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I Nonexistence or Feynman diagram of a decay

  1. Apr 14, 2016 #1
    Hi everyone!

    In an exam on particle physics there has been a problem where for a number of decays we were asked to either reason their non-existence (i.e. name a conservation law that it contradicts to) or draw a Feynman diagram. However, with one of those decays I have a problem:

    [itex]\gamma + p \to \pi^0 + p[/itex]

    Could someone please tell me the right answer? I know it is not a good style to ask for a complete solution. Of course I would also be happy with a clue or a reference, but I don't see in what way it would be possible to give a clue here without telling the result.
    The problem is: I don't see any conservation law that it would contradict to: Lepton numbers (there are no leptons involved), quark flavor (is respected, because pi0 is up-antiup or down-antidown), electric charge (respected) are the only ones that can be checked without a concrete Feynman diagram.
    On the other hand, I cannot imagine how a Feynman diagram should look like. What the hell shall I do with the photon?
    There is another conservation law here, 'Fermion current' or 'fermion flow' (I am not sure about the English term; the German term was "Fermionenstrom"), of that I don't really know what it means. Perhaps, is this the solution?

    Maybe there is a simple principle that I don't understand yet. The lecture was just a very superficial introductory lecture, no QFT.
     
  2. jcsd
  3. Apr 15, 2016 #2

    ChrisVer

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    Feynman diagrams are quiet artistic...
    You can decide how to do that: for example you can make the photon be emitted by any quark [quarks are charged so they can interact with photons at vertices] and that photon together with your incoming one will produce the [itex]\pi^0[/itex]... like reverting the diagram [itex]\pi^0 \rightarrow \gamma \gamma[/itex]. I believe using gluons could also be a possibility...
     
  4. Apr 15, 2016 #3
    Thanks, that helped me a lot!
    Indeed, I didn't consider coupling the photon with quarks. ;)
     
  5. Apr 15, 2016 #4
    I believe I have something to add. Do I simply have [itex] uud \to uud [/itex] and the [itex]\gamma \to u \overline{u}[/itex] individually? Or should I take one up quark (or down, respectively) from the proton? Because if I would just leave the proton as it is then it would not at all make sense to me to mention it in the decay. And also, when I want to calculate a cross section, the proton would have no effect, since there is no vertex.
    This is probably more a question about definitions/conventions than on physics.
     
  6. Apr 15, 2016 #5

    mfb

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    ##\gamma \to u \overline{u}## is not possible on its own, that violates energy/momentum conservation. You need at least some momentum exchange with the proton. You can also take a quark from there, but that is not necessary.
     
  7. Apr 15, 2016 #6
    Aaah! That's why the diagrams are always drawn with curved lines! I was wondering about that the whole semester.

    Thank you so much! That was a very valuable insight for me!
     
  8. Apr 15, 2016 #7

    ChrisVer

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    I don't understand this question, could you rephrase it? which photon are you referring to?

    No, they are drawn in curved lines because it's nicer... and sometimes the vertices can be seen clearly...
    Feynman Diagrams is just a nice/clear way to illustrate the amplitudes of an interaction ...they don't show a physical picture. As amplitudes adding them and taking the magnitude square gives you the cross section.
     
  9. Apr 15, 2016 #8
    Sorry, I don't know how to make a proper quote.

    As to your first question: I wrote proton, not photon. :) I was wondering whether I can just draw a Feynman diagram leaving the proton's quarks as they are. For me, it made no sense to write down +p on both the left and the right side of the 'equation' if the proton doesn't really participate (and therefore, if the corresponding Feynman diagram would be accepted as a solution to the problem in the exam). But mfb just explained to me that the proton is important for the process because of the recoil.

    I am and was aware that Feynman diagrams don't really show a physical picture. But I think that the professor wanted to indicate the real relations as much as possible throughout the lecture. Okay, this is speculation, I might as well be wrong here. It is not really important, anyway.

    Once again, thank you! I'm glad I asked the question here in the board, it was very insightful!
     
  10. Apr 15, 2016 #9

    ChrisVer

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    It's 1 am already, and at such times I always read protons as photons and the other way around.

    yup

    yup. If you want to put it differently, the curves don't really matter because for each propagator you are integrating through the whole possible points of the vertex, as a result you have all possible curves available (Even straight lines). So curves don't show anything in particular
     
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