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Why is this decay allowed and another not?

  1. Mar 26, 2015 #1
    Consider ##\Sigma^0 \rightarrow \Lambda + \pi^0##. (Not Allowed)

    vertex1.png

    According to griffiths, this strong interaction is not allowed by 'Conservation of Energy'. I'm not sure why, as this simply shows an up and anti-up quark coming together, producing a gluon, where mass of gluon is mass of up and anti-up quark combined.

    Consider ##p + p \rightarrow p + p + p + \bar p##. (Allowed)

    Why is this possible by the strong interaction then?

    vertex2.png
     
  2. jcsd
  3. Mar 26, 2015 #2
    Consider this process in the rest from of the initial ##\Sigma^0## particle. What is the total energy of the initial ##\Sigma^0## particle? What is the minimum possible total energy of a final state containing a ##\Lambda^0## particle and a ##\pi^0## particle?

    This is not a diagram that contributes to ##p + p \to p + p + p + \bar p##. In your diagram, one of the protons is unaffected and might as well not be there. So effectively the process you have drawn is ##p \to p + p + \bar p##, which like your first example is not allowed because of energy conservation. (Also there is a more technical reason why this diagram is not allowed, which is that a single gluon cannot produce a proton-antiproton pair because of color conservation. You need at least two gluons).

    Here is a possible diagram for ##p + p \to p + p + p + \bar p##: pp_to_pppp.png
     
  4. Mar 26, 2015 #3

    Orodruin

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    Also, the reason a proton-proton collision can create an additional proton-antiproton pair is that the pp system is not a bound system and the protons can have a significant kinetic energy in the CoM frame (if they do not have enough, the process is forbidden).
     
  5. Mar 26, 2015 #4
    Put simply, the first one is a (forbidden) decay, whereas the second one is a collision
     
  6. Mar 31, 2015 #5
    That totally clears it up. Thanks!
     
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