Look at the following example:

enter image description here

I have two questions:

  1. The observer $B$ is in rest respect the sources of light, so he sees the two photons emanating by the two light sources coming from the same distance at the same velocity $c$. From his point of view $B$ is reached simultaneously by the two photons. So I think that the conclusion of the example should be the following: $A$ sees simultaneously the two lights, but he calculate that $B$ observes first the light coming from the source 2 and then the other light. Despite this, $B$ sees simultaneously the two lights from his point of view. Is it right?
  2. Suppose that we live in a world where the speed of light respects the transformations of Galilei (it is not constant in every inertial system). In this case what does calculate $A$? Does he see both the photons reaching simultaneously $B$?

The point which the exercise was trying to make is the following:

A says: the two lights flash at the same time
B says: the light 2 flashes first, then light 1

This is precisely the relativity of simultaneity - one cannot judge which light flashed first because it depends on the observer

The point is not that A sees that B observes light 2 first, but B sees the two lights simultaneously, in fact this statement is wrong. Image that this was the case. Let's assume that something terrible happens if B gets hit by two light at the same time, let's say B dies. If the lights hit B at different times, however, nothing happens. But now A would observe that B is safe and sound whereas B in his rest frame seeing the two lights at the same time would die. Dead and not dead at the same time? This is not possible and Schrödinger giggles silently in his tomb. The two lights hit B at different times in any rest frame

How is it possible? As you said for B the light sources are at rest. The crucial point is the Lorentz contraction. In this point the exercise is somewhat imprecise. It states that the separation of the electric devices matches the length of the carriage, but the carriage is in move so it gets contracted. A carriage in rest is thus longer then the separation of the devices. Let's correct the exercise and say that the electric devices are situated at such a distance apart that the Lorentz contracted carriage matches their separation in the rest frame of A. Now we go into the rest frame of B. From his point of view the electric devices move towards him. Therefore B sees their separation even shorter, so as the front of the carriage hits the electric device #2, the rear of the carriage will not have reached device #1 yet. So for B light 2 flashes first, then light #1. Note that this is precisely what A would have predicted for B, but the reason that A gives is that B travels towards the light #2 and away from light #1 so it sees light #2 first. As we have seen for B in its rest frame the reason is Lorentz contraction.

If you are interested in special (and general) relativity I can highly recommend you the Book by Lewis C. Epstein - "Relativity visualized".

  • $\begingroup$ Ok, the point $1)$ is clear now, but what about the point $2)$? $\endgroup$ – Dubious Sep 22 '13 at 9:21
  • $\begingroup$ Well, I guess the result would be the same as above. I think one should then ask the question what happens with the Lorentz contraction? Does it still exist? The constant speed of light and the Lorentz contraction are not independent effects, given one of the one can argue that the second must be given too. Abandoning both you will end up with classical Newtonian theory. $\endgroup$ – Stan Sep 22 '13 at 10:11
  • $\begingroup$ Ok, maybe you want to consider the following situation: let the special relativity be still valid, but leave the light alone and consider something else, something that has mass. Let us say small rubber balls are shot at B instead of light. $\endgroup$ – Stan Sep 22 '13 at 10:27
  • $\begingroup$ Assume A observes that these balls fly with the same velocity and hit B at the same time. As before B sees the ball #2 fired first. But now because the velocity also transforms and A observed the two balls having the same velocity relative to B, B will observe the balls having different velocities, this will compensate the delay of #1 and both balls hit B simultaneously, as it should be. Again the conclusion is the same, one cannot tell whether the balls are fired at the same time. Additionally here we need the transformation law for velocities of massive objects. $\endgroup$ – Stan Sep 22 '13 at 10:37
  • $\begingroup$ @Stan Suppose we have two observers A and B and they are at rest. Observer A observes two objects falling from height H (A has same distance between the two objects). Does observer B will measure different times for the duration of falling of the two objects ? (because the two object are not in the same location therefore a finite time interval must exist for the information of their position to reach B). It troubles me because if it was only one object there would be no problem. $\endgroup$ – ado sar Jul 18 at 10:06

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