Solving Curl A in Spherical Coordinates: Tips & Hints

AI Thread Summary
The discussion focuses on solving the curl of vector field A in spherical coordinates, specifically addressing the equations for the components B_r and B_θ. Participants suggest that assuming A_r and A_θ are zero simplifies the problem, leaving only B_r and B_θ to solve. A substitution of A_φ as r^α f(θ) is proposed as a potential solution technique. The conversation highlights the need for vector identities or tricks to simplify the calculations further. Overall, the thread emphasizes collaborative problem-solving in vector calculus within spherical coordinates.
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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!
 
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You can assume that nothing depends on \phi, and the third equation is satisfied by A_r = A_\theta = 0. That leaves <br /> \begin{split}<br /> B_r &amp;= \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (A_\phi \sin \theta) \\<br /> B_\theta &amp;= -\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.
 
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