I have seen this question about the law of reflection in special relativity, and it is shown that, if you have a mirror moving in the opposite direction to the mirror's surface normal, the law of reflection doesn't apply anymore. For example, consider the mirror in the $yz$ plane, so that velocity and normal are both on the $x$-axis. But what about the situation where the mirror is moving perpendicularly to the mirror's surface normal? Consider for example the mirror in the $xz$ plane, with the normal on the $y$-axis and the velocity on the $x$-axis. I have done some calculations, and in my solution, the law of reflection does apply in this case. Am I right?

PS: I hope that this is the correct format to ask this question.

EDIT: This is the procedure I have used:

  1. I looked at the first answer to the question that I have linked before.
  2. I applied the same method that is explained there, but with the new projections of velocities.

My results are (using $c = 1$):

$$ \cos(\theta_i) = \frac{\cos(\theta_i')\sqrt{1-v^2}}{1 + v\sin(\theta_i')} \\ \sin(\theta_i) = \frac{\sin(\theta_i') + v}{1 + v\sin(\theta_i')} $$

$cos(\theta_r)$ and $\sin(\theta_r)$ could be derived from the above equations by replacing $\theta_i \to \theta_r$ and $\theta_i' \to \theta_r'$.

In the primed system of reference, $\theta_i' = \theta_r'$, so $\theta_i = \theta_r$. Is this correct? If it isn't, where am I wrong?

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    $\begingroup$ I see with pleasure that someone appreciates my questions :o) and is also Italian like me. $\endgroup$ – Sebastiano Sep 7 at 21:00

I have done some calculations, and in my solution, the law of reflection does apply in this case. Am I right?

Yes, you're right. However, I offer you to use the following equation that relates the reflected angle to the incident one as measured in $S$ moving at an arbitrary velocity $v$ with respect to the mirror's rest frame ($S^\prime$): (See this article.)

$$\cos \theta_r=\frac{-2\frac{\boldsymbol{\vec{v}_n}}{c}+(1+\frac{\boldsymbol{\vec{v}_n}^2}{c^2})\cos \theta_i}{1-2\frac{\boldsymbol{\vec{v}_n}}{c}\cos \theta_i+\frac{\boldsymbol{\vec{v}_n}^2}{c^2}},$$

where $\boldsymbol{\vec{v}_n}$ is always the velocity vector projection onto the mirror's normal. In your example, when the mirror is set in motion perpendicular to its normal, we have $\boldsymbol{\vec{v}_n}=0$, and thus:

$$\cos \theta_r=\cos \theta_i \rightarrow \theta_r=\theta_i$$

Remember that the light-clock (known as Einstein's light-clock), which is used for deriving the familiar time dilation equation ($t=\gamma t^\prime$), is a good example for your special case: The incident and reflected angles are measured equally by the lab observer as long as the mirror (light-clock) moves perpendicular to its normal.


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