## 3 Extensive Air Showers: The Heitler-Matthews Model
### Hadronic Showers
Assuming that at \(\varepsilon_{c}^{\pi}\), all pions decay, the number of muons is \(N_{\mu}=N_{\pi^{\pm}}=(N_{ch})^{n_{c}}\), where \(n_{c}\) is the number of interaction lengths required for the charged pion's interaction length to exceed its decay length  
\[n_{c}=\frac{\ln(E_{0}/\varepsilon_{c}^{\pi})}{\ln(\frac{3}{2}N_{ch})}=0.85\ { \rm lg}\bigg{(}\frac{E_{0}}{\varepsilon_{c}^{\pi}}\bigg{)}. \tag{6}\]  
Thus, the total energy is divided into two channels, hadronic and electromagnetic  
\[E_{0}=E_{em}+E_{h}=\varepsilon_{c}^{em}\cdot N_{e}+\varepsilon_{c}^{\pi}\cdot N _{\mu}. \tag{7}\]  
This equation represents _energy conservation_, apart from a fraction of a few percent of the primary energy spent in the neutrino component. The relative magnitude of the contribution from \(N_{\mu}\) and \(N_{e}\) does not depend on the details of the model, but only on the respective critical energies, the energy scales at which e.m. and hadronic multiplication ceases. An important conclusion of this description of the hadronic cascades is that the energy is given by a linear combination of muon and electron sizes. This result is insensitive to fluctuations in the division of energy between the hadronic and e.m. channels and independent on the mass of the primary particle. The number of muons is given by  
\[\ln N_{\mu}=\ln N_{\pi^{\pm}}=n_{c}\ln N_{ch}=\frac{\ln(E_{0}/\varepsilon_{c}^{ \pi})}{\ln(3/2N_{ch})}\cdot\ln\big{(}N_{ch}\big{)}=\beta\cdot\ln\bigg{(}\frac{E _{0}}{\varepsilon_{c}^{\pi}}\bigg{)}. \tag{8}\]  
Following [33], we can estimate \(\beta=\frac{\ln\left(N_{ch}\right)}{\ln\left(3/2N_{ch}\right)}=0.85\) for \(E_{0}\) in the range \(10^{14}\)-\(10^{17}\) eV, obtaining  
\[N_{\mu}=\bigg{(}\frac{E_{0}}{\varepsilon_{c}^{\pi}}\bigg{)}^{\beta}=\bigg{(} \frac{E_{0}}{\varepsilon_{c}^{\pi}}\bigg{)}^{0.85}\sim 9900\bigg{(}\frac{E_{0}}{ 10^{15}\text{eV}}\bigg{)}^{0.85}. \tag{9}\]  
Including inelasticity in the Heitler model [33] changes the parameter \(\beta\)  
\[\beta=\frac{\ln\left(N_{ch}\right)}{\ln\left(3/2N_{ch}\right)}\rightarrow\frac {\ln[1+N_{ch}]}{\ln\Big{[}(1+N_{ch})/(1-\frac{1}{3}\kappa)\Big{]}}\approx 1- \frac{\kappa}{3\ln(N_{ch})}=1-0.14\kappa. \tag{10}\]  
The elasticity for the most energetic meson in pion-air interactions yields \((1-\kappa)\) between 0.26 and 0.32, resulting in \(\beta=0.90\). The electronic size can be calculated by inserting the expression (9) for the muon size in the energy conservation relation (7)  
\[\frac{E_{em}}{E_{0}}=\frac{E_{0}-N_{\mu}\varepsilon_{c}^{\pi}}{E_{0}}=1- \bigg{(}\frac{E_{0}}{\varepsilon_{c}^{\pi}}\bigg{)}^{\beta-1}. \tag{11}\]  
The e.m. fraction is 66% at \(E_{0}=10^{15}\) eV, increasing to 83% at \(10^{18}\) eV for proton-induced showers. Therefore, the number of electrons at a maximum shower for proton-induced showers is  
\[N_{e}=\frac{E_{em}}{\varepsilon_{c}^{em}}=\frac{E_{0}}{\varepsilon_{c}^{em}}- \frac{\varepsilon_{c}^{\pi}}{\varepsilon_{c}^{em}}\bigg{(}\frac{E_{0}}{ \varepsilon_{c}^{\pi}}\bigg{)}^{\beta}\approx\frac{E_{0}}{\varepsilon_{c}^{em }}=N_{e|_{max}}^{p}. \tag{12}\]  
The approximation is justified at high energies when the fraction of energy transferred to muons is small [39]. In the framework of the _superposition model_, each nucleus is taken to be equal to \(A\) individual single nucleons, each with energy \(E_{0}/A\) and each acting independently. The shower resulting from the interaction of the primary nucleus \(A\) can be treated as the sum of \(A\) proton-induced independent showers all starting at the same point. Thus, while a proton creates one shower with energy \(E_{0}\), an iron nucleus of the same total energy is expected to create the equivalent of 56 proton showers, each with reduced energy \((E_{0}/56)\). The average properties of showers are well reproduced by this model, though the fluctuations are clearly underestimated and can be studied only with detailed Monte Carlo simulations of the intra-nuclear cascade.