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I am trying to understand the implications of c being relative to the frame of reference. Is the following analysis correct?

N is on a slow train moving 3 m/s and M on the platform next to N both see a distance super nova ahead of the train. The supernova is an event that happened a million light years away, hence a million years ago.

Do they see it at (almost) the same time? Being slow moving, M & N's clocks can be easily synchronized to within milliseconds.

Common experience says yes they do see it at about the same time.

HOWEVER, Within a frame of reference,

Distance = Velocity * Time

So if a one particle P was to travel from the super nova to us as at 0.1c and another particle Q was to travel towards us at 0.1c + 1m/s, then over a million light years Q would arrive long before P even though the speed difference is very small.

M & N are slow, but there will be a tiny difference in their perceived value of C. he super nova is very far away, which will amplify even a tiny difference in velocity as measured in different frames. The speed of the super nova is irrelevant, this is not an emitter model.

Question, is the super nova closer in N's frame of reference than in M's. If not, how can M and N observe it at the same time?

Likewise if the super nova was behind the train, it would be further in N s frame than M's.

(I am trying to make sense of Einstein's train example, which I suspect he fudges.
https://www.bartleby.com/173/9.html

His example as stated makes perfect sense if there is an aether, but not if c is relative to the frame of reference of the observer. I suspect other things are going on, like length reduction or time shifting.)

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closed as unclear what you're asking by Aaron Stevens, ZeroTheHero, Alfred Centauri, Dale, Jon Custer Aug 17 at 23:32

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    $\begingroup$ If it's a slow train, wouldn't you use Newtonian mechanics & Galilean relativity instead and so the answer would be "no, the supernova is not closer in N's frame of reference than M's"? $\endgroup$ – Allure Aug 17 at 9:18
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Well, the question is a bit ill-posed. If they see the supernova at the time they are next to each other, then obviously they do see it at the same time, by definition (provided that they synchronize their clocks at the time when they pass near each other). Otherwise, the one closer to the supernova at the moment of arrival of the light will trivially see it first.

In other words, to be able to answer you have to specify the relative position of M and N at the instant of explosion.

Another option is that you mean something else, and you're asking "do M and N see the same instant in the evolution of the supernova, when they pass next to each other? Or they see the supernova at different times of its life?". In this case, since N is approaching the supernova faster than M (whatever the supernova speed relative to M would be), N will label the same events at the supernova with earlier times than M, hence you can say that N sees before M, even at the instant in which they are next to each other. Moreover, this effect should be modest even if the relative velocity of M and N is small, provided that the supernova is very far away. To see this, let's consider an event at the supernova place, with coordinates $(x_s,t_s)$ in M reference frame. Then N will see this at time ${t'}_{s} = \gamma t_s - \gamma \beta x_s $. Now, $\gamma$ is of order 1 if N is moving slowly in M frame, but $\beta x_s$ can also be of order 1 if $x_s$ is big enough.

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  • $\begingroup$ "the supernova at different times of its life" makes it sound like the change in coordinates of the supernova in the moving frame leads to a real physical effect. It is true that both the time and position of the supernova in N's inertial frame are different from M, but that doesn't mean that N sees something different. They get the exact same light at the exact same time $\endgroup$ – octonion Aug 17 at 19:54
  • $\begingroup$ I meant that the times on the clocks on N and M when they receive the light from some given event happened at the supernova, can be different. The fact that we choose and fix an event at the supernova means that we cannot fix M and N to be at the same distance from the supernova when they receive the light! Only one event at the supernova will shine on M and N when they are at the same place. All the other events in the life of the supernova will arrive at different times to M and N and give different readings on their clocks (even if we talk about "local" readings in the respective frames). $\endgroup$ – Massimo Frigerio Aug 18 at 9:26
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    $\begingroup$ Well, I think the supernova is intended to be an example of a single spacetime event. It has no lifetime in this simplification. But even so, M and N see the same instant of the supernova when they are in the exact same location, and when they are in different locations the only difference is due to the ordinary tiny effect of the finite speed of light. There is no modest effect. It is a moot point anyway since this question was closed for some strange reason. $\endgroup$ – octonion Aug 18 at 15:30
  • $\begingroup$ The supernova actually exploded millions of years ago. Yes, an event. It obviously only happened once. $\endgroup$ – Tuntable Aug 19 at 0:21
  • $\begingroup$ I am asking whether they see it at the same time. That is the question. In relativity, nothing is obvious. And yes their clocks are synchronized they are only moving slowly. $\endgroup$ – Tuntable Aug 19 at 0:22
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First of all, since the train is said to be slow, Galilean Relativity should be used. Secondly, it's safe to assume that the specification "M is next to N" is to be interpreted in the moment when they see the supernova and not when the supernova explodes: otherwise the problem would have said that they are close when the supernova exploded "in the M/N reference frame".

This said, yes, they observe it at the same time.

The problem seems a bit too easy though, so it is possible that I am misunderstanding some detail.

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  • $\begingroup$ Well, electrons moving slowly down a wire have an effect (magnetism), so slow moving is not that simple. In this case C is relative to M and N, and is a 3 m/s (say) difference between them. That is a huge time difference over the distance to the super nova. $\endgroup$ – Tuntable Aug 19 at 0:29

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