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Homework Help: Meson decay

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    a) η decays into two photons, but not three. Through which interaction does the decay occur. What is C(η)?

    b) ρ0 decays into [itex]\pi^{+}\pi^{-}[/itex] but [itex]\eta^0[/itex] does not. Why is that?

    c) [itex]\omega[/itex] decays through electromagnetic interaction into [itex]\pi^0[/itex][itex]\gamma[/itex]. C([itex]\omega[/itex])?

    d) [itex]b_1(1235)[/itex] decays almost solely into [itex]\omega\pi[/itex] Reason what the isospin, parity ,C, and spin from b might be.

    2. Relevant equations
    conservation laws
    spin either 0 or 1 for mesons - is that correct ? I know that there are technically only to constituents but the flavour wave functions have weird structures.
    3. The attempt at a solution

    Would someone be so kind and help me through this? I still struggle to get my head into this.

    a) is the only one I am able to solve:
    the c parity has to be multiplied. the photon has a c parity of -1 and therefore C is 1 here.
    b) I am not sure whether I am allowed to look up the [itex]J^{PC}[/itex] value. I guess so, otherwise it's not possibly to solve ?
    The roh has 1++ and the eta has 0-+
    Using the formulas above I get l=1 for eta and l=0 for rho.
    The Parity is calculated via [itex](-1)^l\cdot P_1\cdot P_2[/itex]
    from the data booklet I get 0- for the pion thus parity(pion=-1) (is there a way to prove the parity by the way?)
    How do I get to L([itex]\pi^{+}\pi^{-}[/itex])?

    With c([itex]\gamma[/itex])=-1 and ([itex]\pi^0[/itex])=1 it should be c([itex]\omega[/itex])=-1*1=-1

    Omega has 1-- and Pion 0-+. The spin follows therefore 1 and 0.
    c(b)=c(omega)*c(pi)= -1.
    Again I have no idea how to calculate the angular momentum between the final products.
  2. jcsd
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