What Is the Coulomb Gauge in Electromagnetic Theory?

[Greg Bernhardt]

Definition/Summary

A particular choice of "gauge fields" in which the vector potential has no divergence.

Equations

$$\nabla\cdot\vec A=0$$

Extended explanation

As explained elsewhere, one introduces "gauge fields" in order to identically solve two of the four Maxwell equations.

When written in terms of the gauge fields the remaining Maxwell equations (in Gaussian units) are given by:
$$\nabla\cdot\left(-\nabla\phi-\frac{1}{c}\frac{\partial \vec A}{\partial t}\right) =4\pi\rho\;,\qquad (1)$$
where $\rho$ is the charge density; and
$$\nabla\times\nabla\times\vec A = 4\pi\frac{\vec j}{c} +\frac{1}{c}\frac{\partial}{\partial t}\left(-\nabla \phi-\frac{1}{c}\frac{\partial \vec A}{\partial t}\right)\;,\qquad (2)$$
where $\vec j$ is the (charge) current density.

In the Coulomb gauge, equation (1) reduces to
$$-\nabla^2\phi=4\pi\rho\;,\qquad (3)$$
which is the same as the usual electrostatic equation for the scalar potential. Thus, in this gauge, charges
apparently interact through an instantaneous Coulomb potential just like in electrostatics. Of course, the instantaneous nature of the interaction is canceled by the interaction of the currents via the vector potential and there is no violation of causality.

The Coulomb gauge is very useful in condensed matter theory where the charges are non-relativistic for the most part and the main contribution to their interaction is most conveniently described as via an instantaneous Coulomb potential.

The downside to the Coulomb gauge is that equation (2) become rather cumbersome:
$$-\nabla^2\vec A = 4\pi\frac{\vec j}{c} -\frac{1}{c} \frac{\partial}{\partial t} \left( \nabla\phi+\frac{1}{c}\frac{\partial \vec A}{\partial t} \right)\;,\qquad (4)$$
where $\phi$ is now consider to be known as a functional of $\rho$ via the solution of equation (3).

From equation (4) we also see that in the non-relativistic limit we can ignore the vector potential. This is
typically the case for condensed matter systems as mentioned above.

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!

 

[fresh_42]

We also recommend our Insight Blog Article:
https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/

 

[vanhees71]

Thanks for quoting my DC-conducting wire Insight, but I'm not so sure, whether it's reallly providing much more insight to the Coulomb gauge. I don't know, if it makes sense to write a bit more on the solutions of the Maxwell equations in Coulomb gauge in this old thread? If so, I can provide something along the lines of

https://arxiv.org/abs/2006.11598

 

[vanhees71]

I think this issue is important enough to justify a more detailed answer even to an old thread.

The Maxwell equations (in Heaviside-Lorentz units) read
$$\begin{alignat}{2}
& \vec{\nabla} \times \vec{E} +\frac{1}{c} \partial_t \vec{B}=0,\\
& \vec{\nabla} \cdot \vec{B} = 0,\\
& \vec{\nabla} \times \vec{B} + \frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j},\\
& \vec{\nabla} \cdot \vec{E}=\rho.
\end{alignat}$$
These are the fundamental Maxwell equations with $\rho$ and $\vec{j}$ being the total charge and current densities.

Eqs. (1) and (2), the homogeneous Maxwell equations admit the introductio of a scalar and a vector potential. We start with (2): According to Helmholtz's fundamental theorem there exists a vector potential such that
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Plugging this in (\ref{1}) we get
$$\vec{\nabla} \times \left (\vec{E} +\frac{1}{c} \partial_t \vec{A} \right)=0 \; \Rightarrow \; \vec{E}+\frac{1}{c} \partial_t \vec{A}=-\vec{\nabla} \Phi,$$
where $\Phi$ is the scalar potential. So we have
$$
\begin{equation}
\vec{E}=-\frac{1}{c} \partial_t \vec{A} -\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.
\end{equation}
$$
Now for a given physical situation, described by $\vec{E}$ and $\vec{B}$, the potentials are not uniquely determined, i.e., although the homogeneous Maxwell equations are automatically fulfilled by introduction of the potentials, the potentials are not uniquely determined by the physical situation, i.e., an entire class of solutions for the potentials will describe the same physical situation. That's called gauge invariance.

Indeed if $\Phi$ and $\vec{A}$ describe a given electromagnetic field, $(\vec{E},\vec{B})$, for any scalar field $\chi$ also
$$
\begin{equation}
\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+\frac{1}{c} \partial_t \chi
\end{equation}
$$
describe the same physical situation. As we shall see, we can use this gauge freedom to impose useful constraints on the potentials, that (at least partially) "fix the gauge", and help to simplify the equations.

The equations for the potentials follow from the inhomogeneous Maxwell equations (3) and (4) by plugging in (5) and using $\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}$:
$$\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}+\frac{1}{c^2} \partial_t^2 \vec{A} + \frac{1}{c} \partial_t \vec{\nabla} \Phi=\frac{1}{c} \vec{j}, \quad -\Delta \Phi -\frac{1}{c} \vec{\nabla} \cdot \vec{A}=\rho.$$
Now we make use of the "gauge freedom" and impose useful gauge constraints. The idea of imposing the socalled Coulomb-gauge condition,
$$\begin{equation}
\vec{\nabla} \cdot \vec{A}=0,
\end{equation}$$
is to decouple the scalar potential from $\vec{A}$ in the last equation. Then it reads as for the potential in electrostatics, which is the origin of the name "Coulomb gauge" for this choice of the gauge constraint:
$$\begin{equation}
-\Delta \Phi=\rho.
\end{equation}$$
This equation can be solved via the Green's function of the Laplace operator as in electrostatics, although $\rho$ may be time dependent:
$$
\begin{equation}
\Phi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(t,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.
\end{equation}
$$
From the point of view of relativity that's a bit strange since it says that the contribution from the source, $\rho$, at place $\vec{x}'$ is instantaneous at the (maybe far distant) position $\vec{x}$. This reminds us that the potential is not an observable, because it is gauge dependent and subject to the quite arbitrary gauge condition.

Now also the equation for $\vec{A}$ simplifies. For the given scalar potential (8) we get
$$\Box \vec{A} = \frac{1}{c} \vec{j}_{\text{perp}}, \quad \vec{j}_{\perp} = \vec{j}-\partial_t \vec{\nabla} \Phi.$$
Here we introduced the d'Alembert operator $\Box=1/c^2 \partial_t^2 - \Box$, i.e., we have an inhomogeneous wave equation, and for causality reasons we use the retarded propagator of the d'Alembert operator to solve it. The result is
$$
\begin{equation}
\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}_{\perp}(t-|\vec{x}-\vec{x}'|/c,\vec{x}')}{4 \pi c |\vec{x}-\vec{x}'|}=\int_{\mathbb{R}} \mathrm{d} t' \int_{\mathbb{R}^3} \delta(t-|\vec{x}-\vec{x}'|/c-t') \frac{\vec{j}(t',\vec{x}')}{4 \pi c |\vec{x}-\vec{x}'|}.
\end{equation}
$$
This formally looks as if it were a field which is causally connected with its source $\vec{j}_{\perp}$ since the source at place $\vec{x}'$ contributes to $\vec{A}$ at place $\vec{x}$ at the retarded time $t-|\vec{x}-\vec{x}'|/c$, i.e., the "signal" seems to propagate with the speed of light, as expected for an electromagnetic wave. However, that's not really the case, because $\vec{j}_{\perp}$ contains the non-local contribution from the term involving $\Phi$.

To see that, nevertheless (9) leads to a valid solution for $\vec{A}$ we need to show that it fulfills the Coulomb-gauge constraint $\vec{\nabla} \cdot \vec{A}=0$. To see that apply $\vec{\nabla}$ under the integral in the 2nd form. Since the expression the operator acts on depends only on $|\vec{x}-\vec{x}'|$ you can change $\vec{\nabla}$ to $\vec{\nabla}'$ and then do a partial integration in the spatial integral. The crucial point is that
$$\vec{\nabla}' \cdot \vec{j}_{\perp}(t',\vec{x}') = \vec{\nabla}' \cdot [\vec{j}(t',\vec{x'})- \partial_{t'} \vec{\nabla}' \Phi(t',\vec{x}')=\vec{\nabla}' \cdot \vec{j}(t',\vec{x}')+\partial_{t'} \rho(t',\vec{x}')=0,$$
due to the continuity equation, which describes (local) charge conservation, which is a necessary integrability constraint for the solvability of the Maxwell equations. This demonstrates the important connection between local gauge symmetry and charge conservation. The latter follows from the field equations of motion alone with no need to use the matter-equations of motion.

Using similar techniques, you can also prove that the observable fields, $\vec{E}$ and \vec{B}##, are given in terms of retarded integrals over local sources only, as it should be for a local relativistic field theory.

For further details, see

https://arxiv.org/abs/2006.11598

PS: Obviously the equation numbering doesn't work anymore. Was there some "update" of the LaTeX software?