Type II supernovae explode because general relativity is wrong?

There is a major open problem in how type II supernovae are able to explode when their core collapses.

In the photo (middle right), we have the bright type II supernova SN 1987A in the Large Magellanic Cloud. The photo was taken with the ESO Schmidt Telescope.

https://en.wikipedia.org/wiki/SN_1987A

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

The leading hypothesis of a type II supernova is that the collapse creates a huge number of high-energy neutrinos, and an unknown mechanism makes the neutrinos to interact with the outer layers of the star.

The interaction then blows the outer layers to space.

Another hypothesis is that some kind of a shock blows the outer layers away.

https://arxiv.org/abs/astro-ph/0612072

H.-Th. Janka et al. (2006) discuss various models.

Is there a connection to dark energy?

In this blog we have remarked that the only large collapse/explosion which we can monitor in detail is the expansion of the universe, and it does not follow the FLRW model derived from the Einstein field equations.

Dark energy spoils things. The expansion seems to be speeding up, though it should slow down.

Similarly, the other example that we have of a large collapse/expansion, the type II supernova, fails to follow the path predicted by general relativity and particle physics.

In our blog we in May 2024 tentatively proved that the Einstein equations do not have a solution for a collapse or expansion. This opens the possibility that the hypothetical correct theory of gravity could explain dark energy. It might also explain the type II supernova explosion.

A longitudinal gravitational wave?

Let us model gravity with the traditional tense rubber membrane, on which we have weights resting.

If a set of weights "collapses" toward each other, and they collide, a longitudinal, circular wave will form, and carry some energy out.

If longitudinal waves are banned, like in electromagnetism, the no wave can form in a spherically symmetric collapse.

          --------             --------    rubber sheet

                     \         /

            v <--   • --- •   --> v

                     ring of 

                     weights

Let us assume that the initial state is an expanding ring of weights sliding on the rubber sheet.

                     bulge

                      ____

         ------ • --       -- • ------   rubber sheet

        v <---                 ---> v

In this blog we suggested on May 22, 2025 that retardation causes a "bulge" which can explain dark energy. Could it be that this phenomenon necessarily creates longitudinal gravitational waves?

Above we have a model of the expanding universe. The rubber membrane has kinetic energy upward. This kinetic energy should go to acceleration of the ring of weights outward. The kinetic energy of the membrane explains why the expansion speeds up.

If the bulge rises above the rubber sheet (membrane) level, we are in trouble. That would enable superluminal communication. We cannot allow that. Could it be that the expansion of the universe must be fine-tuned in such a way that all the kinetic energy of the membrane is spent to accelerate the masses outward?

If we run the process backward, the bulge never rises above the membrane level, since it starts to from the membrane level and starts descending. In the reversed process, there is no superluminal problem.

         -----                ----- tense rubber membrane

                 \          /

                   ••••••  weights

           --->   stop  <---

But what happens if the reverse process (collapse) suddenly stops and a neutron star is formed? Where does the kinetic energy of the membrane moving downward go?

If we would allow longitudinal waves, it would go to them.

What about a "transactional" model? Retardation must eventually be "settled" to ensure energy conservation. The energy could go to the kinetic energy of the weights. But what law would decide who gains the energy and how?

When matter is around, gravitons couple to it, and probably gain a mass. An analogue for electromagnetism is plasma. In plasma, longitudinal waves do happen. This implies that longitudinal gravitational waves probably are present inside a collapsing mass, and probably also close to it.

The plasma analogue may help us. There are no longitudinal electromagnetic waves in empty space. If we have an exploding ball of electrons, then longitudinal waves can happen inside the ball, but they cannot escape outside the ball.

Could it be that the outer layers of a star must absorb the outgoing longitudinal gravitational waves, and that causes the explosion outward?

             )       )       )      •   •   •   •  

      longitudinal         masses

      wave

      (stretched x metric)

      ---------------> x

Let us analyze a longitudinal wave which would periodically stretch distances in, say, the x direction. Suppose that there is some density of mass in space.

Is there a mechanism which would allow the masses to absorb and re-emit the longitudinal wave? A plasma, of course, temporarily absorbs some of the energy of an electromagnetic wave. But if the masses are sitting still, they will not gain any kinetic energy in the longitudinal wave. This means that masses will not allow longitudinal gravitational waves to propagate.

A plasma contains both positive and negative electric charges. But a mass density only contains positive gravity charges. These configurations are very different.

We do not see a mechanism which would allow longitudinal waves in gravity.

The energy of the curved metric inside a collapsing mass shell

Let us have a massive mass shell which collapses. Because of retardation, clocks inside the shell will tick "surprisingly fast". This because they do not yet know that the gravity potential is low.

That is, the metric inside the sphere will not be flat. People often assume that it would be flat, even if the collapse is a dynamic process, and clocks inside cannot "know" what the gravity potential is supposed to be right now.

Mass tends to move to the direction where clocks tick slower. In this case, mass would move toward the shell. Since the metric inside the shell can do work, it must "contain" energy.

Note that this breaks Gauss's law. Gauss's law would imply that there is no force inside the shell. Also, this breaks the Einstein field equations. Einstein says that the metric inside the shell should be flat. But it is not flat because a clock at the center ticks faster than clocks close to the shell.

What happens if the mass shell suddenly stops contracting? The metric inside the shell contains energy. Where does that energy go? It cannot escape as longitudinal gravitational radiation. Because of spherical symmetry, it cannot escape as transverse radiation either.

Since energy goes to the gravity field inside the shell, the shell will contract somewhat slower than in Einstein gravity.

Let us analyze a simpler configuration.

                 ● ---> a                a <--- ●

                M                                   M

Two masses M are accelerated toward each other, and quickly stopped. What does the retarded gravity field between them do to the masses?

The stopping force F gains energy from the system. If the stopping is done slowly, then the deformation energy of the gravity fields should make F to gain more energy.

Another way to interpret this is to consider the gravity field inside the sphere as a "spring" which is deformed by the accelerating collapse of the sphere. In the spring interpretation, the energy of the spring will push the sphere outward, when the collapse stops. In this model, there will be a shock which will make the sphere to expand. If the shock outward is passed to the outer layers of the collapsed star, that can explain the supernova explosion.

If the core collapse ends up as a neutron star, then there necessarily is some kind of a shock as the pressure outward wins the gravity. There is some kind of a "bounce back".

Our hypothetical "gravity shock" adds to this. But is this enough to explain the explosion?

The entropy of the deformed gravity field is low. The energy in it does not easily end up as heat. The collapse to a neutron star generates a lot of heat to the neutrons.

Energies involved in a supernova explosion

https://users-phys.au.dk/jcd/explosion/ott09/Ott_Aarhus_September2009.pdf

Christian David Ott (2009) writes about the "supernova problem". The gravitational energy released in a neutron star collapse is

       ~ 300 Bethe = 300 * 10⁴⁴ J.

The initial bounce back of the neutron star only has ~ 1.2 Bethe of kinetic energy. Almost all energy escapes in neutrinos as the neutron star cools. The collision of the matter, as it forms the neutron star, is almost perfectly inelastic.

But the energy required to throw the outer layers into space is ~ 12 Bethe. The energy of the bounce, 1.2 Bethe, is too small.

We have to calculate what is the energy of the retarded gravity field in the collapse.

The collapsing body is a sphere, not a shell. We have to figure out what happens in the case of a sphere.

How much is the energy of the retarded gravity field?

Hypothesis. The energy stored in the retarded gravity field may be approximately the "extra energy" of the mass inside the forming neutron star, because the mass "does not know yet" that its gravity potential has fallen.

How much does retardation distort the potential at the center?

The radius of a neutron star ~ 10 km.

The velocity of the surface at the bounce back, which happens at r = 10 km, is ~ 0.4 c.

We picked the value 0.4 c from some literature.

We assume a standard retardation rule: an observer "sees" the field as if the source of the field would have moved at a constant velocity. The surface of the collapsing neutron star is accelerated. Thus, the retarded view sees the "current" radius of the star larger than it actually is.

We assume that the density of collapsing matter is constant, and it is a free fall.

            --------------------------------------------->

                                           signal

     surface

           | ---> 0.32 c     |                            × center

           r = 16 km         r = 10 km

Suppose that the radius r was 16 km when a signal to the center left. The signal travels at the speed c. The speed of the surface is 0.32 c, and it will increase to 0.4 at r = 10 km.

The average speed of the surface between r = 16 km and r = 10 km is 0.36 c.

When the signal arrives at the center, the speed has grown to 0.4 c, and r = 10 km. The retarded view sees the radius too large by an amount

       (16 km / c)  *  (0.36  -  0.32) c

       = 640 meters.

The gravity potential at r = 10 km for a neutron star of 1.5 solar masses is

       -G M / r  = -6.7 * 10⁻¹¹  *  3 * 10³⁰ / 10⁴

                      = -2 * 10¹⁶

                      = -0.2 c².

The retarded view at the center sees the gravity potential

       640 m / 10 km  =  6.4%

too high.

The energy in the retarded gravity field would then be something like

       10³⁰ kg  *  0.2 c²  *  6.4%

        = 13 * 10⁴⁴ J

        = 13 Bethe.

We have assumed above that 0.5 solar masses (10³⁰ kg) is located in the volume which sees the potential too high.

We see that the energy of the retarded gravity field might cause a bounce back strong enough, so that the outgoing shock wave can explain how the outer layers of the star are blown to space.

But is there some reason why neutrino radiation does not carry away this 13 Bethe? The bounce, presumably, does not affect elementary particle reactions in the star. Then neutrinos should not carry away the energy.

Gravity almost certainly is retarded, and that breaks the Einstein field equations

        ● ---> a                      O 

       M                               clock of an observer

    static                            static

Suppose that we initially have a static large mass M and a static clock some distance away.

Let us start accelerating the mass M toward the clock. It would be very surprising if the rate of the clock would slow down before the clock "knows" that we started accelerating M.

Suppose that M is very far away from the clock. Suppose that the rate of the clock would change instantaneously in the static frame when we move M. We could observe the change with another clock at some distance from the first clock. We could send signals superluminally. This leads to all the paradoxes in time travel. Obviously, the gravity field must be retarded, in its effect to clock rates.

Since gravity must be retarded for a single mass M, it probably is retarded also for a collapsing shell of mass. If the collapse is accelerated, a clock at the center will tick faster than a clock close to the collapsing shell.

That implies that the metric inside the shell is not flat. But since the inside is empty of matter, the Einstein field equations imply that the metric must be flat. We broke the Einstein field equations.

If the metric were flat inside the shell, could we implement superluminal communication? Probably not. The configuration inside the shell would then be like a movie projector whose frame rate we can change.

What about a nonuniform spherical shell?

Suppose that we can represent the shell as a perfectly uniform shell S plus local deviations S'.

Then the gravity field of S would not be retarded, while that of S' would be retarded.

Let us assume that initially the shell is static and the metric is g inside the shell. Since the inside of the shell is empty, the Ricci tensor R is zero there.

Let the shell then start to collapse. If we assume that the metric associated with S is instantaneously updated inside the shell, then the metric of time, g₀₀, due to S, gradually slows down as the shell contracts.

We probably can adjust the metric which is due to S', so that the Ricci tensor stays zero inside the shell. Tie the metric due to S' to the (proper) time determined by S. Basically, slow down the frame rate of the movie projector.

But the definition of S is ambiguous. Suppose that the shell is thick and dynamic, so that mass moves around in the shell. How should we define S?

Suppose that the shell develops in such a way that all the mass gets concentrated to one side. Why should we assume that a part of the metric gets instantaneously propagated to a certain spherical volume, while another part is retarded?

We showed that defining a non-retarded gravity field is fraught with problems. Our conclusion is that the gravity field must be retarded, and the Einstein field equations are broken inside a collapsing shell.

We have not seen any literature which would analyze the metric inside a collapsing shell. Authors simply assume that the metric stays flat, and the metric of time instantaneously propagates to the entire volume.

In this blog we showed in May 2024 that the Einstein field equations probably have no solution for any dynamical system. Our new result says that the Einstein field equations break reasonable retardation rules in the very simple case of a collapsing shell.

The energy stored in the retarded gravity field

Above we used an ad hoc assumption that the retarded gravity field stores the same energy as how much the "retarded potential" of the mass differs from the final, static potential.

     ____                               ____  tense rubber

             •  -->              <-- •          membrane

                 _____•_____

          

            • = m mass element

This assumption may be the right one for a rubber membrane model of gravity. The mass m at the center in the diagram is too high because it does not "know" yet that the masses on the circumference have fallen lower as they are accelerated.

***  WORK IN PROGRESS  ***

Electron propagator controls bremsstrahlung in the Feynman diagram: how does it know how to do it?

In the November 10, 2025 post we realized that the electron propagator governs the form of the bremsstrahlung wave in the Feynman diagram.

                                k  =  (δ, δ)

                        ~~~~~~~~~~~~~~~  real photon

        p'         /        

        e-  ------------------------------------  p  =  p' + q - k

                                   |                  

                                   | q

                                   | 

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

In the diagram, we have set c = 1. Then the absolute value |δ| of the spatial momentum of the real photon k is the same as its energy δ.

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

The electron propagator measures "how far" is the electron from being on-shell. For an on-shell particle,

       E²  =  P²  +  m²,

where E is its total energy and P is its spatial momentum. In the propagator formula above p is the 4-momentum, and

       p²  =  E²  -  P².

The denominator above is zero for an on-shell particle. That is, there is a pole. The pole is formally removed by the i ε term.

The numerator is a 4 × 4 matrix. The components of p multiply gamma matrices. The term m above is actually m times the 4 × 4 identity matrix I. The numerator is not zero.

Suppose that the electron is off-shell by an energy δ. That is, it has δ "too little" energy, compared to its spatial momentum P:

       p²  -  m²  =  (E  -  δ)²  -  (P  -  δ)²  -  m²

                        =  -2 E δ  +  δ²  -  δ²

                        =  -2 E δ,

where we have assumed that the spatial momentum of the electron P is normal to the momentum δ of the photon. Also, if the electron is not very fast, then |P| is a lot smaller than E, and we can ignore the cross term of P and δ.

We see that the probability amplitude of a real photon, |δ| << |P|, is governed by the electron propagator, and is

         ~  1 / |δ|.

Can a scalar electrically charged particle exist?

The electron propagator is derived from the Dirac equation. How is it possible that the Dirac equation "knows" what kind of an electromagnetic wave will be born if the electron is pushed by the momentum q?

The Dirac equation is kind of a "square root" of the wave equation. This probably explains how it can know about the behavior of the electromagnetic field.

Would the propagator of a scalar charged particle work? The deminator looks like the one for the electron, but the numerator is very different.

The Peskin and Schroeder textbook on QFT (1995) gives the bremsstrahlung formula above for an electron. For a scalar charged particle, the numerators on the right would not contain the 4-momenta p and p'. The polarization ε* of the photon should be coupled to p and p' with a separate mechanism. The propagator of the scalar particle would control the spectrum of bremsstrahlung.

The photon can be emitted from the particle either before the particle scatters from Z+ or after. The propagator in the first case is

       ≈  -1 / (2 E δ)

and in the second case

       ≈  1 / (2 E δ).

The probability amplitudes cancel each out almost completely. It looks like the scalar particle propagator does not "know" what kind of an electromagnetic wave is created by pushing the particle.

This may explain why there are no charged scalar particles.

The Dirac equation under an electric field: it really does not need to "know" anything

                      e-  ~~~  --->         

                              ^

                              |   E

                              

                       

                              |   E

                              v

                    <---  ~~~  e-

Suppose that two wave packet electrons pass by each other so far that their distance is much larger than the size of the packet. Then we can approximate the electric field E of each electron at the other electron.

https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Quantum_Mechanics_III_(Chong)/05%3A_Quantum_Electrodynamics/5.02%3A_Dirac's_Theory_of_the_Electron

       p  →  p  +  e A(r, t).

Since the field E disturbs the Dirac equation, the electron wave probably is off-shell. The Dirac equation understands this because the electron in it is minimally coupled to the field E.

But how is this related to the Green's function of the Dirac field?

Could it be so that we actually derive the properties of the electromagnetic field from the Dirac equation? Then there would be no mystery of how the Dirac equation "knows" the properties of the electromagnetic field.

Yes. We derive the interaction of an electron with the field from the Dirac equation. The macroscopic interaction of a charge and the electromagnetic field is not given to us beforehand. We derive the form of the bremsstrahlung wave from the Dirac equation, using quantum field theory.

Quantum field theory is primary. From it we derive macroscopic Maxwell's equations.

However, this is not entirely satisfactory. We showed on November 25, 2025 that Feynman diagrams miscalculate several classical limits. The classical equation is more robust.

The Dirac propagator as a first "derivative" of the scalar propagator

                  #

                  #=======  sharp hammer

                  V

     

        _____        _____  tense rubber membrane

                 \ • /  pit

                  e- weight

                  x location

          

                        x + δ new location

We can imagine that a pointlike electron builds its electric field by repeatedly hitting the Klein-Gordon equation with a sharp hammer. We assume that the electron is initially static.

Charge must be conserved. We can move the electron, but not create or destroy it. The Dirac equation conserves charge.

If the electron is accelerated sideways, then the pit in the rubber membrane will be deformed. The form of the rubber membrane is then approximately the following:

1.   the static pit

2.   minus the last hammer hit at x to cancel the last hit at the old location x

3.   plus a new hit to the new location x + δ.

In a sense, the deformation from the initial static pit is a "derivative" of the hammer hit operation with respect to a position change of the hammer hit.

Above is the position space propagator (Green's function) for the scalar Klein-Gordon equation. It is the "hammer hit".

The position space propagator is SF for the spin 1/2 electron. We obtain it from the propagator GF of the Klein-Gordon equation by taking a "derivative".

Since the electron is initially static,

       p  =  (m, 0).

Let us assume that m is very small. Then the propagator GF is almost like that for the electromagnetic field.

Peskin and Schroeder (1995) calculate bremsstrahlung from the Feynman diagram. It turns out that since the electron is initially static, p does not contribute anything.

It turns out that the p' term calculates correctly the classical bremsstrahlung spectrum for low-frequency photons k. The value does not depend on the mass of the electron m, only on the momentum p'.

What did we show? That, in a sense, the Dirac propagator is a "derivative" of the photon propagator. This is a (vague) qualitative explanation for the fact that the Dirac propagator "knows" what the bremsstrahlung waveform is like.

Quantum gravity and gravitational bremsstrahlung: the problem with various propagators

Above we suggested that an electrically charged scalar particle might break macroscopic Maxwell's equations. Could there be similar problems with gravitational waves?

Bremsstrahlung depends on the propagator of the particle. It would be strange if gravitational waves for different propagators would be different.

The Higgs particle is a scalar particle with a rest mass. W and Z bosons have a rest mass.

It does not sound reasonable that the gravitational wave from such a particle would be different, depending on the propagator of the particle.

Conclusions

We were able to find a (vague) explanation how the Dirac equations "knows" what form do the low-frequency components of bremsstrahlung take. The reason is that the Green's function of the Dirac equation is a "derivative" of the Green's function for the Klein-Gordon equation.

Why is it a "derivative"? Because the Dirac equation is a "square root" of the Klein-Gordon equation.

This explanation will not work for gravitational waves as bremsstrahlung. Various particles have different propagators. We have to study how we can couple gravity to them.

The Dirac equation conserves charge and energy. The Klein-Gordon equation conserves energy. Energy is the charge in gravity. Can we use a "gravity propagator" for all particles, such that it would differ from the standard propagator of the particle? Gravity does not care if energy is converted to another form.

Feynman diagrams miscalculate the box diagram and bremsstrahlung in the classical limit

In our blog post on November 10, 2025 we showed that Feynman integrals miscalculate several simple processes in the classical limit.

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

                             |              |  p - k       virtual

                             | q + k    |                 photons

               e+   ----------------------------------

                                k = arbitrary 4-momentum

                                       in the loop

Let us assume that we have been able to calculate the Feynman integral for the above loopy diagram – possibly renormalizing the result.

Let us upscale the system, making the electron mass me N²-fold and the elementary charge e N-fold, where N is a large positive number. Then 4-momenta scale by the factor N² and the coupling constant e² / (4 π) scales by the factor N².

We consider a small part of the Feynman integral:

       d⁴k  [photon and electron propagators]

             * [coupling constant for 4 vertices].

1.   The 4-momentum volume element d⁴k scales by (N²)⁴ = N⁸.

2.   The coupling constants e² / (4 π) scale by (N²)⁴ = N⁸.

3.   A photon propagator scales by 1 / (N²)² = N⁴.

4.   An electron propagator scales by 1 / N².

5.   The combined scaling of the propagators is 1 / N⁴ * 1 / N² * 1 / N⁴ * 1 / N² = 1 / N¹².

6.   The total scaling of the integral is

       N⁸  *  N⁸  /  N¹²  =  N⁴.

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

                            | q

        e+ ----------------------------

The scaling of the simplest tree level scattering diagram is from the coupling constants (N²)² and from the photon propagator 1 / (N²)². That is, the scaling is 1.

The scaling of the loopy diagram is N⁴. The loopy diagram would dominate for large N. This only makes sense if the integral of the loopy diagram is zero.

https://dspace.library.uu.nl/bitstream/handle/1874/24911/consoli_79_One-loop%2520corrections.pdf

M. Consoli (1979) calculates box diagram integrals (Figure 7. (a) and (b) in the paper). Consoli does not mention that the integrals would be zero.

We conclude that the classical limit of the Feynman box integral is nonsensical. This suggests that the integral is nonsensical also if we use the normal electron mass me and charge e. What is the problem? Apparently, the Feynman integral is not the right way to analyze the "fine details" of a particle orbit in a Coulomb field. The simplest tree level diagram works very well, but a loop with two virtual photon exchanges is not sensible

the Feynman integral in QED exaggerates the high-energy spectrum of bremsstrahlung, in the classical limit?

        m, q

           • ----------_____ 

                                   ----->  

                                        

                    ● M, Q

Let us have a massive, large negative charge q pass a very massive large positive charge Q.

Let the position vector of q be z(t), where t is the time coordinate. Let t be zero when q is closest to Q.

Intuitively, the spectrum of electromagnetic radiation which the negative charge q will emit, depends on the Fourier decomposition of the acceleration vector:

       a(t) = d²(z(t)) / dt².

https://math.stackexchange.com/questions/318804/relation-between-function-discontinuities-and-fourier-transform-at-infinity

All derivatives of a(t) obviously exist, are continuous and tend to zero as t goes to (negative) infinity. Furthermore, the derivatives behave "nicely" when |t| is small. The theorem in the link states that then the Fourier transform â(k) of a(t) satisfies:

       |â(k)| * |k|ⁿ → 0

for all n > 1. That is, high frequencies |k| are suppressed extremely fast, probably exponentially.

But the Feynman formula, which is used to calculate bremsstrahlung, suppresses high |k| quite slowly. It is suppressed by the electron propagator, and the photon propagator in the Feynman diagram (q = e-, Q = Z e+).

https://sites.ualberta.ca/~gingrich/courses/phys512/node101.html

We can assume that the particle e- has quite a lot of kinetic energy and passes Ze+ from a relatively large distance. The kinetic energy would allow e- to send very high-frequency bremsstrahlung photons – quantum mechanics does not block that.

The Feynman diagram does not understand how sharp is the turn in the path, caused by q

We wrote about this on November 10, 2025. In the bremsstrahlung diagrams above, the momentum change q can be abrupt or slow, depending on how large Ze is. In the classical limit, the bremsstrahlung is drastically different for large and small values of Z. The Feynman integral does not understand anything about this.

We know from experiments that Feynman integrals do calculate practical applications of bremsstrahlung correctly. The electron e- in them is very far from a classical particle. For a quantum particle, the sharpness of the turn in the electron path does not matter.

Conclusions

Feynman diagrams and integrals miscalculate the classical limit in many basic processes.

On the other hand, for a strictly quantum particles, Feynman diagrams work well – we know that from experiments.

Quantum gravity: vertex correction, propagator, coupling

In QED, the vertex correction may be due to almost-bremsstrahlung which cannot escape as a real photon.

https://arxiv.org/abs/hep-th/0211072

Bjerrum-Bohr, Donoghue, and Holstein (2002) calculate some kind of vertex correction diagrams for gravity.

What is the energy density of the gravity field?

         ^  E  

         |

  

                      -----------  +       capacitor 

                      -----------  -        plates

If we have a static electric field E, we can extract energy from it locally by using capacitor plates which have opposite charges.

                        • m

                 -------------    scaffolding

                |              |

                       ●

                       M

To extract energy from a static gravity field, using a small mass m,  we must use a scaffolding which encloses the mass M. Then we can lower a small mass m and extract energy. Energy cannot be extracted locally.

This is the old problem about where is the energy of a gravity field located. Gravitational waves certainly carry energy. But is there energy in a static gravity field and what is the energy density, if not zero?

Electromagnetic/gravitational waves: a quantitative model

           F    M/Q                   M/Q    -F

         ------> ●                        ● <------

                              a                      d

                     -----------------    ----------------

                     accelerate     decelerate

                     time t             time t'

1.   First we accelerate to the right a particle of a charge Q or a mass M, with a constant force F for a time t. The particle moves a distance a.

2.   Then we decelerate it with a constant force -F. It stops after a distance d, after some time t'.

3.  The work lost,

       W  =  F (a  -  d)

escaped in radiation. The momentum

       p  =  F (t  -  t')

escaped in radiation.

Since gravitational waves carry 16X the energy of analogous electromagnetic waves, the difference a - d has to be 16X for gravity.

                \   rubber plate

                  \

                  ● ----> F

                  /

                /

Let us analyze the "effective inertia" of the particle in the process. Let us first look at a "rubber plate" model of the field of the particle. In phase 1, the far field of the particle does not have time to react. The effective inertia of the particle may be reduced because of that.

But the effective inertia also increases because the rubber plate in the diagram pulls the particle to the left.

                                \   rubber plate

                                  \

                     -F <----- ●

                                  /

                                /

In phase 2, the bending of the rubber plate actually helps -F to pull the particle to the left. The effective inertia is reduced. This explains why d < a.

How do we obtain a 16X radiated energy?

A.   If the rubber plate has a zero mass and resists stretching very much, then it carries away a lot of energy.

B.   If the rubber plate has mass, and and is easily stretched, then it carries away a lot of energy.

In the case of gravity, A is a more beautiful option. Assigning mass-energy to a static gravity field does not look nice.

The effects of A and B on the effective inertia of an electric charge

Let us estimate the effects of A and B of the previous section when an electric charge Q is accelerated. The effective inertia of the particle is reduced because the far field does not follow it, but the electric field lines resist stretching, and add effective inertia.

Let a mildly relativistic electron pass a proton at a distance R. In the previous blog post we calculated from the Larmor formula that the radiated energy is

       W  =  4/3  *  1 / (4 π ε₀)³  *  e⁶ / c⁴  *  1 / R³

                         * 1 / me².

The transit past the proton lasts a time

       t  ~  2 R / c.

The pulling force feels the inertia of the electron, and it also must do the work to create the radiation.

Let us assume that the electron is static and that the proton flies past it in the time t. A typical acceleration of the electron is

       ae  ~  1 / (4 π ε₀)  *  e² / R²  * 1 / me.

The distance r which the electron moves is very crudely

       r  ~  1/2 ae t²

           = 2  *  1 / (4 π ε₀)  *  e² / R²  *  1 / me

              * R² / c²

           = 2  *  1 / (4 π ε₀)  *  e² / c²  *  1 / me.

The force required to create the radiation is

       FW  ~  W / r

              =  1/2 * 4/3 * 1 / (4 π ε₀)² * e⁴ / c²

                  * 1 / R³ * 1 / me.

Let us compare this to the typical pulling force by the proton on the electron:

       Fe  =  1 / (4 π ε₀)  *  e² / R²,

       FW / Fe  ~  2/3 * 1 / (4 π ε₀) * e² / c² * 1 / R

                               * 1 / me 

  

                   ≈  2/3 re / R.

We see that when R is roughly re, the creation of radiation involves a force roughly the same as the pulling force of proton.

The far field of the electron does not have time to follow the electron in the process. This reduces the effective inertia of the electron by

       ~  re / R.

We see that the the inertia effect of the radiation is roughly of the same magnitude as of the electric field lagging behind. What is the role and the relation of these two effects?

              mildly relativistic

              proton+

                       ● ---->

                    ○------------- • -------------○  far field

                                      e-

If the "spring" connecting the far field to the electron is very loose, then the the effective inertia reduction from the far field lagging behind dominates. 

Question. Is the inertia of the far field relevant at all? The electron communicates with it through the electromagnetic field. Could it be that the inertia of the far field can be ignored? Is the inertia of the electron always me, even if the far field does not have time to react?

In a hydrogen atom, the effective mass of the electron is not reduced by its far field, which, presumably, cannot keep up with the rapid orbit of the electron around the proton.

On October 22, 2025 we presented a hypothesis that the vertex correction in QED comes from almost-bremsstahlung which almost can escape from the electron, but cannot become a real photon because it does not have h f of energy. Then we can say that the vertex correction in QED does not have anything to do with the far field, but is solely about bremsstrahlung. Also, there would be no purely classical vertex correction.

The vertex correction in gravity

                                       k

                                 ~~~~~~~      graviton

                      p      /                     \     p + q

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

                                       |

                                       |  q

                                       |  photon or graviton

                                       |

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

The virtual graviton k probably couples to the mass me of the electron, and to the kinetic energy of the electron. Then the integral looks much like the QED vertex correction. Can we renormalize it?

The propagator for the graviton should be the impulse response of the gravity field to a Dirac delta disturbance. What is it like?

Since gravity is nonlinear, we do not even know if there exists a solution which corresponds to a Dirac delta disturbance of the field.

The scattering process above is dynamic. In May 2024 we tentatively proved that the Einstein field equations have no dynamic solutions at all.

To make progress, we must try to estimate what the "correct" propagator for the graviton might be like.

The coupling constant for gravity, and the graviton propagator

In a pure momentum exchange, like a planet passing by a star, the graviton propagator, and the associated coupling constants, look like the photon propagator in QED. That is, Newton's force is like Coulomb's force:

       F  ~  m M / r².

But in an emission of gravitational waves, the power is 16-fold compared to electromagnetism. The coupling constant should be tuned to reflect this?

In this blog we have had the idea of "teleportation" to describe a gravitational wave. If we have a binary black hole, there are large dynamic changes in the metric around it. One can use the changes in the metric to drain energy from the gravity field at a 16-fold pace, compared to the analogous electric field.

A gravitational wave "teleports" the metric to a distant place, attenuating it by 1 / R², where R is the distance, and only preserving components which are normal to the radius between the observer and the binary black hole, since there are no longitudinal waves. In the teleportation interpretation, the propagator is similar to electromagnetism, but the coupling is complex, since the metric is complicated close to a binary black hole.

Scattering processes are short-distance phenomena. We do not need teleportation in them.

Hypothesis. The Green's function for the gravity field is a "hammer hit" to each of the 16 components of the metric tensor separately. Each component is capable of carrying away the energy of a hammer hit to the analogous electromagnetic field of the analogous electric charge. This explains why gravitational waves carry 16 times the energy of equivalent electromagnetic waves. The propagator for each component is like the one for the photon.

The following could be a counterargument to the hypothesis: off-diagonal components of the metric g are not independent. E.g., g₀₁ = g₁₀. But it could be that such a component receives a double share of the hammer energy.

What about a hammer hitting just the component g₀₀, i.e., creating a transient mass? The problem may be that the homogeneous gravity differential equation, that is, the vacuum equation, does not have solutions for such a configuration. That is, there is no Green's function.

A hammer strike in QED is allowed to break the rules: the 4-momentum can be arbitrary. Could it be that the hammer strike in gravity can do the same?

Corollary. Why does gravity imitate the Coulomb force in a system of orbiting masses? Because the only component relevant to the orbital motion is the metric of time. Its propagator is just like the photon propagator.

What is a Green's function for a field with many interconnected components?

The Einstein field equations tie together the 16 components of the metric tensor. The equation in vacuum can be called the homogeneous equation. How do we disturb the homogeneous equation with a Dirac delta function, and what is the Green's function for it?

For a scalar field this is clear. But what about a matrix equation?

Worldline quantum field theory of Jakobsen et al. (2021)

https://arxiv.org/abs/2101.12688

Jakobsen, Mogull, Plefka, Steinhoff (2021) calculate the classical bremsstrahlung in gravity.

The authors of the paper have developed a worldline quantum field theory, to be able to calculate the bremsstrahlung waveform perturbatively.

The authors introduce retarded propagators. This looks promising!

https://arxiv.org/abs/2010.02865

Above is another paper by Mogull, Plefka, and Steinhoff (2021).

The setting is that a gravitational wave h is emitted by a particle, or absorbed by a particle.

              h                   particle

         ~~~~~~

         ~~~~~~              • z(τ)  

         ~~~~~~           

             --->                 τ proper time

                                    z location

We may assume that the particle is initially static at the origin (0, 0, 0) of the spatial coordinates. Its position vector is denoted by z(τ), where τ is the proper time of the particle.

The authors calculate a Fourier decomposition of the metric perturbation (gravitational wave) hμν.

Let us assume that the particle only moves on the y axis. Then the second Fourier decomposition above is simply the Fourier decomposition of the real-valued function y(τ).

Above is the position space propagator for the graviton.

The authors give a strange "propagator" for the position of the particle. The propagator should be used to calculate the probability amplitude of a particle moving from one state to another. Why would the amplitude be ~ τ₁ - τ₂ ? Let us take that at a face value and check how the authors will use this "propagator".

The idea in the "interaction action integral" Spm^int probably is the following: let the metric perturbation h be a nice sine wave. If the position of the particle, z(τ) is another nice sine wave, their interaction can probably be calculated in a simple way. If the particle initially is static, z(τ) ≡ 0, andit meets a nice metric perturbation sine wave h, then the particle will start oscillating according that sine wave h.

This sounds like a classical way of calculating gravitational waves.

Classical bremsstrahlung: very different from what Feynman diagrams calculate

Let us analyze the following process

        e- -------------------____

                                           --->  v

                                  |

                                  |  q

                                  |

                                 ● Z+

The momentum exchange q tells us how "curved" the path of the electron e- is.

Let us keep q constant. We vary Z+. If Z is small, then the electron must pass close to Z+ and e- makes a sharp turn in its path.

Let us then make Z+, say, 10-fold. The electron must pass Z+ at a 10-fold distance for the momentum exchange q to be the same. The turn in the path of e- is 10 times smoother.

It is clear that the spectrum of bremsstrahlung has to be quite different in the case where Z+ is 10-fold. Its total energy is only 1/100 because the acceleration of e- is 1/10. It contains very little high frequencies.

Does the Feynman diagram understand this?

                                k

                        ~~~~~~~~~~~~~~~  real photon

                      /

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

                               |

                               | q

                               |

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

                               Z e² = coupling

The coupling to Z+ depends on the value of Z+. If Z+ is 10-fold, the cross section is 10-fold.

Otherwise, the Feynman diagram is essentially identical to the case when Z+ was not 10-fold.

Did we uncover a huge error in Feynman diagrams?

The error requires that the system is classical. But in most cases, the electron passes Z+ so close (~ 10⁻¹⁵ m) that the system is very far from classical. If it would radiate according to the classical path, it would radiate 1 GeV photons.

We found yet anothee case where a Feynman diagram badly miscalculates the classical limit of the system. However, we know from empirical data that it calculates the quantum behavior roughly right. In the quantum world, it does not matter much if Z+ is 10-fold.

Is the Mogull et al. (2021) paper classical?

The authors stress that their method calculates the classical bremsstrahlung (= gravitational wave).

Let us check if that really is the case.

In quantum gravity, we are interested in the Feynman way of calculating bremsstrahlung. It is very far from classical.

The electron propagator is actually the Fourier transform of the 1 / r potential in 2 dimensions?

The photon propagator seems to be related to the Fourier transform of the Coulomb potential 1 / r in 3 dimensions. The transform is

       ~ 1 / k²,

where k is the wave number. The Green's function is the impulse response of the homogeneous wave equation in 3 dimensions to a Dirac delta impulse. Like hitting a 3-dimensional rubber membrane with a sharp hammer.

          -----       -----  potential

                 \   /

                   • e-

When we "hit" an electron, what do we do? We rapidly change its location. In the rubber membrane analogy, the electron has dug itself a pit. When we move the electron rapidly, it makes a V-shaped pit. If the V-shaped pit still has the 1 / r form, but is elongated in some spatial direction, then it looks like a 2-dimensional 1 / r potential, whose Fourier transform is

       ~ 1 / |k|.

In bremsstrahlung, the electron e- receives a hit from its attraction to the nucleus Z+.

                                k

                        ~~~~~~~~~~~~~~~  real photon

                 1  /        

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

                                   | 2

                                   | 

                                   | q

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

The spectrum of bremsstrahlung is controlled by the electron propagator. In the diagram above, the electron propagator between the vertices 1 and 2 determines how much probability amplitude each k receives.

Conclusions

Let us close this long blog post which meandered into diverse topics. We will write more blog posts about quantum gravity.

But first we have to understand quantum electrodynamics better. We need an intuitive model. Especially interesting is the idea we came up last above: is the electron propagator, in some sense, the impulse response (Green's function) of the wave equation to a "line" or "push" impulse, where the line is a short path of the electron? The impulse response is in two spatial dimensions around the line.

The ordinary photon propagator is the impulse response of the wave equation to a point impulse.

In the Feynman diagram for bremsstrahlung, it is the electron propagator which governs the form of the bremsstrahlung wave. As if the electron propagator would actually be a different type of a photon propagator.

If we think of a charged scalar particle (no spin as in the electron), the particle is nothing but its electric field. Then it makes sense to claim that the propagator of the particle is actually a concept of its electric field.

Intuitive vacuum polarization model in QED – no vacuum polarization in gravity?

Let us, finally, try to construct an intuitive vacuum polarization model in QED, such that the model is not just hand waving.

                                             |  q momentum

                                             v      gained by e-

                         v ≈ c                 

                    e-  • --->

                                      R = minimum distance

                     /  |   \

                   |     |     |      E strong field,

                     \  |   /          dense energy

     

                         ● proton+

When e- passes close to the proton, the total energy of the electric field E drops, but the field between the two particles becomes stronger, its energy grows, and this growing energy creates a repulsion between the particles. The repulsion, of course, cannot cancel the overall attraction, but the repulsion is significant.

Vacuum polarization can reduce the energy of the field between the particles, and make the attraction stronger.

Vacuum polarization adds a new degree of freedom to the system. The field between the particles can use this freedom to reduce the total energy of the system. It is a good guess that if the energy W can escape to the Dirac field, then the total energy of the system electric field & Dirac field is reduced by W.

When the electron passes the proton at a distance R, the spatial size of the extra strong field E close to the proton is ~ R, and the time is ~ R / c. It is a smooth disturbance pulse ("bump") of those dimensions. The wavelength of the momentum exchange q is ~ 861 R.

Semiclassical analysis, with point particles e- and proton+

Let us first analyze this semiclassically in the zone of the size R. We Fourier decompose the bump. The most important components have the wavelength ~R.

The bump is the sum of the field of the electron and the proton. When they are close, the energy density of that part of the electric field E is double the sum of the individual fields.

    mildly relativistic

    e- •  --------------------------------------

                          | 

                          | q virtual photon

                          |

                       /     \  k + q  virtual electron-

                       \     /  positron pair e- e+

                          |  

                          | q virtual photon

                          |

       ● ---------------------------------------

  proton+

  static

The transient electric field E of the bump disturbs the Dirac field, through the coupling. We can construct the disturbance with Green's functions of the Dirac field.

We can imagine that q above designates the bump in E. The bump q hits the Dirac field through the coupling. The hit will create Fourier components k according to the propagator of the Dirac field. Because the bump is smooth, destructive interference will mostly cancel out |k| > |q|.

If q creates a virtual pair, then q will be absorbed to the pair, and the energy of the field E will weaken accordingly.

If the pair is long-lived, then the virtual pair will survive for the whole transit of the electron past the proton. The pair will reduce the energy of the field E for a long time. The pair will not escape as a real pair, because the electron e- is only mildly relativistic and cannot donate 1.022 MeV. The pair eventually must annihilate and give q back to E.

A good guess is that a long-lived virtual pair increases the attraction of the electron and the proton as if an extra momentum exchange q would happen between the electron and the proton.

A long-lived pair is called "almost-bremsstrahlung".

For q = 0, no almost-bremsstrahlung will happen. In the standard vacuum polarization calculation, this is represented by the integral

       Π₂(0).

If q ≠ 0, then there will be almost-bremsstrahlung:

       Π₂(0)  -  Π₂(q²).

In the vertex correction we saw that bremsstrahlung appeared as a missing part of the electric form factor F₁(q²) integral. The integral F₁(0) describes the process when the electric field of the electron can keep up with the movement of the electron. Bremsstrahlung is the field which "broke free".

Another way to view q ≈ 0: then R is large, and the transit of the electron past the proton lasts for a long time. Any virtual pair created by the bump will have plenty of time to annihilate, and they will not be able to reduce the average energy of E much.

The role of the coupling constants in the pair loop. We can understand why there is the (small) coupling constant e² / (4 π) in the first vertex which creates the virtual pair from q. But, since the pair necessarily has to annihilate as the electron leaves the proton, why is there the second coupling constant e² / (4 π) at the second vertex? Why is the coupling constant not 1?

One of the reasons might be time symmetry. The process must look the same if we reverse time.

Above we assumed natural units. In them, the coupling constant is e² / (4 π) ≈ 1/137.

The bump in E is at least crudely modeled by a single hit with a blunt hammer. The bump of the increasing E hits the Dirac field, and this hit, at least crudely, is like a hit with a blunt hammer. This partially explains the curious feature of Feynman diagrams that they seem to model complex processes with a single hammer strike. Why not 100 strikes?

Alternatively, the smooth orbit of the electron e- past the proton is a "blunt" disturbance, which can be modeled with a single hit of a blunt hammer.

Quantum magnification hypothesis

Quantum magnification hypothesis. In QED, the Feynman diagram of waves "imitates" the classical process with a

       861 X = 2 π / α

"magnification".

Above we analyzed the classical electric field E between the point particles e- and proton+. If the scattering angle of the electron is large, then

       R  ~  re  =  2.8 * 10⁻¹⁵ m.

That is 1/861 of the electron Compton wavelength λe = 2.4 * 10⁻¹² m.

The Fourier decomposition of the bump E has wavelengths which are ~ 1/861 of the electron Compton wavelength. Quantum phenomena only can have a resolution of ~ the Compton wavelength. Therefore, the wave representation of the process in a Feynman diagram must have waves ~ 861 times longer than the classical description. Is this a problem? Does a calculation with a 861 X magnification produce the same results as the classical calculation?

We know that the simplest Feynman diagram calculates elastic scattering probabilities which agree with classical formulae with point charges. At least in that basic case, the magnification works.

Bremsstrahlung. Let us have a mildly relativistic electron. Let us calculate from the Larmor formula the total energy W it loses if it passes at a distance R from a proton.

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

There,

       a  =  1 / (4 π ε₀)  * e² / R²  * 1 / me.

The time that the power P radiates is

       t  =  2 R / c.

The total radiated energy is

       W  =  4/3  *  1 / (4 π ε₀)³  *  e⁶ / c⁴  * 1 / R³

                        * 1 / me²

             ≈ 2 * 10⁻⁵⁷  *  1 / R³

joules. The mass-energy of an electron is 511 keV ≈ 10⁻¹³ J.

If R = re = 2.8 * 10⁻¹⁵ m, then

       W  ≈  10⁻¹³ J.

We conclude that the cross section of a mildly relativistic electron losing most of its kinetic energy in bremsstrahlung, classically, is very crudely π re², or

       0.3 * 10⁻²⁸ m²  =  0.3 barn.

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

The presumably correct empirical cross section is found from a Wikipedia article. It is something like 

       0.0008 barn.

We see that quantum mechanics efficiently prevents the electron from radiating when R << λe.

Quantum mechanics suppresses the production of high-energy real photons, but not virtual photons? The simple elastic scattering of an electron from a protons obeys the classical cross sections. But bremsstrahlung is severely suppressed when the minimum distance (impact factor) R is much less than the electron Compton wavelength λe.

Quantum mechanics allowed the momentum exchange q to be classical. The photon in q is virtual, it is pure spatial momentum.

Classical limit of the bump E in vacuum polarization

In the electron-proton scattering, let us make the charges N-fold and the masses N²-fold, where N is a large number. The new particle is a "heavy electron". We do not change the ordinary electron in any way. The heavy electron is a new particle, like a muon.

The heavy electron will still track the same classical path. This time, the bulge in the electric field E is classical. We can analyze the process in a localized spatial volume. We do not need to resort to the "momentum space" of Feynman diagrams.

In this semiclassical treatment, obviously, high 4-momenta |k| are strongly suppressed by destructive interference. The time-dependent electric field E disturbs the Dirac field. We can construct the impact response with Green's functions of the Dirac field. Destructive interference cancels high 4-momenta. It is like hitting a tense rubber membrane with a blunt hammer.

Feynman diagrams do not prohibit macroscopic masses or charges. In this classical limit, we know that destructive interference removes ultraviolet divergences.

The classical limit also demonstrates why Feynman diagrams can do with a single hit or the "sharp hammer", or a single application of a Green's function. When the heavy electron comes close to the proton, the change in the field E can be crudely approximated with a single hit of a blunt hammer. Generally, processes which last a limited, short time, often can be modeled with a single hammer hit.

Moving from the classical limit back to the quantum realm

                                             |  q momentum

                                             v      gained by e-

                         v ≈ c                 

                    e-  • --->

                                         R = minimum distance

                     /  |   \

                   |     |     |         E strong field,

                     \  |   /            the "bump"

     

                         ● proton+

The crucial question is what happens if we switch from the heavy electron back to the normal electron. Does the system still behave in a classical way?

When the mildly relativistic electron passes the proton, the process lasts a time

       t  =  R / c.

Let us calculate from the energy-time uncertainty relation how much energy W the system is allowed to "borrow" for such a short time.

       W t  =  h / (4 π)

   =>

       W  =  1 / (4 π)  *  h c / R.

The energy required to create a photon of  wavelength R is

       W'  =  h f  =  h c / R.

We see that we can "almost" borrow enough energy to describe a bump of a size R with a real photon, for the duration of the transit. This suggests that vacuum polarization, which is due to virtual pairs, could be replicated in the classical form, even though we have far too little energy to create real photons which would build the bump. We can borrow enough energy?

Peskin and Schroeder (1995) formula (7.91) has the vacuum polarization correction in QED. If we scale me and q by a factor N², the integral on the right does not change. But scaling the charge of the electron by N adds a factor N² to α = e² / (2 π) (in natural units).

In bremsstrahlung, the Planck constant h has a dramatic effect for the electron, because the electron cannot emit a wave such that h f is larger the kinetic energy of the electron. Apparently, no such dramatic effect is present in vacuum polarization.

Superlinear vacuum polarization as "almost-bremsstrahlung"

We can imagine that a pointlike electron constantly hits a tense rubber sheet with a sharp hammer. The formed pit is the electric field of the electron. Since the electric field interacts with the Dirac field, these hits indirectly also hit the Dirac field, forming some kind of (static) vacuum polarization.

If the electron is free and not under an acceleration, the system is static. The electron is surrounded by an electric field and, presumably, a nonzero Dirac field.

                        v ≈ c

                  e- • ---->

                                               |   q momentum

                                               v       change

                      R

                      ●

                     proton+

When the electron flies past a proton, the hammer hits no longer form a static configuration. The electron gains a momentum q.

A new hammer hit fails to "absorb" everything from the previous hit. The failed part escapes as electromagnetic bremsstrahlung.

In the case of the Dirac field, the failed part is "almost-bremsstrahlung" of a virtual pair. This is the "superlinear polarization" we have been talking about. If the virtual pair does not possess 1.022 MeV, it cannot escape as a real pair – but it can linger for the duration of the electron transit.

The momentum gain q reflects the maximum loss W of the energy of the combined electric field E as the the electron flies by. These are roughly linear if we keep R constant:

       q  ~  W.

                        e-  • ---> 

                             |  q

                             |

                             O  virtual pair

                             |

                             |  q

                             ●

                             proton+

A formed virtual pair means that some electric field energy escaped to a new "degree of freedom". How much energy? Since the field energy fell by W as the electron came close to the proton, we can guess that the virtual pair drains the same energy W.

At the beginning if this blog post we noted that if an energy W can escape to a new degree of freedom, then the total energy of a system often falls by the same amount W. The falling total energy means that the attraction between the electron and the proton is stronger. The magnitude of the added attractive force comes from the relation q ~ W.

Almost-bremsstrahlung hypothesis. The increased attraction between the electron and the proton comes from the electric field energy escaping, for a while, to a virtual pair. This hypothesis explains why the correction makes the attraction stronger between charges of a different sign, and the correction is:

       Π₂(0)  -  Π₂(q²).

If the charges are of the same sign, the escaping energy makes the repulsion weaker.

Does the almost-bremsstrahlung interpretation solve all ultraviolet divergences in any vacuum polarization?

If in some quantum field theory,

       Π₂(0)  -  Π₂(q²)

ultraviolet diverges, then we have a problem. Classically, it would mean that an infinite amount of energy escapes, for a while, into a new degree of freedom, if the particle absorbs a momentum q. The energy in the almost-bremsstrahlung is infinite and the attractive force becomes infinite.

The same problem could appear as infinite bremsstrahlung if we just have a force field surrounding a particle. No vacuum polarization is required to get an ultraviolet divergence if we ignore the fact that the energy must come from somewhere.

These problems would arise from the fact that the energy of the force field of a point particle is, formally, infinite. The acceleration caused by q can then, potentially, extract an infinite energy.

An obvious solution to the problem is to claim that the force field of the particle cannot contain more mass-energy than the mass of the particle. Does that work?

Another possible solution is to appeal to destructive interference. But in the classic model, the field close to a point particle is a problem. Its Fourier decomposition contains very short wavelengths.

The true "gravity charge" is always positive in a Feynman diagram?

Classically, any disturbance in a field always carries a positive energy. Feynman diagrams allow arbitrary negative energies in a virtual particle, but is this "negative energy" just a formal tool, and does not represent genuine negative energy which would have a negative gravity charge?

A sharp hammer hit (Green's function) always puts positive energy into the disturbed field, even if a Fourier component would formally carry a negative energy.

A static field contains energy, even if the Fourier decomposition is time-independent, and formally has no energy.

In general relativity, people often require that the energy density in space is everywere ≥ zero. Otherwise, we could arrive at time paradoxes. This is called an energy condition.

In a Feynman diagram, the 4-momentum really tells us about a net transfer of 4-momentum during the entire process. The "gravity charge" of a line does not need to be the formal amount of energy specified in the line.

In QED, in the vacuum polarization loop, the electron may carry a positive energy W, and the positron a negative energy -W. But classically, polarization of a material always requires positive energy.

If the gravity charge is always positive, is vacuum polarization possible in gravity at all?

In QED, the electric field of colliding charges pumps energy into the Dirac field, creating electric polarization. What would be an analogous process in gravity? We would need energy to create a positive energy particle and pull it away from a negative energy particle? Opposite gravity charges, presumably, repulse each other. Why would we need energy to pull them apart?

Note that a collision of electric charges can produce a real electron and a real positron. Vacuum polarization can be seen as an "almost-production" of a pair. But we do not believe that the corresponding process is possible in gravity. If a particle has a negative energy and a negative gravity charge, the repulsion from masses could lift its energy close to zero, but not to a positive territory. And we not believe that a real particle can have a negative energy.

Hypothesis: no vacuum polarization in gravity. All particles, virtual or real, have a positive "true" energy and a positive gravity charge. There is no vacuum polarization in gravity.

If the hypothesis is true, it solves all problems with vacuum polarization in gravity. It also removes the awkward possibility that we could use quantum mechanics to create a zone of a negative gravity charge.

The hypothesis bans the existence of any virtual particles in empty space. In this blog, we have repeatedly claimed that there cannot exist virtual particles in empty space. Such particles would break unitarity of quantum mechanics, because they would introduce an unknown causal factor into quantum mechanical processes.

Always positive gravity charge and arbitrary loops in Feynman diagrams of gravity

Above we argued that the gravity charge always has to be positive for a virtual particle, even for a particle which formally has a negative energy in a Feynman diagram.

              ~~~~~~~~~~~~~~~~~~~  graviton

                            |         |

                            |         |  virtual gravitons

                            |         | 

              ~~~~~~~~~~~~~~~~~~~  graviton

       --> t

Loops with gravitons are known to ultraviolet diverge in quantum gravity. Let us analyze what our hypothesis would imply for them.

Is the diagram above sensible at all? If a graviton scatters from another graviton (denoted by the first vertical virtual graviton), how could it scatter a second time?

Conclusions

We found an intuitive model for QED vacuum polarization, and an explanation why the renormalized Feynman integral may calculate it correctly.

Vacuum polarization effects have been measured experimentally in high-energy electron-positron scattering experiments, like the LEP of CERN. The precision of the experiments probably is not very good. They do agree with QED predictions.

In low-energy experiments, vacuum polarization probably has a very small effect. We do not know if such experiments have tested vacuum polarization.

In nature, ultraviolet divergences are a concern only in connection with gravity. Our hypothesis of no vacuum polarization in gravity may solve the problems. But let us do more research. What is bremsstrahlung and the vertex correction like in gravity?