first sent away beta version

[lectures/latex.git] / posic / thesis / d_tersoff.tex

diff --git a/posic/thesis/d_tersoff.tex b/posic/thesis/d_tersoff.tex
index cb2e96b..9a5e76b 100644 (file)
--- a/posic/thesis/d_tersoff.tex
+++ b/posic/thesis/d_tersoff.tex
@@ -3,7 +3,7 @@
 
   \section{Form of the Tersoff potential and its derivative}
 
 
   \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{ .}
 \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{ .}


@@ -22,7 +22,7 @@ with
 \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}
 \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}
 \begin{equation}
 f_C(r_{ij}) = \left\{
   \begin{array}{ll}


@@ -109,7 +109,7 @@ In the following all the necessary derivatives to calculate $\nabla_{{\bf r}_i}
 
   \section{Implementation issues}
 
 
   \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 $V_{jki}$
 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 $V_{jki}$


@@ -118,7 +118,7 @@ In simulation, however, it is not practical to evaluate all three potential
 derivatives for each $(ijk)$ triple.
 The $V_{jik}$ and $V_{jki}$ potential and its derivatives will be calculated
 in subsequent loops anyways.
 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 ${\bf r}_k$
 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 ${\bf r}_k$


@@ -128,9 +128,7 @@ This poses a more convenient method to obtain the forces
 keeping in mind that all the necessary force contributions for atom $i$
 are calculated and added in subsequent loops.
 
 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 ${\bf r}_j$}

 
 \begin{eqnarray}
  \nabla_{{\bf r}_j} V_{ij} & = &
 
 \begin{eqnarray}
  \nabla_{{\bf r}_j} V_{ij} & = &


@@ -147,8 +145,8 @@ the following relations are valid:
  \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}
  \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)
 \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)


@@ -157,7 +155,7 @@ Now having a look at $b_{ij}$.
      \frac{\cos\theta_{ijk}}{r_{ij}^2}{\bf r}_{ij}
 \end{eqnarray}
 
      \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 ${\bf r}_k$}

 
 The derivative of $V_{ij}$ with respect to ${\bf r}_k$ just consists of the
 single term
 
 The derivative of $V_{ij}$ with respect to ${\bf r}_k$ just consists of the
 single term


@@ -172,14 +170,14 @@ vanish.
  \nabla_{{\bf r}_k} f_A(r_{ij}) &=& 0 \\
  \nabla_{{\bf r}_k} f_C(r_{ij}) &=& 0
 \end{eqnarray}
  \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} &=&
 \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}
  \nonumber \\
  &=&\frac{1}{r_{ij}r_{ik}}{\bf r}_{ij} -
  \frac{\cos\theta_{ijk}}{r_{ik}^2}{\bf r}_{ik}


@@ -187,9 +185,8 @@ Now look at $b_{ij}$, not only angle important here!
 
   \subsection{Code realization}
 
 
   \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 Erhard 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 $i$.
 \begin{itemize}
  \item Difference in sign of the attractive part.
  \item $c$, $d$ and $h$ values depend on atom $k$ in addition to atom $i$.


@@ -198,11 +195,14 @@ There are slight differences comparted to the original potential by Tersoff:
  \item The exponent of the $b$ term is constantly $-\frac{1}{2}$.
 \end{itemize}
 These differences actually slightly ease code realization.
  \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{}
 
 \begin{figure}
 \renewcommand\labelitemi{}
 \renewcommand\labelitemii{}
 \renewcommand\labelitemiii{}

+{\small
+\fbox{\begin{minipage}{\textwidth}

 LOOP i \{
 \begin{itemize}
  \item // nop (only used in orig. Tersoff)
 LOOP i \{
 \begin{itemize}
  \item // nop (only used in orig. Tersoff)


@@ -214,7 +214,7 @@ LOOP i \{
  \item LOOP j \{
        \begin{itemize}
         \item $\zeta_{ij}=0$
  \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 calculate: $r_{ij}$, $r_{ij}^2$
         \item IF $r_{ij} > S_{ij}$ THEN CONTINUE
        \item


@@ -267,7 +267,9 @@ f_C(r_{ik})dg(\theta_{ijk})\nabla_{{\bf r}_k}\cos\theta_{ijk}\big)$
  \item \}
 \end{itemize}
 \}
  \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}
 
 \end{figure}