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Setting: Let's consider a moving mirror at constant speed in z-direction. A light ray of frequency $\omega$ falls onto the mirror. Its wave vector $k$ forms an angle $\theta$ with the vector normal to the mirror.

I would now like to find the angle of reflection with respect to the vector $\vec{e}_z$ normal to the mirror by using the wave four-vector $k^{\mu}=(\omega/c,\vec{k})$. First I wanted to transform the four-vector to a system $K'$ where the mirror is at rest. I know that the transformed position-vector can be written as $$\begin{Bmatrix}x^{{0}^{'}}=\gamma(x^0-\beta x_\parallel)\\x_\parallel'=\gamma(x_\parallel-\beta x^0)\\x_\perp'=x_\perp \end{Bmatrix}$$ The k-vector in the K-system looks like this $$\begin{Bmatrix}k^0=\omega/c=k\\k_\parallel=k_z=k*cos(\theta)\\k_\perp = k_y=k*sin(\theta)\end{Bmatrix}$$ My problem is now how to transform the k vector into the primed system K'. It should look like this: $$\begin{Bmatrix}k'^0=k \gamma(1-\beta *cos(\theta))\\k'_\parallel=k \gamma(cos\theta -\beta)\\k_\perp'=k_\perp = k*sin(\theta)\end{Bmatrix}$$ The main question is: How do the different cosines come to be?
Note: I know that a similar, but broader question already has been asked here with the title "law of reflection for a moving mirror". But the given references do not provide any insight into this specific problem.

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  • $\begingroup$ Hint: how do you transform four-vectors from one frame to another? $\endgroup$ – udrv Jul 29 '16 at 18:56
  • $\begingroup$ I have to apply a Lorentz transformation by multiplying the 4 Vector with a transformation matrix $\Lambda$. $\endgroup$ – Quasar Jul 29 '16 at 20:00
  • $\begingroup$ And if you have the transform from $(x^0, x_{||})$ to $(x^{0'}, x'_{||})$, what is the transformation matrix? $\endgroup$ – udrv Jul 29 '16 at 20:07
  • $\begingroup$ I think it should be something like $$\Lambda =\begin{Bmatrix}\gamma && \gamma \vec{v}^T/c \\ \gamma \vec{v}/c && 1+ \gamma^2 \vec{v} \vec{v}^T/c^2(1+\gamma)\end{Bmatrix} $$ $\endgroup$ – Quasar Jul 29 '16 at 20:17
  • $\begingroup$ You are complicating yourself way too much. Just look at the first set of equations in your question: It takes position four-vector $x$ into position four vector $x'$. What is this transformation in matrix form? $\endgroup$ – udrv Jul 29 '16 at 20:36
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Take the same transformations as given for the position vector x. They show the "standard" Lorentz-transformation.
Then apply these to: \begin{Bmatrix}k^0=\omega/c=k\\k_\parallel=k_z=k*cos(\theta)\\k_\perp = k_y=k*sin(\theta)\end{Bmatrix} This gives \begin{Bmatrix}k^{{0}^{'}}=\gamma(k^0-\beta k_\parallel)\\k_\parallel'=\gamma(k_\parallel-\beta k^0)\\k_\perp'=k_\perp \end{Bmatrix} Using the relations given for k in the K-system we get the desired result.

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