# Find spherical polar components of $\frac{\partial \textbf{F}}{\partial \phi}$

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 $$B=-\cos(\theta)$$.

• Can you show your calculations that result in both $B>0$ and $B<0$? – md2perpe Sep 21 '18 at 16:33
• @md2perpe I added it – fazan Sep 21 '18 at 16:48
• Why do you assert that equality? – md2perpe Sep 21 '18 at 17:43
• 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. – fazan Sep 21 '18 at 18:00
• So $\mathbf{e}_\phi = [\sin(\phi), \cos(\phi), 0]$? That doesn't look correct. Shouldn't it be $\mathbf{e}_\phi = [-\sin(\phi), \cos(\phi), 0]$? – md2perpe Sep 21 '18 at 18:07