Orthogonality of a Spherical Coordinate System

Posted February 2nd, 2008 by Structure

$x=r \cos \theta sin \phi$
$y=r \sin \theta \sin \phi$
$z = r \cos \phi$

$\chi(r,\phi,\theta)=(r \cos \theta sin \phi, r \sin \theta \sin \phi,r \cos \phi)$

$\frac{\partial \chi}{\partial r}=(\cos \theta sin \phi, \sin \theta sin \phi, \cos \phi) = e_r$

$\frac{\partial \chi}{\partial \phi}= r\frac{\partial}{\partial \phi}(\cos \theta sin \phi, \sin \theta sin \phi, \cos \phi) = r \frac{\partial e_r}{\partial \phi} = r e_{\phi} = r (\cos \theta cos \phi, \sin \theta cos \phi, -\sin \phi)$

$\frac{\partial \chi}{\partial \theta} = r \frac{\partial}{\partial \theta} e_r = r(-\sin \theta sin \phi, sin \theta \sin \phi, 0) = r \sin \theta(-sin \theta, \cos \theta, 0) = r \sin \phi e_{\theta}$

$< e_r,e_r > = 1$

$\frac{\partial}{\partial \phi}< e_r,e_r > = < \frac{\partial e_r}{\partial \phi},e_r > + < e_r,\frac{\partial e_r}{\partial \phi} > = 2 < \frac{\partial e_r}{\partial \phi},e_r > = 2 < e_{\phi},e_r >=0$

$\frac{\partial}{\partial \theta}< e_r ,e_r >= 2< \frac{\partial e_r}{\partial \theta},e_r > = 2<\sin \phi \cdot e_{\theta},e_r>=2 \sin \phi <e_{\theta}, e_r> = 0$