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Solution of the minimization problem.
In quaternion algebra, the rotational minimization problem may now be phrased as follows:
\begin{displaymath}
m({\bf q}) = \sum_\alpha \omega_\alpha
\Vert{\bf Q}{\bf R}^...
...}_\alpha{\bf Q}^T - {\bf R}_\alpha\Vert^2
\stackrel{!}{=} Min.
\end{displaymath} (4.60)

Since the matrix ${\bf Q}$ representing a normalized quaternion is orthogonal this may also be written as
\begin{displaymath}
m({\bf q}) = \sum_\alpha \omega_\alpha
\Vert{\bf Q}{\bf R}^{(0)}_\alpha - {\bf R}_\alpha{\bf Q}\Vert^2.
\stackrel{!}{=} Min.
\end{displaymath} (4.61)

This follows from the simple fact that $\Vert{\bf A}\Vert = \Vert{\bf A}{\bf Q}\Vert$, if ${\bf Q}$ is normalized. Eq. (4.61) shows that the target function to be minimized can be written as a simple quadratic form in the quaternion parameters [55],
$\displaystyle m({\bf q})$ $\textstyle =$ $\displaystyle {\bf q}\cdot{\bf M}{\bf q},$ (4.62)
$\displaystyle {\bf M}$ $\textstyle =$ $\displaystyle \sum_\alpha \omega_\alpha {\bf M}_\alpha.$ (4.63)

The matrices ${\bf M}_\alpha$ are positive semi-definite matrices depending on the positions ${\bf r}_\alpha$ and ${\bf r}^{(0)}_\alpha$:
\begin{displaymath}
\left.
\begin{array}{lll}
M_{\alpha,11} &= &
x_\alpha^2 + y...
...pha y_{0\alpha} -2z_\alpha z_{0\alpha}
\\
\end{array}\right\}
\end{displaymath} (4.64)

The rotational fit is now reduced to the problem of finding the minimum of a quadratic form with the constraint that the quaternion to be determined must be normalized. Using the method of Lagrange multipliers to account for the normalization constraint we have

\begin{displaymath}
m'({\bf q},\lambda) = {\bf q}\cdot{\bf M}{\bf q}
- \lambda({\bf q}\cdot{\bf q} - 1) \stackrel{!}{=} Min.
\end{displaymath} (4.65)

This leads immediately to the eigenvalue problem
$\displaystyle {\bf M}{\bf q}$ $\textstyle =$ $\displaystyle \lambda{\bf q},$ (4.66)
$\displaystyle {\bf q}\cdot{\bf q}$ $\textstyle =$ $\displaystyle 1.$ (4.67)

Now any normalized eigenvector ${\bf q}$ fulfills the relation $\lambda = {\bf q}\cdot{\bf M}{\bf q} \equiv m({\bf q})$. Therefore the eigenvector belonging to the smallest eigenvalue, $\lambda_{min}$, is the desired solution. At the same time $\lambda_{min}$ gives the average error per atom.

The result of RBT analysis is stored in a new trajectory file that contains only RBT motions.


next up previous contents
Next: Parameters Up: Theory and implementation Previous: Quaternions and rotations.   Contents
pellegrini eric 2009-10-06