UPDATE August 21, 2026: The Feynman vertex correction in literature seems to reduce the scattering cross section. That contradicts the classical vertex correction. At the classical limit, the classical vertex correction should yield the right result. Is the Feynman vertex correction erroneous?
On October 22, 2025 and earlier, we noted that Feynman diagrams double-count overlapping probabilities for emission of soft photons. This seems to be the source of infrared divergences. The Feynman vertex correction uses suspicious methods to eliminate infrared divergences.
Classically, bremsstrahlung should affect the (non-elastic) scattering cross section because the electron loses energy when it accelerates. For large turns, bremsstrahlung increases the cross section. The electron spirals toward the proton. For smaller turns?
Since the Feynman diagram method is an extremely crude way to calculate scattering, it is no surprise if it fails at the classical limit.
Or is the problem that the sign of the correction is flipped in the literature? We have suggested that the integral value counts those virtual photons which are absorbed quickly, and do not affect the scattering amplitude.
Let us continue our analysis. What values of k reduce the integral value when q is not infinitesimal?
----
On October 22, 2025 we were able to analyze the QED vertex correction to some extent. We continued our analysis on August 5, 2026.
Our hypothesis is that the formula
F₁(0) - F₁(q²)
calculates the "long-lived" virtual photons which are "detached" from the electron for a long time during the scattering process.
e- • ----------------
\
● v
proton+
The classical vertex correction is due to the fact that the far electric field of the electron does not "follow" the electron as the electron makes a sharp turn. The far field is "detached". The inertial mass of the electron is reduced, and it will pass the proton closer, gaining more momentum. This increases the scattering amplitude.
The question: why would the Feynman diagram and the integral calculate this classical process?
In the link we have a paper which calculates the QED vertex correction. The paper first handles the ultraviolet divergence in electron self-energy. Then it proceeds to calculate the vertex correction. The infrared divergence is handled by assuming a small photon mass λ.
The self-energy integral diverges for large virtual photons |k|. Our hypothesis in this blog is that those k do not exist at all. They are wiped out by destructive interference.
Similarly, we expect large |k| in the vertex correction to be wiped out by destructive interference.
Classical action
If we study the classical behavior through an action, we accept off-shell electrons, and so on. Any history, or a path, is legal in an action. We have to find a locally extremal value for the action integral, in order to find a legal physical history.
Note that since the electric field of a classical electron can be bent or distorted, a classical electron does not always obey the energy-momentum relation (we set c = 1):
E² = p² + mₑ².
The field may be missing some energy, or it can have too much energy. A classical electron can be off-shell.
From this point of view, the Feynman vertex correction diagram describes also a classical electron.
The Feynman vertex correction integral
-k virtual photon
~~~~~~~~
/ \
/ A B \
e- -----------------------------------------
p k + p | k + p + q
|
|
| q virtual photon
proton+ -----------------------------------------
The Feynman integral F₁(0) corresponds to an infinitesimal value of q. The integral is not zero. In QED, that integral is considered the "null case", which does not increase the scattering amplitude of the electron. This makes sense in the classical context, too. The big question is that, when we let q grow, why does the change in the integral value describe the added scattering amplitude?
Recall that the electron propagator very crudely is:
1 / how much off-shell is the electron.
Let the spatial momentum in p be p' and in q let it be q'.
----> p'
|
v p' + q'
If q is infinitesimal, then the most off-shell we get from a value of k which points to the direction of p'. Both internal electron lines are a lot off-shell.
Such a value of k might reduce the value of the integral. But it does not reduce the integral value "maximally", because the formula is
reduced value * reduced value.
We could reduce more, if p' and p' + q' would point to different directions. Then we would have
unreduced value * reduced value.
We see that the integral becomes smaller when q increases. That makes sense.
We can interpret the smaller integral value as saying that, of the virtual photons produced by the hammer strike as the electron approaches the proton, fewer are absorbed "easily" by the electron. All those virtual photons must eventually be absorbed, but the process may be more complicated, and slower.
Let us compare q infinitesimal and q such that p' + q' makes a 90 degree turn.
The integral value may decrease for k pointing to the direction of p' + q'.
^ -k
p |
e- • ------
\ | p' + q'
\ v
v
●
proton+
Analyze which k contribute to the change of the Feynman integral value when q grows from an infinitesimal value
The paper at the link gives the Feynman integral:
The small mass λ of the photon cuts infrared divergences. We can set λ = 0.
For any q, setting k ~ 0 makes both internal electron lines almost on-shell. We get something like
1 / ε³
as the denominator, since also the vertex photon line is almost on-shell.
If q is not infinitesimal, then setting k ~ q makes the second electron line almost on-shell, making the absolute value of the integral larger? Since the total value of the integral is smaller when q grows from infinitesimal, that suggests that the values for k ~ -q will be smaller. That is, virtual photons which carry momentum to the opposite direction from q will "break free". This aligns with our classical analysis: the electron will move closer to the proton.
The vertex correction in the scalar φ³ theory
Dirac gamma matrices γⁱ complicate the analysis of the vertex correction integral. Let us look at the integral in a "scalar" theory. In the southampton.ac.uk paper, there is a simpler theory:
in which the propagator is simply the Klein-Gordon field propagator:
The term i ε is used to handle singularities. We can ignore it. Let us switch i to 1, to ease the analysis.
If k = (E, k'), where E is the energy and k' is the spatial momentum, the the propagator is
1 / (E² - k'² - m²).
The propagator becomes +- infinite when the scalar electron is on-shell. The pole does not contribute much to the integral because it is almost perfectly symmetrically +infinite and -infinite.
k virtual photon
~~~~~~~~
/ \
/ A B \
scalar e- -----------------------------------------
p -k + p | -k + p + q
|
|
| q virtual photon
proton+ -----------------------------------------
Let us assume that q is pure spatial momentum. That is, q only turns the velocity vector of the scalar electron. Let q turn the velocity of the scalar electron by 90 degrees.
scalar e- • ------------------
p \
● |
| p + q
v
If k is pure energy, then the integral value remains the same regardless of q.
Let k be pure spatial momentum. The propagator for the photon line ~ 1 / k² suggests that we should concentrate on values
|k| < |q|.
Let q be infinitesimal. Then the two electron propagators are identical. Their product is positive for any k:
+ + +
+ • +
+ + +
The dot in the diagram denotes k = 0. The "+" or "-" in the diagram tell the sign of the product of the two electron propagators when k points to various directions.
If q is not infinitesimal, the sign diagram looks like this:
+ - -
- • +
- + +
The diagram shows that contributions to the integral change a lot with various k, compared to q infinitesimal. The upper left corner and the lower right corner retain their value.
e-
• --->
/
/ acceleration
v
●
proton+
The acceleration of the electron is mostly toward the lower left corner. Spatial momentum waves k are not well absorbed if they move to the direction of the acceleration.
This can be understood from the sharp hammer and the rubber membrane model. First, the hammer hits at a constant location. Waves are perfectly "absorbed". The wave system is static. The hammer does not need to do any work.
------------ wave front
| k
v ------------
If the hammer starts to accelerate down, the waves with the momentum vector k pointing up or down cause disturbance to the hammer, because the hammer moves relative to the wave fronts of those waves. The wave system is no longer static. The hammer has to do work.
We finally found a connection between the Feynman integral and the sharp hammer.
The classical vertex correction is calculated from the energy of the electric field "lagging behind": why is a -k containing spatial momentum associated with that?
Let us imagine that the electron is a small machine which is embedded into a hard solid material. The machine keeps hammering the material into every direction. A single hammer strike would send a sound wave, but constant hammering makes a permanent squeezed region into the solid.
hit ____
^ / \
| | | sound wave
e- • |
|
^ |
\____/
If we analyze a single hammer strike, the momentum given by the hammer to the solid returns back to the electron when the sound waves reach behind the electron and push it.
Sound waves with a longer wavelength may take longer to be reabsorbed by the electron.
Suppose that the electron is accelerated downward. If there is a delay in absorbing momentum from an upward hit, then the electron may move further down. Here we have another type of a classical vertex correction.
On September 29, 2021 we wrote about our hypothesis that the mass-energy of the electric field of the electron actually is zero, and that the field imitates the inertia of mass-energy through exerting forces onto the electron. This would resolve the paradox of the infinite field energy of a point charge.
The Wikipedia article contains a drawing of a mechanical device, which, in a periodic motion, appears to have a negative inertia. A force field can imitate the effect of inertia.
This may be a way to connect the momentum temporarily lost in hammering, to the inertia of the far field of the electron.
Absorption is destructive interference
Let us have a hammer hitting a tense rubber membrane at very short intervals. The hammer sends an impulse response, or Green's function, into the membrane. A hammer strike makes a sharp pit into the membrane.
For radio waves, the transmitter makes "hammer strikes" which send a nice sine radio wave.
Conclusions
We may have, finally, found the connection between the sharp hammer and Feynman integrals.
Next we will post about the classical bremsstrahlung correction in the scattering probability, and show that it approximately cancels the classical vertex correction.
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