The Big Bang of our matter as an identical fermion explosion
This page is a sequel of paragraph Remnant particles in the fermion explosions at page 4 of THE EXPANSION OF THE UNIVERSE.
Suppose the extra fermion explosion that is meant to yield matter, consists of red u-quarks of spin +1/2. (2.1)
Take in mind three neigboring u-quarks, e.g. in a tetrahedron.
| +1/2 | i | +2/3 | u | |
| +1/2 | i | +2/3 | u | |
| +1/2 | i | +2/3 | u | (2.2) |
Here i is the quaternion unit that is set equal to the color red. Red = i, green = j, blue = k, etc.
10^-23 sec ATB (After the Bang) one of the three u-quarks can be expected to emit a spin +1 gluon, shifting the quark from red to e.g. green. This disrupts the identical fermion explosion. (2.3)
[ spin, color, electric charge, type of particle ]
+1/2 i +2/3 u --------- --------- --------- --------- +1 -k 0 gl -1/2 j +2/3 u
-1/2 j +2/3 u +1/2 i +2/3 u +1/2 i +2/3 u (2.4)
gl = gluon, u = taste of quark, only quarks have taste. The table at the right lists our 3 quarks after the reaction.
color quark * color gluon = i * ( i/-j ) = i * ( j * i ) = i * ( -k ) = j (2.5)
In (2.5) an expression between parentheses is a gluon and when outside parentheses it is the quark, see page 1 and page 2 of the storyline QQD and see the popup frame about Quaternions.
The gluon is subsequently absorbed by one of the two remaining u-quarks.
+1/2 i +2/3 u +1 -k 0 gl --------- --------- --------- --------- +3/2 j +2/3 u
-1/2 j +2/3 u +3/2 j +2/3 u +1/2 i +2/3 u (2.6)
This is according to the same equation as (2.5), but now describing an absorption.
2 * 10^-23 sec ATB the spin +3/2 green u-quark can convert to a spin +1/2 blue u-quark by emitting a spin +1 gluon of color -i.
+3/2 j +2/3 u --------- --------- --------- --------- +1 -i 0 gl +1/2 k +2/3 u
-1/2 j +2/3 u +1/2 k +2/3 u +1/2 i +2/3 u (2.7)
j * ( j/-k ) = j * ( k * j ) = j * ( -i ) = k (2.8)
Since the colors of the three quarks are i, j and k, the conglomerate forms a bound state, a composite, and releases the binding energy, the condensation energy. (2.9)
From now on the BB is driven by the condensation energy. (2.10)
About 10^-10 sec ATB a W+ boson is emitted by one of the two spin +1/2 u-quarks, e.g. [ +1/2, k, +2/3, u], changing it in a spin -1/2 d-quark.
+1/2 k +2/3 u --------- --------- --------- --------- +1 0 +1 W -1/2 k -1/3 d
-1/2 j +2/3 u -1/2 k -1/3 d +1/2 i +2/3 u (2.11)
The list of (2.9) now is a proton p+ (we assume a first-generation ground-state particle to be formed).
The W+ decays in positron e+ and electron neutrino 
e.
+1 0 +1 W --------- --------- --------- --------- +1/2 +1 +1 e +1/2 +1 0 e
-1/2 j +2/3 u -1/2 k -1/3 d +1/2 i +2/3 u +1/2 +1 +1 e +1/2 +1 0 e(2.12)
For the color of W+, e+ and e see (4.35) and (4.42) at page 4 of QG.
As (4.11) in paragraph W+, W-, Z0 and neutrinos at the same page says only spin +1/2 neutrinos and spin -1/2 antineutrinos exist
. So the spin +1/2 e from (2.12) can exist.
Where does [ +1, -i, 0, gl ] from (2.7) go? Suppose it meets a fourth quark [ +1/2, i, +2/3, u ] from the extra fermion explosion. There can be 4 quarks in a tetrahedron, all at same distance to each other.
+1/2 i +2/3 u +1 -i 0 gl --------- --------- --------- --------- +3/2 +1 +2/3 (?)
-1/2 j +2/3 u -1/2 k -1/3 d +1/2 i +2/3 u +3/2 +1 +2/3 (?) +1/2 +1 +1 e +1/2 +1 0 e(2.13)
The 4th quark would end up as a quark of color +1, a colorless quark, and in this website that is a positron, provided the reaction is allowed. The colorlessness makes the particle to be able to stay alone. Since the particle X has electric charge, the spin +3/2 can become +1/2 by emitting one photon, but the electric charge +2/3 doesn't match with the positron. So what to do with this?
In Minus sign go wild is suggested that all this kind of things have to be taken seriously. However, Minus sign go wild is considered to be a previous attempt now. So maybe it is the best to say that [ +3/2, +1, +2/3, e ] is not allowed to be created. Our gluon [ +1, -i, 0, gl ] is not allowed to do this in (2.11). (2.14)
Maybe 3 * 10^-23 sec ATB the gluon [ +1, -i, 0, gl ] meets a likewise created particle.
| +1 | -i | 0 | gl | |
| +1 | -i | 0 | gl | |
| --------- | --------- | --------- | --------- | |
| +2 | -1 | 0 | X | (2.15) |
X is not a gluon because of the spin +2.
Summarizing, if we start with 6 particles [ +1/2, i, +2/3, u ] from the extra fermion explosion, we end up with 2 protons, 2 positrons, 2 electron neutrinos and 1 particle [ +2, -1, 0, gl ]. (2.16)
It is a remarkable resemblance with the proton-neutron-electron world we are used to, but not close enough. So I think it's better to start over again.
What about the electric charge?
In a fermion explosion of solely quarks [ +1/2, i, +2/3, u ] the electric charge, +2/3 per u-quark, mounts up to a very huge quantity. Subsequently the law of conservation of electric charge ensures that this hugh electric charge continues to exist. (2.18)
Observation, as far as I know, show a universe that is more or less electrically neutral. Although I don't know in how far this has been established for interstellar and intergalactic clouds of ions. *a) In matter like in planets the sum of positive electric charge of all nuclei is more or less precisely canceled by the negative electric charge of electrons. (2.19)
So I presume only an extra fermion explosion of [ +1/2, i, +2/3, u ] quarks alone will never do to yield the matter universe we observe around us.
Oké, let's start over again with an extra fermion explosion of red d-quarks of spin -1/2, denoted as [ -1/2, i, -1/3, d ]. Take in mind 3 neigboring d-quarks, e.g. in a triangle, in doing so making sure they have equal relative distance.
[ spin, color, electric charge, type of particle ]
| -1/2 | i | -1/3 | d | |
| -1/2 | i | -1/3 | d | |
| -1/2 | i | -1/3 | d | (2.20) |
After 10^-23 sec ATB (After The Bang) one of the three d-quarks emits a spin -1 gluon, shifting the quark from red to e.g. green. This disrupts the identical-fermion explosion.
-1/2 i -1/3 d --------- --------- --------- --------- -1 -k 0 gl +1/2 j -1/3 d
+1/2 j -1/3 d -1/2 i -1/3 d -1/2 i -1/3 d -1 k 0 gl (2.21)
Again the table at the right shows the state of all particles under consideration after the reaction.
The gluon is subsequently absorbed by one of the two remaining d-quarks.
-1/2 i -1/3 d -1 -k 0 gl --------- --------- --------- --------- -3/2 j -1/3 d
+1/2 j -1/3 d -3/2 j -1/3 d -1/2 i -1/3 d (2.22)
color quark * color gluon = i * ( i/-j ) = i * ( j * i ) = i * ( -k ) = j (2.23)
In (2.23) an expression between parentheses is a gluon and when outside parentheses it is the quark, see page 1 and page 2 of the storyline QQD and see the popup frame about Quaternions.
The spin -3/2 green quark can convert to a blue d-quark by emitting a spin -1 gluon of color -i.
-3/2 j -1/3 d --------- --------- --------- --------- -1 -i 0 gl -1/2 k -1/3 d
+1/2 j -1/3 d -1/2 k -1/3 d -1/2 i -1/3 d -1 -i 0 gl (2.24)
j * ( j/-k ) = j * ( k * j ) = j * ( -i ) = k (2.25)
Our three quarks have color i, j and k now. They can form a composite, a spin -1/2 proton-like particle with one negative electric charge. The binding energy, the condensation energy, is released. From now on, when a gluon from the outside wants to enter the composite and change its zero color end state into a colored end state, it first has to pay the composite a lot of energy to unlock
its bound state. For the moment the condensation energy is released only as kinetic energy of the quarks in the nucleon-like particle and of the particle as a whole. It will be only after 10^-20 sec ATB that the first photons are emitted and the kinetic energy can take the shape of gamma rays, see (2.28). (2.26)
The fermion explosion is disrupted now. From now on the BB is driven by the condensation heat. (2.27)
Typical force reaction times:
(2.28)
W = Weak force; P = Photon, the electromagnetic force; G = Gluon, the strong force.
About 10^-10 sec ATB a W- boson is emitted by one of the two spin -1/2 d-quarks in (2.24), e.g. [ -1/2, k, -1/3, d ], changing it in a spin +1/2 u-quark.
-1/2 k -1/3 d --------- --------- --------- --------- -1 0 -1 W +1/2 k +2/3 u
+1/2 j -1/3 d +1/2 k +2/3 u -1/2 i -1/3 d -1 0 -1 W (2.29)
The list at the right of (2.29) has become a spin +1/2 neutron n0 (we assume a ground-state first-generation particle to be formed). By this time the gluon [ -1, -i, 0, gl ] from (2.24) must have disappeared long
ago, therefore it is not on the list in (2.29). Keep it in mind, we refer to it later.
The neutron is stable when in a nucleus, so it must represent a relative low energy state. Therefore we assume the W-reaction in (2.29) not to cost energy but instead to yield some extra condensation energy
.
The W- decays in electron e- and anti-electron neutrino 
e.
-1 0 -1 W --------- --------- --------- --------- -1/2 -1 -1 e -1/2 -1 0 e
+1/2 j -1/3 d +1/2 k +2/3 u -1/2 i -1/3 d -1/2 -1 -1 e -1/2 -1 0 e(2.30)
For the color of W-, e- and e see (4.35) and (4.42) at page 4 of QG.
(4.11) in paragraph W+, W-, Z0 and neutrinos at the same page says only spin +1/2 neutrinos and spin -1/2 antineutrinos exist
. So the spin -1/2 e from (2.30) can exist. Mark that if we would have started with an identical fermion explosion of [ +1/2, i, -1/3, d ] that spin +1/2 e would have appeared in (2.30) and those neutrinos don't exist.
Where does the gluon [ -1, -i, 0, gl ] from (2.24) go? It cannot go to one of the three quarks that now are bound in the n0, since then it must supply the binding energy before further reaction can take place. Suppose it meets a fourth d-quark [ -1/2, i, -1/3, d ] from the extra fermion explosion. There can be 4 quarks in a tetrahedron, all at same distance to each other.
| -1/2 | i | -1/3 | d | |
| -1 | -i | 0 | gl | |
| --------- | --------- | --------- | --------- | |
| -3/2 | +1 | -1/3 | e | (?) (2.31) |
The 4th quark would end up as a quark of color +1, a colorless quark, and thus free to go where it likes. In this website the colorless +1 quark is a positron, provided the reaction is allowed. The spin -3/2 can become -1/2 by emitting one photon, but the electric charge -1/3 doesn't match with the positron. So what to do with this?
Minus sign go wild as being a previous attempt, suggests that [ -3/2, +1, -1/3, e ] is not allowed to be created and our gluon [ -1, -i, 0, gl ] is not allowed to do this here. (2.32)
So I think that 3 * 10^-23 sec ATB the gluon [ -1, -i, 0, gl ] meets a likewise created particle. For the moment we have two neutrons under consideration.
| -1 | -i | 0 | gl | |
| -1 | -i | 0 | gl | |
| --------- | --------- | --------- | --------- | |
| -2 | -1 | 0 | X | (2.33) |
The resulting particle X can stand alone, as far as the colors are concerned. It is no gluon, gluons have spin 1. It cannot get spin -1 by emitting a photon because the black glueball has no electric charge. However, it can split into two glueballs.
| -2 | -1 | 0 | X | |
| --------- | --------- | --------- | --------- | |
| -1 | -1 | 0 | gl | |
| -1 | +1 | 0 | gl | (2.34) |
The white glueball [ -1, +1, 0, gl ] can be absorbed by any quark any time, as far as the colors are concerned. Only the -1 spin might be an objection. So let's assume the white glueball is absorbed by one of the spin +1/2 quarks in one of the two neutrons. We neglect the second neutron from now on. Our list of particles now has become:
| +1/2 | j | -1/3 | d | |
| +1/2 | k | +2/3 | u | |
| -1/2 | i | -1/3 | d | |
| -1/2 | -1 | -1 | e | |
| -1/2 | -1 | 0 | e | |
| -1 | -1 | 0 | gl | (2.35) |
In the black glueball [ -1, -1, 0, gl ] we recognize the Dark Matter candidate from what I called the free mesonic gluon in paragraph Other colorshifts and mesons
at page 3 of the storyline QCD; as well as in the column at the right at page 2 of THE EXPANSION OF THE UNIVERSE.
In a little less than 1000 sec ATB half of the number of neutrons will have decayed by a W- into protons p, electrons e and anti-electron neutrinos e, by the spin -1/2 d-quark becoming an u-quark: (2.36)
| -1/2 | i | -1/3 | d | |
| --------- | --------- | --------- | --------- | |
| -1 | 0 | -1 | W | |
| +1/2 | i | +2/3 | u | (2.37) |
+1/2 j -1/3 d +1/2 k +2/3 u -1/2 i -1/3 d --------- --------- --------- --------- +1/2 j -1/3 d +1/2 k +2/3 u +1/2 i +2/3 u -1/2 -1 -1 e -1/2 -1 0 e
+1/2 j -1/3 d +1/2 k +2/3 u +1/2 i +2/3 u -1/2 -1 -1 e -1/2 -1 -1 e -1/2 -1 0 e -1/2 -1 0 e -1 -1 0 gl (2.38)
In the right part of the table (2.38) we ended up with a spin +3/2 proton. If the quark [ +1/2, j, -1/3, d ] would have emitted the W-, seemingly the most natural choice, then the spin of the W- would have been +1 and its decay would have yield an e- of spin +1/2 and a e of spin +1/2, and the latter doesn't exist. Later the spin +3/2 proton emits a spin +1 gamma photon and it spin becomes +1/2. In (2.39) we chose the [ +1/2, i, +2/3, u ] to emit the photon.
| -1/2 | i | -1/3 | d | |
| -1/2 | i | -1/3 | d | |
| -1/2 | i | -1/3 | d | |
| --------- | --------- | --------- | --------- | |
| +1/2 | j | -1/3 | d | |
| +1/2 | k | +2/3 | u | |
| -1/2 | i | +2/3 | u | |
| -1/2 | -1 | -1 | e | |
| -1/2 | -1 | -1 | e | |
| -1/2 | -1 | 0 | e | |
| -1/2 | -1 | 0 | e | |
| -1 | -1 | 0 | gl | |
| +1 | 0 | 0 | ph | (2.39) |
(2.39) shows that we started with 3 particles [ -1/2, i, -1/3, d ] from the extra fermion explosion and ended up with 1 proton, 2 electrons, 2 anti-electron neutrinos, 1 black glueball and a photon. This is close enough to the universe we see around us, including the Dark Matter (mind the capitals). However, because of the two electrons instead of just one, the electric charge piles up to a huge negative value that subsequently is conserved, an excess of 1 electric charge per created proton. Should we start all over again?
The Standard Model
(2.40)
The quarks u and d are the only first-generation particles with baryon number not equal to zero. So, when we would have started with an extra fermion explosion of one of the other particles, then still quarks could have been formed but only in quark-antiquark pairs. On the spot each quark would be accompanied by its antiquark and that's not what we want. We want an aera with matter only. (Antimatter in other areas is supposed to origin from its own extra fermion explosion, without any matter.)
So d-quarks it will be, in the extra fermion explosion that is yielding our matter universe as we see it around us. D-quarks here and d-quarks elsewhere, in the antimatter regions (underlining means anti-).
There is no place for the mentioned excess of electrons in planets or stars. Can they roam between the stars? I guess no, streams of such electrons would cause huge magnetic fields that are not observed between the stars and between the galaxies, aren't they?
So I need a second extra fermion explosion of positrons, in such a way that it yields one positron per three d-quarks. These positrons then can annihilate with electrons to gamma rays, which add still more to the condensation energy that drives the BB. As is argued in *e) in the column at the right of page 4 of THE EXPANSION OF THE UNIVERSE, two different fermion explosions right through each other should be allowed as long as the temperature is zero K, as indeed is the case before the disruption of the fermion character of the explosion. But I don't see how to finetune these two independent explosions in such a way. This suggests it will rather be one extra fermion explosion of one particle that decays in three d-quarks and one positron.
| -1/2 | i | -1/3 | d | |
| -1/2 | i | -1/3 | d | |
| -1/2 | i | -1/3 | d | |
| +1/2 | +1 | +1 | e | |
| --------- | --------- | --------- | --------- | |
| -1 | -i | 0 | gl | (2.41) |
This then would be the particle that, with innummerable copies of it on the same spot, would yield our matter universe (however not the vacuum, the vacuum particles).
How does an area of antiquarks in our BB looks like when observed from their own backward time evolving area of space? In their frame they see an implosion towards the BB. Their hadrons exist in an utterly unordered state, caused by a universe's lifetime of reactions. Then, as they observe it, the hadrons spontaneously decouple and form an indentical quark cloud (identical antiquark cloud as we should observe it) that contracts to what they would call a BC, a Big Crunch (fermion implosion). Provided the BC that ends the antimatter area in their frame is a kind of backward evolving copy of the BB in our matter region. I guess it is, but we can't know until we observed it, see Observing Dark Galaxies in the column at the right at page 2 of THE EXPANSION OF THE UNIVERSE.
Since they come from the future it seems reasonable to assume they originate in a second BB somewhere in our far future.
Then in the BB as we observe it, the quarks in an area here are treated a tiny little different than the antiquarks in another area there. Quarks might form a fermion explosion until 10^-23 sec had elapsed or mutual distance had reached 0.9 fm. Antiquarks are just an unorderly bunch that are backward time evolving reaching our BB. (5.6)
Maybe drawing is involved. Discussion about drawing is not yet settled, see e.g. paragraph The time border and the gluon of color 1 at page 3 of QG. (5.7)
