curl of what yields $(0,s^{-1},0)$ in cylindrical coordinates?

Question

In cylindrical coordinates $(s,\theta,z)$, what function $\mathbf{A}$ has the property $$\nabla\times \mathbf{A} = (0, \frac{1}{s} , 0) $$

I know generally that $$\nabla\times \mathbf{A} = \left(\frac{1}{s}\frac{\partial A_3}{\partial \theta}-\frac{\partial A_2}{\partial z}, \frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial s} , \frac{1}{s} \left( \frac{\partial}{\partial z}(s\cdot A_2) - \frac{\partial A_1}{\partial \theta}\right)\right) $$

Is there some better way of solving for $\mathbf{A}$ other than a lot of ugly equations?

Answer 1

Then you can use a Green's function to solve this PDE.

The relevant Green's function is

$$\mathbf G(\mathbf r) = \frac{\mathbf r}{4\pi r^3}$$

The function $\mathbf A$ can then be found by convolution with the Green's function. Let $\mathbf B = s^{-1} \hat \theta$. The solution is then

$$\mathbf A(\mathbf r) = \int_{M} \mathbf B(\mathbf r') \times \mathbf G(\mathbf r - \mathbf r') \, dV'$$

Edit: there was a surface integral here, but I realized I had it written down wrong. I'll have to find a source for this, for while the volume integral is pretty simple, the surface integral looks, to me, to be a bit complicated, and you probably don't want to do that. I'll try to verify the form of that integral. Choosing all space as your integration region safely ignores the surface integral, but you may not be able to do this easily with $1/s$ in your vector field.

Answer 2

There is a path integral formula to construct the inverse of the exterior derivative operator for closed differential forms, for curl in three dimensions: $$ \newcommand{\A}{\mathbf{A}}\newcommand{\ps}{\boldsymbol{\psi}} \newcommand{\p}{\mathbf{p}} \mathscr{R}(\ps) = -(\p - \p_0)\times \int^1_0 \ps\big(\p_0 + t(\p- \p_0)\big)t\,dt,\tag{1} $$ where $\p = (x,y,z)$ and $\ps$ has to be divergence free. We have that $\mathscr{R}(\ps) $ satisfies: $$ \nabla \times \mathscr{R}(\ps) = \ps. $$

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The vector field you gave, $(0,1/s,0) = (0,\dfrac{1}{\sqrt{x^2+y^2}},0)$, is not divergence free, there does not exist any vector potential $\A$ satisfying: $$ \nabla \times \A = (0,\frac{1}{s},0). $$ However, we can just modify a little bit to make your question a well-posed problem, we can move $1/s$ to the $z$-component, and it is divergence free: $$ \nabla \cdot \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right) = 0, $$ and $$ \nabla \times \A = (0,0,\frac{1}{s}) $$ has a solution (not unique though).

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Let $\ps$ be the above vector, and in (1) we take $\p_0$ as the origin we have: \begin{align} \A = &\mathscr{R}(\ps) = -\p\times \int^1_0 \ps(t\p)t\,dt \\ &= -(x,y,z)\times \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right) \\ &= \left(-\frac{y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}},0\right).\tag{2} \end{align}

We can check: \begin{align} \nabla \times\A &= \left(0,0, \frac{\partial }{\partial x}\Big(\frac{x}{\sqrt{x^2+y^2}}\Big) + \frac{\partial }{\partial y}\Big(\frac{y}{\sqrt{x^2+y^2}}\Big)\right) \\ &= \left(0,0, \frac{\sqrt{x^2+y^2} - x\frac{x}{\sqrt{x^2+y^2}}}{x^2+y^2} + \frac{\sqrt{x^2+y^2} - y\frac{y}{\sqrt{x^2+y^2}}}{x^2+y^2}\right) \\ & = \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right). \end{align}

Or you can rewrite (2) in cylindrical coordinates: $$ \A = (-\sin\theta, \cos\theta,0), $$ and use the curl formula you gave.