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Synchronized clocks in special relativity

  1. Dec 1, 2006 #1

    Bos

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    This is sort of a long post but I have to explain the following situation in order to ask the question so bare with me. Lets say two people, person A and person B, are on opposite ends of a train moving at a constant velocity and another observer is at rest on the platform outside (most of you have heard this I'm sure). Person A is at the front of the train and person B at the back. They have agreed to set their wrist watches to 12:00 as soon as the light from a light bulb in the middle of the train reaches them. So, the light bulb is switched on and they both set their clocks to 12:00 simultaneously because from their perspective, the light had to travel an equal distance from the middle to reach both of them, and therefore they both set their clocks at the same time. However, person C on the platform outside who was standing still watching claims that person B set his clock before person A. This is because from his perspective, person B (in the back) is moving toward the light while person A (in the front) is moving away from it and therefore, the light had to travel less distance to reach person B. Since light is constant in all frames of reference, person C will indeed claim that it took the light longer to reach person A than it did to reach person B. Both claims are equally valid and both are justified in their reasoning, so both answers are equally correct.
    This all makes perfect sense to me and thats all great. But here's my question, finally. Since person C sees the light reach person B before it reaches person A, person C will conclude that person B set his clock BEFORE person A. So to person C, or anyone else on the platform, when it is say 12:05 on person B's clock, it will only be 12:03 on person A's clock (the precise number depends on the length and speed of the train but that's not relevant to the point being made). Again, this all makes sense to me. But now what if the train were to immediately stop and person A and B were to get out and confront person C. Now that all of their perspectives are equal, they should all agree upon the readings of both clocks (they have to if they're all standing there looking at them). But how is it that suddenly when the train stopped, it put the clocks back on equal footing from person C's perspective. If person C saw that the two clocks were different (12:05 and 12:03) when the train was moving, than even if the train slowed down and eventually stopped, wouldn't person C still claim that since person A and person B both slowed down and stopped at the same rate, their clocks also slowed down at the same rate and hence would still not be synchronized? What is it about the stopping of the train that puts person C's perspective on equal footing with person A and person B? Clearly I'm missing something, please help! Again I'm very sorry for the long post but this things been buggin me for a while.
    Thanks.
     
  2. jcsd
  3. Dec 1, 2006 #2
    Longwinded question.. I can't find which person's frame your train "immediately stop"s in.
     
  4. Dec 1, 2006 #3

    Bos

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    I don't think that matters. It stops in everyone's frame of course, it just may be that person A and B perceive to stop at a different time than person C perceives it to stop, but the same question applies.
     
  5. Dec 1, 2006 #4

    Galileo

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    Instead of imagining the train stopping, consider the (perhaps conceptionally simpler) case where passengers A and B jump out of the train and onto the platform, in person C's inertial frame.
    If we imagine A and B jumping out of the train simultaneously we see immediately where the problem lies. If they jump out of the train at the same time as seen from the platform frame, they will not jump at the same time as seen from the train frame.

    For example: If they jump out simultaneously as seen from the train's frame, then they jump when their watches reaches, say 13:00 and ofcourse both their watches will read 13:00 when they compare them on the platform.
    What happens from C's point of view is that B's watch is a little ahead of A's, so B jumps out earlier than A. A therefore enjoys time dilation a bit longer, the exact amount such that their watches both read 13:00 when they hit the platform.
     
  6. Dec 1, 2006 #5

    Bos

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    It certainly helps to look at it that way, thanks. But I still have a little problem with your explanation. You say: What happens from C's point of view is that B's watch is a little ahead of A's, so B jumps out earlier than A.

    Why does B jump out earlier just cause his watch is ahead of A's? The only reason their watches aren't synchronized is because of the fact that light traveled less distance to B. When they jump out of the train it has nothing to do with their watches anymore. I would think that if anything, C would see A jump out first, since he's in the front of the train.
     
  7. Dec 1, 2006 #6
    the platform observer will see one end of the train stop at a different time to the other end. don't assume a solid thing like a train acts like you expect in the relativistic scale
     
  8. Dec 1, 2006 #7

    Bos

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    right that makes sense. But if the platform observer sees the front of the train stop first then that means that person B (in the back) will experience the time dilation slightly longer, which would only further deviate the two times, right?
     
  9. Dec 1, 2006 #8
    well is the train travelling toward or away from the observer?

    will the deceleration involved affect things
     
  10. Dec 1, 2006 #9

    JesseM

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    You could have them jump out in any way you want, Galileo was probably assuming you wanted both ends of the train to stop "simultaneously" in the train's own frame. If A and B both jump out simultaneously in the frame of the platform instead, then in the platform's frame (which is now also their frame) their watches will remain out-of-sync by the same amount as when they were riding on the train (as seen in the platform's frame).
     
  11. Dec 1, 2006 #10

    Bos

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    Right but how could they remain out of sync if, according to A and B, they were never out of sync in the first place, so why should they be when they jump out simultaneously?
     
  12. Dec 1, 2006 #11

    Galileo

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    It has. If A and B agree to jump at the same time in their system, then it means they both jump when their watches point at the same time (again, say 13:00), because their watches are perfectly synchronized as viewed from their frame.
    In C's frame it's B's watch that's ahead of A's, so B's watch will reach 13:00 sooner than A's.

    One thing that has really helped in such problems in the past is to just get the Lorentz transformations out and apply them to the situation. In that case you can see exactly what and when things are happening in the different frames.

    Could you say why? I don't understand.
     
  13. Dec 1, 2006 #12

    JesseM

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    Because if they jumped at the same time in the platform's frame, that means that in the train's frame they didn't jump at the same time, A jumped first, and B's clock continued to run at the normal rate until he jumped while A's ran at a slowed-down rate after jumping, causing them to get out-of-sync.
     
    Last edited: Dec 1, 2006
  14. Dec 1, 2006 #13

    Bos

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    Ahhh, thank you. that really cleared things up for me a lot.

    So if we go back and apply it to the train stopping instantly instead of them jumping out, does that mean that from C's perspective, he would see B (in the back) stop first?
     
  15. Dec 1, 2006 #14

    Bos

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    I'm still a little unclear as to how it works out if they don't jump out and the train just stops like my initial question. Is it simply that one end is moving differently relative to the other, but even so, from C's perspective wouldn't the front obviously stop before the back which would therefore mean that person B's clock is still running slower. This would only further deviate their clocks, right?
     
  16. Dec 1, 2006 #15

    Bos

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    sorry to be a pain but i just wanna understand it fully. I do fully understand the situation in which they jump out but not when the train just stops.
    thanks again.
     
  17. Dec 4, 2006 #16

    Ich

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    If the ends are supposed to stop when the light ray reaches them, the back end would stop before the front end.
     
  18. Dec 4, 2006 #17
    I assume that the portion of the train travelling away from 'C' is actually travelling faster than the portion coming towards 'C', So if the train stopped nearly instantaneously 'A' would decelerate faster than 'B' as both ends of the train will stop together (Otherwise the train will be a different length to when it started, and all the commuters will be stressed or strained - Maybe this is why train commuters are always grumpy, must be Einsteins fault) which I assume would equate to a different perception of Time during which the clocks would synchronise.
     
  19. Dec 4, 2006 #18

    JesseM

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    When you say the train stops "instantly", do you mean both ends stop at the same time in the frame of the platform? If so the train had better be fairly elastic, because this will mean the train's length when it stops will be the same as its length when moving, which was shrunk in the platform's frame due to length-contraction; in other words, its new rest length will be its old shrunken length as measured in the platform's frame.

    Other than that, there's no reason anything would be different here than in the case where the observers jump out. In the inertial frame where the train was originally at rest before it stopped relative to the platform, the front end of the train accelerates before the back end, so during the time-interval where the front end is moving alongside the platform while the back end is still at rest in this frame, the front end's clock will be ticking slower and causing the two clocks to become out-of-sync.
     
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