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%\subsection{The description of the target nucleus and fermi motion \editor{Gunter}}

The nucleus is constructed from $A$
nucleons and $Z$ protons with nucleon coordinates $\mathbf{r}_i$ and momenta
$\mathbf{p}_i$, with $i = 1,2,...,A$.
We use a common initialization Monte Carlo procedure, which
is realized in the most of the high energy nuclear interaction models:
\begin{itemize}
\item Nucleon radii $r_i$ are selected randomly in the nucleus rest frame 
according to nucleon density $\rho(r_i)$.
For heavy nuclei with $A > 16$ \cite{GLMP91.BC} nucleon density is
\begin{equation}
\label{NIS1.BC}\rho(r_i) =
 \frac{\rho_0}{1 + \exp{[(r_i - R)/a]}}
\end{equation}
where
\begin{equation}
\label{NIS2.BC}\rho_0 \approx \frac{3}{4\pi R^3}(1+\frac{a^2\pi^2}{R^2})^{-1}.
\end{equation}
Here $R=r_0 A^{1/3}$ \ fm and $r_0=1.16(1-1.16A^{-2/3})$ \ fm and $a
\approx 0.545$ fm.  For light nuclei with $A < 17$ nucleon density is
given by a harmonic oscillator shell model \cite{Elton61.BC}, e. g.
\begin{equation}
\label{4aap6.BC} \rho(r_i) = (\pi R^2)^{-3/2}\exp{(-r_i^2/R^2)},
\end{equation}
where $R^2 = 2/3<r^2> = 0.8133 A^{2/3}$ \ fm$^2$.
 To take into account nucleon
repulsive core it is assumed that internucleon distance $d > 0.8$ \ fm;

\item The nucleus is assumed to be isotropic, i.e. we place each nucleon
using a random direction and the previously determined radius  $r_i$.

\item The initial momenta of the nucleons $p_i$ are randomly choosen between $0$
and $p^{max}_F(r)$, where
the maximal momenta of nucleons (in the local Thomas-Fermi
approximation \cite{DF74.BC}) depends from
the proton or neutron density $\rho$ according to
\begin{equation}
\label{NIS5.BC} p^{max}_F(r) = \hbar c(3\pi^2 \rho(r))^{1/3}
\end{equation}


\item To obtain momentum components, it is assumed that nucleons are distributed
isotropic in momentum space;  i.e. the momentum direction is chosen at random.

\item The nucleus must be centered in momentum space around
$\mathbf{0}$, \textit{i. e.}
 the nucleus must be at rest, i. e. $\sum_i {\bf p_i} = \bf 0$;
 To achieve this, we choose one nucleon to compensate the sum the remaining
nucleon momenta $p_rest=\sum_{i=1}^{i=A-1}$. If this sum is larger than maximum
momentum  $p^{max}_F(r)$, we change the direction of the momentum of a few
nucleons. If this does not lead to a possible momentum value, than we repeat the
procedure with a different nucleon having a larger maximum momentum
$p^{max}_F(r)$. In the rare case this fails as well, we choose new momenta for
all nucleons.

This procedure gives special for hydrogen $^1$H, where the proton has momentum
$p=0$, and for deuterium $^2$H, where the momenta of proton and neutron are
equal, and in opposite direction.


\item We compute energy per nucleon $e = E/A = m_{N} + B(A,Z)/A$,
where $m_N$ is nucleon mass and the nucleus binding energy $B(A,Z)$ is given
by the tabulation of \cite{nucleus_binding}:
 and find the effective mass of each nucleon $m^{eff}_i = 
\sqrt{(E/A)^2 - p^{2\prime}_i}$.
\end{itemize}