Can a static electric field have a vector potential field?

Question

We know very well that static electric fields have scalar potential fields, and magnetic fields can't have scalar fields if there is free current, as $$\nabla\times \vec{B}=\mu_0\vec J$$ $$\Rightarrow \int\nabla\times \vec{B}\cdot d\vec{a}=\oint\vec{B}\cdot d\vec{l}=\int\mu_0\vec J\cdot d\vec{a}$$Since the surface integral of $\vec{J}$ does not have to be zero, that means if we summing up $\vec{B}\cdot{d\vec{l}}$ from point $a$ to point $b$; the result depends not only on end points $a,b$ but also on how you sum it up. Therefore, we do not have a scalar potential for static magnetic field, which is defined as $$\int_{a}^{b}\vec{B}\cdot{d\vec{l}}$$ that is required to be independent of path chosen.

But I have just wondered: can we give static electric field a vector potential field? How? If we can't, how?

---

My attempt:

$$\nabla\times (\nabla\times \vec A_m)=0$$ $$-(\partial_x^2+\partial_y^2+\partial_z^2)\vec A+ \begin{bmatrix} \partial_x \\ \partial_y \\ \partial_z\\ \end{bmatrix} \ \nabla \cdot \vec A=0$$

Then we have this for the $x$ direction:

$$(\partial_x^2+\partial_y^2+\partial_z^2) A_x = \partial_x \nabla \cdot \vec A$$

I do not know how to solve it, except for requiring $\nabla \cdot \vec A=0$, we have $$(\partial_x^2+\partial_y^2+\partial_z^2) \begin{bmatrix} A_x \\ A_y \\ A_z\\ \end{bmatrix} \ =0 $$

Then I can go no further. (I have looked up the three dimensional Laplace partial equation, but I am not sure what does the solution means, especially it will appear in each terms of the vector)

Furthermore: can we define a vector potential field for a non-static eletirc field? $^1$

---

$1:$ I think it has something to do with the Maxwell equations that having each other partial t E/M field in the "curl M/E equations", but I am not quite sure about that.

Answer 1

It looks like you're trying to find a vector field $\vec{A}_m$ such that $\vec{E} = \nabla \times \vec{A}_m$. This is only possible in regions of space that are charge-free: the divergence of the curl of a vector field is always zero, so we necessarily have $$ \frac{\rho}{\epsilon_0} = \nabla \cdot \vec{E} = \nabla \cdot (\nabla \times \vec{A}_m) = 0. $$ Conversely, such a vector field cannot be defined in a region of space where $\rho \neq 0$. So $\rho = 0$ is a necessary condition for this $\vec{A}_m$ to exist.

What's more, $\rho = 0$ is not a sufficient condition for $\vec{A}_m$ to exist. Even if $\rho = 0$ everywhere in a region of space $\mathcal{R}$, we can still run into problems due to the topology of $\mathcal{R}$. To see this, consider a situations where we have a uniform ball of charge with total charge $Q \neq 0$, and let $\mathcal{R}$ be all of space except the volume of the ball (so $\rho = 0$ everywhere in $\mathcal{R}$.) Let $\mathcal{S}$ be any closed surface (lying in $\mathcal{R}$) that encloses the charge. Then we have \begin{align*} \frac{Q}{\epsilon_0} &= \int_\mathcal{S} \vec{E} \cdot d\vec{a} &&\text{(by Gauss's Law)} \\ &= \int_\mathcal{S} (\nabla \times \vec{A}_m) \cdot d\vec{a} &&\text{(assuming $\vec{A}_m$ exists)} \\ &= \oint_\mathcal{\partial \mathcal{S}}\vec{A}_m \cdot d\vec{l}, && \text{(Stokes' Theorem)} \end{align*} where $\partial \mathcal{S}$ is the boundary of the surface $\mathcal{S}$, according to Stokes' Theorem. But $\mathcal{S}$ is a closed surface; it has no boundary. Thus, the integral vanishes, and $Q$ must be zero. This is a contradiction.

The reasons for this are a little long to go into here, but basically $\vec{A}_m$ will exist if and only if every surface that's equivalent to a sphere lying in your charge-free region $\mathcal{R}$ can be smoothly contracted down to a point while remaining in $\mathcal{R}$. If you're interested in this, I encourage you to look into the subjects of differential forms and homology theory; it's a very elegant way of looking at mathematical physics (and electrodynamics in particular.)