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Feynman Diagrams!

  1. Aug 20, 2012 #1
    Feynman Diagrams! Please Help!

    Hi guys!

    I've been asked to draw a scattering and annihilation Feynman diagram for the following scenario:

    e+e- → VeVe (The second Ve is an anti-neutrino just couldn't figure how to put the bar on! :P

    I've tried it and I'm convinced the mediating operator is a photon via the electromagnetic force but I don't see where it can produce two neutrinos :S Unless it's a different mediator?

    Anyways any help would be appreciated!

    thanks guys!

    Karly xoxo
     
  2. jcsd
  3. Aug 20, 2012 #2

    tiny-tim

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    Hi Karly! :smile:
    But can neutrinos feel the electromagnetic force? :confused:
     
  4. Aug 20, 2012 #3
    Re: Feynman Diagrams! Please Help!

    No because they have no charge :) do both sides have to be affected by the mediator? That might be a stupid question I'm sorry but I don't know

    If they do then this must be via a W boson?

    Am I right?

    K xoxo
     
  5. Aug 20, 2012 #4

    tiny-tim

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    any vertex a fermion (in this case, a neutrino) is joined to must have a boson that "mediates" a force that the fermion feels
    W or Z (or both) … you decide! :wink:
     
  6. Aug 20, 2012 #5
    Re: Feynman Diagrams! Please Help!

    But the total charge on the LHS is zero, as is the charge on the RHS so surely it's a Z boson?

    Or am I being silly?

    Karly xoxox
     
    Last edited: Aug 20, 2012
  7. Aug 20, 2012 #6

    tiny-tim

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    i don't like this LHS RHS way of looking at it :redface:

    this is an exercise on Feynman diagrams, so look at the vertices

    what happens if the two electrons are joined to the same vertex?

    what happens if the two electrons aren't joined to the same vertex? :wink:
     
  8. Aug 20, 2012 #7
    Re: Feynman Diagrams! Please Help!

    Well surely if both the electron and the positron are joined to the same vertex then they have annihilated into a mediator, no? And if they are not then they have simply scattered off each other via the exchange of a mediator?

    God I hate this, I know I'm being really stupid but thank-you for your help <3

    K xoxo
     
  9. Aug 20, 2012 #8

    tiny-tim

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    yes and yes :smile:

    but i still think it would be easier if you stuck to considering each individual vertex …

    in the first case (the stick-man diagram), there are two ingoing fermions at one vertex, and two outgoing fermions at the other vertex, and in each their charge adds to 0

    in the second case (the H-diagram), there is one ingoing fermion and one outgoing fermion at each vertex, and in each case their charge difference is not 0

    sooo … ? :smile:
     
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