The spin reasoning. It is always told that the graviton has spin +2 or -2 and that the Higgs particle has spin 0. Let's for the moment propose the vacuum that is gravity to consists of a superposition of a glue2ball field gl gl (spin +2) and a glue2ball field gl gl (spin -2). The vacuum that is gravitation is taken to consist of gravitons made of two gluons each. The vacuum that is the hadronic Higgs field consists of a glue2ball field gl gl and a glue2ball field gl gl (spin 0). The vacuum that is the hadronic Higgs field is taken to consist of Higgs particles, also made of two gluons.
Here “gl” means “gluon” and a glue2ball is a glueball made of 2 gluons; here “” means “spin +1” and “” means “spin -1”.
I tried to work out the concept of gluNons in page 7 of the storyline NETFORCES IN QCD.
Suppose one of the two gluons from the graviton gl gl absorbs a graviton gl gl . Three gluons merge easier than two gluons, as is argued at page 7 of the NET FORCES IN QCD storyline.
( gl , spin +1) + ( gl gl , spin -2) --> ( gl , spin -1) (3.1)
The other gluon from the pair gl gl remains unaffected. The disappeared graviton leaves an empty spot at the place it had occupied. So the two gravitons change into one empty spot and one Higgs particle of same volume (is assumed) as the graviton:
gl gl + gl gl --> gl gl + empty spot (3.2)
Then the Higgs particle gl gl is absorbed at the coupling of some particle in the course of renormalization, leaving another empty place there.
So finally 2 gravitons converted to 2 empty spots and 1 Higgs absorption. (3.3)
This process rules out the possibility of taking e.g. the gl gl as vacuum particle while the other, gl gl , could be real. We need them both as vacuum particles.
If two gravitons gl gl and gl gl would just swap a gluon, one gets gl gl and gl gl . Now there are two Higgs particles in one single strike to be absorbed in the course of renormalization.
So then 2 gravitons convert to 2 empty spots and 2 Higgs absorptions. (3.4)
If this yields a heavier particle then, is this mechanism a candidate for one of the extra generations of elementary particles? One absorption at the spot isn't enough, the mechanism should work for quite a time all over the track of the heavy particle, isn't it? And where is the third generation?
In (3.1) and (3.2) graviton 1, to give them names, had absorbed vacuum marble graviton 2 and in doing so acquires the energy of graviton 2 added to its own energy. It is not precisely the Higgs mechanism because it had absorbed a graviton and not a Higgs particle. So it is in doubt whether the absorbing graviton gains mass. But it has caused one empty spot that will be filled in from the outside and this is one bit of gravitation. A vacuum locally acquiring energy, a local excited state of the vacuum, is not yet defined, maybe even not possible. So we conclude this reaction will not take place in empty space. But it does occur in the neighborhood of a quark ready to give that quark its mass. (3.5)
Then there doesn't have to be a separate Higgs field in empty space. There don't have to be two fields, a gravitational field AND a Higgs field. When a Higgs vacuum marble emerges, it is absorbed immediately thereafter. The hadronic vacuum then is one single grid of gravitons. The Higgs field, the Higgs particle, only does exist as a short-living intermediate state between the gravitational field and any coupling anywhere. The link between space and matter. (3.6)
The quaternion consideration. A vacuum marble like ( i -i ) consists of two gluons, i and -i. We start with two neighboring vacuum marbles, e.g.
( i -i ) ( i -i ) (3.7)
Now we assume the right gluon of the first vacuum marble ( -i ) to absorb the entire second vacuum marble ( i -i ). The left gluon of that first vacuum marble is unaffected.
If we rename ( i -i ) ( i -i ) as ( i a ) ( b c ) then there are 6 multiplication orders: abc, acb, bac, bca, cab, cba. In gluon-gluon reactions there is no preferred multiplication order and so the 6 possible outcomes superpose. In this case they all give same outcome -i, so the superposed possibilities merge to one possibility again.
-i * i * -i = -i * -i * i = i * -i * -i = i * -i * -i = -i * i * -i = -i * -i * i = -i (3.8)
The result is, as far as the colors are concerned, that the first vacuum marble is unchanged and the second vacuum marble is absorbed, leaving behind a hole in the vacuum, in accordance with the spin consideration.
(Quaternion multiplication has the associative property. As long as you don't change the order of multiplication, it doesn't matter whether you first multiply the last two gluons and then multiply by the first gluon, or multiply the first and second gluon and then multiply with the third one.)
The vacuum now is a superposition of ( i -i ), ( j -j ), ( k -k ) and ( 1 1 ) from (2.8), each in spin state gl gl or gl gl . So 8 fields altogether. (3.9)
Suppose, at the spot of a quark in a baryon two gravitons from the vacuum - four gluons altogether - couple as follows.
(gl gl ) graviton 1
(gl gl ) graviton 2
(gl gl ) Higgs particle -->
(gl ) (gl ) two independent gluons (3.10)
Two gluons of opposite spin merge, one gluon from graviton 1 and one gluon from graviton 2. Then the remaining two gluons, also of opposite spin, merge too. There are two possibilities for this, | | and X (one above the other or crosswise). To end up with spin1 gluons we need to assume “one spin from one gluon from graviton 1 to annihilate with one spin from one gluon from graviton 2”. This is thought to take place at the location of a quark, the quark mediates this spin conversion. Take in mind a baryon, three quarks together. If this conversion also detaches the gluons from each other, then you have two independent gluons of opposite color. Then one gluon can go to the second quark and the other to the third quark.
So 2 gravitons disappear from the vacuum, reducing it by their volume, and 2 gluons appear. Vacuum converts into matter. (3.11)
The quaternion approach of Higgs mechanism 1.
( i -i ) is a vacuum particle consisting of the gluons i and -i.
In the Feynman-diagram of the baryon it must be something like the following (quaternion parts only). In the baryon with start state quark of color i, quark of color j and quark of color k (k-quark not shown) a vacuum particle ( -k k ) appears.
i * j
= i * -k * k * j
= i * -k * j * -k
This is quark i that absorbs a gluon -k and quark j that absorbs the other gluon -k. Mark when it would have been quark i that emits a gluon -k and quark j that absorbs the gluon -k, then the equations are precisely the same. See page 2 of QQD.
= j * -i
or likewise leading to end state
= -j * i
See also page 3 of QQD, especially paragraph Color conservation.
When the electron absorbs from the leptonic Higgs field, it gains mass. But here there are two particles, quark i and quark j, that absorb from the (hadronic) Higgs field. So we assume both particles to gain mass.
Finally 2 gravitons convert into 1 Higgs particle at the spot of a quark (nearly) and the Higgs particle converts to 2 separate gluons.
This might be all there is to the exchange of gluons in the baryon. It might be a quark never emits a gluon, they only absorb gluons, in pairs made on the spot. For every gluon pair that is absorbed, there disappears the volume of one graviton from the vacuum.
(Originally in the NET FORCES IN QCD storyline, page 2, is stated: “It is as if the gluon takes away the color of the emitting quark in its upper half AND THEN A COLOR-ANTICOLOR-PAIR OF ITS DESTINATION APPEARS, a so-called correlated pair of particles, a correlated pair of colors cyan-red”. So in the definition of the gluon reaction itself in QCD the emergence of a kind of vacuum particle has been there from the start.)
Maybe the processes superpose. The QCD view where two colors in the baryon swap by a gluon going from one quark to the other superposes with the vacuum marble absorption process just described. But mark, the reaction of gluon exchange between quarks would be pure then, without absorption from the vacuum. When there is no marble absorption from the vacuum, there is no Higgs field absorption nor gravitation either, anyway not in this site. So pure gluon exchange between quarks would be massless and gravity-less - the Einstein assumption that E=mc mass is proportional to gravitational mass (that is the mass causing the gravitational field) remains fulfilled. And according to the conjecture proposed in paragraph Quark mass at page 5 of NET FORCES IN QCD, it would be outside of time either. That is: taking no time at all in our frame of reference. Which for practical use means it is a virtual particle.
Decisive might be an energy consideration. If you have choice OR to provide the energy for an intermediating gluon, albeit only for a short time, OR you can get it for free from the vacuum, just being there to be absorbed, then the choice seems to be easy. The lowest energy one will be chosen most often.
So the possibilities do superpose, but the gluon-interchange reaction from QCD will contribute only little and would have no contribution to the elapse of time. (3.13)
There is concluded the Higgs mechanism prevails over direct gluon exchange, but the direct gluon exchange was in QCD the very mechanism by which quarks attract each other. So we have to conclude the Higgs mechanism not only provides mass but also forms the attraction between quarks. (3.14)
Consider table (25), the outcome of nr 2, ( k ) ( k ) + ( -k ) ( -k ). Two gluons appeared, where do they go? We start with a baryon with colors i, j and k, to denote as baryon ( i j k ). When ( k ) ( k ) emerge at the quark with color k then one ( k ) arrives at the quark of color i and the other ( k ) at the quark with color j. Then ik = -j and jk = i, mark we had to right-multiply the quark by the gluon, see 7b) in Resemblances and differences, page 9 of QQD. The quarks in the baryon then have colors -j, i and k. Then -j * i * k = -1, so this is a permitted state. When i and k would exchange a black glueball then ( -j i k ) converts to ( -j -i -k ) which is an antibaryon. Application 4 at page 7 of storyline QQD, argues baryons usually might be in a quark-antiquark mix of net color white.
So, as far as the colors are concerned, this reaction converts a baryon into an antibaryon. While the subsequent reaction will turn the antibaryon back into a baryon. Note this alternation doesn't affect the electric charge of the baryon, nor its taste, spin or mass. It's only a color-thing and colors, unlike the four other properties, cannot be observed.
Multiplication order in -j * i * k is not defined so all orders superpose, always yielding 1 or -1.
The possibilities for a second set of separated vacuum particles ( k ) ( k ) or ( -k ) ( -k ) next to arrive at baryon ( -j i k ) are:
In each of the sets A up to F the first column ( i -j k ) is the baryon - there is no specific order of i, -j and k in the baryon. The second column is (k) and (k), or (-k) and (-k), the gluons that react with the baryon.
A and B are baryons again. In A: when the end states -j and -i would interchange a black glueball, they would convert to j and i, resulting in baryon ( i j k ). So ( -i -j k ) holds as a baryon state.
In A up to F there is absorbed 1 vacuum particle (Higgs mechanism), the vacuum is reduced by the volume of 1 vacuum particle (gravity).
(Black glueball exchange can also be mimicked by the absorption of a ( -1 -1 ) vacuum particle, that yields the same effect. Alas, according to (8) the ( -1 -1 ) is no part of our forward time evolving vacuum.)
In C the ( k ) ( k ) gluon pair is assumed to emerge at the quark of color i. The gluons then go to quark -j and k respectively. This results in a meson-like composition ( i -i ) and a quark of color -1. We know white gluons, but what is a black quark?
The absorption of the ( k ) ( k ) is the Higgs mechanism, so both the -i (formerly -j) and the -1 (formerly k) acquire mass.
Suppose the baryon is a proton and the i and -i are an u- and d-quark. Then our -1 is an u-quark of spin +1/2 (or -1/2). There are two possibilities: it escapes or it doesn't. When it doesn't, the color -1 quark maintains its spin 1/2, taste u and electric charge +2/3, no reason to assume otherwise. Because -1 doesn't glue, the -1 quark is no longer bound by color, only by electric charge. So I expect it to enlarge its distance and form a halo around the ( i -i ) meson, a zero-color quark-cloud around the remaining two quarks, like the electron does around the nucleus. Well, I guess it would work a lot better when the i and -i were both u and the d circles around them. So let's do it that way.
When distance of d to the two u quarks starts to increase, it becomes increasingly difficult to provide the black d quark with mass by the mechanism described. Two separate gluons (former vacuum particle) have to go to two quarks but when one of them is at large distance the gluon that has to reach for it, has to travel too long a distance. The black d quark has color value -1 and is not attracted by the two u quarks by color force. We concluded in (27) that attraction between quarks IS the Higgs mechanism. So the Higgs mechanism doesn't work between the black d quark and the two u quarks. So the black d quark is massless. It gains lightspeed and so will not be kept in orbit. It escapes after the short transition time.
A particle of -1/3 electric charge, as the d is, is never observed, only integer charges are. Massless electric charge has never been observed. So the charge must somehow have been redistributed. The easiest way is to assume it had transferred its charge to one of the two u's, -1/3 + 2/3 = +1/3, leading to the meson of charge +1/3 +2/3 = +1 electric charge. The escaping particle then has no electric charge.
Could the escaping -1 be a black glueball? The black glueball is massive according to (8). Our quarks have spin 1/2, the black glueball as well as the meson have integer spin. So it seems no, it cannot be the black glueball. Unless there is an extra particle conveying the missing +1/2 or -1/2 spin - the neutrino does such a thing.
Do I recognize the black quark particle? Spin 1/2, electric charge zero, massless, color black (that is -1, not zero, color never is zero). It resembles the neutrino. But no, the neutrino has mass.
Mind the mechanism of charge transference is yet unknown.
For the described Higgs mechanism to work, it needs TWO particles. Sole quarks cannot be given mass. So sole quarks are massless. (3.16)
There seems to be a rule that electrically charged particles have mass. So the sole quark from (3.16) has electric charge zero. (3.17)
Before we forget, let's state as a rule:
Fractional electric charges are never observed, only integer charges can exist observably. (3.18)
Is it for sure sole quarks cannot be given mass? In the neighborhood of a supposed sole quark we start again with a situation as in (3.10) and (3.9).
( i -i ) as graviton 1 and
( i -i ) as graviton 2.
But instead of adding we just swap one spin from graviton 1 with one spin from graviton 2. According to (3.20) in paragraph Color conservation at page 3 of QQD, the swap must be performable by some kind of gluon exchange. One (white glueball with spin up) emitted by the -i would take away the spin of the -i. Arriving at i or -i the spin would cancel to zero.
( i -i 0 )
( i 0 -i ) or ( i -i 0 )
The spin 0 particles are no gluons no more. So this is not a permitted transition. We need a simultaneous second emitted and absorbed by the same particles. This leads to
( i -i )
( i -i ) or ( i -i )
which are two Higgs particles ready to give mass to our sole quark.
When the -i had emitted a gluon i and a gluon -i simultaneously, this would have done too. We recognize ( i -i ) as a vacuum particle. Also the two simultaneous is recognized as vacuum particle ( 1 1 ). Normally the energy for these emissions is not available. The energy can be borrowed, but from what? The sole quark? Then the first thing our sole quark have to do in its sole existence, is to pay back the lent energy: its mass according to E = mc. Instead of lending two particles it always will be easier to get it for free. Since it are vacuum particles we can just take them from the vacuum, in doing so causing two empty spots. Then we recognize this as Higgs mechanism 1.
From observation we insert now that sole quarks are not observed. We conclude that somehow Higgs mechanism 1 does not work. Although it is not clear yet WHY not. (3.19)
Now also a mechanism for the mass of the -1 gluon appears. The color of each gluon consists of two Pauli matrices (see pop-up frame Quaternion units as Pauli matrices) which are two quarks (this is worked out in Four quarks in the shell of page 5 of NET FORCES IN QCD). But as you see, the -1 gluon consists of at least 4 and maybe more likely 6 Pauli matrices. That are 2 or 3 gluons respectively, a glu2on or glu3on. Between two or more colors together the Higgs mechanism can work and so it will. There is no fundamental -1 gluon. The -1 as particle in forward time vacuum alone, exists as massive composition only. (3.20)
(3.20) would also mean gluNons are massive for all N larger than 1. As a general conclusion we draw:
sole color cannot be given mass (3.21)
Somehow for sole gluons Higgs mechanism 1 doesn't work either, like it doesn't for quarks, see (3.16).
So GluNons should be massive for all N larger than 1. What when N = 2 and the colors are opposite as they are in the vacuum particles? The three vacuum particles ( i -i ), ( j -j ) and ( k -k ) are regarded as a particle and an antiparticle. Then one color goes forward in time while the other goes backward, one color has forward vacuum around it, while the other has a tiny parcel of backward time evolving vacuum around it, with a tiny time border between the vacuums (tiny at the Earth surface). One particle emits towards the field, while the other absorbs from the field. The total absorption from the Higgs field is zero. When the colors of one vacuum particle had approached each other within the volume of the time border around the anticolor, then the vacuum particles ( i -i ), ( j -j ) and ( k -k ) are said to coincide massless. (3.22)
Take in mind a vacuum particle ( i -i ). For convenience, look at the vacuum particle as if it is a meson with quarks of colors i and -i, but then gluons of color i and -i instead of the quarks. The i is at B and the -i is at D, see spacetime diagram below. When Higgs mechanism 2 is at work as described in table (3.12), and a pair of gluons like ( k ) ( k ) or ( -k ) ( -k ) appears at X with the intention to give the i and the -i mass, then i absorbs one ( k ) while -i emits the other ( k ) as it is observed by us, forward people. So, as long as i and -i remain within their time borders, the composite of the two remains massless.
Emitting towards the Higgs field is gaining mass too - backward time evolving mass.
Between the two gluons i and -i in our vacuum particle it looks to us as if just one gluon goes DXB from the -i to the i (instead of ( k ) going to i and the other ( k ) going to -i). But it isn't, not completely. In fact both ( k ) are created at the time border and both go from X at the time border to their goal, as observed in their respective local frames. (3.23)
SbXbU is the time border, hidden behind the letters S, X and U there is another b.
Mind the opposite time arrows at both sides of the time border. Chosen is to represent the particles as they are locally observed. The ( k ) at their side is by them observed as ( -k ). The -i at D as we observe it, would have been denoted as an i at D.
Regard the storyline FORWARD BACKWARD TIME DIRECTION, especially page 6 and 7 - but it's difficult to grasp the idea when starting at page 6 or 7. One better starts at page 3.
The vacuum particles ( i -i ), ( j -j ) and ( k -k ) don't have a time arrow, unlike their constituent gluons. In contrast to of course ( 1 1 ) and ( -1 -1 ) that ARE the time arrow. (3.24)
The 1's in ( 1 1 ) and the -1's in ( -1 -1 ) have same time direction. So Higgs mechanism 2 works between them. It is nice that the mass of vacuum particle ( -1 -1 ) is explained now but now it is a problem how ( 1 1 ) stays massless. (3.25)
Let's invoke spin 0 gluons, as they must exist according to paragraph Building vacuum directly from spin 0 gluons at page 5 of this storyline. We can imagine ( 1 1 ) gluon pairs to be forbidden, because being massive. Then remains sole color 1 spin 0 gluons to make the vacuum from. But this trick then can be applied to the sole color -1 spin 0 gluon too, and that's not what we want.
For a pair of gluons to remain massless, the signs in front of the gluons have to be opposite. One gluon has to be a particle (i, j, k or 1) while the other gluon is an antiparticle (-i, -j, -k or -1). The colors don't have to be equal, only the sign in front has to be opposite.
Each of the figures in page 3 of NET FORCE IN QCD describe a pair of gluons merging to one gluon. Oké, if so then the remaining gluon is massless. But two gluons merge less easy than three do. If the gluons first form a pair, a glu2on, then the pair will be massive in fig. (3.2), (3.5), (3.9), (3.10), (3.11) and (3.12). Only the fig.'s (3.3) and (3.4) will yield a massless glu2on.
I really wonder whether the W+, W- and Z0 are in fact two massive gluons orbiting each other while the orbital impulse momentum is 1. And a neutrino, is that a sole quark? Given a tiny mass because of Higgs mechanism 1?
This provides a way to understand the existence of gluon-gluon reactions a little better. To me it was always a little uneasy to imagine how two lightspeed particles can react. Neither of the particles has time elapsing, so there seemed to be no frame where the reaction can take its space and time. So far as I can see, we only had the “quantum mechanical way of thinking”. At a certain moment the state of two lightspeed gluons (call them gluon 1 and gluon 2) superposes with the state of a gluon number 3 (the merger). As it comes to observation then in some cases the two lightspeed gluon state will be chosen again, in the other cases the merger state will be the one. That's without specifying a mechanism how the two-gluon state passes to the merger state. Maybe the following provides such a mechanism.
Gluon 1 has no mass, neither has gluon 2. But when gluon 1 and gluon 2 are at about 0.9 fm mutual distance, then color force is at maximum (see The proton at page 5 of NET FORCES IN QCD). For one single moment the gluons form a pair. There are two colors together and Higgs mechanism 2 sees its chance to work. The gluons didn't see each other, but the vacuum does. When the reacting gluons have color of opposite sign then the pair still remains massless. But if not opposing then for one moment both gluons of the pair get mass. For that single moment both gluons have no lightspeed no longer. They have time to react and so they do. The resulting merger gluon number 3 is massless again and immediately gains lightspeed.
Mind the gluon is pictured as a ring in paragraph “The proton”, page 5, NET FORCES IN QCD. Then there is one orientation that has the largest chance for Higgs mechanism 2 to work: when the rings are parallel, centered around the same axis and at a mutual distance of about 0.9 fm.
If the two gluons 1 and 2 don't see each other, why a gluon from the vacuum and gluon 1 then will? The vacuum consists of the superposition of gluon fields of all possible states, amplitudes, phases and velocities in all possible directions. Among them is the state that is identical to gluon 1. Those then would see gluon 1 and merge with it. I hope.
This allows us to state the rule:
Two gluons of opposite sign do not react (3.26)
They will pass each other by or stay put to each other and form a composition.
Paragraph Meson exchange from page 5 in NET FORCES IN QCD says: “Let us assume the virtual color-anticolor pairs in the shield of a quark can be gluon pairs as well as quark-antiquark pairs.” If so then the quark and its antiquark of one pair are each others opposite, as well as two gluons can be each others opposite. So to fit in with everything else, we are forced to assume that gluons, despite their composition of two quarks of opposing time direction, still have a time direction. The photon has no time direction, but the gluon has. (3.27)
Then the conjecture proposed in paragraph Quark mass at page 5 of NET FORCES IN QCD has to be revised, I guess. The gluon is massless and has a time direction. But since it travels at lightspeed, time is standing still on the gluon. It will never actually do that one step forward in time. Will this do?
According to the reasoning in The time border at page 7 of FORWARD BACKWARD TIME DIRECTION, -i, -j, -k and last but not least -1 exist as respectively i, j, k and 1 in vacuum of backward time direction and are supplied by a time border. But there is no pair of Pauli matrix multiplication yielding -1. The entire -1 gluon goes backward in time and is enveloped by a time border, so the two Pauli matrices it consists of react dark. So 2 Pauli matrices with backward time multiplication rules, that does work, see column at the right. But then the Pauli matrices of -i, -j and -k (all antigluons in baryons and mesons) should react dark.
Changing the order of the two Pauli matrices that compose a gluon, is changing between gluon i and antigluon -i, or j and -j, or k and -k, see pop-up frame Quaternion units as Pauli matrices. So as a rule we can state that
if you have two Pauli matrices in multiplication, then changing multiplication order is the same as changing multiplication rules (from bright to dark, or from dark to bright). (3.28)
I have no physical interpretation for multiplication order of only two factors, nor for quaternion units, nor for Pauli matrices. I do have an interpretation for changing multiplication rules. So it seems clear the last prevails. The mentioned state changes then don't superpose since changing multiplication order is not recognized as a state change (lacking interpretation multiplication order).
(Moreover, a disadvantage of multiplication order as time order of interaction, see (3.40) in paragraph Physical interpretation of multiplication order at page 3 of QQD, is that multiplication order that yield 1 in one frame might become a different multiplication order that yields -1 in another frame. That is, for lower-than-lightspeed particles.)
So it might be that changing multiplication order of the Pauli matrices constituting one color is forbidden because forward time vacuum (bright multiplication rules) cannot be changed in backward time vacuum (dark multiplication rules) just like that. An i is not allowed to change into a -i just by changing multiplication order of its constituting Pauli matrices.
Suppose that the Pauli matrices not only determine color but also determine spin. Then it will occur that when one multiplication order is forbidden, because yielding e.g. -i, that then also one spin is forbidden, e.g. spin down. This is the hallmark of the weak force.