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 th7e 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 kinetic 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? Probably not.
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.
The classical vertex correction is the same as the bremsstrahlung correction? No
We have been saying that the far field of the electron does not have time to react, and that the electron can save kinetic energy because it does not need to accelerate the far field. Could this lead to non-conservation of energy unless the exact same energy is spent on generating bremsstrahlung?
When opposite charges approach each other, the energy released can be calculated either from the Coulomb force, or by integrating the energy density of the joint electric field of the charges.
If the fields would update infinitely fast, then the Coulomb force calculation would always give the exact same result as the field energy integration.
Energy conservation requires:
Ekin-i + Vi + Ekin-o + Vo = constant.
E denotes the kinetic energy of the electron electric field, i denotes inner and o outer, and V denotes the integral of the joint electric field energy density.
Suppose that we are able to save the energy in Ekin-o and that energy goes to Ekin-i.
We can then harvest the entire kinetic energy quickly from the electron, and also harvest the extra Vo. To rescue energy conservation, the electron must lose the extra Vo as bremsstrahlung.
But is the extra Vo the same as Ekin-o?
Let us assume that the electron and the proton have the same absolute charge, and we move them very close to each other.
Let R be the distance from the electron and the proton, and R is much larger than their distance from each other.
The energy of the dipole field outside R is
~ 1 / R³.
But the mass-energy of the electron electric field (and if it moves, also magnetic) outside R is
~ 1 / R.
These formulae are very different. The far dipole field cannot give the kinetic energy to the electron far field. Thus, we cannot prove that the bremsstrahlung correction must equal the vertex correction.
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.
The electron is equivalent to its inner electric field?
All the mass of the electron seems to be in its electric field. This suggests that we should treat the electron somewhat like we treat the electromagnetic field.
We can construct the outer electromagnetic field of the electron from hammer strikes: Green's functions. Maybe we should construct the electron in a similar way, as hammer strikes which create the inner field close to the electron?
This is a classical model which might explain what does the propagator of the electron mean classically.
The Poynting vector tells us the energy and the momentum flows in the electromagnetic field of the electron. What is the role of the point particle electron, if all its mass is in its field? The electron is where the charge lies.
The charge tends to go to the lowest potential of the field. If the mass of the charge is zero, it goes there very quickly. This shows that the field is primary, the charge obeys the orders of the field.
The electron propagator is the propagator of its inner field? What is the inner field propagator like?
In our previous blog post we introduced the "scalar electron" whose propagator is the same as the electromagnetic propagator.
The Dirac equation aims to describe a relativistic particle with mass. It would not be a surprise if the inner field propagator is the electron propagator.
When a hammer hits a rubber membrane, the membrane can go "off-shell": it is bent and does not form a natural sine wave. The hit also makes the hammer off-shell – the hammer is bent, too.
A hit produces a spectrum of Fourier components in the electromagnetic field. It also produces a spectrum of waves in the inner field of the electron. Could this be a classical explanation for the Feynman integral?
If k is a Fourier component, then the outer electromagnetic field gets the hit k, and the inner field the hit -k.
What about the fact that the inner field is local, occupies a very small volume?
k virtual photon
~~~~~~~~
/ \
/ A B \
scalar e- ------------------------------------------------
p -k + p | -k + p + q
|
|
| q virtual photon
proton+ ------------------------------------------------
When the scalar electron approaches the proton, the outer field and the inner field of the scalar electron start to clash: the inner field wants to go closer to the proton, the outer field is not interested.
Both parts of the field will go off-shell: they are bent relative to their undisturbed state.
In the Feynman diagram, the line k and the electron line merge again. This might indicate an "easy absorption" of the off-shell k wave?
Eventually all the off-shell k have to be absorbed by the electron. That may be a more complex, possibly a non-perturbative process.
If |k| is very large, then k has to be absorbed very quickly. The momentum exchange q in the diagram is misleading: the exchange of q happens gradually. For a large |k|, the real q is infinitesimal: only a very small momentum exchange dq happens before the absorption. Thus, the correct value for the Feynman integral for large |k| is identical to the infinitesimal q case. The integral value does not change. This is yet another way to explain why large |k| can be discarded.
Absorption of k: q treated non-perturbatively
k virtual photon
~~~~~~~~
/ \
/ A B \
scalar e- -----------------------------------------------
p -k + p | -k + p + q
|
|
| q virtual photon
proton+ -----------------------------------------------
1. In the diagram, the line A is born from a hammer strike. We can imagine that q is a "catalyst" of that "spontaneous" hammer strike.
The spectrum of the strike is the product of the propagators of the outer field (k) and the inner field (the electron).
2. We can imagine that q is absorbed non-perturbatively. No hammer strike there.
3. The line B ends in a hammer strike which the photon k wields on the inner field. The spectrum of that strike does not need the outer field propagator (for k), but it needs the inner field propagator. We assume that the propagator is symmetric: converting an on-shell electron to the off-shell state -k + p + q has the same probability as the inverse reaction.
In this model, the propagator of B is not associated with the absorption of q. This allows us to treat q non-perturbatively, which takes us closer to a classical model.
Let us assume this: various k are born from the movement of the electron along a path whose length is the wavelength of k. If the wavelength of k is short, then the electron can absorb k very quickly, so that k does not affect the scattering amplitude in any way. We can concentrate on k which have a long wavelength.
We are still far away from understanding how the crude hammer srike model is able to calculate the sophisticated, non-perturbative inertial mass reduction for the electron. Could it be a lucky coincidence?
Suspicious sign flips in the Feynman integral: Feynman rules are wrong?
The scalar electron propagator is
1 / (E² - p'² - mₑ²),
where we have denoted the energy in p by E and the spatial momentum in p by p'. We assume c = 1.
Let k and q be pure spatial momentum ,and |k| small. If
|-k + p'| > |p'|,
then the electron in the line A is off-shell, and the propagator for A has a large negative value. It may be that
|-k + p' + q| < |p' + q|,
which means that the propagator for B has a large positive value.
If we turn the direction of k, the product of the propagators flips the sign. Feynman integrals interpret that the phase of the electron wave changes by 180 degrees, if the sign flips.
It is reasonable that the phase of the electron can drastically change through tuning a tiny virtual photon k?
The phase of a wave flips in classical mechanics if it is reflected from a solid wall. How can a tiny k constitute a solid wall?
The solid wall is the momentum exchange q with the proton?
A sign flip in an individual Fouriet component is not that dramatic, if the entire wave does not flip its sign (be phase shifted 180 degrees).
Electron self-energy: no phase change is possible?
k
~~~~~~
/ \
e- -------------------------------------->
p
In our blog we have remarked that in the self-energy diagram, a virtual photon k can make the phase of the electron to flip 180 degrees if it flips the sign of the electron propagator. That is absurd. How can a particle on its own flip its phase, without interacting with the outside world?
Feynman diagrams are based on the impulse response of the field equation, that is, the Green's function. But what decides if the impulse has a positive sign or a negative sign?
Classical analogue. Let us investigate a classical analogue.
___________________________________ string 2
| |
_ | |
_ / \________|___________|______ string 1
\__/ bar A bar B
The wave in the tense string 1 interacts through two vertical bars A and B, with the tense string 2. We assume that the interaction transfers some of the energy of the wave temporarily to string 2, but the same energy returns back to string 1.
Can the phase change?
The interaction at A creates a small wave to string 2. The original wave loses some of its amplitude.
If the wave in string 2 would travel slower or faster than in string 1, then string 2 would absorb some mass-energy displacement (mass times distance) from the temporary wave in it. That should not be possible for a virtual photon. There is no mass which would permanently change place in empty space.
We can assume that the wave in string 2 travels at the same speed as in string 1. In our example, the wave in string 2 will have the same phase as in string 1. When the wave in string 2 is absorbed back to string 2, the phase has not changed at all.
We can imagine that the field of an electron is an elastic plate which it carries attached to itself. However the electron interacts with its plate, its momentum does not change. If the electron and the plate return to their originl state, an outside observer cannot know anything about a possible interaction earlier within the electron & plate system. This suggests that the phase cannot change.
Let us then analyze the impulse responses in the classical example.
String 1 at bar A feels that the bar resists the movement of the string. We could say that the impulse is negative to string 1.
When string 1 absorbs the wave back at B, we can say that the impulse is positive.
The partial wave in string 1, caused by the impulse, does have a 180 degree phase shift relative to the original wave. It destroys a part of the original wave. We can call it a "negative" wave.
But when the wave in string 2 is absorbed back to string 1, it destroys the negative wave. No phase change to the original wave.
Waves in spacetime. Let us then investigate a virtual photon emission and absorption using waves in a volume of spacetime:
------ e- scattered scattered
\ \ ------ Dirac wave
\ \ (absorbed k and
\ \ returned to original)
k photon
\ \ e- scattered Dirac
\ \ wave (absorbed -k)
----------------
\ \ ---------------- e- original
\ \ ---------------- Dirac wave
\ \ k photon
^ t
|
-----> x
The first scattering can be understood backward in time: the photon k and the scattered Dirac wave arrive from the future, and the Dirac wave scatters from it (absorbs k backward in time) and becomes the original Dirac wave.
Run forward in time, we see the original Dirac wave emitting k, scattering from the photon it itself created.
In the second scattering, the scattered Dirac scatters from k which it itself emitted, and returns to the original wave.
When the Dirac wave scatters from a photon, it may well be that it gets a 180 degree phase shift relative to the original Dirac wave. Since there are two scatterings, the phase shift is canceled.
Feynman rules seem to forget that both scatterings cause the same phase shift.
Let us return to the vertex correction diagram. For classical waves, it is not possible that a small interaction with anything can flip the phase of a big wave. The Feynman rule which causes sign flips is incorrect for classical waves.
The error in the Feynman sign rule for the electron propagator? What is the error in the Feynman rule? Strings 1 and 2 cause an impulse on each other at bar A. The "negative wave" in string 1 has a 180 degree phase change. The Green's function associated with the impulse has positive and negative signs for various Fourier components in string 1. A negative sign marks a phase change.
But when the wave in string 2 is absorbed back to string 1, the absorption destroys the negative wave. The original wave is preserved intact.
Question. Can we allow a phase change for an individual Fourier component of the electron wave, as long as the sum of Fourier components does not have a phase change?
We must allow a phase change, e.g., in the double-slit experiment – but there the wave interacts with the wall.
In the vertex correction, the electron wave interacts with the proton. There is the momentum exchange q.
Hypothesis. No phase change in a Fourier component of the electron wave is possible if the electron does not interact with anything.
Anyway, the question in this special case is not relevant, because destructive interference cancels any change to the wave function. If the electron interacts with something, then a Fourier component can change its phase.
What does the vertex correction calculate in QED?
The "waves in spacetime" diagram does make sense, if we correct possible sign errors in Feynman rules. What does the Feynman integral calculate in the vertex correction?
k virtual photon
~~~~~~~~
/ \
1 / A B \ 2
scalar e- -----------------------------------------------
p -k + p | -k + p + q
|
|
| q virtual photon
proton+ -----------------------------------------------
If q is infinitesimal, then the diagram essentially is the self-energy diagram. The integral calculates something which does not have any effect. Virtual photons k have no effect on anything.
Let |k| be relatively small. Then the we could imagine that k and the line A originate from a hammer strike by the electron. It is enough to hit the electromagnetic field at relatively large time intervals, to keep the field roughly constant ~ 1 / r far away. Small |k| are associated with the far field.
1. The strike at the position marked by 1 in the diagram means an impulse both to the electromagnetic field and the Dirac field. The propagator for k tells how "big" is the Fourier component k in the "spectrum" created by the impulse.
2. The counter-impulse creates a wave in the Dirac field at 1. If we assume that the impulse is very large, then we can assume that the entire electron wave is recreated at 1. The electron propagator tells how big is the component -k + p.
3. The off-shell electron passes close to the proton, and the entire wave turns its direction, so that its momentum becomes -k + p + q. This is non-perturbative.
4. The second hammer strike at 2 creates a Fourier component p + q at a probability amplitude which is the propagator of -k + p + q. We assume a symmetry of the propagator from on-shell => off-shell and to the opposite direction.
5. The amplitude of the electron in A exchanging a virtual photon k with the electron in B is the square of the coupling constant times the propagator for k.
The "flux" through the process above. The flux of the electron wave is the product of the propagators for -k + p and -k + p + q. Let us denote a propagator by G.
The propagator of the photon k tells how big is that Fourier component. The product
G(-k + p) * G(-k + p + q) * G(k) * e²
tells the probability that the photon k interacts with both the initial electron and the final electron wave.
Electron scattering from a photon
photon ~~~~~ ~~~~~~~~~
\ /
e- ---------------------------------------
A
There is only one electron propagator in the diagram above, for A. Why would we have two propagators in the vertex correction diagram, if we assume that q is non-perturbative, or q = 0?
A possible reason is that the in the above diagram, the electron can spontaneously decay into the on-shell state at the end of A. It can do so on its own, by radiating a photon.
In the vertex diagram, the electron absolutely needs the photon k to return back to an on-shell state.
The Feynman vertex diagram does describe the "far electric field" of the electron "detaching"?
Let us, once again, look at the sharp hammer model of the electric field. Let the electron initially be static.
| \ \ k from hit 2
| \ \
| \ \
| destructive interference
| \ \
| \ \
| \ \ k from hit 1
|
^ t •
| e-
-----> x
The Fourier components from the hits of the hammer will have a perfect destructive interference from earlier hits, with the exception of k which have pure spatial momentum. The electric field is static.
/
/ --> a acceleration
/
|
If we start to accelerate the electron, say, to the right, then a part of the k from early hits are left "dangling" – they are not completely destroyed by interference.
Maybe we can interpret this that the far field of the electron is "detached"? For large |k|, destructive interference is essentially complete, even though the electron is accelerating. The detachment only happens for small |k|.
Once the far field is detached, it forms an electromagnetic wave which is independent from the electron. The electron can later absorb that wave again. The Feynman diagram and the integral describe the reabsorption process.
Let F(q²) be the integral. In QED,
F(0)
is "renormalized" to be zero. Only the difference
F(q²) - F(0)
matters.
In the sharp hammer model, destructive interference totally cancels F(0). No waves k are sent at all.
In the sharp hammer model, we are interested in that part of various k that is not destroyed by interference. If the electron makes a sharp turn around a proton, and |k| ~ |q|, or |k| is larger, then a large part of k will survive destructive interference. This is manifest in the electromagnetic radiation sent by the electron. The spectrum of the radiation tells us what size |k| survived interference.
We still need to understand why does the Feynman integral describe the classical vertex correction right. On October 22, 2025 we calculated for a hydrogen atom that the Feynman integral does, very roughly, yield the correct value for the classical effect. We also observed that since both effects scale ~ q², the Feynman value will agree with the classical value for a wide range of q.
Let the electron pass the proton at a distance such that the momentum exchange is roughly q. Why should we sum the probability amplitude of the Feynman vertex correction to the elastic scattering amplitude? Certainly, we will sum overlapping probabilities that way. An electron which goes at the right distance from the proton will receive the momentum q anyway. It may also take part in the vertex correction. We are double counting probabilities.
Only if we knew that the vertex correction diagram will take the electron a lot closer to the proton, then we would know that the probabilities do not overlap, and summing the amplitudes is correct.
The Feynman bremsstrahlung diagram calculates bremsstrahlung roughly correctly. It is a good hypothesis that the vertex correction diagram correctly describes the absorption of a part of the "detached field" back to the electron. The other part flies off as electromagnetic radiation.
When the electron accelerates to a certain direction, it has to "drag" its far field along. It is a good hypothesis that the k absorbed by the electron later will drain from the electron the momentum requires to accelerate its far field to move along the electron. That is, the k that the electron absorbs, slow down the speed of the electron. This means that the electron moved "too fast" toward the proton as it passed the proton. The vertex diagram, thus, does describe some more electrons scattering from the proton, with the momentum exchange q.
The process described in the previous paragraph can be interpreted as an inertial mass reduction of the electron, as the far field does not follow it instantly. This is the connection between the classical vertex correction and the Feynman vertex diagram.
However, if the mass reduction is very small, then the vertex diagram will have a lot of overlapping probability with elastic scattering. Does the Feynman integral take into account this?
The Feynman integral has an infrared divergence, which is handled by assigning the photon a small fictitious mass λ. Thus, there is no unique Feynman vertex correction. It can be tuned to almost any value by manipulating λ. In this blog we have noted that the Feynman integral does not understand the fact that many small photons k are emitted during the same flight past the proton. That is the origin of the infrared divergence.
The Feynman integral must be tuned. We already calculated on October 22, 2025 that through suitable tuning, the integral will match the classical vertex correction.
We have come quite close to showing that the Feynman vertex diagram actually calculates the classical vertex correction. We also explained why it may be able to calculate the classical process.
Classical vertex correction always has mass reduction – why is the probability amplitude of the Feynman vertex correction so small?
Classically, when an electron passes a proton, the inertial mass of the electron is reduced for every such electron. The Feynman vertex correction integral has a probability amplitude which is much smaller than the elastic scattering amplitude. Vertex correction has an additional photon line and three additional propagators. Therefore, its amplitude is much smaller.
How can the Feynman vertex correction calculate the same effect as the classical process, if the Feynman diagram claims that the mass reduction only happens in rare cases?
Our Quantum Magnification Hypothesis from November 4, 2025 explains this. Quantum mechanics has problems describing spatially small processes (high resolution) since a small detail involves a small wavelength, which requires a lot of energy, and that energy is not available.
To overcome the resolution problem, quantum mechanics in most histories ignores the small detail, but in rare cases (often 1/861 of all cases) allocates a lot of energy to create the detail.
In the case of the vertex correction, |k| ~ |q| is a very large virtual photon. It cannot occur in every fly-by of the electron.
We can now explain why QED sums the vertex correction amplitude to the elastic scattering amplitude, and does not need to worry about double counting of probabilities. The few cases covered by the vertex correction diagram have a very large inertial mass reduction to the electron. Then the electron will come much closer to the proton, and the history studied in the vertex correction (which exchanges the momentum q) is unlikely to overlap with elastic scattering histories which exchange the momentum q. The impact factor b in a vertex correction history is much larger than the impact factor in elastic scattering histories.
In the classical vertex correction, every electron has a small inertial mass reduction.
In the QED vertex correction, most electrons have no inertial mass reduction, but a few electrons have a very large inertial mass reduction.
Discussion
We did not prove that the Feynman vertex correction integral calculates the classical process – but we argued heuristically that it does just that. Our numerical calculation on October 22, 2025 proves that numerically, our claim is true.
What implications does our observation have?
In the classical limit, the classical vertex correction must be the "true" nature of the process.
We can argue that the regularization and renormalization of the ultraviolet divergence in the QED vertex correction are not needed, and are actually wrong for a classical process. The QED ultraviolet divergence comes from ignoring destructive interference, which cancels large |k|.
The infrared divergence in the classical vertex correction is just the fact that the electron during every fly-bys creates an infinite number of virtual photons k. The divergence is the result of a wrong interpretation of Feynman diagrams. Feynman diagrams can calculate overlapping probabilities.
At the beginning of this blog post we calculated that the classical vertex correction and bremsstrahlung roughly cancel each other's effect in the scattering amplitude.
The sign of the vertex correction
We still have to analyze why the Feynman diagram gets the sign of the vertex correction wrong, after regularization and renormalization. Literature says that the correction reduces the scattering amplitude, while we know that the classical vertex correction increases it.
Peskin and Schroeder (1995) write about the vertex correction regularization and renormalization. Let us check how do they arrive at the overall sign of the correction.
The amplitude of scattering depends on Γ^μ( p', p ). It comes from the sum of diagrams: the elastic scattering diagram, the one-loop vertex diagram, vertex diagrams with many loops.
The electric form factor F₁(q²) is used in the calculation of electron-proton scattering.
Peskin and Schroeder use the Pauli-Villars regularization to remove (regularize) the ultraviolet divergence in the Feynman integral of the one-loop vertex diagram. Alternatively, we could set a sharp cutoff |k| < Λ. They denote by δF₁ the contribution from the one-loop diagram (first order correction).
Above, the authors calculate the electric form factor from the formula:
F₁(q²) = 1 + [ δF₁(q²) - δF₁(0) ].
The idea is that, for q infinitesimal, F₁(q²) should be 1. It is "renormalized" to 1.
But why do they subtract δF₁(q²) - δF₁(0), and not δF₁(0) - δF₁(q²)?
Destructive interference
Let us use the sharp hammer model for a while. We study the electron in the hydrogen atom, just like on October 22, 2025.
The electromagnetic wave, bremsstrahlung, in the classical treatment gives us a good view about which k are destroyed by destructive interference.
--> 0.008 c
e- • -----------
\
v
● r
proton+
The electron turns 90 degrees around the proton. The radius is r = 5 * 10⁻¹¹ m. The momentum exchange
q = 4,000 eV * c,
whose de Broglie wavelength is the length of the electron orbit, 2 π r.
A typical bremsstrahlung photon should have the wavelength
2 π r / α = 137 * 2 π r,
where α is the fine structure constant.
Destructive interference will destroy almost all photons whose wavelength is shorter. If the electron does not make a full circle around the proton, then the Fourier decomposition of the bremsstrahlung wave will contain many photons of longer wavelengths – actually an infinite number.
Let k be the 4-momentum of a typical bremsstrahlung photon. A reasonable cutoff, because of destructive interference, might be 2 |k|.
The cutoff is a small fraction of |q|, only
2 * |q| / 137.
This is reasonable, because it takes a lot of time to exchange the momentum q. The orbit is smooth. If the electron would be "mildly relativistic", its velocity 1/2 c, or so, then a reasonable cutoff would be |q|.
The classical vertex correction is, obviously, very much associated with bremsstrahlung, because both processes involve the far electric field of the electron "detaching".
We conclude that, in the classical correction, the typical relevant k in the hydrogen atom case has |k| ~ |q| / 137.
The Feynman way of calculation is that a small k in every fly-by of the electron is replaced with a very large k in a small fraction of fly-bys. Therefore, the typical relevant k in the Feynman integral has |k| ~ |q|.
The renormalized Feynman vertex correction integral
The integral seems to employ the "quantum magnification" principle: if every fly-by of the electron in the classical case involves a small |k|, switch that to a model in which most fly-bys do not have any k, but a few fly-bys have very large |k|. This works if the correction is linear in |k|. The classical vertex correction is linear in the mass reduction.
But why does the Feynman integral calculate
δF₁(q²) - δF₁(0) ?
If a certain k is absorbed back to the electron at the end of the process, it contributes to the integral (amplitude) behind δF₁(q²).
If q² < mₑ², then the integral above has a negative value. Fewer k are absorbed back if q ≠ 0. This contradicts the classical vertex correction which increases the scattering amplitude.
The correct way to calculate the vertex correction, employing the Feynman integral to the classical process is this:
1. Cut off the large |k| which are destroyed by interference.
2. Calculate δF₁(q²). That increases the scattering amplitude.
Why would the difference δF₁(q²) - δF₁(0) give anything meaningful?
It can give meaningful results, if the difference, for some reason, has roughly the same absolute value as item 2. That seems to be the case, since the classical vertex correction and the QED vertex correction have roughly the same absolute value, though the sign differs.
Let F be the integral not destroyed by interference. Let q grow from an infinitesimal value. If the "disturbance" from q reduces F to a half, then the difference and item 2 have the same absolute values. This seems to be the case. This probably reflects the fact that bremsstrahlung and the vertex correction are like "mirror images" of each other. Let us investigate.
The far field of the electron is the same regardless of the momentum exchange q it is going to experience close to the proton.
Let the fly-by last a fixed time t. We vary the momentum exchange q.
The Fourier transform of a bump function decays exponentially for large |k|. This suggests that we can ignore k frequencies which have a cycle time much shorter than t.
--> p
--> 0.008 c
e- • -----------
\ | q
v v
● r
proton+
Let us assume that we have removed those k which are destroyed by interference. Let us assume that |q| << mₑ and that the spatial momentum of the electron is |p| ~ |q|, just as in the case of the hydrogen atom on October 22, 2025. A reasonable cutoff might be
|k| < 2 |q|.
The "interesting" part of the Feynman integral for δF₁(q²) is those k for which |k| ~ |q|.
The Feynman integral for δF₁(q²) might be seen as "projecting" various k to p + q, which is the spatial momentum of the electron after scattering.
If p = 0, then we project various k to q. The "uncertainty", or the standard deviation, in the distribution of k is very crudely ~ |q|.
The "average" of k is zero.
The probability density function of the normal distribution at 0 is roughly 1.5X its value at 1 standard deviation.
This heuristic argument may explain why
|δF₁(q²) - δF₁(0)| ~ |δF₁(q²)|,
once we have removed those k which are destroyed by interference.
The above relation would explain why the renormalized QED vertex correction, on the left, has the wrong sign, but its absolute value is roughly correct.
Another way to explain why the above relation holds when we remove k destroyed by interference: the Feynman integral for |k| < 2 |q| is significantly "disturbed" by q ≠ 0 because q affects in a major way how much off-shell the electron goes from k if k is purely spatial momentum. But a significant portion of the integral for the value q = 0 still remains. It is a good guess that the integral value is halved when we let q grow from 0.
Numerical values of δF₁(q²)
Let us try to find values for the Feynman integral δF₁(q²) with a reasonable cutoff, like |k| < 2 |q|.
There probably are no such values calculated in the literature. We would need to use a numerical Monte Carlo software package to do the calculation.
--> p spatial momentum vector
e- • -----------
\
v
|
v p' spatial
momentum
● vector
proton+
^ y
|
------> x
Let us use the Klein-Gordon propagator for a "scalar electron", set the initial spatial momentum of the scalar electron p along the x axis, and the final momentum p' = -|p | along the y axis. Then
q = (-p, -p).
Let k be pure spatial momentum. The interesting part of the Feynman vertex integral is:
I(q²) =
|k| < 2|q|
∫ d⁴k 1 / ((p - k)² - p²)
0
* 1 / ((p' - k)² - p'²)
* 1 / k².
That is, k modulates how much the scalar electron propagators are off-shell.
Let p = (1, 0). Let p' be either (1, 0) or (0, -1).
Let us check what is the value of the product on the right when k is (-1, 0), (1, 0), (0, -1), or (0, -1).
For p' = p, then the product is always positive, and we graphically have:
1
1/9 1
1
For p' = (0, -1) we have:
1/3
1/3 -1
-1
Let us then try values k = (-1/2, 0), etc. For p = p':
64
64/25 64/9
64
For p' = (0, -1):
64/5
64/5 -64/3
-64/3
For very large |k|, the product values are
≈ 1 / k⁶,
regardless of p'.
These few test values suggest that the integral value changes a lot when we turn p' to point to the negative y direction. The integral might even turn negative.
If we would turn p' less, then the integral, obviously, would remain positive.
In our toy model with a scalar electron, the integrals |δF₁(q²)| and |δF₁(0)| are of the same order of magnitude. In this toy model, the hypothesis
|δF₁(q²) - δF₁(0)| ~ |δF₁(q²)|
might be true, though we did not prove that.
Summary
Let us recapitulate why the QED vertex correction roughly calculates the classical vertex correction. It is not a random coincidence that the Feynman integral roughly matches the classical number. The Feynman integral is "intended" to calculate the classical phenomenon.
1. The classical vertex correction reduces the effective mass of the electron during every fly-by past the proton. The reduction is small.
2. The Feynman integral uses the "quantum magnification principle" and transforms this to no effect during most fly-bys, but a very large effect during a few fly-bys.
3. When the electron absorbs the k at the end of the Feynman diagram, it must give up kinetic energy and momentum to the far field of the electron, to get the far field up to the current velocity vector of the electron (after absorbing q). This implies that the electron had "too much momentum" toward the proton before absorbing k. The electron came significantly closer to the proton than if we would assume that the far field is rigidly "attached" to the electron.
4. In the a vertex history, the electron moved a lot closer to the proton. Therefore we can add the amplitude to the simple elastic scattering amplitude. The vertex correction makes the scattering amplitude larger, not smaller as the standard QED claims.
5. In the classical calculation using the Feynman integral, we first remove large |k|, because destructive interference cancels them exponentially. Then we calculate the Feynman one-loop vertex integral denoted by
δF₁(q²).
The standard QED regularizes the Feynman integral of the vertex correction. Then it calculates
δF₁(q²) - δF₁(0).
That is, it calculates the change of the integral from its value δF₁(0) at an infinitesimal q. In the section above, we argued that the change of the integral value may well have roughly the same absolute value as the relevant integral δF₁(q²) itself:
|δF₁(q²) - δF₁(0)| ~ |δF₁(q²)|.
On October 22, 2025 we numerically checked that the QED vertex correction for small q², after a reasonable infrared cutoff, has roughly the same absolute value as the classical vertex correction. The sign is wrong, though.
The infrared divergence in the QED vertex correction is a result of a misinterpretation of the Feynman diagram: the Feynman integral calculates overlapping probabilities. There always are an infinite number of infrared photons involved, just as in the classical vertex correction.
We do not need the "long-lived virtual photons" hypothesis
In our previous blog post we presented a hypothesis that
δF₁(0) - δF₁(q²)
represents "long-lived virtual photons", which cause the vertex correction. But in the Summary above, we do not need that hypothesis. Instead, we assume that after removing those k which are destroyed by interference, what remains of δF₁(q²) is roughly equal to δF₁(0) - δF₁(q²). The hypothesis of long-lived virtual photons has a major weakness: we do not have any mathematical formula which would explain how those long-lived virtual photons behave.
Conclusions
If the above logic holds up, this is a major breakthrough in quantum electrodynamics. The "true" vertex correction is a classical process. Regularization is not needed because destructive interference exponentially cancels high |k| in a classical process.
The infrared divergence is not real. The classical process involves an infinite number of infrared photons in every scattering.
Renormalization is not needed, either. The correction is from the dynamics of the electric field of the electron. It has nothing to do with the charge of the electron. Just like bremsstrahlung is from the dynamics of the electric field.
We showed that the divergence problem in the QED vertex correction are flaws in the numerical approximation method. The problems are not in the physical system. People trying to find a deep physical meaning in the divergences (in renormalization flow, effective field theory, string theory, etc.) made a fundamental error: confusing bad mathematics with physics.
In this blog post we also argued that the electron self-energy integral is nonsensical. We are not sure if the sign rules of Feynman diagrams need to be changed.
