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Use natural units $c=1$ and the Minkowski metric with signs $(+,-,-,-)$. In the ideal centre-of-momentum frame, assume a known initial four-momentum $P=(\sqrt{s},0)$, no unaccounted radiation and exact visible reconstruction. DefineThe missing four-momentum in electron-positron collisions is a subtraction of measured quantities. It is not itself a particle detector hit. With $M_{e\mu }^2=(p_e+p_{\mu }{)}^2$, expansion gives the useful identityFor any collection of future-directed massless invisible four-momenta $q_i=E_i(1,{\hat{\mathbf{n}}}_i)$,Thus a positive missing invariant mass excludes exactly one massless invisible particle under the stated assumptions. It does not exclude one massive invisible particle, identify a neutrino flavor, or determine the number of massless invisible particles beyond that lower bound. Several collinear massless particles can even have zero combined invariant mass. Detector resolution and unaccounted radiation invalidate a literal event-by-event interpretation of these ideal inequalities.

For an original exact example in natural units, choose total energy $6\mathrm{GeV}$ and visible massless four-momenta, in $\mathrm{GeV}$,Two possible invisible massless four-momenta are $(2;-1/2,-1/2,\sqrt{14}/2)$ and $(2;-1/2,-1/2,-\sqrt{14}/2)$. Splitting either into two positive collinear parts gives a three-particle interpretation with identical recoil. Alternatively, one invisible particle of mass $\sqrt{14}\mathrm{GeV}$ has that recoil. This example tests the ambiguity of the subtraction, without asserting any of these alternatives is a tau lepton decay.

Choose $m_{\tau }=2\mathrm{GeV}$ and $\sqrt{s}=6\mathrm{GeV}$ purely for illustration. Then each parent has $E_{\tau }=3\mathrm{GeV}$, $|{\mathbf{p}}_{\tau }|=\sqrt{5}\mathrm{GeV}$ and $\beta =\sqrt{5}/3$. Treat all six daughters as massless, including the charged Electron and antimuon, and generate each three-body particle decay with constant squared decay amplitude. Choose the parent direction isotropically and the two decays independently; this omits the physical production angular law and spin correlations.

The Lorentz-invariant phase space factorizes through $t=q^2$ into a two-body decay to $\ell +q$, followed by a two-body decay of $q$:After angular integration the first factor contributes $1-t/m_{\tau }^2=x$, while the massless second factor is constant. Since $|dt|=m_{\tau }^{2}dx$, the normalized chosen density is $f(x)=2x$. Drawing $x=\sqrt{U}$ with $U$ uniform on $[0,1]$ implements this density. Isotropic directions, followed by a Lorentz boost for the invisible pair and another for the parent, conserve the complete four-momentum of each generated event.

Define transverse acoplanarity by $\alpha =\pi -\Delta \varphi$, where $\Delta \varphi \in [0,\pi ]$ is the angle between the two visible momentum projections perpendicular to the beam. A fully visible two-body event with zero initial transverse momentum has $\alpha =0$ and $M_{\mathrm{miss}}^2=0$. The toy tau lepton events spread away from that point. These observables are complementary: an angular imbalance asks about directions, while the missing invariant mass also uses the visible energies and longitudinal momenta.

A Lorentz-invariant phase space density supplies allowed configurations and their geometric measure. A physical weak interaction decay amplitude reweights them and correlates spins. Consequently the plotted density is a lesson about conservation and available configurations, not a predicted experimental rate or a fit to the discovery sample. The massless charged-daughter approximation is also unsuitable near the muon rest-energy threshold.