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Appendix B

From Fig. (1.5) it follows that

\begin{displaymath}\hat{r}_{i+1}=\sin \vartheta \cos \varphi _{i}\hat{e}^{x}(i)+...
...sin \varphi _{i}\hat{e}^{y}(i)+\cos \vartheta \hat{e}
^{z}(i).
\end{displaymath} (1.50)

Using this in
$\displaystyle \hat{e}^{x}(i+1)$ = $\displaystyle -\frac{1}{\sin \vartheta }\hat{r}_{i}+\frac{\cos
\vartheta }{\sin \vartheta }\hat{r}_{i+1}$ (1.51)
$\displaystyle \hat{e}^{y}(i+1)$ = $\displaystyle \frac{1}{\sin \vartheta }\hat{r}_{i}\times \hat{r}
_{i+1}$ (1.52)
$\displaystyle \hat{e}^{z}(i+1)$ = $\displaystyle \hat{r}_{i+1}$ (1.53)

one easily derives
$\displaystyle \hat{e}^{x}(i+1)$ = $\displaystyle \cos \vartheta \cos \varphi _{i}\hat{e}
^{x}(i)+\cos \vartheta \sin \varphi _{i}\hat{e}^{y}(i)-\sin
\vartheta \hat{e}^{z}(i)$ (1.54)
$\displaystyle \hat{e}^{y}(i+1)$ = $\displaystyle -\sin \varphi _{i}\hat{e}^{x}(i)+\cos \varphi
_{i}\hat{e}^{y}(i)$ (1.55)
$\displaystyle \hat{e}^{z}(i+1)$ = $\displaystyle \sin \vartheta \cos \varphi _{i}\hat{e}
^{x}(i)+\sin \vartheta \sin \varphi _{i}\hat{e}^{y}(i)+\cos
\vartheta \hat{e}^{z}(i)$ (1.56)

from which the matrix $\underline{M}(\varphi _{i})$ may be read off.



W.J. Briels