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A pion is made out of a pair of up and/or down quarks. A neutron or proton is three up or down quarks. So naively I'd expect a pion to be about 2/3 the mass of a nucleon.

In fact it's less than 1/6 the mass. Why is it so much lighter? Is there some reason why a lot less energy is contained in the gluon field of a pion than in the gluon field of a nucleon?

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Related: physics.stackexchange.com/q/9663/2451, physics.stackexchange.com/q/31407/2451 and answers therein. See also Wikipedia. –  Qmechanic May 4 '13 at 0:29
There was a very good post recently about the structure of protons and neutrons at profmattstrassler.com/articles-and-posts/… but, of course a Pion could be equally as complicated so your question about why it isn't is a good one. –  Brandon Enright May 4 '13 at 0:33
Well, this is one of the hints that the quark masses make up a very small part of the masses of the light hadrons. Then, in a very qualitative and hand-wavy way you can argue that because the gluons can interact with each other the glue mass scales faster than the number of quarks. After that you have to ask a theorist. –  dmckee May 4 '13 at 1:08
@dmckee: Nice! If gluons didn't interact, then the gluon field would scale like the number of quarks $n$, and its energy would scale like $n^2$. Presumably the exponent is even higher than 2 because of the gluon-gluon interactions...? –  Ben Crowell May 4 '13 at 1:21
As I said, that's all hand waving. Maybe the lQCD people can shed some light on the detailed mechanism. I'm at the limits of my weak little experimental brain. –  dmckee May 4 '13 at 1:36

1 Answer 1

Let me start off by saying that the "current" masses for the light quarks (up/down) are very small and about 1-4 MeV. The pions have masses of about 140 MeV and the proton/neutron have masses of about 1 GeV.

As you start at high energies and move to lower energies, the strong nuclear force increases in strength... till about $ \lambda_{QCD} \sim 250 MeV$ where the coupling constant blows up to "infinity" and you get confinement for the quark degrees of freedom and they form a condensate and also pick up a "constituent" mass of ~350 MeV each. That means that all the effective matter particles we observe must be colour neutral, with baryons having a mass ~350x3 MeV. Since the quarks condense, the $U(2)_L \times U(2)_R$ flavour symmetry among the light quarks gets broken down to a "diagonal" vector-like $SU(2)_V$ known commonly as isospin.

Note: This process is known as chiral symmetry breaking and it occurs non-perturbatively (since we're in a strongly coupled regime). So it's not yet completely understood.

The pions are (pseudo)Goldstone bosons of the broken axial $SU(2)_A$ symmetry. Since the up/down quarks have slightly different masses, the symmetry is not exact (hence the "pseudo"). But since their "current" masses are so damn light compared to the effective "constituent" masses generated by chiral symmetry breaking, the approximate symmetry is a very good approximation. This means the pions are kinda massless (compared to the mass of the baryons, i.e. the "constituent" masses). I don't have a slick explanation to get the pion mass.

HTH. The conceptual aspects are explained quite well, in more detail, at http://physics.stackexchange.com/a/17214/3998.

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Assigning the quarks a constituent mass a bit over $2m_\pi$ doesn't really help here (for all that it does a passable job of explaining the baryon spectrum): it simply leaves the question of why the pion is too light by a factor of four. –  dmckee May 4 '13 at 6:37
Well, if one works out the nonlinear sigma model, then the pion mass comes out propotional to the sum of current masses of the two quarks (and the condensate vev, divided by ${f_{\pi}}^2$ where $f_\pi$ is the pion decay constant). But like I said, I don't understand it well enough to explain the big picture without getting bogged down by the details. –  Siva May 6 '13 at 15:59

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