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Constrained Molecular Dynamics Method

The idea of using constrained MD to compute the PMF is simple: by using a Lagrange multiplier to fix the distance between two solutes, we are able to compute the required constraint force between two solutes. After a simple integration, modified by the volume entropy term, we obtain the PMF.

Starting with Lagrangian formulation of classical dynamics, we introduce a modified Lagrangian $L^{\prime}$


\begin{displaymath}
L^{\prime}=\sum_{i}\frac{1}{2}m_{i}\dot{\vec{R}_{i}}^{2}-U(\{\vec{R}_{i}\})-\lambda[(\vec{R}_{1}-\vec{R}_{2})^{2}-c^{2}]
\end{displaymath} (27)

where $\lambda$ is the Lagrange multiplier, $\vec{R}_{1}$ and $\vec{R}_{2}$ are the coordinates of the carbon atom in methanes. Let:


\begin{displaymath}
\sigma_{1}(\vec{R}_{1},\vec{R}_{2})=\vert\vec{R}_{1}-\vec{R}_{2}\vert^{2}-c^{2}=R_{12}^{2}-c^{2}.
\end{displaymath} (28)

From Lagrange's equation


\begin{displaymath}
\frac{d}{dt}\frac{\partial L}{\partial\dot{\vec{R}}_{i}}-\frac{\partial L}{\partial\vec{R}_{i}}=0,
\end{displaymath} (29)

we obtain the equation of motion


\begin{displaymath}
m_{i}\ddot{\vec{R}}_{i}=-\frac{\partial U}{\partial\vec{R}_{...
...artial\sigma_{1}}{\partial\vec{R}_{i}}=\vec{F}_{i}+\vec{G}_{i}
\end{displaymath} (30)

The constraint force $\vec{G}_{i}$ is given by


\begin{displaymath}
\vec{G}_{1}=-\lambda\cdot2(\vec{R}_{1}-\vec{R}_{2})
\end{displaymath} (31)


\begin{displaymath}
\vec{G}_{2}=+\lambda\cdot2(\vec{R}_{1}-\vec{R}_{2}).
\end{displaymath} (32)

By its construction, the following condition holds at any $t$:


\begin{displaymath}
\vert\vec{R}_{1}(t)-\vec{R}_{2}(t)\vert=c
\end{displaymath} (33)

Using Eq. (14), Verlet algorithm, we obtain
\begin{displaymath}
\vec{R}_{1}(t+\Delta t)=-\vec{R}_{1}(t-\Delta t)+2\vec{R}_{1...
...ac{\Delta t^{2}}{m_{1}}\lambda[(\vec{R}_{1}(t)-\vec{R}_{2}(t)]
\end{displaymath} (34)


\begin{displaymath}
\vec{R}_{2}(t+\Delta t)=-\vec{R}_{2}(t-\Delta t)+2\vec{R}_{2...
...c{\Delta t^{2}}{m_{2}}\lambda[(\vec{R}_{1}(t)-\vec{R}_{2}(t)].
\end{displaymath} (35)

Subtracting the last two equation, let


\begin{displaymath}
\vec{R}_{2}(t+\Delta t)-\vec{R}_{1}(t+\Delta t)=\overrightarrow{A}+\lambda\overrightarrow{B}
\end{displaymath} (36)

where

\begin{displaymath}
\overrightarrow{A}=2\left[\vec{R}_{2}(t)-\vec{R}_{1}(t)\righ...
...m_{2}}\vec{F}_{2}(t)-\frac{(\Delta t)^{2}}{m_{1}}\vec{F}_{1}(t)\end{displaymath}


\begin{displaymath}
\overrightarrow{B}=-2\frac{(\Delta t)^{2}}{m_{2}}\left[\vec{...
...elta t)^{2}}{m_{1}}\left[\vec{R}_{2}(t)-\vec{R}_{1}(t)\right].
\end{displaymath} (37)

Taking square on both sides, we get


\begin{displaymath}
c^{2}=A^{2}+2\lambda A\cdot B+\lambda^{2}B^{2}.
\end{displaymath} (38)

This tells us how to solve $\lambda$ from the known $\vec{A}(t,t-\Delta t)$ and $\vec{B}(t)$:


\begin{displaymath}
\lambda=\frac{-2\overrightarrow{A}\cdot\overrightarrow{B}\pm...
...w{A}\cdot\overrightarrow{B})^{2}-4B^{2}(A^{2}-c^{2})}}{2B^{2}}
\end{displaymath} (39)

There are two roots for Eq. (39), and we shall use the smaller one. The larger root corresponds to the swap of two atoms, not what we want in a dynamics. Once one solves $\lambda$, the constraint force on the first atom is given by Eq. (31).


next up previous contents
Next: Results and Discussions Up: Free Energy Calculations Previous: Umbrella Sampling   Contents
Je-Luen Li 2007-07-17