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?

https://sites.ualberta.ca/~gingrich/courses/phys512/node103.html

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.

https://accelconf.web.cern.ch/p07/papers/thpmn080.pdf

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.

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?

https://physics.stackexchange.com/questions/172965/in-an-electron-positron-annihilation-in-what-direction-are-the-photons-released

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.

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

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.

https://web2.ph.utexas.edu/~vadim/Classes/2008f/QED.pdf

                            ---------------------------  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.

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

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

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

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

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.

https://en.wikipedia.org/wiki/Spin-1/2

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.

The spin of a scalar field is 0, after all

Let us analyze in more detail what happens when we rotate a symmetric device which disturbs a scalar field. The device is assumed symmetric with respect to the axis of rotation.

                    ^   rotation

                    |

                   ● ---------- ●    symmetric device

                                    |

                                    v

Classically, the device outputs into the scalar field energy and angular momentum, but no linear momentum because the device is symmetric.

Definition of spin and orbital angular momentum

If angular momentum is input to a field, but the associated linear momentum is zero, then we may call it "spin" angular momentum. An example is when we rotate a disk. The disk spins but does not move in another way.

If we input angular momentum relative to a point x, but there is a point y relative to which the angular momentum is zero, then we may call it "orbital" angular momentum. An example is when someone riding in a carousel throws a grain of sand tangentially.

Circularly polarized photons: the frisbee model

Let us assume that the rotating device is an electric dipole and and analyze a single photon. The electromagnetic field is not a scalar field, but it is a good analogy.

If the frequency of the rotation is f, then the energy of a photon is

       h f,

where h is the Planck constant.

Let us have a mass

       m = h f / c²

moving in a circular path at the speed of light. Let it do one cycle in a time 1 / f. The radius is

       r = c / (2 π f)

and the angular momentum is

      J = c m r

        = h / (2 π)

        = ħ.

We can imagine that the field of the rotating dipole loses the energy h f and the angular momentum ħ to the free electromagnetic field from the distance r from the system.

The absorption process is this in the reverse.

The photon itself can be imagined as a "frisbee" which moves at the speed of light and carries the energy and the angular momentum lost by the rotating system.

The frisbee model does not properly explain the photon, though. Since the frisbee moves at the speed of light, it cannot rotate at all - if it would, then the edge would move at more than the speed of light. This is a paradox.

The classical limit of waves made in a scalar field

We finally found an argument which shows that a rotating device outputs into a scalar field orbital angular momentum. Thus, the spin is 0.

                   

                        ● ---------- ●    absorbing device

                                     ---------> 

                                absorbed orbital

                                angular momentum

                                

                    ^   rotation

                    |

                   ● ---------- ●   emitting device

                                    |

                                    v

Let us have a rotating device which creates pressure waves in a gas or a liquid. We use a similar, freely floating, device to absorb the waves.

Suppose that the absorbing device would start rotating around its axis when it absorbs a single quantum of the pressure wave.

The classical limit would then be that the absorbing device starts to rotate macroscopically under a strong wave.

But we have already argued that a classical pressure wave mostly moves any obstacle back and forth to the direction of the wave source.

The emitting device does output angular momentum to the field. Freely floating absorbing devices start to circle around the source of the wave. They receive orbital angular momentum with respect to the emitting device. But they do not receive spin angular momentum.

Conclusions

Waves in a solid can be transverse shear waves or longitudinal pressure waves. Transverse waves can easily make a small freely floating absorbing device to rotate around its axis. Transverse waves carry spin angular momentum.

Longitudinal pressure waves cannot easily turn a small absorbing device around its axis. They give up their angular momentum as orbital angular momentum. Pressure waves are scalar.

We found an explanation of the spin 0 and spin 1 in classical physics.

Let us next study the following two questions:

Can we find a classical model for the spin 1/2 of the electron?

Can the concept of a shear wave clarify how in photon absorption, the absorber harvests the entire energy and angular momentum of the photon from a circularly polarized wave? If the absorber is at an oblique angle from the source, how can it absorb the entire angular momentum?

Waves in a scalar Klein-Gordon field are longitudinal, in a sense

In our July 5, 2022 blog post we suggested that no field can have a quantum whose spin or helicity is 0. That is, a quantum of the field can always carry angular momentum.

https://en.wikipedia.org/wiki/Klein%E2%80%93Gordon_equation

Let us first assume that the scalar field in the Klein-Gordon equation is real.

A Klein-Gordon wave in a drum skin is longitudinal, in a sense

We can generate a circularly polarized electromagnetic wave by rotating an electric dipole. The helicity of the photon is then 1 or -1. Can we do something similar with a Klein-Gordon wave?

The Klein-Gordon field is much like a drum skin (though three-dimensional). We can press the drum skin with our finger. The pressing finger is a "charge" of the field.

                                ^

                              /

                            ● metal sphere

                          /

                        v   movement

 

                 ^

                 |

                ● finger

                 |

                 v    movement

If we move our finger back and forth on the skin, we generate a wave into the skin. If we put a metal sphere on the drum skin at some distance from the finger, the sphere will roll back and forth. The sphere is an antenna which absorbs a part of the dipole wave.

We can easily create waves where the back and forth movement of the sphere is to the direction of the wave source. But making a wave where the metal sphere moves normally to the direction of the wave source seems to be hard, or impossible.

   

               radiated

               energy

           

                   ● spot 1

                                              destructive

                                              interference

                   ● spot 2

We can approximate the wave by imagining that a finger alternately presses the spots 1 and 2 in the diagram. The wave is formed by the interference of the simple circular waves originating from the spots.

Normal to the line spot 1 - spot 2, the waves have a 180 degree phase shift, and almost total destructive interference.

Most of the energy is radiated upward or downward.

The drum skin wave is longitudinal in the sense that the metal sphere moves back and forth to the direction of the wave source. This is a major difference from electromagnetic waves which are always transverse.

Waves in a solid

Sound waves in a solid can be either longitudinal pressure waves or transverse shear waves.

Now we see that scalar field waves correspond to pressure waves. Transverse waves may be similar to electromagnetic waves.

Suppose that we disturb a solid by moving a point inside it along a circulat route. We input energy and angular momentum in both the transverse waves and the longitudinal waves.

Suppose that have a small device which harvests energy from vibrations of the solid. We believe that the device can absorb the entire energy and angular momentum in a single quantum of the transverse wave. The classical wave corresponding to a single quantum "collapses" and is absorbed entirely.

The collapse is not a classical process but purely a quantum process.

Question. Can a small device absorb the entire pressure wave, too? That is, can it absorb the whole energy and angular momentum in the scalar wave?

If the answer to the question is yes, then the quantum of a scalar wave can carry spin angular momentum. Its spin is not 0.

We have not found a theoretical argument which would resolve the question. Maybe it is possible to find the answer empirically?

Conclusions

It is easy to create a circularly polarized electromagnetic wave. Such a wave will move a freely floating electric charge along a circular path. Transverse waves can be used in this way to carry "spin" angular momentum into a small absorbing device, the charge.

Waves in a scalar field are longitudinal. They will move a charge back and forth.

The above are results for a classical field.

However, the collapse of the entire field energy to a small absorbing device is a quantum process.

It is not clear to us what happens in the collapse in a scalar field. Does the entire angular momentum get absorbed into the device? Does it make the device to spin or does it input linear momentum to the device so that it gains "orbital angular momentum" with respect to the emitting system.

We should analyze both the momentum and the angular momentum in the field. If the system which disturbs the field is symmetric, then, classically, the field will only receive angular momentum, not momentum at all relative to the laboratory frame. If the field is entirely absorbed by a device, that device cannot receive orbital angular momentum because then it would also receive momentum!

However it is possible that the field does receive momentum from the emitting device if the field contains just one quantum. It is like throwing a grain of sand from a carousel. In that case, the absorbing device will catch the grain. But if the absorbing device is not on the path of the grain, how can it catch it?

In the next blog post we will analyze the momentum aspect in the process.

The Higgs field, gauge transformations, and the Coulomb gauge

The Coulomb gauge is the "natural" way to define the scalar potential and the vector potential around electric charges.

Let use denote a 4-potential with A.

We get an equivalent physical system if we add the gradient of any sufficiently smooth function g to A.

Adding a gradient is called a gauge transformation because it retains the physics of the system as is.

A real-valued Higgs field which gives the mass for a gauge boson

Let us then introduce a real-valued Higgs field φ whose vacuum expectation value is different from zero. To achieve such a vacuum expectation value, we assume that there is a "potential" on φ of the form

       V(φ) = k (φ - v)²,

where k is a constant and v is the vacuum expectation value.

https://www.theorie.physik.uni-muenchen.de/lsfrey/teaching/archiv/sose_09/rng/higgs_mechanism.pdf

We once again use the framework of Kien Nguyen (2009).

Let us write

       φ = v + h.

We couple the Higgs field φ to the 4-potential A of a gauge boson field.

In the new lagrangian, there is a term

       e² v² A².

This term gives the mass to the gauge boson field A. We ignore the terms of type

       e² v h A²

and

        e² h² A².

The part of the lagrangian for h is of the the same form as for a massive Klein-Gordon field:

https://physics.stackexchange.com/questions/52949/what-is-the-lagrangian-from-which-the-klein-gordon-equation-is-derived-in-qft

The mass m is ~ sqrt(k).

An excitation of the field h is the Higgs boson.

What did we accomplish with the real-valued Higgs field?

We were able to give a mass to the gauge boson of the field A.

The real-valued Higgs field is a very well-behaving massive Klein-Gordon field. Its equation is linear.

The potential of the Higgs field is the familiar quadratic (harmonic) potential, not the complicated Mexican hat potential of the ordinary Higgs field.

There are no defects in the real-valued Higgs field.

There is no self-interaction in the real-valued Higgs field. The self-interaction in the ordinary Higgs field comes from the quartic potential φ⁴.

There is no symmetry breaking process.

The real-valued Higgs field seems to be much better than the ordinary complex-valued Higgs field. What is the problem with the real-valued field?

The problem is that when we do a gauge transformation of the gauge boson field A, then the value of the lagrangian changes. The physics is not conserved.

This is the reason why Peter Higgs among others devised a complex-valued Higgs field which can be transformed along a gauge transformation of A. The goal was to keep the value of the lagrangian the same. In a gauge transformation, the complex value of the Higgs field is rotated just in the right way to conserve the value of the lagrangian.

Let us analyze the whole process by which Peter Higgs gave a mass to the gauge boson.

The introduction of a scalar Higgs field whose vacuum expectation value is an almost constant v ≠ 0, is somewhat ad hoc, but it is required to give A a mass.

What is the value in devising a complex Higgs field, its potential function, its transformation (rotation), and writing a new lagrangian for those? If we can find simple formulae for the complex field, fine. But do we learn anything from the fact that we were able to find such formulae?

It is not clear to us if finding the details for the complex-valued Higgs field system teaches us anything. As if the the whole task would be unnecessary.

Defining the real-valued Higgs field gauge transformation from the Coulomb gauge

Suppose that we are able to do the physics that we want in the Coulomb gauge and using the real-valued Higgs field.

Can we define the gauge transformation of our real-valued field in a trivial way? We do not try to calculate any complicated formulae for the transformed fields nor find a special lagrangian which works with these fields. We simply define that the gauge transformed system must describe the exact same physical behavior as the Coulomb gauge system does with the real-valued Higgs field.

Let the lagrangian of the system in the Coulomb gauge be

        L₀(φ, A).

Let the gauge transformation for A be f. We let the gauge transformation for φ be the identical mapping.

We define the lagrangian L for an arbitrary gauge by the following formula

       L(φ, A) = L₀(φ, f(A)),

where f is the mapping which transforms A to the Coulomb gauge.

Now we have a lagrangian L which trivially is invariant under a gauge transformation. The precise mathematical formula of this lagrangian may look ugly, but is there anything fundamentally wrong with it?

Is there a rule that a lagrangian should have a simple arithmetic formula in all gauges, not just in the Coulomb gauge?

Trade-off of a simple arithmetic formula versus nice properties of a field

Let us again analyze what Peter Higgs and others did at the beginning of the 1960's.

Superconductivity and condensed matter physics inspired the idea of using a charged Higgs field φ to give a mass to gauge bosons in a field A.

The Higgs field must have an almost constant vacuum expectation value v ≠ 0 to do the trick. A simple way to accomplish that is to introduce a potential V whose minimum is at v.

The remaining problem is to make the Higgs field to respect the gauge invariance of A. Higgs and others found out that a complex-valued φ works if its gauge transformation is rotations around the origin of the complex plane, and the potential is the Mexican hat.

Higgs assumed that "symmetry has been broken" and the field has collapsed to be almost precisely equal to a fixed value v.

We can then study the behavior of the field close to this value v. We are able to eliminate the angle parameter from the complex value of φ and restrict the value of φ to vary over the real axis, in the vicinity of the real value v. We define φ = v + h, where all are real.

This endeavor allowed Peter Higgs and others to define the real-valued Higgs field which we suggested in an earlier section of this blog post! 

That is, Peter Higgs and others built a machine which, when assuming symmetry breaking and analyzing the system close to v, behaves like the simple model we introduced above.

Certainly one can build other machines which, when restricted close to v, behave like the simple real-valued Higgs model. Why is the standard Higgs model good? Its formula is very simple. That is its main virtue.

However, the standard Higgs model has three ugly features:

1. It assumes symmetry breaking to a random minimum value v in the valley of the Mexican hat. That kind of behavior has never been observed in other fundamental fields.

2. The Mexican hat potential may allow defects to form. We have not observed defects in other fundamental fields. If defects exist, they are not matter which consists of particles. That would be very strange.

3. The quartic φ⁴ potential means that there is self-interaction in the Higgs field. The probability distribution of the momentum of a single particle may spread as time progresses, even though the particle does not interact with anything! That may contradict quantum mechanics.

Our own real-valued Higgs field avoids all those three ugly features. The drawback of our model is that its arithmetic formula is ugly. The lagrangian L is defined through the Coulomb gauge. The arithmetic formula may be very complicated in another gauge.

Should we favor a simple formula for the Higgs lagrangian, or beautiful properties of the field?

In this blog we think that beautiful properties of the field are more important.

Symmetry breaking and unification in the real-valued Higgs model

Symmetry breaking and unification of the elecromagnetic and weak fields at high energies is something which may be deduced from the Mexican hat potential.

In the real-valued Higgs model, with the quadratic potential, at high temperatures the mass of bosons W and Z will behave erratically. Is there some kind of a unification?

If W and Z particles are very dense, the hamiltonian will attain its minimum value at φ = 0. That would be a unification.

If the density is lowered, then φ slides to its value v. Should we call this symmetry breaking? In some sense, it is.

Conclusions

We showed that a real-valued Higgs field has beautiful properties - but we then have to define the lagrangian through the Coulomb gauge. The arithmetic formula of the lagrangian in other gauges may be terribly complicated.

We believe that beautiful properties of the field are more important than what the arithmetic formula looks like.

The vacuum in the lowest energy state contains a Higgs field whose value is v everywhere. This is ugly compared to other fields whose value is 0 everywhere. Is there a way to remove this ugliness? In this blog we have claimed that the vacuum is truly empty. It contains nothing. The fact that the Higgs field has the value v does not look nice.

In our blog our view is that matter is excitations of fields, and the excitations can be quantized, giving us "particles". The real-valued Higgs model conforms to this. There are no defects.

The notion of a scalar field with a spin zero particle is ugly since other fundamental fields that we know are not scalar fields. Could it be that a scalar field is a vector field in a disguise? In earlier blog posts we noted that the simple φ² probability or energy density is not Lorentz covariant for a massive Klein-Gordon field.

Is "self-interaction" allowed in quantum mechanics?

Scalar fields can be Lorentz covariant. We were wrong to suspect that scalar fields break special relativity. See the update which we wrote to the July 3, 2022 blog post.

However, we found another problem in the Higgs quartic potential. The φ⁴ term in the Higgs potential means that the wave equation is not the massive Klein-Gordon equation.

The Klein-Gordon equation is linear. There is no "self-interaction" of a particle.

In the update to our June 29, 2022 post we asked if a particle can be scattered by itself.

Does self-interaction break basic principles of quantum mechanics when it disperses a wave packet?

Suppose that we prepare a wave packet for a particle. We are able to set a certain probability distribution for the position and the momentum of the particle.

Let us have a wave equation with a self-interaction. The self-interaction may spread the wave packet very quickly, faster than the packet would spread under a linear wave equation.

That is, the uncertainty in the position and the momentum of the particle grows faster than it should.

Is this allowed in quantum mechanics?

There is no external interaction to the wave packet. Conservation of momentum means that we know the state of the particle better than what the self-interacting wave equation claims.

https://en.wikipedia.org/wiki/M%C3%BCnchhausen_trilemma

In our blog we do not like Baron Munchausen tricks. If a particle can disturb itself, that constitutes such a trick.

Can the spin of any particle be 0? Probably no

Let us consider the case of a photon. If we wave a charge back and forth with our hand, we emit energy to the electromagnetic field, but no total angular momentum in the z direction.

Photons, helicity and energy and angular momentum conservation

           z

           ^

           |

           |

           |        <----- + ----->

                   charge moves

                  back and forth

  https://physics.stackexchange.com/questions/154468/difference-between-spin-and-polarization-of-a-photon

According to the link, the system creates sums of two helicity states

      J =  |+1 ħ ⟩

           +

             | -1 ħ ⟩

where J is the angular momentum to the direction of the photon momentum.

            z

            ^

            |

            |

            |        + -------- -

                  rotating dipole

We may also create photons by rotating a dipole from the middle. Let the axis of the rotation be to the positive or negative z direction. For a single photon the helicity is

      J = +-1 ħ.

Is any photon that moves to the direction of the z axis an eigenstate of the helicity?

A photon moving at the right angle to the z axis probably is the sum |+1 ħ ⟩  + | -1 ħ ⟩.

But now we face a dilemma: if we have a rotating antenna which absorbs a photon, how can we simultaneously conserve energy and angular momentum?

     z

     ^ 

     |                                   + ------  -

     |                                 absorbing

     |        +  ------  -

     |        emitting

Suppose that we have an emitting rotating dipole whose axis points to the z direction, and an absorbing dipole with the axis to the z direction. The receiving dipole is placed at an arbitrary direction from the emitting one.

The dipoles rotate at approximately the same rate.

If a photon jumps between the dipoles, it carries the angular momentum +-ħ and the energy hf. How can we explain this transfer if the helicity of the photon is parallel/antiparallel to its momentum?

A possible solution: if we treat the electromagnetic field as a classical field, then the emitting dipole certainly can speed up the rotation of the absorbing dipole. Now, if a transition conserves energy and angular momentum, then it is allowed. In this description we did not refer to a photon at all, and neither to its helicity.

If the axis of the absorbing dipole is not parallel to the emitting dipole, then a simple transition is not possible. Several photons must be absorbed and emitted. Classically this corresponds to waves scattered from the absorbing dipole.

Can a scalar field exist at all: is it possible that the quantum always has the spin 0?

Question. Is there any field where J could be zero always? If we disturb the field with a rotating movement, is it possible that the angular momentum is carried away by J = 0 particles?

                                                 ^

                                                 |

                                                 •  particle

                                      ^

                                      |

                    ●------------●    rotating "antenna"

                     |

                     v

        • particle

        |

        v

                               <----- r----->

We use a motor to rotate the antenna. The axis of the rotation points out from the screen.

If J = 0, then the angular momentum must be carried by particles which depart (tangentially) from some radius r from the center of the rotating movement.

We could say that the particles carry away "orbital angular momentum" rather than spin angular momentum.

The model with two emitted particles is ugly.

Let us try another model.

  

     z

     ^ 

     |                                   ● ------ ●

     |                                 absorbing

     |          ● ------ ●

     |          emitting

Let us assume that we have an emitting rotating system and an absorbing rotating system, like we had for the electromagnetic field.

The systems interact through a classical field. The rotation can be transmitted between the systems through this field.

Suppose that the emitting system loses a quantum of energy and angular momentum, and that quantum shows up in the absorbing system.

We say that a quantum of the classical field carried energy and angular momentum. The quantum had a "spin" that differed from zero.

In this model, the "spin" of the quantum can never be zero. It can always carry angular momentum from one rotating system to another rotating system.

Conjecture. The spin of all bosons differs from zero. The Higgs boson cannot have the spin zero.

Question. Fermions have half-integer spins. Consequently, the spin cannot be zero. Can we find an explanation to this?

Conclusions

What is the spin of a boson? The spin question is intertwined to the fundamental question: what is the quantum of a classical field? It is the "quantization" problem of fields.

In this blog we have claimed that we must work with classical fields and only "quantize" the start and end states of a process. We are not allowed to assume that intermediate phases in the process consist of well-defined quanta. This is in the spirit of quantum mechanics: do not assume "hidden variables" in intermediate states.

Thus, the question of what is the spin of a boson, is somewhat vague. A boson only exists as an intermediate state. However, we conjecture that all bosons can carry angular momentum. The "spin" of a boson cannot be 0.

How to make the Higgs field Lorentz covariant?

UPDATE July 6, 2022: Our reasoning about the oscillation frequency of the "hump" is incorrect. We did not take into account that simultaneity is different for a moving observer.

In several sources it is proved that solutions of the massive Klein-Gordon equation are Lorentz invariant. That means that a solution in the laboratory frame is a solution in a moving frame, too, if we map spacetime points with the Lorentz transformation.

To determine the energy of a solution, we have to integrate over a spatial volume. Length contraction makes the volume smaller in a moving frame, which may mean smaller energy. How can we handle this? A moving particle should have more energy than a static particle.

In the case of an electric field which is normal to the movement of the frame, we below were able to reconcile things. It required transforming the electric field to the moving frame. In the case of a scalar field, there is no such transformation.

https://alpha.physics.uoi.gr/foudas_public/APP/Lecture5-Klein-Gordon-Dirac.pdf

The lecture above by C. Foudas (2007) solves this mystery. The probability density of the massive Klein-Gordon field is not the square of the wave function but a more complex formula involving derivatives! That allows the energy density to be correct in the moving frame. C. Foudas notes on page 6 that the integral of the probability density ρ(x) over a spatial volume V has to be constant.

We still have to check if there is a problem with the Mexican hat form of the Higgs potential. The massive Klein-Gordon equation assumes a simple harmonic potential. Is there a problem with the complicated Mexican hat potential?

----

Let us try to repair the Higgs model. We cannot approve a breach of Lorentz covariance. Higgs must obey the laws of special relativity.

Let us have a volume where the 4-potential A of the electroweak field is non-zero in the Coulomb gauge. It is kind of a "hump" in the potential.

https://www.theorie.physik.uni-muenchen.de/lsfrey/teaching/archiv/sose_09/rng/higgs_mechanism.pdf

Let us use the framework of the Kien Nguyen paper (2009).

Above, v is the vacuum expectation value of the Higgs field. The charge density of the Higgs field is e. The real variables h and Χ (khi) vary a little around zero.

In the scalar field Higgs model, v and e determine the mass term 

        1/2 e² v² = 1/2 m²

of the field A.

How should we Lorentz boost this interaction to a fast-moving frame?

The Lorentz transformation of the electromagnetic field

Let us look at the electric field when we do a Lorentz boost.

https://www.feynmanlectures.caltech.edu/II_26.html

In the formulae, the boosted frame moves to the direction of the x axis. The speed of light c is assumed to be 1. E is the electric field.

    z

    ^

    |            |     |     |

    |            |     |     |    electric field lines

    |            |     |     |

    |

     ---------------------------> x

                  -----------> v moving frame

Let us have a static electric field in the direction of the z axis.

Let the Lorentz factor

       γ = 1 / sqrt(1 - v² / c²)

be 1.01 for the moving frame. The energy in the moving frame includes the kinetic energy, which is 1% of the energy in the static frame.

The volume of the electric field contracts 1%. Its strength grows by 1%. The energy density is

       ~ E²

The energy of the field grows by 1% in the boosted frame. That is the correct value.

An electric field moving to the direction of the field lines

                  E field strength

            +|      |-

            +|      |-

            +|      |-

              -----------> moving frame

Let us then imagine a capacitor where the electric field points to the x direction. We Lorentz transform the field to a frame which moves to the x direction.

The electric field E is the same in the moving frame and the laboratory frame. The volume between the capacitor plates is length-contracted in the moving frame. We have a problem now: the total energy of the field between the plates seems to be less in the moving frame!

Question. Is it so that the formula

       ~ E²

does not calculate the field energy correctly in the moving frame?

The solution to the Question may be that the extra energy is in the field which escapes from the edges of the capacitor plates. Suppose that the capacitor is moving to the x direction. Then the magnetic field, and consequently, the Poynting vector is almost exactly zero between the plates.

But the Poynting vector is not zero for the field which escapes from the edges. The Poynting vector there can explain the energy flow from the left to the right in the field. This might make up for the missing energy.

The famous 4/3 problem may complicate the analysis.

In our blog we have suggested that the energy of a static electric field is zero, and all the mass-energy is concentrated at the charges. That would solve the 4/3 problem. That would also mean that the energy density formula ~ E² is not right in all cases.

The Lorentz transformation of arbitrary matter fields

Suppose that we have 1 kg of iron. It is described by a bunch of fields: electromagnetic, quark, gluon, and so on.

The Lorentz transformation of all these fields is necessarily very complicated.

If we want to calculate the total energy of the whole system in a moving frame, we measure its mass when it is static, and multiply it by the Lorentz factor γ of the moving frame.

Could we do the same with the Higgs field?

If someone disturbs the Higgs field and pumps energy and momentum into it, we do have a preferred coordinate frame where the excitation of the Higgs field is approximately static.

We could use that frame to do calculations and treat the Higgs field as "almost" scalar. The Mexican hat "potential" would be applicable in the preferred frame.

A detailed analysis of a "hump" in the Higgs field, and a moving frame

We again use Kien Nguyen's (2009) framework. Above is the lagrangian for a small radial displacement h from the minimum potential of the Higgs field.

https://physics.stackexchange.com/questions/52949/what-is-the-lagrangian-from-which-the-klein-gordon-equation-is-derived-in-qft

The lagrangian is for the massive Klein-Gordon wave equation.

Let us use the letter ψ instead of h.

https://en.wikipedia.org/wiki/Klein%E2%80%93Gordon_equation

What does a "hump" look like in the Higgs field approximated with the h displacement variable?

Let us assume that the hump spans a large spatial volume. Then spatial second derivatives, the term 𝝯² above, are small.

The equation is like for a harmonic oscillator: the acceleration of the displacement ψ is proportional to -ψ.

The energy density, or the "potential" of the field, is proportional to ψ².

                   ψ = 1

                  ______     

       _____/            \____ 0 

                      V

        -----------> moving frame

Let us assume that in the laboratory frame at the time t = 0, the value of ψ is the maximum = 1 in a certain volume V. The value of ψ will then oscillate like a harmonic oscillator.

Let us have another frame which moves very fast relative to the laboratory frame along the x axis. An observer in this frame will see ψ to oscillate slowly between 1 and -1. The maximum 1 is not attained simultaneously at every point in V.

Furthermore the volume V appears smaller for the moving observer.

Suppose that the frequency of the oscillation of ψ in the laboratory frame is very slow, and the volume V is very elongated along the x axis.

The moving observer will see the "hump" in much the same way as the laboratory observer, except that the notion of simultaneity is different for the moving observer.

The edges of the volume V move rapidly in the eyes of the moving observer.

Now we notice a contradiction: the oscillation of the middle portion appears faster in the laboratory frame than in the moving frame!

The wave equation of ψ is the same in both frames because ψ is a scalar field. The oscillation of the middle portion should have the same frequency in both frames.

This means that the Lorentz covariance is breached by the system.

Length contraction makes V to appear smaller in the moving frame, and the energy of the hump smaller. That is another breach.

Claim. A scalar field which is described by the massive Klein-Gordon equation is not Lorentz covariant.

In our analysis we did not refer to the properties of the Higgs field at all, except that it is described by the massive Klein-Gordon equation.

A massive scalar field does work well in non-relativistic contexts.

A scalar field is defined as a field whose value stays the same if we change the frame. Now we see that the concept does not make sense for a massive field if we allow relativistic speeds.

The Higgs model may work quite well in non-relativistic cases

A scalar field is easy to work with. Suppose that we can divide the system into independent parts, each of which is non-relativistic under its own coordinates. Then we can use the current Higgs scalar field model in each part.

In the interaction between the parts we have to take into account possible large kinetic energies, length contraction, time dilation and so on. In this solution we do not try to define a Lorentz covariant general Higgs field.

Conclusions

Special relativity works with 4-vectors. Based on this fact we may suspect that a scalar field cannot be Lorentz covariant.

One of the consequences is that the Higgs field cannot be scalar. The spin of the Higgs boson must be 1, not 0.

Why the LHC only observed spin 0 Higgs particles? Spin value 0 photons are created by a charge which moves back and forth. Spin 1 photons are produced in a circular movement. It might be that the processes in the LHC are back-and-forth, not circular.

Defining a Lorentz covariant massive field may be complicated. We need to study literature to find out how it can be done.