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How does quantum electrodynamics actually explain HOW reflection occurs on a microscopic scale?

Note that Feynman's QED lecture series/book is not sufficient, as he only assumes that light DOES reflect ('turn around and go back') in order to expound his path integral theory. My question is why does light have the propensity to turn around in the first place.

Is it just the absorption and re-emission of photons, and if so, why does it happen so uniformly (i.e. on a shiny thing, entire scenes are reflected near-perfectly). In essence, why are flat things shiny? Are all the molecules arranged at exactly the same angle?

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Related: physics.stackexchange.com/q/32483/2451 and links therein. –  Qmechanic Aug 25 '12 at 16:27

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the process is coherent--- the same photon is bouncing off all the atoms at once, and you only get constructive interference when the angle of reflection is equal to the angle of incidence. The condition is that the surface is smooth on the scale of the wavelength of light, so that the light can excite each atom independently, and coherently add up all their contributions. This is Feynman's explanation in QED, I think you just misunderstood it as saying that reflection is assumed--- he just assumed rescattering, and then shows you that it happens in the reflected direction preferentially.

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Thanks a lot for that- very enlightening. However, what IS 'bouncing', or 'scattering' (sorry, that was what I meant by reflection rather than the whole process)? Why does it happen? Is it the absorbtion and reemission of the scattered photon (of course of all atoms)? –  Alyosha Aug 26 '12 at 12:42
    
@user11657: It is the absorption of a photon followed by it's later emission, together with the counterintuitive process, also quantum mechanically allowed, of it's emission then later absorption. The emission can happen virtually, then the later absorption supplies the missing energy. –  Ron Maimon Aug 26 '12 at 22:22
    
Why is it that in specular reflection, all wavelengths are reflected, whereas in diffuse reflection, absorption of some wavelengths occurs before emission, so that only some wavelengths are reflective? I assume it's because the final sentence of your last comment does not occur readily in rough surfaces (why?). –  Alyosha Oct 5 '13 at 14:17
    
User @Pete asks "Forgive me for possibly adding a layer of misunderstanding but I was under the impression that QED absorption and re-emission was an entirely random process. If this is the case then how can we predict reflective angles? In addition, refraction through transparent media shows us that interaction with matter does affect the path of light. Again, reemission would have difficulty explaining how we can predict these angles so accurately." –  WetSavannaAnimal aka Rod Vance Feb 6 at 22:26
    
@Pete No, it is measurement in QED (and the rest of quantum mechanics) that is random. Reflexion arises from the coherent, unitarily (and wholly deterministically) evolving pure quantum state of one photon as it interacts with all the matter of the reflector. It actually becomes a pure quantum superposition of free photon and excited matter states - with all of the matter in its field of influence being part of the quantum superposition. The one photon's state propagates following Maxwell's equations, and so the Fresnel equations apply perfectly well. And they can be derived by the ... –  WetSavannaAnimal aka Rod Vance Feb 6 at 22:32

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