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The on-shell measure $d^{3}p/(2E)$ is Lorentz invariant. The product measure together with the four-dimensional Dirac delta distribution enforces four-momentum conservation. For equal masses in two-body scattering, $d{\Phi }_2=\sqrt{1-4m^2/s}d\Omega /(32{\pi }^2)$.

Let ${\mathbf{p}}_1$ and ${\mathbf{p}}_2$ be the incoming and proton momenta. The other incoming atom is at rest, so its total energy is $m_1{c}^2$. By four-momentum conservation, the tritium atom hasThe specified perpendicular emission gives ${\mathbf{p}}_1\cdot {\mathbf{p}}_2=0$. Therefore the relativistic energy-momentum relation impliesDividing by $c^4$ givesHere $E_1,E_2$ are total energies, not just kinetic energies. This identity is conditional on the collision admitting the stated emission: the calculated energies must satisfy the positive-energy mass-shell conditions, in particular $E_2\ge m_2{c}^2$ and $E_3\ge m_3{c}^2$. No nonrelativistic approximation has been used.

Four-momentum conservation gives $P={\sum }_i{Q}_i$, while the mass-shell condition gives $P\cdot P=M^2{c}^2$. ThereforeIn the parent rest frame, $M{c}^2={\sum }_i{E}_i$, and each relativistic energy obeys $E_i\ge m_i{c}^2$. Thus