In 4 quarks in the shell at page 5 of QCD is proposed how each gluon consists of two quarks massless coinciding. In Colors as quaternions
quarks are presented as spin 1/2 particles with the quaternion units i, j, k, -i, -j, -k as colors. The conjecture is proposed that there should exist spin 1/2 particles of color +1 and -1 and that these very well could be the leptons. If we accept the conjecture, does the electron have color 1 or -1? (Quaternion units 1 and -1, is meant here.) I am unable to make calculations, but I can estimate some results.
e- = electron, e+ = positron, ph = photon, 
e = electron neutrino, 
e = anti electron neutrino, μ = muon neutrino, μ = anti muon neutrino, τ = tau neutrino, τ = anti tau neutrino
= spin up = spin +1/2, 
= spin down = spin -1/2
p+ = proton, n0 = neutron, u = anti-u-quark, d = anti-d-quark, ECR = electric charge redistribution, ph = neutrinophoton
Electrons as Baryons
1) Suppose the electron color is -1. The electron then is an antiparticle. The electron is dragged into the nucleus, the wave function of the electron in the nucleus is not entirely zero. It is normalized at zero in the nucleus, but the original wave function has some value there. The electron can react easily with one of the colors of the quarks there:
i * -1 = -i
j * -1 = -j
k * -1 = -k
The electron and the quark merge via the strong nuclear force to one quark of opposite color and spin 1 or 0. A half spin can be carried off by an anti electron neutrino via the weak nuclear force. The electric charge becoms +2/3 -1 = -1/3 (merging with u-quark) or -1/3 -1 = -4/3 (merge with d-quark).
However, the anticolor of that single quark can by no means be combined with the two colors of the remaining quarks in the same baryon. It will lead to a colored end state of the baryon and that is forbidden. (1.1)
So for the quark there is no other way than to re-emit the electron again long before the electromagnetic force or the weak nuclear force can perform any action. So no, any electron will not be absorbed by any quark in the nucleus, which fits in with every-day observation. Compare the black gluon at fig 3.9 at page 3 of QCD.
(Does a problem arise when within the same duration of 10^-23 sec two electrons are absorbed by one quark? The first e- turns the quark color into an anticolor, the second e- turns it back to the original color. If the two electrons have opposite spin, then there is no need to emit e. But the electric charge is a problem: +2/3 -2 = -4/3 electric charge when the u-quark absorbed the two e-, -1/3 -2 = -7/3 for the d-quark. No, this will not occur.)
So the electron very well might have color -1 and be an antiparticle. Let's assume this.
Color electron = -1 (1.2)
2) The positron. When the electron has color -1 then the positron has color +1. The positron is a particle. When in the nucleus absorption of the positron by a quark is inevitable. The colors become:
i * 1 = i
j * 1 = j
k * 1 = k
and this gives zero net color end states. A half spin can be carried away by an electron neutrino via the weak nuclear force.
When the d-quark absorbs the positron, the electric charge becomes -1/3 +1 = +2/3, a neutron (udd) changes in a proton (uud), or a proton (uud) changes in a delta++ (uuu) but the last possibility will cost a lot of energy. When the u-quark absorbs the positron one gets electric charge +2/3 +1 = +5/3 and together with two d-quarks in a neutron this yields still a unitary electric charge. It seems not forbidden. Although in a proton u +5/3 electric charge together with another u +2/3 and one d -1/3 yields a proton with double electric charge, and that seems forbidden. Still I hesitate to restrict to only u +2/3 and d -1/3 from the very first moment, there may be a mechanism - yet unknown - of electric charge redistribution at work.
I suppose the repulsion between the electrically positive nucleus and the also positive positron makes the contribution of the wavefunction in the nucleus very small. Most likely the positron will annihilate with an electron long before being observed in the nucleus.
We conclude
Color positron = +1 (1.3)
3) What if the electron has color 1? Things would immediately go awfully wrong. Each of the three quark can absorb the color 1 easy, the neutrino carries away a spin 1/2.
When the u-quark absorbs the -1 electric charge, one gets +2/3 -1 = -1/3, a proton converts in a neutron, or a neutron converts to the delta-minus (ddd) which cost more energy. When the d-quark in a proton absorbs the -1 electric charge, one gets -1/3 -1 = -4/3 and together with two u-quarks this still yields a unitary electric charge (zero, that is). It seems not forbidden. When the d-quark in a neutron absorbs the -1 electric charge, this yields one d with -4/3 and one d with -1/3. This also seems not forbidden, but the introduction of different d-quarks makes situation more complicated. well, let's suppose there allways is some mechanism of electric charge redistribution at work to maintain u-quarks at +2/3 and d-quarks at -1/3.
As final result virtually all electrons would be absorbed by the nuclei within a second and that has no resemblance with observation. No, the electron cannot have color 1.
1), 2) and 3) are in nice agreement with the considerations in "The baryon number and lepton number of a black hole" at page 6 of the storyline EXPANSION OF THE UNIVERSE.
W+, W-, Z0 and neutrinos
The standard model says that, except for gravitation, neutrinos can only couple to the W+, W- or Z0 particle. Neutrinos do not couple to any of the other particles, including themselves. The standard model says
Spin +1/2 neutrinos and spin -1/2 antineutrinos exist (2.1)
In earlier times was thought that the neutrinos were massless and so the neutrinos propagate at light speed. The picture then is strange but not immediately leading to contradiction. But since the neutrinos turn out to have some mass, things are different.
W1 --> <-- W2
W3
Suppose, we have 1) a standing still neutrino and 2) it has a definite spin direction. There are two W-bosons W1 and W2 approaching it from both sides and a third W-boson W3 approaching from below. (2.2)
In the frame of W1 the approaches W1 and when arriving it couples to W1 and so must have had spin +1/2.
In the frame of W2, that same would have spin -1/2, because has same spin but now travels in opposing direction, so the spin is observed as -1/2. And when would have arrived a little earlier at W2 than at W1, it could and would not couple to W2.
In the frame of the , when W1 arrives at it couples with W1 and if W2 would have arrived a little earlier W2 would not have coupled to . In the view where a with spin -1/2 doesn't exist (call this view B), this sentence would have been: In the frame of the
Which is nonsense, since a few moments later it had to exist again for the meeting with W1., when W1 arrives at it couples with W1 and if W2 would have arrived a little earlier would not have existed.
In the frame of W3, since has spin +1/2 for W1 and spin -1/2 for W2, is in a superposition of spin +1/2 and spin -1/2. In view B, as far as W3 is observing, is in a superposition of existing and not-existing.
W1, W2 and W3 don't agree about the existing or not. (2.3)
Therefore, instead of (2.1), I would presume:
spin -1/2 neutrinos and spin +1/2 antineutrinos do exist but never occur in observations (2.4)
Moreover, it would be better to state that
each (resp. ) has a certain chance between 0 and 1 to be in the spin -1/2 state (resp. spin +1/2 state) as observed by certain W or Z and then the reaction is forbidden (2.5)
This chance should be given as a factor in front of the wavefunction, call it forb
for the moment, from forbidden
. (2.6)
Neutrinophotons
In paragraph Meson exchange at page 4 and especially Four quarks in the shell at page 5 of the storyline QCD, is argued the gluon consists of a quark and an antiquark, massless coinciding. At page 5 of NET FORCE IN QED is also argued the photon consists of an electron and a positron massless coinciding. Then the row can be pursued by a neutrino and an antineutrino massless coinciding to a spin 0 or spin 1 particle. A kind of photon, but not originating from electric charges and thus without internal electric or magnetic fields. A new type of radiation? This particle needs a new name. My proposal would be neutrinophoton.
The neutrinophoton would be able to decay again, but it has the choice to decay to an electron neutrino e and an anti electron neutrino e, or to a muon neutrino μ plus anti muon neutrino μ, or tau neutrino τ plus anti tau neutrino τ, depending on available energy. The neutrinophoton can convert between neutrino types, is the conjecture. (3.1)
Is there a difference between a photon made of an electron and a positron, and a photon made of μ and μ? Since mass is the only difference between the generations, and since in the photon the particle absorbs mass at the same rate as the antiparticle emits mass, so they coincide massless, I suppose there is no difference observable. So the photon should be able to convert between the e+ e-, the mu+ mu- and the tau+ tau- pairs. Sufficiently energetic photons can decay into a mu+ mu- pair or a tau+ tau- pair.
ph → e+ e- → ph → μ+ μ- → ph → τ+ τ- → ph (3.2)
But wait - the spin of e.g. the electron neutrino is +1/2 or -1/2, and -1/2 is forbidden for a neutrino. To form a spin 1 neutrinophoton, the spins of the neutrino and the antineutrino have to align:
+1 = spin neutrinophoton = spin e + spin e = +1/2 +1/2 = +1 or
-1 = spin neutrinophoton = spin e + spin e = -1/2 -1/2 = -1
And that means according to (2.4) and (2.5) that in both cases the reaction with one of the neutrinos is forbidden. Neutrinophotons cannot be formed. So no, we have no new type of radiation. We never observed neutrinophotons, so we won't miss them.
But spin 0 neutrinophotons can exist according to the presented ideas so far. When e and e approach each other within their time borders, they can coincide massless. So we have another massless spin-0 field.
e + e = spin 0 neutrinophoton (3.3)
Ordinary photons with electromagnetic fields in it couple to electric charges by P. (P
= electromagnetic force, see table in the column at the right.)
Neutrinophotons don't have such fields and don't couple to electric charges by P.
The conjecture (3.1)
ph → e e → ph → μ μ → ph → τ τ → ph
thus only holds for spin 0 neutrinophotons. (3.4)
We never observed spin 0 neutrinophotons. So let's take our chance and invoke our old habit to dump them in the vacuum as a field of vacuum particles, meant to become another Higgs field from which neutrinophotons can be absorbed, leaving a hole in the vacuum that is filled in from the outside: the act of gravitation. (3.5)
Analogous to (3.3) there can be:
e + e = spin 0 neutrinophoton (3.6)
but as said, reaction with this state is forbidden. Recall that no IS a spin -1/2 , and no IS a spin +1/2 . Each neutrino has the factor forb to be in the forbidden state or to be in the normal state.
Neutrinophotons might be CTL's, Closed Time Loops, just as is supposed for photons and gluons. See CTL considerations at page 8 of SR.
Start previous attempt
1) the forbidden neutrinophotons (3.6) are absorbed by the vacuum, forming another Higgs field. As will be stated at the next page, the Higgs field where we absorb our mass from, consists of neutrinophotons.
2) the neutrinophotons from (3.3) are not absorbed by the vacuum but decay into the real particles e and e .
It remains a question why (3.3) are real particles ready to be observed, and why (3.6) is absorbed by the vacuum.
When e and e happen to meet each other I see no reason why they shouldn't swap spin, coincide to a vacuum particle and being absorbed by the vacuum, thus making one bit of expansion of the universe. However, this kind of meetings are rare.
The e and e are locked up in vacuum particles. If they would be able to swap spin too, they would be able to depart as real particles. However, they coincide massless and thus have lightspeed. Time is standing still in the vacuum particle. So there must be a real particle available in the neighborhood speeding down the vacuum particle in order to swap the spins after which the vacuum particle can decay into two real particles. Maybe the available energy plays a role too: in pure vacuum the energy to create two neutrinos could be delivered by the vacuum contracting toward the hole the creation left behind.
Most probably the rate of e → e is (very) different from the rate of e → e .
End of previous attempt
The color of W
In Mark the = pairs (...) at page 5 of QCD is argued that W has a quark composition. We go on with that view. In W the quark must have color +1 and the antiquark must have color -1, that is the difference with the 
-meson in which its quarks have colors i and -i, or j and -j, or k and -k.
How the W is created? The first way is described in the paragraph Mark the = pairs (...). For that to occur you need 4 quarks to appear in the shell of a real quark.
Maybe it can be done with 2 quarks in the shell too. Start with a proton p+ (= uud) consisting of three quarks of color i, j and k. One of the two u-quarks in the p+ is in an excited state. Then in the shell of the excited u-quark a quark pair dd emerges. It is a special dd pair in the sense that its colors are respectively +1 and -1, instead of i and -i, etc. (In fact it are no d-quarks anymore.) The excess energy of the u-quark now is used for the mysterious electric charge redistribution (ECR) to take place: one positive electric charge goes from the excited u-quark onto the d-quark of the just-appeared quark pair dd.
Now we make another proposition:
The difference between the u-quark and the d-quark is the electric charge (4.1)
Of course there shall be difference in mass too, but the conjecture is that the mass difference, the rate at which is absorbed from the Higgs field, is a direct consequence from the difference in electric charge.
If so, then the exited u-quark from which the electric charge had been taken away, is changed into a d-quark. The p+ has become a n0. And the d-quark in the newly made dd pair has become an u. Then the two quarks u and d coincide and form the W+ particle.
Likewise in the n0. One of the d-quarks is excited. In its shell a quark pair uu emerges. One negative electric charge goes from the excited d-quark to the u-quark of the new pair, the excited d-quark has become an u-quark, the u-quark in the new pair has become a d-quark and the new pair now is the W- particle.
The color of the excited quark that is converted in the p+ or n0 cannot have changed since that would have left the p+ or n0 in a colored end state, which is forbidden. The only thing that happened was the unknown ECR. If ECR couples to the charges it leaves and arrives at, then ECR must have been a particle. ECR is not allowed to change color then. For ECR there is only one possible color that doesn't change quark color: +1.
1 * i = i * 1 = i
1 * j = j * 1 = i
1 * k = k * 1 = i
If ECR couples, the color of ECR is +1 (4.2)
Is ECR a particle of color +1, electric charge depending on situation, spin 0, mass 0 and no taste?
We have now W in the p+ and n0. What about W and the meson? The standard model says that a - meson (lepton nr 0, spin -1, quark composition du) can spontaneous convert to W- (spin -1) that converts to e- (lepton nr 1, spin -1/2) plus e (lepton number -1, spin -1/2). Mind the spin of e must be -1/2. Ditto for the + meson: + (spin +1) --> W+ (spin +1) --> e+ (spin +1/2) e (spin +1/2) (4.3)
In the + = ud the u has color i, j or k and the d then respectively has color -i, -j or -k. When the color of u changes in +1 and the color of d changes into -1, then the + has changed into a W+. However, the mass of W is much larger than the mass of the -meson. This indicates the colors i, j, k, -i, -j, -k are indeed massless, while the conjecture is that color +1 might be massless, but color -1 has mass, see paragraph Building vacuum from gluons: the hadronic vacuum. The colors of the quark and the antiquark in the -meson i -i, j -j and k -k superpose, but those three are very distinguishable from 1 -1.
The two quarks in W do coincide only partly massless. As a consequence W does not move at lightspeed c, time is elapsing on W. The two quarks of W can merge, resulting in a particle with color -1, the dangerous color.
Suppose that, for the change --> W to occur, the (e.g.) i and -i from the + and the 1 and -1 from the W have to couple. Then the multiplied colors before coupling should equal the multiplied colors after coupling. Before: i * -i = -i * i = 1, after: 1 * -1 = -1 * 1 = -1, and 1 doesn't equal -1. So we conclude the colors certainly don't couple this way.
Suppose we see the event of changing to W as happening in one single point in spacetime. At that point the colors multiply as
i * -i * 1 * -1 = -1
which is the dangerous color. Once again we conclude it most probably doesn't happen like this.
At page 2 and page 3 of QQD and in Four quarks in the shell
at page 5 of QCD, one can see that colors do couple and when merging yield another color. That is a big difference with the particles in the photon and W.
In the photon the e+ and e- keep separated. They never merge, not at their creation, not at their absorption, not during their flight in between, see The color of the photon at page 5 of NET FORCE IN QED. Let's assume the colors in W keep separated too, never merging, not coupling, but instead only coincide, partly massless in this case.
The color of the photon = 0 (4.4)
The two colorless quarks that make up W, emerge from the shell of a quark, coincide partly massless and that is the W particle. When at destination the W-quarkpair is added to the shell of one of the destination particles. In between the colors add: 1 -1 = 0.
The color of W = 0 (4.5)
W and the photon have much in common. Both are a composite of one color +1 particle and one color -1 antiparticle that stay together but don't merge. Both W and the photon have spin 1. The coupling constant of P, the photon coupling, is nearly the same as the coupling constant of the weak force, W that is. But in the photon the two particles are equal (opposite but equal) while in the W- one particle is a d-quark with color +1 and the other particle is the u-quark with color -1. The particles in W differ in electric charge and mass.
Z0 can decay into W+ and W-, so the color Z0 = color W+ * color W- = 1 * 1 = 1.
The color of the neutrino
Now for the meson and the decay of W in an electron and a neutrino. When the meson is created or decays, its quark colors multiply: i * -i = j * -j = k * -k = +1. (5.1)
We always used the rule that the color of a state does not change just from itself, see (3.26) at paragraph Color conservation at page 3 of QQD.
When W decays as W- --> e- e or W+ --> e+ e (let's just stick to the names as used in the Standard Model), the color product of these two particles must multiply to +1 too. We know color e+ = +1 and color e- = -1. There are only two possibilities for this multiplication:
color W+ = color e+ * color e = +1 * +1 = +1
color W- = color e- * color e = -1 * -1 = +1
Color e+ = +1 and color e = +1 (5.2)
Color e- = -1 and color e = -1 (5.3)
e+ and e are particles, e- and e are antiparticles.
The color of the neutrinophoton
Just like the photon consists of an e+ and e- that don't merge, so the neutrinophoton is supposed to consist of an e and an e that don't merge. Their color charges sum up:
color e + color e = +1 -1 = 0
The color of the neutrinophoton = 0 (6.1)
Just like the photon. I would propose that the neutrinophotons are absorbed and emitted in the same way as is described for photons at page 5 of NET FORCE IN QED. Neutrinophotons never merge, they are emitted from the shell of the neutrino and after their flight they are absorbed between the e e pairs in the shell of the absorbing neutrino.
Filling in the vacuum marbles
1) Can the vacuum marbles be filled in with one spin 0 photon per marble?
The procedure from eq. (2.1) at page 2 of QG is:
The wavefunction F of a real particle in the vacuum
times a vacuum particle
must equal F, so the vacuum particle must be 1 (the quaternion 1): F * 1 = F.
The color of the photon is zero, see paragraph The color of the photon at page 5 of NET FORCE IN QED.
So, following same procedure, for our color equation we get: F * 0 = 0, which is not what we want. The solution used in paragraph The color of the photon
is that the e- and the e+ in the photon don't merge at all. Let's apply such a solution also to the spin 0 vacuum photon and e.g. the electron: the spin 0 photon is absorbed by the e-, but without merging. This corresponds with the description of photon absorption and emission in When an electron in an atom emits a photon, where the positron is coming from?
at page 5 of NET FORCE IN QED.
Colors that stay together but do not merge sum up, see (3.03) at page 3 of QQD.
So our equation becomes:
F + 0 = F (7.1)
Then a vacuum marble can be filled in with one single spin 0 photon per marble, in doing so forming the what we call 2nd Higgs field.
2) Can a vacuum marble be filled in with one neutrinophoton per marble?
Spin 1 neutrinophotons don't exist, only spin 0 neutrinophotons do, so the neutrinophoton spin automatically is zero. Neutrinophotons have no electric fields in them, they don't especially couple to electric charges. The color of the neutrinophoton is zero. Internally it consists of two colorless colors
+1 and -1, just like the photon. Therefore the neutrinophoton couples by G (the strong force, from Gluon) but doesn't glue. For gluing one needs colors like i, j, k, -i, -j, -k to be present and they are not.
So for the vacuum made of neutrinophotons also holds F + 0 = F. And every particle with mass from the Standard Model can couple by G to a neutrinophoton in order to absorb it, leaving a hole in the neutrinophoton vacuum.
So it seems yes, a vacuum marble can be filled in with one neutrinophoton per marble and in doing so forming the so-called 1rst Higgs field. That is, as long as absorption of those marbles doesn't lead to a coupling - the absorbed particle does not merge with the absorber.
A previous attempt to form the three generations
First I thought, maybe the mass of the muon is formed when at each Higgs field absorption two spin 0 photons are absorbed from the vacuum, as long as energy for this is available, that is: during the lifetime of the muon. The tauon then might absorb three spin 0 photons at each Higgs field absorption, as long as it lives. And at four spin 0 photon Higgs field absorptions the threshold mass of the lightest quark would have been crossed. From then on mass absorption can be done by the gluon pair absorptions as is used for quarks. Four spin 0 photons would have more energy than one vacuum spin 1 gluon pair. Vacuum photon absorption shifts to another Higgs field.
There is another possibility for this. Follow me to the next page.