So when a solid object hits another solid object, it bounces off because the collision causes a compression and expansion of the objects which push the objects apart. But what are the mechanics of a photon bouncing off of a surface? What pushes the photon off of the surface it strikes?
Actually, the photon is aborbed and re-emitted by atoms at the surface of the material. Those not re-emitted contribute to the heating of the material. That’s why black (non reflecting) objects get hotter than white (highly reflective) objects when exposed to light.
BTW, I assume you mean “visible light”. X-Rays, for example, are also light (EM radiation) and obviously doesn’t “bounce” as well as the lower frequency visible light.
So then how does the atom know how to re-emit the light at the correct angle of reflection? How would the atom be able to “remember” the angle of incidence to correctly spit out the photon at the correct angle? It would seem that if it just absorbed/re-emitted the photon, the re-emitted photon would be traveling in an almost random direction from the original photon.
What would probably happen is the X-ray would ‘knock’ the elctron out of it’s orbital making it a free electron then it would be Compton scattering which is an exchange of momentum between the free electron and the X-ray with the new frequency being dependent on the angle of scattering.
This is a good question. I would think that since photons have momentum, somehow the law of conservation of momentum comes into play, but I’m not sure on how.
Pick up a good book on optics, such a Hecht’s or Jenkins and White. To see a rigorous treatment, look in Born and Wolf’s massive Bible of Optics, or Jackson’s book on Electromagnetic Radiation. Even in the case of the “wave model”, momentum (which can be carried by a wave) hasto be conserved.