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An isotropic two-photon decay of a massive parent with laboratory Lorentz factor $\gamma >1$ has the displayed normalized opening-angle probability density, with $\beta =\sqrt{1-{\gamma }^{-2}}$ and $2\arcsin (1/\gamma )<A<\pi$. To derive it in units $c=1$, the laboratory photon energies are $E_{1,2}=\gamma M(1\pm \beta z)/2$, where $z$ is uniform on $[-1,1]$. The diphoton invariant mass gives ${\sin }^2(A/2)=1/[{\gamma }^2(1-{\beta }^2{z}^2)]$. Apply the change of variables for probability densities to both signs of $z$. Its minimum-angle peak is an inverse-square-root singularity. At $\gamma =1$, the probability distribution is concentrated at $A=\pi$.

The diphoton invariant mass follows by adding the two photon four-momenta: $M^2{c}^4=2E_1{E}_2(1-\cos \theta )$. ThusFor correlated diphoton mass resolution, at fixed angle, $dM/M=(d{E}_1/E_1+d{E}_2/E_2)/2$. If the two quoted fractional errors are independent one-standard-deviation uncertainties,An unknown common energy-scale error would be correlated: with correlation coefficient $\rho$, the fractional variance is $(0.1^2+0.1^2+2\rho 0.1^2)/4$. Fully correlated scale errors give 10%, rather than 7.1%. A mass consistent with a candidate is not by itself evidence establishing its identity.

For two photons, four-momentum conservation and their vanishing rest masses give the diphoton invariant massTaking a logarithmic differential,For small mutually independent measurement errors, propagation of uncertainty consequently givesPrinted-formula correction. The source's left-hand side, if it means $(\Delta M/M{)}^2$ as its description suggests, is missing the factor $1/4$ on the right. Its right-hand side instead equals $\mathrm{Var}(M^2)/M^4$ to first order. A concrete counterexample is a fixed opening angle and an exactly known $E_2$: because $M\propto \sqrt{E_1}$, a small $1\%$ fractional error in $E_1$ produces a $0.5\%$ fractional error in $M$, not $1\%$. Likewise, interpreting $\Delta M^2$ literally as a small change in $M^2$ would make the unsquared left-hand side incompatible with the squared terms. Correlated errors require the covariance terms as well.

The space angle is determined by the two polar angles and azimuthal separation:At ${\theta }_1={\theta }_2=45^{\circ }$ and $\Delta \varphi =180^{\circ }$, $\cos \alpha =0$ and $\alpha =90^{\circ }$. The angular coefficient is therefore one. Equal contribution of opening-angle uncertainty and the combined two independent $1\%$ photon energy errors requiresFor independent equally precise single-photon polar directions, ${\sigma }_{\alpha }^2={\sigma }_{{\theta }_1}^2+{\sigma }_{{\theta }_2}^2$ at this geometry, soAn azimuthal error contributes only at second order about $\Delta \varphi =\pi$. The corresponding invariant-mass resolution from the two energy errors alone is $0.707\%$; with an equally large angular contribution it is $1.00\%$. The common missing factor of four cancels when equating the two contributions, so the angular estimates remain valid despite the printed error.

By four-momentum conservation, $p_{\pi }=p_1+p_2$. With the $(-,+,+,+)$ Minkowski metric, the neutral pion has $p_{\pi }^2=-M_{\pi }^2$, while each photon has $p_i^2=0$ and $|{\mathbf{p}}_i|=E_i$. Taking the Minkowski norm givesThe half-angle identity therefore yieldsThis is a measurement of the diphoton invariant mass using two photon energies and their opening angle, independent of the neutral pion's laboratory velocity.