I just wish to understand why the following reasoning fails.

Suppose observer $B$ (moving reference frame) is moving at a relative velocity $v$ from observer $A$ (stationary reference frame) along the $x$-axis. Observer $B$ shines a ray of light along the $y$-axis, and the ray travels $1m$ before hitting a wall stationary relative to $B$.

From $B$'s point of view, the ray took $t_B=c^{-1}\ \text{seconds}$ to reach the wall, while from $A$'s point of view the ray took $$t_A=\frac{1}{\sqrt{c^2-v^2}}\ \text{seconds}$$

to reach the wall. The above formula can be derived by drawing a right-angle triangle with hypotenuse $ct_A$ and the other sides given by $1m$ and $vt_A$. Then one only needs to solve for $t_A$ in the equation


This yields, however, that

$$t_B=\gamma ^{-1}t_A$$

Which step in the derivation is wrong?

  • $\begingroup$ Who measures the distance to the wall to be 1m, A or B? Is the wall stationary relative to A or to B? For the other one, the wall will not be stationary, but you seem to be ignoring that fact. $\endgroup$ Aug 14, 2021 at 23:11
  • $\begingroup$ @MariusLadegårdMeyer The wall is stationary relative to $B$. I edited to make that explicit. $\endgroup$
    – Sam
    Aug 14, 2021 at 23:14
  • $\begingroup$ Then in what sense is B in a "moving reference frame" and A is in a "stationary reference frame"? It seems that you have just switched the names around in the standard derivation for time dilation... $\endgroup$ Aug 14, 2021 at 23:18
  • $\begingroup$ @MariusLadegårdMeyer I've always imagined myself to be $A$ and that another person, $B$, shines the ray. I'm realizing now that if the roles are switched (if I'm the one who shines the ray) the equation comes out as it should. $\endgroup$
    – Sam
    Aug 14, 2021 at 23:26
  • $\begingroup$ @MariusLadegårdMeyer Yet this confuses me. I thought that if $B$ is moving at a speed $v$ relative to me, $A$, then any event that lasts $t_A$ from my perspective will last $\gamma t_A$ from $B$'s perspective, yet this appears to be false, right? Because in the above posts I described an event that lasts $t_A=c^{-1}$ seconds from my perspective, but $\gamma ^{-1} t_A$ seconds from $B$'s perspective. $\endgroup$
    – Sam
    Aug 14, 2021 at 23:26

1 Answer 1


Your derivation is correct and one further step gives $$t_A =\gamma t_B,$$ completing the derivation. Though the $t_B$ is not quite a proper time, it is half of a different proper time, which would hold if the wall had a mirror which reflected the light back to the emitter of $B$.

Times are always dilated relative to the proper time. A time is defined between two events in spacetime that are timelike-separated, and it turns out that when things are objectively time-separated in relativity they are not objectively space-separated, so there is always a reference frame that thinks both happened at the same point in space, and they measure the proper time—everyone else measures something longer.

You have concocted a scenario where the two events are null-separated, this technically means that the events are objectively both space and time separated, but those separations can be driven arbitrarily close to 0 by choice of reference frame. But, you have mercifully forbidden us from accelerating in the $y$-direction, so thankfully these details can be avoided, and to the easiest way to accomplish this is to make the problem one-dimensional by reflecting the light back at the $x$-axis using the Parity transform, adding these two null-vectors creates a timelike four-vector.


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