Solving Curl A in Spherical Coordinates: Tips & Hints

[phos19]

Homework Statement

Let $$ \vec{B} =\dfrac{1}{4 \pi} \dfrac{-3}{r^4} ( 3\cos^2{\theta} - 1) \hat{u_r} + \dfrac{1}{4 \pi} \dfrac{1}{r^4} ( - 6 \cos{\theta} \sin{\theta} ) \hat{u_{\theta}} $$ (spherical unit vectors)


Find $\vec{A}$ such that $\vec{B} = \nabla \times \vec{A}$

Relevant Equations

(The $\vec{B}$ is divergenceless !)

I've tried writing the curl A (in spherical coord.) and equating the components, but I end up with something that is beyond me:

\begin{equation}
{\displaystyle {\begin{aligned}{B_r = \dfrac{1}{4 \pi} \dfrac{-3}{r^4} ( 3\cos^2{\theta} - 1) =\frac {1}{r\sin \theta }}\left({\frac {\partial }{\partial \theta }}\left(A_{\varphi }\sin \theta \right)-{\frac {\partial A_{\theta }}{\partial \varphi }}\right)&\\B_{\theta}= \dfrac{1}{4 \pi} \dfrac{1}{r^4} ( - 6 \cos{\theta} \sin{\theta} ) ={}+{\frac {1}{r}}\left({\frac {1}{\sin \theta }}{\frac {\partial A_{r}}{\partial \varphi }}-{\frac {\partial }{\partial r}}\left(rA_{\varphi }\right)\right)&\\B_{\varphi}= 0={}+{\frac {1}{r}}\left({\frac {\partial }{\partial r}}\left(rA_{\theta }\right)-{\frac {\partial A_{r}}{\partial \theta }}\right)&\end{aligned}}}
\end{equation}

Is there a "trick" to solve this , or maybe some vector identity to simplify the problem ?
Any hints are greatly appreciated , thanks!

 

[pasmith]

You can assume that nothing depends on $\phi$, and the third equation is satisfied by $A_r = A_\theta = 0$. That leaves $$\begin{split}<br /> B_r &= \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (A_\phi \sin \theta) \\<br /> B_\theta &= -\frac{1}{r} \frac{\partial}{\partial r} (r A_\phi) \end{split}$$ and now substituting $A_\phi = r^\alpha f(\theta)$ will solve the problem.