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Modification of Law of Reflection from a (moving) mirror in Special Relativity

  1. Nov 24, 2009 #1
    Hi everyone

    A few weeks ago, I had worked out the relations between the angle of reflection and angle of incidence in case of reflection from a plane mirror,

    (1) moving toward the incident ray
    (2) moving normal to the incident ray

    (PS -- This is not homework.)

    The way I did it was (for the first case, as an example)

    1. Assume the angle of incidence is [itex]\theta[/itex] in the lab frame.
    2. The velocity of the mirror is

    [tex]\mathbf{v}_{m} = -v_{m}\sin\theta\hat{x} + v_{m}\cos\theta\hat{y}[/tex]

    3. The velocity of the light ray in the lab frame is,

    [tex]\mathbf{v}_{l} = c\sin\theta\hat{x} - c\cos\theta\hat{y}[/tex]

    4. Find the velocity of the incident light ray in the mirror frame,

    [tex]\mathbf{v}_{i;m} = \frac{\mathbf{v}_l + (\gamma-1)\frac{\mathbf{v_l}\cdot\mathbf{V_m}}{V_m^2}\mathbf{V}_m -\gamma\mathbf{V}_m}{\gamma\left(1-\frac{\mathbf{v_l}\cdot\mathbf{V_m}}{c^2}\right)} = c\cos i \hat{x} - c\sin i \hat{y}[/tex]

    where [itex]i[/itex] is the angle of incidence in the frame of the mirror.

    5. Use the law of reflection in the frame of the mirror, to write the velocity of the reflected ray as

    [tex]\mathbf{v}_{r;m} = c\cos i \hat{x} + c\sin i \hat{y}[/tex]

    6. Transform this velocity back to the lab frame, using the inverse of formula 4 (with i --> r, for the reflected ray of course).

    7. From this transformed velocity, find the angle of reflection in the lab frame.

    The original question was to find the modification to Snell's Laws for a moving mirror. I thought the above procedure is a correct way of doing it. I was able to show that for low velocities, Snell's law does also hold in the laboratory frame (we have assumed in step 5 that it always holds in the frame of the mirror -- which seems reasonable to me).

    But I came across this short article: http://home.c2i.net/pb_andersen/pdf/aberration.pdf [Broken], and it seems to suggest that I made some mistake.

    What is going wrong? Is the procedure I outlined above correct?

    Does Snell's Law hold even in the laboratory frame for a moving mirror?

    Would appreciate inputs/insights.

    Thanks in advance,
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 24, 2009 #2


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    You can check your answer by working out the problem normally in the rest frame of the mirror, then Lorentz transforming to the lab frame.

    Edit: Oops, I guess you already did that. You can check your answer here http://arxiv.org/abs/physics/0605057 .
    Last edited: Nov 24, 2009
  4. Nov 25, 2009 #3
    We had a go at working this out in this thread --> https://www.physicsforums.com/showthread.php?t=352427&highlight=reflection+mirror

    I think it is generally accepted that in this special case, the reflected light ray will return along the same path as the incident light ray for any relative velocity of the mirror.

    Also, in the frame of the light source, a mirror with motion components that are parallel and normal to the incident light ray, can be treated as if it only had the parallel motion component as far as working out the angle of reflection is concerned.
  5. Nov 25, 2009 #4
    If the plane mirror (dielectric of index of refraction n) is moving perpendicular to the incident light, and the reflection is observed at a velocity such that the light is incident at Brewster's angle, will light of one polarization be reflected and the other refracted? Suppose the mirror is stationary, and the observer is moving at a velocity such that the angle of incidence is Brewster's angle?
    Bob S
  6. Sep 21, 2010 #5
    If refraction was considered, could it be that incident,reflected and refracted were not in plane, depending on the velocity of the mirror ?
  7. Sep 23, 2010 #6
    I looked at your diagrams in the other thread. Very interesting.
    It appears from the cases you considered that there is no essential difference between reflected light and light emitted at the same angle. Do you think this would be a general equivalence???
    Obviously beaming and aberration are interchangable being essentially the same phenomenon, is it possible that reflection is also basically a consequence of the conservation of momentum wrt light and comp[letely equivalent??
    This also seems to suggest that Doppler shift would apply equally to reflected and emitted light what do you think???

    It would have been nice if you had included a case where the mirrow was actually inclined at a proper 30 deg angle not just a Thomas angle.
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