I was studying Raman scattering and I computed the probability of anti-Stokes scattering (with density of states n, at the numerator) over Stokes (n+1 at denominator). The densities are based on the Bose-Einstein distribution because we are talking about phonons on a crystal.

At nearly 1000 K of temperature the probability of the two events are the same, as we see in the plot (fraction value = 0.5). The curve seems to have a raising behaviour after 1000 K, meaning that anti-Stokes effects are more probable over Stokes' ones.

Anti-Stokes scattering gives us back from the material a photon with higher frequency, thus, higher energy, at cost of a phonon (or more) being absorbed from the crystal lattice. Absorbing a phonon diminishes the "amount of movement" of the crystal atoms, decreasing their thermal velocities that, at over 1000 K is still pretty high. With some SERS effect, the intensity of the phenomenon could be large.

Does that mean that, under specific selection conditions, light is cooling down the material?

anti-Stokes/Stokes efficiencies


1 Answer 1


The specific Raman-scattering process you mention will not have a cooling effect. At high temperatures, the two cross-section will even out (i.e. the relative weight will asymptotically approach $1/2$ from below) and it will not produce any cooling.

On the other hand, it's perfectly possible to use specially-engineered states of light (typically laser light) to cool down matter; this is generally known as laser cooling, and nowadays there's a significant array of different ways to achieve it.

Among those ways there is what's known as Raman sideband cooling, but that requires you to specifically pump on the lower Raman sideband (without pumping the upper band, in contrast to your method) such that a portion of the Raman fluorescence will go on the upper sideband, which will remove energy from the material.


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