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Decay angle

  1. Sep 22, 2009 #1
    Hi,

    I have a question about decay angle. For example,
    Let us consider the decay

    [tex]\Lambda \rightarrow p + \pi^{-}[/tex]

    Here, [tex]\Lambda[/tex] and [tex] p [/tex] are spin 1/2 and [tex]\pi^{-}[/tex] is spin 0. So, spin angular momentum is conserved. So, should not the cosine of angular distribution of [tex] p [/tex] ( or [tex] \pi^{-} [/tex] ) in center of mass frame of [tex] \Lambda [/tex] be flat?

    If the spin angular momentum is not conserved ( like in [tex] \rho \rightarrow \pi^{+} \pi^{-} [/tex] ) then total angular momentum will be conserved because pions system has orbital angular momentum. So, the cosine of angular distribution of one of the pions in center of mass frame of [tex] \rho [/tex] is not flat.

    So, if the spin angular momentum is conserved and there is no orbital angular momentum then should not the cosine of angular distribution of one of the decay product ( two particle decay) in mother reference frame be flat?

    Thanks.
     
  2. jcsd
  3. Sep 23, 2009 #2

    clem

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    Because parity is not conserved in hyperon decay, the final state is a mixture of orbital angular momentum L=1 and L=0. In each case L+S=1/2 so total angular momentum is conserved. The cos\theta dependence of the angular distribution is evidence that parity is not conserved in this decay.
     
  4. Sep 23, 2009 #3
    Thanks clem.

    So, angular dependence comes from non-zero orbital momentum of the final particles, and given by spherical harmonics? In case of zero orbital momentum, angular distribution should be flat? Is the distribution integrated over the z-components of the momentum (m for given J)?
     
  5. Sep 23, 2009 #4

    clem

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    The angular distribution is summed over m_J. m_L and m_s are summed over, weighted by
    Clesch-Gordan coefficients (to make an eigenstate of J). It turns out that pure L=1 or pure L=0 for the orbital angle momentum gives a flat distribution. It is the interference between the two that gives a cos\theta dependence.
     
  6. Sep 23, 2009 #5
    Now, I am confused.

    If parity is conserved (gives L=0) then angular distribution should be flat. If parity is not conserved and L=1, then cos(theta) term will come in the spherical harmonics and as a result in the amplitude. So, even for pure L=1 state, should not be there angular dependence?

    What about weak decay (parity does not conserve)?

    Thanks.
     
  7. Sep 23, 2009 #6

    clem

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    It turns out that the pure state J=1/2, L=1 still has no angular dependence.
    The decay Lambda--> p + pi is a weak decay. Only weak interaction violate parity conservation. All beta decays are also weak and violate parity conservation.
     
  8. Oct 2, 2009 #7
    Sorry for the offtopic, could you please point me to the source of information I can read about the angle dependence on the spin, angular momentum etc.

    Thanks
     
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