more fixes

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

diff --git a/posic/thesis/d_tersoff.tex b/posic/thesis/d_tersoff.tex
index 9a5e76b..8492963 100644 (file)
--- a/posic/thesis/d_tersoff.tex
+++ b/posic/thesis/d_tersoff.tex
@@ -38,9 +38,9 @@ For a three body potential, if $V_{ij}$ is not equal to $V_{ji}$, the derivative
 \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} & = &
@@ -155,7 +155,7 @@ The contribution of the bond order term is given by:
      \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