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Jackson [1] has an interesting presentation of the transverse gauge. I’d like to walk through the details of this, but first want to translate the preliminaries to SI units (if I had the 3rd edition I’d not have to do this translation step).

### Gauge freedom

The starting point is noting that \( \spacegrad \cdot \BB = 0 \) the magnetic field can be expressed as a curl

\begin{equation}\label{eqn:transverseGauge:20}

\BB = \spacegrad \cross \BA.

\end{equation}

Faraday’s law now takes the form

\begin{equation}\label{eqn:transverseGauge:40}

\begin{aligned}

0

&= \spacegrad \cross \BE + \PD{t}{\BB} \\

&= \spacegrad \cross \BE + \PD{t}{} \lr{ \spacegrad \cross \BA } \\

&= \spacegrad \cross \lr{ \BE + \PD{t}{\BA} }.

\end{aligned}

\end{equation}

Because this curl is zero, the interior sum can be expressed as a gradient

\begin{equation}\label{eqn:transverseGauge:60}

\BE + \PD{t}{\BA} \equiv -\spacegrad \Phi.

\end{equation}

This can now be substituted into the remaining two Maxwell’s equations.

\begin{equation}\label{eqn:transverseGauge:80}

\begin{aligned}

\spacegrad \cdot \BD &= \rho_v \\

\spacegrad \cross \BH &= \BJ + \PD{t}{\BD} \\

\end{aligned}

\end{equation}

For Gauss’s law, in simple media, we have

\begin{equation}\label{eqn:transverseGauge:140}

\begin{aligned}

\rho_v

&=

\epsilon \spacegrad \cdot \BE \\

&=

\epsilon \spacegrad \cdot \lr{ -\spacegrad \Phi – \PD{t}{\BA} }

\end{aligned}

\end{equation}

For simple media again, the Ampere-Maxwell equation is

\begin{equation}\label{eqn:transverseGauge:100}

\inv{\mu} \spacegrad \cross \lr{ \spacegrad \cross \BA } = \BJ + \epsilon \PD{t}{} \lr{ -\spacegrad \Phi – \PD{t}{\BA} }.

\end{equation}

Expanding \( \spacegrad \cross \lr{ \spacegrad \cross \BA } = -\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } \) gives

\begin{equation}\label{eqn:transverseGauge:120}

-\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA } + \epsilon \mu \PDSq{t}{\BA} = \mu \BJ – \epsilon \mu \spacegrad \PD{t}{\Phi}.

\end{equation}

Maxwell’s equations are now reduced to

\begin{equation}\label{eqn:transverseGauge:180}

\boxed{

\begin{aligned}

\spacegrad^2 \BA – \spacegrad \lr{ \spacegrad \cdot \BA + \epsilon \mu \PD{t}{\Phi}} – \epsilon \mu \PDSq{t}{\BA} &= -\mu \BJ \\

\spacegrad^2 \Phi + \PD{t}{\spacegrad \cdot \BA} &= -\frac{\rho_v }{\epsilon}.

\end{aligned}

}

\end{equation}

There are two obvious constraints that we can impose

\begin{equation}\label{eqn:transverseGauge:200}

\spacegrad \cdot \BA – \epsilon \mu \PD{t}{\Phi} = 0,

\end{equation}

or

\begin{equation}\label{eqn:transverseGauge:220}

\spacegrad \cdot \BA = 0.

\end{equation}

The first constraint is the Lorentz gauge, which I’ve played with previously. It happens to be really nice in a relativistic context since, in vacuum with a four-vector potential \( A = (\Phi/c, \BA) \), that is a requirement that the four-divergence of the four-potential vanishes (\( \partial_\mu A^\mu = 0 \)).

### Transverse gauge

Jackson identifies the latter constraint as the transverse gauge, which I’m less familiar with. With this gauge selection, we have

\begin{equation}\label{eqn:transverseGauge:260}

\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA} = -\mu \BJ + \epsilon\mu \spacegrad \PD{t}{\Phi}

\end{equation}

\begin{equation}\label{eqn:transverseGauge:280}

\spacegrad^2 \Phi = -\frac{\rho_v }{\epsilon}.

\end{equation}

What’s not obvious is the fact that the irrotational (zero curl) contribution due to \(\Phi\) in \ref{eqn:transverseGauge:260} cancels the corresponding irrotational term from the current. Jackson uses a transverse and longitudinal decomposition of the current, related to the Helmholtz theorem to allude to this.

That decomposition follows from expanding \( \spacegrad^2 J/R \) in two ways using the delta function \( -4 \pi \delta(\Bx – \Bx’) = \spacegrad^2 1/R \) representation, as well as directly

\begin{equation}\label{eqn:transverseGauge:300}

\begin{aligned}

– 4 \pi \BJ(\Bx)

&=

\int \spacegrad^2 \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\

&=

\spacegrad

\int \spacegrad \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

+

\spacegrad \cdot

\int \spacegrad \wedge \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\

&=

-\spacegrad

\int \BJ(\Bx’) \cdot \spacegrad’ \inv{\Abs{\Bx – \Bx’}} d^3 x’

+

\spacegrad \cdot \lr{ \spacegrad \wedge

\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

} \\

&=

-\spacegrad

\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

+\spacegrad

\int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

–

\spacegrad \cross \lr{

\spacegrad \cross

\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

}

\end{aligned}

\end{equation}

The first term can be converted to a surface integral

\begin{equation}\label{eqn:transverseGauge:320}

-\spacegrad

\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

=

-\spacegrad

\int d\BA’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}},

\end{equation}

so provided the currents are either localized or \( \Abs{\BJ}/R \rightarrow 0 \) on an infinite sphere, we can make the identification

\begin{equation}\label{eqn:transverseGauge:340}

\BJ(\Bx)

=

-\spacegrad \inv{4 \pi} \int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

+

\spacegrad \cross \spacegrad \cross \inv{4 \pi} \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

\equiv

\BJ_l +

\BJ_t,

\end{equation}

where \( \spacegrad \cross \BJ_l = 0 \) (irrotational, or longitudinal), whereas \( \spacegrad \cdot \BJ_t = 0 \) (solenoidal or transverse). The irrotational property is clear from inspection, and the transverse property can be verified readily

\begin{equation}\label{eqn:transverseGauge:360}

\begin{aligned}

\spacegrad \cdot \lr{ \spacegrad \cross \lr{ \spacegrad \cross \BX } }

&=

-\spacegrad \cdot \lr{ \spacegrad \cdot \lr{ \spacegrad \wedge \BX } } \\

&=

-\spacegrad \cdot \lr{ \spacegrad^2 \BX – \spacegrad \lr{ \spacegrad \cdot \BX } } \\

&=

-\spacegrad \cdot \lr{\spacegrad^2 \BX} + \spacegrad^2 \lr{ \spacegrad \cdot \BX } \\

&= 0.

\end{aligned}

\end{equation}

Since

\begin{equation}\label{eqn:transverseGauge:380}

\Phi(\Bx, t)

=

\inv{4 \pi \epsilon} \int \frac{\rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’,

\end{equation}

we have

\begin{equation}\label{eqn:transverseGauge:400}

\begin{aligned}

\spacegrad \PD{t}{\Phi}

&=

\inv{4 \pi \epsilon} \spacegrad \int \frac{\partial_t \rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’ \\

&=

\inv{4 \pi \epsilon} \spacegrad \int \frac{-\spacegrad’ \cdot \BJ}{\Abs{\Bx – \Bx’}} d^3 x’ \\

&=

\frac{\BJ_l}{\epsilon}.

\end{aligned}

\end{equation}

This means that the Ampere-Maxwell equation takes the form

\begin{equation}\label{eqn:transverseGauge:420}

\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA}

= -\mu \BJ + \mu \BJ_l

= -\mu \BJ_t.

\end{equation}

This justifies the transverse in the label transverse gauge.

# References

[1] JD Jackson. *Classical Electrodynamics*. John Wiley and Sons, 2nd edition, 1975.