The Ehrenfest paradox breaks conservation of momentum in special relativity?

The Ehrenfest paradox concerns the measured circumference of a rigid rotating disk. The circumference appears lengthened in the co-rotating frame, whereas the circumference is 2 π r in an inertial frame.

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

Let us introduce a linear version of the paradox.

                      

                       movement ---->

                weights connected with springs

           ... ●/\/\/\●/\/\/\●/\/\/\●/\/\/\● ...

           ... ●      ●       ●       ●       ● ...

                                      weights

                       <----  movement

We have two very long rows of weights. Initially they are static in the laboratory frame. There is some energy stored in the lower row weights. We use that energy to push the upper row weights to the right. The lower row starts to move left. Then we stop the movement and return the energy to the lower row.

During the movement, the springs in the upper row are stretched in the co-moving frame of the upper row.

The upper row will move slower than the lower row because in the upper row, in the laboratory frame, some of the added energy becomes potential energy in the springs - it is not kinetic energy of the weights.

After the movement stops, the newtonian center of mass of the system has moved.

Should we give up the newtonian center of mass theorem? Or is there a way to save it?

Possible solution of the paradox. Let the springs be force fields. The fields cannot have zero mass. We need to make the weights lighter in the upper row, in order to have the same mass in both rows. When we use a force between the two rows to accelerate the two rows, the mass of the force field in the upper row appears as reduced, and we do more work to accelerate the upper row than the lower row. The extra work will eventually end up as potential energy in the force field.

If the solution is right, then the inertial mass of a force field must appear reduced when we accelerate the particle carrying that field.

Radiation from the accelerated force field complicates things.

Suppose that we try to minimize mass reduction by making the rows very long and by accelerating very slowly. But minimization probably fails, because, e.g., the electric field strength |E| for a long row of charges is

       ~ 1 / r,

where r is the distance from the row. Total field energy is

       ~ ∫ 1 / r²  *  r dr,

which diverges.

Why does QED vertex correction after regularization calculate correctly the classical effect of reduced mass?

This is continuation to our post on March 18, 2021.

First mystery: why a simple photon propagator captures the complex classical Coulomb scattering?

There is a mystery in QED: why does the very simple photon propagator calculate correctly classical Coulomb scattering in 3 spatial dimensions? QED does not do it correctly if the dimension differs from 3. The classical process is quite complex. Why does such a simple propagator work? It is as if the propagator somehow would be equivalent to the equally simple Coulomb law of force in 3 dimensions.

Second mystery: why the simple vertex correction diagram captures the complex classical mass reduction?

Another mystery is why the simple vertex correction diagram of QED correctly computes the effect of reduced mass, after the Feynman integral is regularized and renormalized. This mystery may be easier to solve than the first one.

                                 virtual photon

                                 energy E, momentum q

                                 ~~~~~~

 momentum k     /                   \

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

                                      | virtual photon

                                      | energy 0,

                                      | momentum p

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

Let us try to solve the classical problem through some kind of wave formalism. In wave formalism, particles are not restricted to their paths of newtonian mechanics. They can zigzag and take strange routes. It is like using the Huygens principle where any point in space can act as a new source of waves. The interference of waves eventually determines where the particle will end up.

A classical lagrangian and path integral resemble a Feynman diagram

Wave formalism is probably equivalent to a lagrangian method of calculation and a path integral.

              ●  "field of the electron"

               |

               |  spring

               |   

         e-  ●--------> 

                               Z+ ● 

Consider the following classical lagrangian problem. The electron is connected to a smallish extra mass through a spring. The extra mass plays the role of the mass of the field of the electron.

We are free to choose the path of the electron and the extra mass as we like, but energy and momentum must be conserved. Note that if we tune the path of the extra mass, we can steal momentum and energy from the electron.

We are studying a classical lagrangian. The interaction between the electron and the extra mass forms a loop in the classical path integral. The electron may give a packet of energy and momentum to the extra mass through the spring, and later take it back.

Is it possible that we get a divergence in the classical path integral? If we allow any energy and momentum to circulate in the loop, that very much resembles a divergent Feynman diagram. How to eliminate divergences in the classical case?

Is there a limit on how much momentum q can the extra mass exchange with the electron? Momentum and energy must be conserved. That puts some limits on E and q. We might approximate the process by assuming a "collision" which transfers some energy and momentum to the extra mass, and another "collision" later. A vertex in a Feynman diagram depicts such a collision.

In a Feynman diagram, the energy of a particle can be negative. In a classical system that is banned. How to reconcile these? If a classical particle moves back in time, then its energy is negative. Or maybe we can allow negative energies in a classical path integral, and that does not affect the final result? We are looking for extreme values of action. Allowing negative energies might not change those extremes? In tunneling there is negative kinetic energy. We have to allow negative energies. The importance of tunneling probably is minor in a typical scattering experiment.

The potential V in the spring complicates things. Let us define that the spring is part of the extra mass. Then the extra mass can flexibly receive various amounts of energy regardless of its momentum. The extra energy is stored in deformation of the spring.

Note that the electron does not satisfy the energy momentum relation

        E² = p² + m_e²

in the classical process if the extra mass is not rigidly fixed to the electron. The effective mass m of the electron is less than m_e. The electron is "off-shell" in the classical process. Also the extra mass, the "photon", is off-shell because it does not move at the speed of light. This is an example of how particles in classical physics can be off-shell, or "virtual".

The reduced mass of the electron and the Larmor formula: classical versus quantum

In the blog post March 2, 2021 we calculated that the anomalous magnetic moment of the electron can be explained if the lagging field of the electron in the zitterwebegung orbit reduces the mass of the electron by the mass of the static field half a wavelength, or π "radians" away.

In the March 15, 2021 post we calculated that the Larmor formula of electromagnetic power dissipation is explained by the mass of the field more than 3/4 radians away using its inertia to pump the maximum power out of the oscillation. Classical bremsstrahlung is Larmor radiation in the case where the acceleration is not periodic.

The Larmor radiation seems to be a "stronger" process than mass reduction. Classically, both phenomena are, of course, always present if an electron is in accelerating motion.

In quantum mechanics, bremsstrahlung creates large quanta only in a small number of fly-bys of the electron. Most fly-bys are elastic, or only radiate soft photons. Classically, the situation is very different. Every fly-by creates also high-frequency radiation.

What about mass reduction? It is always present classically. Is it always present in quantum mechanics? At least in zitterbewegung mass reduction is always present.

The Feynman diagram: the classical interpretation as impulses

       electric field

       of the electron

          | 

          |   impulse

          ● e-  --->

          |

          |

                                          ● Z+

Let us think of the electron as a particle and its static electric field as a field.

When the electron approaches the nucleus, the nucleus pulls on the electron. The electron makes some kind of an impulse into its own static electric field. In the rubber plate model, the electron tugs the rubber plate as the electron accelerates toward the nucleus.

The impulse response in the electric (more precisely, electromagnetic) field can probably be, at least, approximated by a massless Klein-Gordon Green's function.

A Green's function is the response to a Dirac delta impulse. The delta impulse contains too much very high-frequency waves, in contrast to a realistic classical impulse. But if we do some kind of a cutoff of high frequencies, it might approximate the classical process.

The impulse response contains lots of off-shell waves. Typically the waves do not satisfy the energy-momentum relation

       E² = p².

Classically, these are deformations which cannot escape to infinity as well-formed sine waves.

Bremsstrahlung is generated from the on-shell spectrum of the impulse response. Mass reduction is the result of the off-shell spectrum.

Let us think of the rubber plate model. When the electron accelerates, it stretches the rubber plate. When the electron decelerates, the rubber plate will exert various forces on the electron. The nearest parts of the plate will accelerate the electron, but very far parts are still gaining speed and their effect is to decelerate the electron.

If we assume that no waves can escape to infinity, then all the off-shell (and maybe also on-shell) waves have to be absorbed back by the electron. This is clearly the Feynman vertex correction diagram.

We now have a classical explanation for the virtual photon carrying E and q in the diagram. Various E and q combined (integrated) form the impulse response of the field when the electron is accelerated by the nucleus. A "virtual photon", with certain E and q, is a Fourier component of the true classical response.

We still need a classical explanation for the virtual electron and its propagator. We need to figure out a classical way to treat the electron as a wave.

How to model a classical electron as a wave - what is the classical limit of the Dirac equation?

Note the following: it makes sense to assume that all the inertial mass of the electron is in the particle itself. The mass of its static field is zero. The apparent reduced mass of the electron is a result of the forces which its far field exerts on the electron.

The assumption above would explain why the far field does not explicitly appear in the Feynman diagram. Only the impulse which the electron sends to its field is relevant.

             k

        e-  --->

                           k

                      e-  --->

         k

    e-  --->

Let us have a random flux of classical electrons whose momentum is k. If they are free, then we can describe the classical process as a plane wave. The wavelength is determined by the Planck constant, but the wavelength is not relevant in this simple process of linear motion.

Classical scattering of electrons from the field of a nucleus looks like bending of waves in an optically dense material. It is not too big a surprise that the Schrödinger equation (slow electrons), or the Dirac equation (fast electrons), or a Feynman diagram calculates classical scattering correctly, at least in cases where tunneling is unimportant.

A classical wave representation for the electron must, of course, conserve the number of particles and keep the particle number >= 0 everywhere. It should be Lorentz covariant. The Dirac equation satisfies these.

Question. Is the Dirac equation the simplest wave equation which has a positive definite conserved probability density and is Lorentz covariant?

The Dirac equation is

      i / c  h-bar γ^0 dψ / dt

   + i h-bar γ^μ dψ / dμ

   - I m_e c ψ

   = 0,

where we sum over the values μ = 1, 2, 3 (meaning the x, y, z spatial coordinates), and I is the 4 × 4 identity matrix.

What is the classical limit of the Dirac equation? If we let m_e to grow large, or h-bar to become very small, then we have to shorten the wavelength of the solution. The spin of the electron, 1/2 h-bar, becomes insignificant in the classical limit.

Question. Has anyone proved that the limit of the Dirac equation or the Schrödinger equation really is classical physics?

Answer. The Ehrenfest theorem, to some extent, shows that the limit of the Schrödinger equation is classical mechanics:

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

Mario Bacelar Valente (2012) claims that QED is "upgraded" from classical physics:

https://arxiv.org/abs/1201.5536

If we in the Schrödinger equation let the wavelength to become shorter, that will reduce tunneling - as expected if the limit of the Schrödinger equation is classical physics.

Let us think about waves of visible light. The wavelength only matters when the wave meets an obstacle whose feature size is of the order of the wavelength or less. For example, if we have an acromatic lens, it bends all wavelengths in the same way if the wavelength is much smaller than the lens diameter.

In many phenomena, the "classical limit" of a very short wavelength is essentially the same as the phenomenon with substantially longer waves. This observation explains why electrons in many cases seem to follow classical paths.

What is an "impulse" to a classical electron wave representation?

We have no problems imagining in our mind what an impulse to a drum skin or to an electromagnetic field does and what the effect looks like.

But we are used to imagining the classical electron as a particle, not a wave. What would an impulse mean in this case?

Let us write a new blog post about this central problem.

How do we know that coupled classical wave equations have solutions that are stable and smooth?

Short answer: we do not know.

Our goal in this blog is to show that divergences in Feynman integrals are an artifact of the approximation method which is used. A proper approximation method does not need any regularization nor renormalization.

We believe that divergences in Feynman integrals are not an indication of new physics at high energies or short distances. Thus, we believe that the concept of an effective field theory is wrong in this case.

However, there is a much harder mathematical problem in nonlinear differential equations in general. It is almost impossible to prove anything about them. Coupled fields typically form a nonlinear differential equation.

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

Sergiu Klainerman and his coauthors have been able to prove existence theorems for solutions in certain cases. Typically, one assumes that the fields differ from flat very little (small data or small perturbation), and one can then show the existence of a solution for at least a short period of time.

Sidenote: In this blog we have the hypothesis that Einstein equations do not have solutions for any realistic case, because the equations are too "strict". They are not flexible, like the equations of a rubber sheet are.

https://press.princeton.edu/books/hardcover/9780691632551/the-global-nonlinear-stability-of-the-minkowski-space-pms-41

Demetrios Christodoulou and Sergiu Klainerman in 1994 were able to prove the stability of the Minkowski space under "small" perturbations. We believe that under large perturbations, no solution exists. Sidenote ends.

If it turns out impossible to prove the existence and smoothness of solutions for coupled field equations, is that an indication of new physics lurking at very short distances?

No. Even if we discretize physics at very short distances, we still face the problem of proving that large amounts of extreme frequencies cannot appear in a physical process. How to prove that if we input visible light into a physical process, no 1 GeV quanta are produced? The proof may be even harder if we assume that physics is discrete at the Planck length.

What about discretizing physics at many scales, like in quantum mechanics? A wave is represented by a quantum whose size depends on the frequency? That is a better idea, but we face the problem of accelerating emitters whose wave does not seem to conform to simple quantization.

It may be that nonlinear differential equations in nature are well-behaved. We just lack the mathematical ability to prove that. Then there is no need for new physics at all.

This discussion has similarity to various incompleteness theorems proved by Kurt Gödel. For example, we cannot prove the consistency of the Peano arithmetic. Some day, a correct proof of 0 = 1 may appear in a mathematical journal.

Is Gödel's incompleteness an indication of new physics lurking behind the veils? Maybe the universe is a finite state machine and no true Peano arithmetic is needed at all?

It may be that the Peano arithmetic is consistent. Then there is no need for new physics in that case.

The philosophy of effective field theories is that at short distances some new physical law saves us from mathematical uncertainty. The problem in this philosophy is that we cannot know if our inability to prove something is an indication of any physical problem.

If there were mathematical certainty that our theory is inconsistent at short distances, then we would know that there must be new physics. For example, mini black holes of the Planck mass certainly present challenges to field theories.

Conclusions

It may be that coupled field equations are well-behaved in general. We just cannot prove that. Then mathematical problems in them do not require any new physics at short distances, or at any distance.

The problem of divergences in Feynman integrals is probably much easier to solve than the general problem of well-behavedness of nonlinear differential equations.

How to solve a classical wave equation with a source? Can we use Klein-Gordon propagators?

Do the various propagators of the Klein-Gordon equation really produce a Green's function which is zero for times t <= 0, and also zero outside the light cone of the impulse?

https://en.wikipedia.org/wiki/Green%27s_function

A Green's function is an impulse response. The response should be after the response, not before it.

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

The Feynman propagator produces a function which is not zero for t < 0. Times t < t tell the probability amplitude for a particle traveling back in time. It is not a Green's function in the above sense. How is its use justified?

What about the retarded propagator? Wikipedia says that it is zero for t <= 0 and also outside the light cone.

Let us check what is known about various propagators.

https://www.imperial.ac.uk/media/imperial-college/research-centres-and-groups/theoretical-physics/msc/current/qft/handouts/qftretardedpropagator.pdf

Tim Evans (2018) explains why the retarded propagator has to be integrated with a contour at the infinity in the lower complex plane, if t > 0. That gives the sum of residues of the poles. If t < 0, then the contour includes no poles and the integral is zero.

Continuity of a Green's function is not well-defined because the Dirac delta source at t = 0 and x = 0 is infinite. 

The contour integral in the retarded propagator goes to zero when t ---> +0. That is reasonable.

How to solve the response of a classical wave equation to an impulse?

In a blog posting a month ago we wrote that ultraviolet divergences in classical problems are avoided by destructive interference. We also wrote that infrared divergences are avoided because the Fourier decomposition of a typical impulse contains little low frequencies.

                           __

                          |    |   hammer

                          |    |=========

                          |__|

                            |

                            v

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

                     drum skin

Consider the example where we hit a drum skin with a hammer whose head is not sharp. A drum skin approximately satisfies the massless Klein-Gordon equation (= the ordinary wave equation).

Let us try to solve the behavior of the drum skin with Green's functions. The response to a sharp Dirac delta impulse is the Green's function (given as its Fourier decomposition):

       G(E, p) = 1 / (2 π)^4  *  1 / (E^2 - p^2 ),

where the wave is

       exp(-i (E t - p • x)).

Above, E is the energy of the wave, p is the spatial momentum of the wave, t is the time, and x is the spatial location.

G(E, p) goes to infinity when E and p go to zero.

Question. If there is a "smooth" impulse to the massless Klein-Gordon equation, does there exist a solution? Equivalently, if we write a "smooth" source to an inhomogeneous massless Klein-Gordon equation, does there exist a solution?

Can we form a solution from the Green's function? If we set E = 0 in the Green's function, then in an n-dimensional space, let us look at the integral

        ∞

       ∫ 1 / r^2  *  r^(n - 1) dr,

      0

which decides absolute convergence.

If n >= 2, then the integral diverges for large r.

If n <= 2, then the integral diverges for small r.

The Green's function, when put to the massless Klein-Gordon equation, has a Dirac delta source. A Dirac delta source, and the associated solution with a Green's function, seem to have problems with divergence.

Let us put a unit ball source.

https://math.stackexchange.com/questions/489391/fourier-transform-of-the-indicator-of-the-unit-ball

Daniel Fischer (2013) writes that the Fourier transform for n = 4 is

       ~ 1 / p^2 * J_2(p).

The Bessel function J_2 has an asymptotic behavior 1 / sqrt(p).

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

Let us put a gaussian source.

https://math.stackexchange.com/questions/565928/fourier-transform-on-mathbb-rn-of-gaussian-function

Daniel Fischer (2013) confirms that an n-dimensional gaussian has a Fourier transform of the form:

       ~ exp(-π p^2).

These sources look much better with respect to diverging behavior. Note that for small |p|, the Fourier transform is just like for the Dirac delta because exp(-π p^2) is then roughly 1.

The contour integration trick does not work for the gaussian, because

        exp(-π E^2)

has a very large value if E is imaginary and |E| is large. The contour running at infinity should have very small function values. We can probably prove the convergence of the integral directly.

Time asymmetry

There is a serious problem in Green's functions. They are typically symmetric in time. But we want a solution where there is no wave in the past. That is, the initial value is a zero field in the past. The wave is created by the source. Can we still utilize Green's functions?

Maybe we can, for the following reason:

                               |   impulse

                               v

                            

                  |<---- Δx ----->|

      ---------●--------------------------------  tense string

          support

                 

Suppose that you press with your finger to make a wave in a tense string. The wave will spread to the left and the right. How to remove the left wave? Attach the string to a fixed support at the position of the left edge of the impulse area.

The wave which spreads to the right may be relatively unchanged if the impulse is very quick, despite the fixed point on the left. By a "quick impulse" here we mean that the wave can reach the ball during the impulse, but does not have time to proceed much farther. The width Δx of a quick impulse should be roughly

       Δx = c Δt,

where c is the speed of the wave and Δt is the time of the impulse.

If the impulse is slow, then the wave to the right in the fixed case is much weaker than in the free case because the fixed point prevents the finger from pressing the string lower.

If we want the field to be zero in the past, we "fix" it to zero there. The source will create a wave to the future direction, and the wave might be somewhat similar to a Green's function.

We should calculate with a computer the waveform in various cases.

How does a string react to a Dirac delta impulse?

Does a Dirac delta impulse make sense at all? It contains an infinite force. The result may be strange.

Let us analyze the Dirac delta response in 1 + 1 dimensions.

       d^2 ψ / dt^2  -  d^2 ψ / dx^2 = δ(t , x).

Let us approximate the Dirac delta impulse with a box impulse that is Δx wide and Δt long.

               Δx

              ____

         __/         \__  tense string

                 ^

                 |  impulse

We assume that ψ is zero before the sudden  impulse. The impulse forces the second time derivative to jump up suddenly in a narrow area. Around that area, ψ has to satisfy the homogeneous equation. The string is concave downward there.

Suppose that we have a solution with a box size A. We shrink both sizes of the box by a factor 1/2. We get a new solution by shrinking the width and the time of the string deformation by the same factor 1/2. Second derivatives grow 4-fold.

We conclude that nothing strange happens. The string behaves reasonably with a narrow box of impulse.

The wave going backward in time is useful in some configurations

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

                                | virtual photon p

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

                           \          /

                             ~~~               

                     virtual photon q

            -----------------------------------> t

In the vertex correction diagram, when the electron absorbs the virtual photon q, we can also imagine that the electron emits it backward in time.

If there is input which matches the supposed emission backward in time, then emitting backward in time is no problem. This is probably the logic behind the Feynman propagator.

Conclusions

A major question of classical physics is why large amounts of very high-frequency or very low-frequency waves do not arise in wave phenomena. When calculating classical wave processes we instinctively apply cutoffs which remove such frequencies.

If lots of extreme frequencies would arise, then we would have infrared and ultraviolet divergences in classical physics, too.

There is probably no proof that extreme frequencies cannot arise. Coupled classical fields are nonlinear systems, and usually it is impossible to prove anything about nonlinear systems. The Clay Institute Millennium problem about the smoothness of Navier-Stokes is one example.

Above we gave some heuristic arguments why classical physics behaves well. Impulses are smooth, not Dirac delta impulses, and their Fourier decomposition contains very small amounts of extreme frequencies. The gaussian is a prime example of a smooth impulse.

Our analysis suggests that the divergences of Feynman integrals are an artifact from using Dirac delta impulses. The divergences are removed if we switch to more realistic impulse forms, for example, gaussians.

Why does QED after regularization calculate correctly the vertex correction?

We believe that the QED vertex correction, or the vertex function, is due to the elastic nature of the classical static electric field of the electron.

          ---------------------------------    rubber plate

                   ^     ||||/     

                   |       ||   hand

                   v

           oscillation

We have coined the name "rubber plate model" to describe the elasticity of the electric field. It is easy to imagine in one's mind how an elastic rubber plate behaves when we hold it in our hand at one point and wave it. If we make the rubber plate to oscillate with our hand in a cyclic fashion, the effective mass of the plate is reduced because the far parts of the plate oscillate in a different phase from our hand.

Why QED is able to calculate the classical process correctly - after infinities are removed with regularization?

                   virtual photon q

                        ~~~~~~~

                     /                     \

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

                               | virtual

                               | photon

                               | p

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

In the above diagram the electron scatters from the nucleus Z+, p contains spatial momentum, and q is arbitrary 4-momentum.

The virtual photon q is due to the interaction of the electron with its own elastic electric field. The far field does not have time to react to the acceleration caused by the nucleus Z+. The net effect is that the effective mass of the electron is reduced in its collision with the nucleus.

Let p be fixed.

QED uses propagators to calculate the complex-valued weight of the path for each value of q. It then sums the weights. The sum is infinite, and infinities are cut off in some systematic way - regularized. Miraculously, the final result is correct and agrees with empirical measurements.

How is this procedure related to the elastic electric field of the electron?

QED summarizes the smooth approach and receding phases of the electron into a simple line in the diagram: the virtual photon p.

Before the fly-by, the electron gives a Dirac delta impulse to its own electric field in the first vertex. The propagation of that impulse is calculated with the photon propagator. The electron itself moves under its own propagator.

After the fly-by, the electron absorbs the components of the impulse.

How is this procedure related to the elastic interaction of the electron with its own field?

Why can the smooth interaction with the nucleus Z+ be summarized into a single photon line p?

Classical analysis of the fly-by: the non-relativistic case

   y

   ^

   |                  ●  Z+ nucleus     

   |

   |    e- -------->

   |

    -------------------------------> x

Let the electron fly along the x-axis. Let the velocity of the electron be substantially less than the speed of light.

The nucleus first accelerates the electron along the x-axis, then decelerates it. Simultaneously, the nucleus pulls the electron up. The momentum upward becomes p.

The reduced mass of the electron allows the nucleus to accelerate the electron faster. We assume no radiation out. The momentum p is determined exactly by the integral of the upward force over time.

A lighter particle is deflected more. But how does that affect the momentum p? A lighter particle takes a route which comes closer to the nucleus Z+ but the particle moves faster.

http://williamsgj.people.cofc.edu/Scattering%2520Theory.pdf

Let us have a potential

       V = C / r.

According to H. Friedrich (2013), the scattering angle is

       Θ(b) = 2 arccos(1 / sqrt(γ^2 + 1)),

where b is the impact parameter,

       γ = C / (2 E b),

and E is the particle kinetic energy at the start.

Let γ be small. Then

         arccos(1 - γ^2 / 2) = γ.

If we reduce the mass m by 1%, E is 1% less and the deflection angle is 1% more. The momentum p is not affected by the reduced mass, if γ is small. The deflection angle is 2 γ.

Let the particle be very light. Its final upward momentum p cannot be more than m v, where v is its initial velocity. We conclude that if γ is not small, then reducing the mass of the non-relativistic particle reduces p.

If the particle does a head-on collision with the nucleus, then its final momentum q along the x-axis is reduced with a reduced mass because the particle moves faster and has less time to collect momentum.

The effect of mass reduction is to reduce the scattering when the electron flies very close to the proton. Thus, it reduces the cross section for large deflection angles.

The electron fly-by: the relativistic case

What happens if the speed of the electron is close to the light speed? Reduced mass allows it to come closer to the proton but does not affect the speed much. Let us calculate the magnitude on an electron which flies at a distance

       r = λ_e / (2 π) = 4 * 10^-13 m

from the nucleus. Let the kinetic energy of the electron be m_e c^2.

The mass reduction is

       ~ r_e / r = 1/137.

Let us assume that the electric force only affects the electron in an equilateral triangle whose side is r. The acceleration:

       a = k e^2 / (r^2 * 2 m_e)

           = 10^27 m/s^2.

The distance which the electron has come closer to the proton when it leaves the equilateral triangle:

       s = 1/2 a t^2

          = 1/2 a r^2 / c^2

          = 1/2 k e^2 / (2 m_e c^2)

          =  7 * 10^-16 m.

During its journey in the triangle, the electron comes 1/550 closer which adds ~ 1/800 to the momentum.

The mass reduction would increase the acceleration a by 1/137. The effect on the momentum is small and positive, only

       1/800 * 1/137 = 10^-5.

How much does reduced mass affect the speed of the electron?

The electron gains

       1/2 * 1 / 137 m_e c^2

in kinetic energy while it is in the triangle.

The Lorentz factor is

       γ = sqrt(1 / (1 - v^2 / c^2)).

The electron speed is 1/1400 larger in the triangle. Reducing the mass would increase the speed by a ratio

       1/1400 * 1/137 = 0.5 * 10^-5.

The effect on the momentum is small and negative. For a relativistic electron, mass reduction seems to have little effect in Coulomb scattering.

Analysis of the Feynman diagram

In the Feynman diagram, we calculate the response to a Dirac delta impulse which the electron makes to its own field before meeting the nucleus.

The impulse contains waves of any 4-momentum q, weighted by the photon propagator.

We know that the Feynman formula should calculate the effect of mass reduction, where the reduction depends on p. Usually, the mass reduction is quite small.

The electron propagators become infinite when q = 0. The Feynman formula favors electrons which are almost on-shell.

The regularization probably subtracts the q = 0 case from a q != 0 case. That may make the integrals finite.

The electron throws part of its energy and momentum (q) away before the approach, and catches them back as it recedes. The reduction (or increase) in the electron package alters scattering probability amplitudes.

Why does it alter them in the way which gives the correct result after regularization?

We sum all packages. This sounds like a path integral for a lagrangian.

The classical path and lagrangian density for the process

                               package of energy E

                               and momentum q

                                       ~~~~~~~

                                     /                   \

momentum k  e- --------------------------------->

                                              |

                                              |  momentum p

                                              |

                                             ● Z+

Let us analyze the problem in classical physics. The electron makes a very complex wave into its own field. In principle, we should take at least a billion volume elements of space and calculate the action for all the "paths" the field can take in those billion elements.

In the Feynman diagram, all this complex process is summarized into a simple process where the electron throws a package of energy E and momentum q to itself before meeting the nucleus.

The Feynman formula for the diagram probably calculates some kind of classical action for the simplified path.

A classical lagrangian is typically of the form

       L  =  T  -  V,

where T is the kinetic energy and V is the potential energy.

                     ● field

                   /

                 /       rubber

               /       band

             ● e-

Let us imagine that the field of the electron is another particle which is attached to the electron via a rubber band. The electron exchanges spatial momentum and energy with the other particle.

It makes sense to describe the process as the electron sending a package of 4-momentum to the other particle and later taking that package back. It is a simplified way to describe the complex process.

What would a suitable lagrangian density be for the path? The first guess is, of course, that the Feynman integral formula calculates the action for the path. However, divergences in the Feynman formula mean that something is wrong with it. The regularized version of the formula probably is the correct way to calculate the classical process.

The Feynman formula includes arbitrary momentum q in the package. That does not look right. The momentum q should be

       q = E / E_e * k,

where E_e is the total energy of the electron and k its initial momentum. Maybe wrong momenta cancel each other in the integral?

Momentum p determines which package size is favored. Other sizes should have a small weight. The mass-energy of the electric field is the origin of the packages. That part of the field which does not have time to react to the acceleration of p, is thrown as the package.

We will study this in detail in coming blog postings. There has to be a logical reason why a Feynman diagram calculates classical processes right. What is it?

The photograph model: the Feynman path integral is a way to add wave nature to a classical action optimization problem

Consider a classical system and write a lagrangian for it.

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

To find the classical behavior of the system, find the path q which has an extreme value of the action S (usually a minimum). The action is the integral of the lagrangian density L over the path q:

      S = ∫ L dt

           q

How to add wave nature to this?

Use

       exp( i / h-bar ∫ L dt )

                              q

and calculate sums over paths. That is the Feynman path integral.

Paths q which have an extreme value of the action will have constructive interference. Other paths will have destructive interference.

We converted a problem of classical mechanics into a Feynman path integral problem.

The wavelength of the wave nature is determined by the Planck constant h. When h approaches zero, the wavelength approaches zero. The constructive interference in the path integral is concentrated to sharper and sharper points. In the limit, the Feynman path integral reduces to the classical optimization problem.

Our photograph model of quantum field theory is clarified by the above explanation. At the low level, it is classical physics. The wavelength, which is determined by the Planck constant, adds a wave nature over the classical world, through the Feynman path integral.

We have been wondering how Feynman diagrams can calculate classical things, like the reduced mass of the electron under acceleration. Above we have a probable explanation. Feynman diagrams are an approximate way of calculating Feynman path integrals. And Feynman path integrals are built on classical physics.

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

Paul Dirac's 1933 paper seems to contain similar ideas.

This still does not explain details, like why a virtual photon line in a Feynman diagram has a certain propagator and a certain coupling constant. Maybe they calculate some type of classical action approximately right and therefore reproduce the correct classical behavior? We will look into this specific question in coming blog postings.

Does the rubber plate model of the electron electric field explain bremsstrahlung?

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

Wikipedia seems to claim that bremsstrahlung from a quantum process is much larger than the classical Larmor calculation would suggest. The Gaunt factor g_ff,Born, which measures the quantum/classical ratio (or does it?), may easily be 5 or larger.

In Wikipedia, if we multiply the electron mass and its charge by 10^6, then Wikipedia claims that quantum emission of, say, 500 eV photons is 10X larger than in the classical approximation. An electron of that mass should behave like a classical particle. The Wikipedia claim about the Gaunt factor has to be incorrect.

Electromagnetic radiation is a result of the reduced mass of the electron when the electron is under acceleration. If there is just a single jerk on the electron, instead of cyclic oscillation, then the reduced mass during the jerk is probably less than in regular oscillation. Therefore, the Larmor formula may underestimate the radiation in bremsstrahlung. But this can hardly explain a 5-fold difference.

http://stribor.phyast.pitt.edu/REPORTS/pitt-554/pitt-554.pdf

Looking at literature reveals that the classical approximation describes bremsstrahlung quite well.

O. I. Obolensky and R. H. Pratt (2005) provide Figure 8 which shows good agreement, except for 2 - 30 eV.

http://faculty.wcas.northwestern.edu/yoram/teaching/23astron441CP2013/00web/01papers/rlChap5.pdf

George B. Rybicki and Alan P. Lightman (2004) explain the Gaunt factor. It turns out that the Gaunt factor in Wikipedia is determined by impact parameters b_min and b_max:

        ln(b_max / b_min).

The impact parameter b_max is classical: to produce a wave of frequency f the bypass has to happen in time

       ~ 1 / f.

This is definitely a classical thing.

The parameter b_min is the larger of:

1. the value of b where the straight-line approximation breaks down;

2. the de Broglie wavelength of the electron.

The formula has a logarithmic divergence if we allow b_min to be arbitrarily small. In most cases the authors recommend setting b_min to the de Broglie wavelength.

Suppose that b_max is much larger than b_min.

Classically, the contribution of small b is negligible, which shows that the divergence is an artifact of a wrong formula. Thus, we can cut off at some small fixed b, for example at the de Broglie wavelength. Therefore, b_min is essentially a classical thing, too.

The Gaunt factor is mostly classical, not quantum.

Bremsstrahlung: classical or quantum?

A radio transmitter does many cycles and attains a steady state of power dissipation. Bremsstrahlung is an effect which is caused by an electron doing "half a cycle" as it scatters from a nucleus.

                    ●  Z+

      e-  ------>

Let an electron with 1 MeV kinetic energy pass a nucleus.

Two months ago we blogged about the "length scale problem". The electron may make a very sharp turn at a distance ~ 3 * 10^-15 m from the nucleus. Classically, the wave would then contain frequencies of up to 1 GeV energies. Does the length scale problem appear if the distance is large? No. The length scale problem only concerns distances which are less than the de Broglie wavelength of the electron. We cannot draw the path of the electron with a precision higher than the de Broglie wavelength.

We suggested that destructive interference of various paths in the path integral wipes away very large frequencies. Constructive interference strengthens frequencies which are compatible with the electron kinetic energy.

The rubber plate model (or any classical model) cannot explain the interference pattern. But we believe that the underlying process is classical.

If we increase the masses and the charges of the particles, we approach the classical limit. In the classical limit, the production of the bremsstrahlung wave is very probably a classical process.

We can appeal to the fact: quantum mechanics is affected by classical phenomena, unless those phenomena would break laws of quantum mechanics. In the hydrogen atom, the electron cannot crash into the nucleus, even though classical physics would suggest that happening. But classical corrections are visible in the hydrogen spectrum.

An electron passing a nucleus would classically radiate, and radiation does not break the rules of quantum mechanics. We conclude that the classical phenomenon probably is the "cause" of bremsstrahlung.

Thus, the rubber plate model probably explains bremsstrahlung at the low, classical level. The sharp turn of the electron at the nucleus makes the elastic electric field of the electron to vibrate. That vibration is bremsstrahlung.

Why does the Feynman diagram calculate bremsstrahlung correctly?

The Feynman diagram is

                        bremsstrahlung

                        ~~~~~~~~~~  real photon q

                      /

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

                              | virtual

                              | photon

                              | p

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

and the other diagram where the real photon q is emitted after the encounter with the nucleus Z+.

Let us keep p almost fixed and small. Classically p tells the distance how close the electron came to the nucleus. Classically, this determines a complex electromagnetic wave pattern which the electron sends.

Since the acceleration is toward the nucleus, the classical electron sends a dipole wave where most of the energy is aimed to the direction of the electron velocity, and behind it.

The dipole wave is weak and it contains relatively high frequencies (the length scale problem).

Let us concentrate on the wave which the electron sends directly forward: q is parallel with the velocity v of the electron. If q is small, then p gets a small contribution which is at a right angle to p.

Note that the 4-momentum p contains no energy. It is just spatial momentum. The propagator of p is little changed by the contribution from q.

We get a result:

The spectrum of q is determined by the electron propagator.

The spectrum is not a classical phenomenon. The classical wave contains a small amount of energy for every passing electron, and is dominated by very high frequencies if the electron comes very close to the nucleus.  The quantum output has no photon at all for most electrons, but a few electrons lose a large amount of energy to a single quantum.

Our photograph model of quantum field theory claims that the photograph tries to depict the classical process as well as it can, but is limited by the poor resolution of the photograph. The classical process drains a little bit of energy from every electron. The photograph model then decides to put the burden on a few electrons: they lose a full quantum of energy while most electrons lose nothing.

The electron propagator measures how much off-shell the electron is after the quantum of energy is confiscated from the electron. The more it is off-shell, the smaller the probability amplitude.

Thus, the electron propagator favors low-energy photons.

The quantum behavior is determined by the path integral. We cannot explain the quantum behavior from classical physics. It is the interference of various paths which generates the quantum behavior.

An infrared divergence in the low-frequency photons in bremsstrahlung

Since the electron propagator favors low-frequency photons, it overestimates their number from the process. If we sum over all low frequencies, we may get an infinite sum.

http://scipp.ucsc.edu/~haber/ph218/Soft_Photons_Amita.pdf

Amita Kuttner (2016) studies the reaction

       e+  e-   ---->   μ+  μ-  γ,

that is, a collision of a high-energy pair produces a muon pair and a photon.

Kuttner writes that the infrared divergence in bremsstrahlung is canceled by another infrared divergence in the vertex correction of the process

       e+  e-   ---->   μ+  μ-.

The claim is strange. How can a correction to another process cancel the infinite energy of the photons in the first process?

https://web.pa.msu.edu/courses/2017spring/PHY955/Lectures/lecture.April17.renorm.pdf

The linked page claims that the infrared divergence in bremsstrahlung occurs in classical physics, too. The page suggests the same recipe as Amita Kuttner: let the infrared divergence in the vertex correction of elastic scattering cancel the infinite energy emitted in soft photons.

We need to investigate this. Is there really a divergence in classical bremsstrahlung?

Our photograph model would remove the quantum infrared divergence in this way: since the classical wave, which is produced by the passing electron, contains finite energy in very long wavelengths (actually, very little energy), then the quantum "photograph" of the process will contain at most that energy in very long wavelengths.

We need to find out the classical spectrum. Is there enough energy in very long wavelengths, so that every passing electron will send soft photons? Is there a classical infrared divergence? Is the classical calculation in Wikipedia correct?

The classical process in nonlinear. Usually it is impossible in mathematics to prove anything about nonlinear differential equations. We need to use heuristic arguments to show that the emitted energy in the classical process is finite.

The rubber plate model of the electron electric field and the Larmor formula

In our blog post on February 18, 2021 we calculated (but made calculation errors) that we can explain the Larmor formula if we assume that the electron elastic electric far field uses its inertia to pump out the maximum possible power output from an oscillating charge.

The accurate calculation should be done numerically, but we believe that the rubber plate model is right in this. Joseph Larmor in 1897 derived his formula from classical electromagnetism. Electromagnetic radiation is not a quantum phenomenon, and the rubber plate model sits well in the setup as a general model of wave phenomena.

The problem is general in classical physics: what kind of power output does oscillation of an elastic medium produce? It is often hard to calculate, but we do get qualitative understanding of the process when we know we are dealing with an elastic medium.

We should check Larmor's original calculation. Does he assume the electric field to be elastic? Has anyone calculated the power output in a general elastic medium?

Power output for a tense long rope

The power output is easy to calculate in the case of a tense long rope. Let the end of the rope do circular motion of a radius r at the speed of v. Let the mass of the rope be m per unit of length. Let the wavelength be L.

The wave energy of one wavelength is

       E = L m v^2.

The radial acceleration is

       a = v^2 / r.

The person rotating the rope at the end must do the work E per one rotation of length 2 π r. He must use a force

       F = L m v^2 / (2 π r).

If we write

       F = M a,

we get

       M = L / (2 π)  * m.

In this case, the inertial mass which pumps energy from the movement is the mass of

        1 / (2 π)

of the wavelength of the rope. That is, the mass of one radian of the wave in the rope.

The result sounds reasonable. If the wavelength is 6.28 meters, then it is the inertia of one meter of rope which is pumping the maximum energy from the person rotating the rope.

The mass of the rope is uniform per length, and infinite for an infinite rope.

The elastic electric field: the electromagnetic wave gets "detached" from the static electric field at the distance of sqrt(3) / 2 radians

The mass-energy of the electric field of a charge has a formula 1 / R^2 per unit of the radius R.

At larger distances, the electromagnetic wave lives on the combined electric and magnetic fields. The situation is quite different from a tense rope whose mass per meter quickly approaches zero.

Also, the tension in the rope model should be larger for small R.

Maybe we can approximate the generation of an electromagnetic wave with a rope whose mass per length for small R is 1 / R^2, but constant for large R? The tension is larger for small R and constant for large R.

The power drain is one radian of the wave in the large R zone pumping the maximum energy using its inertia.

The Larmor formula corresponds to the far field more than 3/4 radians away using its whole inertia to pump out the maximum energy.

Let us use one radian of the wave as the unit of length. Let initially the mass of the rope per unit of length be 1 / R^2. The total mass M of the rope farther than 3/4 radians away would be

       M = 4/3.

We get the following model: at the distance of R >= sqrt(3) / 2 radians we let the mass of the rope per unit length be constant. For smaller R, the rope still has mass density 1 / R^2.

The rope model indicates that the electromagnetic wave is "detached" from the static electric field of the charge already at a distance of sqrt(3) / 2 radians of the electromagnetic wave.

https://www.rfcafe.com/references/electrical/near-far-field.htm

In radio antenna technology, the field > 1 radian away is considered the far field (Tom Lecklider, 2005).

We get a more detailed qualitative rubber plate model. The far field starts already at the distance 1 radian, or even closer.

The electromagnetic wave compared to the transverse static electric field

Let us do more calculations. Let an electron do a circle whose radius is r, at a speed v. Then

      a = v^2 / r.

The transverse component of the static electric field at a distance R is

       E = e r / (4 π ε_0 R^3).

The Larmor formula says that the power output is

       P = e^2 a^2 / (6 π ε_0 c^3).

The energy density of the power output is

       e^2 a^2 / (24 π^2 ε_0 c^4 R^2).

That corresponds to a transverse dynamic electric field

       E' = e a / (sqrt(12) π ε_0 c^2 R).

E' = E if

       v^2 / (r sqrt(12) c^2 R) = r / (4 R^3)

<=>

       v^2 / (sqrt(12) c^2) = r^2 / (4 R^2)

<=>

       v / (sqrt^2(12) / 2 * c) = r / R

<=>

       R = r / v * 0.9306 c

           = 0.1481 *  2 π r / v * c.

The transverse static electric field is equal to the transverse dynamic electric field at the distance of 0.1481 wavelengths. That is 0.9306 radians. The value is close to the radio technology far field distance which is 1 radian.

Determination of the "lagging mass" in the Larmor formula

Let us do the February 18, 2021 calculation again, this time correctly. We want to find m such that

       P = e^2 v^4 / (6 π ε_0 c^3 r^2)

           = m v^3 / r = m a v = F v

<=>

       m c^2 = e^2 v / (2 π r c * 3 ε_0)

                   = e^2 / (3 ε_0 λ).

The energy density of the electron static field at a distance R is

       E_d(R) = e^2 / (32 π^2 ε_0 R^4).

The radial density is

       e^2 / (8 π ε_0 R^2).

The energy of the field farther than R is

       E(R) = e^2 / (8 π ε_0 R).

Let us solve R from m c^2 = E(R). We get

       R = 3 / (8 π)  λ,

or 0.1194 wavelengths, or 3/4 radians.

The energy of the static electric field outside the classical radius r_e is only HALF of the mass-energy m_e of the electron

On February 18, 2021 we were thinking that the mass-energy of the static field of the electron outside the classical radius r_e is m_e. But it is only 1/2 m_e.

The potential of an electron at the distance r_e from a proton is -m_e c^2.

Thus, the classical radius r_e, in a sense, corresponds to the distance of two unit charges, but is 2X "too large" for the mass-energy of the electric field of a unit charge.

Conclusions

The far field of an oscillating charge turns out to start surprisingly close, only at 3/4 ... 1 radians from the charge, where the wavelength of the electromagnetic wave is denoted as 2 π radians.

The electromagnetic wave lives on the oscillating magnetic and electric fields and does not need the static electric field of the charge. We can say that the wave "breaks off", or is detached from the static field, at the distance of 1 radian.

We can explain the Larmor formula by saying that the static electric field farther than 3/4 radians uses its entire inertia to pump the maximum energy out of the oscillation acceleration.

Which phenomena in QED are quantum, which classical?

There is a very easy way to determine which phenomena are "quantum" and which are classical.

If the formula does not depend on the Planck constant h, then the phenomenon is probably classical.

"Quantum" phenomena are wave phenomena which depend on the wavelength of the particle. The wavelength is determined by the Planck constant.

Let us evaluate which common phenomena are quantum and which classical:

1. Electron orbitals in hydrogen are quantum.

2. The electron spin and the Dirac equation magnetic moment are quantum. Zitterbewegung is quantum.

3. The anomalous magnetic moment correction factor to the Dirac equation value is classical, if we accept that the zitterbewegung circle radius is a quantum phenomenon determined by h.

4. The Lamb shift is classical, but the electron orbit itself is determined by h.

5. Coulomb scattering is classical.

6. The electron-positron annihilation cross section is classical.

7. Compton scattering is quantum. It depends on the wavelength of the photon.

8. Generation of radio waves by an accelerating large charge is classical, though the energy packet size is quantum.

9. Bremsstrahlung is at the low level classical like item 8, but the energy packet size dictates the output, and therefore bremsstrahlung is quantum.

Question. Coulomb scattering is classical, but standard QED believes that it is affected by vacuum polarization. Is the hypothetical vacuum polarization classical?

The rubber plate model of the electron electric field explains the Lamb shift?

We showed last week that we can explain the anomalous magnetic moment of the electron with the rubber plate model. The far field of the electron does not have time to react to the very fast spinning of the electron charge in the zitterbewegung loop. That reduces the effective mass of the electron.

We believe that the Lamb shift is caused by the same phenomenon, but it is harder to calculate an approximation, since the acceleration depends on the distance of the electron from the proton.

Furthermore, the zitterbewegung movement complicates things. How should we handle the assumed light speed orbit of the electron?

Let us assume that the electron on the 2s-orbital is on a very narrow elliptical path whose focus is the proton.

The speed of the electron close to the proton is the speed of light, if we assume zitterbewegung to be real motion.

The reduced mass might follow a similar formula to the calculation of the anomalous magnetic moment:

       m_eff = (1 - r_e / R) m_e,

where r_e is the classical radius of the electron, R is the distance where light can reach during one "cycle" of the accelerated movement, and m_eff is the reduced mass.

The formula is nonsensical when R < r_e. Let us ignore that for now.

Reduced mass reduces the momentum

      p = m v,

but how much? If the mass is 1% less, then the momentum at a certain distance r from the proton is in the non-relativistic case 0.5% less because the electron has less time to collect momentum.

A 0.5% reduction in momentum in non-relativistic movement requires a 1% reduction in the effective potential V, if the mass were not reduced.

Then

       V_eff = (1 - r_e / R) V.

The effect raises the energy of both 2s and 2p, but since 2p has zero probability density at the nucleus, the effect is larger for 2s.

                 --->

     _________________ 

   /     

  |   ● proton

   \__________________ e- at distance r

                <---

What is the "cycle" in our case? The electron does a back-and-forth movement close to the proton in its infinitesimally narrow elliptical orbit.

If the electron starts at a radius r, the cycle might be 2 r. But is that right? It really is just half a cycle because the electron does not return to the same state.

Should we multiply the 2 r by 2?

Let us not multiply.

The Darwin term

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

The Darwin term is the result of smearing the potential by the radius

       λ_e / (2 π) = 4 * 10^-13 m,

where λ_e is the Compton wavelength of the electron.

We know that

       λ_e = 137 r_e.

Then

       r_e / (λ_e / 2 π) = 4.6%.

Let us check the numerical values of the Darwin term and the Lamb shift.

       Darwin = 90.57 μeV

       Lamb    = 1.057 GHz * h / e

                      = 4.371 μeV

The Lamb shift is 4.8% on top of the Darwin term.

It is probably not a coincidence that the percentages are that close.

A very crude calculation

Let us calculate what is the average correction to the 2s potential relative to the 2p potential, when we take into account the reduced mass of the electron.

We start by calculating the effect for r < 5 * 10^-12 m.

Zitterbewegung does not have too much an effect on the electron at this distance, because the zitterbewegung radius is only 4 * 10^-13 m.

The speed of the electron (ignoring zitterbewegung) is roughly 0.03 c at a distance

       r = 5 * 10^-12 m

from the nucleus.

The potential there is roughly -300 eV.

The effective mass reduction: when the electron moves the distance 2 r near the nucleus, its field gets a message at the speed of light, of the moving electron 67 r away:

       R = 67 r.

We can reduce the electron mass by the mass-energy of the far field > 67 r away. The reduction is

       1 / 67 * 300 eV = 4.5 eV,

which is 10^-5 of the electron mass.

The effective potential is reduced by the factor 10^-5 from the value -300 eV.

https://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html

In the link there are formulae for the radial wave functions for 2s and 2p by Jim Branson, 2013.

The 2s electron wave function squared at r = 0.1 is 0.4 if r is given in units of the Bohr radius. That is the probability density.

The volume of the sphere r < 5 * 10^-12 m is

       4 * 10^-3 cubic units

and the integrated probability

       1.6 * 10^-3.

Thus, the average correction of the potential for the 2s orbital is

    ΔE = 1.6 * 10^-3 * 10^-5 * 300 eV

 

          = 4.8 μeV.

The speed of the electron is ~ sqrt(1 / r). The effect of the potential is ~ 1 / r^2. The integrated probability is ~ r^3. The contribution grows as sqrt(r) when r is very small.

http://hyperphysics.phy-astr.gsu.edu/hbase/hydwf.html

The contribution from

       r = 5 * 10^-11 m

is

       ΔE = 0.8 μeV,

because the probability density of 2s there is only 0.045 and the probability density of 2p is already around 0.02

The contribution from

       r = 7 * 10^-11 m ... 20 * 10^-11 m

is negative because 2p has there larger probability density than 2s. The probability density of 2p minus 2s is 0.02.

The contribution is

       ΔE = -1.3 μeV.

The contribution from

       r > 20 * 10^-11 m

is only

       ΔE = 0.1 μeV,

because there the integrated probability grows slowly, and 2s and 2p have roughly equal probability densities.

Zitterbewegung affects the contribution of very short distances. The contribution of

       r = 5 * 10^-13 m

would be

       ΔE = 1.8 μeV

if zitterbewegung has no effect. If we put a "cutoff" at the zitterbewegung radius 4 * 10^-13 m, then the contribution is

       ΔE = 0.9 μeV.

The contribution from 5 * 10^-14 m might be 0.6 μeV. It is not clear what happens at distances shorter than the classical radius 3 * 10^-15 m.

The sum of the contributions, if we put the "cutoff" at the zitterbewegung radius, is

       ΔE = 5.3 μeV,

which is close to the Lamb shift 4.4 μeV.

https://dspace.mit.edu/bitstream/handle/1721.1/72773/31741267-MIT.pdf%3Fsequence%3D2

In his back-of-the-envelope calculation, Hans Bethe in 1947 did put a cutoff at the the zitterbewegung radius h-bar / (m_e c) (see the linked document by Philip O. Koch (1956), page 18).

We made a lot of assumptions in the calculation. The calculated figure could vary 4-fold with different assumptions.

Considerations about classical versus quantum phenomena

The mass reduction is a classical phenomenon. Generally, quantum mechanics is aware of classical physics - quantum mechanics does not ignore classical phenomena. Therefore, the mass reduction should be visible in the hydrogen spectrum. Since the order of magnitude of this classical phenomenon agrees with the Lamb shift, it is probable that it is the same effect.

If we make the particles very heavy and increase their charges in the right proportion, then the "super-electron" passes the "super-proton" along the same orbit as the ordinary electron. The superparticles will behave according to classical physics, if we believe in the correspondence principle. Mass reduction will happen. How does QED handle such superparticles? There is no upper limit for the mass and energy in Feynman diagrams. They should calculate the right approximation for the classical case.

Conclusions

We believe that the Lamb shift can be explained with the rubber plate model of the electron electric field. The field is elastic and the effective mass of the electron is reduced close to the nucleus, because of the large acceleration and speed.

Therefore, the Lamb shift is a classical effect which affects the Schrödinger equation solutions of the hydrogen atom.

Since the effect is classical, there is no need for regularization or renormalization of the electron mass. Hans Bethe in 1947 regularized the self-energy of the electron by calculating the difference between a free electron and the bound electron.

The rubber plate model has turned out to be fruitful: it qualitatively describes the reaction of the electron to its own field, and can explain how a radio transmitter works. It can explain the electron anomalous magnetic moment. It may explain the vertex correction in QED.