In electrodynamics, the scar potential $\phi$ and the vector potential $\textbf{A}$ satisfy the equations $$\frac{\partial}{\partial t}(\boldsymbol{\nabla}\cdot\textbf{A})+\nabla^2\phi=-\frac{\rho}{\epsilon_0}$$ and $$\boldsymbol{\nabla}\Big((\boldsymbol{\nabla}\cdot\textbf{A})+\frac{1}{c^2}\frac{\partial\phi}{\partial t}\Big)+\frac{1}{c^2}\frac{\partial^2\textbf{A}}{\partial t^2}-\nabla^2\textbf{A}=\mu_0\textbf{J}.$$ Radiation gauge implies that one can change from $(\phi,\textbf{A})\to(\phi^\prime,\textbf{A}^\prime)$ such that $\phi^\prime=0$ as well as $\nabla\cdot\textbf{A}^\prime=0$. If $\nabla\cdot\textbf{A}\neq 0$, we can choose a function $\chi(\textbf{r},t)$ such that $$\nabla^2\chi=-\nabla\cdot\textbf{A},\tag{1}$$ we get, $\nabla\cdot\textbf{A}^\prime=0$ because $\textbf{A}^\prime=\textbf{A}+\nabla\chi$.

If one is in vacuum, $\rho=0$. In that case only, one can choose $\phi^\prime=0$, by demanding $$\frac{\partial\chi}{\partial t}=\phi\tag{2}$$ since $\phi^\prime=\phi-\frac{\partial\chi}{\partial t}$.

For radiation gauge, we require both $\nabla\cdot\textbf{A}=\phi=0$. But for that choice of gauge, (1) and (2) has to be satisfied simultaneously. What is the guarantee that we can always find a $\chi(\textbf{r},t)$ satisfying both (1) and (2)?


3 Answers 3


$\nabla\cdot \mathbf{A} = 0$ is known as the Coulomb (or transverse) gauge. It's also called the radiation gauge. Note that, for boundary conditions at infinity, the Coulomb gauge is complete - there is no ambiguity left for making further modifications to the gauge fields that don't violate $\lim_{r\rightarrow \infty} A^\mu(r, t) = 0$ (because any function that satisfies $\nabla^2 f(\mathbf{x})=0$ and $\lim_{r\rightarrow \infty} f(\mathbf{x}) = 0$ is identically zero).

$\phi = 0$ is a different gauge called the Weyl gauge. The residual gauge symmetries aren't enough to make it compatible with the Coulomb gauge. This gauge is unspoilt by any gauge transformation that satisfies $\frac{\partial f}{\partial t} = 0$, leaving $$\mathbf{A}'(\mathbf{x}, t) = \mathbf{A}(\mathbf{x}, t) + \nabla f(\mathbf{x}).$$ True, it might be possible to use the residual freedom in the Weyl gauge to satisfy the Coulomb gauge condition at some particular time, but it will only be satisfied for that instant, and not in general.

  • 1
    $\begingroup$ What is $\nabla\cdot\textbf{A}=\phi=0$ gauge called? Is it not the radiation gauge? In Ryder's book on QFT, while quantizing the free electromagnetic field, he referred the $\nabla\cdot\textbf{A}=\phi=0$ as both the Coulomb and radiation gauge. @Sean E. Lake $\endgroup$
    – SRS
    Feb 24, 2018 at 9:23
  • $\begingroup$ @SRS $\nabla\cdot\mathbf{A}=0$, the Coulomb gauge, is also called the radiation gauge. Setting $\nabla\cdot\mathbf{A}=\phi=0$ is not possible because that also sets the gauge invariant quantity $\mathbf{E}_{\mathrm{div}} = -\nabla\phi - \dot{\mathbf{A}}_{\mathrm{div}}=0$, where $\mathbf{A}_{\mathrm{div}}$ is the divergenceful part of $\mathbf{A}$ in its Helmholtz decomposition. Now, that quantity may happen to be zero because $\rho =\nabla\cdot\mathbf{J}=0$, but that's not part of the gauge choice. $\endgroup$ Feb 24, 2018 at 15:40
  • 2
    $\begingroup$ In vaccuum, I don't see the problem. In your first approach, if you achieve Coloumb gauge $\nabla\cdot\vec{A}=0$, then you get automatically $\varphi=0$, since the Maxwell equation is telling you $\Delta\varphi=0$ and the only solution vanishing at infinity is zero. The same can be seen in your second approach: If $\varphi=0$, Maxwell's equation is telling that $\partial_{t}(\nabla\cdot\vec{A})=0$ which shows that the remaining gauge freedom can be used to achieve Coloumb gauge, since we just solve the time-independent Poisson equation $\Delta f(\vec{x})=\nabla\cdot A$. $\endgroup$
    – B.Hueber
    Mar 24, 2023 at 13:40
  • 2
    $\begingroup$ In other words, the Coloumb gauge condition $\nabla\cdot\vec{A}=0$ implies (in vacuum!) what is usually called radiation gauge, i.e. $\nabla\cdot\vec{A}=0,\varphi=0$. Of course, this gauge only makes sense in vacuum, but that is also when it is usually appplied, i.e. in order to study radiation. $\endgroup$
    – B.Hueber
    Mar 24, 2023 at 13:46

This is not always possible - there is no such guarantee. To see this, assume that the transformation from $(\mathbf A,\phi)$ to the radiation gauge is possible, and calculate \begin{align} \nabla^2\phi +\frac{\partial}{\partial t}\nabla\cdot\mathbf A & = \frac{\partial}{\partial t}\nabla^2 \chi-\frac{\partial}{\partial t}\nabla^2 \chi \\ & = 0. \end{align} There is no guarantee that this was the case originally, and if there are charges present then this is obviously false in general - the radiation gauge is built to describe radiation and it only makes sense in vacuum.

I'm not really sure what the precise conditions are on $(\mathbf A,\phi)$ that will guarantee that a radiation gauge will exist, but frankly I don't think they are that important. The radiation gauge is a restricted gauge and it only applies to a limited set of fields, which are roughly-speaking "radiation-like". If the fields turn out not to have a radiation gauge, then it's because they weren't radiation-like enough to begin with. Most physicists, I suspect, are OK with that kind of fuzzyness - and if you do care about strict delimitations, then you're more likely to care about the Lorenz gauge to begin with.

  • $\begingroup$ Am I correct in my understanding of radiation gauge? In particular, the condition 2? @Emilio Pisanty $\endgroup$
    – SRS
    May 23, 2017 at 12:54
  • $\begingroup$ @EmilioPisanty I think that apart from the vacuum, the gauge $\phi=\vec{\nabla}.\vec{A}=0$ is a possible choice for linear dielectrics for which $\vec{D}=\epsilon\vec{E}$. No inconsistencies arise. $\endgroup$ Nov 22, 2018 at 11:14

Equation (1) implies using (2) that

$$\renewcommand{\vec}[1]{\boldsymbol{#1}}\Delta\partial_t\chi = -\underbrace{\nabla\cdot\partial_t\vec{A}}_f. \tag{I}$$

This is Poisson's equation for $\psi=\partial_t\chi$. If there is a solution, then equation (2) trivially gives $\chi$ by integrating $\psi$ with respect to time. So it boils down to find a solution to (I): this will be highly dependent on the boundary conditions as any treaty on elliptic PDE will tell you. Some classic boundary conditions:

  1. $\psi=0$ on the boundary $\partial\Omega$ of the domain $\Omega$ in which one is working (zero to fix the arbitrary additive constant);
  2. $\nabla\psi\cdot\vec{n}=g$ on $\partial\Omega$ and $\int_\Omega\psi dV=0$ (the latter to fix the arbitrary additive constant);
  3. $\nabla\psi\cdot\vec{n}+\alpha\psi=0$.

But then for any smooth function $\varphi$,

$$\int_\Omega \nabla\psi\cdot\nabla\varphi dV=-\int_\Omega\varphi\Delta\psi dV+\int_{\partial\Omega}\varphi\nabla\psi\cdot\vec{n}dS.$$

Thus if $\psi$ is a solution of (I), we have for each of the considered boundary conditions,

  1. $\int_\Omega \nabla\psi\cdot\nabla\varphi dV = \int_\Omega f\varphi dV$,
  2. $\int_\Omega \nabla\psi\cdot\nabla\varphi dV = \int_\Omega f\varphi$,
  3. $\int_\Omega \nabla\psi\cdot\nabla\varphi dV + \int_{\partial\Omega}\alpha \psi\varphi = \int_\Omega f\varphi dV$.

All these problems have the same structure: $a(\psi,\varphi)=L(\varphi)$ where $a$ is a bilinear form and $L$ is a linear form. In French, we call that formulation the weak formulation and a quick googling seems to show the wording is used in English too (I am sorry but I learned those math in French and I have never used them since!).

This is the domain of the Lax-Milgram-Lions theorem which guarantees the existence, and unicity, of a solution under broad conditions for $a$ and $L$: there exist a Hilbert space for the functions under consideration, and

  • a is continuous, and $|a(\varphi,\varphi')|\le c\|\varphi\|\|\varphi'\|$ for some $c>0$;
  • $a(\varphi,\varphi) \ge d\|\varphi\|^2$ for some $d > 0$ (we say that $a$ is coercive in French);
  • $L$ is continuous.

I would say that in physics, those conditions are fulfilled in practice, save perhaps the odd unrealistic case.


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