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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?

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  • $\begingroup$ Is there a point to asking a question you already had the answer too (which you posted at the exact same time you asked the question)? This is the second time. Are you hoping to provide answers for other readers? You should always check if the questions have been asked before. Locking in your own answer assumes at the onset you have the best answer, and this closes the possibility that another better answer will appear. Thanks. $\endgroup$
    – joseph h
    Dec 13, 2022 at 7:23
  • $\begingroup$ @josephh 1) Both questions have been cases of rubberducking -- I didn't have the answer when I started writing the question, but the act of writing it led me to the solution. 2) I didn't find existing questions before I started writing my questions. Of course, I always have trouble with this site's search engine, so please flag either as duplicate if I just missed something. 3) Your point about locking answers is well-taken -- I've un-accepted my answer on the other question, for now at least. $\endgroup$
    – kc9jud
    Dec 13, 2022 at 7:35
  • $\begingroup$ Of course, if this kind of sharing of answers is frowned upon here, I will be happy to delete my questions and answers! $\endgroup$
    – kc9jud
    Dec 13, 2022 at 7:38
  • $\begingroup$ Hi. I don't think it's frowned upon by most members. Anyone can answer their own question and should if they have found an answer after writing the question. I just thought it was unusual that your question/answer were posted at the exact same time. Thanks. $\endgroup$
    – joseph h
    Dec 13, 2022 at 7:53
  • $\begingroup$ @josephh Ah, yes I just used the "Share your knowledge Q&A style" checkbox on the "Ask Question" interface $\endgroup$
    – kc9jud
    Dec 13, 2022 at 7:59

1 Answer 1

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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.

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