Four quarks in the shell

Read the book of Feynman, QED - The strange theory of light and matter. And Scientific American of June 1980, Gerard 't Hooft, Gauge theories of the forces between elementary particles. In Four quarks in the shell knowledge of renormalization theory is necessary at that level.

A little knowledge about quaternions is necessary. The quaternion approach of strong force colors is worked out in the storyline QQD.

Four quarks in the shell

This yields a new view on the color interaction itself, between the quarks in a baryon or antibaryon. We go back to the superposition of innumerable quark-antiquark pairs, shielding the naked color of the quark. The color coupling constant is about 1. Therefore 4 quarks appearing in 2 quark antiquark pairs, all within their time borders, all seeing each other, count with same importance as 2 quark antiquark pairs superposed to each other at same mutual distance. (The superposed two quark pairs don't see each other, don't react with each other.)              (a)

We just suppose the 4 quarks to appear within their time borders, at the Earth surface within 10^-19 m or 10^-4 fm, (EXPANSION OF THE UNIVERSE, page 2, paragraph The calculation of the time border). Quarks have maximum attraction at about 0.9 fm, so the 4 quarks hardly attract each other. They don't form pairs under strong force attraction.

Emerging pairs always consist of a quark and an antiquark with opposite taste, color and spin. Electric charge already is in the taste, e.g. when the taste is u then electric charge is +2/3 times the electron charge. When the taste is d then the electric charge is -1/3. When the tastes are opposite, the electric charges are so too.

Impulses don't have to be opposite. If one quark from the pair has a small impulse and the other quark has a large impulse in a different direction, when massless coinciding the impulse sum determines the direction and the energy of the resulting lightspeed gluon. It is the direction of the frame of reference in which the quarks do have opposite impulse.

All 4 quarks appear within their time borders. We assume the two pairs to emerge simultaneous within a length of time of 10^-23 sec. A, B, C and D are quarks, A B is one pair, C D is the other pair.

A        C

B        D

The double pair forms 2 gluons (2 times 2 quarks massless coinciding) which can be done in 3 ways: AB CD, AD BC and AC BD (denoted as II column pairs, X crosswise pairs and = row pairs).

A and B have opposite spin and so do C and D. Suppose A and C have spin +1/2. When combining II or when combining X, the double pair can form two spin 0 gluons: quark spins +1/2 -1/2 = 0. In II the two spin 0 gluons necessarily are colorless. But in X they might form a pair of colored spin 0 gluons.              (b)

In fact, as will be described just below, in 18 from 25 cases the two gluons in X have color and in 7 cases they are white-white or black-black.

Colorless and tasteless spin 0 gluon pairs - if formed - will be absorbed by the vacuum, enlarging the vacuum with their volumes, see page 3, 4 and 5 of NEG and page 2 of QG. There is vacuum everywhere, so this will be set into action immediately.

The gluons don't react with the vacuum gluons at the place they are, mutual distance being too small. Instead they react with the ring of gluons at a distance of 0.9 fm mainly. Somewhere on the ring it presses itself between the other vacuum particles.

The energy of the spin 0 gluon is converted into a tiny parcel of space. The ground state real quark - I mean the real quark in which shield all this is taking place - cannot afford to loose this energy. So, if this takes place, immediately thereafter the vacuum re-emits the spin 0 gluon pairs.

When combining = the 4 quarks form two spin 1 gluons:

quark spins +1/2 +1/2 = +1 and -1/2 -1/2 = -1.              (c)

A single gluon emerging consists of two quarks massless coinciding. The two quarks have opposite spin and color, yielding sole white spin 0 gluons. White gluons don't glue. So 4 quarks making up 2 gluons is the only possibility for colored spin 1 gluons to appear.

The single gluons see the core of the quark in which shield they are emerging, since they shield the core by amplifying its strength. This seeing goes by means of gluons and we assume it are spin 1 gluons. Then every gluon swaps the spin. Then spin 1 gluons might be formed. This cost (at least) one cycle of time, which is enough for the constituting quarks to move 3 fm apart. At the Earth surface this is large enough to move far out of the reach of each others time borders. So here on Earth sole emerging quark pairs will not form single spin 1 gluons at a reasonable rate.

And what about the colors? Let's go into quaternions. And use the gluon table. Regard the 2 pairs again. Each pair consists of a color and an anticolor. For each pair that emerges, there are 4 possibilities: and and . (And the fourth pair? Is it ?) The result per pair is to be taken as the application of both colors one after the other. One has to multiply the colors with each other and in quaternions multiplication order makes a difference. So which order is to be taken? Set e.g.:

A     C  
B =      and D =

= i * -i = = j * -j = 1 = , so AB as well as CD will form a white gluon. Multiplication order is not important. But how for the other combinations?

= -j * i = k =
= i * -j = -k =

= j * -i = k =
= -i * j = -k =

= i * j = k =
= j * i = -k =

= -i * -j = k =
= -j * -i = -k =

We didn't need to worry. It are all combinations of i and j - with or without a minus sign in front - and the multiplication always will yield k, one with a minus sign in front and the other without. All possible arrangements and orders of the quarks of color i and j (with or without a minus sign in front) yield the gluon pair k -k. Two different pairs of color-anticolor that appear, always yield two times the third possible color-anticolor pair.

What if the two pairs of quarks are the same?

A     C     = i * i = -1 = = i * -i = 1 =
B =      and D =   = -i * -i = -1 = = -i * i = 1 =

The particle and antiparticle in a pair that emerges in the shell cancel each other out. When do colors cancel each other? + = i + -i = 0, the colors and anticolors happen to add up to zero. But in quaternions we don't add colors, we multiply them. Then = i * -i = 1 and indeed 1 is the neutral element with respect to multiplication. Similar for j and k. But black and white multiply to = 1* -1 = -1. Therefore we take = -1 * -1 = 1 = and = 1 * 1 = 1 = to be the colorless color pairs that appear in the shell, instead of . See also QG page 2 paragraph Filling in the vacuum marbles.

So when AB and CD is the same pair of colors the result is a superposition of the black pair and the white pair.

A few other reactions:

A =
B =
C =
D = or
= i * 1 = i =
= -i * 1 = -i =
= i * -1 = -i =
= -i * -1 = i =

= 1 * 1 = 1 =
= 1 * 1 = 1 =
= 1 * -1 = -1 =
= 1 * -1 = -1 =
= -1 * -1 = 1 =
= -1 * -1 = 1 =

As you see, the pair doesn't occur.

And last but not least there is taste. In the 1st generation there are two tastes available for hadrons: u and d. There u has always +2/3 electron charge, d has always -1/3, anti-u = u has -2/3 and anti-d = d has +1/3 electric charge. As to speak, charge is in the taste. There are 4 possible u and d distributions for the quark pairs AB and CD that emerge in the shell. Keep in mind we assigned spin +1/2 to A and C, and spin -1/2 to B and D.

A C   u u   d d   u d   d u  
B D   u u   d d   u d   d u              (d)

In the first, and similar in the second pair of pairs, II-pairs as well as = are possible, there the tastes cancel out. But X would form charged gluons and that are no gluons. Moreover, X are particle-particle pairs or antiparticle-antiparticle pairs, in which massless coinciding is not possible.              (e)

In the third and fourth pair of pairs situation is even worse: only II can form gluons (colorless spin 0 gluons only) and X and = would yield charged gluons.

The = pairs in the 1st and 2nd scheme are the only possibility to form colored spin 1 gluons, the gluons that mediate the strong force.

Mark the = pairs in the 3rd and 4th scheme are interesting: they form pairs of spin 1 particles of unit charge and the particles can have color but may be white or black too. IF u and d have different mass - and I think they have, at least because of the difference in electric charge - THEN the Higgs field absorption of the one do not cancel the Higgs field emission of the other precisely. The quarks cannot coincide massless, the particle will have some remnant mass. The quarks are within their time borders; they will not separate because the presence within the time border gives some mass reduction. Separation means the quarks have to be supplied by their complete mass; that thus might be huge. We know little about the quark mass, see paragraph Quark mass at the previous page. Admit, the particle has some resemblance to the W+ W- particle. But well, why W has mass 80385 MeV while the meson consisting of the same quarks then, has mass 140 MeV? So this is not that easy.

Finally one of the colored spin 1 gluons formed from the 4 quarks in the shell, can be absorbed by the real quark in which shield all this is happening, changing its real color. The other color then can be reabsorbed too (yielding no change at all) OR escape to another real quark. This can happen in mesons as well as in baryons.

Also possible is that both gluons leave the real quark where they are born, each of them going to a different quark - in baryons only, baryons have 3 quarks. The mother quark then doesn't change color, but the other two do.

And now for the chances

The 4 quarks in the shell emerge in 2 particle-antiparticle pairs, one pair is AB and the other is CD. From (a) there is a chance of 1 out of 2 for emerging 4 quarks all seeing each other within their time borders.              (chance 1 = 1/2)

And a chance of 1 out of 2 for the pairs just to superpose, despite they appear at precisely the same spots.              (chance 2 = 1/2)

A C   u u   d d   u d   d u  
B D   u u   d d   u d   d u              (d)

The best way to proceed now is to start with the taste. We are in chance 1. When written down as in (d) there is a chance of 1 out of 2 for the first or second pair-of-pairs to form.              (chance 3 = 1/2)

And there is a chance of 1 out of 2 for the third or fourth pair-of-pairs to be formed, yielding colorless spin 0 gluon pairs only.              (chance 4 = 1/2)

(For convenience we assumed the u is as likely to appear as d, which might be wrong)

When in chance 3 (1st and 2nd pair-of-pairs in d), there is a chance of 1 out of 2 for II-pairs, yielding sole colorless spin 0 gluons.              (chance 5 = 1/2)

And a chance of 1 out of 2 for =pairs to be formed.              (chance 6 = 1/2)

And no chance for X-pairs to be formed.

We now shift to spin and color. Set A at spin +1/2 and B at spin -1/2. When in chance 6, there is a chance of 1 out of 2 for C spin +1/2 and D spin -1/2.              (chance 7 = 1/2)

And a chance of 1 out of 2 for C spin -1/2 and D spin +1/2.              (chance 8 = 1/2)

As long as we have no reason to assume otherwise, we assign an equal chance to the 5 possible color-anticolor pairs to emerge. In the scheme at the right we see 25 possible combinations, 18 have color and glue, and 7 are white-white or black-black pairs from which the gluons don't glue.

18/7 = 2.57, 7/18 = 0.39, 25/18 = 1.39, 18/25 = 0.72, 25/7 = 3.57, 7/25 = 0.28.

In chance 7 there is a chance of 18 out of 25 for a pair of colored spin 1 gluons, the particles that make up the strong force.              (chance 9 = 18/25)

When in chance 8, there is a chance of 7 out of 25 for a pair of colorless spin 1 gluons (black-black or white-white).              (chance 10 = 7/25)

When 4 quarks in 2 pairs emerge within their time borders, the chance for two colored spin 1 gluons (two opposite colored spin 1 gluons) is:

Chance 1 x chance 3 x chance 6 x chance 7 x chance 9

= 1/2 * 1/2 * 1/2 * 1/2 * 18/25 = 18/400 = 0.045 or about 5 percent.

This percentage is making up the entire strong force. For spin 1 colored gluons, what we interpreted so far as one reaction in one cycle of time, up until now taken as 10^-23 s, must be about 1/0.045 = 22 reactions, 22 cycles of time - from which only one of them is yielding colored spin 1 gluons. One strong force cycle of time then must be rather 0.5 * 10^-24 s. In that time a light speed gluon covers only 0.15 fm.


Scientific American june 1980, Gerard 't Hooft, Gauge theories of the forces between elementary particles, for the model used in Four quarks in the shell.