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Proton Collision Feynman Diagram

  1. Apr 8, 2015 #1
    Supposed to represent a relativistic proton colliding with a stationary proton, leading to changes in the momentum of both and the production of a neutral pion. The pion then decays into two photons.

    No clue if this is right. I've never drawn anything much more complicated than electron-positron annihilations. Any help is appreciated.

    ImageUploadedByTapatalk1428543224.371581.jpg
     
  2. jcsd
  3. Apr 8, 2015 #2

    Simon Bridge

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    You need to make sure that your particles travel in time ... one of the product protons, for eg, travels in the -space direction and backwards in time while one of the u's appears to be stationary in the center of mass frame of the initial protons. Hmmm... don't interactions involving triangles usually sum to zero?
    When you go ##u \to u + \gamma## ... are you saying that the particle spontaneously emits a photon? And why is the second u travelling backwards in time?

    You are basically drawing a position-time graph ... you've been doing those since secondary school - you now how to do those.
     
  4. Apr 8, 2015 #3

    Vanadium 50

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    I'm sorry, but what you are drawing is not a Feynman diagram. It doesn't conserve charge, and there's an arrow on the pi0.
     
  5. Apr 9, 2015 #4
    Draw each fundamental fermion as a separate line with an arrow. Don't draw hadrons as single lines. Make sure the arrows point "backwards" for antifermions.

    I've redrawn the π0 decay since you pretty much got it right, other than the single line for the meson itself. The oval won't be necessary on your diagram, and I didn't care for space and time axes.
     

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  6. Apr 9, 2015 #5

    mfb

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    @Simon Bridge: the pion decay is fine, apart from the drawing of the pion as mentioned before.

    The proton/proton collision process does not make sense. Proton plus proton cannot give pion+Z. And the "space" axis is questionable at best. You don't know the directions the particles will fly to, and you don't want to fix it in a sketch like this.
     
  7. Apr 9, 2015 #6
    Any better?
     

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  8. Apr 9, 2015 #7

    mfb

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    That is an unlikely, but possible process.
     
  9. Apr 9, 2015 #8
    From NASA:
    "In gamma-ray astronomy, "particle-particle collision" usually means a high-energy proton, or cosmic ray, strikes another proton or atomic nucleus. This collision produces, among other things, one or more neutral pi mesons (or pions). These are unstable particles that decay into a pair of gamma rays."
    I have found several other sources as well that indicate it is not a completely rare outcome from a high energy p-p collision. Maybe I'm wrong though.
     
  10. Apr 10, 2015 #9
    Pions from p-p collisions are more likely to come from the decay of high mass baryon states (deltas etc. etc.) than from the kind of process you've drawn.
     
  11. Apr 10, 2015 #10

    mfb

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    I did not say pion production was rare, but pion production from two photons is. A production via the strong interaction is much more likely.
    If the energy is sufficient most pions come from hadronization itself, not from hadron decays afterwards.
     
  12. Apr 10, 2015 #11
    ??
     
  13. Apr 10, 2015 #12

    ChrisVer

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    I'd go with what mfb said in his last post... it's a strong interaction that will give the pion...
     
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