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Modified Twin Paradox in Special Relativity

  1. Jul 23, 2013 #1
    The problem statement
    Suppose the universe is a 2-torus, thus there exists a spatial path that you will go back to where you start eventually. Let Observer A and Observer B travel on this closed path. Observer B is travelling at a constant speed of 0.9c relative to Observer A, where c is the speed of light.

    At time t0, Observer A and Observer B are in the same spatial location and they synchronize their clocks. Then they travel apart. At time t1>t0, Observer B meets Observer A and compares their clock.

    Paradox: from the view point of Observer A, Observer B is travelling at 0.9c, thus B's clock should run slower than A's clock; from the view point of Observer B, Observer A is travelling at 0.9c, thus A's clock should run slower than B's clock. Therefore when A and B compare their clock at t1, there seems to be a contradiction.

    Question: resolve this "paradox", draw the space-time diagrams for both Observer A and Observer B.


    My attempt
    The lecturer has talked about the Twin Paradox, and I know that the acceleration and deceleration leads to the clock discrepancy. However here I think both Observer A and B are in inertial reference frames, so the clock should not be affected by acceleration. But I just can't find any loopholes in the problem. My experience told me that, the paradoxes in special relativity are always caused by clock synchronization (correct me if I am wrong). Is the answer to this paradox due to the fact that, it is impossible for A and B to synchronize their clocks at t0? Why? Or my attempt is wrong and the solution lies on other aspects?

    Thank you for your help!!
     
  2. jcsd
  3. Jul 23, 2013 #2

    Dick

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    Other aspects is the answer. In the Twin Paradox saying acceleration and deceleration leads to the discrepency is a little misleading. You have the same paradox even if the change of velocity is instaneous. It's just the change of frames that count. You really have to follow the hint and draw the space-time diagram. There are obvious problems with synchronizing clocks when the spatial coordinate is periodic. It takes some thought. The torus isn't quite Lorentz invariant. There is a preferred frame where the radius is maximal. Think about it.
     
  4. Jul 23, 2013 #3
    Thank you very much for your reply. But I think in this problem Observer B is always travelling at speed of 0.9c, so there shouldn't be any change of velocity. Can you explain a bit more on "There are obvious problems with synchronizing clocks when the spatial coordinate is periodic"?
     
  5. Jul 23, 2013 #4
    How come there is a preferred frame? I thought the geometry of 2-torus is flat, and every point on 2-torus has infinite radius, or am I mistaken?
     
  6. Jul 23, 2013 #5

    Dick

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    Draw a space time diagram like the problem suggests. Of course, there is no change of velocity in this case. Think about it this way. Light signals can travel around the space multiple times. This means A can see multiple copies of B. He can only synchronize with one of them, not all of them.
     
  7. Jul 23, 2013 #6

    Dick

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    I think what they mean is not really a 'torus' but a 'cylinder'. Just the spatial coordinate is wrapped in a circle. If you shoot a light signal to the right it will travel all of the way around the universe and hit you from the left. What I mean by preferred frame is that there are experiments you can do to detect whether you are moving. So there is an absolute rest frame.
     
    Last edited: Jul 23, 2013
  8. Jul 23, 2013 #7

    TSny

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    For some discussion of this problem see this thread and this thread. They contain links to published papers on the topic.
     
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