The Feynman diagram for Thomson/Compton scattering seems to depict an electron "absorbing" a photon, then "moving" a short distance, and then "emitting" a photon.
photon
~~~~~~
/
e- ------------------------------------------
/ virtual
~~~~ electron
photon
The traditional classical description of the events is quite different: the field of the incoming photon shakes the electron.
| | |
| | | ^ \ scattered wave
| | | | \
| | | ● e-
| | | |
| | | v
incoming electron
vertically moves up
polarized and down
wave
When the electron is shaking, it is in an accelerating movement, and produces a wave. The processes are simultaneous.
The rubber plate model of photon absorption / emission
^ waves are created
| in rubber plate
--------●-----------------------------------
| electron
v moves up and down
The field of the incoming photon exerts a force on the electron. In this blog we have written several times about momentum conservation in emission of electromagnetic waves. Namely, the electron must give up a lot of spatial momentum to provide the energy for the photon. But the photon can only take away a small fraction of that momentum. Where does the excess momentum go?
The rubber plate model solves the mystery. The static electric field of the electron is like an elastic rubber plate attached to the electron. For abrupt movements of the electron, the inertial mass of the electron appears to be small, since the field does not have time to react to the movement of the electron. Some, or all, of the mass of the electron is in its static electric field.
The electron in this temporary state appears as off-shell: it has absorbed more energy relative to the absorbed momentum than what would be allowed for a perfectly rigid object of mass m_e.
The excess energy is subsequently emitted as waves in the rubber plate.
Momentum is, of course, conserved in the rubber plate model. The rubber plate has a backreaction on the electron. The backreaction solves the mystery of the momentum flow.
The rubber plate model is close to the Feynman model.
We notice the following thing: to produce new waves, the electron has to be off-shell. A perfectly rigid system of the electron & the field would not produce any waves.
Special relativity makes a perfectly rigid field impossible. There is always retardation.
What is the classical analogue of the electron propagator?
Geoffrey V. Bicknell gives the total cross section for low-energy photons (Thomson scattering) as
σ_T = 8/3 * π r_0^2,
where r_0 is the classical electron radius. That is, the cross section is 8/3 times the area of the classical silhouette of the electron.
The total cross section for high-energy photons (Compton scattering) is
σ = π r_0^2 / x * ( ln(2 x) + 1/2 ),
where x = E / (m_e c^2) >> 1 and E is the energy of the photons.
The Thomson scattering formula follows from the classical calculation for a small dipole, as can be seen from the lecture notes of Steven Errede (2015).
The propagator controls the amount of power which flows through the scattering system to the scattered flux. Since the power classically is constant regardless of the photon frequency, there is no classical analogue for the electron propagator.
The 1 / r^2 potential of the "electron force"
We can qualitatively explain the Compton cross section if we assume that the photon is a rotating small electric dipole which is not traveling - it only rotates.
The electron moves in the dipole potential 1 / r^2 and is scattered. If we double the energy of the electron, then for "significant" scattering it has to come closer, within a 1 / sqrt(2) distance from the dipole.
The analysis is as for a pair annihilation, since the probability amplitude for the Feynman diagram is the same as in pair annihilation.
Alternatively, we may imagine that the photon scatters from a 1 / r^2 potential of the electron.
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