Vector potential of uniform electric field in Coulomb gauge

Question

In the Coulomb gauge, fixed by $\nabla \cdot \mathbf{A}(\mathbf{x},t) = 0$, we have that a vector potential for a constant, uniform magnetic field $\mathbf{B}$ is $$\mathbf{A}(\mathbf{x},t) = -\frac{1}{2} \mathbf{x} \times \mathbf{B}.$$ If we further set the scalar potential $\phi(\mathbf{x})=0$, we have that the electric field $\mathbf{E}$ is also zero.

Can we also define a vector potential in Coulomb gauge where the magnetic field is zero and the electric field is uniform and constant, while still having a vanishing scalar potential?

Answer 1

We need to find a vector potential $\mathbf{A}(\mathbf{x},t)$ such that \begin{align} -\frac{\partial \mathbf{A}}{\partial t} &= \mathbf{E}, \tag{1}\label{1} \\ \nabla \times \mathbf{A} &= \mathbf{B} = 0, \tag{2}\label{2}\\ \nabla \cdot \mathbf{A} &= 0, \tag{3}\label{3}\\ \nabla^2 \mathbf{A} - c^2 \frac{\partial^2 \mathbf{A}}{\partial t^2} &= -\mu_0 \mathbf{J} = 0. \tag{4}\label{4} \end{align} Equations \eqref{1} and \eqref{2} are simply the electric and magnetic fields in terms of the vector potential (with $\phi(\mathbf{x}) = 0$). \eqref{3} is the gauge condition. \eqref{4} is Maxwell's equation for the vector potential in the Coulomb gauge, again with $\phi(\mathbf{x})=0$.

From \eqref{1} we have that $\mathbf{A}(\mathbf{x},t)$ must be at least linear in $t$, while \eqref{2} and \eqref{3} will be satisfied if $\mathbf{A}(\mathbf{x},t)$ is independent of $\mathbf{x}$. Finally, the second term in \eqref{4} will vanish if $\mathbf{A}(\mathbf{x},t)$ is at most linear in $t$.

Therefore, we can write $$\mathbf{A}(\mathbf{x},t) = - t \mathbf{E}$$ for the vector potential of a uniform electric field.