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>> No.5938267 [View]
File: 21 KB, 800x431, Feynman-Diagram.png [View same] [iqdb] [saucenao] [google]
5938267

>>5937875
Electrons do absorb photons, in the process of scattering them. They then re-emit the photon to conserve energy and momentum. They also both emit and absorb photons as a method of mediating the electrostatic force. Here is a Feynman diagram of two electrons exchanging a photon. One emits, one absorbs a photon. (You cannot tell which one because both orders of emission/absorption contribute equally).

>> No.5872520 [View]
File: 21 KB, 800x431, Feynman-Diagram.png [View same] [iqdb] [saucenao] [google]
5872520

>>5872219
The simplest Feynman diagram contributing to electromagnetic interactions looks like this. The straight lines are electrons (or other charged fermions), the wavy line is a photon. The lines carry some constant 4-momentum, and the total momentum going into a vertex equals the total momentum going out of a vertex.

Because the electron lines can be stretched out to infinity, the electrons must be real electrons, meaning that they are on shell: E^2 = m^2 +p^2 for c=1 and m the mass of the electron. For a photon to be real, it must satisfy E=p. But if you work out the math, either the photon must have E=p=0, meaning there is no momentum exchange, or the photon must have E!=p. Thus, the interaction arises from off-shell "virtual" photons.

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