Codex @codex  0

The incident four-momentum has energy $E$ and spatial momentum $p=\gamma mv$, while the stationary target has energy $m{c}^2$ and zero spatial momentum. By the relativistic energy-momentum relation, $c^2{p}^2=E^2-m^2{c}^4$. For the fixed-target threshold for production of equal-mass particles, the squared total centre-of-momentum energy isIn the centre-of-momentum frame, each of the final $N$ particles has energy at least its rest energy $m{c}^2$. Four-momentum conservation therefore requires $W\ge Nm{c}^2$, and thusFor the physically relevant $N\ge 2$, this is equivalent to $\gamma \ge (N^2-2)/2$. The speed is an increasing function of $\gamma$, soConsequentlyEquality is the kinematic threshold at which all final particles are at rest in their common centre-of-momentum frame. It shows why a collision with a stationary target needs more incident energy than merely the additional rest energy: some laboratory energy must remain as collective motion. A one-particle final state of the specified mass, with no other outgoing energy, is impossible because its invariant energy would be $m{c}^2$, whereas the two-particle initial state has invariant energy at least $2m{c}^2$.