3.3.3.1 Energy Loss of a Point Charge in a Free Electron Gas

A solution of this problem was already given by Lindhard [50]. The energy loss $\frac{dE}{d\vec{r}}$ of the projectile is determined by the force resulting from the electric field $\vec{E}(\vec{r},t)$ which is generated in the plasma by the penetrating charged particle with a charge $Z_1$ moving with the velocity $\vec{v}$. For his solution Lindhard applied the linear response theory.

$$\frac{dE}{dR} = \frac{\vec{v}}{\vert\vec{v}\vert}\cdot \vec{E}(\vec{r},t)\cdot \rho_0(\vec{r},t)$$ (3.100)

$$\rho_0(\vec{r},t) = Z_1\cdot e\cdot \delta(\vec{r}-\vec{v}\cdot t)$$ (3.101)

$$\mathrm{div} \left(\epsilon(\vec{r},t)\cdot \vec{E}(\vec{r},t)\right) = 4\cdot \pi\cdot \rho_0(\vec{r},t)$$ (3.102)

$\epsilon(\vec{r},t)$ is the linear dielectric response function of the medium on the charge.

This set of differential equations can be solved by Fourier transformation [50] ( $v = \vert\vec{v}\vert$).

$$\frac{dE_l}{dR} = \frac{Z_1^2\cdot e^2}{\pi\cdot v}\cdot Im\left\... ...int\limits_{kv}^{-kv} d\omega\cdot \frac{\omega}{\epsilon_l(k,\omega)} \right\}$$ (3.103)

$$\frac{dE_{tra}}{dR} = \frac{Z_1^2\cdot e^2}{\pi\cdot v}\cdot Im\l... ...k^2\cdot v^2}}{k^2-\frac{\omega^2}{c^2}\cdot \epsilon_{tra}(k,\omega)} \right\}$$ (3.104)

The contributions to the energy loss by the transversal electric field $\vec{E}_{tra}$ and the longitudinal electric field $\vec{E}_l$ are treated separately, which requires the definition of a transversal $\epsilon_{tra}(k,\omega)$ (Fourier transform of $\epsilon_{tra}(\vec{r},t)$) and a longitudinal $\epsilon_l(k,\omega)$ (Fourier transform of $\epsilon_l(\vec{r},t)$) dielectric response function. $\epsilon_{tra}(k,\omega)$ and $\epsilon_l(k,\omega)$ are defined in Fourier space via the scalar potential $\Phi(k,\omega)$ and vector potential $\vec{A}(k,\omega)$, describing the electromagnetic field as shown in (3.105) and (3.106), and the Maxwell equations.

$$\vec{E} = - \mathrm{grad}\Phi(\vec{r},t) - \frac{1}{c}\cdot \frac{\partial \vec{A}(\vec{r},t)}{\partial t}$$ (3.105)

$$\vec{B} = \mathrm{rot}\vec{A}(\vec{r},t)$$ (3.106)

$\epsilon_l(k,\omega)$ is defined by

$$\epsilon_l(k,\omega)\cdot k^2 \cdot \Phi(k,\omega) = 4\cdot \pi\cdot \rho_0(k,\omega)$$ (3.107)

while $\epsilon_{tra}(k,\omega)$ is defined by

$$(k^2-\frac{\omega^2}{c^2}\cdot \epsilon_{tra}(k,\omega))\cdot \vec{A}_{tra}(k,\omega) = \frac{4\cdot \pi}{c}\cdot j_{0tra}(k,\omega)$$ (3.108)

The transversal field and the transversal current density are characterized by

$$\vec{k}\cdot \vec{A}_{tra}(k,\omega) = 0$$ (3.109)

$$\vec{k}\cdot \vec{j}_{0tra}(k,\omega) = 0$$ (3.110)

The total energy loss is a sum of the longitudinal and the transversal contribution, respectively. The transversal contribution is negligible since the particle velocity is non-relativistic [50]. This requirement holds for typical ion implantation conditions.

Lindhard derived an expression for the longitudinal dielectric response function $\epsilon_l(k,\omega)$ by applying a self-consistent treatment of the electron gas. He calculates the charge density $\rho_0$ and the current density $j_0$ induced in the plasma by the forces resulting from the scalar potential $\Phi$ and the vector potential $\vec{A}$. These induced sources produce an electro-magnetic field which acts on the electron gas. Therefore the electro-magnetic field must be self-consistent with the induced sources. A quantum-mechanical calculation of this problem results in

$$\begin{split}\epsilon_l &= 1+\frac{2\cdot m^2\cdot \omega_0^2... ...\omega+i\cdot \frac{\gamma}{\hbar}\right)} \right\} \end{split}$$ (3.111)

with

$$\omega_0^2 = \frac{4\cdot \pi\cdot e^2\cdot \varrho}{m}$$ (3.112)

being the classical resonance frequency of the gas. $f(E_n)$ is the electron density distribution function, $m$ is the electron mass, $E_n$ is the energy, $k_n$ the wave vector of the electron in the nth state and $\gamma$ is a small damping factor. By applying this dielectric response function to (3.103) Lindhard derived the energy loss in an electron gas described by the Fermi distribution in the ground state with a maximum energy $E_0 = \frac{m\cdot v_0^2}{2}$.

$$\frac{dE}{dR} = \frac{4\cdot \pi\cdot z_1^2\cdot e^4}{m\cdot v^2}\cdot\varrho\cdot L$$ (3.113)

$$L = \frac{6}{\pi}\int\limits_0^{\frac{v}{v_0}}u\cdot \;du\cdot \i... ...z\cdot \frac{f_2(u,z)}{(z^2+\chi^2\cdot f_1(u,z))^2 + (\chi^2\cdot f_2(u,z))^2}$$ (3.114)

$$\begin{split}f(u',z) &= \frac{1}{2}+ \frac{1}{8\cdot z}\cdot ... ... \right\}\cdot \ln\left(\frac{z+u'+1}{z+u'-1}\right)\end{split}$$ (3.115)

$$u' = \frac{\omega+i\cdot \frac{\gamma}{\hbar}}{k\cdot v_0}$$ (3.116)

$$u = \frac{\vert\omega\vert}{k\cdot v_0}$$ (3.117)

$$\chi^2 = \frac{e^2}{\pi\cdot \hbar\cdot v_0}$$ (3.118)

Assuming a small damping factor $\gamma$ and a small particle velocity $v$, (3.113) can be simplified to an electronic stopping equation which is proportional to velocity.

$$\frac{dE}{dR} = \frac{4\cdot z_1^2\cdot e^4\cdot m^2}{3\cdot \pi\... ...cdot \int\limits_0^1\frac{z^3\cdot dz}{\left(z^2+\chi^2\cdot f_1(0,z)\right)^2}$$ (3.119)

$f_1(u,z)$ is the real part of (3.115), while $f_2(u,z)$ is the imaginary part.