I know a baryon is only stable when it contains a quark of each color. And as far as I know, the gluon essentially changes the color of a quark and moves onto the next, and this is what holds the particles together. But in the process of the gluon moving from one quark to the next, wouldn't the baryon have two quarks of the same color, making it unstable? Or does the gluon move instantaneously, or is the baryon not unstable enough to decay before the gluon reaches the next quark? Or... essentially, how does this process actually work?


3 Answers 3


The idea that baryons contain three quarks is a significant oversimplification wrong. It works for some purposes, but in this case it causes way more confusion than it's worth. So you should stop thinking of baryons as groups of three quarks and start thinking of them as excitations in quantum fields - and in particular, excitations in all the quantum fields at once. Quark fields, gluon fields, photon fields, and everything. These excitations propagate through spacetime and convert among each other as they go, and in a baryon, the propagation and mutual conversion happen to sustain each other so that the baryon can exist as a coherent particle for a while.

One of the conditions required of all these excitations in fields is that they be a color singlet, which is the strong interaction's version of being uncharged. There's a simple intuitive justification for this: just as an electrically charged particle will tend to attract oppositely charged particles to form neutral composites (like protons and electrons attracting each other to form atoms), something which has the charge associated with the strong interaction (color charge) will attract other color-charged particles to form neutral composites (color singlets).

Now, if you literally only had three quarks, the only way to make them a color singlet is to have one be red, one be green, and one be blue.1 (Or the anticolor equivalents.) But with all the complicated excitations that make up a baryon, there are all sorts of ways to make a color singlet. You could have three red quarks, a green-antired gluon, and a blue-antired gluon. Or two red quarks, two green quarks, an antiblue antiquark, a blue-antired gluon, and a blue-antigreen gluon. Or so on; the possibilities are literally infinite.

The point is that you don't actually have to have a quark of each color in the baryon at all times. Only the total color charge in the baryon matters.

Given that, it should seem reasonable that gluons change the color of quarks whenever they are emitted or absorbed, in a way that keeps the total color charge the same. For example, a blue quark could absorb a green-antiblue gluon and become a green quark.

1I'm glossing over some quantum-mechanical details here; specifically, a color singlet wavefunction needs to be an antisymmetrized linear combination, like $\frac{1}{\sqrt{6}}(rgb - rbg + gbr - grb + brg - bgr)$, not just $rgb$. But as long as you don't worry about which quark is which color, for purposes of this answer it's safe to ignore this.

  • $\begingroup$ Makes sense, but does the gluon-bouncing idea still work? I assume that in order to change the quarks color (or even change the structure of the baryon) that would require a gluon, or something, to be absorbed in the process, just like in your example with the blue quark and the green-antiblue gluon. If so, wouldn't that make the gluon-bouncing idea infeasible, since the gluon would just be absorbed instead of "reflected" so to speak? $\endgroup$
    – Vedvart1
    Commented Jun 8, 2017 at 4:25
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    $\begingroup$ Reflection is really just absorption followed by emission. Gluons are constantly being absorbed and emitted, so I suppose you could say that in some cases they're reflected. $\endgroup$
    – David Z
    Commented Jun 8, 2017 at 4:37
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    $\begingroup$ I'm sure you know this David Z, but this answer is an oversimplification of the way color works. There are eight color charges not three and we can't ignore quantum mechanical superpositions. If we have say a red up quark, a blue down quark, and a green strange quark, that is not a pure singlet. Part of it is charged. Similarly any particular color combination like red and antired is not a pure singlet and has a charged component. $\endgroup$
    – octonion
    Commented Jun 8, 2017 at 4:46
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    $\begingroup$ @octonion Indeed I know this is an oversimplification, but IMO an appropriate one for the level of this question. $\endgroup$
    – David Z
    Commented Jun 8, 2017 at 4:52
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    $\begingroup$ +1 for "neutral composites" - great way of intuitively introducing colour singlets by analogy to electrical charge. $\endgroup$
    – Myles
    Commented Jun 9, 2017 at 11:06

The model you are thinking about is really rudimentary and cannot explain the dynamics of Quantum ChromoDynamics, QCD .

In this link there is a better exposition of what a proton is, within QCD.

enter image description here

You may have heard that a proton is made from three quarks. Indeed here are several pages that say so. This is a lie — a white lie, but a big one. In fact there are zillions of gluons, antiquarks, and quarks in a proton. The standard shorthand, “the proton is made from two up quarks and one down quark”, is really a statement that the proton has two more up quarks than up antiquarks, and one more down quark than down antiquarks. To make the glib shorthand correct you need to add the phrase “plus zillions of gluons and zillions of quark-antiquark pairs.” Without this phrase, one’s view of the proton is so simplistic that it is not possible to understand the LHC at all.

In the over all counting balance the proton is color neutral and it is the huge number of gluon exchanges that generate the potential that keeps the three valence quarks tied up as a proton. Because it is a quantum dynamic framework the mathematics is not simple and has to be approximated by lattice QCD .

But in the process of the gluon moving from one quark to the next, wouldn't the baryon have two quarks of the same color, making it unstable

The stability is ensured by the innumerable exchanges between all those constituents. Color transfer from one quark/antiquark is compensated by the color remaining behind, the overall remaining color neutral.

how does this process actually work

It is essentially a quantum mechanical process, could be imagined by a summation of an infinite number of Feynman diagrams of this type exchanging gluons

gluon exchange

The over all colors do not change adding up to color neutral for the whole bag.

To be fair one should explain that there do exist distributions within the proton that show the existence of valence quarks and sea quarks and gluons in the parton distributions, but that is another story.

parton distributions

Figure 2: Overview of the CTEQ6M parton distribution at Q = 100 GeV

The enhancement on large x is the "proof" of valence up and down quarks.

  • 3
    $\begingroup$ The LHC site mentions that in the "sea of quarks" that makes up a proton, after accounting for all the quark-antiquark pairs, you end up with a the net result of the simplified 3-quark proton. Is this just an approximation that usually works, and if so why is it so precise, i.e. why not 0 net quarks or 6? And if it isn't an approximation, why is it so specific given the unimaginable number of quarks in there; what physical property makes this balance so exact? $\endgroup$
    – Vedvart1
    Commented Jun 8, 2017 at 5:18
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    $\begingroup$ @Vedvart1 Sea interactions conserve (quarks-antiquarks)/3, the "baryon number". In theory you could go up to 4 valence quarks + 1 valence antiquark, but that causes meson emission. (I'm simplifying 2 things by neglecting pentaquarks and baryon number conservation violation, both of which are rare, especially at low energies.) $\endgroup$
    – J.G.
    Commented Jun 8, 2017 at 5:48
  • $\begingroup$ @J.G. Why do they conserve that number then? Is it just something we've found experimentally that we essentailly accept as an axiom of sorts? If there is something causing this, do we know what it is? $\endgroup$
    – Vedvart1
    Commented Jun 8, 2017 at 5:50
  • $\begingroup$ Yes , baryon number conservation, charge conservation are basic observational laws incorporated in the standard model of particle physics, with a number of more esoteric quantum number conservations. The model calculations incorporate all conservation laws. $\endgroup$
    – anna v
    Commented Jun 8, 2017 at 5:54
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    $\begingroup$ @Verdvarti Further to anna v's explanation, in the Standard Model baryon number conservation violation can occur only non-perturbatively, but we think its true occurrence is higher based on the Sakharov conditions' explanation of nature's matter-antimatter asymmetry. $\endgroup$
    – J.G.
    Commented Jun 8, 2017 at 6:13

A short answer that's useful when thinking of individual quarks works:

The gluon also has color. There are 8 different forms, to go with the 3 colored quarks.

So if you start with a red and blue pair of quarks, the red changes to blue and you have two blue quarks and a gluon that’s red and anti-blue. Then the second quark absorbs it, cancels out the blue/anti-blue and is left with red itself.

Now the exchange between quarks is with virtual gluons, so don’t worry about it knowing where it’s going ahead of time. It’s the same issue you have in explaining how photons carry the momentum beween charged objects.

  • $\begingroup$ Could you explain exactly what virtual particles are? From what I've gathered, they're just particles from a process with one input and two outputs, regardless the cause. $\endgroup$
    – Vedvart1
    Commented Jun 9, 2017 at 3:48
  • $\begingroup$ No, I won't address what should be a new physics question in the comments. But look: it's already been asked here. Others are mentioned and linked from there. $\endgroup$
    – JDługosz
    Commented Jun 9, 2017 at 9:37
  • $\begingroup$ Could it be explained like RQ + BQ >>> BQ + BQ + RABG >>> BQ + RQ? Being RQ = red quark, BQ = blue quark, RABG = red anti-blue gluon. $\endgroup$
    – Ender Look
    Commented Jun 14, 2017 at 2:29
  • $\begingroup$ @EnderLook that looks right. $\endgroup$
    – JDługosz
    Commented Jun 14, 2017 at 2:38

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