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Simultaneity of relativity

  1. Oct 22, 2008 #1
    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!

    Someone tried to answer this once before by saying that instead of the train stopping, just consider person A and B to just jump out simultaneously (according to them on the train). Then since person B's clock is ahead of A's (according to C), B jumps out first while A still enjoys time dilation a bit longer. In fact, the exact amount in order to synchronize their clocks according to C. But I don't get this because if A is enjoying time dilation longer since he's on the train, wouldn't that further deviate the two clocks, since now A's clock is still moving slower relative to C's but B's is speeding up relative to A's. Wouldn't A's clock have to speed up while B's slows down in order for their clocks to synch up again? Apologies for the long post.
  2. jcsd
  3. Oct 22, 2008 #2


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    Yes, that is what happens. A and B would only be synchronized again if they choose to repeat the synchronization procedure, now in C's frame.
  4. Oct 22, 2008 #3
    Right but my question was that since from C's perspective their clocks are desynchronized when the train stops, then what do A and B percieve their clocks to be. According to their perspecitves, their clocks should still be synchronized, since they were before the train stopped and should still be after, since the train is stopping at the same rate according to both of them. So...when they get out, how do all perspectives (A,B and C) all agree? and clearly they must agree if they're standing there looking at the watches.
  5. Oct 22, 2008 #4
    But their clocks are NOT equal!!! That's the whole point....A&B agree, C sees things differently just as when the train was moving...

    You are correct in your original description up to
    ...then you "run off the tracks"...things are never equal again...

    the whole point of relativity is that things ARE different...frames of reference are everything.
  6. Oct 22, 2008 #5
    no i understand that relativity implies that all reference frames are different. But the reason i say "now that all of their perspectives are equal" is because when the train has stopped and the three of them (A,B,C) are standing next to each other motionless, they should all agree on everything, since nobody is moving relative to each other. Clearly this is true because if they are all standing there motionless, looking at the two watches, they will obvioulsy be the same because nobody is moving relative to each other. They all have the same frame of reference. So, like i said, at what point does A and B's clocks sych up, from C's frame of reference?
  7. Oct 22, 2008 #6
    Your problem is you need to spell out in detail exactly how you got the two to jump from the train “simultaneously”. Not so much for us but for yourself.

    Obviously, the example you have given is flawed – A & B were on opposite ends of the train so in order to for all three to be standing next to each other B would have had to have waited to jump until he caught up with where A jump off to be with C (Not the same time or simultaneous).

    So they jump based on clock time & C put out a number of clocks on a very long C frame platform so A & B could jump at a pre arranged synchronized time on the C clocks. But the A and B clocks will not be on the same time if they do. (Not simultaneous according to them)

    So instead A & B only rely on their own watches to synchronize their jumps – but upon landing one can clearly see the time in the local C frame clock and looking down the Platform and see the other landing late.
    While the other can also see the local C frame clock and looking down the Platform and see the other landing early. (Not simultaneous according to what they see).

    SO what math can you offer to help them to jump “simultaneously”?

    OR as an alternative – consider the Einstein point of “Simultaneity” that regardless of what any one frame may think of their own frame time synchronizations – it cannot assume to know what is simultaneous with the exception of things that literally happen at the same place as well, (Like when A meets C happens simultaneously with C meeting A.)
  8. Oct 22, 2008 #7
    Hey thanks so much you really cleared it up for me. I think the essential point that i was missing is that if indeed they both were to jump out at the location of C, then clearly A (at the front) would have to jump earlier than B (at the back). So, B would therefore enjoy time dilation a bit longer before he jumped out to meet A and C on the platform, which in turn, would put there clocks back in synch according to C. Is this more or less correct?
  9. Oct 22, 2008 #8
    No, not as stated, but if they got on the train in the same manner one at a time from the same start point it should work. but then their watches would not have been in synch while on the train. We are talking about getting A & B in sync with each other they will never read the same as C.

    The real point to is understnad “Simultaneity”.
  10. Oct 22, 2008 #9
    i think i follow. thanks again.
  11. Oct 22, 2008 #10
    Hello curiousBos.

    I hope i interpret what you say correctly.

    A and B are at opposite ends of a train and synchronixze their clocks in the train frame. As long as they remain at reat relative to each other their clocks will emain in synch. While they are moving relative to C their clocks will appear out of synch to C. When they slow, still at rest relative to each other, they will come to a stop in the same reference frame as C and because of this C will see the clocks of A and B in sych, as will A and B see each ohers clock still in synch. The reason for C now seeing the clocks of A and B in synch with each other whereas they were not before is just a consequence of the relativity of simultaneity. A spacetime diagram would elucidate this.

    If they jump from the train simultaneously in the train frame, then they are still at rest relative to each other and the scenario is essentially the same. I did not interpret you as saying that each jumped of the train when level with C but please correct me if i am wrong as this does complicate things a bit. In the case that they do this they would not be jumping from the train simultaneously in the train frame.
  12. Oct 22, 2008 #11
    hey matheinste,
    you wrote: "The reason for C now seeing the clocks of A and B in synch with each other whereas they were not before is just a consequence of the relativity of simultaneity. A spacetime diagram would elucidate this. "

    Thats all fine... but i'd like to know the "consequence of the relativity of simultaneity."

    I think I'm confused again. If A and B's clocks are synched at lets say 12:00, according to them. According to C, B's clock is lets say 1 minute ahead of A's. So according to C, when B's clock reads 12:05, A's reads 12:04. The confusion comes when they want to jump out. If A and B agree to jump out at 12:10, then C would see B jump 1 minute before A. But then that would only further deviate the two clocks, because it would be A who would enjoy time dilation for a minute longer, according to C. But then when they land, A and B's clock would be even more than a minute off according to C, yet according to A and B, they would still have to be synched, since according to them, they jumped at the same time (12:10). I do understand that A and B are in DIFFERENT frames of reference, but only spatially. since they are not moving relative to each other, shouldnt they always agree? I think this is where my confusion lies.
  13. Oct 22, 2008 #12
    Hello curiousBos.

    A and B are in the same frame of reference as each other. Spatial speration does not affect this. Being in the same frame of reference just means that they are not moving relative to each other. If your wording was meant to say that A and B are in a different frame from C then please ignore what i have have just said.

    The consequences of the relativity of simultaneity in this case is that for C the event "A's clock reading 12:10" will not be simultaneous with the event "B's clock reading 12:10". So as you say C will not see A and B jump at he same time in C's frame.

    The transition from train speed into C's frame is not instantaneous and C will see, as A and B slow down, their clock times becoming closer and closer to to each other, finally becoming synched when they halt in C's frame. All of this is of course allowing for the time delay of light.

  14. Oct 22, 2008 #13


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    No. Only observer "D", who doesn't stop*, thinks that the two clocks are always showing the same time.

    *) More specifically: his velocity is constant, and equal to the velocity that A and B had when they synchronized their clocks.

    In this case, you can explain it with the fact that A and B will agree that in their new rest frame, they didn't set their clocks to 12:00 at the same time.

    If you instead had considered the more difficult scenario in which the train takes a much longer time to slow down, you would have had to explain it with the fact that what a clock measures is the integral of sqrt(-dt^2+dx^2) along the curve in spacetime that represents its motion. In C's frame the train keeps getting less Lorentz contracted as it's slowing down, so B is slowing down faster than A. The contribution from dt^2 to the integral will be the same along both paths (the world lines of A and B), but the contribution from dx^2 will be greater along A's path.

    I also have to suggest that you use the Return key a lot more often when you write a long post. It makes it easier to read.
  15. Oct 22, 2008 #14
    Actually these will not be essentially the same;
    it will run afoul the same problem seen in the Bell uniform Acceleration Space ships paradox.

    When they “jump from the train” assume in sync with each other on train time they will land at a some distance apart from each other but will it be longer or short than a train stopped in the C frame – SR cannot tell us this as it depends on which frame C or Train we assume is using “shorter units of distance”. But if they allow the train to stop we know the distance of one full train length will not be the same as whatever happens when they jumped. Factoring the time correctly will have the same problem and most important of all – they is no such thing as the two of them jumping from the train “simultaneously” on “in synch” with either the Train frame or the C frame – and neither of those can be considered simultaneously.

    Sorry curiousBos
    I’m sure this can only be more confusing as it is more advanced.
    IMO for you to make the progress you are looking for right now you need to avoid using assumptions like “According to C, B's clock is lets say 1 minute ahead of A's.”

    Instead build some real examples using speeds time and distances accounting for all of those in each frame. Some simple SR math and being sure you keep your frames right it is not that bad if use simple speeds. I recommend using 0.6c Gamma= 1.250 OR 0.8660254c Gamma = 2.000.
    IMO you will be better off if you do not introduce the crutch of using “spacetime diagrams” to soon.

    Also if and when you advance to needing speeds that are doubles pick from 0.27c, 0.5c, 0.8c depending on how many doubles you need. Gammas there are 1.038 1.155 1.667

    Nothing like working the math in detail to see the point IMO.
    good luck.
  16. Oct 22, 2008 #15
    hey fredrik,

    Oh... so you're saying that the problem is in my question. i had asked what causes A and B's clocks to synch again when they land (according to C).... but the answer is that A and B's clocks ARE desynchronized when they land (according to A,B, and C). That makes sense to me, so i hope its correct. thanks
  17. Oct 22, 2008 #16
    p.s. sorry about the return key thing.
  18. Oct 22, 2008 #17
    Hey thanks for the detailed info. Yea actually I do animation for a living and just love to read about phyics on my spare time. So once i start getting into the math of it all i get confused. But i do enjoy thinking deeply about the problems. I appreciate the explanations. More questions to come!
  19. Oct 22, 2008 #18
    Hello RandallB.

    I see your point. Probably technically accurate, i have yet to get onto acceleration in SR, but perhasps a little complicated for the questioner ( and me ).

  20. Oct 22, 2008 #19


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    Yes, that's what I'm saying.
  21. Oct 22, 2008 #20


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    My opinion (and it's just an opinion) is the exact opposite. Nothing like a spacetime diagram to see the point.
  22. Oct 22, 2008 #21
    it is impossible to be exactly synchronized with an event in both time and space. for to be so would mean that an event occuring at simultaneity with another is in fact the same event occuring in different time frames at the same spot - which is impossible for a body in motion since it has its own time and space frames.
    thus, person A and person B, by this reckoning, should never land at person C's position at the same time - which would imply 'occupying time and space at a moment and location which is true for A B and C and hence, A B and C are the one and the same person - impossible. Either there has to be a change in time or a change in space for person A B and C to be independant and discrete. To be syncronized with C would mean occupying 'your own' space in C's time frame.
    As such, one would have to be ejected out of the train before another - in this instance, A first then C but both at the same spot. The time taken for B to reach the 'point of ejection' of A is that time dilation factor - I think.
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