveler's going away, and when he's coming back. Is that an error too? I hear lots of people talk like that's true.
     
  15. May 28, 2012 #14

    PeterDonis

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    I assume you're referring to the "Time Gap Objection" page...

    http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gap.html

    ...which says

    The "accounting error" is basically changing the "zero point" of time; the outbound reference frame has a "zero" of time that is about 13 years and 8 months earlier than the "zero" point of time in the inbound reference frame. So when you switch frames, you have to switch zeros of time as well. That adds 13 years and 8 months to Terence's clock as seen by Stella.

    Why do you think the two have to be connected? "Ages slower" refers to the *rate* at which time "flows" in one frame compared to the other; it doesn't say anything about where the "zero point" of time is set. On the Time Gap Objection page, it says:

    So Terence does "age slower", according to Stella, on *both* legs, in this interpretation, even though he ends up older at the end (because of the change in "zero point" of time).

    I suspect that's because they're trying to use a different interpretation than the "Time Gap" one. For example, check out the Doppler Shift Analysis page in the FAQ.

    The key is that there is not a single unique "right answer" for most of these questions; how you answer them depends on how you interpret various observer-dependent quantities. The only question that has to have a unique answer is, how do the two twins' ages compare *when they meet again*. That answer is unique because both twins are at the same location at the same time, so none of the ambiguities in interpretation come into play.
     
  16. May 29, 2012 #15
    I'm just saying that if the traveling guy says that his home-staying twin's total aging while he is gone is the sum of 3 parts, then if one of the 3 parts is an error, the other 2 parts can't be right either. Because the sum is right.
     
  17. May 29, 2012 #16

    PeterDonis

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    Ah, I see. You are interpreting "error" as "wrong" :wink:; but the FAQ really means "error" as in "something that has to be compensated for by adding in an additional term". In this case, what has to be compensated for is the difference in the zero of time between the two frames, i.e., the "error" is a "correction" you have to apply to convert one frame's time into the other's. The term "accounting error" is an unfortunate choice of words; the FAQ is not trying to say that the 13 year 8 month difference is an incorrect value; all 3 parts, as given in the FAQ, are correct values, and are added together correctly to give the correct final answer. It's just that one of them, the 13 years 8 months, doesn't correspond (in the Time Gap version, where the turnaround is instantaneous) to anything "observed" by Stella; it's just a change in zero of time that she has to apply when she changes inertial frames.
     
  18. May 30, 2012 #17
    so has anyone actually flown a clock in space at relativistic speeds(or really fast speeds at least - I know clocks can be very accurate tools nowadays) and then collected it to see if it's fast or slow?
     
  19. May 30, 2012 #18

    Nugatory

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    Yes, many times. Look at the FAQ on experimental support, a stcky thread at the top of this sub forum.
     
  20. May 30, 2012 #19
    OK, so it's right for the traveler guy to say that the homey girl gets a lot older when he turns around. It all comes out right. Thanks.
     
  21. May 31, 2012 #20
    For the problem weve been talking about, we can figure out how much older the homey girl got while the traveling guy was turning around by figuring out how much extra she had to age so that she would be the right age when he gets back. But is there any way to figure it out before he gets back? Can he figure it out right after he finishes turning around?
     
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