 Solution

Let $P=m_p{c}^2$ denote the proton rest energy and $W=220\mathrm{GeV}$ the desired particle's rest energy. For an antiproton of total energy $E_{\bar{p}}$ striking a stationary proton, the incoming squared centre-of-mass energy isFor production of the stated single particle, four-momentum conservation requires $s=W^2$. ThusThe numerical value uses $P=0.938272\mathrm{GeV}$. It is a total incident energy; the corresponding kinetic energy is $E_{\bar{p}}^{\mathrm{fixed}}-P\simeq 25.7902\mathrm{TeV}$. Much of the laboratory energy supplies motion of the outgoing system rather than its invariant mass.

For equal and opposite beam momenta, the laboratory is the centre-of-mass frame. Each beam has the same total energy and their sum is $W$, givingThese are the fixed-target production threshold and collider kinematic comparison for the specified final state. Extra final particles or restrictions from conserved quantum numbers would require a separately specified reaction.