Law of reflection of a moving mirror in special relativity (velocity perpendicular to the normal)

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?

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

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.