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So I've been trying a fairly simple thought experiment and I'm not able to wrap my head around an aspect of special relativity.

Here is how it goes:

A train is moving at steady relavisitic velocity V past a platform. There are two observers on the train and two on the platform at equal measured distances. Lets call the ones on the train T1 and T2 at the front and back of the train respectively. Similarly P1 and P2 are on the platform with P1 at the front and P2 at the back. Assume T1,T2 have synchronized atomic clocks and optically triggered cameras. So do P1 and P2 (syncrhonized with each other but not necessarily with T1,T2)

Assume that they believe that the speed of light is always measured the same for all inertial observers.

Now the events are:

  • A bulb is lit on the train exactly between T1 and T2
  • T1 and T2 both photograph their clocks triggered by the light beam from the bulb
  • P1 and P2 do the same

As far as my understanding of SR goes, now both pairs of observers have matching timestamps in their photographs. They would have also measured the exact same speed of light since the time and distance travelled is the same for each pair.

But they will not accept each others photographs because, as far as P1/P2 are concerned, T1 was rushing away from their light beam and T2 was rushing toward their light beam.

If light travels independent of its source and medium, then it has gone at c-V towards T1 and c+V towards T2, hence the timestamps in the photos cant possibly match - unless the clock with T2 went slow (or T1 went fast) or the front half of the train shortened (or back half lengthened)

The symmetric situation occurs with T1/T2 not able to believe how P1 and P2 can have the same timestamp photo without the one clock being off or one half of the platform being of a different length

Now I have read similar examples in many books and the fact that observers in different frames of reference cannot agree on what time something happened. I've also read that relativistic time dilation and space contraction resolves this paradox.

However my understanding is that for the train reference frame, the platform is shortened, and for the platform reference frame the train is shortened. The same applies to the observation of clocks across the reference frames.

To simplify we can replace "platform" with another train moving in the opposite direction (so as to avoid the asymmetry we see in the twin paradox)

Given all this, how does uniform length contraction of the train w.r.t the platform explain the paradox of simultaneity?

If D/T = c for the lightbeams, what D and T do the platform observers have to see for the forward and backward moving beam, in order for the photos to match? Since everyone and their dog have to agree on c being the same....

If I knew nothing else, my logical answer would be that from the viewpoint of P1/P2 the rear half of the train has to lengthen (or the rear clock slow down), or the front half of the train has to shorten (or the front clock speed up).

But this means somehow there is an asymmetry in "approaching" and "receding" from a reference frame - this is at odds with the basic idea that the whole train appears length contracted or that all clocks on the moving train are slowed w.r.t the platforms clocks.

How does this work?

rep_movsd
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    To a seasoned instructor of SR, it is very easy to spot where your problem lies. You have not correctly stated a singular event for the times and positions to be aligned. If you did, you would have discovered that it is impossible to have T1 and T2 and P1 and P2 agree about where the centre of them are, and other issues of this type. As such, they will disagree. This is the kind of situation whereby instructors will remind you that the topic is called relativity for good reasons. – naturallyInconsistent Jul 20 '23 at 15:26
  • @naturallyInconsistent - We could have the train be stopped at the platform and its length and middle agreed upon, and place P1 and P2 in line with T1 and T2 before the train scoots away and comes back. There could very well be a light on the middle of the platform, very close to the trains path that triggers the light on the train right in the middle. Indeed the train appears shorter and the platform apears shorter, but the middle can still be aligned in space and time - no? In any case P can be just one observer, and he still cant believe the photos without D or T changing - right? – rep_movsd Jul 20 '23 at 16:09
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    Related : Special Relativity - Regarding the Simultaneity of Events During the Train Paradox. – Frobenius Jul 20 '23 at 16:13
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    No, it will not work. At the time the train scoots back and at the moment the lamp lights up, T1 and T2 will agree on the time of lighting, but P1 and P2 will point out that the lamp lit up off the centre of the ground. – naturallyInconsistent Jul 20 '23 at 16:33
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    rep_movsd: +1 for having explicitly and distinctly named some of the distinguishable constituents (participants, observers) in your problem statement. You may further denote the constituent of the train whom you identify as "exactly between T1 and T2". (For symmetry and completeness, and in order to address the above complaint by @naturallyInconsistent, you should also denote the constituent of the platform who is/remains "exactly between P1 and P2", a.k.a. "the middle between P1 and P2"; provided you understand and agree that such a constituent of the platform is indeed identifiable, too.) – user12262 Jul 22 '23 at 17:38

1 Answers1

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You are overlooking the relativity of simultaneity and misunderstanding the symmetry of time dilation.

Suppose the light flashes when the mid-point of the observers on the train passes the mid point of the observers on the platform.

The pair of observers on the train must each see the light at the same time, since the light has travelled an equal distance to reach them in their frame. Likewise, each of the observers on the platform must see the light at the same time, since the light has travelled an equal distance to reach each of them in their frame.

However, in the frame of the train, the light reaches the forward observer on the platform before it reaches the rearward observer on the platform. Conversely, in the frame of the platform the light reaches the forward observer on the train after it reaches the rearward observer on the platform.

You are right in supposing that the train seems length contracted in the frame of the platform, and that the platform seems length contracted in the frame of the train. However, you are wrong if you think that time dilation means that all clocks in the train are time dilated relative to all clocks on the platform- time dilation doesn't work that way. What time dilation means is that a single clock on the train will seem time dilated relative to two clocks which it passes in turn on the platform. But two separated clocks on the train will seem time contracted, not time dilated, relative to a single clock on the platform which they pass in turn. That's the crucial point to understand, and it arises not because clocks in one frame tick more slowly than clocks in another, but because clocks in one frame are out of synch compared with clocks in the other. So if a single clock in one frame is passed by two clocks in the other, they will seem time contracted because the leading clock is out of synch with the trailing clock in the frame of the stationary clock. And that is the point you are overlooking in your interpretation of the thought experiment.

Marco Ocram
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  • This clarifies things a bit, but what is the reasoning that the platform observers will apply so that the photos have the same timestamp?

    For D/(c-V) and D/(c+V) to produce the same value.

    Where does the asymmetry come?

     – rep_movsd Jul 21 '23 at 08:49
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    The asymmetry comes from the relativity of simultaneity. If I am moving relative to you, then my planes of simultaneity are tilted relative to yours, the tilt being upwards in my direction of travel. At a given time in my frame, clocks in your frame are increasingly ahead of mine in my direction of travel, and increasingly behind mine in the reverse direction- the discrepancy increases with distance. – Marco Ocram Jul 21 '23 at 11:07
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    Imagine the train and platform are both 2 light-seconds long, that you are the observer at the front of the train and the light flashes at t=t'=0 when the centre of the train passes the centre of the platform. In your frame, the light will reach you and the observer at the rear of the train at t=1. In the platform frame, the light will reach the observers at each end at t'=1. From your perspective, the light had to pass the observer at the nearer end of the platform before it reached you, so that had to be at some time less than t=1s... – Marco Ocram Jul 21 '23 at 11:16
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    ...Conversely, the light reached the observer at the far end of the platform after passing the observer at the rear of the train, so that had to be some time after t=1. – Marco Ocram Jul 21 '23 at 11:17
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    Hopefully you can now see that t'=1 is less than t=1 at the nearer end of the platform, and t'=1 is greater than t=1 at the farther end of the platform, so the time in t' is actually out of synch from the perspective of the train. It is that loss of synchronisation which causes the time dilation effect. – Marco Ocram Jul 21 '23 at 11:20