# Derivation of Lorentz Factor yielding $\gamma ^{-1}$ as opposed to $\gamma$

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?

• 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. Commented Aug 14, 2021 at 23:11
• @MariusLadegårdMeyer The wall is stationary relative to $B$. I edited to make that explicit.
– Sam
Commented Aug 14, 2021 at 23:14
• 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... Commented Aug 14, 2021 at 23:18
• @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.
– Sam
Commented Aug 14, 2021 at 23:26
• @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.
– Sam
Commented Aug 14, 2021 at 23:26

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 $$B$$.
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.