Sunday, August 21, 2022

James Webb results agree with our Minkowski & newtonian cosmology?

UPDATE September 5, 2022: A major problem of the Milne model is how to explain the "first acoustic peak" in the cosmic microwave background (CMB) angular power spectrum. 


A. Benoit-Levy and G. Chardin (2011) suggest that it might be associated with matter/antimatter zones in the universe, and not with baryon acoustic oscillations.


The Hubble constant value can be derived by looking at the angular diameter of these CMB acoustic baryon oscillations in the sky. Andrei Cuceu et al. (2020) quote a value of 67.4 km/s per megaparsec. That is smaller than the value 74 km/s per megaparsec which is measured from standard candles. Could it be that the first acoustic peak is not associated with baryon acoustic oscillations?

----

https://m.slashdot.org/story/403711

The first pictures from the James Webb telescope reveal surprisingly old and surprisingly well-developed galaxies.

"Galaxy formation models may now need a revision, as current ones hold that gas clouds should be far slower to coalesce into stars and galaxies than is suggested by Webb’s galaxy-rich images of the early universe, less than 500 million years after the big bang. “This is way outside the box of what models were predicting,” says Garth Illingworth of the University of California (UC), Santa Cruz."



We wrote about our Minkowski & newtonian model on January 31, 2022. The expansion of the universe happens at a constant speed, just like in the Milne model.

In the Milne model the cosmic microwave background which we observe now was born much farther away from us than in the ΛCDM model. The same for the first galaxies which we observe now. The apparent angular diameter of the first galaxies is smaller in the Milne model than in the ΛCDM model.


         \                         /   now
           \                     /
             \                 /
               \             /
                 \         /   O  first galaxies
                   \     /
                     \ /   Big Bang

           Milne model



         \                         /  now
           \                     /
              \               /   O first galaxies
                 \         /
                     \ /  Big Bang

                  ΛCDM
                  
                  
In the Milne model the first galaxies have a larger redshift and their apparent angular diameter is smaller than in ΛCDM.

James Webb supports the Milne model.

The electron propagator in Thomson scattering is classical

Our analysis in the previous blog post brought up the problem what process, or wave, does the electron propagator really describe in a Feynman diagram.

We know that the photon propagator, for an unknown reason, models Coulomb scattering.


The electron attached to its electric field: the "rubber string" model


Classically, the electron possesses a static electric field. We may imagine that the lines of force are made of rubber.

If we shake the electron, waves will propagate along the rubber strings. The waves are electromagnetic wave.

We may interpret that the rubber mesh controls the movement of the pointlike particle electron. The electron is a kind of an oscillating mass attached to the rubber mesh.


Classical Thomson scattering


                                        electric
                                    lines of force
                                             |
              |    |    |               --- ● ---  
                                             |

           laser beam       electron e-


The laser beam shakes the electron. New electromagnetic waves propagate in the lines of force of the electron.

We may imagine that the laser beam gives small impulses to the electron at very short time intervals. The response of the electron and the rubber mesh is an impulse response.

Recall that a propagator is a Fourier component of the impulse response of a wave equation. For example, if we hit a drum skin with a sharp hammer, the Fourier decomposition of the wave is the set of propagators.

The impulses to the electron have a cycle determined by the frequency of the laser. Constructive interference strengthens the output at this frequency. Destructive interference wipes out other frequencies.

The electron will act as a radio transmitter and send an electromagnetic wave to many directions. It scatters the incoming laser light.

The propagator in this classical treatment tells us how effective the electron is in outputting energy to the scattered "channel", or scattered wave.

We can, in principle, calculate with a computer how the electromagnetic field behaves, and what is the value of the propagator.

The classical Feynman diagram looks like this:


        laser                              scattered wave
        ~~~~~~                       ~~~~~~~~~~~
                           \              /
  e-  ---------------------------------------------------
                            virtual
                           electron 


It is essentially the same as the quantum electrodynamics (QED) Feynman diagram.

The "virtual electron" is the propagator. We may imagine that it represents the electron in the mesh after an impulse hit the electron.


QED Thomson scattering


As we wrote above, the Feynman diagram is essentially the same as in the classical process. Numerical results from Feynman formulae agree with the classical treatment.

Note the following thing: the propagator in the QED diagram is from the impulse response of the Dirac equation. We were able to connect the Dirac equation to a classical process.

The process in Thomson scattering is non-relativistic. The Dirac equation in that case is equivalent to the Pauli equation, or the Schrödinger equation. The propagator for the Schrödinger equation seems to be complicated.


What is a "virtual" electron and how does an electron "absorb" a photon


In the classical interpretation, a virtual electron means the system electron & its electric field where the system has been disturbed by an impulse.

The absorption of a photon means that the system goes to a disturbed (excited) state.

Classically, the electron is a point particle. It cannot have excited states on its own. An excited state has to be the electron in an interaction with something else, in this case its own electric field.

When an excited electron emits a photon, that means that the oscillation of its electric field moves farther from the electron and starts a life of its own.

A propagator does not make much sense for a point particle. But a propagator for the system the electron & its field makes a lot of sense.

Question. How can we extend the classical propagator to an "electron field"? That is, we would not have a point particle but some kind of a field. This might show the connection between the Dirac equation and the classical model.


It is not clear if we can define an electron field in a reasonable way. If the field simply describes the position and the phase (in a path integral) of a single electron, then there is no obvious interaction between different parts of the field.

If the electron is either in the zone A or the zone B of space, there is no interaction between the zones A and B. This is very different from a drum skin where A and B always interact. In a drum skin, a wave equation is natural, but it is not natural for mutually exclusive histories.

The Dirac equation is used in Feynman diagrams to calculate the electron propagator. That is, a single electron is interacting with something else. Then it makes sense to consider the system the electron & its field.

The Dirac equation does have a conserved probability current and it does predict the magnetic moment of the electron. How can we explain these if the equation only describes an interacting electron?

Maybe the electron is a wave phenomenon, after all? The spin of an electromagnetic wave probably is a wave phenomenon.

But then we face the problem how to attach the electric field to the electron.


The Dirac equation describes the system the electron & its field?


The electron propagator in the Dirac field describes something which is off-shell, or not in its ground state.

If the QED propagator is able to calculate something similar as the classical propagator, then it is natural to assume that the QED propagator describes the combined system the electron and its field. We cannot remove the electric field from the electron. It makes sense that the Dirac equation describes the entire system.

However, it is not clear how the classical system gives rise to the Dirac equation. How do we explain zitterbewegung and the magnetic moment?


Conclusions


The analogue of a photon propagator is a sharp hammer hitting a drum skin (where the skin is actually the three-dimensional space).

The Fourier decomposition of the associated - 1 / r potential has the familiar formula

       ~ 1 / p²,

where p is the 4-momentum of the Fourier component.


              |
            _|_  
           |__| -->     ● e-

       hammer


The analogue of the electron propagator might be a hammer hitting the electron. The electric field of the electron moves a little bit. That is like adding a dipole where a positron e+ is put to the old position of the electron and a new electron is put to the new position.

The dipole potential is roughly

       ~ 1 / r²,

and the Fourier decomposition is roughly

       ~ 1 / |p|,

where p is the 4-momentum of the component.

The formula 1 / |p| is similar to the Feynman electron propagator if we set E = 0 and m = 0, where E is the energy of the electron and m is its mass.

Sunday, August 14, 2022

Pair annihilation: the virtual electron is a carrier of a "force"

In our blog post February 16, 2021 we discussed the Feynman diagram of pair annihilation. It looks like scattering where the virtual electron carries (repulsive) momentum.


      momentum -p
             e+ ---------------- ~~~~~~~~~~  photon -k
                                     |
                                     |  virtual
                                     |  electron p - k
                                     |
              e-  --------------- ~~~~~~~~~~  photon k
      momentum p


Suppose that the positron and the electron approach from opposite directions at the same speed. The momenta are p and -p.

The momenta of the outgoing photons are k and -k.

The produced photons cannot carry away all the momentum of the particles. The electron gives the excess momentum p - k to the positron in the form of a virtual electron.


The virtual electron as a carrier of a "force"


The annihilation process looks like a scattering event where there is an approximate repulsive 1 / r² potential between the electron and the positron.

The Fourier transform of a 1 / r potential is

       ~ 1 / q².

The propagator of the photon is of this form.


The Fourier transform of a 1 / r² potential is of the form

       ~ 1 / |q|,

where q is the "momentum" of the Fourier component.

The propagator of the electron is very crudely of the form 1 / |q|.

Our blog post on February 16, 2021 shows that the potential in annihilation is actually even steeper than 1 / r², according to the cross sections calculated from Feynman diagrams.


We have been wrong in our attempts to explain annihilation by the electric attraction between the pair


The momentum transfer in the Feynman annihilation diagram does not happen through a photon.

Maybe it is not possible to make a (semi)classical model of annihilation using the electric force?


The sharp hammer model again: Huygens


We have explained the static electric field of an electric point charge with a sharp hammer which keeps hitting a "drum skin" at the charge, and makes a depression to the skin.

The Huygens principle is that a wave is absorbed by each point in space, and the point then acts as a new source for the wave.

In the case of the electron and its electric field, the Huygens principle might be something like a sharp hammer hitting a drum skin and creating the electron wave and its electric field.

The hit would produce a "point impulse" both in the Dirac field of the electron and the electromagnetic field.

Now we come to the Feynman diagram principle that all "paths" which conserve energy and momentum are allowed, and their probability amplitudes must be summed.

The idea is that the when the electron wave arrives at a spacetime point x, the wave is "absorbed" and immediately recreated with an impulse which hits both the Dirac field and the electromagnetic field.

Our sharp hammer becomes more versatile: it hits two fields at once. All combinations of responses from the two fields are allowed.

The virtual electron in the annihilation diagram is one component of the impulse response. The photon flying away is another component.

If we hit a drum skin, only sine wave "on-shell" Fourier components can travel over a long distance. Other, "off-shell" components have a short range of the effect. This may explain why the electron and the positron must come close to each other in order to annihilate.


A particle model of electron-positron electric scattering

    
             <------  e+
                         |
                         |  electric attraction
                         |
                         e-  ------->


The particle model is very simple and intuitive. Both particles are treated as point charges with an attractive electric force. The paths are calculated with classical relativistic mechanics. This gives results which, according to literature, are (almost?) exactly the same as when calculated using the simplest Feynman diagram.


A wave model of the scattering

 
                   <------ e+
          |       |       |        |        |        overlapping
                |       |        |        |        | waves
                             e-  ------>


Let us then try to form an intuitive wave model from a Feynman diagram. Let us have an average of one electron and one positron in a cubic meter of space.

We do not know the positions of the particles, and represent them with standard plane wave solutions

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

where u is the spinor.

The Feynman diagram is

         
         e+ ----------------------------------------
                                 |
                                 | virtual photon
                                 |
         e- -----------------------------------------


How can we relate this to the wave diagram above?

If we take seriously the wave interpretation, then the scattering of colliding beams of electrons and positrons is a nonlinear effect. If there were just one beam, then scattering would not happen. In a linear system we would be able to sum the solutions of the two beams to obtain a new solution: there would be no scattering.

Suppose that we try to add a "source" to the wave equation of the electron. Something like

       D(ψ) = f(φ, ψ),

where ψ is the electron wave function, φ is the positron wave function, and D(ψ) = 0 is the Dirac wave function of the free electron.

The source term f(φ, ψ) depends on both wave functions.

But if the wave functions φ and ψ are essentially constant in the cubic meter (save a phase factor), how can the source generate a wave which is significantly scattered, to an angle, say, 90 degrees?

To simulate the scattering of the point particles we should have scattered waves where the cross section is

       ~ 1 / α²

where α > 0 is the deflection angle. How to generate such waves without having a localized disturbance of the field?

That looks hard. In the hydrogen atom model, one particle, the proton is treated as a particle while the electron is treated as a wave.

Our view in this blog has been that particles are the "true" nature, and that any wave phenomena are due to path integrals.


An analysis of the electron-positron scattering Feynman diagram


We showed that a pure wave model of the scattering does not work. What does the Feynman diagram then really calculate?

         
         e+ ----------------------------------------
                                 |
                                 | virtual photon
                                 |
         e- -----------------------------------------


Let us assume that the underlying process really is the classical Coulomb scattering of point particles. But we do not know the precise position of the particles. We have to calculate the path integral for very many possible paths which the particles can take. The path integral is a collective phenomenon of all the possible paths. The path integral is the "wave" associated with the process.

What is the photon in the Feynman diagram? It summarizes the interaction in these very many paths.

Why is the photon propagator

       ~ 1 / p²

the "right" way to calculate the effect of the interaction?

It is the Fourier component of the 1 / r potential, but there is no obvious reason why the component would correctly capture the cross section of classical scattering.


           \     \     \        scattered flux
              ________
             |              |
             |              |  cubic meter
             |_______|
                
                 \     \     \   scattered flux
                

Let us again have that cubic meter where electron and positron beams meet. The scattered fluxes come quite uniformly from the entire volume.

We can model the scattering by assuming that for each centimeter of the path of the electron, a small portion of the electron flux gets scattered to various deflection angles, and a corresponding part of the positron flux gets scattered to the opposite direction. It is like both fluxes would travel in a nonuniform medium.

The force which causes the scattering is the electric force, and the propagator for some reason happens to capture it correctly.

Since we cannot interpret the diagram purely with waves, we conclude that the Feynman diagram really does describe the encounter of two particles.

We may use a wave description for one of the particles, though.

What about using a wave packet description for both of the particles? We can make the packets to pass each other at some short distance. That might work reasonably well if the distance is larger than the wavelength.

Classically, we have two point charges, and their electromagnetic fields are kind of "waves".


The interpretation of the annihilation Feynman diagram


Our analysis of the Coulomb scattering diagram concluded that the photon propagator "for some reason" happens to work in that case.

In the annihilation diagram, the "interaction" is not by the photon propagator, but by the electron propagator. How do we analyze this?



Conclusions


A lot of questions but few answers. The key problem in this blog post is what does the electron propagator model in a Feynman diagram. Is it a wave? Is it a path integral of point-like particles?

We will analyze Thomson scattering in the next blog post. There we are able to connect the electron propagator to a classical process.


Sunday, July 31, 2022

The crank model of pair production

Let us continue to develop the crank model which we introduced in the previous blog post.


Pair annihilation / production: a generator of pairs

                 
                          e-     e+
          |             |      |              |
          |             |      |  ----- axis of rotation
          |             |      |              |
   photon --->                       <--- photon


In the previous blog post we suggested that mainly the spins of the incoming electron and positron are responsible for producing the outgoing to two photons in annihilation.

If we place two rotating charges as above, with opposite spins, they will together act as an approximate (linear) electric dipole transmitter.

The reverse process is pair production. Suppose that we let two extremely strong laser beams collide as above. They might create a continuous flow of pairs.

The production of pairs would then be a non-perturbative classical process.

There are two "cranks" in this case. Turning them with incoming laser beams can be interpreted as a "generator" of pairs, or alternatively, the cranks form a "transmitter" of pairs. The process would be classical.

The classical description of the electron and positron streams would be a classical wave. Individual particles are then a quantum phenomenon, just like a photon is a quantum of a classical electromagnetic wave.

However, we have a problem: how to explain the collision of two very high-frequency photons? How do they give the kinetic energies to the created electron and positron?


Pair annihilation cross section


In our February 8, 2021 blog post we noted that the cross section for the annihilation of slow electron - positron pairs is

       σ ~ 1 / β²

in a simple classical model, while it is

       σ ~ 1 / β

in Feynman diagrams. Here, the speed of the colliding electron and the positron is

      v = β c,

where c is the speed of light.

The disrepancy suggests that we cannot treat a slow electron as a "scalar" point particle, like we did in our simple classical model. That makes sense, since the electron does have a spin.

What about large energies?


Douglas M. Gingrich (2004) gives the cross section for fast pairs:

       σ ~ 1 / E * ln(2 E / m),

where m is the electron mass and E is the kinetic energy of the the electron and the positron. The particles approach from opposite directions.

The result differs from a classical calculation where we try to approximate the production of the photons with the Larmor formula. The cross section is too large. It looks like the electron is not like a classical particle even with large energies.


A. Hartin (2007) gives the cross section for the Breit-Wheeler process of pair production from the collision of two gamma ray photons.

If the energies of the photons are ~ 1 MeV, then the cross section for annihilation, pair production, and Compton scattering is of the order

       σ ~ π r₀²,

where r₀ is the electron classical radius. Does this offer us any insight?

We have earlier calculated that in elastic scattering of electrons and positrons, we can use the classical approximation of charged point particles.

How can we reconcile all this? If we want to build a (semi)classical model of the electron and the photon, it should be able to explain  these properties.


Conclusions


The crank model does not seem to help us.

Let us once again look at the "length scale problem" of bremsstrahlung, which we have discussed several times in the past three years.

The problem is the following. If we model a 1 MeV electron as a classical point particle which zooms past a proton, then the it should pass the proton at a distance of

       ~ 10⁻¹⁵ m

to shed most of its kinetic energy. 

But the wavelength of the bremsstrahlung photon is roughly 1,000 times larger. How can the encounter produce such a photon?

In pair annihilation we have the length scale problem if we assume that the electron is a point particle.

Tuesday, July 26, 2022

The mass-energy of the electron is in the rotation?

For a photon, its energy is purely kinetic energy. If we try to stop a photon, its energy drops to zero.

The electron has a mass and it rotates since it has a nonzero spin. Zitterbewegung suggests that the electron moves around a circular loop at the speed of light.

Maybe the mass-energy of the electron is mainly in its rotation? In our water vortex model that is, indeed, the case.

This has relevance for models of pair annihilation. Maybe the released energy mainly does not come from the attraction between the electron and the positron.

In an earlier blog post we remarked that the formula of the cross section of annihilation suggests at a force which goes as 1 / r³. As if the electric attraction would not be the force which annihilates the pair.

We should develop a model of pair production / annihilation where the spin rotation is the central feature.


The electron and the positron doing the zitterbewegung to opposite directions


                         e+        e-
                         |         |
                         |         |      ------ axis of the orbits
                         |         |


In the diagram, the electron and the positron do a circular orbit which is normal to the screen. They have the opposite spins => they orbit to opposite directions.

                               |         radiation
       (     (    (           |        )     )     )
                               |

                    electric dipole


The system is similar to an electric dipole (not rotating) which radiates to its sides.

The new model differs from the vortex model of our previous blog post. The photons are emitted to the directions of the spin axes.

Let us check the literature. If we control the spins of an annihilating pair, to which direction are the photons emitted?


In the link Ali Moh and Tim Sylvester calculate the direction of emitted photons. The spin directions are not controlled, though.

If we set E = 0 and p = 0 in their formula, the cross section is zero. Apparently, a positronium "atom" is formed.

If E is very small, then the cross section goes as

        cos²(θ),

where θ is the angle from the direction of the colliding electron and positron.


This is somewhat similar to the power density of a dipole antenna:

        sin²(α),

where α is the angle from the dipole axis.


Feynman diagrams destroy information?


       p
       e-  --------------  ---------------  photon p'
                             |
                             |
      e+  --------------  ---------------  photon p''
     -p


Suppose that the electron arrives from the left and the positron from the right. They have opposite momenta p and -p.

We calculate the value of the formula for two photons of momenta p' and p''.

Since Feynman diagrams only do perturbative calculation, they are expected to lose information in the process.

The phase of the photons depends on where we measure them. The Feynman formula cannot state the phase information. It loses the phase information.










                            ---------------------------  photon
                          /               ---------------  photon
                        /               /
       e-  ----------------------------------------


In the clip (18) we have the Feynman formula for the diagram above. The incoming and outgoing electron lines are  the 4-component spinors v and u. The internal electron lines are the propagators.

An internal line electron is not described with a spinor, but with a propagator.

Classical Thomson scattering looks very much the same for an electron and a positron. The Feynman diagram should calculate the exact same probability amplitudes for an electron and a positron.

There is no classical annihilation. One might guess that in a classical annihilation process the phase of the two photons is changed by 180 degrees if we switch the electron and the positron. That is because the switch reverses the direction of the electron - positron dipole.

We are looking for a classical wave model for annihilation. In the ideal case the model is non-perturbative and retains all the information which is fed in. It is a unitary model.


Our "spider" model of pair production


A couple of years ago we tried to build a "spider" model of pair production.


                                O  spider
                              / | \
        -----------------------------------------------
           left string          right string


A spider stands on tense string and uses its legs to rotate the left part and the right part to opposite directions. Our zitterbewegung model above bears a resemblance to the spider model, if we look at is as a pair production model. The "spider" is the two photons.

But how can we explain Compton scattering with the spider model?
  

                 ___                     O  spider
      _____/       \_________/ | \_________
           wave ---> 


We can imagine that a wave in the string progresses through the following "spider mechanism": the spider uses the torque in one leg to work against an incoming wave from the left and uses the energy and the torque which it gains from that to create a new wave which proceeds to to the right.

The process resembles pair production. Did we find the relation between Compton scattering and pair production?


The electron as a rotating structured object and the scattering of a 1 MeV photon: a plasma?



                                   ^  zitterbewegung loop
                                   |
                     ----------● e-

                    r = 4 * 10⁻¹³ m


There may be a fundamental difference between low-frequency photons scattering from an electron and high-energy photons.

Let us imagine that the electron is some kind of a rotating structure whose radius is its Compton wavelength divided by two pi:

       r = 2.4 * 10⁻¹² m / (2 π) 
          =  4 * 10⁻¹³ m.

If the wavelength of the photon is much larger, the electron appears as an essentially point charge to the electromagnetic wave. An electron inside a 500 nanometer laser beam oscillates back and forth quite classically. A low-energy photon does Thomson scattering, which can be calculated with a classical model of a point charge.

However, a photon whose wavelength is less than 1/2 of the electron Compton wavelength (> 1 MeV), can see and distort the internal structure of the electron. When it hits the electron, the electron may enter an excited state.

This high-energy process might have a similarity to pair production.

We suggested some kind of a "plasma" model for pair production in our previous blog post. If the internal structure of the electron is severely distorted by the 1 MeV photon, its state might resemble some kind of a plasma.

The emission of the scattered photon happens from this plasma.

No one has ever observed an electron in a stable or metastable excited state. The only stable state is the ground state. However, when a 1 MeV photon hits the electron, it may enter a very short-lived excited state.

How could we fit the spider model to Compton scattering? If there is no photon, the spider harvests the energy and the torque from the incoming wave and "turns the crank" to create the outgoing wave.

If a photon simultaneously is absorbed, we might consider it as some additional energy and angular momentum which the spider puts to "turning the crank".

What is the photon emission then?

The outgoing photon holds a crank which the electron turns?

Maybe we should drop the spider altogether and imagine that the photons are holding the cranks?


The crank model


The name of this model is not a pun :).

Let us model the electron (Dirac) field with a tense string. A photon can interact with it by "turning a crank".


         photon
         ~~~~~~   __   
                            |  crank
          ----------------------------------------  Dirac field


If a photon arrives, it can donate its energy and momentum to a rotating movement of the string. The crank exerts a torque on the string.

The same process backward is the creation of a new photon. The rotating string exerts a torque on the crank.

Now we realize that the model is similar to a Feynman diagram which describes an absorption or an emission of a photon.

But the Feynman diagram is an abstract description, while we aim at a non-perturbative classical description of the process.


Conclusions


A central idea in this blog post is that the spin of the electron is an important classical feature of the particle. It may be that a half of the mass-energy of the electron is kinetic energy of its spinning motion. The other half might be potential energy in the centripetal force which keeps the orbiting parts of the electron in the orbit.

We will next analyze the crank model. Our goal is to construct a classical, non-perturbative model of pair production/annihilation.

I earlier blog posts we have talked about a "generator" which takes strong laser beams as an input and produces a stream of electron-positron pairs. The generator would be analogous to a rotating electric dipole which produces, in a non-perturbative way, a stream of photons. Cranks might be able to function as a generator.

Sunday, July 17, 2022

The electron is a vortex?

Let us consider waves in water. If we disturb the surface of water and put energy into it, we usually create waves. The energy of a wave moves and spreads out rapidly. Waves have the spin 0. They do not carry spin angular momentum.

A wave consists of hills and valleys in the surface of water.


The vortex model of the electron


















If we disturb water in a special way, then we can create a vortex. A vortex is an almost persistent valley in the surface of water, and it may stay at the same place, in contrast to waves.

Now we see an immediate analogy to electromagnetic waves and to the electron. The electron is a persistent source of the electric field, while the field varies rapidly in an electromagnetic wave.

Can a photon have the spin 0? In a classical electromagnetic wave the answer is definitely yes: an oscillating dipole generates a wave which makes a freely floating test charge to move back and forth.

An electron can only have the spin 1/2 or -1/2. The spin cannot be 0. This is like for a vortex: there has to be angular momentum for the vortex to exist.

The analogy might be this:

1. the electric potential corresponds to the height of the surface of water;

2. the electric charge is the water itself;

3. the magnetic field corresponds to the movement of water, just like it in electromagnetism corresponds to the flow of charge.


This analogy explains why the electron must possess a magnetic field: a vortex must contain a rotating motion of water to exist. It also explains why the spin is always ≠ 0.

If we imagine that the electron moves at the speed of light along the zitterbewegung loop, then the magnetic field at the center is ~ 10⁷ tesla.

Vortices in water do not have an analogy for the positron, though. There exists no persistent hill in the surface of water.

Valleys and hills in a water wave can exist because there is inertia in the mass of water. A vortex is, in a vague sense, a wave which formed a loop: the movement and the inertia of water makes the valley persistent.

According to the analogy, there cannot be a truly "static" charge. A static valley in water would immediately get filled. There has to be dynamic motion of charge to maintain the surface of water uneven.

Our analogy works well for a single charge, but for two charges not so well. Suppose that we have two electrons with opposite spins, how are we supposed to maintain their combined valley in water?


Rotation is the standard way to maintain an excited state in classical physics


The photon is a quantum of a classical electromagnetic wave. The electron may be a quantum of a classical Dirac wave.

How do we create a persistent excited state in classical physics? Especially one which can stay at the same place and does not need to move rapidly.

Let us consider a harmonic oscillator. As it oscillates, its energy swings between kinetic and potential energy.


                                                           ^   rotation
                                                           |
                           ● \/\/\/\/\/\/\/\/\ ●  mass
                   fixed       spring
                  point 


We may also let a harmonic oscillator to rotate around a fixed point. In a round orbit, the kinetic and the potential energy stay constant.

According to the virial theorem, the ratio of kinetic energy and the potential energy is often 1 : 1 when averaged over a long time.

Rotation is a very common way to maintain an excited state in classical physics. An example is the orbit of Earth around the Sun.

All this suggests that rotation might be the way to make the electron a persistent particle.


The gyromagnetic ratio 2


Suppose that the electron moves at the speed of light around the zitterbewegung loop which is normal to the z axis. We can explain the magnetic moment of the electron with this model.

But the z-spin of the electron in this naive model would be 1 ħ.

The virial theorem may explain why the z-spin is only 1/2 ħ. If half of the mass energy is potential energy which does not take part in the spinning, then the angular momentum is just 1/2 ħ.

The naive spin 1 ħ model is not a good model in classical mechanics. If the entire mass-energy of the electron is orbiting, then what keeps it in the orbit? There has to be a force field to do that. What is the energy of that force field?

These classical considerations suggest that the gyromagnetic ratio of the electron should be larger than 1. The virial theorem gives us the guess 2.


Earlier vortex models of the atom



In the 19th century Hermann Helmholtz and Lord Kelvin tried to explain the nature of atoms as vortices in a hypothetical medium.

"Helmholtz also showed that vortices exert forces on one another, and those forces take a form analogous to the magnetic forces between electrical wires."

These scientists realized the same thing as we did: a vortex is a way to build persistent, localized excited states.

In a sense, the hydrogen atom really is a vortex: the electron orbits the proton.


The semiconductor model of the electron


Suppose that an electron jumps from the valence band of a semiconductor to the conducting band. We then have an excess density of electrons in a certain zone while the hole and the lack of electrons is in another zone.

This would be a fine model for the creation of an electron-positron pair, but in this model no rotation is needed. The particles would have the spin 0.

Our vortex model has a hard time explaining what are positron vortices. Is space filled with two fluids: the electron fluid and the positron fluid? If we create a vortex in the electron fluid, do we always create another vortex to the positron fluid? If yes, how would this happen?


                 \       /   vortex
                   \   /
              +           +
               |           |
               |           |        -----------------  axis
               |           |
               -            -
 dipole 1                dipole 2
                   /   \
                 /       \   vortex


We can probably create an electron-positron pair from two photons with opposite spins. It is like having two electric dipoles close to each other and rotating to opposite directions. Can they somehow create vortices?

Assume that the dipoles above rotate around an axis which is horizontal in the diagram. They rotate to the opposite directions.

They might create two vortices as drawn above. If we look from up, the vortices rotate to opposite directions.

Now we have a model for the production of an electron-positron pair. The dipoles represent two photons which are circularly polarized with opposite spins. The vortices are the electron and the positron, with opposite spins.


Pair production in a Feynman diagram


How do we calculate the pair production cross section (= probability) from a Feynman diagram?


     photon               
           -----------   -------------  e+
                         |
                         | virtual
                         | electron
                         |
           -----------   -------------  e-
    photon


We assume that the incoming electromagnetic waves disturb the Dirac field according to the QED lagrangian. This disturbance is happening all the time. If we are able to find a history where momentum and energy are conserved, we can calculate its cross section from coupling constants and propagators for virtual particles (= internal lines).

The calculation is perturbative - it is based on the assumption that only very few photons collide. The calculation treats the electron as a wave where the "structure" of the electron is simply the four-component wave function.

It is not easy to visualize how exactly the spin operator in the Dirac formalism processes the four-component wave and outputs its spin.

It is surprising that a perturbative calculation gives probabilities which are very close to measured ones.

How could we visualize the birth of an electron and a positron as two photons collide? We need to think about that.

The production of a photon from a rotating dipole can be visualized quite easily, at least with a computer calculation and computer graphics. The process is not perturbative. Can we do the same for a production of a pair?

Question. Can we find a vortex model which:

1. gives a non-perturbative description of the birth of a pair, and

2. agrees with the perturbative calculation of a Feynman diagram? 


A particle vortex model


                 photon                  photon
                       ---------             ---------
                                   \        /
                e-    -----------------------------


In Feynman diagrams, Compton scattering is closely related to pair production. In a Feynman diagram, one is allowed to bend an outgoing line so that it becomes an incoming line and vice versa. One must switch e- and e+ if the line is an electron.

In our vortex model it is hard to understand how a photon, that makes an electron to oscillate, could be related to pair production.

   
          +
           |
           |    -----------  rotation axis
           |
          -

           -----------> velocity


A circularly polarized photon is like an electric dipole rotating and moving fast.

Since the electron and the positron possess a spin, one way to describe them is to make two charges of the same sign to orbit each other.


                -  ------------ -     electron


                + ------------ +    positron

                         |
                         |  rotation axis


Pair production can then be described with this diagram:


              photon
        + -   ----------------        ---------  ++  positron
                                   \   /
                                   /   \
        + -   ----------------        ---------  - -    electron
              photon


The sign symbols denote zones of charge density: a photon has both + and - zones which orbit each other. That is, a photon is like a rotating dipole.

An electron contains two (or more) negative charge zones that orbit each other. They cannot be bound by the electric attraction. In the water vortex model they are bound by the pressure which prevents the vortex from coming apart.

One can then describe pair production by two electric dipoles colliding. The ends with the same charge become bound and start to orbit each other.
    
  
          rotation axis
                    |
                    |

                <---->  separation d

              +        +
               |        |
               |        |      -------- rotation axis
               |        |  
               -         -       D = dipole length
                    |
                    |
          rotation axis


In the diagram above, the charges + and + break apart from the photons (rotating dipoles) and form a positron.

The negative charges form an electron. If the length of the dipole

       D = 2 d,

where d is the separation where the dipoles break, then the spin of the electron is 1/2 ħ the spin of the positron is -1/2 ħ, assuming that the spins of the photons were 1 ħ and -1 ħ.

The end result is:


          + ------- +    positron


           - ------- -     electron

                |
                |  rotation axis


The diagrams resemble the vortex diagram in an earlier section.

In this blog we have earlier speculated that the photon consists of a "virtual" electron orbiting a "virtual" positron. The diagram above clarifies this idea: in a photon we have charge zones + and - orbiting each other. These charge zones are not electrons or positrons, though. The electron is two negative charge zones orbiting each other and a positron consists of two positive charge zones.

We can explain the gyromagnetic ratio 2 in the same way as we did in a previous section. A half of the total mass-energy of the electron is potential energy of the charge zone pair. Only the other half rotates.

Question. Can we explain Compton scattering with the new particle vortex model? Why should the probability amplitude be the same as in pair production?


Further lines of study


Let us continue this study in the next blog post.

How can the spin of the electron be 1/2?

A possible solution:

The electron consists of two or more negative charge zones. The potential, where the negative charge zones orbit, is not -1 / r. If the potential differs from -1 / r, then the orbit will precess. If the orbit precesses, the system may return to its original state after two rotations!

That would be a simple way to construct a system whose spin is 1/2.

Feynman diagrams say that Compton scattering is related to pair production. How is that possible?

In Compton scattering, if the photon is very energetic, say close to 1 MeV, the collision is violent. The charge zones of the photon and the electron, for a short time, form a "plasma" where charge zones move in a random and violent way.

The plasma will soon decay into particles. In the case of Compton scattering, it is the electron and a photon again.

The same plasma model may explain pair production. Now we see that Compton scattering can, indeed, be related to pair production.


Conclusions


The vortex model takes the spin 1/2 of the electron seriously. The spin is not something abstract, but there is real mass-energy rotating.

In the usual -1 / r potential, a system returns to its original state in one rotation. The orbit does not precess. The spin is 1.

The spin 1/2 suggests that the hypothetical potential which keeps an electron together is not a -1 / r potential.

If a small photon hits an electron in Compton scattering, it only gently shakes the electron. However, a large 1 MeV photon collides violently and produces a "plasma" which decays. The plasma model may explain why pair production is related to Compton scattering.

Richard Feynman's parton model may be similar to our plasma model. We will check if it is the same idea.

Thursday, July 14, 2022

A rotating electric quadrupole produces spin 2 photons?

Suppose that we have a rod and two equal electric charges attached to its ends. We rotate the rod around the axis at its middle.


An electromagnetic quadrupole wave has the spin 2?



                 ^
                 |         rod
             + ● -------------------- ● +    charge
                                             | 
                                             v      rotation


If the rod is microscopic, it returns to the original state after a 180 degree rotation. It may be reasonable to require that the wave function of the system returns to the original state after a 180 degree rotation.

If the wave function only does one full cycle in a full rotation, then the angular momentum is presumably ħ. Niels Bohr built his atomic model from the assumption that angular momentum is quantified in units of ħ.

If the wave function of the rotating rod does two cycles in a 360 degree rotation, then the angular momentum is 2 ħ.

The rotating rod produces an electromagnetic quadrupole wave. The spin (helicity) of that wave, in some sense, is 2.

Maybe it is possible to extract from the wave a photon with the spin 1? If that is true, then we can split a spin 2 photon into two photons of the spin 1.

In gravity, people usually assume that a rotating quadrupole produces spin 2 gravitons.


Spin 0 and spin 1


An electromagnetic spin 1 wave is generated by a rotating dipole. The dipole has to rotate 360 degrees to return to its original state.

Spin 0 waves are longitudinal waves, or pressure waves. The simplest way to generate them is to have a monopole whose charge varies with time. For example, in a drum skin we can generate waves by pressing it rhythmically with our finger.


Spin 1/2


If the rod system would return to its original state after two full rotations, then, presumably, it would be able to store angular momentum in units of 1/2 ħ.



Question. If we are able to construct a system of electric charges which has the property above, are we able to create photons whose spin or helicity is 1/2?


Above we argued that a quadrupole creates a spin 2 wave. A quadrupole is two dipoles rotating with a 180 degree phase shift.


                              x------● +      dipole

             x = axis of rotation


A dipole has to be rotated 360 degrees to bring it to the original state.

We can reduce the required rotation to 180 degrees by combining two dipoles:


                  + ●------x------● +       quadrupole

              x = axis of rotation


An antenna, which has to be rotated 720 degrees to bring it to its original state, would produce spin 1/2 waves. If we combine two such antennas, then we get the familiar dipole antenna.

Does this cast light on the spin 1/2?

Suppose that we have a torus with a sphere inside. We twist and bend the torus so that it becomes a coil with two loops.

If the sphere has a charge, it can probably absorb angular momentum in units of 1/2 ħ.


         loop 1     loop2
                   |     |
                   |     ●    charge moving in the coil
                   |     |
                    -----
                  bridge


In the diagram we schematically have the two loops. The "bridge" is the two tubes which connect the loops.

If the charge moves in the coil, it creates a main wave which is like from a rotating dipole. Also, it creates another, smaller, back-and-forth dipole wave whose frequency is 1/2 of the main wave.

Could it be that a spin 1/2 boson is this simple? Just two dipole waves superimposed?


Spin 1/2 fermions


The waves which we have studied above probably do not conserve the number of particles. We may be able to extract the energy in many different combinations from a wave of a random form. Also, there is no problem in stacking many identical waves on top of each other. The waves describe bosons.

Fermions, like electrons, contain a fixed charge which is conserved. Their wave model must strictly conserve the number of particles.

Erwin Schrödinger in 1925 discovered his famous equation which works for non-relativistic fermions. Paul Dirac in 1928 was able to make a relativistic equation for electrons and positrons.

What kind of a classical wave model strictly conserves the number of particles? An answer is a model of a gas. The number of molecules does not change in a gas, but a gas does exhibit wave phenomena in the form of sound.

A semiconductor might be a suitable classical wave model for fermions. If we have an extra electron somewhere, the associated excess density of electrons is the "wave". If we have a missing electron, then the missing electron density is the associated wave.


Conclusions


We found simple classical models for spin 0, 1, and 2 bosons. We may have a model for a spin 1/2 boson, too.

It remains unclear what the spin of a boson actually means. Is it a unit of what amount of angular momentum can an absorber harvest from the wave? Or is it always the same fixed amount for one type of a boson field? Can a photon transfer angular momentum which is not parallel to its path?

In our blog we claim that gravity is fundamentally analogous to electromagnetism, and that the spin of the graviton is 1. Our observation that a quadrupole radiates "spin 2" photons may clarify why some people say that the spin of the graviton is 2.