The electron-positron Higgs field

There is no remnant force between the virtual particles in QED. The search is terminated.

# The electron-positron photon Higgs field

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

As said at that page, when an electron and a positron coincide within their time borders, at the Earth surface a distance of about 10^-20 m, the electron absorbs Higgs field from the vacuum and the positron, inside its time border, emits Higgs field at the same rate. Therefore when the electron and the positron are within their time borders, 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.

If we take into account point 3 at page 3 of NEG, the proportionality of inert mass (resistance to change of velocity and the E = mc source of energy) with ponderable mass (mass that causes the gravitational field), then we conclude that the photon must have zero mass. That means that if the electron has mass m then the positron must have had mass -m.       (1.1)

For reasons that are explained at page 4 of QG, the positron is the particle and the electron is the antiparticle. So in fact the positron has mass m and the electron mass -m. But on Earth this renders no effect: the electron only has mass -m within its tiny time border. Outside that it reacts just as normal forward-in-time evolving matter. We observe the electron as having mass m.

Is the composite a particle? We defined its spin, but what about its elapse of time? The electron in the photon goes from A to B. The positron in the photon goes from B to A. Neither the electron nor the positron 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 electron-positron 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 a 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 (10^-20 m) would separate the forward and backward vacuums and then the electron and positron would get mass, for which the energy is lacking.

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, 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 electron positron pairs appear with as much as possible quantum numbers being opposite: charge, elapse of time and spin. So the appearing electron positron 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 spin 0 electron positron pair 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 electron positron pairs as Higgs particles) and as it is needed in QED renormalization (Richard Feynman, Martinus Veltman, Gerard 't Hooft).

I am not sure whether the expression Bose condensate applies here. Meant is a particle like the antimuon absorbs one spin 0 photon from the vacuum field in order to gain mass (Higgs field absorption). This means we have here chosen to fill in the vacuum particles - not earlier specified - with spin 0 photons. The Bose condensate is disturbed now, there is one hole in it. Meant is 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. Is that feasible, for a Bose condensate?

O   NVESTIGATE

We treated the spin 0 electron-positron photon field as a Higgs field. As a previous attemtpt I first thought it was the Higgs field of the leptons, all leptons. As argued in the paragraph The three generations at page 5 of QG I now think it is the Higgs field for the second generation, the entire second generation: the quarks s and c, the muon and the muon neutrino.

Within the theory it is the electron that is the antiparticle, emitting towards the Higgs field. The emission causes shells of vacuum around the quark to expand (the time reversed version of sagging-in). As long as inside the bubble, this is antigravity that nevertheless erases velocity (if there were masses inside). Once outside the bubble, the expanding shells are repulsive gravity indeed.

Does the spin 0 photon sea interfere with the sea of gluon pairs from QUATERNION GRAVITATION page 3? To fill in the vacuum particles with gluon pairs is treated from page 2 of QG. The interaction between photons and gluons is discussed in gl ph at page 5 of QG and I conclude ph and gl don't couple by color force (strong nuclear force). Within the theory this is not a redundant conclusion, see e.g. paragraph The color of the photon below.

Both the gluon sea as well as the e- e+ sea are supposed to exist as fields of all possible velocities. When one field is going to get in motion relative to the other, nothing will change since both fields are velocity invariant and special relativistic invariant.

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

1) The original QED renormalization theory is about virtual electron positron pairs shielding the naked core of a real electron or positron, due to electrostatic attraction and repulsion. In case of the electron, it talked about the positron of a pair going a little nearer to the negative charged core and the electron going a little further away from the core. This assumes interaction taking place, virtual photons going from the core to our pair. The different electron positron 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 electron positron pair that appears in the shield.

Anyway it seems clear the electron and the positron 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 their time borders 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 their time borders (that are 4 virtual particles together). The spin up electron combines with the spin up positron, the spin down electron combines with the spin down positron and two photons of opposite spin leave the electron simultaneous.

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. So spin 1 photons it are. But wait, there is no energy available to create real spin 0 photons send from the core, they have to stay virtual. So the absorption from and enlarging of the vacuum must remain virtual too. 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.

O   NVESTIGATE

Real electrons absorb from the Higgs field to gain mass. We suppose real electrons absorb a spin 0 electron positron pair from the vacuum, an electron-positron photon Higgs particle. This is taken as the absorption from the Higgs field to gain mass. Similar to page 3 of NEWTON EINSTEIN GRAVITATION and page 2 of QUATERNION GRAVITATION, this results in a hole in the vacuum, and then the hole is filled in with the surrounding e- e+ vacuum pairs and cause gravitation. 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?

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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.

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

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## The color of the photon

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. A dangerous color! Every time a photon would hit an e- or e+ it would turn it into its antiparticle, e+ would turn in e-, or e- would turn in e+. This is not observed.

Therefore I conclude and propose here that the e- and the e+ in the photon don't merge. In the photon the e+ and e- always keep separated, they never merged, will never have coupled to each other, 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. The e+ as well as the e- keep all their separate properties. Then the colors don't multiply but add:

color photon = -1 +1 = 0

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

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 their time borders 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.

It would explain also why Two gluons of opposite sign do not react, see (3.60) at page 3 of this storyline. In the vacuum particle ( i   -i ) the i and the -i don't couple. They approach each other, coincide massless and gain lightspeed.

## References

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