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P p_bar annihilation

  1. Nov 4, 2009 #1
    In the first edition of Griffiths' Introduction to Elementary Particles, p. 129, I read:

    "In strong interactions, charge conjugation invariance requires, for example, that the energy distribution of the charged pions in the reaction p + p_bar -> [tex]\pi^+[/tex] + [tex]\pi^-[/tex] + [tex]\pi^0[/tex] should (on average) be identical."

    Griffiths gives reference C. Baltay et al, .Phys Rev Lett 15, 591, (1965). I looked it up. This paper appears to only uses the argument, it does not explain its origin.

    Could someone please explain how C-symmetry makes a prediction about the distribution in pion energies in this case? Does this example hint at a broader principle which makes a statement about the energy distribution of reaction products in the final state of a strong interaction which respects C-symmetry?

    Thanks!
     
  2. jcsd
  3. Nov 4, 2009 #2

    clem

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    The initial state is an eigenstate of C, so the final state will also be one.
    The operator C interchanges pi+ and pi-.
     
  4. Nov 5, 2009 #3
    So, is the principle the following:

    "The strong force respects charge conjugation. Because of this, in any strong force interaction where the initial and final states are their own charge conjugate and where the final state particles differ predominately by charge (the mass difference between the pion flavors is small) the final state particles should (on average) have energy equally distributed between the final elements because the strong force can't really "tell the difference" between the final state types of particles, so on average, equal energy should be given to all final state particles."

    Is that the right general line of reasoning?
     
  5. Nov 5, 2009 #4

    clem

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    There is no mass difference between pi+ and pi-, and the pi0 mass doesn't enter.
    "so on average, equal energy should be given to all final state particles."
    More than that, any measured distribution of pi+ must be the same as for pi-.
    pi0 can have a different distribution.
     
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