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Pion help

  1. Nov 15, 2004 #1
    Hi, i have a particle physics exam in 2 days and am confused on pion decay via the weak interaction, namely:

    pi[-] ->muon + anti-muno neutrino

    Thankyou
    Ray Veldkamp
     
  2. jcsd
  3. Nov 15, 2004 #2

    Tom Mattson

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    What's the question? :confused:
     
  4. Nov 15, 2004 #3
    indeed Tom, you are right, i don't get the point either...
    marlon
     
  5. Nov 15, 2004 #4
    Sorry guys i should have been more specific. In lectures we were shown to draw Feynnman diagrams to represent these weak interaction questions. My question is, how do the two quarks in the pion produce the muon and anti-muon-neutrino via virtual W+/W- exchange?
    In lectures we were told that emission or absorption of the W bosons will only change a particle to its doublet partner, so im not seeing how there can be no quarks after the interaction, unless some kind of annhilation takes place.

    Thanks
    Ray
     
  6. Nov 15, 2004 #5

    Tom Mattson

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    In the Feynman diagram, note that the u-bar quark is drawn as a u-quark going backwards in time. So the incoming d-quark and the outgoing u-quark do in fact react to produce a W boson, which decays into the products you mentioned. I wouldn't call this an "annihilation" of the quarks, since that term is normally reserved for the mutual cancellation of a particle with its own antiparticle to produce photons. In this instance, the quarks are destroyed to produce not a photon, but a W.
     
  7. Nov 16, 2004 #6
    Thanks that makes sense.
    I have one more question, concerning helicity and parity violation of the weak interaction. I understand that fermions participating in weak interactions are emitted and absorbed predominantly in negative helicity states, and we have seen that parity transformations change a negative helicity state to a positive helicity.
    What i don't understand, is that if we consider say an electromagnetic interaction, if a particle is in a negative helicity state, wouldnt a parity transformation change it to a positive, and hence the electromagnetic interaction would violate parity?

    Sorry if thats vague
    Thanks
    Ray
     
  8. Nov 16, 2004 #7

    Tom Mattson

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    Don't forget that the parity of the photon is negative. So if you emit a γ in an EM interaction, and there is a parity change in the material system, the overall parity is conserved. Remember that parity is multiplicative, so πparticleπphoton=(-1)(-1)=1.
     
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