I was in an electromagnetism lecture, where we were looking at the magnetostatic Maxwell’s equations:

$$\begin{align} \nabla\cdot\mathbf{B} &= 0 \\ \nabla\times\mathbf{B} &= \mu_0\mathbf{J} \end{align}$$

I am aware that $\mathbf{B}$ can be expressed in terms of a vector potential, as from the the first equation if we let $\mathbf{B}=\nabla\times\mathbf{A}$, where $\mathbf{A}$ is a general vector field, this would thus satisfy the (no magnetic monopoles) equation at the top, as the divergence of the curl of any vector field yields 0.

However the lecturer states that our choice of $\mathbf{A}$ is not unique and that we should add another term (he called it gauge transformation): $$\mathbf{A}' = \mathbf{A} + \nabla\chi$$ where $\chi=\chi(\mathbf{r})$.

He then states that in order to make $\mathbf{A}$ unique, we must impose the following gauge conditions (Coloumb & Lorentz, respectively);

$$\begin{align} \nabla\cdot\mathbf{A} &= 0 \\ \nabla\cdot\mathbf{A} + \frac{1}{c^2}\frac{\partial V}{\partial t} &= 0 \end{align}$$ where $V$ is a scalar potential.

  • What does it mean to be unique in this context?
  • How do these conditions make our choice of $\mathbf{A}$ unique?

2 Answers 2


Below I'll use Planck units, for which, in particular, $c = \epsilon_{0} = =1$.

In fact, the full system of Maxwell's equations provides the statement that the only two vector components of the EM field $\mathbf E, \mathbf B$ are independent (in general, due to a deep symmetry reason, namely that a massless particle has only two polarizations). Next, if we write EM field in terms of 4-potential $$ A_{\mu} \equiv \left( V, \mathbf A\right)_{\mu} $$ (for example, $A_{0} \equiv V$), $$ \mathbf E = -\frac{\partial \mathbf A}{\partial t} - \nabla V, \quad \mathbf B = \nabla \times \mathbf A, $$ we add new unphysical gauge symmetry transformation $$ \tag 1 V \to V + \frac{\partial \varphi}{\partial t}, \quad \mathbf A \to \mathbf A - \nabla \varphi , $$ under which $\mathbf E, \mathbf B$ is unchanged: $$ \mathbf E \to \mathbf E, \quad \mathbf B \to \mathbf B $$ I.e., $A_{\mu}$ is not unique. Also, it seems that without imposing some conditions 4-potential has 4 components, while this number must be 2, in correspondence to the number of independent components of $\mathbf E, \mathbf B$. Is this a problem? No.

We need to constuct the precise scheme of reduction of number of components, which includes the problem of uniqueness of 4-potential. The gauge symmetry is unphysical, so first we may fix $\varphi$ in $(1)$ by imposing so-called gauge condition, which leaves three independent components of $A_{\mu}$ instead four. It may be, for example, Coulomb gauge condition, $$ \nabla \cdot \mathbf A = 0, $$ or lorentz invariant condition, such as $$ \partial_{\mu}A^{\mu} = \frac{\partial V}{\partial t} + \nabla \cdot \mathbf A = 0 $$ Next, we reduce the number of components from three to two by using first Maxwell equation, which is called Gauss law, since in fact it is the bound, because it doesn't contain first time derivatives of $\mathbf E$ or, correspondingly, second time derivatives of $A_{\mu}$ (it doesn't describe propagated degrees of freedom). For example, for Coulomb gauge this equation takes the form $$ \nabla \cdot \mathbf E = -\Delta V - \frac{\partial (\nabla \cdot \mathbf A )}{\partial t} = -\Delta V = 4 \pi \rho , $$ from which we have the scalar potential as function of charge density. For $\rho = 0$, which is true in the case of magnetostatics, we may simply set $V$ to zero.

So that, we reduce the number of $A_{\mu}$ from four to two, as it must be, and solve the uniqueness problem.

Let summarize: if we have used the Gauss law as the bound, i.e., if we have reduced the number of components of 4-potential from four to three, the only thing that we need to fix it uniquely is to impose such gauge condition, since without doing that 4-potential is determined only up to unphysical transformation.

  • $\begingroup$ That seems to overshoot a bit. $\endgroup$
    – Gonenc
    Commented Feb 7, 2016 at 17:49
  • $\begingroup$ @gonenc : why do you think that? $\endgroup$
    – Name YYY
    Commented Feb 7, 2016 at 17:52
  • $\begingroup$ I just thought that the OP doesn't have the background. $\endgroup$
    – Gonenc
    Commented Feb 7, 2016 at 17:53
  • $\begingroup$ @gonenc : thank you, I'll try to change an answer by adding explanations or by removing some part of an answer. $\endgroup$
    – Name YYY
    Commented Feb 7, 2016 at 17:54
  • $\begingroup$ There's no such thing as $c=1$ units. If you don't put $c$ in E&M equations then you're working with quantities whose dimensions differ from the usual. $\endgroup$
    – DanielSank
    Commented Feb 7, 2016 at 18:11

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\div}[1]{\nabla \cdot #1} \newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}} \newcommand{\curl}[1]{\nabla \times \vec{ #1}}$I assume you know how to solve the Maxwell's equations in vacum. In the end you get an expression of the form:

$$\nabla^2\phi = - \pdv {} t(\div \vec A) \quad \text{and}\quad \nabla(\div \vec A)+ \nabla \left( \pdv \phi t \right) + \pdvt{\vec A} t= \Delta \vec A \tag 1 $$

where I've taken $\varepsilon_0= \mu_0 = c =1$ and $$-\nabla \phi = \vec E + \pdv A t \tag 2 \quad \text{and} \quad \vec B = \curl A$$ as usual and $\Delta$ is the vectorial Laplacian. You know that any $\vec A' = \vec A + \nabla f$ also leaves $\vec B$ invariant. This is why we say $\vec A$ is not unique. Note however that you also need to change $\phi$ in order to leave the Maxwell's equation invariant i.e. you need to take another $\phi'$ with

$$\phi' = \phi - \pdv f t$$

which can be easily seen from Eqn $(2)$. We want to fix this ambiguity and also make the Eqn. $(1)$ look nicer. Thus we choose $\div A = - \left( \pdv \phi t\right)$ so that the Eqn $(1)$ becomes:

$$\nabla^2 \phi = \pdvt \phi t \qquad \pdvt {\vec A} t = \Delta \vec A$$

This is the so-called Lorentz gauge. Note that $\vec A$ is still not too unique i.e. $\vec A' = \vec A + \nabla f$ also solves the equations for any function satisfying $\nabla^2 f = \pdvt f t$. In order to fix $\vec A$ completely you need to impose boundary conditions.

If we had $\pdv \phi t = \pdv A t=0$, then the choice of the divergence of $\vec A$ would be $\div { \vec A} = 0$ which is also called the Coulomb gauge.

This is not as neat as the answer of Name YYY but without knowing about tensors and special relativity I think it should be suffice.


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