Classical vertex and bremsstrahlung corrections in scattering

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

https://en.wikipedia.org/wiki/Larmor_formula

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.

https://physics.stackexchange.com/questions/521102/electric-energy-from-dipole-moment

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?

Electron self-energy: no phase change should be 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.

This is a serious error in Feynman rules. Is it so that the vertex correction integral does not calculate anything sensible?

The electron self-energy calculation is nonsensical in QED, as we have noted before. An emission and an absorption of a virtual photon cannot change anything. The correct integral calculation would tell us that nothing changes.

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.

***  WORK IN PROGRESS  ***

QED vertex correction versus classical vertex correction: contradiction?

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?

https://www.southampton.ac.uk/~doug/qft/aqft3.pdf

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

https://www.southampton.ac.uk/~evans/QFT3/qft3.pdf

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.

https://en.wikipedia.org/wiki/Negative_mass

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.

Many-worlds approach to QED, and other aspects

UPDATE August 20, 2026: literature seems to indicate that the Feynman vertex correction reduces the scattering cross section. That is in contradiction with the classical vertex correction. Is the Feynman vertex correction erroneous?

----

UPDATE August 20, 2026: Our analysis about which k contribute to the vertex correction in QED was way too simplicistic. In the next blog post we make the analysis based on literature.

----

In our previous blog post we noted that the action usually presented for quantum electrodynamics (QED) in unclear. What kind of an electromagnetic field is associated with an arbitratry Dirac field wave?

Also, how does the electromagnetic field give rise to a Dirac field wave in pair production? The Bhabha calculation (1935), and the Feynman diagram seem to create the Dirac field wave from "nothing". Bhabha assumes that a positron with a 4-momentum E₊, p₊ is around, and calculates its scattering first to an off-shell electron, and then to an on-shell electron. Why can Bhabha assume that such a positron is around?

A many-worlds approach might make the strange appearance of the positron understandable. Also, it could make the histories such that they are time-symmetric, and obey a typical action of interacting fields.

Bhabha (1935): positron scatters from Z+ and Z-

           Z+   -------------------------------------------------

                            \  virtual photon

                              \ ______________________ e-

                               |                                        E, p

                               | off-shell electron  

                               |______________________ e+

                              /                                         E₊, p₊

                            /  virtual photon

           Z-    ------------------------------------------------

Bhahba calculates backward in time. He assumes a positron wave meeting a massive charge Z+, scattering into an off-shell Dirac wave, and then that off-shell wave scattering from a massive charge Z- in such a way thaf the wave becomes an on-shell electron e-.

Bhabha calculates probability amplitudes for processes which correspond to a positron E₊, p₊, and an electron E, p.

We then have a set of histories in which a positron with various E₊, p₊ scatters into an electron E, p. But something seems to be missing from a time-symmetric picture. What if the positron wave traveling back in time fails to scatter from Z- and Z+ in the described way? The same for the electron: if we calculate backward in time, most electrons should not scatter from Z+ and Z-, and should not become positrons.

Building histories backward in time: many worlds branch also backward in time

     Z+ -------------------------------------------------

                            \ virtual photon

     e+   __________\

     E₊, p₊                 |

                                | off-shell electron  

                                |__________________ e+

                              /                                    E'₊, p'₊    

                            / virtual photon

       Z- ------------------------------------------------

Above we have another scattering process. This time, the positron moves forward in time. It may scatter from Z+, and later from Z-. But most positrons may not scatter at all. Some might scatter backward in time.

It looks like the traditional way of understanding the diagrams cuts off scattering alternatives, which should exist in a time-symmetric interpretation of the diagram.

And time symmetry is needed to make the diagram to correspond to a unitary development. A typical action describes an unitary process.

The Bhabha approach is to build a history which looks nice. He builds it toward the future (though he lets the positron move backward in time for a while). An even nicer history can be built if we can also build backward in time. For this to succeed, and make sense, we need many worlds.

Our branch is such that the only particles entering the reaction are Z+ and Z-. But in other branches going backward in time from the reaction, e+ or e- are outgoing particles if we look backward in time.

The scattering of a Dirac wave from a heavy charge Z+ or Z- can be calculated in a nice way. It is like classical fields. We can calculate the behavior of the wave in space and time coordinates. There is no need to use the "momentum space" of Feynman integrals.

What is the phase of the Dirac waves in pair production?

Different histories produce pairs with different 4-momenta. Destructive interference cannot happen if the histories can be distinguished with a measurement.

We still have the problem: what are the phases of the produced Dirac waves of the electron and the positron? They must have a phase, or must they? Do we have an absolute way to determine what is the phase of an electron? Electrons obey the Pauli exclusion principle. There cannot be a coherent wave of electrons with the same 4-momentum, at least not in a small spatial volume.

For an electromagnetic wave, we can measure its phase, at least if we have many coherent photons.

A typical classical action determines the phase of all the waves it produces. If we have two plates A and B attached with a rubber block, then a wave in A may produce a wave in B, and the phase of the wave in B is deterministic.

                |        #

                |        # ----------  sharp hammer hits

                v       \/

         ----------------------------  metal plate

Hypothesis. If a pair is produced in a small region of spacetime, compared to the wavelength of the produced Dirac waves, then we can assume that all produced Dirac waves have the same phase at their production point. This is the result of a hit with a "sharp hammer": all the produced waves have constructive interference at the point of the hammer hit.

The hypothesis is true for classical electromagnetic waves produced, when two charges Z+ and Z- pass close to each other. It is a sharp hammer hit to the electromagnetic field.

However, there is no obvious reason why the usual QED action would force the Dirac wave to satisfy this. We again encounter the insufficiency of the action.

Hypothesis of inheriting the phase from other particles. Since the produced pair gets its energy and momentum from the colliding particles, we could define that the phase of the combination e- & e+ "inherits" its phase from the particles which donate the energy and the momentum to the pair. Furthermore, we can demand that the phase of the electron and the positron must be continuous over the path of e+ -> off-shell electron -> e-.

           Z+ ----------------------------------------------------

                                   \ virtual photon

                                     \ ___________________ e-

                                    /                            E, p

                                  / off-shell electron  

                                /______________________ e+

                              /                                  E₊, p₊

                            /  virtual photon

           Z- -----------------------------------------------------

Let us analyze the diagram above. Do we obtain reasonable phases for e-, e+?

Suppose that both e- and e+ contain a total energy E which they inherited from Z+ and Z-. The phase α of E is determined by Z+ and Z-. Does this determine the phases of e- and e+?

Imagine that the path of e+ -> off-shell electron -> e- is a metal cylinder. The cylinder is bent. Let us roll the cylinder slightly clockwise at e+. Then it will roll anti-clockwise at e-. The sum of the angles stays the same. Unfortunately, we were not able to find a unique way to determine the phase of e- and e+. The Hypothesis does not work.

Let us assume that Z+ is static. Then it cannot give any energy to the pair. We could demand that all the energy E in the pair has the phase determined by Z-.

Since Z+ is static, it cannot donate any momentum either. It can only absorb momentum. We could demand that e+ inherits its phase from Z-. This fixes also the phase of e-.

If Z- has a macroscopic mass, then its phase varies extremely rapidly. It is not clear what phase should the virtual photon sent by Z- inherit, or e+ inherit. This does not look too nice.

Suppose that we have a system which initially has a 4-momentum E, p. We ignore the spatial extent of the system. Its phase varies in time by the formula

       exp( -i / ħ * (E t - p • x) ).

Then it emits particles e+ and e-, with 4-momenta E+, p+ and E-, p-. Let us define

       E'  =  E  -  E+  -  E-,

       p'  =  p  -  p+  -  p-,

Let us assume that the emission happens at the time t = 0 at the location x = 0. After that, it is natural to assume that the phase of the rest of the system obeys the formula 

      exp( -i / ħ  * (E' t  -  p' • x) ),

the positron e+ obeys

      exp( -i / ħ  * (E+ * t  -  p+ • x) ),

and the electron e- obeys

      exp( -i / ħ  * (E- * t  -  p- • x) ).

The product of the phases obeys the same formula as before the reaction.

Vacuum polarization

    Z+ --------------------------------------

                              | 

                              | q virtual photon

                              |

                             / \

                           /     \  k + q virtual electron-

                    -k    \     /  positron pair e- e+

                             \ /

                              |  

                              | q virtual photon

                              |

      Z- ---------------------------------------

In the phase inheritance hypothesis in the previous section, the idea is that the phase of the entire system develops according to the old formula, even if it splits into a larger set of particles.

Does the inheritance hypothesis say anything about the phase of the virtual pair particles in the vacuum polarization loop?

The virtual photon 4-momentum |q| might be very small, but |k| may be very large. Why would the phase of k be dependent on q? In the diagram above, we can assign any phase to the 4-momentum k, as long as -k has the opposite phase. Then their product is 1, and they do not affect the product of phases.

As we have noted many times in our blog, the diagram above actually summarizes what does not happen. The Feynman integral is infinite, and if  |q| is infinitesimally small, then regularization removes the integral entirely. We should find reasons why the history above does not happen, even though a superficially similar pair production diagram produces pairs.

Let us apply the reasoning which is used in pair production:

1.   We are allowed to assume that a flux of positrons (of electrons) exists, with a 4-momentum k.

2.   The positron scatters from Z-, acquiring an additional 4-momentum q.

3.   The positron scatters from Z+, donating a 4-momentum k to Z+.

4.   The positron returns to its original state with a 4-momentum k. Therefore, we are allowed to assume that the history "really" happened.

Items 1 and 4 differ drastically from the corresponding principle in pair production. In pair production we have the principles:

1'.   We are allowed to assume a flux of real positrons with a 4-momentum E+, p+ exists.

4'.   The positron becomes a real electron, and escapes. Therefore, we are are allowed to assume that the history "really" happened.

In Feynman integrals, anything that conserves 4-momentum is a history that "really" happens. Divergences prove that that rule is wrong.

Long-lived virtual photons in the vertex correction

                                       k  virtual photon

                                   ~~~~

                                /             \

               e-  -------------------------------------

                              1      | 

                                      | q virtual photon

   proton+   -------------------------------------

In our blog post on October 22, 2025 we suggested that the subtracted probability amplitude of the electric form factor

       F₁(0)  -  F₁(q²)

represents long-lived virtual photons which reduce the effective inertial mass of the electron which scatters from the proton. We have a very good classical analogy for this: the inner field of the electron follows the electron closely, and does not reduce its inertial mass. But the far fields "falls off" for a while, and does not contribute to the inertial mass.

Interpretation of the vertex Feynman diagram. We can interpret the diagram in this way: the "hammer hit" (impulse) at the position 1 always produces the same full spectrum of virtual photons, regardless of q. That is, F₁(0) is the "response" to the hammer hit.

The Feynman integral calculates which of these are absorbed so quickly that they have no effect on the process. The missing part of the integral is those virtual photons which live long and are only later absorbed back to the electron. The long-lived photons reduce the inertial mass of the electron, and contribute to the scattering amplitude.

Let us model the electric field with an elastic rubber plate attached to the electron. When the electron is under an acceleration, energy flows to the rubber plate, to distort and bend the plate. Less mass energy is left in the electron. This is another way to explain the reduced inertial mass.

If the electron is static, the Fourier decomposition of its electric field only contains time-independent waves, whose energy E is zero:

      exp( -i / ħ  *  (E t  -  p • x) ).

We could say that the electric field of a static electron has "zero energy".

Long-lived off-shell pairs in vacuum polarization

              e- --------------------------------------

                                      |  q

                                    / \

                                  /     \               vacuum

                            -k   \     /  k + q    polarization

                                    \ /

                                      | q virtual photon

  proton+ ---------------------------------------

On November 4, 2025 we suggested that in vacuum polarization, the subtracted probability amplitude

       Π₂(0)  -  Π₂(q²)

represents long-lived off-shell pairs which exist for a long enough time, so that they will reduce the energy in the electric field between e- and the proton, and effectively make the attractive force between e- and the proton stronger.

Interpretation of the vacuum polarization diagram. Let us assume that the electron passes the proton at a distance r such that the momentum transfer is q. The value q measures the "disturbance" of the electric field between the electron and the proton. The disturbance q hits the Dirac field "with a hammer". An off-shell pair may be born. The hammer hit is represented by Π₂(0).

The Feynman integral Π₂(q²) calculates those off-shell pairs which annihilate quickly and have no effect, except relaying the momentum q to the proton. That is, those pairs do not affect the scattering amplitude at all.

The remaining off-shell pairs are long-lived, and they add to the scattering amplitude. The difference

       Π₂(0)  -  Π₂(q²)

tells us how much they add. Why do they add to the scattering amplitude? Because they can relay the momentum exchange q between the electron and the proton, even if the particles pass at a distance > r.

A standard principle in interpreting Feynman diagrams is that each different diagram with the same input and the same output adds to the scattering amplitude of a reaction. Why should it be so? If the second diagram "eats the market share" of the first diagram, then that principle does not need to hold.

Another way to explain why Π₂(q²) has no effect on anything: if the electron is moving at a constant velocity, then Π₂(0) represents a "hammer strike" which destructive interference from earlier hammer strikes completely wipes out (except maybe Fourier components with E = 0). As the electron passes the proton, then Π₂(q²) is that part which is destroyed by interference from earlier strikes. That is, Π₂(q²) has no effect on anything.

The Feynman diagram in the case of vacuum polarization describes the complement of the relevant process.

The action of QED for the vertex correction

Above, we wrote about hammer strikes. Let us try to interpret these in the framework of the (unknown) QED action.

If we assume that the electron is very heavy, and its electric field is very strong, then the vertex correction is a classical process. It should obey a classical action of electrodynamics.

Note that we argued on February 25, 2025 that Maxwell's equations fail for accelerated systems. Thus, we do not know the correct action for classical electromagnetism, either.

In the classical action, high frequencies, or large 4-momenta |k|, probably play no role. Destructive interference cancels them.

Why would the Feynman integral correctly calculate the effect of the reduced mass of the electron in the vertex correction?

The Quantum Magnification Principle of November 4, 2025 may explain that. The scattering of the electron happens in a very small volume of spacetime. If we try to model the mass-energy of the far field of the electron with a wave, that wave will have a very short wavelength, and a very large energy. To work around that, quantum mechanics assigns a large mass reduction in a small number of cases. Other cases have no mass reduction.

              e-  -------------------------------------

                                      | 

                                      | q virtual photon

   proton+  -------------------------------------

The bulk of the scattering amplitude comes from the plain Coulomb scattering diagram, above. There is no mass reduction.

In a small number of cases, the Feynman integral for the vertex correction diagram has the mass of the electron greatly reduced. Does the vertex correction diagram (integral) then calculate the added scattering amplitude right? The integral should calculate the difference which the mass reduction makes to the ordinary scattering amplitude. Does it?

The formula

       F₁(0)  -  F₁(q²)

calculates a difference. Could it be that F₁(0) represents the elastic scattering behavior? The subtraction then calculates the impact of adding the disturbance q to the process?

If that is the case, then if we add the integral of the elastic scattering diagram, and the integral of the vertex correction diagram, then we do not count overlapping probabilities twice.

On October 22, 2025 we calculated that the QED vertex correction in a hydrogen atom roughly agrees with the classical vertex correction. But we do not understand the details: how do the Feynman integrals manage to calculate an estimate for the classical vertex correction? In that post, we just presented a hypothesis that it calculates the classical effect.

                                    -k virtual photon

                              ~~~~~~~~

                            /                     \

                          /     A           B    \

               e-  -----------------------------------------

               p          p + k     |      p + k + q

                                        |

                                        |

                                        | q virtual photon

   proton+  -----------------------------------------

When we add the virtual photon k to the elastic scattering diagram, the Feynman integral also gain the electron propagators for the lines A and B as coefficients.

If q is infinitesimal, then A and B are symmetric.

If q is larger, it "displaces" the Feynman integral more.

Let us assume that k contains a large amount of positive energy. Adding the contribution of the Feynman integral is equivalent to saying that then the electron comes closer to the proton, and adds to the scattering amplitude. It is an additional history which contributes to the total amplitude.

The classical vertex correction is increased "Coulomb focusing" from a lower inertia of the electron

          e- • ------------------  

                                       \

            b                           \

                               ●

                    proton+

Let us assume that the electron is mildly relativistic, and scatters from the proton to a large angle. Then, using natural units, 

       |mₑ|, |q|, |k|

can be assumed to be of roughly the same order of magnitude.

Let

       r

be the classical impact factor b which gives the electron a pull q as it passes the proton.

When a classical electron approaches the proton, the attraction to the proton will make the impact factor b less when the electron is close to the proton. This increases the scattering amplitude, and is called Coulomb focusing. The elastic scattering Feynman diagram does not understand Coulomb focusing.

Classically, the inertia of the electron is less than mₑ close to the proton, because the far field of the electron does not have time to react.

Suppose that k takes a large portion of the energy of the electron away. Then, Coulomb focusing will increase the scattering cross section greatly, possibly twice or more. This proves that it may be reasonable to add the Feynman vertex correction integral to the elastic scattering cross section.

For very small |k|, the integral is small. Presumably, the main contribution to the integral comes from values |k| ~ |q|. In the classical analogy, most of the contribution to the scattering amplitude will come from electrons whose impact factor is > r. That is, they will increase the scattering amplitude.

We are interested in the number of electrons which scatter to > |q|. The elastic diagram gives us the first estimate. The vertex correction diagram gives an estimate of the extra amplitude we should add. We can add it because most of that extra amplitude comes from electrons with the impact factor b > r.

Is it reasonable to assume that if the inertial mass of the electron is halved, then Coulomb focusing will double the scattering cross section?

When the electron passes the proton, its angular momentum is conserved relative to the proton. A large scattering angle requires that the total energy of the (relativistic) electron is very crudely double close to the proton. Then its distance to the proton is only b / 2. Coulomb focusing is a very prominent phenomenon.

If the inertial mass of the electron is halved close to the proton, then it will accelerate faster toward the proton. It will pass the proton much closer. It is reasonable to assume that the cross section is double in such a case.

We conclude that for mildly relativistic electrons, adding the Feynman vertex correction integral to the elastic scattering integral makes sense.

Earlier in our blog we have shown that the electron anomalous magnetic moment correction, which Julian Schwinger in 1948 calculated from the vertex correction, agrees with the classical vertex correction if we assume that the electron in zitterbewegung makes a loop whose length is the Compton wavelength.

An mildly relativistic electron is very far from the classical limit. The bulk of the mass-energy in the electric field of the electron is close to its classical radius

       rₑ  =  2.8 * 10⁻¹⁵ m,

but its Compton wavelength is much larger,

       λₑ  =  2.4 * 10⁻¹² m.

We cannot "build" the electric field from waves which are 861 times longer than the most prominent detail in the field. Thus, we cannot expect that the classical vertex correction in this case is anything close to the quantum vertex correction. Classically, an electron scattering to a large angle passes the proton at a distance rₑ. Then the mass-energy of the far field is close to mₑ, and the classical vertex correction could easily be 100% to the scattering amplitude. Looking at the literature, the QED vertex correction seems to be much smaller, on the order of 1/861.

Comparing the classical vertex correction and the QED vertex correction in a hydrogen atom

On October 22, 2025 we calculated that if an electron passes a proton at the speed (0.008 c) and the distance (5 * 10⁻¹¹ m) as in a hydrogen atom, then the classical vertex correction and the QED vertex correction roughly agree. In this case, the electron and its field can, maybe, be considered an almost classical object. But how does the Feynman integral manage to calculate the classical phenomenon?

                                         -k virtual photon

                              ~~~~~~~~~~

                            /                           \

                          /   A                   B    \

               e- -----------------------------------------

               p          k + p    |   k + p + q

                                        |

                                        |

                                        | q virtual photon

   proton+ -----------------------------------------

                    p

           e-  • -----------------

                                        \         

                             ●          v

                        proton+

                                          | p + q

                                          v

The 4-momentum q is sufficient to turn the path of the electron by 90 degrees.

The Feynman integral for vertex correction mostly consists of contributions in which

       |k|  ~  |q|.

If the electron emits such a virtual photon, and the photon survives for a long time, then the emission substantially affects the path of the electron. But why would all such emissions contribute positively to the scattering amplitude?

If q is infinitesimal, then, obviously, the vertex correction integral cannot affect the scattering amplitude. The integral describes a symmetric process in which the electron sends a photon k to an arbitrary direction. The difference

       F₁(0)  -  F₁(q²)

describes how much does the integral change from the q infinitesimal case. Could it be that the entire change in the integral adds to the scattering amplitude?

https://en.wikipedia.org/wiki/Propagator

The Feynman integral contains coefficients from the electron propagator:

       1 / (E² - s² - mₑ²),

where E is the total energy of the electron, s is its spatial momentum, and mₑ is its mass. The denominator essentially calculates how much the electron is off-shell.

The 4-momentum p is on-shell, as well as p + q.

The 4-momentum k takes the electron off-shell. Then the absolute value of the integral grows smaller, since with p or p + q it is infinite.

Let us denote

       k = (E, s).

If s is parallel to the spatial momentum of p + q, then the propagator for k + p + q is significantly smaller than the propagator for k + p? The velocity in k + p + q is significantly larger => it is more off-shell.

If s is antiparallel to the spatial p + q, then the velocity in k + p + q is smaller. It is off-shell. But about as much off-shell as k + p. The integral value changes its sign, though.

The integral grows smaller in the case that k is rougly parallel to spatial p + q? And in this case, the electron moves closer to the proton, so that the impact factor b becomes smaller. This makes sense.

The Feynman vertex correction integral has an infrared divergence when |k| becomes small. This is because of the fact that, to describe the behavior of the far field of the electron, we need photons of an arbitrary wavelength. It is just like in the case of bremsstrahlung: classical bremsstrahlung contains an infinite number of long-wavelength photons. Feynman diagrams do not understand that all these photons are emitted simultaneously, and counts the same probability multiple times. This causes the infrared divergence.

https://www.southampton.ac.uk/~doug/qft/aqft3.pdf

The paper in the link gives for q² << mₑ²:

On October 22, 2025 we already calculated  that the Feynman vertex correction yields the same numerical result as the classical vertex correction in the hydrogen atom case, if we choose the infrared cutoff, the small "photon mass" λ above, in a suitable way.

We thus know that the Feynman vertex correction can be tuned to imitate the classical vertex correction.

The classical vertex correction depends on the mass-energy of the far field of the electron. But the mass cannot directly affect the electron. All effects on the electron must happen through the Poynting vector. Could it be that the Feynman diagram approximates the Poynting vector behavior?

Let us have a sharp hammer which keeps hitting a rubber membrane at short intervals. Initially, the hammer is static. Then we start to accelerate it to the direction of a vector q. Could it be that the momentum which the hammer emits to the opposite direction of q, then can almost "break free", and is only much later absorbed back to the hammer?

Conclusions

Let us close this long post. We first investigated if a many-worlds model can help us to design an action for QED, such that it could explain pair production. Our results were inconclusive.

Next we wandered to study the vertex correction in QED. If the infrared cutoff (photon mass λ) is chosen in a suitable way, the QED vertex correction numerically agrees with the classical vertex correction. But it is still unclear to us why does it agree.

Our next post will be about the vertex correction. Why does QED agree with the classical phenomenon?