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

The variable initial energy $E_0$ replaces $m_{\pi }$ in the kinematic expression before differentiation. The charged-lepton mass remains fixed:Thus the required Jacobian determinant in this one-dimensional change of variables isEquivalently, $E_0=\sqrt{p^2+m_{\ell }^2}+p$ gives $d{E}_0/dp=1+p/E_{\ell }$, so $dp/d{E}_0=E_{\ell }/(E_{\ell }+p)=E_{\ell }/m_{\pi }$, the same result. It is dimensionless in natural units.