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Can massless quarks form massive hadrons?

  1. Oct 31, 2008 #1
    It is clear that the mass of the valence quarks is only a small fraction of the mass of the hadrons (for example, the proton).

    However, I wonder if it would be possible to get massive protons in QCD if the quarks were truly massless.

    On one hand, for massless quarks, the pions, as Goldstone bosons of an exact Chiral symmetry, would be massless.

    On the other hand, the binding energy of any electron-positron bound state would vanish in the limit of zero electron(positron) mass.

    Is there some general argument to exclude in QCD finite masses for systems of masless quarks?
    In the case of glueballs (made of massless gluons), do they have finite mass?
  2. jcsd
  3. Nov 2, 2008 #2


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    QCD is a funny theory - it exibits asymptotic freedom, and (presumably) there is a mechanism of "chiral symmetry breaking" that gives a vev to the quark bilinear of order a few hundred MeV. It is THIS scale that generates masses for the hadrons! This leads to a very amusing statement: if there were no Higgs at all, the W/Z bosons would be 1 GeV, NOT massless (this is a famous qualifying exam question for particle physicists looking to get a PhD!).

    Of course, now you ask - where does this chiral symmetry breaking come from? Answer that, and you are on your way to winning $1,000,000 from the Clay Math Institute!

    So in answer to your question: the mass of the hadrons presumably comes from the (as yet not well understood) QCD dynamics that generate a mass gap - quark mass is a relatively small contribution compared to this. So yes: glueballs are massive. Proof of this will win you lots of $$ ;-)

    Hope this helps!
  4. Nov 3, 2008 #3
    Thank you for your answer. There are a couple of things from it that i do not understand yet:

    What is a "vev" ?

    Why the W/Z bosons, which are associated to weak and electromagnetic interactions, should remain massive if there was no Higgs? What is the link between QCD and W/Z bosons?
  5. Nov 3, 2008 #4


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    vev = "Vacuum Expectation Value" - this is the thing that breaks the symmetry. Take a look at any book or website that describes "Spontaneous Symmetry Breaking" for more details.

    This is subtle, and perhaps I should not have mentioned it if it confuses you. The thing is - the spontaneous chiral symmetry breaking that happens in QCD also breaks the electroweak symmetry (since EW theory is a chiral gauge theory) - it wasn't planned, it was just an accident, but there you are. Therefore it would contribute to the W and Z masses. If they were 1 GeV each, we would have a complete theory of electrweak physics! However, they are 2 orders of magnitude heavier than that, so this cannot be the only source of the symmetry breaking. These ideas are the basis of an alternative theory called "technicolor" which is just like QCD with the scale cranked up to a TeV. Unfortunately, this theory does not describe the precision measurements that we have been making since the 1980's, so it probably isn't correct (at least in its naive form).

    So the short answer is: QCD has nothing to do with W/Z masses in the world that we live in. But it could have.
  6. Nov 10, 2008 #5
    vev is the vacum expectation value as mentioned before,but breaks the symmetry only when taking non zero value
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