I have a question regarding the calculation of the $G$-matrix in the classic "Molecular vibrations" textbook by Wilson (1955, McGraw Hill).
On page 63 it gives an example for the calculation of an off-diagonal term coupling stretching with bending internal coordinates. Unfortunately I get a slightly different result when calculating it by hand.
Assume a non-linear triatomic molecule. Atoms 1 and 2 are on the outside, atom 3 is in the middle.
$s_1$ is the bond between Atoms 1 and 3, $s_2$ is the bond between Atoms 2 and 3, $s_3$ is the angle between atoms 1, 2, and 3.
The $s$-vectors are (page 63) given by:
Bond between Atom 1 and 3: $$ \vec{s}_{11} = \vec{e}_{31} $$ $$ \vec{s}_{13} = -\vec{e}_{31} $$ $$ \vec{s}_{12} = 0 $$ Bond between Atom 2 and 3: $$ \vec{s}_{21} = 0 $$ $$ \vec{s}_{23} = \vec{e}_{32} $$ $$ \vec{s}_{22} = -\vec{e}_{32} $$ Angle over Atoms 1, 3, and 2: $$ \vec{s}_{31} = \frac{\cos(\phi) \vec{e_{31}} - \vec{e}_{32}}{r_{31} \sin(\phi)} $$ $$ \vec{s}_{32} = \frac{\cos(\phi) \vec{e_{32}} - \vec{e}_{31}}{r_{32} \sin(\phi)} $$ $$ \vec{s}_{33} = \frac{ (r_{31} - r_{32} \cos(\phi))\vec{e}_{31} + (r_{32} - r_{31} \cos(\phi))\vec{e}_{32} } {r_{31} r_{32} \sin(\phi)} $$ If I calculate $G_{13}$ I get: $$ G_{13} = \mu_1 \vec{s}_{1, 1} \cdot \vec{s}_{3, 1} + \mu_2 \vec{s}_{1, 2} \cdot \vec{s}_{3, 2} + \mu_3 \vec{s}_{1, 3} \cdot \vec{s}_{3, 3} $$ $$ = \mu_1 \vec{e}_{31} \cdot \frac{\cos(\phi) \vec{e_{31}} - \vec{e}_{32}}{r_{31} \sin(\phi)} - \mu_3 \vec{e}_{31} \cdot \frac{ (r_{31} - r_{32} \cos(\phi))\vec{e}_{31} + (r_{32} - r_{31} \cos(\phi))\vec{e}_{32} } {r_{31} r_{32} \sin(\phi)} $$ (EDIT: in the following equality I wrongly assumed $\vec{e}_{31} \cdot \vec{e}_{32} = 0$) $$ = \mu_1 \frac{\cos(\phi)}{r_{31} \sin(\phi)} + \mu_3 \cdot \frac{ (r_{32} \cos(\phi) - r_{31}) } {r_{31} r_{32} \sin(\phi)} $$ $$ = \mu_1 \frac{\cos(\phi)}{r_{31} \sin(\phi)} + \frac{\mu_3 \cos(\phi)}{r_{31} \sin(\phi)} - \frac{\mu_3}{r_{32} \sin(\phi)} $$ $$ = (\mu_1 + \mu_3) \frac{\cos(\phi)}{r_{31} \sin(\phi)} - \frac{ \mu_3 } {r_{32} \sin(\phi)} $$
The result in the book is instead: $$ G_{13} = -\frac{\mu_3 \sin(\phi) }{ r_{32} } $$
What am I doing wrong?
