I understand how to determine if a force is conservative from \begin{equation} \nabla\times \mathbf{F}=0 \implies \mathbf{F}\text{ is conservative} \end{equation} When $F$ is in cartesian coordinates. What happens if you’re given a force such as \begin{equation} \mathbf{F}=r\sin\theta e^{i\phi}\hat{r} \end{equation} How do you compute the curl and potential?
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The curl of any vector $\vec{F}$ in spherical polar coordinates is: $$ \nabla \times \vec{F} = \frac{1}{r\sin \theta}\left(\frac{\partial (F_\phi \sin \theta) }{\partial \theta} - \frac{\partial F_\theta}{\partial \phi}\right)\,\hat{\boldsymbol r} + \frac{1}{r}\left(\frac{1}{\sin \theta} \frac{\partial F_r}{\partial \phi} - \frac{\partial (r F_\phi)}{\partial r}\right)\, \hat{\boldsymbol \theta} + \frac{1}{r}\left(\frac{\partial ( r F_\theta)}{\partial r} - \frac{\partial F_r}{\partial \theta}\right)\, \hat{\boldsymbol \phi} \ . $$ To "compute the potential" you would have to solve the equation $$ \nabla V = -\vec{F}\ , $$ which, in spherical polar coordinates for a force which only has a $\hat{r}$ component, gives three equations: \begin{eqnarray} \frac{\partial V}{\partial r} & = & -F_r \\ \frac{1}{r}\frac{\partial V}{\partial \theta} & = & 0 \\ \frac{1}{r\sin \theta}\frac{\partial V}{\partial \phi} & = & 0 \end{eqnarray} The potential is not uniquely determined because you can add any scalar field with a zero gradient to $V$.
