Perfectly elastic and inelastic collision of photon

Question

I`m having some trouble understanding (and finding any relevant information) about the elastic and inelastic collision between a photon and a mirror in non-quantum machanical terms. When a photon hits a surface and the collision is inelastic, than the momentum delta I = I, but when it hits it in a perfectly elastic way delta I= 2I. Why? Because of the change in the frequency? If yes, how do I know that it changes to half of its original sum?

Answer 1

Firstly, as @AnubhavGoel has mentioned, I'm assuming you have mixed up the terms "elastic" and "inelastic" here as what you have said is nonphysical. Now by definition in an elastic collision total kinetic energy is conserved. In a photon, $E_k = pc$, so the total momentum of the photon is also conserved.

Secondly, I'm assuming the symbol $I$ you have used defines the momentum in a specific direction rather than the total momentum (I will use $p_x$ to make it a bit more explicit). So in the elastic case if total momentum is conserved $p_x \rightarrow -p_x$, hence $\Delta p_x = 2 p_x$.

In the inelastic case I'm assuming you are referring to Raman scattering where the photon energy is absorbed by matter in the mirror. In this case $p_x \rightarrow 0$, hence $\Delta p_x = p_x$.

However, in general inelastic collision the photon doesn't have to give all its energy to the matter it interacts with as in Compton scattering. In this case the photon is remitted with a longer wavelength and given, $p = \frac{h}{\lambda}$, a lower total momentum, hence $0 \leq \Delta p_x \leq 2p_x$.

I hope this helps.