The energy resolution

Fig. 3 shows the energy distribution of the harder shower ($E_1$) vs. the softer shower ($E_2$) for the $2\gamma$ samples. Showers with separation $D_{\gamma\gamma} \geq 20$ $cm$ are shown on the left, while pairs with $M < 0.5$ $GeV$ are plotted on the right of Fig. 3. Green dots represent the pairs with masses inside the respective meson window. Solid lines represent the case where $E_1 =E_2$. Fig. 3 shows that the statistics of pairs with $E_1 =E_2$ is quite poor and covers a restricted energy range. Consequently, we measured $V(M^2)$ for the set of ($E_1,E_2$) values shown as black dots in Fig. 3. For each dot, the squared mass distribution ($\eta$ sample) or the invariant mass distribution ($\pi ^0$ data) has been created. Energy limits for a dot have been chosen to be the $0.5\sigma_E$ for each $E$, where $\sigma_E$ is deduced from Eq. (4) with $A=0.02$ and $B=0.08$ Gev $^{\frac{1}{2}}$. These distributions, shown in Figs. 4 - 7, are all fitted with a Gaussian over a polynomial background. Fitting errors reflect the statistics of the data sample. Systematic errors are governed by the choice of background function, and they are estimated to be $5\%$ and $10\%$ for the $\eta$ and $\pi ^0$ data respectively. Reducing the energy bins to $0.3 \sigma_E$ did not have a significant impact on the results; it only reduced statistics. From the Gaussian fit the value $V(M^2)/M^4$ has been calculated. In the first approximation, for the $\eta$ data we neglected the spatial contribution to the mass resolution. Following Eq. (3) we formed the set of equations

$$\left(\frac{V(M^2)}{M^4}\right)_k = F_i + F_j,$$ (9)

where $k = 1,\dots,18$ labels calculated values. We try to solve these equations for the set of $F_i$ values, with $i = 1,\dots,11$ numbering the selected photon energies. The least-square solution we obtained is model-independent since we have not presumed the functional form of $F(E)$. It is free of any restrictions on the $F$ values except that $F_i = F_j$ for $i=j$. The corresponding $\sigma_{E}(E_i) = E_i \sqrt{F_i}$ from the free solution is plotted in Fig. 8. These points agree very well with the standard expression for the energy resolution (Eq. 4), with $A=0.037$ and $B=0.080$ GeV $^{\frac{1}{2}}$.