Bremsstrahlung is the prime example of this.
Claim. The electron and its field behaves in the classical way at the scale of "large distances". That is, when an electron scatters from a proton and the minimum distance of the electron to the proton is not small, then the low-frequency (soft) spectrum of bremsstrahlung is essentially classical, though the energy is, of course, divided into individual quanta.
Consider the classical limit of the process. We have a macroscopic charge changing its course in scattering. The energies involved in the process are very large relative to the energy of an individual low-frequency radiated photon. Bohr's correspondence principle says that the bremsstrahlung spectrum is the classical spectrum. The charge will radiate an infinite number of low-frequency photons in a single scattering event.
bremsstrahlung photon
~~~~~~
/
e- ---------------------------------
|
| 4-momentum q
|
---------------------------------
proton
Feynman formulae do not put any limit on the charge and the mass of the scattering particle. The Feynman integral formulae for the simplest bremsstrahlung diagrams assume that the electron radiates only one photon of bremsstrahlung. That is an error. We claim that it radiates an infinite number of soft photons in one scattering event.
We can remove the infrared divergence simply by cutting off the low-frequency spectrum of bremsstrahlung. The probability of emitting a high-frequency photon in bremsstrahlung is low. The overlap of classical probabilities is negligible when the involved probabilities are small. This is the reason why regularization works for infrared divergences.
The number of photons in classical bremsstrahlung
Wikipedia contains a semiclassical calculation "Simplified quantum description" for a nonrelativistic electron. Let the electron velocity be v. We assume that v is at least 100 km/s, which is the thermal velocity at 450 kelvins. The electron scatters from a proton.
We want to determine bremsstrahlung intensity for a frequency f. Wikipedia gives two values for the impact parameter,
b_max = v * 1 / f
and
b_min = h / (m_e v).
There b_min is the de Broglie wavelength of the electron. The bremsstrahlung output for the frequency f is obtained by integrating over the classical output between these impact factors b.
The limit b_max clearly is dictated by classical physics. If b is the impact parameter then the close encounter with the proton lasts for a time
t = b / v.
The classical power output is only significant for frequencies
f < 1 / t = v / b.
We get
b < v * 1 / f = b_max.
The limit b_min is quantum mechanical. Classically, there should be a lot of power output from much smaller impact parameters b. The total kinetic energy of the electron puts a ceiling on the classical output, but that ceiling is nowhere near for b = the de Broglie wavelength.
Let us consider the case of a mildly relativistic electron, whose wavelength is
~ 2.4 * 10⁻¹² m,
or the Compton wavelength of the electron. We probably can build an electron wave packet whose length and width are
~ 10⁻⁹ m = 1 nm.
The diameter of a typical atom is 0.1 nm. We can build a "screen" which forces electrons to fly past the proton at a distance of 1 nm or more. The wave packet presumably produces almost the classical spectrum of bremsstrahlung.
What is the classical spectrum like if our wave packet flies past the proton at a distance b ~ 1 nm?
It is a sudden change of the electron path in a distance of 1 nm. The Fourier transform of the classical electron path probably divides the radiated energy fairly evenly to frequencies
0 Hz ... 3 * 10¹⁷ Hz = c / 1 nm.
Let 2 f be smaller than c / 1 nm. The number of radiated photons in the interval
f ... 2f
is some constant C. That is because the energy output in that range scales with f, and so does the energy of a photon. Let us divide the frequency range into intervals of the form f ... 2f. There is an infinite number of such intervals, and an infinite number of photons are emitted.
Conclusion. Bremsstrahlung from a single scattering event always contains an infinite number of soft photons. That is true classically as well as in quantum mechanics.
The Feynman integral formula for bremsstrahlung assumes that an emitted photon in an interval I is a separate case from an emitted photon in another interval I', and that we can sum the classical probabilities of these events. That is clearly an error. The probabilities overlap.
The Feynman formula probably calculates right the probability of emitting a photon in a small frequency interval df, but there is no sense in summing these probabilities over all intervals.
A crude calculation of the number of photons C in a frequency interval
We assume that a mildly relativistic electron passes a proton.
Then b_min is roughly the Compton wavelength of the electron. We divide possible impact parameters into intervals
b ... 2b.
Let us divide possible photon frequencies into intervals
f ... 2f.
For an interval b ... 2b, let us calculate the number of photons in each frequency interval.
In the picture we have the Larmor formula from Wikipedia. The square of the acceleration a² scales as
1 / b⁴.
The duration of the fly-by scales as b. The cross section of the interval scales as b². The radiated energy is evenly divided on frequencies from 0 Hz to ~ 1 / b. We see that each interval b ... 2b contributes an equal number C of photons to an interval f ... 2f, if b > b_max for the frequency f.
What is C? Let us calculate it in the case b = the Compton wavelength. The photon frequency range is then, given in photon energy, from 0 eV to m_e c² = 511 keV.
b
e- ----------------->
|
| b
● proton
We assume that the electron only radiates when it travels the distance b, roughly at the distance b away from the proton. The acceleration is
a = k e² / (b² m_e) = 5 * 10²⁵ m/s²,
where k is the Coulomb constant, e is the electron charge, and m_e is the electron mass. The radiated power is
P = 0.01 W,
and the radiated energy is
E = 10⁻²² J = 0.7 meV.
In the frequency interval 256 keV ... 511 keV the energy of a single photon is roughly 400 keV. We see that only
C = 2 * 10⁻⁹
photons are emitted in that interval.
Let us calculate how many photons whose wavelength is 1 m ... 2 m are emitted. Then b_max is 2 m and b_min is 2.4 * 10⁻¹² m. About 40 intervals of b contribute. The number of photons is only 10⁻⁷.
The wavelength 1 m corresponds to a photon whose energy is 1 μeV. Suppose that we measure bremsstrahlung of mildly relativistic electrons, and we can detect photons down to the energy of 1 μeV. There are then roughly 40 frequency intervals of photons. Each interval only contains some 10⁻⁷ photons. We conclude that the overlap of probabilities is negligible. We can ignore the infrared divergence of bremsstrahlung.
What part of the Green's function escapes as bremsstrahlung?
In our September 29, 2021 blog post we explained the "sharp hammer" model of the electric field of the electron. The electron keeps hammering the electromagnetic field to create its electric field.
The electron at every point of its path disturbs the electromagnetic field which we model with the massless Klein-Gordon wave equation. The disturbance is a Dirac delta source in the equation. The response is a Green's function of the Klein-Gordon equation.
We showed that for a free electron, there is total destructive interference for a component of the Green's function, if the component contains energy. A component which only contains spatial momentum survives.
Another interpretation is that the free electron totally reabsorbs the Green's function which it sent.
When the electron is scattered by a proton, the electron cannot absorb the Green's function completely. A part escapes as bremsstrahlung.
In this blog we have claimed that high-frequency components of the Green's function are almost completely wiped out by destructive interference. What about low-frequency components?
^ e-
/
/
x
● / \
/ \ k'
e- ------------------
\
\ k
When a free electron at a point x creates a component k' of the Green's function, then destructive interference destroys k' along with a component k of the same 4-momentum as k', but with a 180 degree phase shift relative to k'. Alternatively, we may interpret that the electron reabsorbed k.
Total destructive interference is spoilt when the electron is scattered by the proton (marked with ●).
Looking at the diagram above, it is obvious that if the deflection angle is small, then destructive interference still wipes out most of a long-wavelength k component, but some of the component survives.
The geometry of the diagram is essentially the same for all long-wavelength k. We conclude that the "survival rate" is the same for all such k.
For a very short-wavelength k, destructive interference is almost total. The wave only "sees" a small part of the deflection curve before it is destroyed by interference.
Conclusion. In a typical scattering event, a small part of each long-wavelength on-shell component of the Green's function escapes as bremsstrahlung. Most of the component is still destroyed by interference (alternatively, reabsorbed by the electron).
How to handle infrared divergences
Setting a cutoff for low frequencies probably works in all situations. The infinite tail of photons contains exceedingly low-energy photons. They have a very small impact on events and we can safely ignore them.
There are no infrared divergences in QED, once we recognize that Feynman integral formulae should not sum overlapping probabilities.
Literature contains a strange argument that the infrared divergences in bremsstrahlung and in the vertex function somehow "cancel each other out". They are separate processes with different outputs and cannot "cancel" each other. The argument has to be erroneous. We will analyze it.
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