Pair mass versus collision energy ($M_{ee}{c}^2$ versus $E_{\mathrm{cm}}$)

For a reconstructed Electron–Positron pair, four-momentum conservation givesNeglecting the electron mass compared with the daughter energies, and calling the opening angle $\theta$, this becomesAs an original example, choose $E_{+}=2\mathrm{GeV}$, $E_{-}=3\mathrm{GeV}$ and $\theta =60^{\circ }$. Then $M_{ee}{c}^2=\sqrt{6}\mathrm{GeV}\simeq 2.449\mathrm{GeV}$, although the pair's laboratory energy is $5\mathrm{GeV}$. The pair carries net momentum. Neither its laboratory energy nor the original proton beam energy is its invariant mass; unobserved products carry additional four-momentum.

For equal, oppositely directed incoming beams with equal total energy $E_b$, their total momentum vanishes. Thus the incoming center-of-mass energy isStipulated $1.50\mathrm{GeV}$ beams therefore provide $3.00\mathrm{GeV}$ of incoming center-of-mass energy. This identity is exact for the stated equal-beam geometry. Radiation before annihilation can reduce the energy forming the final state; the apparatus also has a distribution of beam energies. A nominal machine setting is therefore not an exact event-by-event final-state invariant mass.