7 
QQD 
Quaternion applications  Part 2Application 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 everythingoverwhelming 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 b2below 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 b3like 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 pluscolor and 2 minuscolors) 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 b1likes one soon ends up with a population of b3likes, with only a few b1likes. The ratio of pluscolor abundance to minuscolor 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 uquarks 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 socalled “unused gluons” would be used there. and E. g. and would swap i and j.Still application 4: Mesons
O NVESTIGATE NEXT PAGE Up CONTACT 
QUATERNION
