Do photons lose energy after radiation pressure is applied to a perfect reflector?

Question

I was readng Wikipedia article (English one) about radiation pressure because there is something I still cannot figure out. As I understand it, radiation pressure emerges from conservation of momentum. Photons or electromagnetic waves possess momentum and when they are absorbed, reflected or even emitted, the aborber/reflector/emitter experiences a pressure that is proportional to the irradiance (in watts per square meter). OK. In the case of a perfect reflector that does nothing but reflecting incoming radiation, it would seem the reflector is pushed and the radiation is reflected in opposite directions. But the radiation has not lost anything, it has only changed direction. And I do not understand how something can be set in motion (the reflector), which amounts to do work, yet the source of this work does not lose energy. I mean, if the radiation, after a U turn, meets a second reflector, it would have pushed two reflectors a way, yet it would continue in its original direction as if nothing happened...? What am I missing here? Should not photons lose something?

Answer 1

In the case of a perfect reflector that does nothing but reflecting incoming radiation, it would seem the reflector is pushed and the radiation is reflected in opposite directions. But the radiation has not lost anything, it has only changed direction.

This is true only if the reflector has an infinitely high mass. Then the reflector doesn't begin to move.

In reality the reflector has a high but finite mass.

According to $p_{\text{photon}}=\frac{h}{\lambda}$ you can set up the conservation of momentum $$\frac{h}{\lambda_{\text{incident}}}=-\frac{h}{\lambda_{\text{reflected}}}+mv$$ where $m$ is mass of the reflector, and $v$ is the velocity of the reflector after reflection. This means the reflector receives momentum from the incident photon. And thus the reflected photon has a momentum roughly the negative of the incident photon, but not exactly, as we see below.

And according to $E_{\text{photon}}=h\nu$ you can also set up the conservation of energy. $$h\nu_{\text{incident}}=h\nu_{\text{reflected}}+\frac{1}{2}mv^2$$ This means the reflector also receives a tiny bit of energy from the incident photon. And thus the reflected photon has a tiny bit less energy than the incident photon. And thus its frequency $\nu$ is a little less after reflection.

With the help $\nu=\frac{c}{\lambda}$ this becomes $$\frac{h}{\lambda_{\text{incident}}}=\frac{h}{\lambda_{\text{reflected}}}+\frac{mv^2}{2c}$$ and thus after reflection the photon's momentum is a little less and the photon's wavelength $\lambda$ is a little longer.

Answer 2

Your intuition is correct: each photon loses a very small amount of energy when it reflects from a perfect teflector that can move. The reflected photons will have slightly longer wavelengths than their incident counterparts.

Answer 3

There will be a recoil momentum ($2p$ and energy ($2p^2/M$ of the mirror. The photon energy will therefore be $pc-2p^2/M$. For a mirror of one mole of glass (60g) and a photon of 500 nm (~2eV) that is a relative frequency change of $2p/Mc= 2/(6\cdot 10^{23}\cdot 10^9 \approx 3\cdot10^{-32}$. However this is not what will happen, as the mirror will not move as a rigid body. Instead a collective electronic excitation in a very thin layer of metal will absorb the momentum. As a consequence I guess that the momentum will be carried by some $10^9$ electrons and the effect will be a relative frequency change of somewhere around $10^{-10}$, give or take a few orders of magnitude.