Application 4: Baryons
IF kji = 1 is color neutral THEN -k * -j * i = 1 is color neutral too (i, j, k, -i, -j, -k are quark colors, 1 and -1 are gluons). It would violate the conservation of color, but such a law is not yet known. The only law used is the everything-overwhelming urge for color neutrality. And that a color cannot just like that on its own change in another color; to do so is needed a color particle like is used here -1.
The transition from b1 (baryon 1) to b3 is caused by -1 going from i to j (b2) OR by -1 going from j to i (b2 below). The emitting color is divided by -1 and the absorbing color is multiplied by -1. The two possibilities b2 and b2-below cannot be distinguished and superpose. Compare the usual color swap by exchange one gluon out of i, j, k, -i, -j. -k, where the gluon goes from a color to a second color or the other way around.
The transition from b3 to b5 is caused by -1 going from -j to k (b4 up) OR by -1 going from k to -j (b4 below). Also these two possibilities superpose.
But wait, in b4 below, couldn't the -1 escape? -1 itself is colorless and it leaves behind a white antibaryon. Normally, in ground state: no, energy not available. But when energy is sufficient to provide the free -1's mass, then yes. IF this can be, then a baryon disappears and an antibaryon appears. The conservation of baryon number is violated by 2.
We consider now the single -1 gluon emission only. Starting from b1, an ordinary baryon, there are 6 different ways to arrive at a b3-like state (i > j meaning
-1 gluon going from i to j)
i > j, j > i, j > k, k > j, k > i, i > k
and 3 ways to remain in the b1 state:
i > i, j > j, k > k.
All -1 emissions have same chance, and so have all -1 absorptions.
( When an i emits a -1, the -1 is initially nearest to that i, eventually giving it the first chance to absorb. So there might be a little enhanced chance for i > i, j > j and so on. We take this effect negligible. )
At b3 there are 7 ways to stay in a similar state (with 1 plus-color and 2 minus-colors)
-i > k, k > -i, -j > k, k > -j, -i > -i, -j > -j, k > k
and 2 ways to return to the original baryon state:
-i > -j, -j > -i.
The total number of steps is 6 + 3 +7 + 2 = 18, yielding a net number of steps of 6 - 2 = 4 towards the
one + and two - state. So when starting with a population of b1-likes one soon ends up with a population of b3-likes, with only a few b1-likes. The ratio of plus-color abundance to minus-color abundance then is 1 : 2 or 1/3 : 2/3. Is there a connection with the -1/3 and +2/3 electric charge of the d- and u-quarks in a baryon?
According to the representation of the vacuum in QUATERNION GRAVITATION page 2, even when the reactions shown above would not take place, then a baryon state b1 will absorb one (-1 -1) vacuum marble to arrive at state b3. Subsequently it will remain in that state as argued.
Mark that IF we would still work with gluons with an upper and a lower color, instead of a quaternion unit replacing the gluon, that THEN in -k * -j * i = 1 some of the 12 so-called
unused gluons would be used there.
E. g. and would swap i and -j.
Still application 4: Mesons
|i i||i -1 -i||-i -i||i -1 -i||i i|
|i -i||i -1 i||-i i||i -1 i||i -i|