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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 $D$.

doppler_geometry

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

$f_D=\frac{V}{\lambda}(\cos\theta_1+\cos\theta_2)$

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?

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  • $\begingroup$ Your equations are "from Physics." Can you clarify what alternative approach you are asking for? $\endgroup$ – Carl Witthoft Sep 1 '15 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 '15 at 16:05
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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.

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  • $\begingroup$ 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 '15 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 '15 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 '15 at 11:11
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    $\begingroup$ 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 '15 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 '15 at 9:23
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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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