Past exam of the particle and nuclear physics course of the University of Cambridge / 2014 / Paper 1 / 3 / b

Let $q=(m_C^2-m_B^2)/(2m_B)$ and $p=(m_B^2-m_A^2)/(2m_B)$. In the B frame these are the energies of $t$ and $s$. If $\theta$ is their relative angle,The minimum has parallel massless particles; the maximum has antiparallel ones. The isotropic decay allows both limiting directions. The bounds are Lorentz invariant even though the B frame makes their calculation simple.

Since ${\mathbf{p}}_A=-{\mathbf{p}}_s$, the same angle gives $m_{At}^2=m_A^2+2q(E_A+p\cos \theta )$. ThereforeThe minimum now occurs when $s$ and $t$ are antiparallel. The eventwise identity $m_{At}^2+m_{st}^2=m_C^2+m_A^2-m_B^2$ independently checks the opposite ordering of the two endpoints.