One rule of relativity is that the speed of light remains constant from all reference points. However, consider the case of four objects A, B, S, and O. O will be our own reference point. S is held still, and A and B have the same velocity approaching lightspeed. From A’s viewpoint, B would be motionless while S would be moving away from it. This is all simple.

Now, what happens if A and B are both moving at lightspeed in the same direction and neither is ahead of the other in the direction they are moving(From the perspective of something placed directly between them)? (I know that technically nothing could observe from lightspeed, but this is theoretical so be quiet about that. It doesn’t solve the problem to say it has no real-life applications.) A would perceive itself as being held still, but it would perceive S as moving at the speed of light. This presents a problem: if lightspeed is constant for all reference frames, how is S moving at lightspeed in one reference frame but not moving at all in our own? Things get even more complicated when you look at A’s perception of B. If lightspeed is constant for all reference frames, then A must perceive B as far ahead of A, while B must perceive A as far ahead of B. Clearly these cannot be both accurate, unless some sort of quantum shenanigans are going on (considering that neither A nor B can have any mass this seems possible.)

One thing that occurred to me about this is, if relativity treats space and time as the same, it would be reasonable to think of lightspeed as a vector through space time (a.k.a. everything moves through space time at a speed up f c). However, since light does take time to travel through space, this means light must be slightly angled into the time dimension as well. This means it would be possible to have something traveling even faster through space if it did not travel through time at all, though of course a constant position in time would make something unobservbale, which seems to semi-negate the problem. (I wonder if, when something goes into a black hole, it actually gets to this speed through space, so information isn’t deleted, it’s just held still. Might bring this up another time.) (Also this would imply a minimum unit of time)

  • 1
    $\begingroup$ A reference frame going the speed of light is not valid in special relativity, so any paradoxes derived from assuming such a frame exists are meaningless. $\endgroup$
    – Chris
    Commented Dec 29, 2017 at 6:14
  • $\begingroup$ speed of light is constant for all reference frames. What do you mean by lightspeed? $\endgroup$ Commented Dec 29, 2017 at 6:27

1 Answer 1


If A or B have any mass at all, they simply cannot move at light speed, in our reference frame or any other. So anything you ask about an object A or B moving at light speed is simply nonsense. You can only consider what happens if A and B move at a very high fraction of light speed (in some particular reference frame).

The fact that the speed of light is constant does not mean that an object that is moving at near light speed in one reference frame will be moving at near light speed in another reference frame. The speed of light is a constant, not the speed of A, B, or S.

  • 1
    $\begingroup$ I never said they had mass, I just said that’s where the frame of reference is $\endgroup$
    – Notchmath
    Commented Dec 29, 2017 at 6:09
  • $\begingroup$ In that case look to any of the many previous questions about what happens "in a photon's reference frame" for an explanation of why it doesn't make sense to consider a reference frame of a massless particle. $\endgroup$
    – The Photon
    Commented Dec 29, 2017 at 6:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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