4 
NET FORCES IN QCD 
Three gluons emitted simultaneouslyThis page is in the COLORSHIFT APPROACH and is UT OF SE (it's quaternions now) The schemes when all the three quarks of a baryon simultaneously emit one gluon yet have to be investigated. Let’s give it a try. We forget about the colors of the gluons and denote their colorshifts only. The various possibilities are discussed in paragraphs. The first line of a paragraph is given in bold and starts with the 3 quarks of the antibaryon denoted as “+” or “” or “0” in a triangle, followed by the 3 gluons setting off from these 3 quarks. Sole gluons and gluoncomposites are denoted between brackets ( ). When the quark is denoted by +, its color shifts +2/6 along the arc of the colorcircle, because it has emitted a gluon of colorshift 2/6. Likewise when the quark is denoted by , its color shifts 2/6 along the colorcircle because it had emitted a gluon of colorshift +2/6. Denotation by 0 means the emission of a glueball, a white gluon. No color is shifted. Below the bold line are the various possible quarkgluon reactions necessary to restore white in the antibaryon. Before starting with the first paragraph, we have to say something about gluon emission of shifts +2/6 and 2/6. The 8 possibilities of three quarks simultaneously emitting a +2/6 or 2/6 gluon (no zeroshift gluons yet) are: The three triangles of + and  in the second line change into each other by rotating the colorcircle behind them over +2/6 and 2/6 turn. They are different but their treatment in analysis is the same. That holds too for the three triangles in the third line. So from the triangles shown here, we treat only four triangles, in four paragraphs. If one really wants to know the actual colors of the quarks, overlay the colorcircle with a triangle and apply the denoted shift. plus equals means the same as “+2/6 quark”, the magenta quark’s color is shifted over +2/6 to yellow. Then the color is left out, leaving the denotation “+2/6 quark” only  one has to imagine the color of the quark in front of the +2/6 there. From the colors of the quarks in the start state and the applied colorshifts one can construct for oneself the colors of the appropriate gluons  or the ones that are identified with! For the actual applied gluon one has to reconstruct it afterward from the color of the destination quark. In the example the applied gluon is , in socalled timesymmetric representation. (2/6) and (2/6) and (2/6) Once all 3 gluons have being emitted, the antibaryon is left behind white and need no further reactions. But the colored gluons yet have to remain somewhere. So unless the gluons find their place, this group of 3 emissions will not occur. 1.1.a +2/6 quark 2/6 gluon = 0 quark, +2/6 quark 2/6 gluon = 0 quark, +2/6 quark 2/6 gluon = 0 quark. If the three gluons each hit another quark white will be maintained, it doesn’t matter which gluon hits which quark. When we denote +2/6 quark, we mean: “the color of the quark in question, shifted over +2/6”, no matter which color that quark had. We left out the color of the quark here and use the colorshift of the gluon only. The outcome 0 quark means the original color of the quark is restored. (It is shifted over no angle.) Number of possibilities = 6 One can rotate the colorcircle behind the +++triangle over +2/6 or 2/6 turn, but this yields precisely the same quarkcolorstates after gluonemission. We identify gluons of the same shift. For the quarks, gluons of the same colorshift have the same colorstate. But their paths may be different and that counts! An emitted 2/6 gluon here can hit the quark it came from, or one of the two others. This yields 3 different paths. For each of those 3 possibilities a next gluon can hit one of the two remaining quarks, which gives 2 possibilities. Only the last gluon then has no choice anymore. So this reaction has 3 * 2 * 1 = 6 possible ways to happen. 1.1.b +2/6 quark 2/6 gluon 2/6 gluon 2/6 gluon = +2/6 quark, +2/6 quark = +2/6 quark, +2/6 quark = +2/6 quark. If the three gluons all hit the same quark white will be restored. Number of possibilities = 3 The colors of the quarks are different states. Each of the 3 quarkcolors can be the one absorbing all 3 gluons. 1.1.c +2/6 quark 2/6 gluon 2/6 gluon = +2/62/62/6 = 2/6 quark = quark, +2/6 quark 2/6 gluon = quark, +2/6 quark = quark. The denotation +2/6 quark means: “When we overlay the colorcircle with the +++triangle and look at the cyan quark, we see it must be shifted over +2/6” (due to gluonemission). This is a tricky one. These rules are applied here in anticlockwise direction along the colorcircle (cyan, magenta, yellow) which gives an interchange of and . When applied in clockwise direction (cyan, yellow, magenta) one gets: +2/6 quark 2/6 gluon 2/6 gluon = +2/62/62/6 = 2/6 quark = quark, +2/6 quark 2/6 gluon = quark, +2/6 quark = quark, which are 3 yellows and this is forbidden. So the first two gluons may hit the same of any antiquarkcolor, but if the third intents to hit one of the two other quarks, it have to be the one that obeys the order quarkplusgluonplusgluon, quarkplusgluon, quarkplusnothing applied in anticlockwise direction along the colorcircle. Number of possibilities = 9 Each of the 3 quarkcolors can be the one that absorbs only one gluon. This gives 3 possibilities. That gluon can come from each of the 3 quarks. That delivers 3 different gluonpaths and so 3 different possibilities. The rule above then dictates which quark absorbs the remaining two gluons. So there are 3 * 3 = 9 different possibilities for this quarkgluonreaction to occur here. (2/6 2/6) and (2/6) = +2/6 glu2on 2/6 gluon (2/6 2/6) is a twogluoncomposite. Let’s call it a glu2on. One obtain it’s analysis by replacing an occurring “2/6 gluon 2/6 gluon” by “+2/6 glu2on” in the previous paragraph. (A sole gluon could have been denoted as a glu1on, a composite of one gluon. Likewise a glu0on should mean “no gluons at all”, but we don’t denote anything then.) 1.2.a +2/6 quark +2/6 glu2on 2/6 gluon = +2/6 quark. Number of possibilities = 3 The glu2on and the gluon hit the same color, so 3 possibilities for this reaction to occur. 1.2.b +2/6 quark +2/6 glu2on = +2/6+2/6 = 2/6 quark = quark, +2/6 quark 2/6 gluon = quark, +2/6 quark = quark. If the glu2on hits an antiquarkcolor, the gluon has to hit the one that obeys the order quarkplusglu2on, quarkplusgluon, quarkplusnothing applied in anticlockwise direction along the colorcircle. Number of possibilities = 3 Once the glu2on or the gluon has been absorbed, the other particle is dictated where to be absorbed. So the three colors in the antibaryon, the three quarks, are the places where the reaction can start. (2/6 gluon 2/6 gluon 2/6 gluon) = 2/62/62/6 = 6/6 = 0 This threegluoncomposite, a glu3on from now on, is a glueball of three gluons (let’s call this a glue3ball). The glue3ball is white and, provided the antibaryon is left behind white too, it will leave the antibaryon. Glueballs are always white and appear only at excited baryonstates because of the energy involved in their creation. 1.3.a There are no further reactions necessary to restore white since the antibaryon is already left behind white. Number of possibilities = 1 (2/6 gluon) and (2/6 gluon) and (+2/6 gluon) 2.1.a 2/6 quark 2/6 gluon 2/6 gluon +2/6 gluon = +2/6 quark Number of possibilities = 3 There is only one quark having emitted a +2/6 gluon. When all 3 gluons hit that quark, the result is the colors of all 3 quarks being shifted over +2/6, which leaves the antibaryon white. Rotating the colorcircle over +2/6 and 2/6 behind the ++triangle gives the 3 possibilities for the color of the 2/6 quark. So there are 3 possibilities for this reaction to occur. 2.1.b +2/6 quark 2/6 gluon = 0 quark, +2/6 quark 2/6 gluon = 0 quark, 2/6 quark +2/6 gluon = 0 quark. Number of possibilities = 6 Rotating the colorcircle over +2/6 and 2/6 behind the ++triangle gives 3 possibilities for the color of the 2/6 quark. Then there are 2 pairs of paths of the 2 quarks that both emitted a 2/6 gluon to the two +2/6 quarks that subsequently absorb them. So there are 3 * 2 = 6 different possibilities for this reaction. 2.1.c +2/6 quark +2/6 gluon = 2/6 quark = quark, +2/6 quark 2/6 gluon = +2/62/6 quark = quark, 2/6 quark 2/6 gluon = +2/6 quark = quark. Applied in anticlockwise direction (c, m, y) this gives an interchange of and . Applied in clockwise direction (c, y, m) this would yield three identical colors. Number of possibilities = 6 Each of the 3 quarkcolors can be that only one that emits a +2/6 gluon. That 2/6 quark will absorb then a 2/6 gluon which can originate from either of the two +2/6 quarks. This doubles the number of possibilities, which now is 3 * 2 = 6. Once this has been chosen, it is dictated which quark will absorb the remaining 2/6 and +2/6 gluon. 2.1.d +2/6 quark 2/6 gluon 2/6 gluon = +2/6 2/62/6 = 2/6 quark, +2/6 quark +2/6 gluon = +2/6 +2/6 = 2/6 quark, 2/6 quark. In the end the color of each quark is shifted over 2/6. This overall shift leaves the antibaryon always white. Number of possibilities = 6 There are 3 possibilities for choosing the quark that absorbs the two 2/6 gluons. Then there are two possibilities for choosing the quark that absorbs the remaining +2/6 gluon. So there are 3 * 2 = 6 possibilities for this reaction to occur. (2/6) and (2/6+2/6) = 2/6 gluon + glue2ball 2.2.a 2/6 quark 2/6 gluon = 2/62/6 = +2/6 quark. Number of possibilities = 6 The 2/6 quark absorbs one of the two 2/6 gluons and the other 2/6 gluon combines with the +2/6 gluon to a glue2ball. The result is the colors of all 3 quarks being shifted over +2/6, which leaves the antibaryon white. Then the glue2ball leaves the antibaryon. Rotation of the colorcircle behind the ++triangle over +2/6 and 2/6 gives 3 different possibilities for this reaction. The 2/6 gluon that is absorbed by the 2/6 quark can come from either of the two +2/6 quarks, which doubles the number of possibilities once again. So there are 3 * 2 = 6 possibilities for this reaction to occur. (2/62/6) and (+2/6) = +2/6 glu2on +2/6 gluon 2.3.a 2/6 quark +2/6 glu2on +2/6 gluon = 2/6+2/6+2/6 = +2/6 quark Number of possibilities = 3 Rotating the colorcircle over +2/6 and 2/6 behind the ++triangle gives the 3 possibilities for the color of the 2/6 quark that absorbs the glu2on and the gluon. 2.3.b +2/6 quark +2/6 glu2on = +2/6 +2/6 = 2/6, +2/6 quark +2/6 gluon = +2/6 +2/6 = 2/6, 2/6 quark. In the end the color of each quark is shifted over 2/6. This overall shift leaves the antibaryon always white. Compare 2.1.d and substitute 2/6 gluon 2/6 gluon by +2/6 glu2on. Number of possibilities = 6 (2/6 2/6 +2/6) = 2/6 glu3on 2.4.a 2/6 quark 2/6 glu3on = 2/62/6 = +2/6 quark. The antibaryon is restored to white only if the glu3on hits the quark. Otherwise the reaction will not occur. Number of possibilities = 3 Rotating the colorcircle over +2/6 and 2/6 behind the ++triangle gives the 3 possibilities for the color of the 2/6 quark that absorbs the glu3on. The 3paragraphs are completely analoguous to the 2paragraphs, so most of the explanation is left out. (+2/6 gluon) and (+2/6 gluon) and (2/6 gluon) 3.1.a +2/6 quark +2/6 gluon +2/6 gluon 2/6 gluon = 2/6 quark Number of possibilities = 3 3.1.b 2/6 quark +2/6 gluon = 0 quark, 2/6 quark +2/6 gluon = 0 quark, +2/6 quark 2/6 gluon = 0 quark. Number of possibilities = 6 3.1.c 2/6 quark 2/6 gluon = +2/6 quark = quark, 2/6 quark +2/6 gluon = 2/6+2/6 quark = quark, +2/6 quark +2/6 gluon = 2/6 quark = quark. Applied in clockwise direction (c, y, m) this gives an interchange of cyan and yellow. Applied in anticlockwise direction (c, m, y) this would yield three identical colors. Number of possibilities = 6 3.1.d 2/6 quark +2/6 gluon +2/6 gluon = 2/6 +2/6+2/6 = +2/6 quark, 2/6 quark 2/6 gluon = 2/6 2/6 = +2/6 quark, +2/6 quark. In the end the color of each quark is shifted over +2/6. This overall shift leaves the antibaryon always white. Number of possibilities = 6 (+2/6) and (+2/62/6) = +2/6 gluon + glue2ball 3.2.a +2/6 quark +2/6 gluon = +2/6+2/6 = 2/6 quark. Then the glue2ball leaves the antibaryon. Number of possibilities = 6 (+2/6+2/6) and (2/6) = 2/6 glu2on 2/6 gluon 3.3.a +2/6 quark 2/6 glu2on 2/6 gluon = +2/62/62/6 = 2/6 quark Number of possibilities = 3 3.3.b 2/6 quark 2/6 glu2on = 2/6 2/6 = +2/6, 2/6 quark 2/6 gluon = 2/6 2/6 = +2/6, +2/6 quark. In the end the color of each quark is shifted over +2/6. This overall shift leaves the antibaryon always white. Compare 3.1.d and substitute +2/6 gluon +2/6 gluon by 2/6 glu2on. Number of possibilities = 6 (+2/6 +2/6 2/6) = +2/6 glu3on 3.4.a +2/6 quark +2/6 glu3on = +2/6+2/6 = 2/6 quark. The antibaryon is restored to white only if the glu3on hits the +quark. Otherwise the reaction will not occur. Number of possibilities = 3 The 4paragraphs are completely analoguous to the 1paragraphs, so most of the explanation is left out. (+2/6) and (+2/6) and (+2/6) 4.1.a 2/6 quark +2/6 gluon = 0 quark, 2/6 quark +2/6 gluon = 0 quark, 2/6 quark +2/6 gluon = 0 quark. It doesn’t matter which gluon hits which quark. Number of possibilities = 6 4.1.b 2/6 quark +2/6 gluon +2/6 gluon +2/6 gluon = 2/6 quark, 2/6 quark = 2/6 quark, 2/6 quark = 2/6 quark. Number of possibilities = 3 4.1.c 2/6 quark +2/6 gluon +2/6 gluon = 2/6+2/6+2/6 = +2/6 quark = quark, 2/6 quark +2/6 gluon = quark, 2/6 quark = quark. These rules are applied here in clockwise direction along the colorcircle (c, y, m) which gives an interchange of and . If applied in anticlockwise (c, m, y) direction all quarks would be magenta: 2/6 quark +2/6 gluon +2/6 gluon = 2/6+2/6+2/6 = +2/6 quark = quark, 2/6 quark +2/6 gluon = quark, 2/6 quark = quark. The first two gluons may hit the same of any antiquarkcolor, but if the third intents to hit one of the two other quarks, it have to be the one that obeys the order quarkplusgluonplusgluon, quarkplusgluon, quarkplusnothing applied in clockwise direction along the colorcircle. Number of possibilities = 9 (+2/6 +2/6) and (+2/6) = 2/6 glu2on +2/6 gluon 4.2.a 2/6 quark 2/6 glu2on +2/6 gluon = 2/6 quark, Number of possibilities = 3 4.2.b 2/6 quark 2/6 glu2on =2/62/6 = +2/6 quark = quark, 2/6 quark +2/6 gluon = quark, 2/6 quark = quark. These rules are applied here in clockwise direction along the colorcircle (c, y, m) which gives an interchange of and . If applied in anticlockwise (c, m, y) direction all quarks would be of identical color. Number of possibilities = 3 (+2/6 gluon +2/6 gluon +2/6 gluon) = +2/6+2/6+2/6 = +6/6 = 0 4.3.a There are no further reactions necessary to restore white since the antibaryon is left behind white already. The glue3ball leaves the antibaryon. Number of possibilities = 1 There are 4 possibilities for 3 quarks to simultaneously emit one zeroshift gluon (a glue1ball, a glueball consisting of one gluon) and two nonzeroshift gluons. As a result amongst the remaining reactions, except for the glue1ball leaving the antibaryon, are the ones we discussed in the previous page of The Strong Nuclear Force Gluon and quark reactions. (2/6) and (2/6) and (0) = 2/6 gluon 2/6 gluon + glue1ball 5.1.a +2/6 quark 2/6 gluon = 0 quark, +2/6 quark 2/6 gluon = 0 quark. Then the 0 gluon, the glue1ball, leaves the antibaryon. Number of possibilities = 6 The glue1ball can originate from each of the 3 colors. Then the 2/6 gluon can be absorbed by the same +2/6 quark it came from or by the other +2/6 quark (2 different gluonpaths). 3 * 2 = 6 possibilities. 5.1.b 0 quark 2/6 gluon 2/6 gluon = +2/6 quark Then the glue1ball leaves the antibaryon. Number of possibilities = 3 The 0 quark can be each of the 3 colors. (2/6 2/6) and (0) = +2/6 glu2on + 0 gluon 5.2.a 0 quark +2/6 glu2on = +2/6 quark. Then the glue1ball leaves the antibaryon. Number of possibilities = 3 The 0 quark can be each of the 3 colors. (2/6) and (+2/6) and (0) = 2/6 gluon +2/6 gluon + glue1ball 6.1.a +2/6 quark = quark, 2/6 quark = quark, 0 quark = quark. There are no reactions necessary to restore white since the antibaryon is left behind white already. The glue1ball leaves the antibaryon. When the +2/6 gluon and the 2/6 gluon then both hit the same quark, no matter which one, they will cancel out each others shift. Number of possibilities = 3 * 3 = 9 The glue1ball can come from either quark, but the order quark+0 gluon, quark+2/6 gluon, quark2/6 gluon in anticlockwise direction has to be obeyed. (The clockwise order of reactions is treated in 7.1.a.) So 3 possibilities for emission. Then each of the 3 quarks can absorb both gluons giving 3 * 3 possibilities total. 6.1.b +2/6 quark 2/6 gluon = 0 quark, 2/6 quark +2/6 gluon = 0 quark. Then the glue1ball leaves the antibaryon. The other way around would yield three yellows, so that doesn’t work: +2/6 quark +2/6 gluon = 2/6 quark = quark, 2/6 quark 2/6 gluon = +2/6 quark = quark. But this does work in 7.1.b. Number of possibilities = 3 Each quark can be the one emitting the glue1ball. (+2/6 2/6) and (0) = glue2ball + glue1ball 6.2.a Since the antibaryon is left behind white already, the glue2ball and the glue1ball leave the antibaryon. Compare fig. 3.4. Number of possibilities = 3 Each of the quarks can be the one that emits the glue1ball. (+2/6) and (2/6) and (0) = +2/6 gluon 2/6 gluon + glue1ball 7.1.a 2/6 quark +2/6 gluon = 0 quark, +2/6 quark 2/6 gluon = 0 quark. Then the glue1ball leaves the antibaryon. Number of possibilities = 3 Each quark can be the one emitting the glue1ball. 7.1.b 2/6 quark 2/6 gluon = +2/6 quark = quark, +2/6 quark +2/6 gluon = 2/6 quark = quark, 0 quark = quark. Then the glue1ball leaves the antibaryon. Compare Fig. 3.3 of the previous page. The equivalent of 6.1.a doesn’t work here: 2/6 quark = quark and +2/6 quark = quark. The antibaryon is not left behind white already. Number of possibilities = 3 Each quark can be the one emitting the glue1ball. The 8paragraphs are analoguous to the 5paragraphs. (+2/6) and (+2/6) and (0) = +2/6 gluon +2/6 gluon + glue1ball 8.1.a 2/6 quark +2/6 gluon = 0 quark, 2/6 quark +2/6 gluon = 0 quark. Then the glue1ball leaves the antibaryon. Number of possibilities = 6 8.1.b 0 quark +2/6 gluon +2/6 gluon = 2/6 quark Then the glue1ball leaves the antibaryon. Number of possibilities = 3 (+2/6 +2/6) and (0) = 2/6 glu2on + 0 gluon 8.2.a 0 quark 2/6 glu2on = 2/6 quark. Then the glue1ball leaves the antibaryon. Compare Fig. 3.2 and 11. Number of possibilities = 3 (+2/6) and (0) and (0) = +2/6 gluon + glue1ball + glue1ball 9.1.a 0 quark = quark, 0 quark +2/6 gluon = quark, 2/6 quark = quark. The magenta and yellow quark interchange color. Then the two glue1balls leave the antibaryon. Mark 0 quark +2/6gluon = quark and 2/6 quark = quark. There would be 3 magenta’s, which is forbidden. When the meets the cyan first they don’t react. The yellowtomagenta gluon will pass by the cyan quark without interacting with it until it meets the magenta quark. Compare Fig.7. Number of possibilities = 3 Each quark can be the one emitting the +2/6 gluon. But it will be absorbed by the next quark in clockwise direction. 9.1.b 2/6 quark +2/6 gluon = 0 quark. The quark that emitted the gluon reabsorbs it immediately thereafter. Then the two glue1balls leave the antibaryon. Number of possibilities = 3 Each quark can be the one emitting and absorbing the +2/6 gluon. (2/6) and (0) and (0) = 2/6 gluon + glue1ball + glue1ball 10 is analogue to 9. 10.1.a 0 quark2/6 gluon = quark, 0 quark = quark +2/6 quark = quark. The cyan and yellow quark interchange color. Then the two glue1balls leave the antibaryon. When the meets the magenta first they don’t react. Compare fig 3.1. Number of possibilities = 3 Each quark can be the one emitting the 2/6 gluon. But it will be absorbed by the next quark in anticlockwise direction. 10.1.b +2/6 quark 2/6 gluon = 0 quark. The quark that emitted the gluon reabsorbs it immediately thereafter. Then the two glue1balls leave the antibaryon. Number of possibilities = 3 Each quark can be the one emitting and absorbing the 2/6 gluon. (0) and (0) and (0) = glue1ball + glue1ball + glue1ball The three glue1balls leave the antibaryon behind white. This is not the most exciting possibility. Number of possibilities = 1 12 (+3/6) and (+3/6) and (+3/6) Each of the three quarks emits one mesonic gluon. The mesonic gluon +3/6 and 3/6 have identical outcome: they both shift a color to the same anticolor, “its” anticolor. Therefore we take only one of them, the +3/6 colorshift. Most possibilities have no white outcome. Therefore the mesonic gluons contribute little to the superposition: 186 parts to 6 = 186 / 6 = 31. 12.1 3/6 quark +3/6 gluon = 0 quark, 3/6 quark +3/6 gluon = 0 quark, 3/6 quark +3/6 gluon = 0 quark. It doesn’t matter which gluon hits which quark. Number of possibilities = 6 Summary Par  par  nr of possib’s par  par  nr of possib’s par  nr of possib’s
Can we forget about the colors of the quarks too? Just for the moment, imagine the color is in fact the phase of a wavefunction. If a full turn around the colorcircle is precisely one wave then the 3 colors are each 1/3 of this wave, which yields very unequal waveparts. One can distinguish the colors by the difference in shape of the colorwave. But the colors would no longer be equally fundamental. This happens likewise with a circle of 2 waves divided in 3. So it seems the most convenient to regard one full turn around the colorcircle as precisely 3 waves and each color as one wave . If the wave of a color is then the wave of an anticolor is . One anticolorwave consists of the second half of the previous colorwave and the first half of the next colorwave in the colorcircle. When a color starts with a crest followed by a trough then an anticolor start with a trough followed by a crest. That provides a way to always distinguish a color from an anticolor. If a color should be an arbitrary 1/3 of the colorcircle then the anticolor has the same length but its begin and endpoints shifted by an extra 1/6 of a turn. Then a particle and an antiparticle do annihilate to zero. For colorreactions within one and the same hadron little difficulty is expected then. But when two hadrons meet the result becomes unpredictable, depending on the arbitrary shape of the colorwave in each hadron. So I assume a colorwave always has to start at zero, then going to one maximum, back to zero and going to the opposite maximum and finally ending at zero. The colorcircle is closed and offers six points on it that are more special than others: the zeropoints where a color or anticolor begins or ends. There is nothing attached to those points to let some prevail above others. So one can arbitrary choose to divide the circle in 3 colorwaves or in 3 anticolorwaves or even in 6 “halfcolorwaves”! So there is not really a difference between particle and antiparticle then. When 2 baryons meet, one cannot decide whether this is a meeting between 2 quarks or 2 antiquarks or 1 quark and 1 anti. The colors (and anticolors too) are equally fundamental now, but they don’t differ from each other. They can be identified with each other. The colorstates differ only by their order in the colorcircle. One can call an arbitrary wave yellow, the next one coupled to it shall be cyan then while the previous one  if there are 3 of them  must be magenta. One cannot distinguish between (forbidden) and (allowed). This leads nowhere. No, we cannot forget about the colors of the quarks. NEXT PAGE Up CONTACT 
