Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This question has been asked twice already, with very detailed answers. After reading those answers, I am left with one more question: what is color charge?

It has nothing to do with colored light, it's a property possessed by quarks and gluons in analogy to electric charge, relates to mediation of strong force through gluon exchange, has to be confined, is necessary for quarks to satisfy Heisenberg principle, and one of the answers provided a great colored Feynman diagram of its interaction, clearly detailing how gluon-exchane leads to the inter-nucleon force. But what is it?

To see where I'm coming from, in Newton's equation for gravity, the "charge" is mass, and is always positive, hence the interaction between masses is always attractive. In electric fields the "charge" is electric charge, and is positive or negative. (++)=+, (--)=+, so like charges repel (+-)=(-+)=-, so opposite charges attract. In dipole fields, the "charge" is the dipole moment, which is a vector. It interacts with other dipole moments through dot and cross products, resulting in attraction, repulsion, and torque. In General Relativity, the "charge" is the stress-energy tensor that induces a curved metric field, in turn felt by objects with stress-energy through a more complicated process.

So what is color charge?

The closest that I've gotten is describing it through quaternions ($\mbox{red}\to i$, $\mbox{blue}\to j$, $\mbox{green}\to-k$, $\mbox{white}\to1$ , "anti"s negative), but that leads to weird results that don't entirely make sense (to me), being non-abelian.

Since $SU(3)$ is implicated, what part of $SU(3)$ corresponds to, for instance, "red" or "antigreen"? (Like "positive charge" is $+e$, "negative charge" is $-e$). What is the mathematical interaction of red and antired (like positive and negative is $(+e)(-e)=-e^2$), and what happens when you apply that interaction to red and antiblue? (Like how electric charges interact with magnetic dipoles through their relative velocities).

If I had to point to a thing on paper and say "this here represents the red color charge", what would that thing be? Does such a thing even exist?

In short, what is color charge?

I've had abstract algebra and group theory and some intro courses on field theory and QED, but I don't know a lot of jargon, or really a lot of algebra.

Sorry the question's so long. Thanks for the future clarification!

share|improve this question
1  
Scratch my earlier comment, perhaps I see where you are going with this, but I don't think that you have expressed it very well. Are you asking about the mathematical structure of the strong force? –  dmckee Jan 8 '13 at 2:59
    
It may not pay to press to far for an analogy similar to charge in EnM and mass in gravity since in QCD (or any nonabelian gauge theory really) gluon number is not conserved. This means that nearby gluons can change the representation of a state. I don't claim to fully understand the ramifications of this, but see users.ictp.it/~pub_off/lectures/lns007/Strassler/Strassler.pdf. –  DJBunk Jan 8 '13 at 17:54
1  
Also note the gluon propagator just differs from the photon propagator by an identity matrix in color space. SO at high energies, or for a large number of flavors so that the theory doesn't become confining, where the tree level Feynman diagram dominates (a single gluon exchange)-in this limit the theory reduces to just N copies of Coulomb's law, one for each color. –  DJBunk Jan 8 '13 at 18:03
    
Photon number isn't conserved either in QED, though. So the fact that gluon number isn't conserved doesn't represent a radical departure from QED. Unless you meant how gluon number can change even without interactions with the charges? –  David Z Jan 8 '13 at 18:44
    
@DavidZaslavsky Photons don't carry charge though. Gluons carry color charge themselves (in the adjoint rep) so they can change the representations of the state. –  DJBunk Jan 8 '13 at 21:04

2 Answers 2

Try this on for size: color charge is a name for a set of three related charges, which are arbitrarily labeled red, green, and blue. Each of the individual charges works kind of like electromagnetic charge, in that you have positive and negative values: red and antired, green and antigreen, blue and antiblue. So it's kind of like a three-dimensional space of charge, with three independent charges on the axes. Thus the color charge of a particle would be represented by a vector $(c_1, c_2, c_3)$.

The one major difference that makes this not a regular 3D charge space is that an equal combination of all three charges is equivalent to no color charge at all. So you can imagine taking that 3D space of color charge and projecting it on to the 2D plane orthogonal to the neutral-color axis. That is, if you have a particle whose color charge is $(c_1, c_2, c_3)$, that's equivalent to the projection of that vector on to the plane perpendicular to $(1,1,1)$.

Now on to the details. I don't know offhand what the formula equivalent to e.g. Coulomb's law for the strong force would be; there is some very complicated math involved. But qualitatively, I can tell you that color charges always try to stay in neutral groups. (Singlets, in the language of group theory) For example, if you have red, green, and blue particles together, they will be very difficult to break apart. Similarly, red and antired will be difficult to break apart, so you could say they attract each other. If you put a red and an antiblue particle together, they don't form a color neutral pair, so I think there will be a bit of a repulsion, unless the two of them come together with a third particle that has the right color charge to make them group color-neutral (which would actually just be a composite of an antired and a blue). For certain, it's not as simple as just multiplying two numbers.

share|improve this answer
1  
David, I think you should clarify that "color" is a quantized measure, the way spin is, and elementary particle charge is. You do refer to charge but do not stress the quantized nature of color at the particle level. –  anna v Jan 8 '13 at 6:39
    
Your model then would make, for instance, red=$(1,0)$, blue=$(-\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})$, green=$(-\tfrac{1}{2}, -\tfrac{\sqrt{3}}{2})$, white=$(0,0,0)$, and the anti's negative. This seems to work. At least it produces the right results for the known color interactions. It kind of bothers me that it doesn't produce directly a factor like $q_1q_2$ or $\vec{p}_1\circ\vec{p}_2$ that gives the interaction strength, but at least it's abelian. It also bothers me that it doesn't form a group and isn't closed. But it does work. Thanks for answering. –  rboston Jan 8 '13 at 18:33
    
@rboston Remember that this is just an analogy, and you can't expect to get specific quantitative expressions from analogies. Plus, there's no reason to expect the charges to form a group. It's the transformations they are subject to that form the group $SU(3)$. –  David Z Jan 8 '13 at 18:42
    
@annav I don't know about that. As far as I know, quantization of color charge is something that is observed but not theoretically required for the theory to make sense. It's like EM charge, but unlike spin, in that sense. –  David Z Jan 8 '13 at 18:45
1  
It is the unfortunate choice of the concept "color" ,. In contrast with strangeness and charm it gives the impression that one might have a half blue half red quark, for example.whereas we know the charge is either +/-1/3 or+/-2/3 or +/-1 for elementary particles. anyway it is just my opinion. –  anna v Jan 8 '13 at 18:54
up vote 7 down vote accepted

I asked this question a few weeks ago and was dissatisfied with most of the answers I found on the internet, so I eventually managed to procure a copy of Griffiths' excellent text on elementary particles (really, all of his texts are excellent) which includes a section exactly answering my question with what I was looking for. I decided then to answer it myself, in case some other curious person reads this and wants to know.

This is just a very cursory explanation, intended to answer my own question to my own satisfaction.

Griffiths starts by introducing what are basically three copies of EM charge called color charge, and proposes these to be three-element column vectors: $$c_{red} = \left(\begin{array}{c}1\\0\\0\end{array}\right), c_{blue} = \left(\begin{array}{c}0\\1\\0\end{array}\right), c_{green}=\left(\begin{array}{c}0\\0\\1\end{array}\right).$$ These could in principle could take any vector value whatsoever, except for effects of symmetry in the theory and color confinement.

To figure out how these vector charges interact, we turn to the Gell-Mann $\lambda$-matrices, which are to $SU(3)$ what the Pauli matrices are to $SU(2)$. These are listed by Griffiths, but writing matrices would be a pain; you can look them up on Wikipedia.

Griffiths then takes Feynman scattering amplitudes in lowest order for the chromodynamic interaction, and from these develops potentials for various interactions.

For quark-anti-quark, he has $$V_{q\bar{q}}(r) = -f\frac{\alpha_s\hbar c}{r}.$$ This is a long-range force in principle, but it is made short-range due to confinement. It takes the same form as the Coulomb potential. The important thing here is the $f$, which Griffiths calls the "color factor". This color factor is like $q_1q_2$ in electrostatics or $\mathbf{p}_1\circ\mathbf{p}_2$ for dipole-dipole forces, and will depend on the color state of the interacting particles in question. It is calculated by $$f = \frac{1}{4} (c_3^\dagger\lambda^\alpha c_1)(c_2^\dagger\lambda^\alpha c_4),$$ where summation is implied over $\alpha$. Here $c_1$ is charge of incoming quark, $c_3$ charge of outgoing quark, and $c_2,c_4$ charges of incoming and outgoing antiquark.

As an example, Griffiths calculates the interaction between red and anti-blue. $$c_1=c_3=\left(\begin{array}{c}1\\0\\0\end{array}\right), c_2=c_4=\left(\begin{array}{c}0\\1\\0\end{array}\right).$$ Hence $$f = \frac{1}{4}\left[(1,0,0)\lambda^\alpha\left(\begin{array}{c}1\\0\\0\end{array}\right) \right]\left[ (0,1,0)\lambda^\alpha \left(\begin{array}{c}0\\1\\0\end{array}\right)\right] = \frac{1}{4}\lambda^\alpha_{11}\lambda^\alpha_{22}.$$ That is, it involves a sum over products of the 1st diagonal element and 2nd diagonal element o each of the Gell-Mann matrices. By looking at their form, the only matrices with both these elements non-zero are the ones labeled by Griffiths $\lambda^3$ and $\lambda^8$. These lead to $$f = \frac{1}{4}[(1)(-1)+(1/\sqrt{3})(1/\sqrt{3})] = -\frac{1}{6},$$ $$V_{r\bar{b}} = \frac{1}{6}\frac{\alpha_s \hbar c}{r},$$ which is evidently a repulsive force. Griffiths also calculates other interactions. For instance, quark-antiquark singlet interactions, $(1/\sqrt{3})(r\bar{r}+b\bar{b}+g\bar{g})$, which have color factor $f=\frac{4}{3}$ and thus are attractive, explaining confinement of quarks to color-singlet states and the lack of observation for colored states. He also calculates quark-quark interactions, which have a slightly different potential, $$V_{qq}=f\frac{\alpha_s \hbar c}{r}.$$ As an example, he calculates red-red interaction; it has factor 1/3, hence is repulsive.

There is a lot of this in this very wonderful book, but that's enough to satisfy my curiosity of what color charge is and how it works. Hopefully it is helpful to anyone else. Of course, this was highly simplified for the sake of my own simplified brain and no doubt infuriating to pedants in the field, but if you would like a better explanation and understanding, this was all taken from Chapter 8.4 of Introduction to Elementary Particles by David Griffiths, published by Wiley-VCH, Second Revised Edition -- just to cite sources.

share|improve this answer

protected by Qmechanic Jul 27 '13 at 17:13

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.