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].(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Higgs-Z ditau decays

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