From 335354349680c91e46e19a0d63694bc05223394e Mon Sep 17 00:00:00 2001
From: Olivier Coulaud <Olivier.Coulaud@inria.fr>
Date: Wed, 15 Jan 2014 15:11:49 +0100
Subject: [PATCH] Add DL_POLY factors for energy and force
---
.../ScalFmm-PeriodicModel/periodicmodel.tex | 21 ++++++++++---------
1 file changed, 11 insertions(+), 10 deletions(-)
diff --git a/Doc/Src_tex/ScalFmm-PeriodicModel/periodicmodel.tex b/Doc/Src_tex/ScalFmm-PeriodicModel/periodicmodel.tex
index aca699ddd..c4b31e8a3 100644
--- a/Doc/Src_tex/ScalFmm-PeriodicModel/periodicmodel.tex
+++ b/Doc/Src_tex/ScalFmm-PeriodicModel/periodicmodel.tex
@@ -122,11 +122,11 @@ The potential at particle $x_i$ computed by ScalFMM code is given by
$$
V(x_i) = \sum_{j=0,i\neq j}^{N}{\frac{q_j}{\|x_i-x_j\|}}
$$
-and the force on atom $x_i$
+and the force on atom $x_i$ writes
$$
f(x_i) = \sum_{j=0,i\neq j}^{N}{q_j\frac{x_i-x_j}{\|x_i-x_j\|^3}}.
$$
-Finally, the total energy of the system writes
+Finally, the total energy of the system is
$$
U = \frac{1}{2}\sum_{i=0}^{N}{q_i V(x_i)}.
$$
@@ -396,29 +396,30 @@ energy & $E_0 = m_0(l_0/t_0)^2$&$1.6605402 \; 10^{−23}\; Joules $ & $10\; J\,
In internal variables the energy writes
$$
-U = \frac{q_0^2}{4 \pi\epsilon_0 l_0}\sum_{i=0}^{N}{\sum_{j<i}{\frac{q_i q_j}{\|x_i-x_j\|}}} = C_{energy}\sum_{i=0}^{N}{\sum_{j<i}{\frac{q_i q_j}{\|x_i-x_j\|}}}
+U = \frac{q_0^2}{4 \pi\epsilon_0 l_0}\sum_{i=0}^{N}{\sum_{j<i}{\frac{q_i q_j}{\|x_i-x_j\|}}} = C_{energy} \;\sum_{i=0}^{N}{\sum_{j<i}{\frac{q_i q_j}{\|x_i-x_j\|}}}
$$
and the forces write
$$
f(x_i) = -\frac{q_0^2}{4 \pi\epsilon_0 l_0^2} q_i \sum_{j=0,i\neq j}^{N}{q_j\frac{x_i-x_j}{\|x_i-x_j\|^3}}
- = -C_{force} q_i \sum_{j=0,i\neq j}^{N}{q_j\frac{x_i-x_j}{\|x_i-x_j\|^3}}
+ = C_{force} \; q_i \sum_{j=0,i\neq j}^{N}{q_j\frac{x_i-x_j}{\|x_i-x_j\|^3}}
$$
-The Energy conversion factor is $\gamma_0 = \frac{q_0^2}{4 \pi\epsilon_0 l_0}/E_0 = 138935.4835$. The energy unit is in Joules and if you want $kcal mol^{-1}$ unit the the factor becomes $\gamma_0/418.400$.
+The Energy conversion factor is $\gamma_0 = \frac{q_0^2}{4 \pi\epsilon_0 l_0}/E_0 = 138935.4835$. The energy unit is in Joules and if you want $kcal\, mol^{-1}$ unit the the factor becomes $\gamma_0/418.400$.
To compare with the molecular dynamics code, the forces and the energy is multiplied by $C_{force}$ ($C_{energy}$ respectively).
\begin{center}
-\begin{tabular}{|l|c|c|c|c|c|}
+\begin{tabular}{|l|c|c|c|}
\hline
- & \multicolumn{2}{|c|}{Energy} & \multicolumn{2}{c|}{Force} \\
-units& $kcal\, mol^{-1}$& & &\\
+ & Constant & Value & Unit \\
\hline
-$C_{energy}$& & & &\\
-$C_{force}$& -138935.4835/418.400& & &\\
+Energy & $C_{energy}$& 138935.4835/418.4 & $kcal\, mol^{-1}$\\
+Force &$C_{force}$& -138935.4835 & $10\,J mol^{-1}\, \AA^{-1}$\\
\hline
\end{tabular}
\end{center}
+The force unit is the internal DL\_Poly unit.
+
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Conclusion}
A new approach has been proposed to compute a periodic FMM.
--
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