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Twin paradox, again sorry

  1. Oct 28, 2012 #1
    I know the twin paradox has been answered a lot on here, sorry for bringing it up again. I thought i about had it figured out but then i read a few things that confused me again. For starters, its starting to sound like time dilation wont happen unless there is a complete round trip, that somehow the actuall process of turning around and returning to the second twin causes the time difference as well as the velocity. Thats not right is it? I think where i got confused is when they started talking about doppler shift, which i really didnt understand at all, but thats what it looked like they were saying.

    Thanks
     
  2. jcsd
  3. Oct 28, 2012 #2

    D H

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    Ignore relativity for a bit. Suppose you are moving from point A to point B. You are torn between going directly to just get the move done and over with versus a more circuitous route, stopping to see long-lost friends and relatives along the way. Obviously the second route will take longer. You can't pick one single point on that circuitous route as being the cause of that option taking longer because every point on that route contributes.

    This point A to point B analogy applies directly to the twin paradox. The twins follow different paths from point A in spacetime to point B in spacetime. The lengths of these two paths are different, so the time durations measured by their clocks differ. You can't attribute any one point as the cause of this difference because every point contributes to the difference.

    Note well: Time dilation is something that appears to happen to the other twin. Consider the traveling twin. According to her senses and her clock, she's aging normally throughout the trip. Now consider the stay at home twin. According to his senses and his clock, he's aging normally the entire time his twin is gone.

    An easy way to understand the doppler shift explanation is to imagine that the twins regularly communicate with one another. Now look at what each twin perceives as happening to their twin based on these communications. On the outbound leg of the journey, each twin will perceive the other as aging at a slower rate than they are. Just before the traveling twin returns home, each twin will perceive the other as aging at a faster rate than they are. What makes things different is where this switch from aging slower to aging faster happens. For the traveling twin, the switch happens at the turn around point. For the stay at home twin, the switch happens on the receipt of the "I've arrived at the target star! Time to come home" message.
     
  4. Oct 28, 2012 #3
    Ok, i get doppler shift now, thanks. But say the one twin leaves earth then just stops at some point. Then the other twin leaves at a much slower speed and meets up with him, what then? The distance traveled will have been the same.
     
  5. Oct 28, 2012 #4

    ghwellsjr

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    All you have to do to analyze any scenario is pick an Inertial Reference Frame (IRF) and define or determine how each Observer/Clock/Object (OCO) moves according to that IRF. The IRF is a coordinate system which is how you measure distances and times. The speed of each OCO determines its time dilation which is the rate that its clock ticks at. The faster the speed, the slower the tick rate.

    So in your last question, twin A leaves and travels some distance (according to the IRF) at a fast speed and stops. That means his clock will be ticking slower than twin B's while he's traveling and therefore will lose more and more time on it according to the time coordinate of the IRF until he stops at which point it starts ticking at the same rate as the time rate defined by the IRF and it ends up with an earlier time than twin B's clock. But now twin B leaves and travels at a slower rate than twin A traveled which means his clock will be time dilated but not as much as twin A's was. So since they will have traveled the same distance, twin B's clock will not lose as much time as twin A's did and when he gets there, twin A's clock will have an earlier time on it than twin B's clock does.

    Got it? It's real simple.
     
  6. Oct 28, 2012 #5
    Got it! Ive been looking at this for hours now, i think my head was about to explode but you put it in terms that made it click. Thanks
     
  7. Oct 28, 2012 #6

    D H

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    The distance through spacetime is not the same here. The twin who travels slower takes the shorter path through spacetime and hence ages more. Suppose the target star is 4.3 light years away and that the slower twin goes at 0.1 c to reach that star, the faster one at 0.9 c. The fast twin will reach the star in 2.08 years by his clock and then will have to wait 38.25 years for the other twin to arrive. Total time per his clock is 2.08+38.25 years, or 40.33 years. The slow twin will reach the target star in 42.81 years by his clock. The fast twin will be 2.48 years younger than the slow twin.

    Here's another way to reach that same age difference. Rather than having the two twins starting out at the same time, suppose one twin departs on a ship using the best technology available at that time, one capable of reaching 0.1 c and then stopping at the target star. The other twin stays at home, waiting for technology to improve. Technology does improve, and 38.25 years later, the stay at home twin departs on a spaceship that travels at 0.9 c to the target star. The two twins will now arrive simultaneously, with the fast twin being 2.48 years younger than the slow twin when they meet at the target star.
     
  8. Oct 28, 2012 #7
    I get the rest i think, but this part threw me. Wouldn't the slower twin be taking the longer path? The same spatial distance with a longer time.
     
  9. Oct 28, 2012 #8
    The (instantaneous) acceleration at the beginning only allows for the turn around phase later on to happen, but it does not make a difference which of the twins does this instantaneous acceleration.
    Of course it is not just the turn around which makes up for the difference in aging. It is also the distance traveled at vrel. However, both twins consider the other to have been traveling away at vrel.
    Again, it is not the distance traveled which decides WHO aged less or more, but it is a prerequisite for a turn around happening at a distance to occur, which is the decisive factor on who aged less/more.


    A turn around which happens instantly after the acceleration phase. Therefore accelerating instantaneous, then instantly turn around instantaneous and accelerate instantaneous back to be at rest in the initial frame would of course make no difference in aging as the prerequisites, getting the twins to vrel and getting a distance between the two, did not happen.



    Nor does it make a difference which twin, AFTER the turn around phase, accelerates back into the other twin's frame.


    Of course we can just draw the problem in a minkowski diagram and just get the results, but making sense of it in a humanly understandable way is different imo.


    I believe that it would be wiser to change the twin paradox in such a way, that the twins initially move at vrel towards each other.
    When they meet at the sync/reference point, one twin accelerates into the other twin's rest frame instantaneous, while the other twin also accelerates instantaneous into the other twin's former rest frame.
    This would make it more clear that it does not matter which of the two twins did the initial acceleration.
    Same goes for the final acceleration phase.
    It would also make it more clear that until the turn around phase, there is a symmetry between the twins allowing for anyone to do the turn around with same results.
     
  10. Oct 31, 2012 #9
    I will have to add some info on this. Because while above is correct, it can be misleading. After thinking it out, it is true that it is the turn around phase which is the decisive factor on who ages less, but now that i am getting a clearer picture of it, one has to also consider that if twin A does the turn around, he will meet at some space-time point (x,t) with twin B. But if twin B does the turn around, it will be a different space-time point (x2,t2) they meet at.

    And so, while the decisive factor about the outcome of the experiment(who aged less/more) was who of the twins does the turn around, it is the whole trip to that space-time point which makes up for the difference, not just along the x axis, but also along the t axis.

    The turn around twin gets to that point in less time, even though he is considered the traveling one, he seems to choose the faster path to that point.

    If i manage, i will try to code some java visualization of it, not just for the forum, but also for me to wrap my mind better around this. I remember when someone in the forum stated something along the lines that acceleration rotates space-time in a sense and it feels like that might be the case in a certain sense.
     
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