[1211] | 1 | \section{Multifragmentation process simulation.} |
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| 2 | \hspace{1.0em} The GEANT4 multifragmentation model is capable to predict |
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| 3 | final states as result of an highly excited nucleus statistical break-up. |
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| 4 | |
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| 5 | The initial information for calculation of |
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| 6 | multifragmentation stage consists from the atomic mass number $A$, |
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| 7 | charge $Z$ of excited nucleus and its excitation |
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| 8 | energy $U$. At high excitation energies $U/A > 3$ \ MeV the |
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| 9 | multifragmentation mechanism, when nuclear system can eventually breaks |
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| 10 | down into fragments, becomes the dominant. Later on the excited primary |
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| 11 | fragments propagate independently in the mutual Coulomb field and |
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| 12 | undergo de-excitation. Detailed description of multifragmentation |
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| 13 | mechanism and model can be found in review \cite{BBIMS95}. |
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| 14 | |
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| 15 | \subsection{Multifragmentation probability.} |
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| 16 | |
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| 17 | \hspace{1.0em}The probability of a breakup channel $b$ is given by the |
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| 18 | expression (in the so-called microcanonical approach \cite{BBIMS95}, |
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| 19 | \cite{Botvina87}): |
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| 20 | \begin{equation} |
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| 21 | \label{MFS1}W_b(U,A,Z)=\frac{1}{\sum_{b}\exp [S_b(U,A,Z)]} \exp [S_b(U,A,Z)], |
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| 22 | \end{equation} |
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| 23 | where $S_b(U,A,Z)$ is the entropy of a multifragment state corresponding |
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| 24 | to the breakup channel $b$. The channels $\{b\}$ can be parametrized by |
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| 25 | set of fragment multiplicities $N_{A_f,Z_f}$ for fragment with atomic |
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| 26 | number $A_f$ and charge $Z_f$. All partitions $\{b\}$ should satisfy |
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| 27 | constraints on the total mass and charge: |
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| 28 | \begin{equation} |
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| 29 | \label{MFS2}\sum_{f}N_{A_f,Z_f}A_f = A |
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| 30 | \end{equation} |
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| 31 | and |
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| 32 | \begin{equation} |
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| 33 | \label{MFS3}\sum_{f}N_{A_f,Z_f}Z_f = Z. |
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| 34 | \end{equation} |
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| 35 | It is assumed \cite{Botvina87} that thermodynamic equilibrium is |
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| 36 | established in every channel, which can be characterized by the channel |
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| 37 | temperature $T_b$. |
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| 38 | |
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| 39 | The channel temperature $T_b$ is determined by the equation constraining |
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| 40 | the average energy $E_b(T_b, V)$ associated with partition $b$: |
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| 41 | \begin{equation} |
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| 42 | \label{MFS4}E_b(T_b, V)= U+E_{ground} = U+M(A,Z), |
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| 43 | \end{equation} |
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| 44 | where $V$ is the system volume, $E_{ground}$ is the ground state (at |
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| 45 | $T_b = 0$) energy of system and $M(A,Z)$ is the mass of nucleus. |
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| 46 | |
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| 47 | According to the conventional thermodynamical formulae the average energy |
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| 48 | of a partitition $b$ is expressed through the system free energy $F_b$ |
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| 49 | as follows |
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| 50 | \begin{equation} |
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| 51 | \label{MFS5}E_b(T_b, V)= F_b(T_b,V) +T_bS_b(T_b,V). |
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| 52 | \end{equation} |
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| 53 | Thus, if free energy $F_b$ of a partition $b$ is known, we can find |
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| 54 | the channel temperature $T_b$ from Eqs. ($\ref{MFS4}$) and ($\ref{MFS5}$), |
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| 55 | then the entropy $S_b = -dF_b/dT_b$ and hence, decay probability $W_b$ |
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| 56 | defined by Eq. ($\ref{MFS1}$) can be calculated. |
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| 57 | |
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| 58 | Calculation of the free energy is based on the use of the liquid-drop |
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| 59 | description of individual fragments \cite{Botvina87}. The free energy |
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| 60 | of a partition $b$ can be splitted into several terms: |
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| 61 | \begin{equation} |
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| 62 | \label{MFS6}F_b(T_b,V) = \sum_{f}F_f(T_b,V) + E_{C}(V), |
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| 63 | \end{equation} |
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| 64 | where $F_f(T_b,V)$ is the average energy of an individual fragment |
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| 65 | including the volume |
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| 66 | \begin{equation} |
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| 67 | \label{MFS7} F^V_f = [-E_0 - T^2_b/\epsilon(A_f)]A_f, |
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| 68 | \end{equation} |
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| 69 | surface |
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| 70 | \begin{equation} |
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| 71 | \label{MFS8}F^{Sur}_f = \beta_0[(T_c^2 - T^2_b)/(T_c^2 + T^2_b)]^{5/4}A_f^{2/3} |
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| 72 | = \beta(T_b)A_f^{2/3}, |
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| 73 | \end{equation} |
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| 74 | symmetry |
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| 75 | \begin{equation} |
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| 76 | \label{MFS9}F^{Sim}_f = \gamma(A_f - 2Z_f)^2/A_f, |
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| 77 | \end{equation} |
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| 78 | Coulomb |
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| 79 | \begin{equation} |
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| 80 | \label{MFS10}F^{C}_f = \frac{3}{5}\frac{Z_f^2e^2}{r_0A_f^{1/3}} |
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| 81 | [1 - (1+ \kappa_{C})^{-1/3}] |
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| 82 | \end{equation} |
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| 83 | and translational |
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| 84 | \begin{equation} |
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| 85 | \label{MFS11}F^{t}_f = -T_b\ln{(g_fV_f/\lambda^3_{T_b})} + |
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| 86 | T_b\ln{(N_{A_f,Z_f}!)}/ N_{A_f,Z_f} |
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| 87 | \end{equation} |
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| 88 | terms and the last term |
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| 89 | \begin{equation} |
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| 90 | \label{MFS12} E_{C}(V)=\frac{3}{5}\frac{Z^2e^2}{R} |
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| 91 | \end{equation} |
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| 92 | is the Coulomb energy of the uniformly charged sphere with charge $Ze$ |
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| 93 | and the radius $R = (3V/4\pi)^{1/3}= r_0A^{1/3}(1 + \kappa_{C})^{1/3}$, |
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| 94 | where $\kappa_{C} = 2$ \cite{Botvina87}. |
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| 95 | |
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| 96 | Parameters $E_0 = 16$ \ MeV, $\beta_0 = 18$\ MeV, $\gamma = 25$ \ MeV |
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| 97 | are the coefficients of the Bethe-Weizsacker mass formula at $T_b = 0$. |
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| 98 | $g_f=(2S_f+1)(2I_f+1)$ is a spin $S_f$ and isospin $I_f$ degeneracy |
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| 99 | factor for fragment ( fragments with $A_f > 1$ are treated as the |
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| 100 | Boltzmann particles), $\lambda_{T_b} = (2\pi h^2/m_N T_b)^{1/2}$ is the |
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| 101 | thermal wavelength, $m_N$ is the nucleon mass, $r_0 = 1.17$ \ fm, |
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| 102 | $T_c=18$ \ MeV is the critical temperature, which corresponds to the |
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| 103 | liquid-gas phase transition. $\epsilon(A_f) = \epsilon_0[1 + 3/(A_f-1)]$ |
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| 104 | is the inverse level density of the mass $A_f$ fragment and |
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| 105 | $\epsilon_0=16$ \ MeV is considered as a variable model parameter, whose |
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| 106 | value depends on the fraction of energy transferred to the internal |
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| 107 | degrees of freedom of fragments \cite{Botvina87}. The free volume $V_f |
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| 108 | =\kappa V=\kappa\frac{4}{3}\pi r_0^4 A$ available to the translational |
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| 109 | motion of fragment, where $\kappa \approx 1$ and its dependence on the |
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| 110 | multiplicity of fragments was taken from \cite{Botvina87}: |
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| 111 | \begin{equation} |
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| 112 | \label{MFS13}\kappa =[1 + \frac{1.44}{r_0A^{1/3}}(M^{1/3} - 1)]^{3} - 1. |
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| 113 | \end{equation} |
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| 114 | For $M = 1$ $\kappa = 0$. |
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| 115 | |
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| 116 | The light fragments with $A_f < 4$, which have no excited |
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| 117 | states, are considered as elementary particles characterized by the |
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| 118 | empirical masses $M_f$, radii $R_f$, binding energies $B_f$, spin |
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| 119 | degeneracy factors $g_f$ of ground states. They contribute to the |
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| 120 | translation free energy and Coulomb energy. |
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| 121 | |
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| 122 | |
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| 123 | \subsection[Direct simulation of low multiplicity disintegration]{Direct simulation of the low multiplicity multifragment disintegration} |
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| 124 | |
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| 125 | \hspace{1.0em}At comparatively low excitation energy (temperature) |
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| 126 | system will disintegrate into a small number of fragments $M \leq 4$ and |
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| 127 | number of channel is not huge. For such situation a direct |
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| 128 | (microcanonical) sorting of all decay channels can be performed. Then, |
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| 129 | using Eq. ($\ref{MFS1}$), the average multiplicity value $<M>$ can be |
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| 130 | found. To check that we really have the situation with the low |
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| 131 | excitation energy, the obtained value of $<M>$ is examined to obey the |
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| 132 | inequality $<M> \leq M_0$, where $M_0 = 3.3$ and $M_0 = 2.6$ for $A \sim |
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| 133 | 100$ and for $A \sim 200$, respectively \cite{Botvina87}. If the |
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| 134 | discussed inequality is fulfilled, then the set of channels under |
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| 135 | consideration is belived to be able for a correct description of the |
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| 136 | break up. Then using calculated according Eq. ($\ref{MFS1}$) |
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| 137 | probabilities we can randomly select a specific channel with given |
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| 138 | values of $A_f$ and $Z_f$. |
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| 139 | |
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| 140 | |
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| 141 | \subsection{ Fragment multiplicity distribution.} |
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| 142 | |
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| 143 | \hspace{1.0em}The individual fragment multiplicities $N_{A_f,Z_f}$ in |
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| 144 | the so-called macrocanonical ensemble \cite{BBIMS95} are distributed |
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| 145 | according to the Poisson distribution: |
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| 146 | \begin{equation} |
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| 147 | \label{MFS14} P(N_{A_f,Z_f}) = \exp{(-\omega_{A_f,Z_f})} |
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| 148 | \frac{\omega_{A_f,Z_f}^{N_{A_f,Z_f}}}{N_{A_f,Z_f}!} |
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| 149 | \end{equation} |
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| 150 | with mean value $<N_{A_f,Z_f}>=\omega_{A_f,Z_f}$ defined as |
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| 151 | \begin{equation} |
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| 152 | \label{MFS15} <N_{A_f,Z_f}> =g_fA_f^{3/2}\frac{V_f}{\lambda^3_{T_b}} |
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| 153 | \exp{[\frac{1}{T_b}(F_f(T_b,V)-F^{t}_f(T_b,V) - \mu A_f - \nu Z_f)]}, |
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| 154 | \end{equation} |
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| 155 | where $\mu$ and $\nu$ are chemical potentials. The chemical potential |
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| 156 | are found by substituting Eq. ($\ref{MFS15}$) into the system of |
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| 157 | constraints: |
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| 158 | \begin{equation} |
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| 159 | \label{MFS16}\sum_{f}<N_{A_f,Z_f}>A_f = A |
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| 160 | \end{equation} |
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| 161 | and |
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| 162 | \begin{equation} |
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| 163 | \label{MFS17}\sum_{f}<N_{A_f,Z_f}>Z_f = Z |
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| 164 | \end{equation} |
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| 165 | and solving it by iteration. |
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| 166 | |
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| 167 | |
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| 168 | \subsection{ Atomic number distribution of fragments.} |
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| 169 | |
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| 170 | \hspace{1.0em}Fragment atomic numbers $A_f > 1$ are also distributed |
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| 171 | according to the Poisson distribution \cite{BBIMS95} (see |
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| 172 | Eq. ($\ref{MFS14}$)) with mean value $<N_{A_f}>$ defined as |
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| 173 | \begin{equation} |
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| 174 | \label{MFS18} <N_{A_f}> = A_f^{3/2}\frac{V_f}{\lambda^3_{T_b}} |
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| 175 | \exp{[\frac{1}{T_b}(F_f(T_b,V)-F^{t}_f(T_f,V) - \mu A_f - \nu <Z_f>)]}, |
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| 176 | \end{equation} |
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| 177 | where calculating the internal free energy |
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| 178 | $F_f(T_b,V)-F^{t}_f(T_b,V)$ one has to substitute $Z_f |
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| 179 | \rightarrow <Z_f>$. The average charge $<Z_f>$ for fragment having atomic |
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| 180 | number $A_f$ is given by |
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| 181 | \begin{equation} |
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| 182 | \label{MFS19}<Z_f(A_f)> = \frac{(4\gamma + \nu)A_f}{8\gamma + 2[1 - (1 + |
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| 183 | \kappa)^{-1/3}]A_f^{2/3}}. |
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| 184 | \end{equation} |
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| 185 | |
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| 186 | \subsection{ Charge distribution of fragments.} |
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| 187 | |
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| 188 | \hspace{1.0em}At given mass of fragment $A_f > 1$ the charge $Z_f$ |
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| 189 | distribution of fragments are described by Gaussian |
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| 190 | \begin{equation} |
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| 191 | \label{MFS20} P(Z_f(A_f))\sim \exp{[-\frac{(Z_f(A_f) - |
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| 192 | <Z_f(A_f)>)^2}{2(\sigma_{Z_f}(A_f))^2}]} |
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| 193 | \end{equation} |
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| 194 | with dispertion |
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| 195 | \begin{equation} |
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| 196 | \label{MFS21}\sigma_{Z_f(A_f)} = \sqrt{\frac{A_fT_b} |
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| 197 | {8 \gamma + 2[1 - (1 +\kappa)^{-1/3}]A_f^{2/3}}} |
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| 198 | \approx \sqrt{\frac{A_fT_b}{8 \gamma}}. |
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| 199 | \end{equation} |
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| 200 | and the average charge $<Z_f(A_f)>$ defined by Eq. ($\ref{MFS17}$). |
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| 201 | |
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| 202 | \subsection{ Kinetic energy distribution of fragments.} |
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| 203 | |
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| 204 | \hspace{1.0em}It is assumed \cite{Botvina87} that at the instant of the |
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| 205 | nucleus break-up the kinetic energy of the fragment $T^{f}_{kin}$ in the |
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| 206 | rest of nucleus obeys the Boltzmann distribution at given temperature |
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| 207 | $T_b$: |
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| 208 | \begin{equation} |
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| 209 | \label{MFS22} \frac{dP(T^{f}_{kin})}{dT^{f}_{kin}}\sim \sqrt{T^{f}_{kin}} |
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| 210 | \exp{(-T^{f}_{kin}/T_b)}. |
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| 211 | \end{equation} |
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| 212 | Under assumption of thermodynamic equilibrium the fragment have |
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| 213 | isotropic velocities distribution in the rest frame of nucleus. The |
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| 214 | total kinetic energy of fragments should be equal $\frac{3}{2}MT_b$, |
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| 215 | where $M$ is fragment multiplicity, and the total fragment momentum |
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| 216 | should be equal zero. These conditions are fullfilled by choosing |
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| 217 | properly the momenta of two last fragments. |
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| 218 | |
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| 219 | The initial conditions for the divergence of the fragment system are |
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| 220 | determined by random selection of fragment coordinates distributed with |
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| 221 | equal probabilities over the break-up volume $V_f = \kappa V$. It can be |
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| 222 | a sphere or prolongated ellipsoid. Then Newton's equations of motion are |
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| 223 | solved for all fragments in the self-consistent time-dependent Coulomb |
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| 224 | field \cite{Botvina87}. Thus the asymptotic energies of fragments |
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| 225 | determined as result of this procedure differ from the initial values by |
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| 226 | the Coulomb repulsion energy. |
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| 227 | |
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| 228 | \subsection{Calculation of the fragment excitation energies.} |
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| 229 | |
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| 230 | \hspace{1.0em}The temparature $T_b$ determines the average excitation |
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| 231 | energy of each fragment: |
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| 232 | \begin{equation} |
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| 233 | \label{MFS23} U_{f}(T_b) = E_f(T_b) - E_f(0) = \frac{T_b^2}{\epsilon_0}A_f + |
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| 234 | [\beta(T_b) - T_b \frac{d\beta(T_b)}{dT_b} - \beta_0]A^{2/3}_f, |
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| 235 | \end{equation} |
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| 236 | where $E_f(T_b)$ is the average fragment energy at given temperature |
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| 237 | $T_b$ and $\beta(T_b)$ is defined in Eq. ($\ref{MFS8}$). There is no |
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| 238 | excitation for fragment with $A_f < 4$, for $^{4}He$ excitation energy |
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| 239 | was taken as $U_{^{4}He} = 4T^2_b/\epsilon_o$. |
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| 240 | |
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