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) $?
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.