Curl resulting in a potential field

Question

Can we find a vector field $\mathbf{A}$ such that

$\nabla\times\mathbf{A}=q\nabla p$ ,

where $q,p$ scalar functions of the position?

Answer 1

For arbitrary $p$ and $q$ this equation may not have a solution. Taking the divergence in both sides of the equation we see that a necessary condition for the equation to have a solution is that $\vec\nabla\cdot(q\vec\nabla p)=0$, or equivalently that $q\vec\nabla p=\vec\nabla\times\vec B$ for some vector field $\vec B$ (function of the position). In that case any solution to $\vec\nabla\times\vec A = \vec B+\vec\nabla\Phi$, with $\Phi$ some scalar field (and there are always solutions to this equation) will solve your equation. One only needs to find $\Phi$ so this equation is compatible, which requires that $\vec\nabla\cdot(\vec\nabla\times \vec A) =\vec\nabla\cdot \vec B + \nabla^2\Phi=0$. Solving this Poisson equation to find $\Phi$ ($\nabla^2\Phi=-\vec\nabla\cdot \vec B$), and inserting in $\vec\nabla\times\vec A = \vec B+\vec\nabla\Phi$ will give a solution of your original problem.

Answer 2

The answer is in general, "No!"

For if

$\nabla \times \mathbf A = q \nabla p, \tag 1$

then, since the divergence of a curl always vanishes, (1) implies

$\nabla \cdot (q \nabla p) = \nabla \cdot (\nabla \times \mathbf A) = 0; \tag 2$

also,

$\nabla \cdot (q \nabla p) = \nabla q \cdot \nabla p + q \nabla^2 p; \tag 3$

thus a necessary condition for (1) to hold is that $p$ and $q$ together obey the equation

$q\nabla^2 p + \nabla q \cdot \nabla p = 0, \tag 4$

which is clearly not the case for arbitrary $p$ and $q$.