= Two-pion kaon event reconstruction
{title2=$M_{\pi\pi}^2=(E_++E_-)^2-|\mathbf p_++\mathbf p_-|^2$}
A two-track <invariant mass> and a direction test answer different questions. In units $c=1$, assign each track the <pion> <mass> $m_\pi$ and reconstruct its <momentum> before magnetic deflection. With $p_\pm=|\mathbf p_\pm|$ and opening angle $\alpha$,
$$
E_\pm=\sqrt{p_\pm^2+m_\pi^2},\qquad
M_{\pi\pi}^2=2m_\pi^2+2(E_+E_--p_+p_-\cos\alpha).
$$
This follows by expanding the squared sum of the two <four-momenta>. The angle $\theta$ between the reconstructed parent <momentum> $\mathbf P=\mathbf p_++\mathbf p_-$ and the neutral <particle beam> direction $\hat{\mathbf n}$ satisfies
$$
\cos\theta=\frac{\mathbf P\cdot\hat{\mathbf n}}{|\mathbf P|}.
$$
A fully reconstructed two-body <decay channel> gives the parent <mass> and $\theta\simeq0$, within experimental resolution. A three-body <decay channel> with an unseen <neutrino> generally misses both constraints; an accidental correct <invariant mass> alone is insufficient. Notice that the track opening angle $\alpha$ need not be small even when the parent direction angle $\theta$ vanishes.
The 1964 analysis selected $490<M_{\pi\pi}<510$ MeV and remeasured events with $\cos\theta>0.9995$, comparing adjacent <mass> bands.<Evidence for the 2π Decay of the K₂⁰ Meson by James H. Christenson and others>[p.139, Figs.2–3 and remeasurement selection]{ref}
For an original dimensionless example, take $m_\pi=1$, $\hat{\mathbf n}=(0,0,1)$ and $\mathbf p_\pm=(\pm\sqrt3,0,3)$. Each energy is $\sqrt{13}$; the opening angle is $60$ degrees since $\cos\alpha=(9-3)/(9+3)=1/2$. Yet $\mathbf P=(0,0,6)$ points exactly along the beam, and $M_{\pi\pi}^2=4(13)-36=16$. Thus a parent of <mass> $4$ passes both ideal tests. These chosen numbers illustrate <four-momentum conservation>, not detector performance.
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