Subsections

The dot product gradient of the spheroid

The partial derivative of the dot product with respect to the orientational parameter $\mathfrak{O}_i$ is

$\displaystyle {\frac{{\partial \delta_z}}{{\partial \mathfrak{O}_i}}}$ = $\displaystyle {\frac{{\partial}}{{\partial \mathfrak{O}_i}}}$$\displaystyle \left(\vphantom{ \widehat{XH} \cdot \widehat{\mathfrak{D}_{\scriptscriptstyle \parallel}} }\right.$$\displaystyle \widehat{{XH}}$$\displaystyle \widehat{{\mathfrak{D}_{\scriptscriptstyle \parallel}}}$$\displaystyle \left.\vphantom{ \widehat{XH} \cdot \widehat{\mathfrak{D}_{\scriptscriptstyle \parallel}} }\right)$
= $\displaystyle \widehat{{XH}}$$\displaystyle {\frac{{\partial \widehat{\mathfrak{D}_{\scriptscriptstyle \parallel}}}}{{\partial \mathfrak{O}_i}}}$ + $\displaystyle {\frac{{\partial \widehat{XH}}}{{\partial \mathfrak{O}_i}}}$$\displaystyle \widehat{{\mathfrak{D}_{\scriptscriptstyle \parallel}}}$.
(15.183)

Because the XH bond vector is constant and not dependent on the spherical angles its derivative is zero. Therefore

$\displaystyle {\frac{{\partial \delta_z}}{{\partial \mathfrak{O}_i}}}$ = $\displaystyle \widehat{{XH}}$$\displaystyle {\frac{{\partial \widehat{\mathfrak{D}_{\scriptscriptstyle \parallel}}}}{{\partial \mathfrak{O}_i}}}$. (15.184)

The Dpar gradient

The partial derivatives of the unit vector $\widehat{{\mathfrak{D}_{\scriptscriptstyle \parallel}}}$ with respect to the spherical angles are
\begin{subequations}\begin{align}
\frac{\partial \widehat{\mathfrak{D}_{\scripts...
...\\
\sin \theta \cos \phi \\
0 \\
\end{pmatrix}.
\end{align}\end{subequations}



The relax user manual (PDF), created 2019-03-08.