1
$\begingroup$

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

$$(ct_A)^2=(vt_A)^2+1.$$

This yields, however, that

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


Which step in the derivation is wrong?

$\endgroup$
5
  • $\begingroup$ 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

0
$\begingroup$

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.

$\endgroup$

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.