trunk/oopsePaper/pbc.tex

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\begin{document}
\section{\label{Sec:pbc}Periodic Boundary Conditions}

\textit{Periodic boundary conditions} are widely used to simulate truly
macroscopic systems with a relatively small number of particles. Simulation
box is replicated throughout space to form an infinite lattice. During the
simulation, when a particle moves in the primary cell, its periodic image
particles in other boxes move in exactly the same direction with exactly the
same orientation.Thus, as a particle leaves the primary cell, one of its
images will enter through the opposite face.If the simulation box is large
enough to avoid "feeling" the symmetric of the periodic lattice,the surface
effect could be ignored. Cubic and parallelepiped are the available periodic
cells. \bigskip In OOPSE, we use the matrix instead of the vector to describe
the property of the simulation box. Therefore, not only the size of the
simulation box could be changed during the simulation, but also the shape of
it. The transformation from box space vector $\overrightarrow{s}$ to its
corresponding real space vector $\overrightarrow{r}$ is defined by
\begin{equation}
\overrightarrow{r}=H\overrightarrow{s}%
\end{equation}


where $H=(h_{x},h_{y},h_{z})$ is a transformation matrix made up of the three
box axis vectors. $h_{x},h_{y}$ and $h_{z}$ represent the sides of the
simulation box respectively.

To find the minimum image, we need to convert the real vector to its
corresponding vector in box space first, \bigskip%
\begin{equation}
\overrightarrow{s}=H^{-1}\overrightarrow{r}%
\end{equation}
And then, each element of $\overrightarrow{s}$ is casted to lie between -0.5
to 0.5,
\begin{equation}
s_{i}^{\prime}=s_{i}-round(s_{i})
\end{equation}
where%

\begin{equation}
round(x)=\lfloor{x+0.5}\rfloor\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }if\text{
}x\geqslant0
\end{equation}
%

\begin{equation}
round(x)=\lceil{x-0.5}\rceil\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }if\text{ }x<0
\end{equation}


For example, $round(3.6)=4$,$round(3.1)=3$, $round(-3.6)=-4$, $round(-3.1)=-3$.

Finally, we could get the minimum image by transforming back to real space,%

\begin{equation}
\overrightarrow{r^{\prime}}=H^{-1}\overrightarrow{s^{\prime}}%
\end{equation}


\end{document}
Revision:	904
Committed:	Wed Jan 7 15:21:00 2004 UTC (21 years, 4 months ago) by tim
Content type:	application/x-tex
File size:	3836 byte(s)
Log Message:	* empty log message *
#	Content
1	\documentclass{article}%
2	\usepackage{amsfonts}
3	\usepackage{amsmath}
4	\usepackage{amssymb}
5	\usepackage{graphicx}%
6	\setcounter{MaxMatrixCols}{30}
7	%TCIDATA{OutputFilter=latex2.dll}
8	%TCIDATA{Version=5.00.0.2552}
9	%TCIDATA{CSTFile=40 LaTeX article.cst}
10	%TCIDATA{Created=Friday, September 19, 2003 08:29:53}
11	%TCIDATA{LastRevised=Wednesday, January 07, 2004 10:20:42}
12	%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
13	%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
14	%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Standard LaTeX Article">}
15	%TCIDATA{ComputeDefs=
16	%$H$
17	%}
18	\newtheorem{theorem}{Theorem}
19	\newtheorem{acknowledgement}[theorem]{Acknowledgement}
20	\newtheorem{algorithm}[theorem]{Algorithm}
21	\newtheorem{axiom}[theorem]{Axiom}
22	\newtheorem{case}[theorem]{Case}
23	\newtheorem{claim}[theorem]{Claim}
24	\newtheorem{conclusion}[theorem]{Conclusion}
25	\newtheorem{condition}[theorem]{Condition}
26	\newtheorem{conjecture}[theorem]{Conjecture}
27	\newtheorem{corollary}[theorem]{Corollary}
28	\newtheorem{criterion}[theorem]{Criterion}
29	\newtheorem{definition}[theorem]{Definition}
30	\newtheorem{example}[theorem]{Example}
31	\newtheorem{exercise}[theorem]{Exercise}
32	\newtheorem{lemma}[theorem]{Lemma}
33	\newtheorem{notation}[theorem]{Notation}
34	\newtheorem{problem}[theorem]{Problem}
35	\newtheorem{proposition}[theorem]{Proposition}
36	\newtheorem{remark}[theorem]{Remark}
37	\newtheorem{solution}[theorem]{Solution}
38	\newtheorem{summary}[theorem]{Summary}
39	\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
40	\begin{document}
41	\section{\label{Sec:pbc}Periodic Boundary Conditions}
42
43	\textit{Periodic boundary conditions} are widely used to simulate truly
44	macroscopic systems with a relatively small number of particles. Simulation
45	box is replicated throughout space to form an infinite lattice. During the
46	simulation, when a particle moves in the primary cell, its periodic image
47	particles in other boxes move in exactly the same direction with exactly the
48	same orientation.Thus, as a particle leaves the primary cell, one of its
49	images will enter through the opposite face.If the simulation box is large
50	enough to avoid "feeling" the symmetric of the periodic lattice,the surface
51	effect could be ignored. Cubic and parallelepiped are the available periodic
52	cells. \bigskip In OOPSE, we use the matrix instead of the vector to describe
53	the property of the simulation box. Therefore, not only the size of the
54	simulation box could be changed during the simulation, but also the shape of
55	it. The transformation from box space vector $\overrightarrow{s}$ to its
56	corresponding real space vector $\overrightarrow{r}$ is defined by
57	\begin{equation}
58	\overrightarrow{r}=H\overrightarrow{s}%
59	\end{equation}
60
61
62	where $H=(h_{x},h_{y},h_{z})$ is a transformation matrix made up of the three
63	box axis vectors. $h_{x},h_{y}$ and $h_{z}$ represent the sides of the
64	simulation box respectively.
65
66	To find the minimum image, we need to convert the real vector to its
67	corresponding vector in box space first, \bigskip%
68	\begin{equation}
69	\overrightarrow{s}=H^{-1}\overrightarrow{r}%
70	\end{equation}
71	And then, each element of $\overrightarrow{s}$ is casted to lie between -0.5
72	to 0.5,
73	\begin{equation}
74	s_{i}^{\prime}=s_{i}-round(s_{i})
75	\end{equation}
76	where%
77
78	\begin{equation}
79	round(x)=\lfloor{x+0.5}\rfloor\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }if\text{
80	}x\geqslant0
81	\end{equation}
82	%
83
84	\begin{equation}
85	round(x)=\lceil{x-0.5}\rceil\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }if\text{ }x<0
86	\end{equation}
87
88
89	For example, $round(3.6)=4$,$round(3.1)=3$, $round(-3.6)=-4$, $round(-3.1)=-3$.
90
91	Finally, we could get the minimum image by transforming back to real space,%
92
93	\begin{equation}
94	\overrightarrow{r^{\prime}}=H^{-1}\overrightarrow{s^{\prime}}%
95	\end{equation}
96
97
98
99	\end{document}