The electron-positron Higgs field




There is no remnant force between the virtual particles in QED. The search is terminated. How does the picture look now?



The electron-positron photon Higgs field

This is a sequel of paragraph Massless coinciding at page 2 of THE EXPANSION OF THE UNIVERSE.

The vacuum consists of particles. The distance between two neighboring particles is calculated to be 10^-21 m, see Calculation of the radius of a vacuum marble. Within that distance the vacuum cannot distinguish particles to have different locations anymore. Particles closer than 10^-21 m are judged by the vacuum to have identical location.

When an electron e- and a positron e+ coincide within 10^-21 m, one of them absorbs Higgs field from the vacuum and the other, inside its time border, emits Higgs field to the vacuum at the same rate. Therefore when the e- and the e+ are within 10^-21 meter, the composite does not absorb nor emit Higgs particles and is massless and has zero gravitational field. We are going to take this as the photon.

For reasons that are explained at page 4 of QG - page 44 of TONE - the e+ is the particle and the e- is the antiparticle. Therefore we take the e+ to have mass m and then the e- must have mass -m.

But on Earth this makes no difference: the electron only has mass -m within its tiny time border as is calculated in paragraph The calculation of the time border . Outside that it reacts just as normal forward-in-time evolving matter. We observe the electron as having mass m.      (1.1)

Is the composite a particle? What about its elapse of time? The e+ in the photon goes from A to B. The e- in the photon goes from B to A. Neither the e+ nor the e- has entropic development during their photon flight. Besides, according to SR time is standing still on the photon, the photon has a first moment but not a second one. So yes, the composite just is the photon and the photon is a particle.

When an electron in an atom emits a photon, where the positron is coming from? Inside each electron, according to QED renormalization theory, there is a superposition of a horde of e+ e- pairs shielding the naked core. The building up of shielding makes the naked core to approach infinite charge and zero mass. That is, when peeling of layer after layer of shield, the remaining core grows to infinite charge and zero mass. One of such pairs from the shield may sufficiently coincide and form a photon and leave the electron. Mind the e+ e- pairs are superpositions relative to each other. They don't see each other. They don't form an interacting cloud of e+ e- pairs. Besides, there are infinite of them. They won't miss a pair.

Since the two particles as composite have no mass, they immediately gain lightspeed. They force each other along the same path, because the slightest separation (larger than 10^-21 m) would separate the forward and backward vacuums and then the e+ and e- would get mass, for which the energy is lacking.

In this website matter goes forward in time and antimatter goes backward in time. When the e+ in the photon goes from A to B and the e- in the photon goes backward in time from B to A, then can a photon be emitted without being sure of its absorbing destiny at the end of its track? I mean, if there is no absorbing destiny set at the end of the photon's track, where the e- is coming from then? Well, just realize that the e+ e- pair that makes up a photon, is taken out of the virtual cloud shielding the naked core of the electric charge that emits the photon. The fact that the e- is there is enough at that moment, we don't need to know more about the end of the photon's track yet.

In accepted physics, an electron and a positron that approach each other too near, will annihilate each other into two gamma photons. However, when the e+ and e- coincide massless as described above and within a moment the composite speeds up to lightspeed, then time dilation immediately is at maximum, so time is standing still on the pair. There is no need to fear for annihilation, there is no time for that. Besides, to what they should annihilate? When coinciding massless they already are a photon! They try to approach each other in order to annihilate, but before they can do so they become a massless pair, a photon. Well, I guess, this is the annihilation!

Normally, to maintain composition the spin of the particles best align and as soon as they gained lightspeed they stay aligned. But wait, since they will force each other along the same path, spin alignment is no longer needed to keep the composite together. This changes situation. Normally e+ e- pairs appear with as much as possible quantum numbers being opposite: charge, elapse of time and spin. So the appearing e+ e- pair sets out as opposite spin particles. Spin electron + spin positron = 1/2 -1/2 = 0, so why a spin 0 photon shouldn't form?

I think they do. But we observe no spin 0 photons. Where did they go? The usual habit seems to be, when we need in our theory a particle at ground state - our first choice in bringing it to existence - and we don't find it in the real world around us, then we dump it in the vacuum as a Bose Einstein condensate vacuum field. So we do now. We presume a electron positron pair spin 0 vacuum field, a spin 0 photon vacuum field, to cover entire spacetime and thus form the Higgs field as Paul Dirac originally created it (well, he constructed an electron sea, with a positron as a hole in the sea, and he didn't talk about gaining mass) and as Peter Higgs et al. presumed it (he talked about gaining mass but never talked about photons or e+ e- pairs as Higgs particles) and as it is needed in QED renormalization (Richard Feynman, Martinus Veltman, Gerard 't Hooft).

So we have chosen here to fill in the vacuum particles - not earlier specified - with spin 0 photons. I hope the expression Bose condensate applies here, that the vacuum field is a Bose condensate. Meant is an electron or muon or tauon absorbs one spin 0 photon from the vacuum field in order to gain mass (Higgs field absorption *). The Bose condensate is disturbed now, there is one hole in it. Meant then is that the field streams in the hole in order to fill it, dragging along everything floating in it. This is gravitation, a gravitation performed by the spin 0 photon field. After rearrangement the Bose condensate is restored. This is the same mechanism of gravitation as explained at page 3 of NEG, The Higgs field - Part 1. The electron has its own type of gravitational field.

*) In fact it are the positron, the anti-muon and anti-tauon that absorb mass. The electron, muon and tauon within their time border emits toward the Higgs field, as observed from us.

Spin e+ e- = +1/2 +1/2 = +1 and spin e+ e- = -1/2 -1/2 = -1, this is the photon spin as we observe it. I see two main paths how this can come to be.

1) The original QED renormalization theory is about virtual e+ e- pairs shielding the naked core of a real positron or electron. In case of the electron, it talked about the e+ of a pair going a little nearer to the negative charged core and the e- going a little further away from the core, due to electrostatic attraction and repulsion. This assumes interaction taking place, virtual photons going from the core to our pair. The different e+ e- pairs of the shield are superpositions to each other, they don't see each other and don't react with each other at all. But the stages of the described separation process in one single e+ e- pair in its subsequent moments are all part of one single virtual process. Thus each subsequent extra photon from the core, coupling to our pair, diminishes the contribution of the matching Feynman diagram with a factor 10, see previous pages of this storyline. Therefore only the first photons coming in from the core give a significant contribution, resulting in only a small charge separation in an e+ e- pair that appears in the shield.

Anyway it seems clear the e+ and the e- from one pair of the shield do absorb photons from the core. Each photon absorption swaps the spin of the e+ or e-. As soon as they are within 10^-21 m and their spins happen to align, they become a spin 1 (or spin -1) photon and leave the electron at lightspeed. When energy is available the spin 1 photon becomes real, otherwise it stays virtual.

2) Two pairs appear simultaneous and all within 10^-21 m (that are 4 virtual particles together). The spin up e+ combines with the spin up e-, the spin down e+ combines with the spin down e- and two photons of opposite spin leave the electron simultaneous.

1.
Real electrons absorb from the Higgs field to gain mass. The picture sketched at this page is that real electrons absorb a spin 0 electron positron pair from the vacuum, an electron-positron photon Higgs particle, as Higgs field absorption to gain mass. Similar to page 3 of NEWTON EINSTEIN GRAVITATION this results in a hole in the vacuum, and then the hole is filled in with the surrounding e+ e- vacuum pairs (spin 0 photons that are) and cause gravitation.

Real electrons have their own Higgs field. The electron Higgs field and the electron gravitational field is the same field in the sense that both fields consist of the same vacuum particles.

How many absorptions per second there are, for one electron, about 10^18 à 10^20, isn't it? What is the density of the e+ e- vacuum field?

This is the Higgs field for the electron, muon and tauon, according to nowadays insights. The name photonic Higgs field probably suits the best. It is also called the leptonic Higgs field although this name is in fact not quite correct: leptons are the electron-like particles PLUS their associated neutrinos. However, the neutrinos have their own Higgs field made of neutrinophotons. The hadronic Higgs field - or gluonic Higgs field - for the quarks is probably the only constant present Higgs field, the other two Higgs fields should be created on the spot when needed. See the first four paragraphs of page 5 of QG (page 45 in the TONE storyline).
- August 2026

TO   INVESTIGATE

2.
Does the core use spin 1 photons or spin 0 photons or both, to interact with the e+ e- pairs of the shield? Does it make any difference? The core does emit spin 0 photons, I see no reason why not. But the spin 0 photons are absorbed by the vacuum before they can reach the e+ e- pairs, in doing so enlarging the vacuum by their volume (this is the time reversed version of Higgs field absorption and the subsequent act of gravitation). But wait, there is no energy available to do such, to create real spin 0 photons send from the core and enlarge the vacuum. It is difficult to judge, but let's go for the assumption that IF the core emits spin 0 photons, THEN the vacuum would absorb them before the e+ e- pairs do. So interaction core to e+ e- pairs is by spin 1 photons only.

Maybe this is a stupid question. Spin 0 fields are scalar fields, aren't they? You cannot emit and send out a spin 0 particle just as if it is a spin 1 particle, isn't it?
- August 2026

TO   INVESTIGATE

3.
The wavefunction of the electron in the photon goes from A to B. The wavefunction of the positron in the photon goes from B to A. The resulting wavefunction is a standing wave. Its nodes, where the amplitude is zero, are standing-still in space. The amplitude of the crests in between go up and down to the summed-up value of the original wavefunctions.

Any photon of a certain wavelength - gamma rays, visible light, radio signals - can be converted to any other photon (except for the polarization, phase and spin) by moving relative to it . You don't touch the photon, you only change your frame of reference. What if we do so with the composite? The wavefunction of e.g. the electron gains strength while that of the positron looses strength then. Still they do add up to the standing wave, now moving relative to us. The nodes move at constant speed now. How does this work out?

The mass-energy of the electron and of the positron cancel each other to zero energy, see (1.1) above on this page. Still the photon has some energy E = hf, h = constant of Planck, f = frequency. It must be due to the getting-out-of-phase of the wavefunctions of the electron and the positron composing the photon.

I guess this is a stupid question too, due to my lack of knowledge of the physics under consideration. But well, they sketch the problem.
- August 2026

TO   INVESTIGATE

The photon very well might be a CTL, a closed time loop, see CTL considerations at page 8 of SR.


 

The color of the photon

If you are reading TONE for the first time, this paragraph runs ahead on pages yet to come. You might skip it and read it later.


What is the color of the photon? Paragraph Electrons as Baryons at page 4 of QG, argues the color of e+ is +1 and the color of e- is -1.

Take an electron and a positron that annihilate into two photons. If we would interprete this as that the e- and the e+ merge, then the the colors of the e- and e+ multiply to yield the color of the merger: a superposition of -1 * 1 = -1 and 1 * -1 = -1, which is just -1, the dangerous color! Every time a photon would hit an e- or e+ it would turn it into its antiparticle without changing its electric charge. This is forbidden in paragraph Mesons at page 3 of QQD, especially (5.2) and (5.9).

Therefore I conclude and propose here that the e- and the e+ in the photon don't merge (there is no state transition taking place.) In the photon the e+ and e- always keep separated (albeit virtually at the same place), they never merged, will never have coupled to each other by color force, not at their creation, nor at their absorption, nor during their flight in between. They coincide but keep separated. They coincide massless without merging and gain lightspeed and because of that lightspeed the e- and e+ don't even interchange particles to stay put.

And even if the e+ and e- in the photon would merge with a target particle, their colors still won't apply to it, see paragraph What if the two pairs of quarks are the same? at page 5 of QCD, especially (2.6) up to and including (2.8). Merging is prohibited because of the accepted rule Quarks have color and antiquarks have anticolor.

The e- and e+ stay close to each other but they don't see each other. Instead the vacuum sees. The vacuum sees no difference anymore between the location of Higgs absorption and Higgs emission and concludes to masslessness of the composite.

The e+ as well as the e- keep all their separate properties. Then the colors don't multiply but add:

color photon = color e- + color e+ = -1 +1 = 0       (2.1)

and 0 is not a color. (The color quaternion units family is +1, i, j, k, -1, -i, -j, -k and 0 is not amongst them.)

The net color of the photon is zero       (2.2)

When the photon is arriving in the shell of an electron, the e- and the e+ that formed the photon, are no longer bound to each other within 10^-21 m and become again one of the infinite number of e+ e- pairs that renormalization already assumed there to be present.

As I just said: 2 photons (Take an electron and a positron that annihilate into two photons.) While 1 e+ plus 1 e- is just enough to form 1 photon. The other photon must be formed in the shell that surrounds the naked core of the e- or the e+.

At page 1 and 2 of QQD is argued the colors i, j, k, -i, -j, -k glue. The colors 1 and -1 don't glue but nevertheless can couple. But the photon doesn't even couple by color force.

See also discussion about this topic in paragraph Two gluons of opposite sign do not react at page 3 of QG.


 

References

Scientific American june 1980, Gerard 't Hooft, Gauge theories of the forces between elementary particles.