X-Git-Url: https://hackdaworld.org/gitweb/?p=lectures%2Flatex.git;a=blobdiff_plain;f=posic%2Fthesis%2Fd_tersoff.tex;h=b031be2f58ecd016152de51b4e26e136c23450aa;hp=9a5e76b8694259247af052e36bc5550010d780fc;hb=17d5c879c418790a154098e51c524eca183c4d98;hpb=46f67509943dd32ee23123168681f64f252a5ec4 diff --git a/posic/thesis/d_tersoff.tex b/posic/thesis/d_tersoff.tex index 9a5e76b..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{ .} @@ -32,15 +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: $\theta_{ijk}$, $\cos(\theta_{ijk})$,