Commit 33535434 authored by COULAUD Olivier's avatar COULAUD Olivier
Browse files

Add DL_POLY factors for energy and force

parent 27be2040
...@@ -122,11 +122,11 @@ The potential at particle $x_i$ computed by ScalFMM code is given by ...@@ -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\|}} 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}}. 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)}. 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\, ...@@ -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 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 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}} 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). To compare with the molecular dynamics code, the forces and the energy is multiplied by $C_{force}$ ($C_{energy}$ respectively).
\begin{center} \begin{center}
\begin{tabular}{|l|c|c|c|c|c|} \begin{tabular}{|l|c|c|c|}
\hline \hline
& \multicolumn{2}{|c|}{Energy} & \multicolumn{2}{c|}{Force} \\ & Constant & Value & Unit \\
units& $kcal\, mol^{-1}$& & &\\
\hline \hline
$C_{energy}$& & & &\\ Energy & $C_{energy}$& 138935.4835/418.4 & $kcal\, mol^{-1}$\\
$C_{force}$& -138935.4835/418.400& & &\\ Force &$C_{force}$& -138935.4835 & $10\,J mol^{-1}\, \AA^{-1}$\\
\hline \hline
\end{tabular} \end{tabular}
\end{center} \end{center}
The force unit is the internal DL\_Poly unit.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Conclusion} \section{Conclusion}
A new approach has been proposed to compute a periodic FMM. A new approach has been proposed to compute a periodic FMM.
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