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Moving Frame With Many Clocks Pass By

  1. Mar 1, 2015 #1
    I had asked this about 2 years ago - but for the life of me cannot find it. Please feel free to cancel this if you can find my OP. Anyhow my question was this:

    A "train" passes by at significant relativistic speed, along the "x" axis. "Platform" observer peers in windows of train as it passes, only watching the particular clock which is aligned with the observer as it is passing at the same "x" coordinate. The picture seen by the observer is akin to a film of a clock but each moment is actually a different clock from the train. My recollection is the answer I got was that to the observer, the impression of the clocks would be that time passes *faster* by the lorentz factor for the speed. Can someone confirm?
  2. jcsd
  3. Mar 1, 2015 #2


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    It must (that is, platform observers see train clocks running slow - your statement isn't completely clear about who is fast/slow so I'm adding this just for definiteness).

    All clocks on the train are at rest relative to each other, so we can assume they're all synchronized and measure time as experienced by any observer sitting on the train.

    "The train passing through the station" is well-defined without reference to observers, as the interval between events (a) "front of train passes back of platform" and (b) "back of train passes front of platform".

    So we are comparing the time interval between physical events (a) and (b) as measured by an observer sitting on the train, and by one on the platform.

    But the Lorentz transform tells us exactly how the measure of elapsed time between two events changes between observers.

    And of course if you put clocks along the platform, an observer on the train will come to the conclusion that platform clocks run slow by the same factor, the situation is completely symmetric.
    Last edited: Mar 1, 2015
  4. Mar 1, 2015 #3


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    You are measuring t' at x=0, where t' is the time in the moving frame and x is the position in your frame. The Lorentz transforms immediately give you that ##t'=\gamma t##. Your recollection is correct. The situation is symmetrical as usual. Observers on the train watching clocks on the platform would also see them apparently sped up.
  5. Mar 2, 2015 #4
    I am not the OP, but I have always wondered how to visualize what happens when you keep switching reference frames.

    Set up the train moving with v and the station so that at t=t'=0 the nose of the train and the beginning of the station line up(event A).
    Set v so β=.5√3 and γ=2 .

    Now in the IRF of the station, time passes as the beginning of the station lines up with the rear end of the train(event B).

    t time has passed for the station. So reading the clock at the rear of the train from the beginning of the station reads t/γ.

    Now we switch to the train's IRF at time t`=t/γ. Say we are in the rear end of the train. For the train event B has not yet happened, right? Because if we again switch to the station we are now at time t=/γ^2 and we can keep Lorentz transforming, adding more powers, and we get closer to event A at t=0 and t`=0? Each time we Lorentz transform we half the time and in the limit we go to 0 for both?
  6. Mar 2, 2015 #5


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    You can't violate causality that way. If an event has happened, it's happened. Changing frames just means changing your choice of coordinates (that is,the numbers you use to label points in space time), and changing coordinates doesn't change anything, any more than scrolling the map on your phone. That's really what the Lorentz transforms are doing - changing your perspective on space time (from ones that are natural to you to ones that are natural to someone in relative motion) in a way analogous to moving your view of a map.
  7. Mar 2, 2015 #6


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    But the Lorentz transform isn't [itex]t = \gamma t'[/itex]; it is [itex]t = \gamma \left( t' - vx'/c^2 \right)[/itex] and these formulas agree only in the special case [itex]x' = 0[/itex].

    And the inverse isn't [itex]t' = \gamma t[/itex]; it is [itex]t' = \gamma \left( t + vx/c^2 \right)[/itex] and these formulas agree only in the special case [itex]x = 0[/itex].

    It's impossible for both [itex]x[/itex] and [itex]x'[/itex] to be zero (except when [itex]t = 0[/itex]).
  8. Mar 4, 2015 #7
    I think this is a very nice question. And your "impression" about what happens is correct too.

    You can figure it out even without math. Train observers see the platform clock go slower, and all the train clocks always in sync. So if the platform clock and the train clocks are all 0 at some point, on some following point the train clocks will be ahead of the platform clock.

    But for observations that happen on the same spot in space, reference frame should not matter and can not introduce differences. So if the train observers think that a specific train clock is ahead of the platform clock when they are at the same position, the platform observer should agree too. So each next train clock will seem further and further ahead to the platform observer at the moment it is co-located with his clock.
  9. Mar 5, 2015 #8
    I was thinking that for such an observer - if he did not know about relativity of simultaneity - he might conclude that "moving clocks run fast."
  10. Mar 5, 2015 #9


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    Certainly. If you make wrong assumptions then you can make wrong conclusions too.
  11. Mar 5, 2015 #10
    It's more of a naive observation, eh? I think that the issue is ignorance of SR, not incorrect assumptions.
  12. Mar 5, 2015 #11


    Staff: Mentor

    In order to conclude that the clocks in the "moving clocks run fast" not only do you need your proposed observation, you also need the assumption that the moving clocks are all synchronized. Otherwise, since you only observe each clock once, you cannot make any inference about how fast they run. It is the assumption that they are all synchronized that is incorrect and therefore leads to the incorrect conclusions.

    If you don't make incorrect assumptions, then either you will not be able to conclude anything from your observations, or you will observe relativistic effects.
  13. Mar 5, 2015 #12
    I think if the observer had never heard of SR, he would assume that *all* clocks everywhere were synchronized. I can see how that is an assumption.
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