Vector potentials of vector potentials In $\mathbb{R}^{3}$, we have that $\mathbb{div}(A_0)=0 \implies A_0=\mathbb{curl}(A_1)$
for some vector field $A_1$
We can also, using the Lorentz Gauge, choose $A_1$ such that $\mathbb{div}(A_1)=0$, which made me wonder;
What happens if you repeat this process, producing $A_2$ a divergence free potential for $A_1$ and so on. Does this terminate to a fixed point? Does it enter a loop?
 A: Following the prescription of this question, given a vector field $\mathbf A_0$ such that $\nabla \cdot \mathbf A_0=0$, we construct a sequence $\mathbf{A}_0, \mathbf A_1, \mathbf A_2, \ldots$ such that
$$
\begin{array}{cc} 
\nabla \cdot \mathbf A_1 = 0, & \nabla \times \mathbf A_1= \mathbf A_0\\ 
\nabla \cdot \mathbf A_2 = 0, & \nabla \times \mathbf A_2= \mathbf A_1\\ 
\nabla \cdot \mathbf A_3 = 0, & \nabla \times \mathbf A_3= \mathbf A_2\\ 
\vdots & \vdots 
\end{array}$$
Taking the curl of the right column, and using the formula for the double curl, we see that
$$
-\nabla^2 \mathbf A_{n+2} = \mathbf A_n, \quad n=0, 1, 2, \ldots,$$
where $\nabla^2$ denotes the vector Laplacian. This is the Poisson equation and it can be solved by convolution against  the Newtonian potential;
$$\tag{*}
\mathbf A_{n+2}(x)=\int_{\mathbb R^3} \frac{ \mathbf A_n(x-y)}{4\pi \lvert y\rvert}\, dy.$$
Remark. The vector field $\mathbf A_{n+2}$ given by (*) is divergence-free if $\mathbf A_n$ is. The proof is immediate, just take the divergence termwise and interchange differentiation and integration. Thus, (*) has the correct gauge.

To gather information about the limit as $n\to \infty$, (*) is impractical. However, taking the Fourier transform, defined as
$$
\hat{f}(k)=\int_{\mathbb R^3} f(x)e^{-i x\cdot k}\, dx, $$
we see that
$$\tag{**}
\hat{\mathbf A}_{n+2}(k)=\frac{\hat{\mathbf A}_n(k)}{\lvert k\rvert^2}.$$
We conclude that
$$
\hat{\mathbf A}_n(k)=\begin{cases} \frac{\hat{\mathbf A}_0(k)}{\lvert k\rvert^m}, & n=2m, \\ 
\frac{\hat{\mathbf A}_1(k)}{\lvert k\rvert^m}, & n=2m+1.
\end{cases}$$
In particular:

*

*the sequence never reaches a fixed point, except for the trivial cases in which $\mathbf A_0=\mathbf 0$ or $\mathbf A_1 =\mathbf 0$.

*if $\hat{\mathbf A}_0$ is supported in $\lvert k\rvert <1$, then the sequence $\hat{\mathbf A}_n$ for even $n$ blows up. If it is supported in $\lvert k\rvert >1$, it tends to zero. Similar considerations for $\hat{\mathbf A}_n$ for odd $n$.

