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Higgs-Z ditau decays

  1. Feb 24, 2015 #1

    ChrisVer

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    I am having some hard time to understand how the Higgs-Z boson ditau decays can result in different momenta for the pion-products (in the very simple scenario where [itex]\tau \rightarrow \pi \nu[/itex].

    The decay angular distribution is given by:
    [itex]\frac{1}{\Gamma} \frac{d \Gamma}{d \cos \theta} \propto \frac{1}{2} (1+ P_\tau \cos \theta)[/itex]
    Where [itex]P_\tau[/itex] is the average polarization of the sample and [itex]\theta[/itex] is the angle between the momentum of the tau and the pion (I believe in the lab frame). For all left-handed taus [itex]P_\tau = -1[/itex] and for all right-handed taus [itex]P_\tau = +1[/itex].
    By using some kinematics one can find that:
    [itex] \cos \theta = \frac{2x -1 -a^2}{\beta_\tau (1- a^2)} \approx 2x -1 [/itex]
    Where [itex]a= \frac{m_\pi}{m_\tau}[/itex] and so we neglected its square, and [itex]\beta_\tau[/itex] is tau's velocity which can be taken to be [itex]\beta \approx 1[/itex] due to the large mass difference between Z/H and taus.
    So
    [itex]\frac{1}{\Gamma} \frac{d \Gamma}{d x} \propto 1+ P_\tau (2x -1)[/itex]

    From this last relation, how can someone deduce that a right-handed [itex]\tau^-[/itex] (left-handed [itex]\tau^+[/itex]) decays into hard pions while left-handed [itex]\tau^-[/itex] (right-handed [itex]\tau^+[/itex]) decays into soft pions?
     
  2. jcsd
  3. Feb 24, 2015 #2

    mfb

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    Every angle in the lab frame would depend on the tau energy. It could be the angle between pion motion and tau spin or something similar.

    What is x?

    One spin orientation (relative to flight) tends to decay via a forward emission of pions (high energy), one tends towards a backwards emission (low energy).
     
  4. Feb 24, 2015 #3

    ChrisVer

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    Sorry I forgot to define x ... [itex]x= E_\pi / E_\tau[/itex] the fraction of the energy of the tau carried away by the pion.
     
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