# How can two different inertial frames observe two different events?

Consider an apparatus similar to that used in Michelson-Morley experiment kept in a frame (S') moving with a constant speed (v) to the right of an inertial frame (S).

The apparatus consists of a light source, a partially silvered glass (B) in front of the source and two mirrors facing B (at a length L from B) and perpendicular to each other.

Light leaves the source and is reflected as well as transmitted at B. Let $$T_1$$ and $$T_2$$ be the total times to reach the mirrors from B and return back.

According to Special Relativity, the speed of light (c) is constant in both the frames. Thus, for S' $$T_1=T_2=2L/c$$ and for S $$T_1=\frac{2L/c}{1-v^2/c^2}$$ and $$T_2=\frac{2L/c}{\sqrt{1-v^2/c^2}}$$.

From this it is evident that the two beams of light reach back B simultaneously in S' but not in S. But how can the two frames observe two different events - light reaching simultaneously at the same point and light not reaching simultaneously?

• You have accounted for only the Lorentz contraction (as far as I can tell). In frame S not only is the distance between B and the mirror contracted, but also the beam splitter (B) and mirrors are all moving. So the light that goes "up" to the top mirror actually does a triangular path in that frame (IE has to go further) and the light going to the other mirror first has to catch up to it (the mirror is "running away"). – Dast Oct 22 '19 at 16:12

If the arrival of the photons at a specific point (B) is simultaneous from the viewpoint of one inertial observer, it will be simultaneous from the viewpoint of any other inertial observers. Your calculations for the upper mirror is correct. That is, the total time $$T$$ needed for the photon to return to the splitter, as measured in $$S$$, is:

$$T=\frac{2L/c}{\sqrt{1-v^2/c^2}}$$

However, when calculating the time for the photon emitted towards left, you must be noticed that the assigned mirror escapes from the mirror so that the photon has to travel a longer path than the Lorenz contracted $$L/\gamma$$. Therefore, we have:

$$L/\gamma+vT_1=cT_1 \rightarrow T_1=\frac{L/\gamma}{c-v}$$

As the photon reflects back from the left mirror towards the splitter, you must consider that the splitter approaches the photon at $$v$$. Therefore, the traveled path is smaller than $$L/\gamma$$ so that we can write:

$$L/\gamma-vT_1=cT_1 \rightarrow T_1=\frac{L/\gamma}{c+v}$$

The total time is thus calculated to be:

$$T=T_1+T_2=L/\gamma \left( \frac{1}{c-v}+\frac{1}{c+v} \right)=\frac{2L/\gamma}{c^2-v^2}=\frac{2L/c}{\sqrt{1-v^2/c^2}}$$

This time equals to the prior one. That is, both of the photons move parallel and perpendicular to motion direction reach the splitter simultaneously.