Application 4: Baryons

IF kji = 1 is color neutral THEN -k * -j * i = 1 is color neutral too, according to (1.2) and (1.3) and also (2.5) at page 3 of QQD. (1.1)

i, j and k are colors, -i, -j and -k are anticolors. Take in mind i, j and k as the colors of the quarks in a baryon.

The transition of (1.1) can be achieved by a gluon with colorless color -1 going from j (the quark with color j) to k (the quark with color k) or the other way around -1 going from k to j. The electric charge and taste of the quarks remain unchanged by the -1 gluon exchange, so we end up with two quarks with anticolor in the baryon.

a) The i, -j, -k state of the baryon is by no means color neutral as meant in (1.4) at page 3 of QQD, when the three quarks in the baryon stay together but do not merge. (1.2)

b) The paragraph Mesons at page 3 of QQD tries to show that the restriction quarks have color and antiquarks have anticolor is compulsory. (1.3)

We recognize (1.1) as an application of the Minus Sign Go Wild approach at page 3 of QG. Minus sign go wild is considered now to be a previous attempt. So no, despite it is tempting, we don't go on with this.

IF YOU ARE READING TONE FOR THE FIRST TIME, ALL NECESSARY THINGS FOR THE CHAPTERS TO COME ARE GIVEN NOW, YOU CAN SKIP THE REST OF THE PAGE AND READ IT LATER.

What if we still give it a view.

Table (1.4)

B1 is a baryon with quarks of color i, j and k; b2 up to b7 are subsequent states of the baryon. The transition from b1 to b3 is caused by a gluon with color -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 between colors by means of a colored gluon.

Let's regard the chain of reactions b1 up to and including b7 including the spin and taste (type of particle). We choose the baryon to be a proton. According to The spinpuzzle in the column at the right at page 4 of QCD, we start with the spins of the u-quarks aligned. (1.5)

[ spin, color, electric charge, type of particle ]

plus for b2-below, b4-below and b6-below in the the right side of the column [ spin, color ]

In blue are the -1 gluon and the quark that emitted it.

-1/2	i	-1/3	d
+1/2	j	+2/3	u
+1/2	k	+2/3	u	b1
---------	---------	---------	---------		---------	---------
+1/2	-i	-1/3	d		-1/2	i
-1	-1	0	gl		+1	-1
+1/2	j	+2/3	u		-1/2	-j
+1/2	k	+2/3	u	b2	+1/2	k
---------	---------	---------	---------		---------	---------
+1/2	-i	-1/3	d
-1/2	-j	+2/3	u
+1/2	k	+2/3	u	b3
---------	---------	---------	---------		---------	---------
+1/2	-i	-1/3	d		+1/2	-i
+1/2	j	+2/3	u		-1/2	-j
-1	-1	0	gl		+1	-1
+1/2	k	+2/3	u	b4	-1/2	-k
---------	---------	---------	---------		---------	---------
+1/2	-i	-1/3	d
+1/2	j	+2/3	u
-1/2	-k	+2/3	u	b5
---------	---------	---------	---------		---------	---------
+1/2	-i	-1/3	d		-1/2	i
+1/2	j	+2/3	u		+1/2	j
+1/2	k	+2/3	u	b6	-1/2	-k
-1	-1	0	gl		+1	-1
---------	---------	---------	---------		---------	---------
-1/2	i	-1/3	d
+1/2	j	+2/3	u
+1/2	k	+2/3	u	b7

Table (1.6)

As you see, the end state b7 is identical to the starting state b1.

The spins between two subsequent --------- always add up to +1/2 due to spin conservation. (1.7)

The colors between two subsequent --------- always multiply to -1, reflecting color conservation, that is: conservation of colorlessness, IF the colors would merge, which they don't. See (1.2) and (1.3) at page 3 of QQD. OR if the colors of the three quarks one after the other are applied on a passing test particle. The application of the colors of the three quarks on the test particle yields no net color change. It should be impossible for a test particle to graze the baryon at only one or two of its quarks. (1.8)

In b4 below, couldn't the -1 escape? -1 itself is colorless and it leaves behind a black baryon (a baryon with color -1) and black is one of the colorless colors +1 and -1. Normally, in ground state: no, energy not available. But when energy is sufficient to provide the free -1's mass, then yes. However, we forbid quarks with antiquark colors. (1.9)

Starting from b1, the ordinary baryon in table (1.4), there are 6 different ways to arrive at a b3-like state, a baryon of three quarks with two anticolors and one color:

( i > j means gluon with color -1 goes from quark with color i to quark with color 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

When a quark i emits a color -1 gluon so the quark becomes -i, is the -1 initially nearest to the -i? Let's suppose so. Strong force is zero at zero distance, so the -i and the just created -1 don't react. So i > i, j > j, k > k don't contribute here. (1.10)

Call this process 1. Process 1 takes the typical strong force reaction time of 10^-23 sec.

Starting from b3 there are 7 ways to stay in a similar state (1 color and 2 anticolors)

-i > k, k > -i, -j > k, k > -j, -i > -i, -j > -j, k > k

Once again we take the -i > -i, -j > -j, k > k not to work here, mutual distance too small. So starting from state b3 there are 4 ways left to stay in a similar state.

Starting from b3 there are 2 ways to return to the original baryon state:

-i > -j, -j > -i.

This is process 2, also taking 10^-23 sec.

Process 1 leads all present b1-states to b3-states. Process 2 leads B3-states in 2 out of 4 + 2 = 6 possibilities back to the b1-state. So in any step of 10^-23 sec all b1-states become b3-states and 1/3 of the present b3-states become b1-states again. No matter if you start with e.g. 1000 b1's and no b3's *), or 500 b1's and 500 b3's, after about six or seven steps you end up with 250 b1's and 750 b3's and that's the stable division, 1 : 3. One out of 4 is a b1, 3 out of 4 is a b3-like state. A baryon is only 25 % of the time in a normal i, j, k color state and 75 % of the time in a b3-like state, provided the gluon -1 reaction was allowed. (1.11)

*) The other way around, starting with no b1's and 1000 b3, is the result of step one in process 1.

According to the representation of the vacuum at page 2 of QG, a baryon state b1 will absorb one ( -1 -1 ) vacuum particle to arrive at state b3. (1.12)

When minus sign could go wild and b3-states were possible, in the baryon the ratio of color to anticolor abundance then soon is 1 : 2 or 1/3 : 2/3. There could be a connection with the -1/3 and +2/3 electric charge of the d- and u-quarks in a baryon. It is this kind of things that keeps me from throwing away Minus Sign Go Wild just like that. (1.13)

Maybe Minus Sign Go Wild is true for unobserved intermediate states, the so-called virtual particles. The immoral virtual particles, as Feynman called them, in renormalization procedure go backward in time, faster than light and its photons move in small curves wherever required. So maybe more immoral behavior can be expected between the virtual particles under consideration. I am still not completely convinced that Minus Sign Go Wild plays no role at all. (1.14)

In the following previous attempt (1.10) is left out. You still end up with 250 b1-states and 750 b3-states as stable division. It doesn't matter whether (1.10) is true or not. (1.15)

A previous attempt

Starting from b1, the ordinary baryon in table (1.5), there are 6 different ways to arrive at a b3-like state, a baryon of three quarks with two anticolors and one color:

( i > j means gluon -1 goes 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.)

Call this process 1. Process 1 takes the typical strong force reaction time of 10^-23 sec.

Starting from b3 there are 7 ways to stay in a similar state (with 1 plus-color and 2 minus-colors, 1 color and 2 anticolors that is)

-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.

This is process 2, also taking 10^-23 sec.

Process 1 leads a b1-state in 6 out of 6 + 3 = 9 possibilities to a b3-state. Process 2 leads B3-states in 2 out of 7 + 2 = 9 possibilities back to the b1-state. So in any step of 10^-23 sec 2/3 of the present b1-states become b3-states and 2/9 of the present b3-states become b1-state again. No matter you start with e.g. 1000 b1's and no b3's, or the other way around no b1's and 1000 b3's, or 500 b1's and 500 b3's, after about four steps you end up with 250 b1's and 750 b3's and that's the stable division, 1 : 3. One out of four is a b1, the rest is b3-like states. A baryon is only 25 % of the time in a normal i, j, k color state and 75 % of the time in a b3-like state, provided the gluon -1 reaction was allowed.

End of previous attempt

Still application 4: Mesons

1	2	3	4	5

i -i	i -1 i	-i i	i -1 i	i -i

i i	i -1 -i	-i -i	i -1 -i	i i

The lower mesons are no mesons, it are colored end state objects, despite their colorless outcome -1 in multipliction, i * i = -i * -i = -1. See the text below (5.2) in paragraph Mesons at page 3 of QQD (the first 7 alineas).

Mark that if the upper meson nr 1 is u u (the left quark is u with color i and the right quark is u with anticolor -i) that then the upper middle meson nr 3 is at the left the same u-quark with anticolor -i and at the right the same u-antiquark with color i. It are a quark-with-anticolor and an antiquark-with-color, but since overall color composition is maintained, there might be no objection to it. (2.1)