| 417 |
|
macroscopic systems with a relatively small number of particles. The |
| 418 |
|
simulation box is replicated throughout space to form an infinite |
| 419 |
|
lattice. During the simulation, when a particle moves in the primary |
| 420 |
< |
cell, its image in other boxes move in exactly the same direction with |
| 421 |
< |
exactly the same orientation.Thus, as a particle leaves the primary |
| 420 |
> |
cell, its images in other boxes move in exactly the same direction with |
| 421 |
> |
exactly the same orientation. So, as a particle leaves the primary |
| 422 |
|
cell, one of its images will enter through the opposite face.If the |
| 423 |
< |
simulation box is large enough to avoid "feeling" the symmetries of |
| 423 |
> |
simulation box is large enough to avoid \textquotedblleft feeling\textquotedblright\ the symmetries of |
| 424 |
|
the periodic lattice, surface effects can be ignored. Cubic, |
| 425 |
< |
orthorhombic and parallelepiped are the available periodic cells In |
| 425 |
> |
orthorhombic and parallelepiped are the available periodic cells in |
| 426 |
|
{\sc oopse}. We use a matrix to describe the property of the simulation |
| 427 |
< |
box. Therefore, both the size and shape of the simulation box can be |
| 427 |
> |
box. Both the size and shape of the simulation box can be |
| 428 |
|
changed during the simulation. The transformation from box space |
| 429 |
|
vector $\mathbf{s}$ to its corresponding real space vector |
| 430 |
|
$\mathbf{r}$ is defined by |
| 431 |
|
\begin{equation} |
| 432 |
< |
\mathbf{r}=\underline{\underline{H}}\cdot\mathbf{s}% |
| 432 |
> |
\mathbf{r}=\underline{\mathbf{H}}\cdot\mathbf{s}% |
| 433 |
|
\end{equation} |
| 434 |
|
|
| 435 |
|
|
| 437 |
|
the three box axis vectors. $h_{x},h_{y}$ and $h_{z}$ represent the |
| 438 |
|
three sides of the simulation box respectively. |
| 439 |
|
|
| 440 |
< |
To find the minimum image, we convert the real vector to its |
| 440 |
> |
To find the minimum image of a vector $\mathbf{r}$, we convert the real vector to its |
| 441 |
|
corresponding vector in box space first, \bigskip% |
| 442 |
|
\begin{equation} |
| 443 |
< |
\mathbf{s}=\underline{\underline{H}}^{-1}\cdot\mathbf{r}% |
| 443 |
> |
\mathbf{s}=\underline{\mathbf{H}}^{-1}\cdot\mathbf{r}% |
| 444 |
|
\end{equation} |
| 445 |
|
And then, each element of $\mathbf{s}$ is wrapped to lie between -0.5 to 0.5, |
| 446 |
|
\begin{equation} |
| 461 |
|
For example, $\roundme(3.6)=4$, $\roundme(3.1)=3$, $\roundme(-3.6)=-4$, |
| 462 |
|
$\roundme(-3.1)=-3$. |
| 463 |
|
|
| 464 |
< |
Finally, we obtain the minimum image coordinates by transforming back |
| 464 |
> |
Finally, we obtain the minimum image coordinates $\mathbf{r}^{\prime}$ by transforming back |
| 465 |
|
to real space,% |
| 466 |
|
|
| 467 |
|
\begin{equation} |
| 468 |
< |
\mathbf{r}^{\prime}=\underline{\underline{H}}^{-1}\cdot\mathbf{s}^{\prime}% |
| 468 |
> |
\mathbf{r}^{\prime}=\underline{\mathbf{H}}^{-1}\cdot\mathbf{s}^{\prime}% |
| 469 |
|
\end{equation} |
| 470 |
|
|
