 Solution

Momentum identity. From the Lorentz factor, $v^2/c^2=1-{\gamma }^{-2}$. The relativistic three-momentum therefore satisfiesTogether with relativistic energy $E=\gamma m{c}^2$, this gives the energy–momentum relation $E^2-p^2{c}^2=m^2{c}^4$.

Parent in the laboratory. The rest energy is $m_0{c}^2=1\mathrm{GeV}$, while the total relativistic energy is $3\mathrm{GeV}$. ConsequentlyIts momentum magnitude is $p_{0}c=\sqrt{3^2-1^2}\mathrm{GeV}=\sqrt{8}\mathrm{GeV}$.

Daughters in the zero-momentum frame. The centre-of-momentum frame is initially the parent's rest frame. Conservation of momentum gives ${\mathbf{p}}_1^{\ast }=-{\mathbf{p}}_2^{\ast }$. Equal daughter rest masses then imply equal relativistic energies. Conservation of relativistic energy givesTheir speeds are thereforeThe two velocities are opposite, and $p_X^{\ast }c=\sqrt{(1/2{)}^2-(1/3{)}^2}\mathrm{GeV}=\sqrt{5}\mathrm{GeV}/6$. Their total kinetic energy is $m_0{c}^2-2m_X{c}^2=m_0{c}^2/3$.

Invariant of the complete system. Before the decay, in the centre-of-momentum frame,After the relativistic two-body decay,In the laboratory, four-momentum conservation fixes the summed daughter energy to $3\mathrm{GeV}$ and their summed momentum to the parent's ${\mathbf{p}}_0$. Thus the same calculation isThe invariant is formed after summing the daughter four-momenta; adding their separate squared rest energies would omit the relative-motion contribution to the invariant mass of the pair.