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In his popular book on relativity, in chapter IX, "The relativity of simultaneity", Einstein describes an experiment in which a flash happens simultaneously on A and B, as defined by the fact that an observer at the middle point M can see the light coming from A and B at the same moment:

          --train-->

---embankment---A-------M-------B----

Then, he proceeds to say that an observer on the train "is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A".

So Einstein concludes that simultaneity is not absolute. And an explicit calculation using Special Relativity principles confirms this. The conclusion is that:

So the answer is that the observer on the train sees the lightning strike the front of the train at $t' = -\gamma\tfrac{vd}{2c^2}$ and the rear of the train at $t' = \gamma\tfrac{vd}{2c^2}$. The time between the lightning strikes is $\gamma\tfrac{vd}{c^2}$.

By setting $\gamma=1$, we obtain the result in Galilean Relativity (ie: "The time between the lightning strikes is $\tfrac{vd}{c^2}$"), which is the theory of space time before Einstein came out with Special Relativity.

The point is that regardless of whether we are calculating using Galilean Relativity or Special Relativity, the conclusion is that different people on different speed will always have different notion of simultaneity. Since this is a conclusion known to everyone before Einstein, how did Einstein use this insight to derive his Special Relativity?

Truly, I fail to see how can this thought experiment on simultaneity can give Einstein insights into Special Relativity. Anything I miss?

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Graviton
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    In a pre-einsteinian calculation the wave is assumed to take it's speed relative some medium, so though the two observer see the wave fronts with different arrival times they compute the events that triggered the waves to have the same timing in both frames. – dmckee --- ex-moderator kitten May 08 '14 at 05:54
  • @dmckee, would you like to expand this into full answer-- and explain with calculations how Galilean Relativity and Special Relativity differ crucially here? – Graviton May 08 '14 at 05:59
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    @Graviton: if $\gamma = 1$ then $v = 0$ and $t' = t = 0$. So in the limit of zero velocity the lightning strikes are simultaneous in both frames. – John Rennie May 08 '14 at 06:00
  • @JohnRennie, that's not how you interpret it, one can reduce from Lorentz frame to Galilean frame by assuming $\gamma=1$, regardless of speed $v$ – Graviton May 08 '14 at 06:02
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    @Graviton: see my answer - $\gamma$ is a function of $v$ and $c$ and you can't just set it to unity and leave $v$ and $c$ unchanged. – John Rennie May 08 '14 at 06:09
  • @graviton: rereading my answer to your previous question I see how I might have given the impression that $\gamma$ is an independant variable. If so I apologise - we get so used to using $\gamma$ as a shorthand for $1/\sqrt{1 - v^2/c^2}$ that we do it without thinking. – John Rennie May 08 '14 at 06:16
  • If you assume $c$ is infinite then $\gamma=1$ regardless of $v$ (providing it is finite). – bright magus May 08 '14 at 07:58

2 Answers2

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You say:

By setting γ to 1, we obtain the result in Galilean Relativity (ie: "The time between the lightning strikes is $\tfrac{vd}{c^2}$"), which is the theory of space time before Einstein came out with Special Relativity.

But remember that $\gamma$ is not some independant parameter. It's just shorthand for:

$$ \gamma = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}} $$

So you can't just set $\gamma = 1$ without changing either $v$ or $c$ or both.

If you set $v = 0$ then $t = t' = 0$. In this case the events are simultaneous in both frames, but that's not surprising because if $v = 0$ both frames are the same inertial frame.

If you want to use the Galilean limit but keep $v$ non-zero the way to do this is to increase the speed of light to infinity (obviously this is a thought experiment). In that case $\gamma\tfrac{vd}{c^2} = 0$ and again $t = t' = 0$ so the lightning strikes are simultaneous in both frames.

John Rennie
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Graviton,

The simple and short answer is that although Einstein did realize there is a difference whether you are moving towards or away from emitted ray of light, yet he derived his transforms for time and length considering the case in which the train is moving away from the station.

This is the situation he considered (look at the velocity vector):eins

(Comment: Should anyone disagree as to what inertial frame should actually be primed, then it is not relevant for what I wrote: the train was leaving the station in Einstein's thought experiment.)

bright magus
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