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From Einstein's "easy" explanation https://www.bartleby.com/173/9.html

(Yes, I have reviewed 8 other answers to similar questions. Please bear with me.)

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M is on the platform, T (aka M') is on the train with open carriages. Just as T passes M, lightening strikes two points A and B equidistant behind and ahead of both M and T.

For Thunder (not in Einstein's explanation), M will clearly hear both bangs together, and knowing the speed of sound through air and distances to A and B can calculate when the lightening struck. It will also take a little time for the sound to reach T in the middle of the train, during which T will move ahead of M. T will thus hear the front B strike first as he will now be closer to B than A, and the speed of sound is relative to the air. Clear.

For Light let us assume M will also see both flashes at the same time. Let that be the definition of the flashes being Simultaneous. But what about T?

Unlike sound, the speed of light is relative to the frame of reference of the observer. T is equidistant from A and B when the flash occurred, so T should also see both flashes together. Like for thunder, T has moved a little to the right of M before seeing the flashes, but M's relative position is not relevant to T's observation of the flashes.

But Einstein says, without explanation, that T sees B before A?

Question: Would T actually see B before A? If so, why?

There has also been some commentary about what M might think that T observed if M did not understand Relativity. If T had mirrors and M was observing flashes in the mirrors. But that is not what Einstein said. So let us stick to just what M and T actually observe.

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  • $\begingroup$ You have a valid point $\endgroup$ – yuvraj singh Aug 19 at 6:50
  • $\begingroup$ Although i read about is speed of light is frame dependent or it is constant ,that is speed is constant but it is the time which is relative , $\endgroup$ – yuvraj singh Aug 19 at 6:54
  • $\begingroup$ Your question was Would T actually see B before A? ,although light strike both place at same time for M as it is frame of reference ,every thing will decide according M'. $\endgroup$ – yuvraj singh Aug 19 at 7:01
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The speed of light is not instantaneous. I believe what is meant is that by the time that the first pulse of light reaches T, the train will have travelled a certain distance to the right. So the observer on T sees first the pulse of light from the right, then the pulse of light from the left. It's the same reasoning as in the case of sound. This is evident in the frame of reference of M.

The important takeaway is that simultaneity is not necessarily preserved in a transformation of reference frame. In the frame of T, the lightning strikes indeed happen at different times, so that is why they can be observed at different times.

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  • $\begingroup$ Thanks for posting, but please read the question again. $\endgroup$ – Tuntable Aug 19 at 5:35
  • $\begingroup$ Can you please clarify what you need clarification on? $\endgroup$ – DanDan0101 Aug 19 at 5:37
  • $\begingroup$ OK, added: Question: Would T actually see B before A? If so, why. $\endgroup$ – Tuntable Aug 19 at 5:40
  • $\begingroup$ I had also added " Like for thunder, T has moved a little to the right of M before seeing the flashes, but M's relative position is not relevant to T's observation of the flashes." $\endgroup$ – Tuntable Aug 19 at 5:41
  • $\begingroup$ Indeed. Remember that the lightning strikes are simultaneous specifically in frame M. Due to the relativity of simultaneity, in T's frame, event B happens before event A. Einstein tries to show why this is/where it arises from, by considering the same event in M's frame. When doing that, the same basic reasoning applies as when considering something like sound. $\endgroup$ – DanDan0101 Aug 19 at 5:44
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Here are some spacetime diagrams that might help clarify what is going on. I have used my "relativity on rotated graph paper" method to visualize elapsed proper time along worldlines and to visualize the lines of simultaneity for various observers.

The first diagram is drawn in the embankment frame.
The second diagram is drawn in the train frame.

Lightning strikes at events A and B.
From each strike, we draw in
the [green-dashed] light-rays (with speed c) and
the [violet-dotted] sound-rays (modeled with speed $v_{sound}=0.8c$... in the embankment frame) to be received by the train.
The train is traveling with velocity $(3/5)c=0.6c$.

Here are the results (what "M and T actually observe":

  • The central observer M (in the embankment frame) receives the two light-rays at a common event, and the two sound-rays at a later common event.
    The received light-rays have the same speed c.
    The received sound-rays have the same speed $v_{sound}$.

  • The central observer T (in the train frame) receives the rays at distinct events in this order: light-ray hits front ("T sees B's flash"), sound-ray hits front, light-ray hits rear ("T sees A's flash"), sound-ray hits rear.
    Thus, "T sees B before A"
    The received light-rays have the same speed c.
    The received sound-rays have unequal speeds,
    the first greater than $v_{sound}$ and the second less than $v_{sound}$.
    (This is more evident in the second diagram, which is drawn from the train-frame.
    Note that in the train frame, the strike at B occurs before the strike at A.)

In the embankment frame... RRGP-robphy-embankment-frame

In the train frame... RRGP-robphy-train-frame

(Admittedly, I can't quite tease out the issues you have with this thought experiment. Hopefully, this might help clarify the issues.)

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Unlike sound, the speed of light is relative to the frame of reference of the observer.

Einstein is not talking about relativity of speed here, rather this explanation comes from the fact that speed of sound, or light is not instantaneous as @DanDan0101 said. But let's write it in mathematical details so that it becomes more clear.

Let's instead of thunders use a light lamp at position of each thunder (This is not necessary but it helps us to make the problem less abstract) in other words, we place a lamp at A and another lamp at B. M can activate them with a signal, i.e he can send a signal from M to either A or B to turn them on. Assume that M sends two signals to A and B at the same time, so that they turn on simultaneously. From M's point of view, lamps are turned on simultaneously (like thunders in Einstein's explanation), what about T? Well according to T signals are sent at the same time (because they are being sent from the same place, that's M.) but they don't arrive at A and B simultaneously, why? Well let's say that the distance between MB and MA is $D$ and $-D$ respectively, from T's perspective. T is at the rest relative to himself, but T sees that B and A lamps are moving to left with velocity $-v$, right? So from his point of view, the signal that goes to the right, that's B, will arrive at: $$ct_1=D-vt_1 \to t_1=\frac{D}{c+v}$$

While the second signal which goes to the A will arrive at:

$$-ct_2=-D-vt_2\to t_2=\frac{D}{c-v}$$So they won't be turned on at the same time. Note that it's being assumed that signals go to A and B with speed of light, that's $c$ and I didn't deal with light rays that comes from lamps themselves when they are turned on. Why? Because imho Einstein has not defined very nicely where light rays from thunders will be seen.

Edit (answer to comment): Let's define two events $E_1$ and $E_2$ in M's frame such that $E_1$ indicates the moment when M receives the first light ray at $x=0$ and $t=0$ and $E_2$ indicates the moment when M receives second light ray at $x=0$ and again $t=0$, because these events are simultaneous. Now we want to know whether $E'_1$ and $E_2'$ are simultaneous or not, that's, same events but from T's point of view. Are they simultaneous or not? We can see that with Lorentz transformation: $$t'=\gamma(t+vx/c^2)$$ using $x=0$ and $t=0$ we can clearly see that $t'=0$ for both events, that's, from T's point of view these events are simultaneous as well. And it doesn't depend on T's position. However I think it is not what Einstein meant.

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  • $\begingroup$ I think you have provided a good explanation of a different problem. For light traveling from T to A then A's speed relative to T is important. But for light traveling from A to T it is not, because we are not using an emitter model. $\endgroup$ – Tuntable Aug 20 at 0:51
  • $\begingroup$ @Tuntable I think whether we use thunders or lamps, in the end of the day relativity of simultaneity is the same. But my point is, the position of T doesn't matter at all, as you can see T can understand relativity of simultaneity even without lamps sending him a light ray. The problem with Einstein explanation, is that he didn't specify "where" light rays will be seen from T's point of view, in other words, if Einstein were to say $\endgroup$ – Paradoxy Aug 20 at 7:16
  • $\begingroup$ @Tuntable "Light rays will arrive at M simultaneously, from M's point of view, what about T? Then we could see that T would say, they have arrived at M simultaneously too! Although thunders themselves are not simultaneous, their light rays will arrive at M simultaneously" but from Einstein explanation, I am not sure "where" light rays from thunders will be seen. Is it at M? Obviously not, then where? Note that by seeing I mean physical observation. That's why I tried to change the problem a little bit to show that at least thunders themselves are not simultaneous. $\endgroup$ – Paradoxy Aug 20 at 7:19
  • $\begingroup$ I Added "For Light let us assume M will also see both flashes at the same time. Let that be the definition of the flashes being Simultaneous. But what about T?" I see no reason for T not to see the flashes simultaneously. $\endgroup$ – Tuntable Aug 20 at 7:48
  • $\begingroup$ Incidentally, for clarity of language, "Lightening" is the physical phenomenon, which produces the "Thunder" sound (travels at speed relative to the air) as well as the "Flash" (travels at speed relative to the observer). $\endgroup$ – Tuntable Aug 20 at 7:50
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Would T actually see B before A? If so, why.

Yes T will actually see B before A. At the moment when M receives the light from both B and A, T is to the right of M. This means that at that moment T has already received the light from B since the light from B came from the right and so it had to pass T to reach M. Similarly, at that moment B has not yet received the light from A since the light from A came from the left and so it has to pass M to reach T. Since there is a time after the light from B reaches T and before the light from A reaches T then T sees B before A.

Note, although it is not proven here this argument applies in any reference frame. So the phrase “at that moment” is not restrictive.

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