|NET FORCES IN QCD|
When we stick to 3 gluons emitted simultaneously we can allow 1 quark to emit 2 gluons or all 3 gluons. Simultaneously, or one after the other. At least 1 quark doesn’t emit a gluon then.
We can allow 1 gluon to be emitted that subsequently emits a second gluon. Or that first gluon emits the 2 other gluons. Simultaneously, or one after the other. Or a gluon emits a gluon that emits a gluon.
In groundstate there is hardly energy for glueballs to come to existence. If they appear in analysis they have to be reabsorbed before the first measurement. Since all 3 quarks can always do so in the antibaryon, there are 3 different paths of nearly equal length for a glueball to be absorbed, each ending on a different quark. A glueball can also be absorbed by the 1 or 2 remaining gluons, if there are any.
In the antibaryon the number of possibilities that contain glueballs will be multiplied by 3 for each glueball appearing in analysis, plus the number of gluons remaining at that moment.
Two glueball’s can absorb each other, reducing the number of glueballs present by 1.
All these possibilities have to be added to the mentioned possibilities on page 3 of this storyline (“Three gluons emitted simultaneously”).
The last possibility considered on that page, triangle nr 11, contains 3 glueballs and its number of possibilities increases from 1 to 3 * 3 * 3 = 27 (plus more). Not thát boring after all.
Provided the antibaryon is left behind white, the glueball can leave the antibaryon, by example an antiproton. It can reach the quarks of a neighboring antiproton or antineutron in the nucleus and become absorbed by one of them. Due to distance reduction this contribution is smaller than the absorption in the private antibaryon. The antiprotons and antineutrons in other nuclei are at least 10 exp 5 times farther away and these possibilities contribute lesser by the same amount.
Another group of possibilities that count separately are all possible ways for 2 gluons (instead of 3) to be emitted simultaneously by the 3 quarks of an antibaryon. Regard the mentioned 186 possibilities with at least one zero-shift gluon (glue1ball) emitted. The 3 possible glue1balls , and are identified, so leaving out a glue1ball diminish the number of possibilities by 1 (and not by 3). So you can indeed replace 1 glueball by 1 nothing, a no-emission. The triangles 5 up to 11 then count, resulting in a number of extra possibilities of
12 + 15 + 6 + 12 + 6 + 6 + 1 = 58.
All possible ways for maintaining white state after 1 gluon emission, are found by taking the possibilities with at least 2 glue1balls and replacing both balls by no emission. These are the triangles 9, 10 and 11, giving extra possibilities 6 + 6 + 1 = 13.
There is 1 way to emit no gluon at all, the 3 quarks going from nearby points A, B and C without any further emissions to nearby points D, E and F.
The total amount of possibilities of the 3 quarks of an antibaryon going from nearby points A, B and C to nearby points D, E and F and emit and absorb in between 0, 1, 2 or 3 gluons, is 186 + 58 + 13 + 1 = 258. Except for the extra possibilities of glueballs to be absorbed. Or the shifting between gluNon-states.
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