In paragraph Massless coinciding at page 2 of THE EXPANSION OF THE UNIVERSE is introduced that the photon consists of an electron and a positron coinciding massless: within the time border Higgs field absorption from the electron is canceled by Higgs field emission of the positron. The e- and e+ in the photon don't merge, they each keep all their separate properties. Also in that paragraph is introduced a new particle, the neutrinophoton.
At page 5 of NET FORCE IN QED is argued, when an electron in the atom emits a photon, the positron-electron pair needed to form this photon, originates from the shield of virtual particles surrounding the infinite strong electric core of the electron. There is concluded spin 0 photons should exist.
In paragraph Four quarks in the shell at page 5 of QCD, is shown how pairs of quarks, taken out of the shell of virtual particles that surrounds each quark, can form colored spin 0 gluons. I don't think two quarks do really merge into a gluon, that they couple to form a new gluon particle. The quarks in the gluon keep separated and keep their properties. Anyway I think that when a gluon is absorbed by the color of a target quark, the colors of the constituting quark and antiquark in the gluon are applied one after the other onto the target quark and that means multiplication of the colors of the quark and antiquark in the gluon one after the other with the color of the target quark.
For the abbreviations of storyline names, see the Contents page.
e+ positron, e- electron, μ muon, τ tauon, 
e electron neutrino, μ muon neutrino, τ tau neutrino; ph neutrinophoton
Three Higgs fields
In this paragraph my model for the three generations is abandoned because of arguments of reaction time.
In (1.1) up to and including (1.4) at page 3 of NEG is argued that velocity relative to the vacuum is impossible
(virtually impossible) because coupling to standing-still field, absorbing from standing-still field, takes least energy and thus is most abundant.
Mass should be absorbed from the field that is standing still with respect to the mass-absorbing particle because it leads to the lowest energy absorption. We pursue this argument here and state that the Higgs field from which mass is absorbed, is chosen out of the field of all possible velocities, to be the field of lowest energy. (1.1)
Assumed is that the stronger a field is, the faster it reacts and the more energy it contains. Therefore is assumed the spin 0 colored gluon field contains most energy, the spin 0 photon field contains lesser energy and the spin 0 neutrinophoton field is assumed to contain least energy. (1.2)
The abandoned model for the three generations:
START PREVIOUS ATTEMPT
The particles of the 1rst generation (quarks u and d, the e+ e-, and e) absorb their mass from the lowest energy field, the Higgs field that consists of ph, see Neutrinophotons at page 4 of QG. Let's call the neutrinophotonic Higgs field the 1rst Higgs field (belonging to the 1rst generation). (1.3)
The particles of the 2nd generation (quarks s and c, the μ, and μ) absorb their mass from the photonic Higgs field, made of spin 0 photons. Let's call it the 2nd Higgs field (belonging to the 2nd generation). (1.4)
And finally the 3rd generation (quarks b and t, the τ, and τ;) should absorb their mass from the most energetic field, the gluon field made of spin 0 pairs of colored gluons, which can best be called the 3rd Higgs field (belonging to the 3rd generation). (1.5)
STOP PREVIOUS ATTEMPT
Pro's
-- For the stable particles we observe in the world around us, (1.3) suits best with the Standard Model as it is now: a Higgs field virtually without properties and only meant for giving mass. (1.6)
-- The picture The 1rst generation absorbs mass from the 1rst Higgs field, the 2nd generation from the 2nd Higgs field and the 3rd generation from the 3rd Higgs field
should not surprise. The main difference between the three generations is the mass of the particles. The Higgs field gives mass, that's what it does. So one Higgs field for each generation is an obvious possibility. (1.7)
-- Why is not always absorbed from the lowest energy Higgs field, the 1rst Higgs field? The reason might be the reaction time, making the colored reaction always being faster and thus earlier occuring than the colorless color reaction. This would occur only when there is plenty internal energy available within the particles to afford the higher mass absorption. (1.8)
The 10^-23 sec reaction time of colored particle absorption from the vacuum is significant shorter than the reaction time of the colorless colors
with electric charge (which are the e- and e+ that are supposed to be inside the photon) which is at best 10^-20 sec. While the reaction time of colorless colors with electric charge should be on its turn significantly shorter than the reaction time of colorless colors without electric charge (the spin 0 neutrinophoton field that is: vacuum particles only, reaction time unknown, maybe equal to the W or Z reaction time).
Typical reaction times for W (W+ W- and Z0, weak force), P (Photons, electromagnetic force) and G (Gluons, color force). (1.9)
Con's
-- The spin 0 photons are vacuum particles, but it still are just photons, with typical reaction time of 10^-20 up to 10^-16 sec. Imagine that on a real s-quark (s-quarks are second generation) three subsequent 10^-23 sec color reactions are taking place, just one after the other. If the quarks s and c get their mass from the spin 0 photon Higgs field, then the (at best) 10^-20 sec spin 0 photon reaction is still completely involved in giving mass to the first of the three color reactions, while the other two color reactions already have been completed. Since we in (1.8) introduced reaction time as a working mechanism, we have to consider this here too and then this would not work. Likewise the spin 0 neutrinophoton field might be fast enough to give mass to e, but it might be not fast enough in giving mass to d and u, since d as well as u react in 10^-23 sec on color. (1.10)
-- And the trick couple by G but not glue
for general mass absorption from the 3rd Higgs field? This trick is supposed to work only for particles with the coloress colors +1 and -1, like the electron and the positron. And the color of the photon is zero, it doesn't couple to gluons by G, see the previous page. So there is no route for a 10^-20 sec spin 0 photon Higgs particle to react 10^-23 sec when being absorbed by a s- or c-quark that is acquiring mass. When giving mass to the s and c the spin 0 photon will be too late.
In this website so far has been used a reaction time of 10^-23 sec for the color force, including the colorless colors +1 and -1. Maybe I better conjecture, when a reacting particle has color +1 or -1 AND the particles inside the reacting particle have color +1 or -1, that then
The colorless color reaction is as fast as the (at best) 10^-20 sec photon reaction (1.11)
Maybe it IS the photon reaction time, the reaction time of the electromagnetic force.
Because of the contras (1.10) and (1.11) I see no other way than to abandon this previous attempt model for the three generations. Instead I return to the view of the hadronic or gluonic Higgs field (A) made of the spin 0 colorless gluons to be absorbed by quarks d, u, s, c, b and t to give them mass. That is, the spin 0 gluons are colorless, but the quark and antiquark they consist of have color (B). The leptonic Higgs field (made of spin 0 photons) gives mass to the leptons e, μ and τ. Finally the neutrinophoton Higgs field gives mass to the electron neutrino e, the muon neutrino μ and the tauon neutrino τ. (1.12)
(A) Massless coinciding at page 2 of EXPAN, and paragraph Filling in the vacuum marbles at page 2 of QG between (1.2) and (1.3).
(B) (5.13) in paragraph Remnant particles in the fermion explosions at page 4 of EXPAN
As a consequence I cannot explain the three generations anymore.
So a particle first tries to absorb mass from the fastest field, the hadronic Higgs field, if it can afford the energy. The quarks d, u, s, c, b and t succeed because they have color. If not so, then a little later it tries to absorb from the leptonic Higgs field, provided it can afford that. There the e, μ and τ succeed. If it still hasn't enough energy for that, it finally absorbs from the lowest-energy field, the neutrinophoton-field. That's what the neutrinos e, μ and τ do. And then we have all material particles (spin-1/2 particles or fermions) supplied with their mass. W+ W- and Z0 would have a different story, see paragraph The conversion between pi-meson and W at page 5 of QCD (for W) and see e, mu, tau 
e, mu, tau below. (1.13)
Neutrinos are only reacting with W and Z. Then a neutrino absorps a spin 0 ph Higgs boson by W and Z force, isn't it? The slowest field giving mass to the neutrinos after faster fields failed to do so. (1.14)
So available energy determines what Higgs absorption can be afforded. Regard a muon. It is created because there was enough energy to create it. But then it stays a muon all over its track. So the muon's energy must be an internal state. (1.15)
Regard the quark composition uds, with spin 3/2, called the Σ*0 (sigma star null) particle. The u and d are in groundstate while the s is supposed to be in an excited state, e.g. circling around an u-d axis. Then, when it de-excitates and falls back to a lower energy state, it decays into e.g. Δ0 (delta null) = udd with spin 3/2. So internal state can be given meaning in the hadrons. (1.16)
But how could μ be an exited state of e? Both particles have no internal structure. But wait, regard spin. If we observe that (a part of) vacuum itself is mirror-imaged, I assume the shape of a single particle in that part of vacuum is mirror-imaged along with it, not noticing something had changed. But we see its spin has swapped. Spin is connected to space, spin is a structure. Then again, spin e is +1/2 or -1/2 and ditto for μ and τ, so I see no use for spin in this case.
Here STRING THEORY has an advantage: it allows sole electrons that are moving freely in the vacuum, to have excited states. Pointlike electrons and stringlike muons and tauons? Has the muon stringlike properties? In this website there is only one other place where strings are suggested: (1.10) at page 2 of QQD. This also ruins internal excited states as a possible explanation for the three generations. (1.17)
One single particle has its structure, its shape, relative to the vacuum. Take in mind e.g. its spin.
No grid
In this paragraph I want to keep the expansionfield view AND the velocityfield view AND their equivalence. That rules out any process between the particles of the vacuum, like cohering processes.
One velocity all over space constitutes a velocityfield: the vacuum (one specific frame) consists of points (vacuum particles) that all have the same velocity (relative to another velocityfield). All possible velocityfields together form the field of all possible velocities. So far we assumed that the vacuum particles of one velocityfield cohere somehow, in doing so shaping 3-dim space. We assumed also that the different velocityfields are superposed to each other: they don't see each other and thus don't cohere either. (2.1)
Likewise holds for the field of all possible expansions as is built up in paragraph The fermion explosion of the vacuum particles at page 4 of EXPAN, which is the same page as page 7 of NEG, or the next page of TONE that also is its last page. *) There is argued that the Big Bang of the vacuum particles consists of a superposition of one expansion at every point of spacetime. Every expansion has one point standing still to you, the so-called expansion center. The Big Bang consists of a superposition of expansions that all have another point of space as expansion center (as observed from one such expansion center). The different expansions don't see each other. (2.2)
*) If you are reading TONE you very well could read the first two paragraphs of the next page and then return here.
Now there is a problem. Regard Fig (2.6) at page 4 of EXPAN, the left side of the equation. The arrow colors indicate that the velocities of one fermion expansion all originate from different velocityfields that all are superposed to each other. (2.3)
Then again, the fermion expansions were there first. So it is one fermion expansion that coheres
as a Bose condensate, while the different Bose condensates (out of the field of all possible expansions) are superposed to each other. Then one velocityfield (consisting of the same velocity in all its points) consists of velocities that all originate from different superposed states and thus those velocities don't see each other and how points that don't see each other would ever cohere? (2.4)
At the moment of Higgs field absorption we still assume one Higgs field particle is absorbed and a hole in the vacuum particle sea appears. The Bose condensate is disrupted then and for the moment (2.7) (just a few lines ahead) doesn't hold. At this moment I still want to maintain the sagging-in process of gravitation to occur infinite fast
because in that view the gravitation law of Newton is derived (page 6 of NEG) and the vacuum particle density is calculated to be 10^-21 m mutual distance (The radius of the Higgs boson). But I also want to keep the expansionfield view of my Big Bang theory at page 4 of EXPAN. Especially I want to maintain the equivalence of the field of all possible velocities and the field of all possible expansions. This equivalence is too beautiful to let it go. (2.5)
This forces me to conclude that the particles of the vacuum are not held together by any coherence mechanism, nor are vacuums superposed to each other. (2.6)
The vacuum is a Bose condensate without coherence and without superposition of fields (2.7)
However, when a Higgs field absorption takes place, it does so from the field that is standing still with respect to the Higgs field absorbing particle, that's what we assume. Absorbing from a velocityfield or an expansionfield? We don't need the rest of spacetime, we only need the Higgs field absorbing spot. So it seems you may choose freely whether you are in the velocityfield that is standing still to you, or that you happened to be in the center of an expansionfield. It doesn't really matter as long as the Higgs field absorbing spot is standing still to you. (2.8)
Page 4 of EXPAN says that wherever you are, you always are in an expansion center that is standing still relative to you. Then, in the field of all possible expansions, if you are going from expansion center A in space to another point B, this resets the expansion center from A to B. Being in A and having the suiting velocity there changes the expansion center also to B, see paragraphs Velocity and The Globular Cluster
at page 5 of EXPAN. So changing to a different expansion state out of the superposed states of expansion, is the same as changing to another velocity. (2.9)
One Higgs field only
Particle accelerator measurements indicate there is only one complex Higgs field with only one Higgs boson. (Tien jaar Higgs field
, NTvN mei 2022 - Ten years Higgs field, Dutch Magazine of Physics, May 2022.) (3.1)
The 3rd Higgs field, made of colored spin 0 gluon pairs, color force that is, is the underlying force of the Bose condensate of the vacuum, necessary to determine the reaction time of the Bose condensate. (3.2)
It can be that there are only 3rd generation vacuum particles (pairs of colored spin 1 gluons), and then a pair of colored gluons converts at the spot into two spin 0 photons that subsequently are absorbed in order to give an e, μ or τ its mass. Or instead of giving mass, the two spin 0 photons convert at the spot into two spin 0 neutrinophotons to give mass to the neutrinos e, μ and τ. There should be no conversion of a pair of spin 1 gluons directly into two neutrinophotons, I guess. The spin 0 photon Higgs field and the spin 0 neutrinophoton Higgs field as well don't have to exist as separate fields but are locally created from the colored gluon pair Higgs field at the spot at the moment they are needed, compare paragraph The spin consideration at page 3 of QG, especially (1.6) **). Let's adopt this view. (3.3)
**) The spin consideration
at page 3 of QG is a hadronic field feature only. Nor spin 0 photons nor spin 0 neutrinophotons will ever add to the spin 2 of the graviton. Spin 1 photons can align to a spin 2 graviton vacuum particle, but two photons do not cohere. Spin 1 neutrinophotons even do not exist.
In paragraph The Big Bang of our matter as a fermion explosion of d-quarks at page 6 of EXPAN (not a TONE page) is shown that our matter very well might be the result of an identical-fermion explosion of spin -1/2 d-quarks. That page starts with an attempt to derive our matter from an extra fermion explosion of spin +1/2 u-quarks, which leads to a universe that is close to ours, but not close enough. This is very illustrative for that you cannot derive our matter from just any particle in fermion expansion. (3.4)
Because of reaction time, it also could not be the other way around, that the vacuum consists of spin 0 neutrinophotons that on the spot, when internal energy is available, is upgraded to a spin 0 photon field or even a spin 0 colored gluon pair field. The upgrading to spin 0 colored gluon pair field comes after the very first contacting with the vacuum field, which would be the spin 0 neutrinophoton field, and that is supposed to be far too slow for a color reaction where mass is given to a quark. (3.5)
Bose-Einstein condensates
In the past I tried to describe the coherence of the vacuum particles as ordinary interactions mediated by the well-known forces. A good example is the dynamic view of the vacuum as sketched in paragraph The grid: a previous attempt
in the column at the right. It can't be true, for a vacuum it's of no use.
At page 4 of EXPAN is argued that the Big Bang is an identical-fermion explosion, an expanding frame that from its own nature already resembles a Bose condensate, an expanding Bose condensate - such an expression should have meaning then. In such a frame it is still more difficult to imagine any coherence by well-known forces between vacuum particles that all are receding from each other. (4.1)
If one field out of the field of all possible velocities is a superfluid then also one field out of the field of all possible expansions must be a superfluid, an expanding superfluid. Conjectured is there is a corresponding description of the process of gravitation (page 3 of NEG) in the expanding Bose condensate frame. (4.2)
The Bose condensate for spin 0 vacuum photons and spin 0 vacuum neutrinophotons can be understood as an expanding Bose condensate: the Pauli principle causing the fermion expansion from page 4 of EXPAN. An other underlying force for a Bose condensate is missing for spin 0 vacuum photons and spin 0 vacuum neutrinophotons. We need the underlying force to determine the reaction time of the condensate. Moreover, we adopted the view that spin 0 ph as well as spin 0 ph are created at the spot, they don't exist as separate fields. (4.3)
All in all we conclude:
Set the vacuum particles to ONE SINGLE HIGGS FIELD of COLORLESS GLUONS OF SPIN 0 that ALSO is a Bose condensate (4.4)
with the remark that there is color inside the colorless gluons.
See also Paragraph Vacuums that are Bose condensates that are superfluids at page 6 of NEG.
Everything can couple to everything by color force. Examining the Standard Model.
In the course of this website is found that all particles of the standard model have color, although a number of them have the colorless color +1 or -1 (quaternion +1 or -1). The gluon consists of two colored quarks, massless coinciding within their timeborders. The photon consists likewise of an electron and a positron massless coinciding, the e+ and e- being quarks with color +1 and -1 and electric charge. The neutrinos are quarks with color +1 or -1 without electric charge. We always take particles to have color (quaternion value +1, i, j, k) and antiparticles to have anticolor (quaternion value -1, -i, -j, -k).
Regard paragraph Four quarks in the shell
at page 5 of QCD, the paragraph starting with Mark the = pairs in the 3rd and 4th scheme are interesting, nearly at the end of the paragraph. If we indeed take this to be the W+ and W- particle, we in fact have identified all particles from the Standard Model with quarks, particles that somehow feel the strong force G. Except for the Z0, I still didn't found that.
In Four quarks in the shell
W+ has quark composition u d and W- has quark composition u d. (d means anti d quark
, etc.) Our supposed W+ and W- have the freedom to have any of the colors 1, i, j, k, -1, -i, -j, -k. The particles with color i, j, k, -i, -j, -k glue. Let's take the freedom to identify this with the 
meson with mass 140 MeV. Our supposed W+ and W- with colorless color +1 or -1 don't glue but only couple by G; this is very comparable with the W+ and W- from the standard model that only couple but do not mediate a force. Let's take this as the true W from the standard model with mass 80385 MeV.
So everything couples to everything else by G (strong nuclear force), although not every coupling glues. Regard the scheme of the standard model below. How do the non-existing grey lines in the standard model work out? I can't calculate but I can make an estimation. The difference with the standard model is the supposed coupling by G, the rest is the same as in the standard model.
Mark we don't try to identify force fields with each other. Some speculative thoughts about that are in paragraph What IS electric charge? at the previous page.
Let's call a non-existing coupling (according to the standard model) by the two particles the grey line connencts, separated by a small piece of grey line , gl = gluon, q = quark, ph = photon, e = electron or positron. We don't immediately bother which particle absorbs which. At the moment they couple, the particles merge to a merger particle and that's all there is for the moment.
The gluon has color [ i, j, k, -1, -j, -k ] and only will be found in the very neighborhood of quarks.
OR the gluon has the colorless color
+1 or -1 and is a glueball, see the free mesonic gluon in fig 3.9 at page 3 of QCD.
Here [ i, j, k, -i, -j, -k ] means one element out of the set [ i, j, k, -i, -j, -k ]
, etc. So [ i, j, k ] means i, or it means j, or it means k.
When both merging particles have mass the mass of the merger line should not be taken as it is given here. The result is given between ( ).
Color e = +1, color 
e = -1, see color of the neutrino at the previous page.
Mass merger = gl mass + mass = 0 + mass = mass
spin merger = spin gl + spin = -1 +1/2 () = -1/2. Or spin merger = +1 -1/2 () = +1/2
Electric charge merger = 0 + 0 = 0
Color merger = color gl * color = [ 1, i, j, k, -1, -i, -j, -k ] * [ -1, +1 ] = [ 1, i, j, k, -1, -i, -j, -k ].
When color = +1, then the merger is a particle with the same color or anticolor as the gluon it couples with.
When color = -1, then the merger is a particle with the opposite color or anticolor of the gluon it couples with.
When the color of the gluon was [ i, j, k, -i, -j, -k ] then the merger cannot leave the confinement the gluon was in. Moreover, the gluon was part of a process of color interchange, there is no other place for colored gluons, and that gluon now is missing. I suppose the merger particle cannot take over the color task of the missing gluon for several reasons (mass, spin). I am not sure what happens then, but I guess the merger splits again in the original gluon and neutrino.
When the color of the gluon was +1 then the merger is a particle with neutrino mass, spin [ -1/2, +1/2], zero electric charge and same color as the neutrino it coupled with. So the merger is a neutrino. However, it are spin -1/2 and spin +1/2 which do not occur. This reaction will not take place.
When the color of the gluon was -1 then the merger is a particle with neutrino mass, spin [ -1/2, +1/2], zero electric charge and opposite color as the neutrino it coupled with. This means the has become an , or the has become a . The coupling with the gluon changed the spin into its opposite too. This results in existing spin +1/2 and spin -1/2 . So I conclude, when the color -1 gluon is not part of a color interchange process in a hadron, in other words when the color -1 gluon has escaped into outer space, then the color -1 gluon indeed couples with a neutrino to yield the opposing existing neutrino.
gl W+ W- Z0
Mass merger = mass gl + mass [ W+, W-, Z0 ] = 0 + mass [ W+, W-, Z0 ] = mass [ W+ W- Z0 ]
Spin merger = spin gl + spin [ W+ W- Z0 ] = 1 + 1 = 2, or -1 -1 = -2, or 1 - 1 = 0.
Electric charge merger = 0 + [ -1, +1 ] = [ -1, +1 ]
Color merger = color gl * color [ W+ W- Z0 ] = [ 1, i, j, k, -1, -i, -j, -k ] * 0 = 0
The merger particle has mass W+ W- or mass Z0, spin -2, 0 or +2, electric charge +1 or -1 and no color (0 is not a color). Color W = 0, see The color of W at the previous page. So the gluon and the W+, W- and Z0 don't react with the gluon by G.
gl e, mu, tau
The color of e- is -1, the color of e+ is +1, see paragraph Electrons as Baryons.
Mass merger = mass gl + mass [ e, mu, tau ] = 0 + mass [ e, mu, tau ] = mass [ e, mu, tau ].
Spin merger = spin gl + spin [ e, mu, tau ] = 1 - 1/2 = +1/2. Or spin merger = -1 +1/2 = -1/2.
Electric charge merger = 0 +1 = +1 or 0 -1 = -1
Color merger = color gl * color [ e, mu, tau ] = [ 1, i, j, k, -1, -i, -j, -k ] * [ +1, -1 ] = [ 1, i, j, k, -1, -i, -j, -k ]
The merger is a particle of mass [ e, mu, tau ], spin [ -1/2, +1/2], electric charge [ +1, -1 ] and color [ +1, -1 ].
The interaction between the color of e+ e- and the color of the quarks is discussed in paragraph Electrons as Baryons at the previous page. The same reasoning holds for the interaction between e+ e- and the gluons in the hadrons.
Coupling e+ e- with a gluon of color +1 (somewhere in free space) leads to spin swap of the e+ e-, which is a permitted state change.
Coupling e+ with gluon of color -1, would that change it into e-? Which violates electric charge conservation. Or only give a positron with an anticolor? Which will not be observable, but is not was is agreed upon. Ditto for coupling e- with color -1 gluon. I conclude the reaction will not take place. Ditto for mu and tau particles.
gl ph
Mass merger = gl mass + ph mass = 0 + 0 = 0
Spin merger = spin gl + spin ph = 1 + 1 = 2. Or spin merger = 1 - 1 = 0.
Electric charge merger = 0 + 0 = 0
Color merger = color gl * color ph = [ 1, i, j, k, -1, -i, -j, -k ] * 0 = 0
The color of the photon (at page 5 of NET FORCE IN QED) is zero and the color of the merger is zero. Zero is not a quaternion unit, is not a color. So I conclude ph and gl don't couple by G.
Color e = +1, color e = -1, see color of the neutrino at the previous page.
Mass merger = (mass + mass quark)
Spin merger = 1/2 + 1/2 = 1. Or 1/2 - 1/2 = 0.
Electric charge merger = 0 +2/3 = +2/3. Or 0 -1/3 = -1/3.
Color merger = [ +1, -1 ] * [ i, j, k, -i, -j, -k ] = [ i, j, k, -i, -j, -k ]
The merger has more or less one quark mass, spin [ -1, 0, 1 ], electric charge [ -1/3, +2/3 ] and color [ i, j, k, -i, -j, -k ].
In the standard model the neutrino is supposed not to react with a quark directly. The neutrino couples with W+ W- Z0 and W+ W- Z0 couples with a quark, that's the rule. If every particle has a kind of color, as is suggested in this website, then in principle there is a contribution of neutrino's directly coupling to quarks.
The e with its color -1 cannot be absorbed by a quark in a baryon or meson because it would leave the baryon or meson colored, compare 1) in Electrons as Baryons
at previous page. So the e is not observed by the baryon or meson, they doesn't see the e.
The spin +1/2 e has color 1. This, as far as color is concerned, makes the e extremely vulnerable for absorption by any quark via the strong nuclear force, compare 2) in Electrons as baryons
. But then again, when spin merger is 1, where to leave the extra spin +1/2? There is no other way than emit an e again. When spin merger is zero I guess the merger has a need for a spin 1/2, and where to get that? So yes, the e reacts extremely easily with the quarks in the baryon or meson, but no, they don't stay there. This does hold as an observation, the e is seen by the baryon or meson, but it never causes a state change.
Well, that is, two simultaneous neutrinos e and e can do the job: within one and the same 10^-23 sec two e happen to coincide near a quark in a proton or neutron. The two neutrinos then merge by the strong nuclear force to a spin -1 particle of color -1 x -1 = 1 that is immediately absorbed by the quark. The quark then emits the extra spin 1 by one photon. It seems possible, but doesn't sound as an easy process.
Color e = +1, color e = -1, see color of the neutrino at the previous page.
Mass merger = mass + mass ph = mass + 0 = mass
Spin merger = [ -1/2, +1/2 ] + [ -1, +1 ] = [ -1/2, +1/2 ]
Electric charge merger = 0 + 0 = 0
Color merger = [ color ] * [ color ph ] = [ -1, +1 ] * 0 = 0
The merger is a particle with mass, spin [ -1/2, +1/2 ], no electric charge and color 0.
The color of the photon is zero and the color of the merger is zero. Zero is not a quaternion unit, is not a color. So I conclude ph and don't couple by G.
u, d, s, c, b, t e, mu, tau
Mass merger = (mass [ u, d, s, c, b, t ] + mass [ e, mu, tau ] )
Spin merger = [ 1/2, -1/2 ] + [ 1/2, -1/2 ] = [ -1 , 0, 1 ]
Electric charge merger = [ +2/3, -1/3 ] + [ 1, -1 ] = [ +5/3, +2/3, -1/3, -4/3 ]
Color merger = [ i, j, k ] * [ +1, -1 ] = [ i, j, k, -i, -j, -k ]
The merger particle has unknown mass, spin [ -1 , 0, 1 ], electric charge [ +5/3, +2/3, -1/3, -4/3 ] and color [ i, j, k, -i, -j, -k ].
The interaction between the color of e+ e- and the color of the quarks is discussed in paragraph Electrons as Baryons. u, d, s, c, b, t in the baryon or meson always have color [ i, j, k, -i, -j, -k ]. Coupling e+ e- with a quark of color +1 somewhere in free space - such a quark, isn't that just the positron? After ECR? (Electric charge redistribution, obviously necessary to yield an observable particle.)
A spin 1 particle is not a quark, nor an electron. There has to be emitted a spin 1/2.
I conclude the reaction will not take place. Ditto for mu and tau particles.
u, d, s, c, b, t u, d, s, c, b, t
Mass merger = (mass [ u, d, s, c, b, t ] + mass [ u, d, s, c, b, t ] )
Spin merger = [ 1/2, -1/2 ] + [ 1/2, -1/2 ] = [ -1 , 0, 1 ]
Electric charge merger = [ +2/3, -1/3 ] + [ +2/3, -1/3 ] = [ -2/3, +1/3, +4/3 ]
Color merger = [ i, j, k ] * [ i, j, k ] = [ i, j, k, -1, -i, -j, -k ]
The merger particle has unknown mass, spin [ -1 , 0, 1 ], electric charge [ -4/3, -1/3, +1/3, +4/3 ] and color [ i, j, k, -1, -i, -j, -k ]. The color +1 seems to be lacking.
This reaction is already discussed in the standard model. The standard model says that quarks don't couple by G. Quarks and gluons couple, but quarks directly to quarks do not. Gluons couple with each other by G, so why wouldn't quarks do likewise?
A spin 1 particle is not a quark, a spin 1/2 has to be removed by e.g. a neutrino.
We take for granted that the reaction will not take place.
ph ph
Mass merger = mass ph + mass ph = 0 + 0 = 0
Spin merger = [ +1, -1 ] + [ +1, -1 ] = [ -2, 0, +2 ]
Electric charge merger = 0 + 0 = 0
Color merger = 0 + 0 = 0
The merger particle has no mass, spin [ -2, 0, +2 ], no electric charge and no color.
The color of the photon is zero and the color of the merger is zero. Zero is not a quaternion unit, is not a color. So I conclude ph and ph don't couple by G.
e, mu, tau e, mu, tau
The color of e- is -1, the color of e+ is +1, see paragraph Electrons as Baryons.
Mass merger = ( mass [ e, mu, tau ] + mass [ e, mu, tau ] )
Spin merger = [ -1/2, +1/2 ] + [ -1/2, +1/2 ] = [ -1 , 0, 1 ]
Electric charge merger = [ +1, -1 ] + [ +1, -1 ] = [ -2, 0, +2 ]
Color merger = [ -1, +1 ] * [ -1, +1 ] = [ -1, +1 ]
The merger particle has unknown mass, spin [ -1 , 0, 1 ], electric charge [ -2, 0, +2 ] and color [ -1, +1 ].
When e+ and e- coincide massless they form the photon, but then they don't merge.
Could e+ and e- merge? The electric charge -2 and +2 is an integer charge, so not immediately forbidden. However, we find it only at the Δ++ particle with quark composition uuu. Let's suppose it somehow is forbidden. So electric charge 0 is left. Then it only can be the pairs e+ e-, mu+ mu- and tau+ tau- that can merge. The color then always yield +1 * -1 = -1, the dangerous colorless color. The merger will not be absorbed by a p+ or n0 since it would leave them colored which is forbidden. We now have a particle with some mass, spin -1, 0 or +1, no electric charge and color -1. Do we recognize Z0? Did we finally found Z0, our last missing particle?
Color e = +1, color e = -1, see color of the neutrino at the previous page.
Mass merger = ( mass + mass )
Spin merger = [ -1/2, +1/2 ] + [ -1/2, +1/2 ] = [ -1 , 0, 1 ]
Electric charge merger = 0 + 0 = 0
Color merger = [ (color spin +1/2 e), (color spin -1/2 e) ] * [ (color spin +1/2 e), (color spin -1/2 e) ] = [ -1, +1 ] * [ -1, +1 ] = [ -1, +1 ].
Neutrinos don't glue but couple to each other by G, that's the consequence of the view so far. The merger particle has unknown (but maybe small) mass, spin [ -1 , 0, 1 ], no electric charge and color [ -1, +1 ].
A spin 1 particle is not a neutrino, a spin 1/2 has to be removed by... a neutrino. So after all the number of neutrinos doesn't change.
Final conclusion
The provisional final conclusion is that there is no objecting reaction/coupling, and thus no objection, to the view of every elementary particle of the Standard Model having a color.