From Helmholtz’s theorem, any smooth vector field $\mathbf{F}$ that goes to zero at infinite distance can be uniquely decomposed everywhere in the sum of a divergence free component and an irrotational component.

In particular if it's conservative, one can prove that it can be uniquely determined everywhere solving the Poisson equation for its potential: \begin{equation} \Delta\varphi(\mathbf{r})=f(\mathbf{r}) \end{equation} where $f$ is the divergence of $\mathbf{F}$. This equation can be found from the definition of scalar potential provided by Helmholtz’s theorem \begin{equation} \varphi(\mathbf{r})=\frac{1}{4\pi}\int_{\mathbb{R}^{n}}\frac{f(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d\mathbf{r}' \end{equation} I was wondering if something similar can be done (and how) for the vector potential, i.e., if it's possible to uniquely determine a solenoidal vector field by solving a differential equation that involves its vector potential and its curl starting from the Helmholtz’s theorem \begin{equation} \mathbf{A}(\mathbf{r}):=\frac{1}{4\pi}\int_{\mathbb{R}^{n}}\frac{\mathbf{C}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d\mathbf{r}' \end{equation} where $\mathbf{C}$ is the curl of $\mathbf{F}$. So I'm assuming this field goes to zero at infinite distance and it's smooth. In case this is not possible, I would appreciate a proof or a counterexample.

  • $\begingroup$ If you additionally assume that $$ F= \nabla\times G , \quad \nabla\cdot G = 0$$ then when you take the curl again and use the double curl identity $\nabla \times (\nabla\times G) = \nabla(\nabla\cdot G)-\Delta G$ you get $$ -\Delta G = \nabla\times F $$ which is again a Poisson equation so you can solve for $G$. this is called the Biot-Savart law. Without this assumption you still have an equation but its more complicated $\endgroup$ Commented Jul 30, 2021 at 10:48
  • $\begingroup$ @CalvinKhor That's the Coulomb gauge. I was hoping for a more general approach. $\endgroup$
    – Simo
    Commented Jul 30, 2021 at 11:26
  • $\begingroup$ well I have no idea what a Coulomb gauge is but it wasn’t in your question despite being a partial answer hence my comment. That’s the end of what I know so gl $\endgroup$ Commented Jul 30, 2021 at 12:29

1 Answer 1


This is a very good question and it is addressed in some treatments of classical field theory. Specifically, I think the answer you seek is nicely discussed in the excellent discussion in pages 52-54 in Morse and Feshbach's book, "Methods of Theoretical Physics".

It can be found online here, for example:


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .