Doppler broadening upon reflection from liquid interface

Question

I just came across the question "Why are the surfaces of most liquid so reflective?", in which the author asks how the surface of a liquid gives rise to a mirror image, even though it consists of many individually moving atoms/molecules. I understand that the density of charges and therefore the macroscopic polarizability of the liquid is not changed by the movement of the atoms/molecules. How this leads to the refractive index and the Fresnel equations was explained in a recent video by 3Blue1Brown.

However, I wonder what is the effect of the movement of the charges on the frequency of the light. At $T = 300 \, \text{K}$ the velocity of water molecules exhibits a distribution of root-mean-squared $v_\text{rms} = \sqrt{\frac{3 k_B T}{m}} = 645 \, \frac{\text{m}}{\text{s}}$. Light reflected from this ensemble of molecules is the interference of the small contributions of all the individual molecules. As some are moving away from the surface while some are moving towards the surface, all these contributions give rise to different Doppler shifts. The reflection of monochromatic light of frequency $\nu_0$ should therefore exhibit a frequency spectrum of width $\frac{v_\text{rms}}{c} \nu_0$, up to a geometry factor depending on the angle of incidence. For green light ($\nu_0 = 500 \, \text{THz}$), the broadening of $1.1 \, \text{GHz}$, although too narrow to percieve, is large enough to be detected with a Fabry-Pérot interferometer.

Is this Doppler broadening upon reflection actually happening? A short search on Google didn't result in something relevant, but maybe I'm using the wring keywords here. It might also just be irrelevant to most experiments, since we don't use liquid mirrors in standard optics setups.

Answer 1

Classically, waves tend to scatter off things that are bigger than a wavelength and not be affected by things that are smaller. You can see this in the ocean. Waves go by the poles supporting a pier, but are reflected from the shore.

For light, the wavelength is about 5000 angstroms, and molecules are roughly 1 angstrom. light would not be affected if you had a molecule of fluid B in fluid A, or even a tiny droplet. It takes an interface larger than a wavelength.

For an interface to give reflected light a Doppler shift, enough molecules would have to be traveling in the same direction to make the interface move. Since they all travel in uncoordinated random directions, this does not happen.

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However, quantum mechanics makes different predictions when things get very small. Light is photons, which are sort of like particles and sort of like waves. A wave equation predicts where you are likely to find them next. But when they get there, they can interact with individual electrons.

So some photons can scatter off a molecule of B in fluid A. Whether or not many do depends on the properties of B and wavelength. Wavelength is important in a different way than for wave reflection. It corresponds to energy of the photon. If a molecule of B has a transition where an electron gains the energy in a photon, the photon can be absorbed. When the molecule decays, a new photon can be radiated in a random direction. If there is no such transition at the right wavelength, photons go on by without interacting.

This doesn't mean an interface will have a Doppler shift. If B had such a transition, all photons would scatter before getting very far. B would be opaque at that wavelength.

For examples of these classical and quantum mechanical effects, see If we repeatedly divide a colorful solid in half, at what point will the color disappear?.

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There is a quantum mechanical explanation of how waves reflect at an interface. It related to the reason why light travels slower in a fluid than a vacuum. See What really causes light/photons to appear slower in media?.