Three colors applied together
Three colors applied together
In SUMMARY 1 (the 0th page of the storyline NET FORCES IN QCD), Page 1: Colorcharges, eq. (1.1) and (1.4) state
+ + = white and
+ + = white. (3.00)
Mark we placed a color and its anticolor just above each other. “White” here means “zero”, zero color. It is from QCD where quaternion units play no role.
Previous two pages of this storyline QQD - colors are quaternion units now - indicate that when colors couple, they have to be multiplied.
* * = white and
* * = white. (3.01)
Mark we did not place a color and its anticolor right above each other. Well, the green-magenta are, but the others are not. In quaternions white = 1, the 3 colors together (or the 3 anticolors together) must equal 1. When white = 1 then -1 will be black.
What is right, should we sum or multiply?
What if we experiment a little with the original color sum equation? In a baryon we have red + green + blue = white. We did set red = i, green = j, blue = k, white =1. Then red + green + blue = white becomes i + j + k = 1. This is the quaternion q = -1 + i + j + k. Two baryons colliding then should be the multiplication
(-1 + i + j + k) (-1 + i + j + k) = -2 -2i -2j -2k = -2 ( -1 + i + j + k)
What is this, minus two times a baryon? The original two baryons? Why the minus sign?
Try this at the excel document baryoncollision.xls. Two yellow sections next to each other are two colliding baryons. Just fill in the yellow sections of one line, click somewhere else and in the blue section appears the resulting multiplication. (3.02)
Do quarks couple to each other? According to QCD the answer is no: quarks couple to gluons (emission or absorption) and the gluons then couple to other quarks. But two quarks don't couple, anyway not their colors. What a remarkable situation it is! The colors of the quarks from a single hadron can never depart from each other nor will they ever merge to one color.
They say that white (or black) protons have quarks with colors that sum up to zero. Or multiply to 1, or -1. The meaning and purpose of this seems to be that any two neighboring protons or neutrons in a nucleus show no net color to each other - which is a bit of nonsense because when within reach (less than 2 fm) two quarks do attract each other anyway by their colors.
Is there meaning to a gluon meeting a white proton? Imagine a proton and a nearby passing meson. A gluon from one of the two quarks in the meson couples (emission), goes to the proton and then couples again to one of the 3 colored quarks of the proton (absorption). Suppose it is determined from which quark to which quark the gluon is going - to give them names, the gluon goes from quark 1 in the meson to quark 2 in the proton - then according to QCD it does so along all possible routes. That includes the route going to one of the other two quarks in the proton, couple there and change its color (gluon is absorbed), couple again and changing the color back (gluon re-emitted) and then set for the quark of destination. The whiteness of the proton is tested in between then. But then again, why is whiteness compulsory? White or not, the quarks do attract each other anyway, when within range.
Quarks are fermions and so the Pauli principle holds for them. Color is a quantum number and when the colors of nearby quarks (about 1 fm relative distance, not more, not less) are the same, these two quarks have to differ in another quantum number. If not then repulsion is the only option. The three quarks in the hadron differ in color. However, the color of quark 1 and quark 2 may be alike, as well as their spin, electric charge and taste. Oké, then there will be repulsion for the moment. But if not for the quantum numbers all quarks attract each other when within reach. That's how I did understand it so far.
There is another meaning to the zero sum end state of colors in the next paragraph.
From (1.1), (1.2) and (1.4) at page 1 of this storyline we find
Multiplication order does count in quaternions and (3.12) and (3.13) indeed equal 1. But (3.11) and (3.14) equal -1. One is tempted to define a “normal” and a “reversed” multiplication order. Let's define the equations (3.12) and (3.13) with outcome 1 as normal order. Then (3.11) and (3.14) with outcome -1 is the reversed order.
= k j i k j i k j i
= -i -j -k -i -j -k -i -j -k
= i j k i j k i j k
= -k -j -i -k -j -i -k -j -i
So when you have the 3 different colors and want them to yield white, you have to choose your multiplication order from the color normal order, kji = jik = ikj = 1. And if you have the 3 anticolors and want them to yield white, you have to choose their order from the anticolor normal order, -i -j -k = -j -k -i = -k -i -j = 1.
ij = k which equals * = . But also is ji = -k equaling * = , resembling better the cyclic behavior of colors as summed up in SUMMARY 1, page 1 colorcharges eq. (1.3) and (1.5). Color multiplication in normal order * is the right one.
Likewise is -i * -j = k or * = , while -j * -i = -k or * = . Anticolor multiplication in normal order * must be chosen there.
Let's apply this in Fig. 3.2 at page 3 of the storyline NET FORCES IN QCD. We judge ik from (2) as “taken out of the color normal order” as being right. And we see ki from (1) as “color taken out of the reversed order” and thus not to be used here. The canceling out of (1) and (2) in fig. 3.2 is not true then.
Next figure to apply this can be Fig. 3.3 at the same page. And there we meet a problem: where to take k * -j from, or -j * k? It is not found in a normal order, nor in a reversed order.
This might mean we have to stop with this normal order vs reversed order business. It seems better to reformulate the starting theorem. The starting theorem was: red, green and blue applied together (and also cyan, magenta and yellow together) yields white, or 1 that is in quaternions. Instead would be better:
The color product end state of red, green and blue, and also of cyan, magenta and yellow, equals +1 white or -1 black, both are colorless states. (3.15)
We restore the quaternion canceling-out in fig. 3.2 at page 3 of the storyline NET FORCES IN QCD as well as the other quaternion canceling-outs and doublings-in-the-wavefunction at the page.
For 3 colors applied together (baryons) eq. (3.11) and (3.13) from this page superpose, yielding a color sum end state 1 -1 = 0. So the contribution in the wavefunction is zero. The baryon is colorless. (3.16)
For 3 anticolors applied together (antibaryons) eq. (3.12) and (3.14) superpose, yielding a color sum end state 1 -1 = 0 too, no contribution in the wavefunction either from this possibility. The antibaryon is colorless. (3.17)
So this might yield another rule:
The color sum end state of red, green and blue, and also of cyan, magenta and yellow, equals zero. (3.18)
Zero color can never be achieved by quaternion unit multiplication only. Whatever multiplication order you take from no matter which quaternion units, their multiplication always yield a quaternion unit again. Zero is not a quaternion unit.
For baryons in first occasion the sum end state is i + j + k or i -j -k, and so on (for i -j -k, see paragraph Baryons below). So for baryons, zero color sum end state can only be achieved by superposition of wavefunctions.
Let's generalize this to all color compositions: the sum end state of a color composition is always zero. Or anyway try to. (3.19)
This is all not too different from electrostatics between e.g. an electron and a proton. Their electric charges sum up to zero, but when applied together in the law of Coulomb they multiply to -1, in this case, and not to zero.
Without mentioning explicitly so far, we use a kind of color conservation: the color of a fundamental particle (quark, gluon) does not spontaneously change into a different color. An i cannot become j just like that, nor -i. To change the color of a quark or gluon one needs interaction with another color, a color coupling.
This rule can be shortened to “A color on its own stays the same” - but colors are never alone. Only the colorless colors white +1 and black -1 can be. Nevertheless we stick to this rule.
Therefore, when you have a color composition, superposed with it are all other compositions into which it can transform by gluon exchange. (3.20)
i * i
We take this as the start state, a composition of two i colors. Colors are particles (quarks or gluons), in this case we take the i's as quarks and the composition is a meson.
= i * -k * k * i
= i * -k * i * -k
and this last line represents a quark i that emits a gluon -k and the other quark i that absorbs the gluon -k (the quark is always right multiplied by the gluon)
= j * j
So i i can change into j j. Likewise i i can change into k k. (3.21)
i i = j j = k k superpose.
The appearance of -k k can be the absorption of a vacuum particle as presented in The Higgs mechanism 2 at page 2 of the QUATERNION GRAVITATION storyline.
i * i
= i * -1 * -1 * i
= i * -1 * i * -1
-i * -i
So i i can change into -i -i by black glueball exchange, and likewise j j into -j -j and k k into -k -k. (3.22)
Any two colors in composition continually change into their opposite colors and back again by black glueball exchange. In the baryon this has a peculiar result, see application 4 at page 7 of this storyline. In the meson it causes a net color sum end state of zero: +i +i -i -i = 0.
From 1) and 2) follows
i i = j j = k k = -i -i = -j -j = -k -k superpose. (3.23)
The black glueball has mass, according to (8) in paragraph Building vacuum from gluons at page 2 of QG. The transition between i i and j j and k k and also the transition between -i -i and -j -j and -k -k goes relative fast, see Higgs mechanism 2 at page 2 of QG, while the transitions between i i and -i -i, and likewise between j j and -j -j and between k k and -k -k, goes relative slow. When you start e.g. with i i then the other states from (3.23) do superpose immediately, but not immediately with same amplitude.
i * j
Start state present in baryons only; k color not shown.
= i * -k * k * j
= i * -k * j * -k
This is quark i that emits a gluon -k and quark j that absorbs the gluon -k.
= j * -i
or, when the emerging pair is k -k, leading to end state
= -j * i
So i and j can swap color but one of the colors changes sign.
We met such a sign change before, see (2.4), (2.5) and (2.6) at previous page. But there it was half a gluon exchange that makes sign change. While here it is no gluon exchange at all but a vacuum marble absorption, a Higgs absorption, replacing the gluon exchange. This is worked out in Higgs mechanism 2 at page 2 of QUATERNION GRAVITATION (QG). See also paragraph Some preliminary conclusions at page 2 of QG. (3.24)
Is it possible to regard every computation act as a particle reaction? i * -k * k * j changes into i * -k * j * -k by quaternion computation rule; ij = -ji by computation rule. But it can also be seen as a physical swap of colors with a minus sign appearing.
The swapping reaction then is part of the description of the swapping reaction. It is an endless series of disturbances we recognize from renormalization techniques. But in renormalization normally there is a main reaction and added to it are the disturbances. However in this case, as far as I can see, the main reaction doesn't exist, only the series of disturbance does.
One better uses the quantum view of superpositions: ( i * j ) and ( i * -k * k * j ) and ( i * -k * j * -k ) and ( j * -i ) and ( -j * i ) are states of same outcome and thus superpose. One of the possibilities will be chosen according to chances of occurrence (square of amplitude). No one bothers about how one possibility mechanically goes into the other. (3.25)
But then again, do these states superpose simultaneously? Does it take time to go from state ( i * j ) to state ( i * -k * k * j ), does the emergence of the vacuum particle take at least one typical reaction time (10^-23 s)?
j * -i = j * k * -k * -i = j * k * -i * k = i * j
-j * i = -j * -k * k * i = -j * -k * i * -k = i * j
return to the starting state i j.
When the colors of one baryon have the letters i, j and k, regardless the sign in front or the order of multiplication, then the multiplication outcome will always be a real state +1 or -1.
If not i, j and k then there always is a pair i i or j j or k k (regardless their sign in front). Therefore according to the rules from previous paragraph, are superposing:
8 different three-color compositions that cannot transform into one another by gluon exchange and thus not superpose in a baryon. They have to be set initially.
The color product end state of these compositions are:
These end states are the 8 gluons. The compositions are the possible glu3ons (more about it below) that superpose with the gluons, regardless their multiplication order. A spin consideration of these compositions is in paragraph Can 3 gluons change into 1 gluon easily? at page 7 of storyline NET FORCES IN QCD. This is not yet worked out properly.
Regard baryon color composition i i j.
i * i * j = j * i * i = -j and i * j * i = j. In 2/3 of the cases the outcome is -j and in 1/3 the outcome is j. Superposing yields
2/3 * -j - 1/3 * j = 1/3 * -j (3.32)
Physical interpretation of multiplication order
We passed over applying three colors together. There “applying together” does NOT mean “at the same time”. A physical interpretation of multiplication order can be time order of multiplication: start multiplying with this, a little later with that and at last with this - as if it were couplings performed one after the other. Reversing multiplication order then is reversing time order. (3.40)
Then ijk becomes different from jki and kij, despite that ijk = jki = kij. There are 3 times more possibilities to superpose. (3.41)
ijk from (3.11) then means i and j couple first, and then k. And kji from (3.13) then would mean k and j couple first and then i. So far so good, but there is still no time order in e.g. ij, i * j that is; ij and ji have different outcomes, but there cannot be assigned a different reaction time order to it.
(3.11) up to (3.14) shows reversing multiplication order suffices to convert between end state +1 and -1. If multiplication order is physically interpreted as time order of multiplication, then multiplication order can be different in frames of reference that differ a large speed. In one frame there is multiplied i * j first and then * k; and in the other frame j * k first and then i *.
That might be the case with quarks. Quarks have mass and move below lightspeed. But not for gluons. Gluons move at lightspeed and gluon worldlines are the same in all frames. Except the black gluon, of course.
Preliminary conclusion: Multiplication order as time order of occurrence runs into trouble with two-color multiplication. And in three-color multiplication it runs into trouble when the colors don't have lightspeed.
Let's conclude there is no physical interpretation of multiplication order yet. (3.42)
Three gluons applied together
How to visualize ijk, what example shall we take? Take in mind an antibaryon, see the gluon table.
Quarks don't couple, despite they stay apart at least 0.7 fm. I mean, colors can couple to each other when within reach and at 0.7 fm relative distance the strong force is indeed quite strong. So it is quite strange the colors of the quarks don't couple. But oké, they don't.
In fact I can imagine only one example: the glu3on, a composite of 3 gluons, an object that I suppose to exist, see page 7 of NET FORCE IN QCD. When the color product is +1 or -1 we call it a glue3ball. The 3 gluons can merge to 1 gluon (goes relatively easy). When they don't merge and stay a glu3on then the 3 gluons will remain a certain distance to each other, because at zero distance the force between colors is zero and the force is proportional to the distance, see paragraph The proton at page 5 in the storyline NET FORCES IN QCD.
If this would be all, then every single-gluon state (glu1on we call it) is always superposed by all glu3on states that yields the glu1on when their 3 composing gluons merge. In (3.21) from Higgs mechanism 2 at page 3 of QG is suggested that glu3ons are massive. A particle has mass OR it is massless, even the tiniest mass value will do. The glu3on mass would spoil the superposition. When energy for the glu3on mass is lacking, gluons will be massless, except for -1. As soon as energy is available, the massless glu1on plus the energy concentrated in its parcel of space time, superposes with the massive glu3on state. It does so for only one instant. After that single moment the massless glu1on state and the massive glu3on state depart from each other. Is this mechanism candidate for the next generation of elementary particles? Because of its mass, the glu3on will have range. But gluons already have range. (3.50)
Consider gluons i, j and k in one glu3on. Since kji = +1 and ijk = -1 (both are colorless) it's a glue3ball, a colorless gluon.
The k and -k colors shown next to each other in the picture do not represent two different particles but only one particle in two superposing states. Likewise the black and the white colors. From +1 or -1 there is a like chance to convert back to ANY glu3on state. These are only two of them.
And now for the mesons, two quarks together. A multiplication of two different colors like i and j always yield a colored thing, in this case a product end state +k or -k. Reversing multiplication order renders no effect. So different kind of colors are not permitted for the quarks in the meson. Two equal colors multiplication, whatever their sign, always yields +1 or -1.
As you see, according to (3.15) only, the two quarks of the meson may be of opposite color as well as equal color. Then i * i, j * j and k * k from the first column superpose swiftly by absorption from the Higgs field. And so do the members of each of the other columns.
Then i * i and -i * -i superpose by black glueball exchange. This is more slowly because of the mass of the black glueball. So finally the first two columns superpose and the last two columns superpose, leading to one class with outcome +1 (last two columns) and one class with outcome -1 (first two columns). (3.61)
As said in point 2 of paragraph Color conservation at this page, any two colors in composition continually change into their opposite colors and back again by black glueball exchange. In the meson this leads to a net color sum end state of zero, for the +1 class directly via i -i = j -j = k -k = 0 and for the -1 class as a net color over time +i +i -i -i = +j +j -j -j = +k +k -k -k = 0. (3.62)
A previous attempt
The application of colors goes by means of gluons, there is no other way. All gluons have 2 colors, so e.g
* * * = * ( / ) * = *
and for the single blue we have no gluon to transmit it.
(Mind green divided by cyan equals green left-multiplied by anti-cyan, red that is.)
Applying the 3 colors twice then can be done by 3 gluons:
ijkijk = * * * * * * = * ( / ) * ( / ) * ( / ) = * * = j/-i * i/-k * k/-j
= kji = * * * = 1 (3.70)
use the gluon table.
We anyway have ijkijk = kji, which here means “applying the colors twice equals applying the colors in reversed order once”. (3.71)
So a rule like “reversed multiplication order for colors and normal order for anticolors” might be replaced by the rule “apply this or that twice”. The appliance of the 3 colors twice, or the 3 anticolors twice, then always gives -1 * -1 = 1 * 1 = 1
Applying all colors (the 3 colors applied once and the 3 anticolors applied once). When applying (3.12) and (3.13) the outcome is 1:
* * * * * * = k * j * i * -i * -j * -k = 1 (3.72)
Contrary, “walking around the colorcircle”, no matter what direction or starting point, always yield -1.
* * * * * * = k * -i * j * -k * i * -j = -j * -i * -k = -1. Then walking around the colorcircle twice always yields -1 * -1 = 1. (3.73)
Any color multiplication sequence ends up with one out of 1, i, j, k, -1, -i, -j, -k.
i = j = k = (-i) = (-j) = (-k) = -1
-1 * -1 = 1 * 1 = 1
So any factor sequence of colors, when repeated 4 times at most, will always yield 1, the colorless white end state.
Open color is never observed. All color reactions are virtual. When a quark emits a gluon, it sets out as the superposition of the 8 possible colors, 1, i, j, k, -1, -i, -j, -k. Whatever color exchange might have taken place, after 4 times setting out as the same superposition of the 8 possible colors, each exchange occurred 4 times. Yielding net color white. So it seems IF colors are virtual, THEN they practically are confined. Is this all there is to color confinement?
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