\section{Form of the Tersoff potential and its derivative}
-The Tersoff potential 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{ .}
\zeta_{ij} & = & \sum_{k \ne i,j} f_C (r_{ik}) \omega_{ik} g(\theta_{ijk}) \textrm{ ,}\\
g(\theta_{ijk}) & = & 1 + c_i^2/d_i^2 - c_i^2/[d_i^2 + (h_i - \cos \theta_{ijk})^2] \textrm{ .}
\end{eqnarray}
-The cutoff function $f_C$ is taken to be
+The cut-off function $f_C$ is taken to be
\begin{equation}
f_C(r_{ij}) = \left\{
\begin{array}{ll}
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} \\
\section{Implementation issues}
-As seen in the last sections the derivatives of $V_{ij}$, $V_{ji}$ and $V_{jk}$
+As seen in the last sections, the derivatives of $V_{ij}$, $V_{ji}$ and $V_{jk}$
with respect to ${\bf r}_i$ are necessary to compute the forces for atom $i$.
According to this, for every triple $(ijk)$ the derivatives of the three
potential contributions, denoted by $V_{ijk}$, $V_{jik}$ and $V_{jki}$
derivatives for each $(ijk)$ triple.
The $V_{jik}$ and $V_{jki}$ potential and its derivatives will be calculated
in subsequent loops anyways.
-To avoid multiple computation of the same potential derivatives
+To avoid multiple computation of the same potential derivatives,
the force contributions for atom $j$ and $k$ due to the $V_{ijk}$ contribution
have to be considered by calculating the derivatives of $V_{ijk}$
with respect to ${\bf r}_j$ and ${\bf r}_k$
keeping in mind that all the necessary force contributions for atom $i$
are calculated and added in subsequent loops.
-The following symmetry considerations help to obtain the
-
- \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} & = &
b_{ij} \nabla_{{\bf r}_j} f_A(r_{ij}) +
f_A(r_{ij}) \nabla_{{\bf r}_j} b_{ij} \big]
\end{eqnarray}
-Using the equality $\nabla_{{\bf r}_i} r_{ij}=-\nabla_{{\bf r}_j} r_{ij}$
+Using the equality $\nabla_{{\bf r}_i} r_{ij}=-\nabla_{{\bf r}_j} r_{ij}$,
the following relations are valid:
\begin{eqnarray}
\nabla_{{\bf r}_j} f_R(r_{ij}) &=& - \nabla_{{\bf r}_i} f_R(r_{ij}) \\
\nabla_{{\bf r}_j} f_A(r_{ij}) &=& - \nabla_{{\bf r}_i} f_A(r_{ij}) \\
\nabla_{{\bf r}_j} f_C(r_{ij}) &=& - \nabla_{{\bf r}_i} f_C(r_{ij})
\end{eqnarray}
-The pair contributions .... easy.
-Now having a look at $b_{ij}$.
+The pair contributions are, thus, easily obtained.
+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
\nabla_{{\bf r}_k} f_A(r_{ij}) &=& 0 \\
\nabla_{{\bf r}_k} f_C(r_{ij}) &=& 0
\end{eqnarray}
-Now look at $b_{ij}$, not only angle important here!
+Concerning $b_{ij}$, in addition to the angular term, the derivative of the cut-off function has to be considered.
\begin{eqnarray}
\nabla_{{\bf r}_k}\zeta_{ij} &=&
g(\theta_{ijk})\nabla_{{\bf r}_k}f_C(r_{ik}) +
f_C(r_{ik})\nabla_{{\bf r}_k}g(\theta_{ijk}) \\
\nabla_{{\bf r}_k}f_C(r_{ik}) &=& - \nabla_{{\bf r}_i}f_C(r_{ik}) \\
\nabla_{{\bf r}_k}\cos\theta_{ijk} &=&
- \nabla_{{\bf r}_k}\Big(\frac{{\bf r}_{ij}{\bf }r_{ik}}{r_{ij}r_{ik}}\Big)
+ \nabla_{{\bf r}_k}\Big(\frac{{\bf r}_{ij}{\bf r}_{ik}}{r_{ij}r_{ik}}\Big)
\nonumber \\
&=&\frac{1}{r_{ij}r_{ik}}{\bf r}_{ij} -
\frac{\cos\theta_{ijk}}{r_{ik}^2}{\bf r}_{ik}
\subsection{Code realization}
-The implementation of the force evaluation shown in the following
-is applied to the potential designed by Erhard and Albe.
-There are slight differences comparted to the original potential by Tersoff:
+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 $c$, $d$ and $h$ values depend on atom $k$ in addition to atom $i$.
\item The exponent of the $b$ term is constantly $-\frac{1}{2}$.
\end{itemize}
These differences actually slightly ease code realization.
+The respective flow chart is displayed in Fig.~\ref{fig:flowchart}.
\begin{figure}
\renewcommand\labelitemi{}
\renewcommand\labelitemii{}
\renewcommand\labelitemiii{}
+{\small
+\fbox{\begin{minipage}{\textwidth}
LOOP i \{
\begin{itemize}
\item // nop (only used in orig. Tersoff)
\item LOOP j \{
\begin{itemize}
\item $\zeta_{ij}=0$
- \item set $S_{ij}$ (cutoff)
+ \item set $S_{ij}$ (cut-off)
\item calculate: $r_{ij}$, $r_{ij}^2$
\item IF $r_{ij} > S_{ij}$ THEN CONTINUE
\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: $\theta_{ijk}$, $\cos(\theta_{ijk})$,
\item \}
\end{itemize}
\}
-\caption{Implementation of the force evaluation for Tersoff like bond-order
- potentials using pseudocode.}
+\end{minipage}}
+}
+\caption{Flow chart of the force evaluation for Tersoff-like bond order potentials using pseudocode.}
+\label{fig:flowchart}
\end{figure}
projects / lectures / latex.git / blobdiff