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By reflection of photons from a mirror, the mirror must get an impulse of 2p=h/pi.l for every photon. This means that the mirror receives kinetic energy. But the photon has the energy it has before the refection. How does QED explain this? In "Photons and perfect mirror" the question is answered as a consequence of other explanation but I does not see the mechanism which leads to it. I try to clarify this as respond to the answers.

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marked as duplicate by John Rennie, user36790, ACuriousMind, Kyle Kanos, Qmechanic Apr 12 '16 at 0:33

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    $\begingroup$ Possible duplicate of: Photons and perfect mirror $\endgroup$ – John Rennie Apr 10 '16 at 18:41
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    $\begingroup$ See also: Does a reflection still transfer momentum to an mirror?. The point is that unless the mirror has an infinite mass the photon does not have the same energy after the reflection as before it. The reflection process will redshift the photon, though in practice this redshift is undetectably small. $\endgroup$ – John Rennie Apr 10 '16 at 18:43
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/10301/2451 , physics.stackexchange.com/q/83105/2451 and links therein. $\endgroup$ – Qmechanic Apr 10 '16 at 19:47
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    $\begingroup$ Also note that you can get the same basic result (radiation pressure) working entirely classically, from Maxwell's equations. You don't need quantum mechanics, much less QED. $\endgroup$ – knzhou Apr 10 '16 at 22:01
  • $\begingroup$ @JohnRennie 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. $\endgroup$ – Mercury Apr 11 '16 at 22:08
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QED uses Feynman diagrams and they are written in the center of mass. In the case of a photon hitting a mirror, it is hitting the field of electrons in the outer band of the mirroring substance. For reflection there should be elastic scattering in the center of mass so that the phases of the photons are kept and the image emerges intact on reflection. So it should be a QED feynman diagram of a photon scattering elastically off a field.

The kinematics are similar to a ball bouncing off a wall, i.e. elastically scattering. The assumption is that the mass of the wall is practically infinite and the center of mass with respect to the ball+wall, where the elastic scattering happens is at the same (x,y,z) as the center of mass of the wall itself.

For the photon and the collective electric field of the electrons it is scattering off, the center of mass and the laboratory frame are practically the same, as the photon has zero mass and low energy (optical photons). Even the mass of a single electron , (.5MeV, gamma ray energies are of this size) is large enough for the lab frame to be considered the same as the center of mass frame. The effect of the transformation to the lab on the energy of the photon will be very very small.

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  • $\begingroup$ 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. $\endgroup$ – Mercury Apr 11 '16 at 22:08
  • $\begingroup$ 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". $\endgroup$ – anna v Apr 12 '16 at 4:31
  • $\begingroup$ 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) $\endgroup$ – Mercury Apr 12 '16 at 11:47
  • $\begingroup$ 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. $\endgroup$ – anna v Apr 12 '16 at 13:15
  • $\begingroup$ 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? $\endgroup$ – Mercury Apr 12 '16 at 14:03

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