X-Git-Url: https://hackdaworld.org/gitweb/?p=lectures%2Flatex.git;a=blobdiff_plain;f=posic%2Fthesis%2Fd_tersoff.tex;h=b031be2f58ecd016152de51b4e26e136c23450aa;hp=c01a65eaf34d207e955e3b23b4f2275909113b85;hb=17d5c879c418790a154098e51c524eca183c4d98;hpb=fdf1f976b879c9b7403c1d76c9906aa850614862

diff --git a/posic/thesis/d_tersoff.tex b/posic/thesis/d_tersoff.tex
index c01a65e..b031be2 100644
--- 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{ .}
@@ -32,13 +32,13 @@ 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 written down.
+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 ${r}_i$]{\boldmath Derivative of $V_{ij}$ with respect to ${\bf r}_i$}
 
@@ -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}) \\
@@ -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 Erhart 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.