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--> --> --> -->A.Trajectory of incident nucleon by classical mechanics
It is assumed that the mean field provided by the target nucleus stays static during the interaction. The time evolution of the incident nucleon in the mean field is governed by the Newtonian equation of motion $ \vec{F} = \mu\frac{{\rm d}^{2} \vec{r}}{{\rm d}t^{2}}, $ | (1) |
$\overrightarrow {{F_c}} (r) = \left\{ {\begin{array}{*{20}{l}}{\dfrac{{CZzr}}{{{R^3}}}\dfrac{{\vec r}}{r},\quad }&{r < R}\\{\dfrac{{CZz}}{{{r^2}}}\dfrac{{\vec r}}{r},\quad }&{r \geqslant R}\end{array}} \right. $ | (2) |
The nuclear force is written as
$ \begin{aligned}[b] \vec{F_{n}}(r) =& -\frac{\partial V_{n}}{\partial r}\frac{\vec{r}}{r}, \\ V_{n} =& \alpha\frac{\rho}{\rho_{0}}+\beta\frac{\rho^\gamma}{\rho_{0}^\gamma} +\frac{C_{sp}(\gamma_{i}+1)}{2}\frac{\rho^{\gamma_{i}}}{\rho_{0}^{\gamma_{i}}} \delta ^{2} +\frac{|\tau|}{\tau} \frac{C_{sp}}{2}\frac{\rho^{\gamma_{i}+1}}{\rho_{0}^{\gamma_{i}}}2\delta,\end{aligned} $ | (3) |
The following form of the nucleon density was used:
$ \rho_{i}(r) = \dfrac{\rho_{i0}}{1+{\rm exp}\left(\dfrac{r-R_{i}}{d_{i}}\right)}, $ | (4) |
$ \begin{aligned}[b] \rho_{p_{0}}+\rho_{n_{0}} &= \rho_{0},\\ \frac{\rho_{p_{0}}}{\rho_{n_{0}}} &= \frac{Z}{N},\\ \int\rho_{p}(r){\rm d}\vec{r} &= Z,\\ \int\rho_{n}(r){\rm d}\vec{r} &= N,\\\int{r}^2\rho_{p}(r){\rm d}\vec{r} &= Z\langle{r}^2\rangle_{p},\\ \int{r}^2\rho_{n}(r){\rm d}\vec{r} &= N\langle{r}^2\rangle_{n}, \end{aligned} $ | (5) |
Solving the equation of motion, i.e., Eq. (1), with initial conditions depending on impact parameter b and incident energy
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B.Total reaction cross section of the reaction between nucleon and nucleus
Suppose that a nucleon beam with intensity I passes through a nuclear matter with density $ \frac{{\rm d}I}{I} = -\sigma\rho {\rm d}x, $ | (6) |
$ P = \frac{I^{\rm out}}{I^{\rm in}} = {\rm exp}\left [-\int_{-\infty}^{+\infty} \sigma \rho v {\rm d}t\right]. $ | (7) |
$ \sigma_{t} = \int_{0}^{+\infty}2\pi b[1-P(b)]{\rm d}b. $ | (8) |
$\sigma _{pp}^{\rm free} = \sigma _{nn}^{\rm free} = \left\{ {\begin{array}{*{20}{l}}{34{{\left( {\dfrac{{{p_{{\rm{lab}}}}}}{{0.4}}} \right)}^{ - 2.104}},}&{{p_{{\rm{lab}}}}\leqslant 0.4}\\{23.5 + 1000{{\left( {{p_{{\rm{lab}}}} - 0.7} \right)}^4},}&{0.4 < {p_{{\rm{lab}}}}\leqslant 0.8}\\{23.5 + \dfrac{{24.6}}{{1 + \exp \left( { - \left( {{p_{{\rm{lab}}}} - 1.2} \right)/0.10} \right)}},}&{0.8 < {p_{{\rm{lab}}}}\leqslant 1.5}\\{41 + 60\left( {{p_{{\rm{lab}}}} - 0.9} \right)\exp \left( { - 1.2{p_{{\rm{lab}}}}} \right),}&{1.5 < {p_{{\rm{lab}}}}\leqslant 5.0}\end{array}} \right.$ | (9) |
$ \sigma _{np}^{\rm free} = \sigma _{pn}^{\rm free} = \left\{ {\begin{array}{*{20}{l}}{6.3555p_{\rm lab}^{ - 3.2481}\exp \left( { - 0.377{{\left( {\ln {p_{{\rm{lab}}}}} \right)}^2}} \right),}&{{p_{{\rm{lab}}}} \leqslant 0.4}\\{33 + 196{{\left| {{p_{{\rm{lab}}}} - 0.95} \right|}^{2.5}},}&{0.4 < {p_{{\rm{lab}}}} \leqslant 1.0}\\{24.2 + 8.9{p_{{\rm{lab}}}},}&{1.0 < {p_{{\rm{lab}}}} \leqslant 2.0}\\{42,}&{2.0 < {p_{{\rm{lab}}}} \leqslant 5.0}\end{array}} \right.$ | (10) |
$\sigma \rho = \left\{ {\begin{array}{*{20}{l}}{\sigma _{pp}^{\rm free}{\rho _p} + \sigma _{pn}^{\rm free}{\rho _n},\quad {\rm for}\; p\; {\rm reaction}}\\{\sigma _{nn}^{\rm free}{\rho _n} + \sigma _{np}^{\rm free}{\rho _p},\quad {\rm for}\;n\; {\rm reaction}.}\end{array}} \right.$ | (11) |
Third, the in-medium factor proposed by Cai et al. [40] was considered.
$ \sigma \rho = \left\{{\begin{array}{*{20}{l}}{\sigma _{pp}^{\rm med}{\rho _p} + \sigma _{pn}^{\rm med}{\rho _n},\quad {\rm for}\; p\; {\rm reaction}}\\{\sigma _{nn}^{\rm med}{\rho _n} + \sigma _{np}^{\rm med}{\rho _p},\quad {\rm for}\; n\; {\rm reaction},}\end{array}} \right.$ | (12) |
$ \begin{aligned}[b]&\sigma_{p p}^{\rm med} = \sigma_{n n}^{\rm med } = \frac{1.0+0.1667 E_{\rm lab}^{1.05} \rho^{3}}{1.0+9.704 \rho^{1.2}} \sigma_{p p}^{\rm free}, \\ &\sigma_{n p}^{\rm med} = \sigma_{pn}^{\rm med} = \frac{1.0+0.0034 E_{\rm lab}^{1.51} \rho^{2}}{1.0+21.55 \rho^{1.34}} \sigma_{n p}^{\rm free}, \end{aligned} $ | (13) |
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C.Deuteron-induced reaction
Let us assume that the positions of proton and neutron in the deuteron satisfy a Gaussian distribution. Let the center of the deuteron be at the origin of the coordinates; then, the probability density function that the neutron or proton is at point $\begin{aligned}[b] f =& A\exp\left(-\frac{x_{d}^2+y_{d}^2+z_{d}^2}{L}\right)\\ =& A\exp\left(-\frac{r_{d}^2}{L}\right),\end{aligned} $ | (14) |
$ \int f(r_{d}){\rm d}\vec{r_{d}} = 1, $ | (15) |
$ \int {r_{d}}^2 f(r_{d}){\rm d}\vec{r_{d}} = \langle{r}^2\rangle = (2.1421)^2{\rm fm}^{2}. $ | (16) |
The deuteron-induced reaction is equivalent to one neutron and one proton reacting with the target nucleus. Let deuteron be incident with impact parameter b in the z-axis direction, and let us establish coordinates with deuteron center as origin. Then, according to the determined Gaussian distribution function, the probability of the neutron being located at point
$ b_{p} = \sqrt{(b-x_{d})^2+y_{d}^2},\quad b_{n} = \sqrt{(b+x_{d})^2+y_{d}^2}. $ | (17) |
$\begin{aligned}[b] \sigma_{d-ABS} =& \int 2\pi bdb\:\:[1-P_{n}(b_{n})][1\\&-P_{p}(b_{p})] \:\:f {\rm d}x_{d}{\rm d}y_{d}{\rm d}z_{d}, \end{aligned}$ | (18) |
$ \sigma_{p-NEB} = \int 2\pi bdb\:\: P_{n}(b_{n})[1-P_{p}(b_{p})] \:\: f {\rm d}x_{d}{\rm d}y_{d}{\rm d}z_{d}, $ | (19) |
$ \sigma_{n-NEB} = \int 2\pi bdb\:\:[1-P_{n}(b_{n})]P_{p}(b_{p}) \:\: f {\rm d}x_{d}{\rm d}y_{d}{\rm d}z_{d}. $ | (20) |
$ \sigma_{t} = \sigma_{d-ABS}+\sigma_{p-NEB}+\sigma_{n-NEB}. $ | (21) |