I have to find the partial derivatives in spherical form of $\textbf{F}=r[\textbf{e}_{\theta}+\textbf{e}_{\phi}],$ and I managed to find all the others except for the one over $\phi$.

I got this far: $\frac{\partial \textbf{F}}{\partial \phi}=r[\cos(\theta)\textbf{e}_\phi+\cos(\phi)\textbf{x}-\sin(\phi)\textbf{y}]$ , with the last two terms being the partial derivative of $\textbf{e}_{\phi}$ over $\phi$.

Since $\textbf{e}_{\phi} $ is an unit vector, its derivative has to be perpendicular to it, and as such must be of the form $A\textbf{e}_{\theta}+B\textbf{e}_r$, but if I try doing that I get that B must be both positive and negative. Where do I go wrong?

Edit: calculation of coefficients

$[\cos(\phi),-\sin(\phi),0]=[A\sin(\theta)\cos(\phi)+B\cos(\theta)\cos(\phi), A\sin(\theta)\sin(\phi)+B\cos(\theta)\sin(\phi), A\cos(\theta)-B\sin(\theta)]$

From the third components: $A\cos(\theta)=B\sin(\theta)$ we get $A=\frac{\sin(\theta)}{\cos(\theta)}B$.

Looking at the first components, we get $B=\cos(\theta)$, but looking at the second components we get $B=-\cos(\theta)$.

  • $\begingroup$ Can you show your calculations that result in both $B>0$ and $B<0$? $\endgroup$ – md2perpe Sep 21 '18 at 16:33
  • $\begingroup$ @md2perpe I added it $\endgroup$ – fazan Sep 21 '18 at 16:48
  • $\begingroup$ Why do you assert that equality? $\endgroup$ – md2perpe Sep 21 '18 at 17:43
  • $\begingroup$ The left-hand-side is the derivative of $\textbf{e}_\phi$. Since $\textbf{e}_\phi$ is an unit vector, its derivative is perpendicular to it. The polar spherical coordinate system is orthogonal, so a vector perpendicular to $\textbf{e}_\phi$ is a linear combination of $\textbf{e}_\theta$ and $\textbf{e}_r$. Hence, $\frac{\partial \textbf{e}_\phi}{\partial \phi}=A \textbf{e}_\theta+B \textbf{e}_r$. Note that A and B are scalars, but not neccessarily constants. $\endgroup$ – fazan Sep 21 '18 at 18:00
  • $\begingroup$ So $\mathbf{e}_\phi = [\sin(\phi), \cos(\phi), 0]$? That doesn't look correct. Shouldn't it be $\mathbf{e}_\phi = [-\sin(\phi), \cos(\phi), 0]$? $\endgroup$ – md2perpe Sep 21 '18 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.