Abrasion process

In the abrasion process, as implemented by Wilson et al [1] it is assumed that the nuclear density for the projectile is constant up to the radius of the projectile () and zero outside. This is also assumed to be the case for the target nucleus. The amount of nuclear material abraded from the projectile is given by the expression:

$$\Delta _{abr} = FA_P \left[ {1 - \exp \left( { - \frac{{C_T }}{\lambda }} \right)} \right]$$ (26.6)

where F is the fraction of the projectile in the interaction zone, is the nuclear mean-free-path, assumed to be:

$$\lambda = \frac{{16.6}}{{E^{0.26} }}$$ (26.7)

is the energy of the projectile in MeV/nucleon and is the chord-length at the position in the target nucleus for which the interaction probability is maximum. For cases where the radius of the target nucleus is greater than that of the projectile (i.e. ):

$$C_T = \left\{ {\begin{array}{*{20}c} {2\sqrt {r_T^2 - x^2 }... ...{2\sqrt {r_T^2 - r^2 } } & {:x \le 0} \\ \end{array}} \right.$$ (26.8)

where:

$$x = \frac{{r_P^2 + r^2 - r_T^2 }}{{2r}}$$ (26.9)

In the event that then is:

$$C_T = \left\{ {\begin{array}{*{20}c} {2\sqrt {r_T^2 - x^2 } } & {:x > 0} \\ {2r_T } & {:x \le 0} \\ \end{array}} \right.$$ (26.10)

where:

$$x = \frac{{r_T^2 + r^2 - r_P^2 }}{{2r}}$$ (26.11)

The projectile and target nuclear radii are given by the expression:

$$\begin{array}{l} r_P \approx 1.29\sqrt {r_{RMS,P}^2 - 0.84^... ... r_T \approx 1.29\sqrt {r_{RMS,T}^2 - 0.84^2 } \\ \end{array}$$ (26.12)

The excitation energy of the nuclear fragment formed by the spectators in the projectile is assumed to be determined by the excess surface area, given by:

$$\Delta S = 4\pi r_P^2 \left[ {1 + P - \left( {1 - F} \right)... ...\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$3$}}} } \right]$$ (26.13)

where the functions and are given in section 26.7. Wilson et al equate this surface area to the excitation to:

$$E_S = 0.95\Delta S$$ (26.14)

if the collision is peripheral and there is no significant distortion of the nucleus, or

$$\begin{array}{l} E_S = 0.95\left\{ {1 + 5F + \Omega F^3 } \... ... {:12 \le A_P \le 16} \\ \end{array}} \right. \\ \end{array}$$ (26.15)

if the impact separation is such that +. is in MeV provided is in fm$^2$.

For the abraded region, Wilson et al assume that fragments with a nucleon number of five are unbounded, 90% of fragments with a nucleon number of eight are unbound, and 50% of fragments with a nucleon number of nine are unbound. This was not implemented within the Geant4 version of the abrasion model, and disintegration of the pre-fragment was only simulated by the subsequent de-excitation physics models in the G4DeexcitationHandler (evaporation, etc. or G4WilsonAblationModel) since the yields of lighter fragments were already underestimated compared with experiment.

In addition to energy as a result of the distortion of the fragment, some energy is assumed to be gained from transfer of kinetic energy across the boundaries of the nuclei. This is approximated to the average energy transferred to a nucleon per unit intersection pathlength (assumed to be 13 MeV/fm) and the longest chord-length, , and for half of the nucleon-nucleon collisions it is assumed that the excitation energy is:

$$E_X^* = \left\{ { \begin{array}{*{20}c} {13 \cdot \left[ {1... ... {\rm C}_l } & {:C_t \le 1.5{\rm fm}} \\ \end{array}} \right.$$ (26.16)

where:

$$C_l = \left\{ { \begin{array}{*{20}c} {2\sqrt {r_P^2 + 2rr_... ...r > r_T } \\ {2r_P } & {r \le r_T } \\ \end{array}} \right.$$ (26.17)

$$C_t = 2\sqrt {r_P^2 - \frac{{\left( {r_P^2 + r^2 - r_T^2 } \right)^2 }}{{4r^2 }}}$$ (26.18)

For the remaining events, the projectile energy is assumed to be unchanged. Wilson et al assume that the energy required to remove a nucleon is 10MeV, therefore the number of nucleons removed from the projectile by ablation is:

$$\Delta _{abl} = \frac{{E_S + E_X }}{{10}} + \Delta _{spc}$$ (26.19)

where is the number of loosely-bound spectators in the interaction region, given by:

$$\Delta _{spc} = A_P F\exp \left( { - \frac{{C_T }}{\lambda }} \right)$$ (26.20)

Wilson et al appear to assume that for half of the events the excitation energy is transferred into one of the nuclei (projectile or target), otherwise the energy is transferred in to the other (target or projectile respectively).

The abrasion process is assumed to occur without preference for the nucleon type, i.e. the probability of a proton being abraded from the projectile is proportional to the fraction of protons in the original projectile, therefore:

$$\Delta Z_{abr} = \Delta _{abr} \frac{{Z_P }}{{A_P }}$$ (26.21)

In order to calculate the charge distribution of the final fragment, Wilson et al assume that the products of the interaction lie near to nuclear stability and therefore can be sampled according to the Rudstam equation (see section 26.6). The other obvious condition is that the total charge must remain unchanged.