Multifragmentation process simulation.

The GEANT4 multifragmentation model is capable to predict final states as result of an highly excited nucleus statistical break-up.

The initial information for calculation of multifragmentation stage consists from the atomic mass number $A$, charge $Z$ of excited nucleus and its excitation energy $U$. At high excitation energies $U/A > 3$ MeV the multifragmentation mechanism, when nuclear system can eventually breaks down into fragments, becomes the dominant. Later on the excited primary fragments propagate independently in the mutual Coulomb field and undergo de-excitation. Detailed description of multifragmentation mechanism and model can be found in review [1].

Multifragmentation probability.

The probability of a breakup channel $b$ is given by the expression (in the so-called microcanonical approach [1], [2]):

$$W_b(U,A,Z)=\frac{1}{\sum_{b}\exp [S_b(U,A,Z)]} \exp [S_b(U,A,Z)],$$ (32.1)

where $S_b(U,A,Z)$ is the entropy of a multifragment state corresponding to the breakup channel $b$. The channels $\{b\}$ can be parametrized by set of fragment multiplicities $N_{A_f,Z_f}$ for fragment with atomic number $A_f$ and charge $Z_f$. All partitions $\{b\}$ should satisfy constraints on the total mass and charge:

$$\sum_{f}N_{A_f,Z_f}A_f = A$$ (32.2)

and

$$\sum_{f}N_{A_f,Z_f}Z_f = Z.$$ (32.3)

It is assumed [2] that thermodynamic equilibrium is established in every channel, which can be characterized by the channel temperature $T_b$.

The channel temperature $T_b$ is determined by the equation constraining the average energy $E_b(T_b, V)$ associated with partition $b$:

$$E_b(T_b, V)= U+E_{ground} = U+M(A,Z),$$ (32.4)

where $V$ is the system volume, $E_{ground}$ is the ground state (at $T_b = 0$) energy of system and $M(A,Z)$ is the mass of nucleus.

According to the conventional thermodynamical formulae the average energy of a partitition $b$ is expressed through the system free energy $F_b$ as follows

$$E_b(T_b, V)= F_b(T_b,V) +T_bS_b(T_b,V).$$ (32.5)

Thus, if free energy $F_b$ of a partition $b$ is known, we can find the channel temperature $T_b$ from Eqs. ($\ref{MFS4}$) and ($\ref{MFS5}$), then the entropy $S_b = -dF_b/dT_b$ and hence, decay probability $W_b$ defined by Eq. ($\ref{MFS1}$) can be calculated.

Calculation of the free energy is based on the use of the liquid-drop description of individual fragments [2]. The free energy of a partition $b$ can be splitted into several terms:

$$F_b(T_b,V) = \sum_{f}F_f(T_b,V) + E_{C}(V),$$ (32.6)

where $F_f(T_b,V)$ is the average energy of an individual fragment including the volume

$$F^V_f = [-E_0 - T^2_b/\epsilon(A_f)]A_f,$$ (32.7)

surface

$$F^{Sur}_f = \beta_0[(T_c^2 - T^2_b)/(T_c^2 + T^2_b)]^{5/4}A_f^{2/3} = \beta(T_b)A_f^{2/3},$$ (32.8)

symmetry

$$F^{Sim}_f = \gamma(A_f - 2Z_f)^2/A_f,$$ (32.9)

Coulomb

$$F^{C}_f = \frac{3}{5}\frac{Z_f^2e^2}{r_0A_f^{1/3}} [1 - (1+ \kappa_{C})^{-1/3}]$$ (32.10)

and translational

$$F^{t}_f = -T_b\ln{(g_fV_f/\lambda^3_{T_b})} + T_b\ln{(N_{A_f,Z_f}!)}/ N_{A_f,Z_f}$$ (32.11)

terms and the last term

$$E_{C}(V)=\frac{3}{5}\frac{Z^2e^2}{R}$$ (32.12)

is the Coulomb energy of the uniformly charged sphere with charge $Ze$ and the radius $R = (3V/4\pi)^{1/3}= r_0A^{1/3}(1 + \kappa_{C})^{1/3}$, where $\kappa_{C} = 2$ [2].

Parameters $E_0 = 16$ MeV, $\beta_0 = 18$ MeV, $\gamma = 25$ MeV are the coefficients of the Bethe-Weizsacker mass formula at $T_b = 0$. $g_f=(2S_f+1)(2I_f+1)$ is a spin $S_f$ and isospin $I_f$ degeneracy factor for fragment ( fragments with $A_f > 1$ are treated as the Boltzmann particles), $\lambda_{T_b} = (2\pi h^2/m_N T_b)^{1/2}$ is the thermal wavelength, $m_N$ is the nucleon mass, $r_0 = 1.17$ fm, $T_c=18$ MeV is the critical temperature, which corresponds to the liquid-gas phase transition. $\epsilon(A_f) = \epsilon_0[1 + 3/(A_f-1)]$ is the inverse level density of the mass $A_f$ fragment and $\epsilon_0=16$ MeV is considered as a variable model parameter, whose value depends on the fraction of energy transferred to the internal degrees of freedom of fragments [2]. The free volume $V_f =\kappa V=\kappa\frac{4}{3}\pi r_0^4 A$ available to the translational motion of fragment, where $\kappa \approx 1$ and its dependence on the multiplicity of fragments was taken from [2]:

$$\kappa =[1 + \frac{1.44}{r_0A^{1/3}}(M^{1/3} - 1)]^{3} - 1.$$ (32.13)

For $M = 1$ $\kappa = 0$.

The light fragments with $A_f < 4$, which have no excited states, are considered as elementary particles characterized by the empirical masses $M_f$, radii $R_f$, binding energies $B_f$, spin degeneracy factors $g_f$ of ground states. They contribute to the translation free energy and Coulomb energy.

Direct simulation of the low multiplicity multifragment disintegration.

At comparatively low excitation energy (temperature) system will disintegrate into a small number of fragments $M \leq 4$ and number of channel is not huge. For such situation a direct (microcanonical) sorting of all decay channels can be performed. Then, using Eq. ($\ref{MFS1}$), the average multiplicity value $<M>$ can be found. To check that we really have the situation with the low excitation energy, the obtained value of $<M>$ is examined to obey the inequality $<M> \leq M_0$, where $M_0 = 3.3$ and $M_0 = 2.6$ for $A \sim 100$ and for $A \sim 200$, respectively [2]. If the discussed inequality is fulfilled, then the set of channels under consideration is belived to be able for a correct description of the break up. Then using calculated according Eq. ($\ref{MFS1}$) probabilities we can randomly select a specific channel with given values of $A_f$ and $Z_f$.

Fragment multiplicity distribution.

The individual fragment multiplicities $N_{A_f,Z_f}$ in the so-called macrocanonical ensemble [1] are distributed according to the Poisson distribution:

$$P(N_{A_f,Z_f}) = \exp{(-\omega_{A_f,Z_f})} \frac{\omega_{A_f,Z_f}^{N_{A_f,Z_f}}}{N_{A_f,Z_f}!}$$ (32.14)

with mean value $<N_{A_f,Z_f}>=\omega_{A_f,Z_f}$ defined as

$$<N_{A_f,Z_f}> =g_fA_f^{3/2}\frac{V_f}{\lambda^3_{T_b}} \exp{[\frac{1}{T_b}(F_f(T_b,V)-F^{t}_f(T_b,V) - \mu A_f - \nu Z_f)]},$$ (32.15)

where $\mu$ and $\nu$ are chemical potentials. The chemical potential are found by substituting Eq. ($\ref{MFS15}$) into the system of constraints:

$$\sum_{f}<N_{A_f,Z_f}>A_f = A$$ (32.16)

and

$$\sum_{f}<N_{A_f,Z_f}>Z_f = Z$$ (32.17)

and solving it by iteration.

Atomic number distribution of fragments.

Fragment atomic numbers $A_f > 1$ are also distributed according to the Poisson distribution [1] (see Eq. ($\ref{MFS14}$)) with mean value $<N_{A_f}>$ defined as

$$<N_{A_f}> = A_f^{3/2}\frac{V_f}{\lambda^3_{T_b}} \exp{[\frac{1}{T_b}(F_f(T_b,V)-F^{t}_f(T_f,V) - \mu A_f - \nu <Z_f>)]},$$ (32.18)

where calculating the internal free energy $F_f(T_b,V)-F^{t}_f(T_b,V)$ one has to substitute $Z_f \rightarrow <Z_f>$. The average charge $<Z_f>$ for fragment having atomic number $A_f$ is given by

$$<Z_f(A_f)> = \frac{(4\gamma + \nu)A_f}{8\gamma + 2[1 - (1 + \kappa)^{-1/3}]A_f^{2/3}}.$$ (32.19)

Charge distribution of fragments.

At given mass of fragment $A_f > 1$ the charge $Z_f$ distribution of fragments are described by Gaussian

$$P(Z_f(A_f))\sim \exp{[-\frac{(Z_f(A_f) - <Z_f(A_f)>)^2}{2(\sigma_{Z_f}(A_f))^2}]}$$ (32.20)

with dispertion

$$\sigma_{Z_f(A_f)} = \sqrt{\frac{A_fT_b} {8 \gamma + 2[1 - (1... ...a)^{-1/3}]A_f^{2/3}}} \approx \sqrt{\frac{A_fT_b}{8 \gamma}}.$$ (32.21)

and the average charge $<Z_f(A_f)>$ defined by Eq. ($\ref{MFS17}$).

Kinetic energy distribution of fragments.

It is assumed [2] that at the instant of the nucleus break-up the kinetic energy of the fragment $T^{f}_{kin}$ in the rest of nucleus obeys the Boltzmann distribution at given temperature $T_b$:

$$\frac{dP(T^{f}_{kin})}{dT^{f}_{kin}}\sim \sqrt{T^{f}_{kin}} \exp{(-T^{f}_{kin}/T_b)}.$$ (32.22)

Under assumption of thermodynamic equilibrium the fragment have isotropic velocities distribution in the rest frame of nucleus. The total kinetic energy of fragments should be equal $\frac{3}{2}MT_b$, where $M$ is fragment multiplicity, and the total fragment momentum should be equal zero. These conditions are fullfilled by choosing properly the momenta of two last fragments.

The initial conditions for the divergence of the fragment system are determined by random selection of fragment coordinates distributed with equal probabilities over the break-up volume $V_f = \kappa V$. It can be a sphere or prolongated ellipsoid. Then Newton's equations of motion are solved for all fragments in the self-consistent time-dependent Coulomb field [2]. Thus the asymptotic energies of fragments determined as result of this procedure differ from the initial values by the Coulomb repulsion energy.

Calculation of the fragment excitation energies.

The temparature $T_b$ determines the average excitation energy of each fragment:

$$U_{f}(T_b) = E_f(T_b) - E_f(0) = \frac{T_b^2}{\epsilon_0}A_f... ...beta(T_b) - T_b \frac{d\beta(T_b)}{dT_b} - \beta_0]A^{2/3}_f,$$ (32.23)

where $E_f(T_b)$ is the average fragment energy at given temperature $T_b$ and $\beta(T_b)$ is defined in Eq. ($\ref{MFS8}$). There is no excitation for fragment with $A_f < 4$, for $^{4}He$ excitation energy was taken as $U_{^{4}He} = 4T^2_b/\epsilon_o$.