| 285 |
|
\end{equation} |
| 286 |
|
The difficulty is selecting points $\mathbf{r}^N$ such that they are |
| 287 |
|
sampled from the distribution $\rho_{kT}(\mathbf{r}^N)$. A solution |
| 288 |
< |
was proposed by Metropolis et al.\cite{metropolis:1953} which involved |
| 288 |
> |
was proposed by Metropolis \emph{et al}.\cite{metropolis:1953} which involved |
| 289 |
|
the use of a Markov chain whose limiting distribution was |
| 290 |
|
$\rho_{kT}(\mathbf{r}^N)$. |
| 291 |
|
|
| 481 |
|
simulation box on an infinite lattice in Cartesian space. Any given |
| 482 |
|
particle leaving the simulation box on one side will have an image of |
| 483 |
|
itself enter on the opposite side (see Fig.~\ref{introFig:pbc}). In |
| 484 |
< |
addition, this sets that any given particle pair has an image, real or |
| 485 |
< |
periodic, within $fix$ of each other. A discussion of the method used |
| 486 |
< |
to calculate the periodic image can be found in |
| 484 |
> |
addition, this sets that any two particles have an image, real or |
| 485 |
> |
periodic, within $\text{box}/2$ of each other. A discussion of the |
| 486 |
> |
method used to calculate the periodic image can be found in |
| 487 |
|
Sec.\ref{oopseSec:pbc}. |
| 488 |
|
|
| 489 |
|
\begin{figure} |
| 557 |
|
\mathcal{O}(\Delta t^4) |
| 558 |
|
\label{introEq:verletBack} |
| 559 |
|
\end{equation} |
| 560 |
< |
Adding together Eq.~\ref{introEq:verletForward} and |
| 560 |
> |
Where $m$ is the mass of the particle, $q(t)$ is the position at time |
| 561 |
> |
$t$, $v(t)$ the velocity, and $F(t)$ the force acting on the |
| 562 |
> |
particle. Adding together Eq.~\ref{introEq:verletForward} and |
| 563 |
|
Eq.~\ref{introEq:verletBack} results in, |
| 564 |
|
\begin{equation} |
| 565 |
< |
eq here |
| 565 |
> |
q(t+\Delta t)+q(t-\Delta t) = |
| 566 |
> |
2q(t) + \frac{F(t)}{m}\Delta t^2 + \mathcal{O}(\Delta t^4) |
| 567 |
|
\label{introEq:verletSum} |
| 568 |
|
\end{equation} |
| 569 |
|
Or equivalently, |
| 570 |
|
\begin{equation} |
| 571 |
< |
eq here |
| 571 |
> |
q(t+\Delta t) = |
| 572 |
> |
2q(t) - q(t-\Delta t) + \frac{F(t)}{m}\Delta t^2 + |
| 573 |
> |
\mathcal{O}(\Delta t^4) |
| 574 |
|
\label{introEq:verletFinal} |
| 575 |
|
\end{equation} |
| 576 |
|
Which contains an error in the estimate of the new positions on the |
| 579 |
|
In practice, however, the simulations in this research were integrated |
| 580 |
|
with a velocity reformulation of the Verlet method.\cite{allen87:csl} |
| 581 |
|
\begin{equation} |
| 582 |
< |
eq here |
| 582 |
> |
q(t+\Delta t)= q(t) + v(t)\Delta t + \frac{F(t)}{2m}\Delta t^2 |
| 583 |
|
\label{introEq:MDvelVerletPos} |
| 584 |
|
\end{equation} |
| 585 |
|
\begin{equation} |
| 586 |
< |
eq here |
| 586 |
> |
v(t+\Delta t) = v(t) + \frac{\Delta t}{2m}[F(t) + F(t+\Delta t)] |
| 587 |
|
\label{introEq:MDvelVerletVel} |
| 588 |
|
\end{equation} |
| 589 |
|
The original Verlet algorithm can be regained by substituting the |
