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The Coulomb or Transverse Gauge

Let us return to the equations of motion:

$\displaystyle \nabla^2\phi + \frac{\partial (\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$})}{\partial t}$ $\textstyle =$ $\displaystyle -
\frac{\rho}{\epsilon_0 }$ (8.55)
$\displaystyle \nabla^2\mbox{\boldmath$A$}+ - \frac{1}{c^2}\frac{\partial^2 \mbox{\boldmath$A$}}{\partial t^2}$ $\textstyle =$ $\displaystyle - \mu_0 \mbox{\boldmath$J$}
+ \mbox{\boldmath$\nabla$}\left (\mb...
...\cdot \mbox{\boldmath$A$}+ \frac{1}{c^2}\frac{\partial \phi}{\partial t}\right)$ (8.56)

There is another way to make at least one of these two equations simplify. We can just insist that:

\begin{displaymath}
\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$}= 0.
\end{displaymath} (8.57)

It isn't so obvious that we can always choose a gauge such that this is true. Since we know we can start with the Lorentz gauge, though, let's look for $\Lambda$ such that it is. That is, suppose we've found $\phi,\mbox{\boldmath$A$}$ such that:
\begin{displaymath}
\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$}+ \frac{1}{c^2}\frac{\partial \phi}{\partial t} = 0
\end{displaymath} (8.58)

As before, we propose:

$\displaystyle \phi'$ $\textstyle =$ $\displaystyle \phi - \frac{\partial \Lambda}{\partial t}$ (8.59)
$\displaystyle \mbox{\boldmath$A$}'$ $\textstyle =$ $\displaystyle \mbox{\boldmath$A$}+ \mbox{\boldmath$\nabla$}\Lambda$ (8.60)

such that
\begin{displaymath}
\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$}' = \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$}+ \nabla^2\Lambda = 0.
\end{displaymath} (8.61)

If we substitute in the Lorentz gauge condition:
\begin{displaymath}
\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$}= -\frac{1}{c^2}\frac{\partial \phi}{\partial t}
\end{displaymath} (8.62)

we get:
\begin{displaymath}
\nabla^2\Lambda = - \mbox{\boldmath$\nabla$}\cdot \mbox{\bol...
...}\frac{\partial \phi'}{\partial t} = g(\mbox{\boldmath$x$},t)
\end{displaymath} (8.63)

As before, provided that a solution to the equations of motion in the Lorentz gauge exists, we can in principle solve this equation for a $\Lambda$ that makes $\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$A$}= 0$ true. It is therefore a legitimate gauge condition.

If we use the Coulomb gauge condition (which we are now justified in doing, as we know that the resulting potentials will lead to the same physical field) the potentials in the Coulomb gauge must satisfy the equations of motion:

$\displaystyle \nabla^2\phi$ $\textstyle =$ $\displaystyle -
\frac{\rho}{\epsilon_0 }$ (8.64)
$\displaystyle \nabla^2\mbox{\boldmath$A$}+ - \frac{1}{c^2}\frac{\partial^2 \mbox{\boldmath$A$}}{\partial t^2}$ $\textstyle =$ $\displaystyle - \mu_0 \mbox{\boldmath$J$}
+ \frac{1}{c^2}\mbox{\boldmath$\nabla$}\frac{\partial \phi}{\partial t}$ (8.65)

The potential $\phi$ is therefore the well-known solution
\begin{displaymath}
\phi(\mbox{\boldmath$x$}) = \frac{1}{4\pi\epsilon_0 } \int \...
...{\vert\mbox{\boldmath$x$}- \mbox{\boldmath$x$}_0\vert} d^3x_0
\end{displaymath} (8.66)

that you probably originally saw in elementary introductory physics and solved extensively last semester using the Green's function for the Poisson equation:
\begin{displaymath}
G(\mbox{\boldmath$x$},\mbox{\boldmath$x$}_0) = - \frac{1}{4\pi\vert\mbox{\boldmath$x$}- \mbox{\boldmath$x$}_0\vert}
\end{displaymath} (8.67)

that solves the ``point source'' differential equation:
\begin{displaymath}
\nabla^2G(\mbox{\boldmath$x$},\mbox{\boldmath$x$}_0) = \delta(\mbox{\boldmath$x$}- \mbox{\boldmath$x$}_0)
\end{displaymath} (8.68)

In this equation one uses the value of the charge density on all space as a function of time under the integral, and then adds a source term to the current density in the inhomogeneous wave equations for the vector potential derived from that density as well.

There are several very, very odd things about this solution. One is that the Coulomb potential is instantaneous - changes in the charge distribution instantly appear in its electric potential throughout all space. This appears to violate causality, and is definitely not what is physically observed. Is this a problem?

The answer is, no. If one works very long and tediously (as you will, for your homework) one can show that the current density can be decomposed into two pieces - a longitudinal (non-rotational) one and a transverse (rotational) one:

\begin{displaymath}
\mbox{\boldmath$J$}= \mbox{\boldmath$J$}_\ell + \mbox{\boldmath$J$}_t
\end{displaymath} (8.69)

These terms are defined by:
$\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$J$}_\ell$ $\textstyle =$ $\displaystyle 0$ (8.70)
$\displaystyle \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$J$}_t$ $\textstyle =$ $\displaystyle 0$ (8.71)

Evaluating these pieces is fairly straightforward. Start with:

\begin{displaymath}
\mbox{\boldmath$\nabla$}\times (\mbox{\boldmath$\nabla$}\ti...
...bla$}\cdot \mbox{\boldmath$J$}) - \nabla^2\mbox{\boldmath$J$}
\end{displaymath} (8.72)

This equation obviously splits into the two pieces - using the continuity equation to eliminate the divergence of $\mbox{\boldmath$J$}$ in favor of $\rho$, we get:
$\displaystyle \nabla^2\mbox{\boldmath$J$}_t$ $\textstyle =$ $\displaystyle - \mbox{\boldmath$\nabla$}\times (\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$J$})$ (8.73)
$\displaystyle \nabla^2\mbox{\boldmath$J$}_\ell$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\nabla$}(\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$J$}) = -\mbox{\boldmath$\nabla$}\frac{\partial \rho}{\partial t}$ (8.74)

(which are both Poisson equations).

With a bit of work - some integration by parts to move the $\mbox{\boldmath$\nabla$}$'s out of the integrals which imposes the constraint that $\mbox{\boldmath$J$}$ and $\rho$ have compact support so one can ignore the surface term - the decomposed currents are:

$\displaystyle \mbox{\boldmath$J$}_t$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\nabla$}\times (\mbox{\boldmath$\nabla$}\times \i...
...\boldmath$J$}d^3x_0}{4\pi\vert\mbox{\boldmath$x$}- \mbox{\boldmath$x$}_0\vert})$ (8.75)
$\displaystyle \mbox{\boldmath$J$}_\ell$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\nabla$}\frac{\partial }{\partial t}\left(\int \f...
..._0\right) = \epsilon_0 \mbox{\boldmath$\nabla$}\frac{\partial \phi}{\partial t}$ (8.76)

Substituting and comparing we note:
\begin{displaymath}
\frac{1}{c^2} \mbox{\boldmath$\nabla$}\frac{\partial \phi}{\partial t} = \mu_0 \mbox{\boldmath$J$}_\ell
\end{displaymath} (8.77)

so that this term cancels and the equation of motion for $\mbox{\boldmath$A$}$ becomes:
\begin{displaymath}
\nabla^2\mbox{\boldmath$A$}- \frac{1}{c^2}\frac{\partial^2 ...
...{\boldmath$A$}}{\partial t^2} = - \mu_0 \mbox{\boldmath$J$}_t
\end{displaymath} (8.78)

only.

In the Coulomb gauge, then, only the transverse current gives rise to the vector potential, which behaves like a wave. Hence the other common name for the gauge, the transverse gauge. It is also sometimes called the ``radiation gauge'' as only transverse currents give rise to purely transverse radiation fields far from the sources, with the static potential present but not giving rise to radiation.

Given all the ugliness above, why use the Coulomb gauge at all? There are a couple of reasons. First of all the actual equations of motion that must be solved are simple enough once one decomposes the current. Second of all, when computing the fields in free space where there are no sources, $\phi = 0$ and we can find both $\mbox{\boldmath$E$}$ and $\mbox{\boldmath$B$}$ from $\mbox{\boldmath$A$}$ alone:

$\displaystyle \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle - \frac{\partial \mbox{\boldmath$A$}}{\partial t}$ (8.79)
$\displaystyle \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$A$}$ (8.80)

The last oddity about this gauge is that it can be shown - if one works very hard - that it preserves causality. The transverse current above is not localized within the support of $\mbox{\boldmath$J$}$ but extends throughout all space just as instantaneously as $\phi$ does. One part of the field evaluated from the solution to the differential equations for $\mbox{\boldmath$A$}$, then, must cancel the instantaneous Coulomb potential and leave one with only the usual propagating electomagnetic field. This is left as a homework problem.


next up previous contents
Next: Poynting's Theorem, Work and Up: Potentials Previous: The Lorentz Gauge   Contents
Robert G. Brown 2014-08-19