= Pair mass versus collision energy
{title2=$M_{ee}c^2$ versus $E_{\rm cm}$}
For a reconstructed <Electron>–<Positron> pair, <four-momentum conservation> gives
$$
M_{ee}^2c^4=(E_++E_-)^2-c^2|\mathbf p_++\mathbf p_-|^2.
$$
Neglecting the <electron mass> compared with the daughter <energies>, and calling the opening angle $\theta$, this becomes
$$
M_{ee}^2c^4\simeq2E_+E_-(1-\cos\theta).
$$
As an original example, choose $E_+=2\,\mathrm{GeV}$, $E_-=3\,\mathrm{GeV}$ and $\theta=60^\circ$. Then $M_{ee}c^2=\sqrt6\,\mathrm{GeV}\simeq2.449\,\mathrm{GeV}$, 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> is
$$
E_{\rm cm}=\sqrt{(E_b+E_b)^2-c^2|\mathbf p+(-\mathbf p)|^2}=2E_b.
$$
Stipulated $1.50\,\mathrm{GeV}$ 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>.
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