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Muon decay & quark mass

  1. Sep 5, 2011 #1

    I was looking at the quark masses and had this doubt regarding the muon decay: Why can't the muon decay into up and down quarks (plus neutrino). I know the explanation is because muon doesn't exhibit hadronic decays because its mass isn't big enough [itex]m_\mu=107MEV[/itex]. But I just saw that the up and down quark masses are [itex]m_u=2.4MEV[/itex] and [itex]m_d=4.8MEV[/itex]. Being much smaller than the mass of the muon, why isn't this hadronic decay possible.

  2. jcsd
  3. Sep 5, 2011 #2


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    salparadise, The quark masses you're quoting are the current or bare mass. You can't produce quarks singly, they must appear in a colorless combination, and when they do the mass they carry is the constituent or dressed mass, which is considerably heavier. See these terms on Wikipedia.
  4. Sep 5, 2011 #3


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    Indeed the decay is a decay to [itex]\pi^-[/itex], and a good question is why does the pion decay to muon instead of the inverse, muon to pion? Or, why is the mass of the pion barely higher than the muon?
  5. Sep 5, 2011 #4


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    Fundamentally, muons are leptons (as are tauons) so they cannot decay into quarks irrespective of mass.
  6. Sep 5, 2011 #5
    The dominant decay of the tau is actually hadronic; so, you might want to rethink that.
  7. Sep 5, 2011 #6
    That's because quarks and gluons cannot get more than about 10-15 m (1 fermi) away from each other -- that's quark confinement.

    That means that they get a wavefunction wavelength of about that amount, implying a total mass of a few hundred MeV -- more than the muon mass.

    This accounts for the masses of most of the light-quark hadrons, with the partial exception of pions and kaons. In their case, the masses are ~ sqrt(mquarks*mQCD). But even then, the up and down quarks are massive enough to make pions more massive than muons.
  8. Sep 5, 2011 #7
    I understand. So, the threshold energy for having hadron production in a process that has only leptons as input particles is the pion mass. This is minimum energy required for producing a new pair of quarks.

    I believe you're confusing strong interaction with weak interaction. It is true that leptons do not interact strongly via the exchange of gluons. But hadrons do feel weak interaction, which is responsible for tau decays into hadrons, via the exchange of the W boson.

    Interesting to see your estimation for the mass of a bounded system of quarks, obtained from their separation wavelength. This follows the same argument used by Yukawa to estimate pion mass from bounded nucleon systems. But, QCD is much more complicated than OPE, and you estimate the total system mass, not just that of the bosons that mediates it... although, by comparison with the undressed valence quarks mass, I see that most of the pions mass must come from gluons and sea of quarks, .
  9. Sep 5, 2011 #8
    And valence-quark kinetic energy. For light quarks in hadrons, their kinetic energy >> their rest masses.
  10. Sep 6, 2011 #9


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    You're right. My bad.
  11. Sep 8, 2011 #10
    Might also be useful to think in terms of Feynman diags. The u and d quarks will play a role at higher loop orders. The muon goes to W + neut. Now the W can go to a u d pair but they do not have enough energy to hadronise so the pair just exist as a virtual loop which go back to W which then decays leptonically.

    So the u and d quark do play a role, just as virtual particles at higher loop orders.
  12. Sep 8, 2011 #11

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    I'm not sure it's helpful to discuss loop effects, as tree level calculations are good to half a percent here.
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