# Relativistic Transform Law for wave-vector [closed]

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, peterh, David Z
If this question can be reworded to fit the rules in the help center, please edit the question.