Irrotational vector Field in Spherical coordinates If the vector field $ \overrightarrow{A} = A_{\varphi} (r, \theta) \widehat{e}_{\varphi} $ is irrotational, then what will be the value of $ A_{\varphi} (r, \theta) $
My work:
For irrotational case, curl of the vector must be zero.
$curl \ A = \dfrac{1}{r^2 \sin \theta } \begin{vmatrix}
 \widehat{e}_{r} & r \widehat{e}_{\theta} & r \sin \theta \widehat{e}_{\varphi} \\ 
\frac{\partial }{\partial r} & \frac{\partial }{\partial \theta} & \frac{\partial }{\partial \varphi} \\ 
0 & 0 & r \sin \theta A_{\varphi} (r, \theta)
\end{vmatrix}$
$curl \ A =  e_r \left [ r \cos \theta A_{\varphi} +  r \sin \theta \frac{\partial A_{\varphi}}{\partial \theta} \right ] - r \widehat{e}_{\theta} \left [ ( \sin \theta A_{\varphi} + r \frac{\partial \sin \theta A_{\varphi}}{\partial r}   \right ]$
Is this correct? How to proceed further? I have evaluated as follows not sure if that is correct.
$r \cos \theta A_{\varphi} +  r \sin \theta \frac{\partial A_{\varphi}}{\partial \theta} = 0 $
$ \frac{\partial A_{\varphi}}{\partial \theta} = - \cot \theta A_{\varphi} $
$\int \frac{d A_{\varphi}}{w_{\varphi}} = -\int \cot \theta \ d \theta $
$ \ln A_{\varphi} = -\ln (\sin \theta) + K(r) \ ........... \ (1)$
$ \sin \theta A_{\varphi} + r \frac{\partial \sin \theta A_{\varphi}}{\partial r} = 0 $
$ \ln A_{\varphi} = -\ln  (r) + K (\theta) \ .......... \ (2)$
Combining $ (1) $ and $ (2) $
$ ln A_{\varphi} = -(\ln  (r \sin \theta)) $
$ A_{\varphi}(r, \theta) = \dfrac{1}{r \sin \theta }$
This is the result I am getting. Could you please advise where I am wrong, What would be the value of
$ A_{\varphi}(r, \theta) $?
 A: You almost have done it.
Curl $\nabla \times \vec A$ in spherical coordinates can be defined by the following equation
$$
\nabla \times \vec A = \frac1{r \sin\theta}
\left[
\frac{\partial}{\partial \theta}\left( A_\phi \sin \theta\right) - \frac{\partial A_\theta}{\partial \phi}
\right]\hat r 
+
\frac1r\left[
\frac1{\sin\theta}\frac{\partial A_r}{\partial \phi} - \frac\partial{\partial r} \left(r A_\phi\right)
\right]\hat \theta
+
\frac1r\left[
\frac\partial{\partial r}\left(rA_\theta\right)-\frac{\partial A_r}{\partial \theta}
\right]\hat \phi.
$$
Assuming that vector field has the following form $\vec A = A_\phi(r,\theta)\hat \phi$, curl can be described as
$$
\nabla\times\vec A = \frac{1}{r\sin\theta}\left[\sin\theta\frac{\partial A_\phi}{\partial \theta}+ A_\phi\cos\theta \right]\hat r - \frac 1r\left[ 
r\frac{\partial A_\phi}{\partial r} + A_\phi
\right]\hat \theta.
$$
Next, applying the condition $\nabla \times\vec A = \vec 0$ leads to set of the equations,
since to obtain zero vector all vector elements should be equal to zero
$$
\begin{cases}
\displaystyle\sin\theta\frac{\partial A_\phi}{\partial \theta}+ A_\phi\cos\theta = 0\\[1ex]
\displaystyle r\frac{\partial A_\phi}{\partial r} + A_\phi = 0
\end{cases}
$$
I hope this will help.
