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Let a plane mirror in motion with a speed $\vec{\beta}=\vec{v}/c$ along $x$ axis. Reflecting surface extends on the $y/z$ plane. Let a light ray reflecting on the mirror surface. The incident wave vector forming an angle $\theta_i$ with the normal to the surface. In this set-up i'm trying to get the Einstein's result described on [1]:

$ \begin{eqnarray} \cos\left(\theta_r\right)=\frac{(1+\beta^2)\cos\left(\theta_i\right)-2\beta}{(1+\beta^2)-2\beta\cos\left(\theta_i\right)} \end{eqnarray} $

where $\theta_i$ and $\theta_r$ are the incident and reflecting angles views by laboratory system.

The reflecting's law $\theta_i$ = $\theta_r$ is valid in the mirror's referance system. For first i find an expression for $\theta_r$.

$\frac{\omega}{c}\left(\begin{array}{c}1\\\cos(\theta_r)\\\sin(\theta_r)\\0\end{array}\right) = \frac{\omega'}{c}\gamma\left(\begin{array}{c c c c}1 & +\beta & 0 & 0 \\ +\beta & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0& 0& 0& 1 \end{array}\right)\left(\begin{array}{c}1\\\cos(\theta_r')\\\sin(\theta_r')\\1\end{array}\right)$

so

$ \begin{eqnarray} \cos(\theta_r) = \frac{\beta + \cos(\theta_r')}{1+\beta\cos(\theta_r')} \end{eqnarray} $

Now to find $\cos(\theta_r')$ i can use the transform law for $\theta_i$

$\frac{\omega'}{c}\left(\begin{array}{c}1\\\cos(\theta_i')\\\sin(\theta_i')\\0\end{array}\right) = \frac{\omega}{c}\gamma\left(\begin{array}{c c c c}1 & -\beta & 0 & 0 \\ -\beta & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0& 0& 0& 1 \end{array}\right)\left(\begin{array}{c}1\\\cos(\theta_i)\\\sin(\theta_i)\\1\end{array}\right)$

and i find that

$ \begin{eqnarray} \cos(\theta_i') = \frac{\cos(\theta_i)-\beta}{1-\beta\cos(\theta_i)} \end{eqnarray} $

on the mirror referance system i can use $\cos(\theta_i') = -\cos(\theta_r')$ and so

$ \begin{eqnarray} \cos(\theta_r') = \frac{\beta-\cos(\theta_i)}{1-\beta\cos(\theta_i)} \end{eqnarray} $

using the expression for $\cos(\theta_r')$ in $\cos(\theta_r)$ i can reach the result

$ \begin{eqnarray} \cos(\theta_r) = \frac{\beta + \frac{\beta-\cos(\theta_i)}{1-\beta\cos(\theta_i)}}{1+\beta\frac{\beta-\cos(\theta_i)}{1-\beta\cos(\theta_i)}} = \frac{2\beta - (\beta^2+1)\cos(\theta_i)}{(1+\beta^2)-2\beta\cos(\theta_i)} \end{eqnarray} $

....there is a minus sign! Why? What are the errors? I wrong the boost direction?

[1] Einstein, Albert. "Zur elektrodynamik bewegter körper." Annalen der physik 322.10 (1905): 891-921.

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closed as off-topic by Kyle Kanos, peterh, David Z Jun 12 '17 at 2:04

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