# Relativistic mirror

A (very long) mirror is moving from left to right at 0.5c and we point a laser at it perpendicularly as it passes in front of us. (it glides in its own plane, like the windows of a train)

How will the laser bounce off the mirror? will it follow the original path back to the source, in the same manner as if the mirror was at rest? or will it reflect at an angle?

Clarification: let's consider the mirror as being front surface, to avoid discussing about light traveling through a layer of glass or anything similar. The point of the question is simply about reconciling the fact that different reference frames point to different results.

EDIT: The reason I asked the question is because from the source's perspective it would appear the movement of the mirror is irrelevant. However, for the reflected ray we can also consider the mirror as being the source, which means the light would go straight in the mirror's reference frame, meaning oblique for the outside observer.

• this is not homework, what makes it look that???
– Jony
Jul 25 at 20:17
• @PM2Ring yes, it's a front surface mirror. Please see my edit: the glass layer is not the point of the question.
– Jony
Jul 25 at 20:23
• This should not have been closed. Jul 25 at 21:54
• physics.stackexchange.com/questions/203691/… is valid here. The fact that the speed is relativistic should not matter. Jul 26 at 5:40
• In our frame the light will go forth and back perpendicular to the mirror. In the frame of the mirror the light will reach the mirror in an angle and be reflected in an angle of the same size but another direction, and return to the source. Jul 26 at 20:16

I think the clearest way to see this is with a double Lorentz transformation. In the lab frame the mirror moves at $$v$$ in the $$z$$ direction, and the laser beam in in the $$x$$ direction. The k-vector of the light is thus $$\mathbf{k}= \begin{pmatrix}\omega/c\\k\\0\\0\end{pmatrix}$$ where $$\omega$$ is the angular frequency and $$k$$ the wavevector, with $$\omega/c=k$$.

We now do a LT to the rest frame of the mirror: $$\mathbf{k'}=\begin{pmatrix}\gamma&0&0&-\gamma\beta\\ 0&1&0&0\\ 0&0&1&0\\-\gamma\beta&0&0&\gamma \end{pmatrix}\begin{pmatrix}\omega/c\\k\\0\\0\end{pmatrix}=\begin{pmatrix}\gamma\omega/c\\k\\0\\-\gamma\beta k\end{pmatrix}.$$ In this frame the mirror is at rest, so we have the angle of relection equals the angle of incidence; in other words the normal component of $$k$$ changes sign: $$\mathbf{k'}\to\mathbf{k'_r}=\begin{pmatrix}\omega/c\\-k\\0\\-\gamma\beta k\end{pmatrix}.$$

Finally we LT back to the lab frame: $$\mathbf{k_r}=\begin{pmatrix}\gamma&0&0&\gamma\beta\\ 0&1&0&0\\ 0&0&1&0\\\gamma\beta&0&0&\gamma \end{pmatrix}\begin{pmatrix}\gamma\omega/c\\-k\\0\\-\gamma\beta k\end{pmatrix}=\begin{pmatrix}\omega/c\\-k\\0\\0\end{pmatrix}.$$ But this is just a normal reflection just as though the mirror is not moving.

Laser device has a mirror, which is a source of wave-fronts.

In the laser frame each wave-front is parallel to the laser mirror, because the creation of the wave-front occurs simultaneously at different parts along the mirror. Waves move perpendicular to the wave-fronts. So waves move perpendicular to the laser mirror.

In the long mirror frame different parts of the wave-front are created at different times inside the laser, which means that the wave-front is not parallel to the laser mirror, which means that the wave does not move parallel to the laser mirror.

In the laser frame different parts of a wave-front are absorbed at the same time by the long mirror, which means that they are later emitted at the same time, which means that the emitted wave-front is parallel to the long mirror, which means that the wave moves perpendicular to the long mirror.

In the long mirror frame different parts of a wave-front are absorbed at different times by the long mirror, which means that they are later emitted at the different times, which means that the emitted wave-front is not parallel to the long mirror, which means that the wave does not move perpendicular to the long mirror.

• This attributes a width to the laser beam, which is ok. But if we were to consider the laser as a series of individual single photons (of if instead of a laser source there was a photon gun shooting just a single photon) this analysis wouldn't make sense because the photon doesn't have a width. Or maybe it's incorrect to still consider it a "photon" when switching between the two reference frames?
– Jony
Jul 26 at 18:15
• @Jony If a photon is to have a certain direction at birth, then its birthplace must be uncertain, which means its birth time is uncertain in some frames. All possible birthplaces and all possible birth times must be correlated so that only at a certain direction the sum of all possible paths is large. ... Or probability-wave-front is tilted. Jul 27 at 3:09