If Navier-Stokes allows simulation of a Turing machine, then Gödel may make existence of a solution undecidable

Let us consider the Navier-Stokes equation of perfect fluid, with no atoms or other type of a cutoff at very short distances.

Solutions of the Navier-Stokes equation easily develop turbulence. Turbulence is a fractal-like phenomenon.

Can we harness turbulence to do complex digital calculations, like on a Turing machine?

If yes, then we might get analogues of the Gödel incompleteness theorem for solutions of the Navier-Stokes equation.

Suppose that a proof of a contradiction from the Peano axioms is equivalent to proving that a certain solution of the Navier-Stokes equation develops a singularity. Then it would be an undecidable problem if a singularity appears.

https://terrytao.wordpress.com/2007/03/18/why-global-regularity-for-navier-stokes-is-hard/

Terence Tao in his 2007 blog post refers to a possible connection of problems of complexity theory (e.g., P = NP) to solutions of the Navier-Stokes equation. Turbulence develops complex, pseudorandom structures. (See his note "6. Understanding pseudorandomness".)

If we can build a digital computer from turbulence, then complexity theory will pop up.

Real fluid has a cutoff at the atomic scale. We do not expect a turbulence-based Turing machine to have any relevance in the real physical world.

Quantum fields probably have a cutoff at the scale of the Planck length, because mini black holes may turn up. It is unlikely that we can harness microscopic quantum fields to make a Turing machine, but this deserves further thought.

A swarm of virtual quanta: a possible solution to the renormalization problem of QED

UPDATE October 27, 2019: If there is just a single vacuum polarization loop, the loop length is 2, and the propagator for large k is roughly 1 / k^2. Since k can have any value in R^4, the integral diverges badly, like k^2.

But if we have a swarm of virtual photons, and some of the virtual photons create virtual pairs, and the pairs recombine in such a way that the loop length for a momentum loop is 6 or more, then the propagator for large k is 1 / k^6 or less. Then the integral converges.

If we could modify the Feynman method in such a way that the momentum loop length is always at least 6, we would get rid of the ultraviolet divergence.

---

Let us assume that two electrons are colliding in a particle accelerator.

         ^                   ^
             \     p      /
              |~~~~|
             /             \
         e-                 e-

The simplest Feynman scattering diagram contains just one virtual photon which is exchanged between the electrons. The photon carries a four-momentum p.

The propagator of a photon is

       i g_μν / p^2.

Let us think about a classical scattering of two electrons. If the distance is larger than the Compton wavelength 2 * 10^-12 m, then classical Coulomb scattering is a good approximation.

In the classical scattering, the electrons move in curved paths. If we want to use the quantum mechanical particle interpretation, the electrons exchange a large number of virtual photons on their path.

Consider then the vacuum polarization loop for a single virtual photon:

         p           p + k
   r_0 ~~~~~O~~~~~ r_3
                       -k     

The photon starts from a spacetime point r_0 and ends up at r_3. The electron-positron pair are born at r_1 and annihilate at r_2. The contribution in the phase of the Feynman diagram is something like:

       exp(i *[(r_1 - r_0) • p
                    + (r_2 - r_1) • (p + k - k)
                     + (r_3 - r_2) • p
               ])
       =
       exp(i (r_3 - r_0) • p).

We see that there is a perfect constructive interference if we let r_1, r_2, and k vary.

The constructive interference causes the integral over all k to diverge badly.

We have in our blog stressed that momenta k >> p should have a negligible contribution to natural phenomena where a momentum p is the input. That is, if the process is fuzzy at a length scale L, phenomena with a length scale << L should not affect much.

But in the vacuum polarization loop, high k contribute greatly.

A possible solution is to require that in the scattering, the electrons exchange a very high number of low-momentum virtual photons.

Then the corresponding virtual electrons and positrons form large "swarms", where they can annihilate with any of a large number of opposite charges. There is no longer perfect constructive interference because an electron of a momentum p + k can annihilate with a positron of an arbitrary momentum k', and the positron was not born at the same spacetime point as the electron.

We conjecture that there is an almost perfect destructive interference for high momenta k. Only if k is of the same order of magnitude as p, is the contribution considerable.

Let the electrons pass by at a distance L.

We conjecture that the classical limit of the process is that the EM field as well as the reaction to it, the vacuum polarization electron field, are fuzzy at the length scale L. The fields have little contribution from high momentum k planar waves.

Our suggestion has an obvious problem: how do we model the creation of an energetic real photon or a real electron-positron pair in a collision? We should allow a high momentum p photon to produce a 1.022 MeV pair.

So far we have not found in literature practical examples of how large a cutoff Λ one should use in a collision of a momentum p, so that the Feynman formulas predict the outcome well. Is 2 × p a suitable cutoff? Or should it be much larger?

The Feynman diagram with just one virtual photon is a perturbation diagram where we approximate the perturbation as a single Dirac delta impulse on the other field. That sounds like a very crude way to calculate a solution for the QED lagrangian (whatever the correct lagrangian is). The divergence in the Feynman integral might be an artifact of the very crude approximation.

If we calculate with a swarm of virtual quanta, we might be closer to solving the fields non-perturbatively. That is, closer to the correct solution. If large momenta k have a very low weight in the correct solution, then there is no divergence problem. Then the solution to divergences is to calculate correct, non-perturbative solutions.

What is the bare charge of an electron?

The scattering experiment in our previous blog post can be interpreted as a way to measure the electric repulsion between two electrons, and therefore, their charge.

Some people think that the electron is surrounded by a cloud of virtual electron-positron pairs which screen part of the negative bare charge of the electron. For an unknown reason, there is less screening if real electrons come close to each other. That would explain the stronger coupling constant. But this does not help much because we have to refer to an "unknown reason".

We in this blog have the philosophy that only measurable quanta exist as "particles" and the rest is classical fields which obey classical field equations.

We do not know the correct QED lagrangian. Let us assume that Feynman diagrams indirectly describe the lagrangian correctly, whatever it is.

The trigger to create a virtual pair in the one-loop Feynman diagram is the exchange of the large momentum virtual photon between the colliding electrons.

The reaction in the electron field is to the momentum exchange. That suggests that the reaction is a dynamic phenomenon.

Suppose that we somehow attach two electrons very close to each other. They exchange a lot of momentum in a second. Is there a similar virtual pair loop present in this case?

We need to analyze the Feynman formulas for the loop. What kind of a reaction do the formulas describe in the electron field?

Scattering of two electrons is a classical field phenomenon - is renormalization really needed?

In last fall we promised in this blog that we will show that the divergences in the Feynman loop diagrams are an artifact, which is a result of a wrong integration order.

Thus, no renormalization is needed if the scattering amplitudes are calculated in the correct way.

Our analysis of of the QED lagrangian brought this question up again.

Scattering of two electrons with low energy

Let us consider the scattering of two electrons which possess much less than 1.022 MeV of kinetic energy. Let us assume the bounce is symmetric.

    ^                ^
      \            /
        \        /
         |~~~| virtual photon
        /         \
      /             \
   e-                e-

There is electric repulsion between the two electrons which makes them to bounce off each other. The virtual photon marks the repulsion. The virtual photon is not a particle in any way.

An analogue for the repulsion is a spring:

  e-  ---|\/\/\/\/\/\/\/\/|---  e-

The electrons push each other with a rod containing a spring.

What is analogous to the virtual electron-positron loop in the Feynman diagram for the virtual photon?

       ~~~~O~~~~
             loop

An analogue is that the spring can give way not just in the spring /\/\/\ part but also in the straight rod parts ---.

If we want to calculate the bounce very precisely, we need to take into account all other degrees of freedom where the kinetic energy of electrons can be stored temporarily, not just the electric field.

In the case of the spring, the straight rods could store a little energy. It is similar for a virtual electron-positron loop: it can make the repulsive potential between the two electrons a little less steep. The loop will store some energy for the time when the electrons pass by. The loop will return the energy back to the kinetic energy of the electrons when they start to recede.

The virtual loop, which is also called vacuum polarization, might create a temporary charge distribution like this:

       -   e-  ++  e-    -

The positive charge density between the electrons makes the repulsion a little weaker in the bounce.

But the running coupling constant makes the electron repulsion stronger at short distances. What can cause that?

We remarked in our earlier post about a radio transmitter that if an EM wave is created by a disturbance of the EM field, and the wave is guaranteed to get absorbed soon again by another disturbance, then the wave may store a lot of momentum relative to the energy. A free plane wave in an EM field stores less momentum per energy.

The same is probably true for a short-lived wave in the electron field. We would need a correct QED lagrangian to analyze this in detail. The temporary field is born by interaction from the rapidly changing electric field between the bouncing electrons. We should show from the correct QED lagrangian that the electron field, indeed, is disturbed by the rapidly changing electric field, and stores some energy and momentum for a short time.

Let us assume that a temporary wave in the electron field is able store a little energy and considerable momentum. The temporary field does not need the 1.022 MeV of energy which would be needed for a real pair.

Is our analysis of the waves fully classical? Where does quantum mechanics enter the picture?

Note that our qualitative analysis did not refer to quanta anywhere except that the colliding electrons in the pictures were assumed to be particles.

A more precise analysis would assume that the electrons are waves obeying the Dirac equation when moving free of interactions.

The waves are smooth. The fuzziness of the waves in space it at least of the order of the Compton wavelength of the electrons.

If we solve the waves fully classically, where does quantum mechanics enter the picture? Maybe only at the measuring device. It will measure particles whose probability distribution can be derived from the classical wave solution.

What is the divergence in the Feynman loop diagram?

The well-known problem in the Feynman diagram formulas is the divergence of the calculation of the loop contribution. If we integrate over all possible momenta carried by the loop, the integral diverges.

How does that divergence show up in our classical analysis?

It does not because when two classical waves of a wavelength λ meet, we do not need to consider detail much smaller than λ in the reaction of the electron field. We can use a cutoff at λ, and intuitively know that finer detail has very little effect.

Using large momenta in the electron field would involve fine spatial detail in the electron field.

A Feynman diagram calculates all possible paths of the bouncing electrons. Electrons are point particles in the Feynman diagram and can come to a very close distance from each other. Very fine detail in the reaction of the electron field to that close encounter does have an effect on the Feynman calculation formula. If we try to calculate an intermediate result after the loop, the result may well diverge.

However, that intermediate result is not what we measure from the experiment. It makes no sense to calculate such.

If we only calculate the end results of the experiment, and if our intuition that fine detail has a vanishing effect is right, then the end results will not diverge.

If we are right, the divergence in Feynman diagrams is just an artifact from a wrong integration order. We must not calculate the diverging intermediate result.

The general problem of divergences in partial differential equations

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

A Millennium problem is to prove the existence and the smoothness of solutions for the Navier-Stokes equation. The problem is in turbulence. Does its infinitely fine detail have a large effect on the solution?

The QED classical wave equations might have a similar problem. We need to prove that no turbulence-like phenomenon can appear. If we cannot prove the existence and smoothness, then the divergence of Feynman intermediate results is a symptom of a real mathematical problem.

If we can prove the smoothness and existence of QED wave solutions, then the divergence is just an artifact.

We have in this blog post shown the connection between the existence of smooth solutions for a classical partial differential equation and the need to renormalize a Feynman diagram calculation.

To recapitulate:

1. If the classical equation has smooth solutions, then renormalization is really not needed. The apparent need for renormalization is a result of trying to calculate nonsensical intermediate results.

2. If the classical equation does not have smooth solutions, then it is not a proper physical model. We need to modify it, for example, by introducing a cutoff, which is equivalent to renormalization with a cutoff.

We need to look at the literature about the existence of smooth solutions for various partial differential equations. Coupled field equations are probably nonlinear in most cases. Little is known about the existence of solutions for nonlinear equations.

Linearized general relativity is not renormalizable. How does that show up when we try to solve the classical Einstein equations?

Conclusions

Renormalization may be unnecessary in QED.

The need for renormalization, or new physics at the Landau pole scale or the Planck scale, is connected to the existence of solutions for the corresponding classical field equations.

If smooth solutions exist, no renormalization nor any new physics is required. In such a case, the concept of an effective field theory is unnecessary.

A drum skin inspired QED lagrangian density

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

The lagrangian density is

       L = ψ-bar (iγ_μ (∂_μ + ieA_μ + ieB_μ) - m) ψ
           - F_μν F^μν.

Wikipedia says that A is the EM field of "the electron itself" and B is the external field.

The first question is how many fields there are in the lagrangian? It looks like each electron has its own field ψ, and also its own EM field A.

Apparently, there is also a free EM field present as well as an external field whose origin may be charges far away.

The basic idea of a lagrangian density is to find a local extreme value in a spacetime integral of that density.

Suppose that we have just a slow-moving single electron in a static external electric field B.

The lagrangian above treats the energy of the combined electric fields A + B as potential energy. The mass of the electron is potential energy, and is constant.

But the potential of the electron in the field A + B is strange. If B is the static potential of other electrons, then the potential in B should have a minus sign in the lagrangian. That is correct in the above formula.

And what is the potential energy of the electron in its own electric field? If the potential energy is negative, then the sign in the above formula is wrong.

The above formula seems to mention the total energy of an electron three times as potential energy:

1. potential energy in its own field,
2. mass,
3. -2 × energy of the electric field of the electron, in the last term.

A drum skin QED lagrangian

How could we write a lagrangian which makes sense?

We had the analogy where a drum skin is pressed with a finger. The static electric field of the electron is the depression in the skin.

The depression is like a negative electric field and the finger is a positive charge sitting at the bottom of the depression.

The variables are the skin depression and the finger position.

       L = eA - F^2,

where A is the skin depression as a positive value, F^2 is the skin deformation energy and e is the downward force on the finger as a positive value.

What if we add another pressing finger? We have to introduce another field A_2 for it. The other finger stays in its own depression.

Note that the pressing force for each finger i only releases energy for its own depression A_i and ignores other depressions. We need separate terms F_i^2 for each A_i.

We can implement a repulsive force between the fingers by adding the energy of the total field F^2. If the fingers move closer, then F^2 grows.

There is a problem, though. Each A_i should be independent of the other A_j. But the term F^2 couples them. What to do? We may multiply the energy associated with each i with a large number M. The private field of each i is then "rigid".

To accommodate drum skin vibrations as well as an external field, we need to introduce yet another field B which is not visible for any of the electrons. Drum skin vibrations live in the global field B.

The total QED lagrangian is

       L = Σ_i M (e A_i(x_i) - F_i^2) - F^2,

where x_i is the position of the i'th electron and F^2 is the positive energy of the total field, which is the sum of all A_i and B. M is the large positive number which makes the static field of each electron rigid.

The mass of an electron is

       M (F_i^2 - e A_i(x_i) + F_i^2.

We should still add kinetic energies to the lagrangian.

The classical lagrangian

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

The lagrangian is

       L = L_field + L_interaction

           = -1/(4μ) F^αβ F_αβ - A J,

where Wikipedia says A J means "many terms expressing the electric currents of other charged fields in terms of their variables".

If there are no pointlike charges but a finite charge density, then the lagrangian above might be similar to the lagrangian of an attractive force, and it would calculate correctly the field inside and around a uniform charge density ball.

But if we have two such charge balls, the lagrangian then attracts them together, which is wrong.

We need to check the literature. Has anyone got the EM lagrangian right?

Is the blowup problem of classical fields related to the renormalization problem of quantum field theory?

https://www.physicsforums.com/threads/validity-of-theoretical-arguments-for-unruh-and-hawking-radiation.978501/page-3#post-6243209

https://terrytao.wordpress.com/2007/03/18/why-global-regularity-for-navier-stokes-is-hard/

There might exist a classical field problem which resembles the renormalization problem. In the Navier-Stokes equation, a Millennium problem is to prove that solutions do not "blow up" because of turbulence.

In a realistic fluid there is a natural scale, the scale of molecules, at which the Navier-Stokes equation stops working. The blowup cannot happen. This sounds like an energy cutoff which is used to eliminate the divergence in renormalization.

The concept of an "effective theory" contains the idea that at very short distances there is new physics which provides the necessary cutoff.

If we try to model an electromagnetic field in a gravitational field, and consider the backreaction of the two fields when they interact, the renormalization problem may appear in the classical fields as a blowup problem. For example, the solution might not be stable under small perturbations.

After all, Feynman diagrams are perturbation calculations. If the perturbations diverge, then a classical solution might be unstable.

Which brings us to the old topic if general relativity has any solutions at all under realistic matter fields.

Christodoulou and Klainerman (1990) proved the "nonlinear stability" of the Minkowski metric under general relativity.

This is a very interesting question: if Feynman diagrams with gravitons diverge, how could Christodoulou and Klainerman show the stability in the (very restricted) case of the Minkowski metric?

In physics, if we are calculating with two fields, we usually ignore the backreaction. If we calculate the behavior of a laser beam which climbs out of a gravitational field, we assume that the backreaction on the gravitational field is negligible. But it might be that a precise calculation shows the the solution is not stable.

Is the magnetic field just a Lorentz-transformed Coulomb electric field?

Suppose that we have a charge which is static in the laboratory frame. We know that it will interact with a moving test charge only through its static Coulomb electric field.

The movement of the test charge is explained solely through the Coulomb interaction. There is no magnetic field.

Suppose that we switch to a moving frame. Clocks which were placed at various spatial coordinates of the laboratory frame appear to go slower, and they are no longer in synchrony when viewed from the moving frame.

If an observer in the laboratory frame measures the test charge to move from a position x_1 to x_2, and measures the elapsed time as t, then the observer in the moving frame will measure a different time interval.

Furthermore, the measured time interval depends on the spatial distance x_2 - x_1. The speed of the charge appears to affect the force on the test charge, when the force is observed in the moving frame. The moving observer interprets that a magnetic field is affecting the path of the test charge. The force depends on the field strength and the speed of the test charge.

We see that in this simple case, we can say that the magnetic force is just an illusion which appears when the test charge path is Lorentz transformed over to a moving frame. A human has problems understanding the transformation and explains the surprising movement with an imagined magnetic force.

The simple model can explain the magnetic field which we see around a wire carrying an electric current.

http://www.feynmanlectures.caltech.edu/II_13.html

Richard P. Feynman calculated the relativistic effect of the electrons moving inside a wire, relative to the protons. He considers a negative test charge moving at the same (very slow) velocity v as the electrons inside the wire.

Feynman writes that there is a magnetic field which will cause the test charge to curve toward the wire. If we move to a frame moving along the test charge, then, he writes, the protons in the wire will appear length-contracted, and the negative test charge, in this moving frame, is pulled by their electric attraction.

Let the density of protons in the laboratory frame be P.

The density of electrons in the laboratory frame is γE, where γ > 1 (because of the length contraction) and E is the density of the electrons in the rest frame of the electrons. If there is no electric Coulomb force in the laboratory frame, then it must be that

        P = γE.

Let us then switch to the moving frame. The density of electrons is E, and the density of protons in this frame is γP. The density of protons is γ^2 times the density of electrons. There, indeed, is a pulling electric force on the negative test charge.

---

Feynman writes that, "of course", we cannot reduce magnetic fields solely to the Coulomb electric field and a Lorentz transformation. He does not specify the reason, though.

When we consider configurations like the radio transmitter of our previous blog post, it is not clear at all how we could get rid of the magnetic field.

If we move a charge back and forth, some of the spherical waves which are produced will reflect back. If we just believe in a Coulomb field which spreads at the speed of light out from the moving charge, what then causes the reflection? There has to be a wave equation acting, if there is a reflection. And it is hard to write a wave equation without a magnetic field.

A clearer picture of how a radio transmitter works

For concreteness, let us assume that we want to send 1 GHz radio waves. Let us assume that we have a device which moves an electric charge back and forth a distance significantly less than 15 cm. The cycle time is one nanosecond.

   <--------- charge --------->

                < 15 cm

We feed mechanical energy to the device, and the device converts the energy into photons which can be observed far away.

In earlier blog posts we have asked the question where does the momentum go? The device moves at a speed considerably less than light. Its kinetic energy is converted into photons which can only carry away a fraction of the momentum lost by the device. Where is the momentum stored?

We earlier introduced the fishing float analogy for the charge. When we move the float, a complex pattern of water waves forms around the float. Waves reflect also back toward the float. The force which water exerts on the float at each moment is a very complex phenomenon. We should make a computer simulation to calculate the force.

The radio transmitter, similarly, has a "self-force" which the field of the charge exerts on the charge itself.

Now we see that some of the momentum which the device lost while accelerating forth, can be reabsorbed by the device when it moves back. The momentum was temporarily stored in the electromagnetic field of the moving charge.

In half a nanosecond, momentum and energy in the field can move at most 7.5 cm away, if they want to be reabsorbed by the device. Thus, it is the immediate vicinity of the antenna which stores the extra momentum.

The Larmor formula estimates the average energy flux from a charge doing a periodic motion. The Larmor formula has been tested empirically on radio transmitters.

If the motion of the charge is non-periodic, then the energy flux is hard to calculate  and there is no reason to assume that the Larmor formula would be correct in such a case.

Quantization of the electromagnetic wave

Far away from our radio transmitter, the EM wave is very close to a plane wave. More precisely, it is a sum of two circularly polarized plane waves that rotate in opposite directions.

A plane wave is easy to quantize. We may imagine that it consists of an integer number of coherent photons. The Fourier decomposition stays almost constant. The photons are long-lived.

The waveform close to the transmitter is much more complex. It can be calculated using classical Maxwell's equations. Does the complex wave consist of photons and "virtual photons"?

We can, of course, calculate a Fourier decomposition for the complex waveform. Since the wave is interacting with the moving charge, the Fourier decomposition changes fast.

Can we meaningfully Fourier decompose the static electric field of the charge? If it is really static, the decomposition is strange. Maybe the decomposition makes sense if the charge moves?

We could interpret that the modes in the Fourier decomposition consist of photons or virtual photons which are being created and absorbed at a fast pace.

A spherical wave will always have some of the wave energy reflected back toward the center.

If a Fourier mode is a plane wave filling the entire space, then every mode is "approaching", as well as "receding" from the center.

We cannot describe the functioning of a radio transmitter just by photons moving away from the transmitter. There are always virtual photons which will be reabsorbed by the transmitter.

Where are the photons and virtual photons born in a radio transmitter?

Where are the photons or virtual photons "born" in space? The Coulomb electric field of the transmitter charge oscillates in a wide area of space. Are the quanta born at the charge? Or are they born farther away from the movements of its static electric field?

Let us consider a simple analogy. We have a drum skin which we disturb by pressing a finger agains it and moving the finger back and forth.

Where are the waves in the drum skin born? The energy comes through the finger.

Away from the finger, the drum skin obeys some kind of a partial differential equation.

As the finger moves back and forth, parts of the skin oscillate up and down. That is probably the main origin of the vertical waves in the skin.

Only part of the energy and the momentum which the finger inputs is carried away as vertical, ring-form waves. The rest is absorbed back into the finger in a backreaction.

Is there conversion from longitudinal waves to vertical waves? Suppose that the drum skin has zero friction.

All energy is at the lowest level transferred as horizontal stretching of the skin. We conjecture that we can ignore longitudinal waves.

Let us return to the radio transmitter.

1. The moving charge acts as a source in Maxwell's equations. It produces disturbance to the EM field.

2. The waves are born at the charge, but much of the energy and the momentum in the waves is reabsorbed back into the charge in a backreaction. Specifically, some of the energy in spherical waves is always reflected back.

3. Some of the waves escape as well-formed spherical waves.

4. Photons as well as virtual photons and longitudinal photons are born at the moving charge.

5. Virtual photons and longitudinal photons are quickly reabsorbed by the moving charge.

6. Virtual photons and longitudinal photons can store a considerable amount of momentum and return it back to the moving charge during the next half-cycle. The magnetic field of the moving charge resists changes in the speed of the charge. Most of the momentum is probably stored in that magnetic field and returned back to the charge when the charge changes the direction.

Is there a unique decomposition of an EM field into virtual and real photons?

We defined above a virtual photon as a quantum of a wave which is quickly reabsorbed to the moving charge in the backreaction.

For a plane wave, we have a unique decomposition into photons: we just specify the wave and the number of photons.

Is there a similar decomposition into virtual photons?

In Feynman diagrams, virtual photons can carry any momentum and energy. They have a "continuous spectrum". That suggests that there is no unique decomposition.

However, the success of Feynman diagrams suggests that virtual photons are, indeed, emitted and absorbed as discrete quanta. If we take that literally, then the decomposition of an EM field is those virtual and real photons which have been emitted but not yet absorbed. We probably cannot measure what exact quanta are present, though.

How can a moving charge know that the virtual photons that it emitted will be reabsorbed soon?

Let us consider an electromagnetic coil. It is well known that a current will create a magnetic field which, in turn, will keep the current running even if the voltage is switched off. There is considerable momentum stored in the field.

How does the coil know that it can store momentum into the magnetic field, and that the momentum will be reabsorbed? If we remove the coil, the momentum cannot linger in thin air.

The solution probably is that we cannot remove the charges which are present in the magnetic field, without making the charges to reabsorb the momentum in the magnetic field. The magnetic field of a moving charge resists changes in the motion of the charge. The momentum stored in the field will always be returned back to the charge.

In an earlier blog post we had the model where the field of a charge is an elastic object attached to the charge. Then it is clear that any momentum stored in the movements of the elastic object will always be returned back to the charge.

Thus, the short-lived electromagnetic waves close to a moving charge are such that they inevitably will get reabsorbed if we try to remove the charge.

Summary

The nature of virtual photons was revealed, and we explained the functioning of a radio transmitter with the new concepts.

Virtual photons correspond to classical waves which the moving charge will reabsorb soon after they were emitted. The reabsorption is the backreaction of the charge on its own field. It is also called the "self-force" of the charge on itself.

The charge can only emit short-lived waves which certainly will get reabsorbed. Virtual photons cannot live for long.

The momentum stored in short-lived waves can be considerable. The stored momentum explains how a radio transmitter can convert kinetic energy of its charges to real photons, even though the real photons cannot carry away all the momentum.

Some of the waves which the moving charge emits will survive to a large distance. Those waves consist of real photons which the charge emitted. Far away, spherical waves look like plane waves, and there is negligible reflection back toward the source of the waves.