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\documentclass{article}%
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\usepackage{amsfonts}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{graphicx}%
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%TCIDATA{OutputFilter=latex2.dll}
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%TCIDATA{Version=5.00.0.2552}
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%TCIDATA{CSTFile=40 LaTeX article.cst}
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%TCIDATA{Created=Friday, September 19, 2003 08:29:53}
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%TCIDATA{LastRevised=Tuesday, January 13, 2004 10:22:03}
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%}
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\newtheorem{theorem}{Theorem}
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\newtheorem{algorithm}[theorem]{Algorithm}
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\newtheorem{axiom}[theorem]{Axiom}
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\newtheorem{case}[theorem]{Case}
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\newtheorem{claim}[theorem]{Claim}
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\newtheorem{conclusion}[theorem]{Conclusion}
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\newtheorem{condition}[theorem]{Condition}
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\newtheorem{conjecture}[theorem]{Conjecture}
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\newtheorem{criterion}[theorem]{Criterion}
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\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
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\begin{document}
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\section{\label{Sec:pbc}Periodic Boundary Conditions}
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\textit{Periodic boundary conditions} are widely used to simulate truly
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macroscopic systems with a relatively small number of particles. The
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simulation box is replicated throughout space to form an infinite lattice.
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During the simulation, when a particle moves in the primary cell, its image in
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other boxes move in exactly the same direction with exactly the same
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orientation.Thus, as a particle leaves the primary cell, one of its images
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will enter through the opposite face.If the simulation box is large enough to
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avoid "feeling" the symmetries of the periodic lattice, surface effects can be
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ignored. Cubic, orthorhombic and parallelepiped are the available periodic
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cells In OOPSE. We use a matrix to describe the property of the simulation
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box. Therefore, both the size and shape of the simulation box can be changed
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during the simulation. The transformation from box space vector $\mathbf{s}$
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to its corresponding real space vector $\mathbf{r}$ is defined by
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\begin{equation}
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\mathbf{r}=\underline{\underline{H}}\cdot\mathbf{s}%
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\end{equation}
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where $H=(h_{x},h_{y},h_{z})$ is a transformation matrix made up of the three
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box axis vectors. $h_{x},h_{y}$ and $h_{z}$ represent the three sides of the
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simulation box respectively.
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To find the minimum image, we convert the real vector to its corresponding
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vector in box space first, \bigskip%
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\begin{equation}
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\mathbf{s}=\underline{\underline{H}}^{-1}\cdot\mathbf{r}%
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\end{equation}
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And then, each element of $\mathbf{s}$ is wrapped to lie between -0.5 to 0.5,
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\begin{equation}
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s_{i}^{\prime}=s_{i}-round(s_{i})
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\end{equation}
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where
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%
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\begin{equation}
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round(x)=\left\{
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\begin{array}
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[c]{c}%
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\lfloor{x+0.5}\rfloor & \text{if \ }x\geqslant0\\
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\lceil{x-0.5}\rceil & \text{otherwise}%
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\end{array}
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\right.
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\end{equation}
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For example, $round(3.6)=4$,$round(3.1)=3$, $round(-3.6)=-4$, $round(-3.1)=-3$.
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Finally, we obtain the minimum image coordinates by transforming back to real space,%
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\begin{equation}
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\mathbf{r}^{\prime}=\underline{\underline{H}}^{-1}\cdot\mathbf{s}^{\prime}%
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\end{equation}
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\end{document} |