\section{Form of the Tersoff potential and its derivative}
-The Tersoff potential \cite{tersoff_m} is of the form
+The Tersoff potential~\cite{tersoff_m} is of the form
\begin{eqnarray}
E & = & \sum_i E_i = \frac{1}{2} \sum_{i \ne j} V_{ij} \textrm{ ,} \\
V_{ij} & = & f_C(r_{ij}) [ f_R(r_{ij}) + b_{ij} f_A(r_{ij}) ] \textrm{ .}
0, & r_{ij} > S_{ij}
\end{array} \right.
\end{equation}
-with $\theta_{ijk}$ being the bond angle between bonds $ij$ and $ik$ as shown in Figure \ref{img:tersoff_angle}.\\
+with $\theta_{ijk}$ being the bond angle between bonds $ij$ and $ik$ as shown in Figure~\ref{img:tersoff_angle}.\\
\\
For a three body potential, if $V_{ij}$ is not equal to $V_{ji}$, the derivative is of the form
\begin{equation}
\nabla_{{\bf r}_i} E = \frac{1}{2} \big[ \sum_j ( \nabla_{{\bf r}_i} V_{ij} + \nabla_{{\bf r}_i} V_{ji} ) + \sum_k \sum_j \nabla_{{\bf r}_i} V_{jk} \big] \textrm{ .}
\end{equation}
-In the following all the necessary derivatives to calculate $\nabla_{{\bf r}_i} E$ are done.
+In the following all the necessary derivatives to calculate $\nabla_{{\bf r}_i} E$ are written down.
- \section{Derivative of $V_{ij}$ with respect to ${\bf r}_i$}
+ \section[Derivative of $V_{ij}$ with respect to ${r}_i$]{\boldmath Derivative of $V_{ij}$ with respect to ${\bf r}_i$}
\begin{eqnarray}
\nabla_{{\bf r}_i} V_{ij} & = & \nabla_{{\bf r}_i} f_C(r_{ij}) \big[ f_R(r_{ij}) + b_{ij} f_A(r_{ij}) \big] + \nonumber \\
& = & \Big[ \frac{\cos\theta_{ijk}}{r_{ij}^2} - \frac{1}{r_{ij} r_{ik}} \Big] {\bf r}_{ij} + \Big[ \frac{\cos\theta_{ijk}}{r_{ik}^2} - \frac{1}{r_{ij} r_{ik}} \Big] {\bf r}_{ik}
\end{eqnarray}
- \section{Derivative of $V_{ji}$ with respect to ${\bf r}_i$}
+ \section[Derivative of $V_{ji}$ with respect to ${r}_i$]{\boldmath Derivative of $V_{ji}$ with respect to ${\bf r}_i$}
\begin{eqnarray}
\nabla_{{\bf r}_i} V_{ji} & = & \nabla_{{\bf r}_i} f_C(r_{ji}) \big[ f_R(r_{ji}) + b_{ji} f_A(r_{ji}) \big] + \nonumber \\
& = & \frac{1}{r_{ji} r_{jk}} {\bf r}_{jk} - \frac{\cos\theta_{jik}}{r_{ji}^2} {\bf r}_{ji}
\end{eqnarray}
- \section{Derivative of $V_{jk}$ with respect to ${\bf r}_i$}
+ \section[Derivative of $V_{jk}$ with respect to ${r}_i$]{\boldmath Derivative of $V_{jk}$ with respect to ${\bf r}_i$}
\begin{eqnarray}
\nabla_{{\bf r}_i} V_{jk} & = & f_C(r_{jk}) f_A(r_{jk}) \nabla_{{\bf r}_i} b_{jk} \\
keeping in mind that all the necessary force contributions for atom $i$
are calculated and added in subsequent loops.
-\subsection{Derivative of $V_{ij}$ with respect to ${\bf r}_j$}
+\subsection[Derivative of $V_{ij}$ with respect to ${r}_j$]{\boldmath Derivative of $V_{ij}$ with respect to ${\bf r}_j$}
\begin{eqnarray}
\nabla_{{\bf r}_j} V_{ij} & = &
The contribution of the bond order term is given by:
\begin{eqnarray}
\nabla_{{\bf r}_j}\cos\theta_{ijk} &=&
- \nabla_{{\bf r}_j}\Big(\frac{{\bf r}_{ij}{\bf }r_{ik}}{r_{ij}r_{ik}}\Big)
+ \nabla_{{\bf r}_j}\Big(\frac{{\bf r}_{ij}{\bf r}_{ik}}{r_{ij}r_{ik}}\Big)
\nonumber \\
&=& \frac{1}{r_{ij}r_{ik}}{\bf r}_{ik} -
\frac{\cos\theta_{ijk}}{r_{ij}^2}{\bf r}_{ij}
\end{eqnarray}
-\subsection{Derivative of $V_{ij}$ with respect to ${\bf r}_k$}
+\subsection[Derivative of $V_{ij}$ with respect to ${r}_k$]{\boldmath Derivative of $V_{ij}$ with respect to ${\bf r}_k$}
The derivative of $V_{ij}$ with respect to ${\bf r}_k$ just consists of the
single term
\subsection{Code realization}
-The implementation of the force evaluation shown in the following is applied to the potential designed by Erhard and Albe \cite{albe_sic_pot}.
+The implementation of the force evaluation shown in the following is applied to the potential designed by Erhart and Albe~\cite{albe_sic_pot}.
There are slight differences compared to the original potential by Tersoff:
\begin{itemize}
\item Difference in sign of the attractive part.
\item
\item LOOP k \{
\begin{itemize}
- \item set $ik$-depending values
+ \item set $ik$-dependent values
\item calculate: $r_{ik}$, $r_{ik}^2$
\item IF $r_{ik} > S_{ik}$ THEN CONTINUE
\item calculate: $\theta_{ijk}$, $\cos(\theta_{ijk})$,