Suppose you have a train moving forward relative to an inertial observer at velocity $v$. Suppose you have a clock 1 at the front of the train and a clock 2 at the back, and, in the frame of the train, you synchronize the clocks via

$$t_2 = t_1 + \frac{l_0}{c}$$

where $t_1$ is the omission time. Using only length contraction, the fact that light moves at a speed $c$ in all reference frames, and time dilation, I need to derive the fact that the train in the rear appears faster by $$\frac{vl_0}{c^2}.$$

So what I've done so far:

Let the coordinates of the observer off the train be denoted $(t^{*},x^{*})$. We have that $\gamma t_1^* = t_1$ and that the apparent length of the train is in fact $\frac{l_0}{\gamma}$.

We emit a light beam from the front to the back. It would take $$\frac{l_0}{\gamma c}$$ time to reach the back, but the back is moving towards the front at a velocity $v$. So in fact, it has to travel less distance.

Now this is where I am confused. I cannot just write $$ct = vt + \frac{l_0}{\gamma}$$ and find $t$, since this would imply light is moving at speed $c+v$.

For some reason, if I just say "well, the back of the train moves $$\frac{v l_0}{\gamma c}$$ in the time it takes the light to move $${l_0}{\gamma}$$", then I appear to get the right relativity of simultaneity constant. Is this somehow the right way to "ignore" relative velocity?


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  • $\begingroup$ You need two events on the rest frame of the spatially moving train having the same time coordinate with different spatial coordinates, i.e simultaneous. Then you use constancy of $c$, or invariance of the interval, to show that the same two events might have different time coordinates in a different frame. $\endgroup$ – user76568 Feb 8 '18 at 17:56
  • $\begingroup$ @Dror How do you use constancy of $c$? we are not allowed to use the invariance of the spacetime interval $\endgroup$ – JohnDD Feb 8 '18 at 18:03
  • $\begingroup$ Just look the first chapter of Landau and Lifshitz "Classical Field Theory " $\endgroup$ – Kiryl Pesotski Feb 10 '18 at 0:52
  • $\begingroup$ I can, ofcourse, write an answer, but this will be a bad reproduction of what is done in that book. It is mathematical, but also with philosophy in there. Strongly recommend to read this book through to anyone who desires to know about relativity! $\endgroup$ – Kiryl Pesotski Feb 10 '18 at 0:54

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