forgot to change ...

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

diff --git a/posic/thesis/d_tersoff.tex b/posic/thesis/d_tersoff.tex
index 9a5e76b..b031be2 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}
 
-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{ .}
@@ -32,15 +32,15 @@ f_C(r_{ij}) = \left\{
     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 \\
@@ -67,7 +67,7 @@ In the following all the necessary derivatives to calculate $\nabla_{{\bf r}_i}
  & = & \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 \\
@@ -95,7 +95,7 @@ In the following all the necessary derivatives to calculate $\nabla_{{\bf r}_i}
  & = & \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} \\
@@ -128,7 +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.
 
-\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} & = &
@@ -138,7 +138,7 @@ are calculated and added in subsequent loops.
  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}) \\
@@ -149,13 +149,13 @@ 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
@@ -185,7 +185,7 @@ Concerning $b_{ij}$, in addition to the angular term, the derivative of the cut-
 
   \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.
@@ -220,7 +220,7 @@ LOOP i \{
        \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: $\theta_{ijk}$, $\cos(\theta_{ijk})$,