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