If the laboratory frame is a different frame, we have to do a Lorentz transformation. The effects of the transformation can be called "Doppler effects".
The part which is not obvious is how a (dipole) wave behaves in a frame change. The natural frame for a wave is the one where the source of the wave is static. How does a moving observer see the intensity of the wave?
An example: Compton scattering
incoming beam
photon E
~~~~>
wave burst source e- wave burst
( ( ( ● ) ) )
<----- v ●
observer
<----------------->
distance d
Let us assume that an incoming photon of the energy E << 511 keV is linearly polarized in the vertical direction.
There, 511 keV is the electron mass-energy m c².
The photon hits a static electron and makes the electron to oscillate up and down. The electron emits a burst of waves.
Simultaneously, the photon pushes the electron and makes it to move.
We are interested in the intensity of the produced classical electromagnetic wave in the direction from which the photon arrived. We assume that the collision is almost head-on and that we observe a photon coming back from the process.
The electron receives an impulse
p = 2 E / c
to the right.
We guess that the best frame to analyze the process classically is the one where the collision is half-way: the electron has received an impulse of p / 2 and is moving to the right at the velocity
v = c E / 511 keV
= α c,
where we denote by α the ratio E / 511 keV.
If we work in that moving frame, then the electron is static relative to the horizontal axis and is oscillating up and down, and the observer is moving to the left at the same velocity
α c.
The observer sees a burst of waves arrive from the electron. How does he see the burst?
Let us consider a hypothetical (wrong) model where the electron stays at the same horizontal position throughout the collision process. Then the observer would also stay static. The observer would see a classical dipole wave emitted by the electron which oscillates up and down. The dipole wave would have the same frequency as the incoming beam. The intensity of the back scattered wave would not depend on α, as long as the intensity of the incoming beam stays constant.
Let us have a very naive observer in the laboratory frame. He uses the hypothetical (and wrong) model to calculate the frequency and the intensity of the back scattered wave. We compare the correct calculation to his naive calculation.
The Doppler effect on the wave emitted by the electron:
Doppler shift of the frequency. The observed frequency is smaller by the factor
1 / (1 + α)
relative to the frequency emitted by the electron.
Doppler effect on the intensity. The intensity goes down as
~ 1 / d²
on the distance d.
We look at the spatial distance between the following events:
1. the collision event, and
2. the observation event at the location of the observer.
If the spatial distance is d in the laboratory frame, then it is
(1 + α) d
in the moving frame.
Also, since the observer is moving away at the speed α c, the energy density of the wave appears to him by a ratio
1 / (1 + α)
smaller than for an observer comoving with the electron.
The end result: the intensity is by a factor
1 / (1 + α)³
less than what the very naive observer in the laboratory frame would calculate.
The Doppler effect on the incoming beam:
Doppler shift of the incoming beam (photon). Since the electron is moving away at the speed α c, the frequency of the incoming beam it sees is smaller by the factor
1 / (1 + α)
relative to the frequency measured in the laboratory frame.
Doppler effect on the incoming beam (photon) intensity. The electron in the moving frame is moving also relative to the light source which produced the incoming beam. The intensity of the incoming beam that the electron sees is reduced by the factor
1 / (1 + α)³
relative to what the very naive observer in the laboratory frame would calculate.
Above we assume that the incoming beam of light is produced by point sources. The analysis might be different if the incoming beam would be a true plane wave. Is it possible to create such a plane wave?
Conclusions:
1. The frequency of the observed photon is
1 / (1 + α)²
of the incoming beam frequency.
2. The intensity of the back scattered wave is
1 / (1 + α)⁶
times a constant, if we vary α but not the intensity of the incoming beam.
3. The cross section for back scattering is
1 / (1 + α)⁴
times a constant if we vary α.
Comparison to Compton scattering in quantum mechanics
At the link, D. H. Delphenich has an English translation of a Walter Gordon 1926 paper where Gordon uses the Schrödinger equation to calculate Compton scattering. Gordon observed that the frequency and the intensity of the scattered wave is the geometric mean of the classically computed values at the start of the transition and at the end of the transition. This corresponds to our analysis above where we used a frame where the collision is "half-way".
Yuji Yazaki (2017) recounts the history the famous Klein-Nishina (1929) formula for Compton scattering.
Paul Dirac in 1926 used Heisenberg methods to calculate Compton scattering.
For the intensity of the scattered beam he obtained the formula
where I₀ is the incoming beam intensity, I is the scattered beam intensity, θ is the scattering angle, and φ is the angle between the electric field polarization and the propagating direction of the scattered wave.
If α = 0, then Dirac's formula is the classical result where the electron oscillates vertically and sends a dipole wave.
Above we analyzed the case where θ is roughly 180 degrees and φ is roughly 0. Our result agrees with that of Dirac.
This is the famous Klein-Nishina formula. The cross section is averaged over all incoming beam polarizations.
If the incoming beam is polarized, the formula is
If the energy of the incoming photon is << 511 keV, then the incoming wavelength λ' is quite close to the scattered wavelength λ, and
2 ≌ λ / λ' + λ' / λ.
The cross section formula agrees with our analysis of back scattering above.
Conclusions
The Doppler effect on the intensity of the scattered beam is tricky to derive. We have to take into account the Doppler effect on the incoming beam, too.
We can approximate classically Compton scattering very precisely for photons << 511 keV, if we assume that the electron is "half-way" through the scattering process with the photon: the electron is already moving at half the speed it will eventually have.
Since the classical approximation is quite precise, then the corresponding Feynman diagram and the integral must describe an essentially classical process.
The classical process is this: if we disturb the electron by making it to oscillate, what happens in its electric field? It makes the field lines to oscillate.
Does the electron propagator describe the behavior of the electric field of the electron under a disturbance?
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