If I imagine a photon being released from the plate below as soon as the plates start moving, shouldn't the photon hit the opposite plate a bit behind the point where it would have hit if the plates were stationary, how does the photon move with the two plates?
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In the primed frame, the pulse of light was emitted at an angle as depicted. Angles Lorentz-transform generally as (see: wiki: Abberation)
- $$ \theta' = \tan^{-1}{\frac{\sin{\theta}}{\gamma(v/c + \cos{\theta})}} $$
which in our case with a right angle between the pulse and the velocity in the unprimed frame simplifies nicely to:
- $$ \theta' = \tan^{-1}{\frac{c}{\gamma v}} $$
Examine the diagram geometrically and pick the same trigonometric function, the tangent of the angle is the opposite leg ($L$) over the adjacent leg ($.5 v \Delta t'$), so:
- $$\theta' = \tan^{-1}{\frac{L}{.5 v \Delta t'}} $$
Substitute $\Delta t' = \gamma \Delta t$ and $\Delta t = 2L/c$ and simplify to obtain eqation 2 from equation 3.
