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If a photon cannot be accelerated or decelerated, how is reflecting them possible at all?

Qmechanic
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Satyajit Sen
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1 Answers1

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This answer adresses the content of the question:

If a photon cannot be accelerated or decelerated, how is reflecting them possible at all?

A photon is a quantum mechanical particle/entity and obeys quantum electrodynamics interaction rules. The reflection is one of the possible interactions of a photon with the electric fields of atoms and molecules, the elastic scattering which only changes the direction of the photon, not the energy, i.e. frequency , as E=h*nu.

anna v
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  • Anna, this seems misleading, if not incorrect. "The reflection is one of the possible interactions of a photon with the electric fields" - (1) Photons do not interact with electric fields; (2) photons do not reflect - scattering of a photon on an electron involves the annihilation and creation operators with the virtual electron mediating the interaction. A popular view that "photons scatter" is an improper mix of quantum and classical approaches. Classical light reflects, but quantum photons get absorbed and re-emitted. – safesphere Mar 14 '19 at 20:57
  • @safesphere All quantum mechanical interactions can be expressed by the scattering matrix to be compared with measurements. Scattering by definition is a+b->a+b if it is elastic, which it has to be for the photon to build up a classical image reflection. It is erroneous , imo, to talk of electrons absorbing and emitting. Atoms, yes,. The scattering on fields happens with virtual particle exchanges as far as writing up the scattering matrix elements to be calculated. On free electrons the virtual particle is an electron, – anna v Mar 15 '19 at 04:37
  • similar to compton scattering .http://hyperphysics.phy-astr.gsu.edu/hbase/scacon.html – anna v Mar 15 '19 at 04:38
  • Again, all this language confuses classical and quantum concepts. To make it simple, can you present a valid Feynman diagram with two photons at the same vertex, one incoming and the other outgoing? If you can, fine. Otherwise "photon scattering" is but a confused simplification by combining two events, absorption and emission, into a non-existing "single event" of "scattering"". – safesphere Mar 15 '19 at 05:13
  • @safesphere you are going against the mainstream mathematics usage and visualization in the quantum framework . Of course there will be a virtual electron positron loop in the feynman diagram, but the end of the calculations gives the scattering of a+b-->a+b for elastic scattering. That is the way it is used in particle physics. Classical and quantum concepts overlap because the classical emerge from the quantum. – anna v Mar 15 '19 at 05:17
  • There is no electron-positron loop in the Compton scattering. The incoming photon is absorbed by the electron that goes off shell and emits a different photon. I don't object your calculations that take advantage of simplifications, but from the quantum standpoint there is no such thing as "photon scattering". It also is prohibited by special relativity. – safesphere Mar 15 '19 at 05:26
  • compton scattering was given as similar to free electon +photon scatter . loops are necessary where quantum number conservations etc enter., that is why loops are needed for photon +field interactions.. in general virtual partilcles carry over the impulse and quantum numbers, because there are no four vertices in QFT. – anna v Mar 15 '19 at 05:56