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I Positron/Proton Scattering Weak Charged Current

  1. Mar 23, 2017 #1
    Hello,

    I'm wondering how you would go about drawing a Feynman diagram for a weak charged current scattering interaction of a positron and a proton. I have attached a diagram of what I have tried but it doesn't conserve lepton number (I think this is a problem, from what I gather non-conservation of lepton number is fringe stuff). It started out with a lepton number of -1 and finishes with +1.

    Gijpo1B.png

    I know it is possible to draw a p/e+ scattering diagram, the question I am looking at asks for one. Would scattering imply that you get e+ and p coming back out, possibly with other particles so long as no conservation laws are violated? Would it still be 'scattering' if one or both of these is missing from the products, say I got a neutron or some mesons coming out instead of a proton?

    Thanks

    edit - oh I'm dumb the first neutrino given off should be an antineutrino. Never mind. Anyway, is this what you'd draw as a diagram or is there something simpler?
     
  2. jcsd
  3. Mar 23, 2017 #2

    mfb

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    While that is a possible diagram (with the fixed neutrino->antineutrino of course), it is extremely unlikely, and it is probably not what the problem asked for. The first half is a charged current interaction already, there is no need to produce an additional electron/neutrino pair.
     
  4. Mar 23, 2017 #3

    Orodruin

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    The first interaction is not correct, the arrow on the neutrino line is in the wrong direction. The outgoing neutrino should be an anti-neutrino.
     
  5. Mar 23, 2017 #4
    Thanks, so yeah I was thinking how unlikely this would be too. But then again, you would be left with three up quarks which can't form a hadron, something else would have to happen, and I'm not sure what.
     
  6. Mar 23, 2017 #5

    mfb

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    uuu can form a hadron.

    It is also possible to make a box diagram with two W in the box and proton plus positron in the final state. That is more scattering-like, but it is negligible compared to the simple photon exchange.
     
  7. Mar 23, 2017 #6
    Ah, so they do. I hadn't realised the delta^++ was considered as such, thanks. re the box diagram - it's fine that it's suppressed, that's what the question is getting at (it's easy to draw scattering for Z^0 and a photon, the latter of which has highest amplitude as you say).
     
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