Home

HW #5: Spherical Coordinates, Duex

This work is borrowed gratefully from Flanders 4.1.

Position in $ \Re^3 $ is given in terms of spherical parameters as:

\[ \vec{r} = &(r\sin{\phi}\cos{\theta}, r\sin{\phi}\sin{\theta}, r\cos{\phi}) \]

with basis $ \{\hat{i}, \hat{j}, \hat{k}\} $. Applying the exterior derivative operator gives:

\begin{align*} d\vec{r} = &(\sin{\phi}\cos{\theta}, \sin{\phi}\sin{\theta}, \cos{\phi})dr \\<br /> &+ (r\cos{\phi}\cos{\theta}, r\cos{\phi}\sin{\theta}, -r\sin{\phi})d\phi \\<br /> &+ (-r\sin{\phi}\sin{\theta}, r\sin{\phi}\cos{\theta}, 0)d\theta \\<br /> = &dr\hat{r} + rd\phi\hat{\phi} + rd\theta\hat{\theta} \\<br /> \end{align*}

with

\begin{align*} \hat{r} &= (\sin{\phi}\cos{\theta}, \sin{\phi}\sin{\theta}, \cos{\phi}) \\<br /> \hat{\phi} &= (\cos{\phi}\cos{\theta}, \cos{\phi}\sin{\theta}, -\sin{\phi}) \\<br /> \hat{\theta} &= (-\sin{\phi}\sin{\theta}, \sin{\phi}\cos{\theta}, 0) \\<br /> \end{align*}

Therefore, (answer part (a))

\begin{align*} d\hat{r} = &(0,0,0)dr \\<br /> &+ (\cos{\phi}\cos{\theta}, \cos{\phi}\sin{\theta}, -\sin{\phi})d\phi \\<br /> &+ (-\sin{\phi}\sin{\theta}, \sin{\phi}\cos{\theta}, 0)d\theta \\<br /> = &\hat{\phi}d\phi + \hat{\theta}d\theta \\<br /> d\hat{\phi} = &(0,0,0)dr \\<br /> &+ (-\sin{\phi}\cos{\theta}, -\sin{\phi}\sin{\theta}, -\cos{\phi})d\phi \\<br /> &+ (-\cos{\phi}\sin{\theta}, \cos{\phi}\cos{\theta}, 0)d\theta \\<br /> d\hat{\theta} = &(0,0,0)dr \\<br /> &+ (-\cos{\phi}\sin{\theta}, \cos{\phi}\cos{\theta}, 0)d\phi \\<br /> &+ (-\sin{\phi}\cos{\theta}, -\sin{\phi}\sin{\theta}, 0)d\theta \\<br /> \end{align*}

Mon, 2007-02-19 06:31 | eric