Timeline for How does QED explain reflection?
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7 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2016 at 15:49 | comment | added | anna v | I think if one will go through the math one will again find that it is the eV values of the photons versus the large mass of the medium that make the difference in wavelengths not detectable. | |
| Apr 12, 2016 at 14:03 | comment | added | Mercury | By reversal of the helicity of a photon the birefringent plate receives an angular momentum of 2h (Beth 1936) which has to show in a redshift of the photon wavelength (by the same logic as in reflection). But as far as know there is no such redshift observed. Maybe someone can shed some light on it? I think a redshift will be more observable then as in reflection? | |
| Apr 12, 2016 at 13:15 | comment | added | anna v | compton: look at the formula here: en.wikipedia.org/wiki/Compton_scattering . The difference in wavelengths is inversely proportional to the mass of the electron. The "free" is a handwave explanation for the conduction band hyperphysics.phy-astr.gsu.edu/hbase/solids/band.html . For birefringent in general if there exists an interaction it is possible, but I do not know details. | |
| Apr 12, 2016 at 11:47 | comment | added | Mercury | If there is a free electron never mind the great difference in E and m there is a very much noticeable redshift (Compton scattering). If the there is some sort of lattice reaction then it seems curious why they say that the elctrons form a free electron gas. However I am more interested do you think there is also a redshift when a photon of helicity 1 passes a birefringent plate and change the helicity to -1. (or vice versa) | |
| Apr 12, 2016 at 4:31 | comment | added | anna v | As John said in the comments and as I say in the answer there is no noticeable redshift because the energy transferred is very small, due to the large "inertia" (in the analogue of ball scattering on wall). Even one electron will pick up a very small part of the photon's energy because of the energy difference ( E=mc^2) for the electron at rest, 500.000eV and the photon at most some hundred eV. And hitting a reflective surface, means hitting a lattice where the electrons are in bands ( not free) and the lattice will react collectively with much larger "inertia". | |
| Apr 11, 2016 at 22:08 | comment | added | Mercury | I still have problems with the answer. It is known that the electrons in metals are free. So if there is an elastic impact between an electron and a photon and the electron is free, why it is not a Compton scattering? Then there would be a noticeable redshift. | |
| Apr 10, 2016 at 19:42 | history | answered | anna v | CC BY-SA 3.0 |