fozed said:
When the bullet is fired it is in the same reference frame as the experiment. It accelerates very quickly, granted (its a futuristic bullet), but it still is at rest the instant the light is at C. The question, I suppose, is what happens during the acceleration. However, other problems arise if you are to say that the light distance does not 'contract' at the same rate as the experiment apparatus. What do you think?
You don't have to worry about acceleration, you could just imagine the bullet coming in at constant velocity from far away and passing by A at the precise moment the light beam reached that point.
Anyway, the easy way to figure this out is just to use the Lorentz transformation to find the coordinates of different events first in the apparatus frame, then in the bullet frame. In the apparatus frame, we can set the coordinates of the events as:
light hits Y, splits into two beams: x=0, y=0, t=0
left beam reaches U: x=-0.5, y=0, t=0.5
right beam reaches V: x=0.5, y=0, t=0.5
left beam reaches A as bullet passes A (assume distance from U to A is 0.2 light seconds): x=-0.5, y=0.2, t=0.7
right beam reaches C: x=0.5, y=0.2, t=0.7
right beam reaches B: x=-0.1, y=0.2, t=1.3
bullet reaches B: x=-0.1, y=0.2, t=1.5
...with x and y in units of light-seconds, and t in units of seconds.
Now, the Lorentz transformation is:
x' = \gamma (x - vt)
y' = y
t' = \gamma (t - vx/c^2)
with \gamma = 1/\sqrt{1 - v^2/c^2}, so with v=0.5c for the bullet frame, \gamma is around 1.1547.
So, the coordinates of all these events in the bullet frame are:
light hits Y, splits into two beams: x'=0, y'=0, t'=0
left beam reaches U: x'=-0.866, y'=0, t'=0.866
right beam reaches V: x'=0.289, y'=0, t'=0.289
left beam reaches A as bullet passes A: x'=-0.981, y'=0.2, t'=1.097
right beam reaches C: x'=0.173, y'=0.2, t'=0.520
right beam reaches B: x'=-0.866, y'=0.2, t'=1.559
bullet reaches B: x'=-0.981, y'=0.2, t'=1.790
You can check my math to see if I made any mistakes, but these numbers indicate that the beam reaches the mirror before the bullet in both frames. In the bullet's frame, you can see that the bullet passes A about 0.577 seconds after the right beam hits mirror C, so the right beam has a head start towards B in this frame. The "paradox" arises from forgetting the different definitions of simultaneity in the two frames, and wrongly supposing that if the beam leaves C at the same moment the bullet passes A in the apparatus frame, then this would be true in the bullet frame as well. Once you notice that the bullet passes A later than the light beam bounces off C in the bullet's own frame, there is no paradox, and both frames agree the light reaches B before the bullet.
