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Shower phenomenology

The distribution of energy deposition in electromagnetic showers generated by high-energy electrons and photons is summarized by the Particle Data Group [1]. In their review is given the following expression for the shower profile as a function of penetration depth $t$ into a material.

\begin{displaymath}
-\frac{dE}{dt} = E_0 b \frac{(bt)^{a-1}e^{-bt}}{\Gamma(a)}
\end{displaymath} (1)

Here $b$ is a constant that depends on the material, but is approximately $(2X_0)^{-1}$ where $X_0$ is the radiation length of the material. The variable $a$ in Eq. 1 depends on the energy of the shower through the relation
\begin{displaymath}
t_{max} = \frac{a-1}{b} = X_0 \left[ \ln{\left( \frac{E_0}{E_c}\right)}
+C_j \right]
\end{displaymath} (2)

where $E_0$ is the incident gamma energy, $E_c$ is the critical energy of the shower medium, and the constant $C_j$ is +0.5 for incident photons and -0.5 for electrons or positrons. The distribution given by Eq. 1 measures the energy deposition via ionization by all shower particles at a given depth, which is not in general identical to the Cerenkov light emission rate. However electrons remain relativistic for most of the length of their track, in which case both the ionization energy loss and the Cerenkov emission rate are proportional to the total $e^{\pm}$ track length.

The quantity of interest in reconstructing the polar angle from cluster measurements in the LGD is the average depth of the Cerenkov emission in the LGD. This is not exactly the quantity $t_{max}$ in Eq. 2 but the two are closely related. From Eq. 1 one can see that the quantity $t_{max}$ in Eq. 2 is just the depth at which the energy deposition rate reaches its maximum. Using Eq. 2 to define $a$ allows us to go back to Eq. 1 and calculate the average deposition depth $t_{ave}$ which one finds is simply $a/b$. This leads to a formula that gives $t_{ave}$ for gamma showers in terms of the shower energy $E_0$ and physical constants.

\begin{displaymath}
t_{ave} = X_0 \left[ \ln{\left( \frac{E_0}{E_c}\right)} + 0.5 +
\frac{1}{b} \right]
\end{displaymath} (3)


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Next: Simulations Up: report Previous: report
Richard T. Jones 2003-02-12