 ### Merge branch 'master' into PB_tensorialFMM

parents 731ae45d 7dcb5cc7
 ... ... @@ -77,19 +77,22 @@ \begin{minipage}{0.4\textwidth} \begin{flushleft} \large \emph{Author:}\\ Berenger \textsc{Bramas} % Your name Berenger \textsc{Bramas} \\% Your name Olivier \textsc{Coulaud} % Your name \end{flushleft} \end{minipage} ~ \begin{minipage}{0.4\textwidth} \begin{flushright} \large \emph{Supervisor:} \\ \textsc{} % Supervisor's Name %\emph{Supervisor:} \\ %\textsc{} % Supervisor's Name \end{flushright} \end{minipage}\\[4cm] {\large \today}\\[3cm] % Date, change the \today to a set date if you want to be precise {\large \today}\\[1cm] % Date, change the \today to a set date if you want to be precise {\Large \textbf{Working note}} \vfill % Fill the rest of the page with whitespace \end{titlepage} ... ... @@ -387,6 +390,16 @@ and the force on atom $x_i$ $$f(x_i) = \frac{q_i }{4 \pi\epsilon_0}\sum_{j=0,i\neq j}^{N}{q_j\frac{x_i-x_j}{\|x_i-x_j\|^3}}$$ The Ewald method is equivalent to consider a surrounding conductor ($\epsilon=\infty$). The method described in this document and implemented in ScalFMM consider a finite set of the original box surrounded by a vacuum with dielectric constant $\epsilon=1$. The results by these two approaches are different \cite{DeLeeuw1980, Heyes1981} and for a very large sphere surrounding all boxes we have the following equation \begin{equation} E_{Ewald} = E_{FMM} -\frac{2\pi}{3V}|\sum_{i=1}^{N}{q_i x_i}|^2. \end{equation} By tacking the gradient of this equation, the relationship between FMM forces and Ewald forces is \begin{equation} F_{Ewald}(x_k) = -\nabla_k E_{Ewald} = F_{FMM}(x_k) + \frac{4\pi}{3V} q_k \sum_{i=1}^{N}{q_i x_i}. \end{equation} \subsubsection{DL\_Poly comparisons} DL\_POLY\_2 uses the following internal molecular units \\ ... ... @@ -428,12 +441,49 @@ Force &$C_{force}$& -138935.4835 & $10\,J mol^{-1}\, \AA^{-1}$\\ The force unit is the internal DL\_Poly unit. The first test consists in a small crystal $4\times 4\times 4$ of NaCl. It is composed of 128 atoms The second tests is a larger crystal $10\times 10\times 10$ of NaCl and have X atoms. The first test consists in a small crystal $4\times 4\times 4$ of NaCl. It is composed of 128 atoms The second tests is a larger crystal $10\times 10\times 10$ of NaCl and have 2000 atoms. The positions and the forces are stored in the \texttt{REVCON} file and the energy in the \texttt{STATIS} file. The Ewald's parameter are chosen such that the Ewald sum precision is $10^{-08}$. \begin{center} \begin{tabular}{|l|c|c|c|c|} \hline Crystal & Cube size & Cut off & reciprocal lattice vector &Energy \\ & $\AA$ & $\AA$ & & $kcal\, mol^{-1}$ \\ \hline Small & 16 & 8 & (9,9,9) & $-2.346403 \,10^{4}$ \\ \hline Big & 40 & 12 & (15,15,15) & $-3.666255 \,10^{5}$ \\ \hline \end{tabular} \end{center} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Conclusion} A new approach has been proposed to compute a periodic FMM. This approach can be used with any kernel and without the need of developing new operators. Moreover, it is possible to compute an important repetition size for a small cost. \bibliographystyle{plain} \bibliography{fmm} \newpage \section*{Appendix 1 } \begin{center} \large DL\_poly CONTROL file \end{center} \begin{verbatim} timestep 1.e-15 steps 1 temperature 0.0 ensemble nve delr 0.5 cutoff 8.0 ewald precision 1.e-8 no vdw traj 1,1,2 print 1 stats 1 job time 3600 close time 10 finish \end{verbatim} \end{document}