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Sigma plus decays

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data
    The question is to draw possible Feynman diagrams of the following decay.
    Sigma(plus) ----> proton + photon



    2. Relevant equations
    Quark content of sigma(plus) is uus
    Quark content of proton is uud


    3. The attempt at a solution
    I know that the decay has to proceed by the weak interaction. After all strangeness is not conserved. So the the strange quark in the sigma plus must first decay into a W(minus) and an up quark. The W- will then produce a down quark plus and an anti-up quark.

    With this I will now have the uud for the proton...which is good but I will have an up quark and an anti up quark left over. These can annihilate each other to form 2 photons but not one (due to conservation of momentum). I can see no way of getting rid of this "extra" photon. Do you think its a typo?
     
  2. jcsd
  3. Sep 15, 2008 #2

    Vanadium 50

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    No, it's not a typo.

    You say they can annihilate each other to form 2 photons but not one (due to conservation of momentum). Do you know what a Feynman diagram is? It is not a step by step description of the process - "first this decays, then these pieces interact", etc. is most assuredly not what a Feynman diagram is supposed to represent, and it sounds like you think it is.

    It's best to get this misconception cleared up sooner rather than later. I'd make an appointment with my professor so he or she can clear this up. It should take an hour or so.
     
  4. Sep 15, 2008 #3
    I see what your saying but the reaction, sigma(plus) ----> proton + photon, is a physical process (at least according to the question) and should obey the physical laws like conservation of momentum. So the Feynman diagram should have a sigma(plus) coming in and proton and a photon going out for the external lines of the diagram. So far, I see nothing wrong with the Feynman diagram except that I get a proton and two photons coming out for the external lines. I can see no way of drawing my diagram to get a single photon coming out as an external line by drawing correct vertices.
     
  5. Sep 16, 2008 #4

    Vanadium 50

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    <sigh>

    Like I said, It's best to get this misconception cleared up sooner rather than later. I'd make an appointment with my professor so he or she can clear this up. It should take an hour or so.

    If you want to keep struggling, and see how permanently you can ingrain in this misunderstanding before removing it, hey, be my guest. The job market is plenty tight as it is.

    By the way, you switched from teh quark to the hadron representation between the first and third messages. In the hadron representation, you have the sigma in, and a proton and photon out. In the quark representation, you have it as you described above, but with one photon, not two.
     
  6. Sep 23, 2008 #5
    I did not have a misconception of what a Feynman diagram. I was simply not seeing the way to draw it correctly and it was an easy clarification when I asked my professor. It appears you did not understand my question.
     
  7. Sep 23, 2008 #6

    Dick

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    As free on-shell particles, u+ubar->gamma can't happen. But it's a perfectly valid internal vertex in a feynman diagram where all of the lines aren't external (and the particles can be off-shell). I don't think that would have been terribly hard to explain and shouldn't take a one hour conference.
     
  8. Sep 25, 2008 #7
    Exactly. That was all I needed to hear and it was a simple fix. I considered doing that vertex but at first it didn't seem right. My professor took all of five seconds to answer my question and it immediately made sense. Thanks.
     
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