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Ignoring for the moment the experimental necessity of eliminating the gluon singlet state, the nine "raw" gluon color components form two quite distinct sets. The first set consists of the six (color,anti-other-color) components:

$r\overline{g}$, $r\overline{b}$, $g\overline{b}$, $g\overline{r}$, $b\overline{r}$, $b\overline{g}$

The second set consists of the three (color,anti-color) components:

$r\overline{r}$, $g\overline{g}$, $b\overline{b}$

Only the first six components are needed to explain color flipping in hadrons and mesons. The second smaller group cannot change quark colors since they are net neutral in color.

So: For gluon exchanges between quarks in hadrons and mesons, is the role of the three gluon color-anticolor components purely attractive or repulsive? That is, are these three components more closely analogous to the photons of the electromagnetic force, which similarly carry no net electric charge?

Popular explanations, even ones by very good physicists, seem usually to skip over the existence of these two distinct groups, and instead describe only the color-changing consequences of the first group of gluon color components.

Finally, are there experimental implications of in the existence of these two groups? For example, would gluons that happen to carry mostly color-anticolor charge generate identifiably different jets from gluons that carry mostly color-anti-other-color charge?

This is a well-studied area, of course, so I suspect I just haven't been looking in the right places.

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Note that, for nuclear forces, based on strong interaction, the attractive/repulsive behaviour depends on distance : The force is attractive at distance of about 1 femtometer (fm), at very short distances less than 0.7 fm, it becomes repulsive. Beyond 2.5 fm. , the force is nor more signigicant. So, it means that you have to make precise calculus, before getting an answer. –  Trimok Jun 3 '13 at 10:26
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2 Answers

up vote 8 down vote accepted

Texts on QCD don't divide the generators of $SU(3)$ – and therefore "bicolors of gluons" – into two groups because this separation is completely unphysical and mathematically artificial (basis-dependent).

Moreover, the number of "bicolors of gluons" i.e. generators of $SU(3)$, the gauge group of QCD, isn't nine as you seem to think but only eight. The group $U(3)$ has nine generators but $SU(3)$ is the subgroup of matrices with the unit determinant so one generator is removed. At the level of gluons or Lie algebra generators, the special condition $S$ means that the trace is zero. So the combinations $$ A(r\bar r) + B (g\bar g) + C(b \bar b)$$ are only allowed "bicolors of gluons" if $A+B+C=0$. Now, in this 8-dimensional space of "bicolors of gluons", there are no directions ("bicolors") that are better than others. For any direction in this space, there exists an $SU(3)$ transformation that transforms this direction into a direction non-orthogonal to any chosen direction you choose. This is true because the 8-dimensional representation is an irreducible one (the adjective "irreducible" means that one shouldn't try to split it to two or several separated collections!). And there doesn't exist any consistent Yang-Mills theory that would only contain the six off-diagonal "bicolors" because the corresponding six generators aren't closed under the commutator.

The actual calculations of the processes with virtual gluons ("forces" between quarks etc.) therefore never divide terms to your two types because this separation is just an artifact of your not having learned group theory. Instead, all the expressions are summing over three colors of quarks, $i=1,2,3$ indices of some kind, and there is never any condition $i\neq j$ in the sums because such a condition would break the $SU(3)$ symmetry.

Now, the $r\bar r,g\bar g,b\bar b$ "bicolors of gluons" (only two combinations of the three are allowed) are actually closer to the photons than the mixed colors. So it's these bicolors that produce an attractive force of a very similar kind as photons – they are generators of the $U(1)^2$ "Cartan subalgebra" of the $SU(3)$ group and each $U(1)$ behaves like electromagnetism. That's why these components of gluons cause attraction between opposite-sign charges and repulsion between the like charges.

The six off-diagonal "bicolors of gluons" (and let me repeat that the actual formulae for the interactions never separate them from the rest – they're included in the same color-agreement-blind sum over color indices) cause neither attraction nor repulsion: they change the colors of the interacting quarks so the color labels of the initial and final states are different. It makes no sense to compare them, with the idea that only momentum changes, because that would be comparing apples and oranges (whether the force looks attractive or not depends on the relative phases of the amplitudes for the different color arrangements of the quarks).

At any rate, particles like protons contain quarks of colors that are "different from each other" so they're closer to the opposite-sign charges and one mostly gets attraction. However, the situation is more complicated than it is for the photons and electromagnetism because of the six off-diagonal components of the gluons; and because gluons are charged themselves so the theory including just them is nonlinear i.e. interacting.

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Maybe an analogy using the singlet state and the triplet state for two spin-1/2 particles might be a good way to expalin this. –  Peter Shor Jun 3 '13 at 12:35
    
Luboš, thanks! Your answers are always highly informative, factually precise, and good at emphasizing the important points in the interpretation of the math. And Peter, nice to see you are still active. Wow, would I love to quiz you on the latest D-Wave discussions... maybe in some other forum. –  Terry Bollinger Jun 9 '13 at 3:14
    
Thanks for your kind words and interest, @Terry, and I join your comments and requests for Peter. –  Luboš Motl Jun 9 '13 at 7:05
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LM gives the more in depth answer that includes these points too, but regarding the primary question of distinguishing cross-color anti-color states and homogeneous color anti-color states, at the simplest level we don't due to these facts:

There is no experimental evidence for any preferred direction in "color - anti-color" space

Therefore, all linearly independent combinations are equally likely (except of course the "colorless" or neutral singlet combination which is also ruled out experimentally, due to no long-range element in the strong force). There is no particular "color -anti-color" combination observed to be favored in any strong interaction process.

Linear superposition in QM. When nature makes no distinction among equivalent states neither can our calculations.

Therefore, we must sum over (linearly superpose) all the "equivalent" states to get the correct result. In a sense the existence of "equivalent" states is an artifact of our calculation methods -- we have to refer to some combination of colors and anti-colors in each element of the calcuation just by writing it down, so the way we make our calculations match nature in this situation is to linearly superpose all the allowed elements, i.e. all allowed linearly independent color-anti-color states.

In other words we cannot favor any of the eight allowed combinations of color and anti-color in our calculations simply because (so far as we've seen) nature doesn't.

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