Below is the diagram of the question I have on the Doppler shift of a light emitted from a stationary light source $S$ at an angle $\theta_1$ toward a transversely moving plane mirror $O$ having velocity as shown $V$ along the plane. The reflected light from the mirror is detected by a stationary detector $D$.


I tried to use Doppler shift equation as derived by Drain, 1980, which essentially for such a setup we would have:


where $f_D$ is the Doppler shift seen by the detector, and $\lambda$ is the wavelength of the light.

If we make use the fact that $\theta_2=(\pi-\theta_1)$ due to mirror reflection, and upon substitution, $\cos\theta_2=-\cos\theta_1$, and mathematically we get $f_D=0$.

I would like to ask if the derivation for Doppler shift is correct. Is there any intuition for this result?

  • $\begingroup$ Your equations are "from Physics." Can you clarify what alternative approach you are asking for? $\endgroup$ Sep 1, 2015 at 11:47
  • $\begingroup$ I have edited to reflect what I would want to know -- the physical intuition behind the result. Thanks. $\endgroup$
    – kuskus
    Sep 1, 2015 at 16:05
  • $\begingroup$ FWIW, this was verified in the mid-70s by an experiment capable of detecting a frequency shift of two parts in $10^{18}$. See iopscience.iop.org/article/10.1088/0022-3735/10/3/017 $\endgroup$
    – PM 2Ring
    May 26, 2021 at 15:50
  • $\begingroup$ @PM2Ring thanks for the reference. the paper is using different setup (i.e. the laser ray is perpendicular to the motion of the mirror). From my recollection on this paper, this is also relates to transverse Doppler effect of the moving train as observed by a stationary observer - as asked by Einstein. $\endgroup$
    – Karsun
    Jul 18, 2021 at 15:36
  • $\begingroup$ @Karsun Sorry, I don't understand your objection. That paper is directly about the setup described in this question. As that paper and the answers below explain, there is no Doppler shift in this setup. $\endgroup$
    – PM 2Ring
    Jul 18, 2021 at 20:37

2 Answers 2


When we think about balls bouncing off walls and light bouncing off mirrors, we assume that the there will be a momentum exchange, but only for components that are perpendicular to the plane.

If the mirror has some velocity component in the perpendicular direction, it affects the interaction. It can add or subtract momentum from the reflected particles. In the case of light, this momentum change will affect the wavelength.

But there is no momentum interaction for components parallel to the mirror. Since there is no interaction, the parallel component of the light's momentum is unchanged and the parallel velocity of the mirror is irrelevant.

  • $\begingroup$ Using ball analogy, the ball has a velocity parallel component to the mirror -- it is not like the ball is dropped vertically. So I am not clear about "But there is no momentum interaction for components parallel to the mirror" point. $\endgroup$
    – kuskus
    Sep 2, 2015 at 6:43
  • $\begingroup$ After striking the mirror, the ball's velocity parallel to the mirror is unchanged. Only the perpendicular component changes. $\endgroup$
    – BowlOfRed
    Sep 2, 2015 at 6:44
  • $\begingroup$ I got it now with the ball analogy. One follow up question: Is there any relationship between frequency shift and the momentum change/interaction in this comparison? Thanks. $\endgroup$
    – kuskus
    Sep 2, 2015 at 11:11
  • 1
    $\begingroup$ Yes. When a ball's momentum changes, the mass is constant and the velocity changes. When a photon's momentum changes, the speed is constant and the energy/frequency/wavelength changes. $\endgroup$
    – BowlOfRed
    Sep 2, 2015 at 14:38
  • $\begingroup$ I made some edit to correct the mixed-up between momentum gain, and loss due to parallel component not the perpendicular component. $\endgroup$
    – kuskus
    Oct 1, 2015 at 9:23

Yes, the derivation is correct. The intuition should be that there should be no Doppler shift, since the source and detector positions are fixed (light path length is not changing).


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