+The function $b_{ij}$ represents a measure of the bond order, monotonically decreasing with the coordination of atoms $i$ and $j$.
+It is of the form:
+\begin{eqnarray}
+b_{ij} & = & \chi_{ij} (1 + \beta_i^{n_i} \zeta^{n_i}_{ij})^{-1/2n_i} \\
+\zeta_{ij} & = & \sum_{k \ne i,j} f_C (r_{ik}) \omega_{ik} g(\theta_{ijk}) \\
+g(\theta_{ijk}) & = & 1 + c_i^2/d_i^2 - c_i^2/[d_i^2 + (h_i - \cos \theta_{ijk})^2] \\
+b_{ij} & = & \chi_{ij} \Big( 1 + \beta_i^{n_i} \Big[ \sum_{k \ne i,j} f_C (r_{ik}) \omega_{ik} \big[ 1 + c_i^2/d_i^2 - c_i^2/[d_i^2 + (h_i - \cos \theta_{ijk})^2] \big] \Big] \Big)^{-1/2n_i}
+\end{eqnarray}
+where $\theta_{ijk}$ is the bond angle between bonds $ij$ and $ik$.
+This is illustrated in Figure \ref{img:tersoff_angle}.
+
+\printimg{!h}{width=8cm}{tersoff_angle.eps}{Angle between bonds of atoms $i,j$ and $i,k$.}{img:tersoff_angle}
+
+In order to calculate the forces the derivation of the potential with respect to $x^i_n$ (the $n$th component of the position vector of atom $i$ $\equiv$ ${\bf r}_i$) has to be known.
+This is gradually done in the following.
+The $n$th component of the force acting on atom $i$ is
+\begin{eqnarray}
+F_n^i & = & - \frac{\partial}{\partial x_n^i} \sum_{j \neq i} V_{ij} \nonumber\\
+ & = & \sum_{j \neq i} \Big( \partial_{x_n^i} f_C(r_{ij}) \big[ f_R(r_{ij}) + b_{ij} f_A(r_{ij}) \big] + \nonumber\\
+& & + f_C(r_{ij}) \big[ \partial_{x_n^i} f_R(r_{ij}) + b_{ij} \partial_{x_n^i} f_A(r_{ij}) + f_A(r_{ij}) \partial_{x_n^i} b_{ij} \big] \Big) \textrm{ .}
+\end{eqnarray}
+For the implementation it is helpful to seperate the two and three body terms.
+\begin{eqnarray}
+F_n^i & = & \sum_{j \neq i} \Big( f_R(r_{ij}) \partial_{x_n^i} f_C(r_{ij}) + f_C(r_{ij}) \partial_{x_n^i} f_R(r_{ij}) \Big) + \nonumber\\
+& + & \sum_{j \neq i} \Big( \partial_{x_n^i} f_C(r_{ij}) b_{ij} f_A(r_{ij}) + f_C(r_{ij}) \big[ b_{ij} \partial_{x_n^i} f_A(r_{ij}) + f_A(r_{ij}) \partial_{x_n^i} b_{ij} \big] \Big)
+\end{eqnarray}
+The cutoff function $f_C$ derivated with repect to $x^i_n$ is