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

by fazan   Last Updated September 21, 2018 18:20 PM

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



Answers 1


solved, I misplaced a sign in $\textbf{e}_\phi$

thanks to @md2perpe for pointing it out

fazan
fazan
September 21, 2018 18:16 PM

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