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1\section{Reaction initial state simulation.}
2
3\subsection{Allowed projectiles and bombarding energy range for interaction
4with nucleon and nuclear targets.}
5\hspace{1.0em}
6The GEANT4 parton string models are capable to predict final states (produced
7hadrons
8which belong to the scalar and vector meson nonets and the baryon (antibaryon)
9octet and decuplet) of reactions on nucleon and nuclear targets
10with nucleon, pion and kaon
11 projectiles. The allowed bombarding energy
12 $\sqrt{s} > 5$ \ GeV is recommended.
13 Two approaches, based on diffractive excitation or soft scattering with
14 diffractive admixture according to cross-section, are considered.
15\hspace{1.0em}Hadron-nucleus  collisions in the both
16approaches (diffractive and parton exchange) are considered
17as a set of the independent hadron-nucleon collisions.
18 However, the string excitation procedures
19in these approaches are rather different.
20
21\subsection{ MC initialization procedure for nucleus.}
22\hspace{1.0em}The initialization of each nucleus, consisting from $A$
23nucleons and $Z$ protons with coordinates $\mathbf{r}_i$ and momenta
24$\mathbf{p}_i$, where $i = 1,2,...,A$ is performed.
25We use the standard initialization Monte Carlo procedure, which
26is realized in the most of the high energy nuclear interaction models:
27\begin{itemize}
28\item Nucleon radii $r_i$ are selected randomly in the rest of nucleus according
29to proton or neutron density $\rho(r_i)$.
30For heavy nuclei with $A > 16$ \cite{GLMP91} nucleon density is
31\begin{equation}
32\label{NIS1}\rho(r_i) =
33 \frac{\rho_0}{1 + \exp{[(r_i - R)/a]}}
34\end{equation}
35where
36\begin{equation}
37\label{NIS2}\rho_0 \approx \frac{3}{4\pi R^3}(1+\frac{a^2\pi^2}{R^2})^{-1}.
38\end{equation} 
39Here $R=r_0 A^{1/3}$ \ fm and $r_0=1.16(1-1.16A^{-2/3})$ \ fm and $a
40\approx 0.545$ fm.  For light nuclei with $A < 17$ nucleon density is
41given by a harmonic oscillator shell model \cite{Elton61}, e. g.
42\begin{equation}
43\label{4aap6} \rho(r_i) = (\pi R^2)^{-3/2}\exp{(-r_i^2/R^2)},
44\end{equation}
45where $R^2 = 2/3<r^2> = 0.8133 A^{2/3}$ \ fm$^2$.
46 To take into account nucleon
47repulsive core it is assumed that internucleon distance $d > 0.8$ \ fm;
48
49\item The initial momenta of the nucleons are randomly choosen between $0$ and
50$p^{max}_F$, where
51the maximal momenta of nucleons (in the local Thomas-Fermi
52approximation \cite{DF74}) depends from
53the proton or neutron density $\rho$ according to
54\begin{equation}
55\label{NIS5} p^{max}_F = \hbar c(3\pi^2 \rho)^{1/3}
56\end{equation}
57with $\hbar c = 0.197327$ GeV fm;
58
59\item To obtain coordinate and momentum components, it
60 is assumed that nucleons are distributed isotropicaly in configuration
61 and momentum spaces;
62
63\item Then perform shifts of nucleon coordinates ${\bf r_j^{\prime}}
64= {\bf r_j} - 1/A \sum_i {\bf r_i}$ and momenta ${\bf p_j^{\prime}}
65= {\bf p_j} - 1/A \sum_i {\bf p_i}$ 
66of nucleon momenta. The nucleus must be centered in configuration space around
67$\mathbf{0}$, \textit{i. e.} $\sum_i {\mathbf{r}_i} = \mathbf{0}$ and
68 the nucleus must be at rest, i. e. $\sum_i {\bf p_i} = \bf 0$;
69
70\item We compute energy per nucleon $e = E/A = m_{N} + B(A,Z)/A$,
71where $m_N$ is nucleon mass and the nucleus binding energy $B(A,Z)$ is given 
72by the Bethe-Weizs\"acker formula\cite{BM69}:
73\begin{equation}
74\begin{array}{c}
75\label{NIS6} B(A,Z) = \\
76= -0.01587A + 0.01834A^{2/3} + 0.09286(Z- \frac{A}{2})^2 +
770.00071 Z^2/A^{1/3},
78\end{array}
79\end{equation} 
80 and find the effective mass of each nucleon $m^{eff}_i = 
81\sqrt{(E/A)^2 - p^{2\prime}_i}$.
82\end{itemize}
83
84\subsection{Random choice of the impact parameter.}
85
86\hspace{1.0em}The impact parameter $0 \leq b \leq R_t$ is randomly
87selected according to the probability:
88\begin{equation}
89\label{NIS11}P({\bf b})d{\bf b} = b d{\bf b},
90\end{equation}
91where  $R_t$ is the target  radius,
92respectively. In the case of nuclear projectile or target the nuclear radius is
93determined from condition:
94\begin{equation}
95\label{NIS12}\frac{\rho(R)}{\rho(0)} = 0.01.
96\end{equation}
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