# Law of reflection for a moving mirror [closed]

An elementary fact that people learn about mirrors is the law of reflection, that the angle of incidence of a light beam striking the mirror (as measured with respect to a normal) equals the angle of reflection.

1. Does this law also hold for a mirror that is moving? Consider a square mirror that is moving at speed v in a direction perpendicular to the mirror. (You can think of the mirror as starting in the xy-plane and moving in the positive z direction of a Cartesian coordinate system.) As the mirror approaches a certain observation point, another person shines a laser beam of frequency w at the mirror so that the beam makes an angle A with the normal to the mirror. (You can think of the beam as lying in the yz-plane.)
2. What angle and frequency will you measure for the reflected light beam? Does the law of reflection still hold?
3. Do your conclusions change if the mirror moves parallel, rather than perpendicular, to its plane (say in the y direction if it starts in the xy-plane)?

Note:

1. The large mirror of the Hubble space telescope is an example of a mirror in motion as it orbits the earth. From your analysis, do you think the users of the Hubble have to take into account the motion of the mirror when measuring properties of its images?
2. I know that this question could be according to the .SE standards be too broad. But I request you not to close this question, I would really appreciate people if this question is answered rather than flagged or closed.

## closed as off-topic by David Z♦Sep 11 '16 at 8:30

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• Can't you always shift to the frame of reference of the mirror, so you only need to deal with the images of moving objects? – DJohnM Dec 28 '14 at 21:51
• I can't tell if the equality between the angles remains, but I can tell something about the direction of the wave-vector. The projection $k_{x,y}$ doesn't change because the mirror remains in the plane $x, y$. But along the vertical direction $z$, while the reflected beam travels a distance $cTcos(A)$, the mirror advances by $vT$ where $T = 1/f$ and $f$ is the light frequency. So within a time $T$, the distance that the beam made with respect to the mirror along $z$ is only $cTcos(A)[1 - \frac {v}{c \ cos(A)}]$ . Thus, $k_z$ became longer by a factor $[1 - \frac {v}{c \ cos(A)}]^{-1}$. – Sofia Dec 28 '14 at 22:22
• Denoting by $B$ the reflection angle, could it be that $tan(B) = k_z / k_{x, y}$ ? – Sofia Dec 28 '14 at 22:27
• The law of reflection breaks down because the frequency that is being reflected by the mirror is not the same as the original frequency, but it's shifted by the Doppler effect. There is a transversal as well as a longitudinal Doppler effect for light (and any combination for other than the two extreme angles). – CuriousOne Dec 28 '14 at 22:36
• I don't see any Doppler in the plane $x, y$ because the mirror remains in this plane. What yes though happens is that, as the mirror comes toward the observer (that sends the beam) and as the beam path is oblique, the impact point of the beam on the mirror undergoes a continuous translation in the plane $x, y$. So, because of it, the whole reflected ray undergoes a corresponding translation. – Sofia Dec 29 '14 at 0:43

Maesumi has already provided references, I would just like to point out how you can solve the problem yourself: the law of reflection most certainly holds in the rest-frame of the mirror. So to find out what happens when it is moving, transform to that frame, apply the law of reflection, transform back.

As correctly pointed out in the comments you will find that the law of reflection does not hold anymore and that the frequency of the incident light changes upon reflection.

I do realize that this is only a qualitative answer, but the quantitative results you can easily work out yourself by applying the recipe provided. This will probably be more insightful too.

The correct solution was given even before the advent of relativity. As soon as Michelson-Morely experiment indicated constancy of speed of light various papers gave simple solutions to this problem. See for example W. Pauli book, Theory of Relativity, comment 166.

This problem is also considered in the following papers. There are different types of derivations, using:

(A) Relativity, Lorentz transformations and velocity addition formula

(B) Maxwell's equations at a discontinuity

(C) Plane geometry or Calculus

1) A. Einstein, Zur Elektrodynamik bewegter Korper {Ann. Phys. (Leipzig)} {17} (1905) 891–-921.

Page 915. Look for something like $\cos \phi''' = - {{(1+ \beta ^2) \cos(\phi) -2 \beta}\over{(1+\beta^2)-2\beta \cos \phi}}$ . He uses $v/V$ for $\beta$. $\phi$ is angle of incidence, $\phi'''$ is angle of reflection, $v$ is velocity of the mirror, $V$ is speed of light, all measured with respect to the same axis normal to the mirror.

http://onlinelibrary.wiley.com/doi/10.1002/andp.19053221004/pdf

https://www.fourmilab.ch/etexts/einstein/specrel/www/

Reprinted in Einsteins Miraculous Year: Five Papers That Changed the Face of Physics, edited by John Stachel (Princeton U.P., Princeton, 1998)

2) J. Ronald Galli and Farhang Amiri, A general principle for light reflecting from a uniformly moving mirror: A relativistic treatment , American Journal of Physics, {80} (2012) 680--683.

http://dx.doi.org/10.1119/1.4720389

http://physics.weber.edu/galli/RelativisticReflection.pdf

3) Aleksandar Gjurchinovski and Aleksandar Skeparovski, Einstein's mirror {Phys. Teach.} {46} (2008) 416--418.

http://scitation.aip.org/content/aapt/journal/tpt/46/7/10.1119/1.2981289

4) Malik Rakhmanov, Reflection of light from a moving mirror

http://arxiv.org/abs/physics/0605100

5) Z. K. Silagadze, Test problems in mechanics and special relativity

https://arxiv.org/pdf/physics/0605057.pdf (pages 5 and 57) This is the shortest physics-based solution. I believe (1) skips the steps of derivation.

6) I also have a paper on the topic A conic section approach to the relativistic reflection formula. The paper uses elementary math (meaning precalculus, Euclidean geometry, properties of hyperbolas, etc). There is also a short calculus-based proof.