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Three particle pion decay

  1. Feb 8, 2010 #1
    hi,

    is the following decay possible?

    [tex]\pi^- \rightarrow u \bar{u} + \mu^- + \bar{\nu}_\mu[/tex]

    my idea: the quark content of [tex]\pi^-[/tex] is [tex]\bar u d[/tex] , the d quark could decay into an up-quark, emitting a [tex]w^-[/tex] boson which decays into the muon and anti-muon-neutrino.

    or is there any problem?

    (normally, you only see this decay: [tex]\pi^- \rightarrow \mu^- + \bar{\nu}_\mu[/tex] listed)

    --derivator
     
  2. jcsd
  3. Feb 8, 2010 #2
    Well, have you already checked to make sure the reaction conserves baryon number, lepton number, strangeness, etc.? If not, make sure the reaction satisfies all of the requirements, because from what you have shown in your post it doesn't look like you've tried that yet.
     
  4. Feb 8, 2010 #3
    as far as I can see, its all conserved:

    -there are no baryons -> baryon number trivially conserved
    -lepptonnumber is 0 on both sides
    - there are no strange quarks involved -> conserved
    - charge is conserved
     
  5. Feb 8, 2010 #4
    I would say that all looks conserved, except maybe for conservation of energy. Check to see if the annihilation of up and antiup quarks produces a nonconservation of energy.
     
  6. Feb 8, 2010 #5
    In charged pion decay to a muon and neutrino, there is only about 33.91 MeV energy available: the muon gets about 4 MeV and the neutrino gets about 30 MeV. There is insufficient energy for a quark-antiquark pair.

    Bob S
     
  7. Feb 8, 2010 #6
    Ya, that's what I was trying to get Derivator to do.
     
  8. Feb 9, 2010 #7

    Vanadium 50

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    At the end of the day, you need to do something with the quark-antiquark pair you created. You have two options - they can bind to form a neutral pion, or they can annihilate to form a photon. The first, as pointed out, is not energetically allowed with a muon. It is, however, energetically allowed with an electron, and the following decay occurs (but very rarely - about 11 times per billion decays):

    [tex]\pi^+ \rightarrow \pi^0 + e^+ + \nu[/tex]

    The decay with a photon is much more common, but of course the only thing that's observable is the final state - we don't (and can't) know whether the photon is from annihilation, or if it's from radiation off the incoming pion or outgoing muon.
     
  9. Feb 9, 2010 #8
    so beta decay, in contrast, is possible, beacuse the electron is way lighter than for example the muon?

    Which energy is avaible in beta-decay to create the up-quark, electron and neutrino? only the invariant-mass of the down-quark. if so, wich mass? the mass of the "naked" current up quark, or the mass of the constituent up-quark?

    -----------------------------------------------------------------
    @Bob S
    back to three-particle pion decay:
    (all masses/energies only approx)

    pion mass: 140 Mev
    mass of both up quark: 6 Mev
    mass of Muon: 106 Mev
    neutrino mass: negligible

    so (140-106-6)Mev=28Mev is left. why should there be any energy problem?
     
    Last edited: Feb 9, 2010
  10. Feb 9, 2010 #9

    Vanadium 50

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    Neither - the energy available is the mass of the parent (which must exceed the mass of the daughters).
     
  11. Feb 9, 2010 #10
    The problem is not the total available energy (33.91 MeV), but the energy available to recoiling particles with mass. Because of the neutrino's (near?) zero mass in pi-mu decay and the requirement to conserve momentum, the recoil muon kinetic energy is only about 4 MeV**, with the neutrino taking away nearly 30 MeV. So there probably isn't sufficient energy available for the additional quark-antiquark pair. A better possibility is to look at the 2-body decay process pi -> electron + neutrino (branching ratio about 1 x 10-4) and the 3-body process pi -> electron + neutrino + gamma (branching ratio about 1.6 x 10-7). Both these decay modes have much more free energy.

    Bob S

    **[added] In 2-body pion -> muon+ neutrino decay at rest, the muon kinetic energy is

    Tmuon = [(Mpic2)2 + (Mμc2)2]/(2Mpic2) - Mμc2 = 4.108 MeV

    assuming massless neutrino and using present best values for pion and muon masses.
     
    Last edited: Feb 9, 2010
  12. Feb 10, 2010 #11
    I understand your calculation for a two body decay. But in a three body decay, why can't I give every particle as much energy as it needs (of course not more than 33.91 MeV in total), so that the sum of there momentum vectors will cancle?

    Suppose, there was no free energy left. The three particle just won't move. That's no problem, since the pion had no momentum, too!
     
    Last edited: Feb 10, 2010
  13. Feb 10, 2010 #12
    Vanadium-50's suggestion in post #7 may be your best option. The u u -bar quark-antiquark pair must (I think) go via a pi-zero with isospin 1. (There is also an electron-neutrino-gamma decay mode?). Using an effective mass of the pi-zero + electron, the recoil kinetic energy is about 60 KeV. As V-50 says, the branching ratio is ~ 11 per billion.

    Bob S
     
  14. Feb 11, 2010 #13
    I see!
     
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