Here is a thought experiment for which I cannot find a good answer or explanation. Let us imagine that in the void and on a (very long) line there are two lights switched off in points A and C. Two observers are waiting in point B, at the same distance of A and C.

The experiment is such as at 12 PM in the reference frame of the line, the lights are switched on and off, so that they are both emitting a photon towards the point B. As the line is very long, one of the observer has enough time to move to a point B' on the line while the two photons are going towards the observers. After he has moved to point B', the observer stops and waits, and so now the two observers are both again at rest with respect to the reference frame of the line.

Finally, at some time, the photons reach point B. They cross at point B, as they were equally distant from this point when they left, and as they travel with constant speed. But they also reach, at some (other?) time point B', where they cross each other, as for the second observer they were equally distant from him when they left and as they travel with constant speed.

So we now have two photons crossing each other at some different places in the same reference frame of the line, which seems very weird given their trajectory. What truly happens?

  • 2
    $\begingroup$ DataXplorer, I implore you, please draw a spacetime diagram of this thought experiment. And this is generally good advice because often, in the process of drawing the diagram, you'll see the answer to your question. $\endgroup$ May 23 '20 at 0:02
  • 1
    $\begingroup$ Do you also believe that if the observer walks past B, he will continue to say that A and B are equally far from him, even though B is between him and A? $\endgroup$
    – WillO
    May 23 '20 at 2:43
  • $\begingroup$ Your error is in assuming that the line has a reference frame. It doesn't. No acceleration is involved here. You simply have one static observer and one moving with a constant speed. They meet in the middle of the line. In their frames photons are emitted at different times. The one who sees them emitted simultaneously will receive them simultaneously as well. See Einstein's Train here, it is the same setup as yours well explained with a nice animation: en.wikipedia.org/wiki/Relativity_of_simultaneity - I've removed the General Relativity tag, because it is not applicable to this case. $\endgroup$
    – safesphere
    May 23 '20 at 6:12
  • 1
    $\begingroup$ @safesphere The "reference frame of the line" is the reference frame that is stationary with respect to points A and B. It is incorrect to say that such a reference frame doesn't exist. If the observer "stops and waits" at B', acceleration is definitely involved (because acceleration is involved in stopping). And the relativity of simultaneity isn't especially useful here, since this entire thought experiment can be done in the non-relativistic regime, with slow-moving billiard balls instead of light, and you'd get the same qualitative behavior that the OP finds "weird". $\endgroup$ May 23 '20 at 6:40
  • $\begingroup$ @probably_someone Sorry, but I believe that your answer and comments are misleading. $\endgroup$
    – safesphere
    May 23 '20 at 7:32

This apparent paradox is resolved by relativity of simultaneity.

Consider observer O at rest relative to points A and C, and observer O' moving towards point A. Both observers are at point B, equidistant between A and C.

Photons are emitted simultaneously from points A and C. This statement, in a question about relativity, should immediately prompt the question "simultaneous in what frame?" For the photons to meet at point B, it must be simultaneous in the rest frame of observer O.

Observer O' is not at rest in that frame. In the rest frame of O', the photons are not emitted simultaneously. Instead, point A emits a photon before point C does. In fact, point A emits a photon before O' even reaches point B. By the time the observers meet at B, in the rest frame of O' point A has already emitted its photon. Point C then emits its photon in that frame some time after observer O' has passed point B.

In the rest frame of observer O, the photon from point A has less distance to travel to reach O', and therefore O' sees it first.

In the rest frame of observer O', the photon from point A travels the same distance as the photon from point C, but A's photon starts the trip earlier, and therefore O' sees it first.

Both frames agree that observer O' sees the photon from point A and then the photon from point C. They just have different reasons for the sequence.


Point B' is not the same distance away from A and C. Therefore, the photons will not reach point B' at the same time. The observer at B' will not see the photons cross; that observer will see one photon moving in one direction, and then, after some time delay, the other photon moving in the other direction.

  • $\begingroup$ If it was the case, this observer would see a photon go faster than the other. They left at the same time for him, from the same distance, and they go to the same speed towards him. Furthermore, he does not know he is moving : for him, the observer staying on point B is moving. $\endgroup$
    – Unknown
    May 22 '20 at 21:59
  • $\begingroup$ @DataXplorer After the observer moves to B', the photons are no longer the same distance from him. $\endgroup$ May 22 '20 at 22:00
  • 1
    $\begingroup$ there is only one point, $B$, on the line $AC$ that is equidistant from $A$ and $C$, so $B'=B$. If you disagree, please provide a diagram. $\endgroup$
    – JEB
    May 22 '20 at 22:04
  • 1
    $\begingroup$ @DataXplorer Well, that's a completely different situation, but there is still no contradiction. In that situation, the moving observer (from the frame of the lights) has no reason to assume that A and C are equidistant from him at the time of measurement. $\endgroup$ May 22 '20 at 22:25
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Unknown
    May 22 '20 at 22:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.