Let us assume that a macroscopic electron passes by a macroscopic proton at a high speed.
p
e- • ------> v
| q
v
b = impact factor v' gained
● proton+ velocity
We have two classical phenomena which affect the scattering amplitude and work to opposite directions:
1. The effective inertial mass of the electron is less because its far field does not have time to react to the pull of the proton. The electron will move a little closer to the proton and obtain more momentum q downward in the diagram.
2. The electron will accelerate downward toward the proton and radiate electromagnetic waves. This "friction" slows down the electron's descent toward the proton. The electron gains less momentum q from the proton.
If the electron moves almost at the speed of light, and gains a very large momentum q from the proton, then it does not need to give kinetic energy to the far field immediately – but the far field will "break free" and is radiated away. At suitable values of v and q, these two effects should be roughly equal, and cancel each other.
The kinetic energy saved by not moving the far field
We assume that v is much less than c. Let us use the values of a real electron, and pretend that they are macroscopic. The fly-by lasts roughly
t = 2 b / v.
Light moves the distance
R = c / v * 2 b
in that time.
The mass of the electron far field at the distance R is
m = rₑ / R * mₑ
= 1 / (4 π ε₀) * e² / c²
= 1 / (4 π ε₀) * e² / c³ * v / (2 b).
The electron gains a velocity v' downward, besides its horizontal velocity v.
The electron temporarily "saves" the energy
1 / (4 π ε₀) * e² / c³ * v / (2 b) * 1/2 v'²,
and can accelerate slightly faster vertically toward the proton, as it passes the proton.
The energy lost to radiation
The acceleration v-dot is
a = v' / t.
The energy
P t = 2/3 * 1 / (4 π ε₀) * e² / c³ * v'² / t
= 1 / (4 π ε₀) * e² / c³ * v / (2 b) * 2/3 v'².
The formula is the same as for the "saved" kinetic energy, except that we instead of 1/2 have 2/3!
Our model was extremely crude. A more precise evaluation may find the values to be exactly the same?
In our blog we have noted that zitterbewegung and the classical vertex correction explain the anomalous magnetic moment of the electron. If the electron cannot radiate, then the vertex correction might exist without the bremsstrahlung correction canceling it.
Comparison to quantum electrodynamics: QED has sign errors in the vertex correction and the bremsstrahlung correction
Various papers about QED claim that the infrared divergences of soft bremsstrahlung and the QED vertex correction cancel each other out in the scattering cross section. We noted in the summer of 2025, that the cancellation really cannot happen. In principle, we can observe soft protons of an infinitesimal energy. They cannot cancel out something which is elastic scattering: the vertex correction.
We have remarked that the QED process in each individual history sends out many soft photons at a time, but the Feynman rules count them as non-overlapping probabilities. This is wrong, and causes the sum of probabilities to diverge.
Anyway, the general idea that the effects of the vertex correction and bremsstrahlung approximately cancel each other out in the scattering cross section, is true classically, and might be true in QED, too.
Various QED papers claim that the vertex correction reduces the scattering cross section. That must be wrong, since classically, it increases the cross section. There is a sign error in QED.
A similar sign error must exist in QED treatments of bremsstrahlung, if it cancels out the QED vertex correction.
Why a sign error? The renormalization procedure in QED corrections is ad hoc. It can easily flip the sign of corrections.
Why does the Feynman integral approximate the classical vertex correction?
We have been working on this question for a long time. In our previous blog post we finally found (?) a connection to the sharp hammer model.
*** WORK IN PROGRESS ***
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