Past exam of the particle and nuclear physics course of the University of Cambridge / 2021 / Section B / 5 / a / Solution

In the pion rest frame, write its four-momentum as $P=(m_{\pi },0)$ and the charged lepton's as $p_{\ell }=(E_{\ell },\mathbf{p})$. By four-momentum conservation, the massless antineutrino has $k=P-p_{\ell }$, with $k^2=0$. ThereforeIts energy is $E_{\nu }=m_{\pi }-E_{\ell }$, and for a massless antineutrino, $E_{\nu }=|\mathbf{k}|$. The final momenta are equal and opposite, soThe expression is positive precisely when the two-body decay is energetically allowed, $m_{\pi }>m_{\ell }$, neglecting the neutrino mass.