What is the Coulomb gauge

1. Jul 24, 2014

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 cancelled 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.

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