Vector field where curl of the field is the field itself

Question

I was curious if there is a possibility of a field where the curl of the field would be the field itself, i.e, $$\nabla \times \vec{A} = \vec{A}$$

I can immediately see that the divergence of such a field is $0$ so it may be written as $\vec{A} = \nabla \times \vec{B}$. This is also sort of equivalent to find the eigenvector of the anti symmetric matrix $$ \begin{bmatrix} 0 & -\frac{\partial}{\partial z} & \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} & 0 & -\frac{\partial}{\partial x} \\ -\frac{\partial}{\partial y} & \frac{\partial}{\partial x} & 0 \\ \end{bmatrix} $$As this is normal, I think eigenvectors should exist.I am asking this because I am curious to see how a field whose rotation at every point is the field itself would look like except the zero vector field.

@BenSteffan I have modified the question and excluded the trivial case — Rishabh Jain, Nov 15 '19 at 13:54
Related: What are the Eigenvectors of the curl operator? I'm not sure if an eigenfunction with eigenvalue 1 exists. — , Nov 15 '19 at 13:59

score 2 · Accepted Answer · 2019-11-16T12:51:58.460

Eigenfunctions of the curl operator, i.e. $\nabla\times \mathbf B=k\mathbf B$ for eigenvalue $k$, are studied in physics as examples of force-free magnetic fields. Kirk T. MacDonald's notes list many solutions. One of the simplest is Lundquist's solution, which in cylindrical coordinates $(\rho,\phi,z)$ is given by

$$ B_\rho = 0, \quad B_\phi = J_1(k\rho), \quad B_z = J_0(k\rho) $$ where $J_0$ and $J_1$ are Bessel functions. The field lines are helices, and since the Bessel functions are oscillatory in $\rho$ there are both left- and righthanded helices, and ones with both positive and negative $B_z$.

In the paper mentioned, the simplest solution is (cos z, -sin z, 0). — Rishabh Jain, Nov 16 '19 at 12:47

Vector field where curl of the field is the field itself

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