Relativity paradox with mirrors and light pulses 
Consider two very short light pulses emitted from the centre (C) of two mirrors A and B (as shown in the diagram).
From the point of view of the lab frame, the apparatus is all moving to the left at velocity v.
Imagine there is also an electron near the centre of the apparatus, which is stationary in the apparatus frame and therefore also moving with velocity v to the left according to the lab frame.
The short light pulses (much shorter than the apparatus length) bounce off mirrors A and B and return and strike the electron.
This situation has similarities with the Michelson-Morley experiment.
According to the frame moving with the apparatus, the pulses take an equal time to bounce off the mirrors and arrive back at C. Therefore the EM waves cancel and there is no net radiation pressure exerted on the electron.
According to the lab frame, the light pulse emitted to the left has less distance to travel overall and so arrives at C before the pulse that was emitted to the right. Therefore the first pulse accelerates the electron by exerting a radiation pressure on it.
Does the electron accelerate or not? :)
(I'm looking for derivations/proofs showing both frames' interpretations)
 A: In the lab frame both pulses arrive at C at the same time. The reason is that the distances traveled are the same (they do not reach A and B simultaneously). The distances of paths CA and BC are equal, the same happens with the paths AC and CB. The distance CAAC is equal to CBBC.
A: If $t_{CA}$ refers to the time it takes in the lab frame for the light to reach C from A, and the same with $t_{AC}$, $t_{CB}$ and $t_{BC}$ then we have:
$t_{CA}=\frac{L/2+v t_{CA}}{c}$
$t_{AC}=\frac{L/2-v t_{AC}}{c}$
$t_{CB}=\frac{L/2-v t_{CB}}{c}$
$t_{BC}=\frac{L/2+v t_{BC}}{c}$
Thus $t_{CA}=t_{BC}$ and $t_{AC}=t_{CB}$
Finally: $t_{CA}+t_{AC}=t_{CB}+t_{BC}$ 
A: Let $E$ be the event "pulse 1 arrives back at point $C$" and let $F$ be the event "pulse 2 arrives back at point $C$".  Then $E$ and $F$ either are or are not two names for the same event, and this has nothing to do with choices of frames.  Therefore if $E$ and $F$ have the same coordinates in one frame (e.g. the apparatus frame), then they must have the same coordinates in any other frame (e.g. the lab frame).  
In other words:  It's obvious in the apparatus frame that the pulses arrive at $C$ simultaneously.  Therefore the pulses arrive at $C$ simultaneously, and all observers must agree on this.
A: Because I like setting the $x$ axis point towards the right, reverse the speed on your figure. In the lab frame, the light moving towards the mirror A reaches the mirror in a time $\Delta t_1$ determined from $v \Delta t_1 + c \Delta t_1 = L/2$ giving the $\Delta t_1 = \frac{L/2}{c + v}$. It is then reflected by this mirror. Though in the lab frame the electron has moved it is still a distance $L/2$ from the A mirror. The light reflected from $A$, according to an observer in the lab frame, will reach the electron in the time $\Delta t_2$ determined by $c \Delta_2 = L/2 + v \Delta_2$ so giving $\Delta t_2 = \frac{L/2}{c - v}$. This gives the total time as $\Delta t_A = \frac{L c}{c^2 - v^2}$.
If you look to see what is happening with the pulse of light moving towards the B mirror, then the time to reach the mirror is $\frac{L/2}{c-v}$ and once reflected the electron has also moved forward so that it is still a distance $L/2$ when the pulse of light begins its reflected journey from the B mirror, but now it takes a time $\frac{L/2}{c + v}$ (because the electron is moving towards the reflected beam).
This gives the same time in the lab frame, specifically because when the light is reflected at mirror A it still leaves the mirror with speed $c$ and similarly for the light being reflected by the mirror B, the speed of light is constant.
