## Motivation.

This revisits my last blog post where I covered this content in a meandering fashion. This is an attempt to re-express this in a more compact form. In particular, in a form that is amenable to include in my book. When I wrote the potential section of my book, I cheated, and didn’t try to motivate the results. My cheat was figuring out the multivector potential representation starting with STA where things are simpler, and then translating it back to a multivector representation, instead of figuring out a reasonable way to motivate things from the foundation already laid.

I’d like to eventually have a less rushed treatment of potentials in my book, where the results are not pulled out of a magic hat. Here is an attempted step in that direction. I’ve opted to put some of the motivational material in problems (with solutions at the chapter end.)

## Multivector potentials.

We know from conventional electromagnetism (given no fictitious magnetic sources) that we can represent the six components of the electric and magnetic fields in terms of four scalar fields
\label{eqn:mvpotentials:80}
\begin{aligned}
\BE &= -\spacegrad \phi – \PD{t}{\BA} \\
\BH &= \inv{\mu} \spacegrad \cross \BA.
\end{aligned}

The conventional way of constructing these potentials makes use of the identities
\label{eqn:mvpotentials:60}
\begin{aligned}
\end{aligned}

applying those to the source free Maxwell’s equations to find representations of $$\BE, \BH$$ that automatically satisfy those equations. For that conventional analysis, see section 18-6 [2] (available online), or section 10.1 [3], or section 6.4 [4]. We can also find such a potential representation using geometric algebra methods that are cross product free (problem 1.)

For Maxwell’s equations with fictitious magnetic sources, it can be shown that a potential representation of the field
\label{eqn:mvpotentials:100}
\begin{aligned}
\BH &= -\spacegrad \phi_m – \PD{t}{\BF} \\
\BE &= -\inv{\epsilon} \spacegrad \cross \BF.
\end{aligned}

satisfies the source-free grades of Maxwell’s equation.
See [1], and [5] for such derivations. As with the conventional source potentials, we can also apply our geometric algebra toolbox to easily find these results (problem 2.)

We have a mix of time partials and curls that is reminiscent of Maxwell’s equation itself. It’s obvious to wonder whether there is a more coherent integrated form for the potential. This is in fact the case.

## Lemma 1.1: Multivector potentials.

For Maxwell’s equation with electric sources, the total field $$F$$ can be expressed in multivector potential form
\label{eqn:mvpotentials:520}
F = \gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } \lr{ -\phi + c \BA } }{1,2}.

For Maxwell’s equation with only fictitious magnetic sources, the total field $$F$$ can be expressed in multivector form
\label{eqn:mvpotentials:540}
F = \gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } I \eta \lr{ -\phi_m + c \BF } }{1,2}.

The reader should try to verify this themselves (problem 3.)

Using superposition, we can form a multivector potential that includes all grades.

## Definition 1.1: Multivector potential.

We call $$A$$, a multivector with all grades, the multivector potential, defining the total field as
\label{eqn:mvpotentials:600}
\begin{aligned}
F
&=
&=
\lr{ \spacegrad – \inv{c} \PD{t}{} } A

\end{aligned}

Imposition of the constraint
\label{eqn:mvpotentials:680}

is called the Lorentz gauge condition, and allows us to express $$F$$ in terms of the potential without any grade selection filters.

## Lemma 1.2: Conventional multivector potential.

Let
\label{eqn:mvpotentials:620}
A = -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF }.

This results in the conventional potential representation of the electric and magnetic fields
\label{eqn:mvpotentials:640}
\begin{aligned}
\BE &= -\spacegrad \phi – \PD{t}{\BA} – \inv{\epsilon} \spacegrad \cross \BF \\
\end{aligned}

In terms of potentials, the Lorentz gauge condition \ref{eqn:mvpotentials:680} takes the form
\label{eqn:mvpotentials:660}
\begin{aligned}
0 &= \inv{c} \PD{t}{\phi} + \spacegrad \cdot (c \BA) \\
0 &= \inv{c} \PD{t}{\phi_m} + \spacegrad \cdot (c \BF).
\end{aligned}

See problem 4.

## Problem 1: Potentials for no-fictitious sources.

Starting with Maxwell’s equation with only conventional electric sources
\label{eqn:mvpotentials:120}

Show that this may be split by grade into three equations
\label{eqn:mvpotentials:140}
\begin{aligned}
\spacegrad \wedge \BE + \inv{c}\PD{t}{} \lr{ I \eta \BH } &= 0 \\
\spacegrad \wedge \lr{ I \eta \BH } &= 0.
\end{aligned}

Then use the identities $$\spacegrad \wedge \spacegrad \wedge \BA = 0$$, for vector $$\BA$$ and $$\spacegrad \wedge \spacegrad \phi = 0$$, for scalar $$\phi$$ to find the potential representation.

Taking grade(0,1) and (2,3) selections of Maxwell’s equation, we split our equations into source dependent and source free equations
\label{eqn:mvpotentials:200}

\label{eqn:mvpotentials:220}

In terms of $$F = \BE + I \eta \BH$$, the source free equation expands to
\label{eqn:mvpotentials:240}
\begin{aligned}
0
&=
\lr{ \spacegrad + \inv{c} \PD{t}{} } \lr{ \BE + I \eta \BH }
}{2,3} \\
&=
&=
+ \spacegrad \wedge \lr{ I \eta \BH }
+ I \eta \inv{c} \PD{t}{\BH},
\end{aligned}

which can be further split into a bivector and trivector equation
\label{eqn:mvpotentials:260}
0 = \spacegrad \wedge \BE + I \eta \inv{c} \PD{t}{\BH}

\label{eqn:mvpotentials:280}
0 = \spacegrad \wedge \lr{ I \eta \BH }.

It’s clear that we want to write the magnetic field as a (bivector) curl, so we let
\label{eqn:mvpotentials:300}
I \eta \BH = I c \BB = c \spacegrad \wedge \BA,

or
\label{eqn:mvpotentials:301}
\BH = \inv{\mu} \spacegrad \cross \BA.

\Cref{eqn:mvpotentials:260} is reduced to
\label{eqn:mvpotentials:320}
\begin{aligned}
0
&= \spacegrad \wedge \BE + I \eta \inv{c} \PD{t}{\BH} \\
&= \spacegrad \wedge \BE + \inv{c} \PD{t}{} \spacegrad \wedge \lr{ c \BA } \\
&= \spacegrad \wedge \lr{ \BE + \PD{t}{\BA} }.
\end{aligned}

We can now let
\label{eqn:mvpotentials:340}
\BE + \PD{t}{\BA} = -\spacegrad \phi.

We sneakily adjust the sign of the gradient so that the result matches the conventional representation.

## Problem 2: Potentials for fictitious sources.

Starting with Maxwell’s equation with only fictitious magnetic sources
\label{eqn:mvpotentials:160}

show that this may be split by grade into three equations
\label{eqn:mvpotentials:180}
\begin{aligned}
-\eta \spacegrad \wedge \BH + \inv{c}\PD{t}{(I \BE)} &= 0 \\
\spacegrad \wedge \lr{ I \BE } &= 0.
\end{aligned}

Then use the identities $$\spacegrad \wedge \spacegrad \wedge \BF = 0$$, for vector $$\BF$$ and $$\spacegrad \wedge \spacegrad \phi_m = 0$$, for scalar $$\phi_m$$ to find the potential representation \ref{eqn:mvpotentials:100}.

We multiply \ref{eqn:mvpotentials:160} by $$I$$ to find
\label{eqn:mvpotentials:360}

which can be split into
\label{eqn:mvpotentials:380}
\begin{aligned}
\end{aligned}

We expand the source free equation in terms of $$I F = I \BE – \eta \BH$$, to find
\label{eqn:mvpotentials:400}
\begin{aligned}
0
&= \gpgrade{ \lr{ \spacegrad + \inv{c}\PD{t}{} } \lr{ I \BE – \eta \BH } }{0,3} \\
&= \spacegrad \wedge \lr{ I \BE } + \inv{c} \PD{t}{(I \BE)} – \eta \spacegrad \wedge \BH,
\end{aligned}

which has the respective bivector and trivector grades
\label{eqn:mvpotentials:420}
0 = \spacegrad \wedge \lr{ I \BE }

\label{eqn:mvpotentials:440}
0 = \inv{c} \PD{t}{(I \BE)} – \eta \spacegrad \wedge \BH.

We can clearly satisfy \ref{eqn:mvpotentials:420} by setting
\label{eqn:mvpotentials:460}
I \BE = -\inv{\epsilon} \spacegrad \wedge \BF,

or
\label{eqn:mvpotentials:461}
\BE = -\inv{\epsilon} \spacegrad \cross \BF.

Here, once again, the sneaky inclusion of a constant factor $$-1/\epsilon$$ is to make the result match the conventional. Inserting this value for $$I \BE$$ into our bivector equation yields
\label{eqn:mvpotentials:480}
\begin{aligned}
0
&= -\inv{\epsilon} \inv{c} \PD{t}{} (\spacegrad \wedge \BF) – \eta \spacegrad \wedge \BH \\
&= -\eta \spacegrad \wedge \lr{ \PD{t}{\BF} + \BH },
\end{aligned}

so we set
\label{eqn:mvpotentials:500}
\PD{t}{\BF} + \BH = -\spacegrad \phi_m,

and have a field representation that automatically satisfies the source free equations.

## Problem 3: Total field in terms of potentials.

Prove lemma 1.1, either by direct expansion, or by trying to discover the multivector form of the field by construction.

Proof by expansion is straightforward, and left to the reader. We form the respective total electromagnetic fields $$F = \BE + I \eta H$$ for each case.

We find
\label{eqn:mvpotentials:560}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= -\spacegrad \phi – \PD{t}{\BA} + I \frac{\eta}{\mu} \spacegrad \cross \BA \\
&= -\spacegrad \phi – \inv{c} \PD{t}{(c \BA)} + \spacegrad \wedge (c\BA) \\
&= \gpgrade{ \spacegrad \lr{ -\phi + c \BA } – \inv{c} \PD{t}{(c \BA)} }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad -\inv{c} \PD{t}{} } \lr{ -\phi + c \BA } }{1,2}.
\end{aligned}

For the field for the fictitious source case, we compute the result in the same way, inserting a no-op grade selection to allow us to simplify, finding
\label{eqn:mvpotentials:580}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= -\inv{\epsilon} \spacegrad \cross \BF + I \eta \lr{ -\spacegrad \phi_m – \PD{t}{\BF} } \\
&= \inv{\epsilon c} I \lr{ \spacegrad \wedge (c \BF)} + I \eta \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } \\
&= I \eta \lr{ \spacegrad \wedge (c \BF) + \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } } \\
&= I \eta \gpgrade{ \spacegrad \wedge (c \BF) + \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } }{1,2} \\
&= I \eta \gpgrade{ \spacegrad (-\phi_m + c \BF) – \inv{c} \PD{t}{(c \BF)} }{1,2} \\
&= I \eta \gpgrade{ \lr{ \spacegrad -\inv{c} \PD{t}{} } (-\phi_m + c \BF) }{1,2}.
\end{aligned}

## Problem 4: Fields in terms of potentials.

Prove lemma 1.2.

Let’s expand and then group by grade
\label{eqn:mvpotentials:n}
\begin{aligned}
\lr{ \spacegrad – \inv{c} \PD{t}{} } A
&=
\lr{ \spacegrad – \inv{c} \PD{t}{} } \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF }} \\
&=
-\inv{c} \PD{t}{\phi} + c \inv{c} \PD{t}{ \BA } + I \eta \lr{ -\inv{c} \PD{t}{\phi_m} + c \inv{c} \PD{t}{\BF} } \\
&=
+ I \eta c \spacegrad \wedge \BF
– c \inv{c} \PD{t}{\BA}
– c I \eta \inv{c} \PD{t}{\BF} \\
+\inv{c} \PD{t}{\phi}
+ \inv{c} \PD{t}{\phi_m} } \\
&=
– \PD{t}{\BA}
– \PD{t}{\BF}
} \\
+\inv{c} \PD{t}{\phi}
+ \inv{c} \PD{t}{\phi_m} }.
\end{aligned}

Observing that $$F = \gpgrade{ \lr{ \spacegrad -(1/c) \partial_t } A }{1,2} = \BE + I \eta \BH$$, completes the problem. If the Lorentz gauge condition is assumed, the scalar and pseudoscalar components above are obliterated, leaving just
$$F = \lr{ \spacegrad -(1/c) \partial_t } A$$.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.

[2] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics, Volume II.[Lectures on physics], chapter The Maxwell Equations. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963. URL https://www.feynmanlectures.caltech.edu/II_18.html.

[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

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

[5] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.

## Potential solutions to the static Maxwell’s equation using geometric algebra

When neither the electromagnetic field strength $$F = \BE + I \eta \BH$$, nor current $$J = \eta (c \rho – \BJ) + I(c\rho_m – \BM)$$ is a function of time, then the geometric algebra form of Maxwell’s equations is the first order multivector (gradient) equation
\label{eqn:staticPotentials:20}

While direct solutions to this equations are possible with the multivector Green’s function for the gradient
\label{eqn:staticPotentials:40}
G(\Bx, \Bx’) = \inv{4\pi} \frac{\Bx – \Bx’}{\Norm{\Bx – \Bx’}^3 },

the aim in this post is to explore second order (potential) solutions in a geometric algebra context. Can we assume that it is possible to find a multivector potential $$A$$ for which
\label{eqn:staticPotentials:60}

is a solution to the Maxwell statics equation? If such a solution exists, then Maxwell’s equation is simply
\label{eqn:staticPotentials:80}

which can be easily solved using the scalar Green’s function for the Laplacian
\label{eqn:staticPotentials:240}
G(\Bx, \Bx’) = -\inv{\Norm{\Bx – \Bx’} },

a beastie that may be easier to convolve than the vector valued Green’s function for the gradient.

It is immediately clear that some restrictions must be imposed on the multivector potential $$A$$. In particular, since the field $$F$$ has only vector and bivector grades, this gradient must have no scalar, nor pseudoscalar grades. That is
\label{eqn:staticPotentials:100}

This constraint on the potential can be avoided if a grade selection operation is built directly into the assumed potential solution, requiring that the field is given by
\label{eqn:staticPotentials:120}

However, after imposing such a constraint, Maxwell’s equation has a much less friendly form
\label{eqn:staticPotentials:140}

Luckily, it is possible to introduce a transformation of potentials, called a gauge transformation, that eliminates the ugly grade selection term, and allows the potential equation to be expressed as a plain old Laplacian. We do so by assuming first that it is possible to find a solution of the Laplacian equation that has the desired grade restrictions. That is
\label{eqn:staticPotentials:160}
\begin{aligned}
\end{aligned}

for which $$F = \spacegrad A’$$ is a grade 1,2 solution to $$\spacegrad F = J$$. Suppose that $$A$$ is any formal solution, free of any grade restrictions, to $$\spacegrad^2 A = J$$, and $$F = \gpgrade{\spacegrad A}{1,2}$$. Can we find a function $$\tilde{A}$$ for which $$A = A’ + \tilde{A}$$?

Maxwell’s equation in terms of $$A$$ is
\label{eqn:staticPotentials:180}
\begin{aligned}
J
\end{aligned}

or
\label{eqn:staticPotentials:200}

This non-homogeneous Laplacian equation that can be solved as is for $$\tilde{A}$$ using the Green’s function for the Laplacian. Alternatively, we may also solve the equivalent first order system using the Green’s function for the gradient.
\label{eqn:staticPotentials:220}

Clearly $$\tilde{A}$$ is not unique, as we can add any function $$\psi$$ satisfying the homogeneous Laplacian equation $$\spacegrad^2 \psi = 0$$.

In summary, if $$A$$ is any multivector solution to $$\spacegrad^2 A = J$$, that is
\label{eqn:staticPotentials:260}
A(\Bx)
= \int dV’ G(\Bx, \Bx’) J(\Bx’)
= -\int dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} },

then $$F = \spacegrad A’$$ is a solution to Maxwell’s equation, where $$A’ = A – \tilde{A}$$, and $$\tilde{A}$$ is a solution to the non-homogeneous Laplacian equation or the non-homogeneous gradient equation above.

### Integral form of the gauge transformation.

Additional insight is possible by considering the gauge transformation in integral form. Suppose that
\label{eqn:staticPotentials:280}
A(\Bx) = -\int_V dV’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \tilde{A}(\Bx),

is a solution of $$\spacegrad^2 A = J$$, where $$\tilde{A}$$ is a multivector solution to the homogeneous Laplacian equation $$\spacegrad^2 \tilde{A} = 0$$. Let’s look at the constraints on $$\tilde{A}$$ that must be imposed for $$F = \spacegrad A$$ to be a valid (i.e. grade 1,2) solution of Maxwell’s equation.
\label{eqn:staticPotentials:300}
\begin{aligned}
F
&=
-\int_V dV’ \lr{ \spacegrad \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
&=
\int_V dV’ \lr{ \spacegrad’ \inv{\Norm{\Bx – \Bx’} } } J(\Bx’)
&=
\int_V dV’ \spacegrad’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V dV’ \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
&=
\int_{\partial V} dA’ \ncap’ \frac{J(\Bx’)}{\Norm{\Bx – \Bx’} } – \int_V \frac{\spacegrad’ J(\Bx’)}{\Norm{\Bx – \Bx’} }
\end{aligned}

Where $$\ncap’ = (\Bx’ – \Bx)/\Norm{\Bx’ – \Bx}$$, and the fundamental theorem of geometric calculus has been used to transform the gradient volume integral into an integral over the bounding surface. Operating on Maxwell’s equation with the gradient gives $$\spacegrad^2 F = \spacegrad J$$, which has only grades 1,2 on the left hand side, meaning that $$J$$ is constrained in a way that requires $$\spacegrad J$$ to have only grades 1,2. This means that $$F$$ has grades 1,2 if
\label{eqn:staticPotentials:320}
= \int_{\partial V} dA’ \frac{ \gpgrade{\ncap’ J(\Bx’)}{0,3} }{\Norm{\Bx – \Bx’} }.

The product $$\ncap J$$ expands to
\label{eqn:staticPotentials:340}
\begin{aligned}
\ncap J
&=
&=
\ncap \cdot (-\eta \BJ) + \gpgradethree{\ncap (-I \BM)} \\
&=- \eta \ncap \cdot \BJ -I \ncap \cdot \BM,
\end{aligned}

so
\label{eqn:staticPotentials:360}
=
-\int_{\partial V} dA’ \frac{ \eta \ncap’ \cdot \BJ(\Bx’) + I \ncap’ \cdot \BM(\Bx’)}{\Norm{\Bx – \Bx’} }.

Observe that if there is no flux of current density $$\BJ$$ and (fictitious) magnetic current density $$\BM$$ through the surface, then $$F = \spacegrad A$$ is a solution to Maxwell’s equation without any gauge transformation. Alternatively $$F = \spacegrad A$$ is also a solution if $$\lim_{\Bx’ \rightarrow \infty} \BJ(\Bx’)/\Norm{\Bx – \Bx’} = \lim_{\Bx’ \rightarrow \infty} \BM(\Bx’)/\Norm{\Bx – \Bx’} = 0$$ and the bounding volume is taken to infinity.

# References

## Motivation

Geometric algebra (GA) allows for a compact description of Maxwell’s equations in either an explicit 3D representation or a STA (SpaceTime Algebra [2]) representation. The 3D GA and STA representations Maxwell’s equation both the form

\label{eqn:potentialMethods:1280}
L \boldsymbol{\mathcal{F}} = J,

where $$J$$ represents the sources, $$L$$ is a multivector gradient operator that includes partial derivative operator components for each of the space and time coordinates, and

\label{eqn:potentialMethods:1020}
\boldsymbol{\mathcal{F}} = \boldsymbol{\mathcal{E}} + \eta I \boldsymbol{\mathcal{H}},

is an electromagnetic field multivector, $$I = \Be_1 \Be_2 \Be_3$$ is the \R{3} pseudoscalar, and $$\eta = \sqrt{\mu/\epsilon}$$ is the impedance of the media.

When Maxwell’s equations are extended to include magnetic sources in addition to conventional electric sources (as used in antenna-theory [1] and microwave engineering [3]), they take the form

\label{eqn:chapter3Notes:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \boldsymbol{\mathcal{M}} – \PD{t}{\boldsymbol{\mathcal{B}}}

\label{eqn:chapter3Notes:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}

\label{eqn:chapter3Notes:60}

\label{eqn:chapter3Notes:80}

The corresponding GA Maxwell equations in their respective 3D and STA forms are

\label{eqn:potentialMethods:300}
\lr{ \spacegrad + \inv{v} \PD{t}{} } \boldsymbol{\mathcal{F}}
=
\eta
\lr{ v q_{\textrm{e}} – \boldsymbol{\mathcal{J}} }
+ I \lr{ v q_{\textrm{m}} – \boldsymbol{\mathcal{M}} }

\label{eqn:potentialMethods:320}
\grad \boldsymbol{\mathcal{F}} = \eta J – I M,

where the wave group velocity in the medium is $$v = 1/\sqrt{\epsilon\mu}$$, and the medium is isotropic with
$$\boldsymbol{\mathcal{B}} = \mu \boldsymbol{\mathcal{H}}$$, and $$\boldsymbol{\mathcal{D}} = \epsilon \boldsymbol{\mathcal{E}}$$. In the STA representation, $$\grad, J, M$$ are all four-vectors, the specific meanings of which will be spelled out below.

How to determine the potential equations and the field representation using the conventional distinct Maxwell’s \ref{eqn:chapter3Notes:20}, … is well known. The basic procedure is to consider the electric and magnetic sources in turn, and observe that in each case one of the electric or magnetic fields must have a curl representation. The STA approach is similar, except that it can be observed that the field must have a four-curl representation for each type of source. In the explicit 3D GA formalism
\ref{eqn:potentialMethods:300} how to formulate a natural potential representation is not as obvious. There is no longer an reason to set any component of the field equal to a curl, and the representation of the four curl from the STA approach is awkward. Additionally, it is not obvious what form gauge invariance takes in the 3D GA representation.

### Ideas explored in these notes

• GA representation of Maxwell’s equations including magnetic sources.
• STA GA formalism for Maxwell’s equations including magnetic sources.
• Explicit form of the GA potential representation including both electric and magnetic sources.
• Demonstration of exactly how the 3D and STA potentials are related.
• Explore the structure of gauge transformations when magnetic sources are included.
• Explore the structure of gauge transformations in the 3D GA formalism.
• Specify the form of the Lorentz gauge in the 3D GA formalism.

### No magnetic sources

When magnetic sources are omitted, it follows from \ref{eqn:chapter3Notes:80} that there is some $$\boldsymbol{\mathcal{A}}^{\mathrm{e}}$$ for which

\label{eqn:potentialMethods:20}
\boxed{
}

Substitution into Faraday’s law \ref{eqn:chapter3Notes:20} gives

\label{eqn:potentialMethods:40}

or
\label{eqn:potentialMethods:60}
\spacegrad \cross \lr{ \boldsymbol{\mathcal{E}} + \PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} } } = 0.

A gradient representation of this curled quantity, say $$-\spacegrad \phi$$, will provide the required zero

\label{eqn:potentialMethods:80}
\boxed{
\boldsymbol{\mathcal{E}} = -\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }.
}

The final two Maxwell equations yield

\label{eqn:potentialMethods:100}
\begin{aligned}
-\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \spacegrad \lr{ \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} } &= \mu \lr{ \boldsymbol{\mathcal{J}} + \epsilon \PD{t}{} \lr{ -\spacegrad \phi -\PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} } } } \\
\end{aligned}

or
\label{eqn:potentialMethods:120}
\boxed{
\begin{aligned}
\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{e}} – \inv{v^2} \PDSq{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }
\inv{v^2} \PD{t}{\phi}
}
&= -\mu \boldsymbol{\mathcal{J}} \\
\end{aligned}
}

Note that the Lorentz condition $$\PDi{t}{(\phi/v^2)} + \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} = 0$$ can be imposed to decouple these, leaving non-homogeneous wave equations for the vector and scalar potentials respectively.

### No electric sources

Without electric sources, a curl representation of the electric field can be assumed, satisfying Gauss’s law

\label{eqn:potentialMethods:140}
\boxed{
\boldsymbol{\mathcal{D}} = – \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}}.
}

Substitution into the Maxwell-Faraday law gives
\label{eqn:potentialMethods:160}
\spacegrad \cross \lr{ \boldsymbol{\mathcal{H}} + \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}} } = 0.

This is satisfied with any gradient, say, $$-\spacegrad \phi_m$$, providing a potential representation for the magnetic field

\label{eqn:potentialMethods:180}
\boxed{
\boldsymbol{\mathcal{H}} = -\spacegrad \phi_m – \PD{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}}.
}

The remaining Maxwell equations provide the required constraints on the potentials

\label{eqn:potentialMethods:220}
\lr{
-\boldsymbol{\mathcal{M}} – \mu \PD{t}{}
\lr{
}
}

\label{eqn:potentialMethods:240}
\lr{
}
= \inv{\mu} q_m,

or
\label{eqn:potentialMethods:260}
\boxed{
\begin{aligned}
\spacegrad^2 \boldsymbol{\mathcal{A}}^{\mathrm{m}} – \inv{v^2} \PDSq{t}{\boldsymbol{\mathcal{A}}^{\mathrm{m}}} – \spacegrad \lr{ \inv{v^2} \PD{t}{\phi_m} + \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}} } &= -\epsilon \boldsymbol{\mathcal{M}} \\
\end{aligned}
}

The general solution to Maxwell’s equations is therefore
\label{eqn:potentialMethods:280}
\begin{aligned}
\boldsymbol{\mathcal{E}} &=
– \inv{\epsilon} \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
\boldsymbol{\mathcal{H}} &=
\end{aligned}

subject to the constraints \ref{eqn:potentialMethods:120} and \ref{eqn:potentialMethods:260}.

### Potential operator structure

Knowing that there is a simple underlying structure to the potential representation of the electromagnetic field in the STA formalism inspires the question of whether that structure can be found directly using the scalar and vector potentials determined above.

Specifically, what is the multivector representation \ref{eqn:potentialMethods:1020} of the electromagnetic field in terms of all the individual potential variables, and can an underlying structure for that field representation be found? The composite field is

\label{eqn:potentialMethods:280b}
\boldsymbol{\mathcal{F}}
=
– \inv{\epsilon} \spacegrad \cross \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
+ I \eta
\lr{
}.

Can this be factored into into multivector operator and multivector potentials? Expanding the cross products provides some direction

\label{eqn:potentialMethods:1040}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
– \PD{t}{ \boldsymbol{\mathcal{A}}^{\mathrm{e}} }
– \eta \PD{t}{I \boldsymbol{\mathcal{A}}^{\mathrm{m}}}
– \spacegrad \lr{ \phi – \eta I \phi_m } \\
+ \frac{1}{2 \epsilon} \lr{ \rspacegrad I \boldsymbol{\mathcal{A}}^{\mathrm{m}} – I \boldsymbol{\mathcal{A}}^{\mathrm{m}} \lspacegrad }.
\end{aligned}

Observe that the
gradient and the time partials can be grouped together

\label{eqn:potentialMethods:1060}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
– \PD{t}{ } \lr{\boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \boldsymbol{\mathcal{A}}^{\mathrm{m}}}
– \spacegrad \lr{ \phi + \eta I \phi_m }
+ \frac{v}{2} \lr{ \rspacegrad (\boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \boldsymbol{\mathcal{A}}^{\mathrm{m}}) – (\boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \boldsymbol{\mathcal{A}}^{\mathrm{m}}) \lspacegrad } \\
&=
\inv{2} \lr{
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} } \lr{ v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}} }

\lr{ v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}} \lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
} \\
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} } \lr{ -\phi – \eta I \phi_m }
– \lr{ \phi + \eta I \phi_m } \lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
}
,
\end{aligned}

or

\label{eqn:potentialMethods:1080}
\boxed{
\boldsymbol{\mathcal{F}}
=
\inv{2} \Biglr{
\lr{ \rspacegrad – \inv{v} {\stackrel{ \rightarrow }{\partial_t}} }
\lr{
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
}

\lr{
\phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
+ \eta I \phi_m
}
\lr{ \lspacegrad + \inv{v} {\stackrel{ \leftarrow }{\partial_t}} }
}
.
}

There’s a conjugate structure to the potential on each side of the curl operation where we see a sign change for the scalar and pseudoscalar elements only. The reason for this becomes more clear in the STA formalism.

## Potentials in the STA formalism.

Maxwell’s equation in its explicit 3D form \ref{eqn:potentialMethods:300} can be
converted to STA form, by introducing a four-vector basis $$\setlr{ \gamma_\mu }$$, where the spatial basis
$$\setlr{ \Be_k = \gamma_k \gamma_0 }$$
is expressed in terms of the Dirac basis $$\setlr{ \gamma_\mu }$$.
By multiplying from the left with $$\gamma_0$$ a STA form of Maxwell’s equation
\ref{eqn:potentialMethods:320}
is obtained,
where
\label{eqn:potentialMethods:340}
\begin{aligned}
J &= \gamma^\mu J_\mu = ( v q_e, \boldsymbol{\mathcal{J}} ) \\
M &= \gamma^\mu M_\mu = ( v q_m, \boldsymbol{\mathcal{M}} ) \\
I &= \gamma_0 \gamma_1 \gamma_2 \gamma_3,
\end{aligned}

Here the metric choice is $$\gamma_0^2 = 1 = -\gamma_k^2$$. Note that in this representation the electromagnetic field $$\boldsymbol{\mathcal{F}} = \boldsymbol{\mathcal{E}} + \eta I \boldsymbol{\mathcal{H}}$$ is a bivector, not a multivector as it is explicit (frame dependent) 3D representation of \ref{eqn:potentialMethods:300}.

A potential representation can be obtained as before by considering electric and magnetic sources in sequence and using superposition to assemble a complete potential.

### No magnetic sources

Without magnetic sources, Maxwell’s equation splits into vector and trivector terms of the form

\label{eqn:potentialMethods:380}
\grad \cdot \boldsymbol{\mathcal{F}} = \eta J

\label{eqn:potentialMethods:400}

A four-vector curl representation of the field will satisfy \ref{eqn:potentialMethods:400} allowing an immediate potential solution

\label{eqn:potentialMethods:560}
\boxed{
\begin{aligned}
&\boldsymbol{\mathcal{F}} = \grad \wedge {A^{\mathrm{e}}} \\
\end{aligned}
}

This can be put into correspondence with \ref{eqn:potentialMethods:120} by noting that

\label{eqn:potentialMethods:460}
\begin{aligned}
\grad^2 &= (\gamma^\mu \partial_\mu) \cdot (\gamma^\nu \partial_\nu) = \inv{v^2} \partial_{tt} – \spacegrad^2 \\
\gamma_0 {A^{\mathrm{e}}} &= \gamma_0 \gamma^\mu {A^{\mathrm{e}}}_\mu = {A^{\mathrm{e}}}_0 + \Be_k {A^{\mathrm{e}}}_k = {A^{\mathrm{e}}}_0 + \BA^{\mathrm{e}} \\
\gamma_0 \grad &= \gamma_0 \gamma^\mu \partial_\mu = \inv{v} \partial_t + \spacegrad \\
\grad \cdot {A^{\mathrm{e}}} &= \partial_\mu {A^{\mathrm{e}}}^\mu = \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}},
\end{aligned}

so multiplying from the left with $$\gamma_0$$ gives

\label{eqn:potentialMethods:480}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \lr{ {A^{\mathrm{e}}}_0 + \BA^{\mathrm{e}} } – \lr{ \inv{v} \partial_t + \spacegrad }\lr{ \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}} } = \eta( v q_e – \boldsymbol{\mathcal{J}} ),

or

\label{eqn:potentialMethods:520}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \BA^{\mathrm{e}} – \spacegrad \lr{ \inv{v} \partial_t {A^{\mathrm{e}}}_0 – \spacegrad \cdot \BA^{\mathrm{e}} } = -\eta \boldsymbol{\mathcal{J}}

\label{eqn:potentialMethods:540}
\spacegrad^2 {A^{\mathrm{e}}}_0 – \inv{v} \partial_t \lr{ \spacegrad \cdot \BA^{\mathrm{e}} } = -q_e/\epsilon.

So $${A^{\mathrm{e}}}_0 = \phi$$ and $$-\ifrac{\BA^{\mathrm{e}}}{v} = \boldsymbol{\mathcal{A}}^{\mathrm{e}}$$, or

\label{eqn:potentialMethods:600}
\boxed{
{A^{\mathrm{e}}} = \gamma_0\lr{ \phi – v \boldsymbol{\mathcal{A}}^{\mathrm{e}} }.
}

### No electric sources

Without electric sources, Maxwell’s equation now splits into

\label{eqn:potentialMethods:640}

\label{eqn:potentialMethods:660}
\grad \wedge \boldsymbol{\mathcal{F}} = -I M.

Here the dual of an STA curl yields a solution

\label{eqn:potentialMethods:680}
\boxed{
\boldsymbol{\mathcal{F}} = I ( \grad \wedge {A^{\mathrm{m}}} ).
}

Substituting this gives

\label{eqn:potentialMethods:720}
\begin{aligned}
0
&=
&=
&=
\end{aligned}

\label{eqn:potentialMethods:740}
\begin{aligned}
-I M
&=
&=
&=
\end{aligned}

The $$\grad \cdot \boldsymbol{\mathcal{F}}$$ relation \ref{eqn:potentialMethods:720} is identically zero as desired, leaving

\label{eqn:potentialMethods:760}
\boxed{
=
M.
}

So the general solution with both electric and magnetic sources is

\label{eqn:potentialMethods:800}
\boxed{
}

subject to the constraints of \ref{eqn:potentialMethods:560} and \ref{eqn:potentialMethods:760}. As before the four-potential $${A^{\mathrm{m}}}$$ can be put into correspondence with the conventional scalar and vector potentials by left multiplying with $$\gamma_0$$, which gives

\label{eqn:potentialMethods:820}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \lr{ {A^{\mathrm{m}}}_0 + \BA^{\mathrm{m}} } – \lr{ \inv{v} \partial_t + \spacegrad }\lr{ \inv{v} \partial_t {A^{\mathrm{m}}}_0 – \spacegrad \cdot \BA^{\mathrm{m}} } = v q_m – \boldsymbol{\mathcal{M}},

or
\label{eqn:potentialMethods:860}
\lr{ \inv{v^2} \partial_{tt} – \spacegrad^2 } \BA^{\mathrm{m}} – \spacegrad \lr{ \inv{v} \partial_t {A^{\mathrm{m}}}_0 – \spacegrad \cdot \BA^{\mathrm{m}} } = – \boldsymbol{\mathcal{M}}

\label{eqn:potentialMethods:880}

Comparing with \ref{eqn:potentialMethods:260} shows that $${A^{\mathrm{m}}}_0/v = \mu \phi_m$$ and $$-\ifrac{\BA^{\mathrm{m}}}{v^2} = \mu \boldsymbol{\mathcal{A}}^{\mathrm{m}}$$, or

\label{eqn:potentialMethods:900}
\boxed{
{A^{\mathrm{m}}} = \gamma_0 \eta \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} }.
}

### Potential operator structure

Observe that there is an underlying uniform structure of the differential operator that acts on the potential to produce the electromagnetic field. Expressed as a linear operator of the
gradient and the potentials, that is

$$\boldsymbol{\mathcal{F}} = L(\lrgrad, {A^{\mathrm{e}}}, {A^{\mathrm{m}}})$$

\label{eqn:potentialMethods:980}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
&=
&=
+ \frac{1}{2} \lr{ -\rgrad I {A^{\mathrm{m}}} – I {A^{\mathrm{m}}} \lgrad } \\
&=
\inv{2} \lr{ \rgrad ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}}) – ({A^{\mathrm{e}}} + I {A^{\mathrm{m}}}) \lgrad }
,
\end{aligned}

or
\label{eqn:potentialMethods:1000}
\boxed{
\boldsymbol{\mathcal{F}}
=
\inv{2} \lr{ \rgrad ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}}) – ({A^{\mathrm{e}}} – I {A^{\mathrm{m}}})^\dagger \lgrad }
.
}

Observe that \ref{eqn:potentialMethods:1000} can be
put into correspondence with \ref{eqn:potentialMethods:1080} using a factoring of unity $$1 = \gamma_0 \gamma_0$$

\label{eqn:potentialMethods:1100}
\boldsymbol{\mathcal{F}}
=
\inv{2} \lr{ (-\rgrad \gamma_0) (-\gamma_0 ({A^{\mathrm{e}}} -I {A^{\mathrm{m}}})) – (({A^{\mathrm{e}}} + I {A^{\mathrm{m}}}) \gamma_0)(\gamma_0 \lgrad) },

where

\label{eqn:potentialMethods:1140}
\begin{aligned}
&=
-(\gamma^0 \partial_0 + \gamma^k \partial_k) \gamma_0 \\
&=
-\partial_0 – \gamma^k \gamma_0 \partial_k \\
&=
-\inv{v} \partial_t
,
\end{aligned}

\label{eqn:potentialMethods:1160}
\begin{aligned}
&=
\gamma_0 (\gamma^0 \partial_0 + \gamma^k \partial_k) \\
&=
\partial_0 – \gamma^k \gamma_0 \partial_k \\
&=
+ \inv{v} \partial_t
,
\end{aligned}

and
\label{eqn:potentialMethods:1200}
\begin{aligned}
-\gamma_0 ( {A^{\mathrm{e}}} – I {A^{\mathrm{m}}} )
&=
-\gamma_0 \gamma_0 \lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \\
&=
-\lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \phi_m – \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}} } \\
&=
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
\end{aligned}

\label{eqn:potentialMethods:1220}
\begin{aligned}
( {A^{\mathrm{e}}} + I {A^{\mathrm{m}}} )\gamma_0
&=
\lr{ \gamma_0 \lr{ \phi -v \boldsymbol{\mathcal{A}}^{\mathrm{e}} } + I \gamma_0 \eta \lr{ \phi_m – v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \gamma_0 \\
&=
\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + I \eta \phi_m + I \eta v \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
&=
\phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta v I \boldsymbol{\mathcal{A}}^{\mathrm{m}}
+ \eta I \phi_m
,
\end{aligned}

This recovers \ref{eqn:potentialMethods:1080} as desired.

## Potentials in the 3D Euclidean formalism

In the conventional scalar plus vector differential representation of Maxwell’s equations \ref{eqn:chapter3Notes:20}…, given electric(magnetic) sources the structure of the electric(magnetic) potential follows from first setting the magnetic(electric) field equal to the curl of a vector potential. The procedure for the STA GA form of Maxwell’s equation was similar, where it was immediately evident that the field could be set to the four-curl of a four-vector potential (or the dual of such a curl for magnetic sources).

In the 3D GA representation, there is no immediate rationale for introducing a curl or the equivalent to a four-curl representation of the field. Reconciliation of this is possible by recognizing that the fact that the field (or a component of it) may be represented by a curl is not actually fundamental. Instead, observe that the two sided gradient action on a potential to generate the electromagnetic field in the STA representation of \ref{eqn:potentialMethods:1000} serves to select the grade two component product of the gradient and the multivector potential $${A^{\mathrm{e}}} – I {A^{\mathrm{m}}}$$, and that this can in fact be written as
a single sided gradient operation on a potential, provided the multivector product is filtered with a four-bivector grade selection operation

\label{eqn:potentialMethods:1240}
\boxed{
}

Similarly, it can be observed that the
specific function of the conjugate structure in the two sided potential representation of
\ref{eqn:potentialMethods:1080}
is to discard all the scalar and pseudoscalar grades in the multivector product. This means that a single sided potential can also be used, provided it is wrapped in a grade selection operation

\label{eqn:potentialMethods:1260}
\boxed{
\boldsymbol{\mathcal{F}} =
\lr{
– \phi
+ v \boldsymbol{\mathcal{A}}^{\mathrm{e}}
+ \eta I v \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \eta I \phi_m
} }{1,2}.
}

It is this grade selection operation that is really the fundamental defining action in the potential of the STA and conventional 3D representations of Maxwell’s equations. So, given Maxwell’s equation in the 3D GA representation, defining a potential representation for the field is really just a demand that the field have the structure

\label{eqn:potentialMethods:1320}
\boldsymbol{\mathcal{F}} = \gpgrade{ (\alpha \spacegrad + \beta \partial_t)( A_0 + A_1 + I( A_0′ + A_1′ ) }{1,2}.

This is a mandate that the electromagnetic field is the grades 1 and 2 components of the vector product of space and time derivative operators on a multivector field $$A = \sum_{k=0}^3 A_k = A_0 + A_1 + I( A_0′ + A_1′ )$$ that can potentially have any grade components. There are more degrees of freedom in this specification than required, since the multivector can absorb one of the $$\alpha$$ or $$\beta$$ coefficients, so without loss of generality, one of these (say $$\alpha$$) can be set to 1.

Expanding \ref{eqn:potentialMethods:1320} gives

\label{eqn:potentialMethods:1340}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
+ \beta \partial_t A_1
+ \beta \partial_t A_1′
&=
\boldsymbol{\mathcal{E}} + I \eta \boldsymbol{\mathcal{H}}.
\end{aligned}

This naturally has all the right mixes of curls, gradients and time derivatives, all following as direct consequences of applying a grade selection operation to the action of a “spacetime gradient” on a general multivector potential.

The conclusion is that the potential representation of the field is

\label{eqn:potentialMethods:1360}
\boldsymbol{\mathcal{F}} =

where $$A$$ is a multivector potentially containing all grades, where grades 0,1 are required for electric sources, and grades 2,3 are required for magnetic sources. When it is desirable to refer back to the conventional scalar and vector potentials this multivector potential can be written as $$A = -\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ -\phi_m + v \boldsymbol{\mathcal{A}}^{\mathrm{m}} }$$.

## Gauge transformations

Recall that for electric sources the magnetic field is of the form

\label{eqn:potentialMethods:1380}

so adding the gradient of any scalar field to the potential $$\boldsymbol{\mathcal{A}}’ = \boldsymbol{\mathcal{A}} + \spacegrad \psi$$
does not change the magnetic field

\label{eqn:potentialMethods:1400}
\begin{aligned}
\boldsymbol{\mathcal{B}}’
&= \boldsymbol{\mathcal{B}}.
\end{aligned}

The electric field with this changed potential is

\label{eqn:potentialMethods:1420}
\begin{aligned}
\boldsymbol{\mathcal{E}}’
&= -\spacegrad \lr{ \phi + \partial_t \psi } – \partial_t \BA,
\end{aligned}

so if
\label{eqn:potentialMethods:1440}
\phi = \phi’ – \partial_t \psi,

the electric field will also be unaltered by this transformation.

In the STA representation, the field can similarly be altered by adding any (four)gradient to the potential. For example with only electric sources

\label{eqn:potentialMethods:1460}

and for electric or magnetic sources

\label{eqn:potentialMethods:1480}

In the 3D GA representation, where the field is given by \ref{eqn:potentialMethods:1360}, there is no field that is being curled to add a gradient to. However, if the scalar and vector potentials transform as

\label{eqn:potentialMethods:1500}
\begin{aligned}
\boldsymbol{\mathcal{A}} &\rightarrow \boldsymbol{\mathcal{A}} + \spacegrad \psi \\
\phi &\rightarrow \phi – \partial_t \psi,
\end{aligned}

then the multivector potential transforms as
\label{eqn:potentialMethods:1520}
-\phi + v \boldsymbol{\mathcal{A}}
\rightarrow -\phi + v \boldsymbol{\mathcal{A}} + \partial_t \psi + v \spacegrad \psi,

so the electromagnetic field is unchanged when the multivector potential is transformed as

\label{eqn:potentialMethods:1540}
A \rightarrow A + \lr{ \spacegrad + \inv{v} \partial_t } \psi,

where $$\psi$$ is any field that has scalar or pseudoscalar grades. Viewed in terms of grade selection, this makes perfect sense, since the transformed field is

\label{eqn:potentialMethods:1560}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&\rightarrow
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } \lr{ A + \lr{ \spacegrad + \inv{v} \partial_t } \psi } }{1,2} \\
&=
\gpgrade{ \lr{ \spacegrad – \inv{v} \PD{t}{} } A + \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi }{1,2} \\
&=
\end{aligned}

The $$\psi$$ contribution to the grade selection operator is killed because it has scalar or pseudoscalar grades.

## Lorenz gauge

Maxwell’s equations are completely decoupled if the potential can be found such that

\label{eqn:potentialMethods:1580}
\begin{aligned}
\boldsymbol{\mathcal{F}}
&=
&=
\lr{ \spacegrad – \inv{v} \PD{t}{} } A.
\end{aligned}

When this is the case, Maxwell’s equations are reduced to four non-homogeneous potential wave equations

\label{eqn:potentialMethods:1620}
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } A = J,

that is

\label{eqn:potentialMethods:1600}
\begin{aligned}
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \phi &= – \inv{\epsilon} q_e \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \boldsymbol{\mathcal{A}}^{\mathrm{e}} &= – \mu \boldsymbol{\mathcal{J}} \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \phi_m &= – \frac{I}{\mu} q_m \\
\lr{ \spacegrad^2 – \inv{v^2} \PDSq{t}{} } \boldsymbol{\mathcal{A}}^{\mathrm{m}} &= – I \epsilon \boldsymbol{\mathcal{M}}.
\end{aligned}

There should be no a-priori assumption that such a field representation has no scalar, nor no pseudoscalar components. That explicit expansion in grades is

\label{eqn:potentialMethods:1640}
\begin{aligned}
\lr{ \spacegrad – \inv{v} \PD{t}{} } A
&=
\lr{ \spacegrad – \inv{v} \PD{t}{} } \lr{ -\phi + v \boldsymbol{\mathcal{A}}^{\mathrm{e}} + \eta I \lr{ -\phi_m + v \boldsymbol{\mathcal{A}}^{\mathrm{m}} } } \\
&=
\inv{v} \partial_t \phi
+ v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
+ I \eta v \spacegrad \wedge \boldsymbol{\mathcal{A}}^{\mathrm{m}}
– \partial_t \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
– I \eta \partial_t \boldsymbol{\mathcal{A}}^{\mathrm{m}} \\
&+ \eta I \inv{v} \partial_t \phi_m
+ I \eta v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}},
\end{aligned}

so if this potential representation has only vector and bivector grades, it must be true that

\label{eqn:potentialMethods:1660}
\begin{aligned}
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} &= 0 \\
\inv{v} \partial_t \phi_m + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{m}} &= 0.
\end{aligned}

The first is the well known Lorenz gauge condition, whereas the second is the dual of that condition for magnetic sources.

Should one of these conditions, say the Lorenz condition for the electric source potentials, be non-zero, then it is possible to make a potential transformation for which this condition is zero

\label{eqn:potentialMethods:1680}
\begin{aligned}
0
&\ne
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}} \\
&=
\inv{v} \partial_t (\phi’ – \partial_t \psi) + v \spacegrad \cdot (\boldsymbol{\mathcal{A}}’ + \spacegrad \psi) \\
&=
\inv{v} \partial_t \phi’ + v \spacegrad \boldsymbol{\mathcal{A}}’
+ v \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi,
\end{aligned}

so if $$\inv{v} \partial_t \phi’ + v \spacegrad \boldsymbol{\mathcal{A}}’$$ is zero, $$\psi$$ must be found such that
\label{eqn:potentialMethods:1700}
\inv{v} \partial_t \phi + v \spacegrad \cdot \boldsymbol{\mathcal{A}}^{\mathrm{e}}
= v \lr{ \spacegrad^2 – \inv{v^2} \partial_{tt} } \psi.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[3] David M Pozar. Microwave engineering. John Wiley \& Sons, 2009.

## Transverse gauge

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

\label{eqn:transverseGauge:20}

Faraday’s law now takes the form
\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}

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

\label{eqn:transverseGauge:60}
\BE + \PD{t}{\BA} \equiv -\spacegrad \Phi.

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

\label{eqn:transverseGauge:80}
\begin{aligned}
\spacegrad \cdot \BD &= \rho_v \\
\spacegrad \cross \BH &= \BJ + \PD{t}{\BD} \\
\end{aligned}

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

\label{eqn:transverseGauge:140}
\begin{aligned}
\rho_v
&=
&=
\end{aligned}

For simple media again, the Ampere-Maxwell equation is

\label{eqn:transverseGauge:100}
\inv{\mu} \spacegrad \cross \lr{ \spacegrad \cross \BA } = \BJ + \epsilon \PD{t}{} \lr{ -\spacegrad \Phi – \PD{t}{\BA} }.

Expanding $$\spacegrad \cross \lr{ \spacegrad \cross \BA } = -\spacegrad^2 \BA + \spacegrad \lr{ \spacegrad \cdot \BA }$$ gives
\label{eqn:transverseGauge:120}

Maxwell’s equations are now reduced to
\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 \\
\end{aligned}
}

There are two obvious constraints that we can impose
\label{eqn:transverseGauge:200}
\spacegrad \cdot \BA – \epsilon \mu \PD{t}{\Phi} = 0,

or
\label{eqn:transverseGauge:220}

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

\label{eqn:transverseGauge:260}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA} = -\mu \BJ + \epsilon\mu \spacegrad \PD{t}{\Phi}

\label{eqn:transverseGauge:280}

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

\label{eqn:transverseGauge:300}
\begin{aligned}
– 4 \pi \BJ(\Bx)
&=
\int \spacegrad^2 \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\int \spacegrad \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
+
\int \spacegrad \wedge \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’ \\
&=
\int \BJ(\Bx’) \cdot \spacegrad’ \inv{\Abs{\Bx – \Bx’}} d^3 x’
+
\int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
} \\
&=
\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
\int \frac{\spacegrad’ \cdot \BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’

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

The first term can be converted to a surface integral

\label{eqn:transverseGauge:320}
\int \spacegrad’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
=
\int d\BA’ \cdot \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}},

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

\label{eqn:transverseGauge:340}
\BJ(\Bx)
=
+
\spacegrad \cross \spacegrad \cross \inv{4 \pi} \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’
\equiv
\BJ_l +
\BJ_t,

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

\label{eqn:transverseGauge:360}
\begin{aligned}
&=
&=
&=
&= 0.
\end{aligned}

Since

\label{eqn:transverseGauge:380}
\Phi(\Bx, t)
=
\inv{4 \pi \epsilon} \int \frac{\rho_v(\Bx’, t)}{\Abs{\Bx – \Bx’}} d^3 x’,

we have

\label{eqn:transverseGauge:400}
\begin{aligned}
&=
\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}

This means that the Ampere-Maxwell equation takes the form

\label{eqn:transverseGauge:420}
\spacegrad^2 \BA – \epsilon \mu \PDSq{t}{\BA}
= -\mu \BJ + \mu \BJ_l
= -\mu \BJ_t.

This justifies the transverse in the label transverse gauge.

# References

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

## Line charge field and potential.

When computing the most general solution of the electrostatic potential in a plane, Jackson [1] mentions that $$-2 \lambda_0 \ln \rho$$ is the well known potential for an infinite line charge (up to the unit specific factor). Checking that statement, since I didn’t recall what that potential was offhand, I encountered some inconsistencies and non-convergent integrals, and thought it was worthwhile to explore those a bit more carefully. This will be done here.

### Using Gauss’s law.

For an infinite length line charge, we can find the radial field contribution using Gauss’s law, imagining a cylinder of length $$\Delta l$$ of radius $$\rho$$ surrounding this charge with the midpoint at the origin. Ignoring any non-radial field contribution, we have

\label{eqn:lineCharge:20}
\int_{-\Delta l/2}^{\Delta l/2} \ncap \cdot \BE (2 \pi \rho) dl = \frac{\lambda_0}{\epsilon_0} \Delta l,

or

\label{eqn:lineCharge:40}
\BE = \frac{\lambda_0}{2 \pi \epsilon_0} \frac{\rhocap}{\rho}.

Since

\label{eqn:lineCharge:60}

this means that the potential is

\label{eqn:lineCharge:80}
\phi = -\frac{2 \lambda_0}{4 \pi \epsilon_0} \ln \rho.

### Finite line charge potential.

Let’s try both these calculations for a finite charge distribution. Gauss’s law looses its usefulness, but we can evaluate the integrals directly. For the electric field

\label{eqn:lineCharge:100}
\BE
= \frac{\lambda_0}{4 \pi \epsilon_0} \int \frac{(\Bx – \Bx’)}{\Abs{\Bx – \Bx’}^3} dl’.

Using cylindrical coordinates with the field point $$\Bx = \rho \rhocap$$ for convience, the charge point $$\Bx’ = z’ \zcap$$, and a the charge distributed over $$[a,b]$$ this is

\label{eqn:lineCharge:120}
\BE
= \frac{\lambda_0}{4 \pi \epsilon_0} \int_a^b \frac{(\rho \rhocap – z’ \zcap)}{\lr{\rho^2 + (z’)^2}^{3/2}} dz’.

When the charge is uniformly distributed around the origin $$[a,b] = b[-1,1]$$ the $$\zcap$$ component of this field is killed because the integrand is odd. This justifies ignoring such contributions in the Gaussing cylinder analysis above. The general solution to this integral is found to be

\label{eqn:lineCharge:140}
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\evalrange{
\lr{
\frac{z’ \rhocap }{\rho \sqrt{ \rho^2 + (z’)^2 } }
+\frac{\zcap}{ \sqrt{ \rho^2 + (z’)^2 } }
}
}{a}{b},

or
\label{eqn:lineCharge:240}
\boxed{
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\lr{
\frac{\rhocap }{\rho}
\lr{
\frac{b}{\sqrt{ \rho^2 + b^2 } }
-\frac{a}{\sqrt{ \rho^2 + a^2 } }
}
+ \zcap
\lr{
\frac{1}{ \sqrt{ \rho^2 + b^2 } }
-\frac{1}{ \sqrt{ \rho^2 + a^2 } }
}
}.
}

When $$b = -a = \Delta l/2$$, this reduces to

\label{eqn:lineCharge:160}
\BE
=
\frac{\lambda_0}{4 \pi \epsilon_0}
\frac{\rhocap }{\rho}
\frac{\Delta l}{\sqrt{ \rho^2 + (\Delta l/2)^2 } },

which further reduces to \ref{eqn:lineCharge:40} when $$\Delta l \gg \rho$$.

### Finite line charge potential. Wrong but illuminating.

Again, putting the field point at $$z’ = 0$$, we have

\label{eqn:lineCharge:180}
\phi(\rho)
= \frac{\lambda_0}{4 \pi \epsilon_0} \int_a^b \frac{dz’}{\lr{\rho^2 + (z’)^2}^{1/2}},

which integrates to
\label{eqn:lineCharge:260}
\phi(\rho)
= \frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}}.

With $$b = -a = \Delta l/2$$, this approaches

\label{eqn:lineCharge:200}
\phi
\approx
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ (\Delta l/2) }{ \rho^2/2\Abs{\Delta l/2}}
=
\frac{-2 \lambda_0}{4 \pi \epsilon_0 } \ln \rho
+
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \lr{ (\Delta l)^2/2 }.

Before $$\Delta l$$ is allowed to tend to infinity, this is identical (up to a difference in the reference potential) to \ref{eqn:lineCharge:80} found using Gauss’s law. It is, strictly speaking, singular when $$\Delta l \rightarrow \infty$$, so it does not seem right to infinity as a reference point for the potential.

There’s another weird thing about this result. Since this has no $$z$$ dependence, it is not obvious how we would recover the non-radial portion of the electric field from this potential using $$\BE = -\spacegrad \phi$$? Let’s calculate the elecric field from \ref{eqn:lineCharge:180} explicitly

\label{eqn:lineCharge:220}
\begin{aligned}
\BE
&=
-\frac{\lambda_0}{4 \pi \epsilon_0}
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 }
\PD{\rho}{}
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0}
\lr{
\inv{ b + \sqrt{ \rho^2 + b^2 }} \frac{ \rho }{\sqrt{ \rho^2 + b^2 }}
-\inv{ a + \sqrt{ \rho^2 + a^2 }} \frac{ \rho }{\sqrt{ \rho^2 + a^2 }}
} \\
&=
-\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 \rho}
\lr{
\frac{ -b + \sqrt{ \rho^2 + b^2 }}{\sqrt{ \rho^2 + b^2 }}
-\frac{ -a + \sqrt{ \rho^2 + a^2 }}{\sqrt{ \rho^2 + a^2 }}
} \\
&=
\frac{\lambda_0 \rhocap}{4 \pi \epsilon_0 \rho}
\lr{
\frac{ b }{\sqrt{ \rho^2 + b^2 }}
-\frac{ a }{\sqrt{ \rho^2 + a^2 }}
}.
\end{aligned}

This recovers the radial component of the field from \ref{eqn:lineCharge:240}, but where did the $$\zcap$$ component go? The required potential appears to be

\label{eqn:lineCharge:340}
\phi(\rho, z)
=
\frac{\lambda_0}{4 \pi \epsilon_0 }
\ln \frac{ b + \sqrt{ \rho^2 + b^2 }}{ a + \sqrt{\rho^2 + a^2}}

\frac{z \lambda_0}{4 \pi \epsilon_0 }
\lr{ \frac{1}{\sqrt{\rho^2 + b^2}}
-\frac{1}{\sqrt{\rho^2 + a^2}}
}.

When computing the electric field $$\BE(\rho, \theta, z)$$, it was convienent to pick the coordinate system so that $$z = 0$$. Doing this with the potential gives the wrong answers. The reason for this appears to be that this kills the potential term that is linear in $$z$$ before taking its gradient, and we need that term to have the $$\zcap$$ field component that is expected for a charge distribution that is non-symmetric about the origin on the z-axis!

### Finite line charge potential. Take II.

Let the point at which the potential is evaluated be

\label{eqn:lineCharge:360}
\Bx = \rho \rhocap + z \zcap,

and the charge point be
\label{eqn:lineCharge:380}
\Bx’ = z’ \zcap.

This gives

\label{eqn:lineCharge:400}
\begin{aligned}
\phi(\rho, z)
&= \frac{\lambda_0}{4\pi \epsilon_0} \int_a^b \frac{dz’}{\Abs{\rho^2 + (z – z’)^2 }} \\
&= \frac{\lambda_0}{4\pi \epsilon_0} \int_{a-z}^{b-z} \frac{du}{ \Abs{\rho^2 + u^2} } \\
&= \frac{\lambda_0}{4\pi \epsilon_0}
\evalrange{\ln \lr{ u + \sqrt{ \rho^2 + u^2 }}}{b-z}{a-z} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\ln \frac
{ b-z + \sqrt{ \rho^2 + (b-z)^2 }}
{ a-z + \sqrt{ \rho^2 + (a-z)^2 }}.
\end{aligned}

The limit of this potential $$a = -\Delta/2 \rightarrow -\infty, b = \Delta/2 \rightarrow \infty$$ doesn’t exist in any strict sense. If we are cavilier about the limits, as in \ref{eqn:lineCharge:200}, this can be evaluated as

\label{eqn:lineCharge:n}
\phi \approx
\frac{\lambda_0}{4\pi \epsilon_0} \lr{ -2 \ln \rho + \textrm{constant} }.

however, the constant ($$\ln \Delta^2/2$$) is infinite, so there isn’t really a good justification for using that constant as the potential reference point directly.

It seems that the “right” way to calculate the potential for the infinite distribution, is to

• Calculate the field from the potential.
• Take the PV limit of that field with the charge distribution extending to infinity.
• Compute the corresponding potential from this limiting value of the field.

Doing that doesn’t blow up. That field calculation, for the finite case, should include a $$\zcap$$ component. To verify, let’s take the respective derivatives

\label{eqn:lineCharge:420}
\begin{aligned}
-\PD{z}{} \phi
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ -1 + \frac{z – b}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ -1 + \frac{z – a}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ 1 + \frac{b – z}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ 1 + \frac{a – z}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\inv{\sqrt{ \rho^2 + (b-z)^2 }}
-\inv{\sqrt{ \rho^2 + (a-z)^2 }}
},
\end{aligned}

and

\label{eqn:lineCharge:440}
\begin{aligned}
-\PD{\rho}{} \phi
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{ \frac{\rho}{\sqrt{ \rho^2 + (b-z)^2 }} }{
b-z + \sqrt{ \rho^2 + (b-z)^2 }
}

\frac{ \frac{\rho}{\sqrt{ \rho^2 + (a-z)^2 }} }{
a-z + \sqrt{ \rho^2 + (a-z)^2 }
}
} \\
&=
-\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{\rho \lr{
-(b-z) + \sqrt{ \rho^2 + (b-z)^2 }
}}{ \rho^2 \sqrt{ \rho^2 + (b-z)^2 } }

\frac{\rho \lr{
-(a-z) + \sqrt{ \rho^2 + (a-z)^2 }
}}{ \rho^2 \sqrt{ \rho^2 + (a-z)^2 } }
} \\
&=
\frac{\lambda_0}{4\pi \epsilon_0 \rho}
\lr{
\frac{b-z}{\sqrt{ \rho^2 + (b-z)^2 }}
-\frac{a-z}{\sqrt{ \rho^2 + (a-z)^2 }}
}
.
\end{aligned}

Putting the pieces together, the electric field is
\label{eqn:lineCharge:460}
\BE =
\frac{\lambda_0}{4\pi \epsilon_0}
\lr{
\frac{\rhocap}{\rho} \lr{
\frac{b-z}{\sqrt{ \rho^2 + (b-z)^2 }}
-\frac{a-z}{\sqrt{ \rho^2 + (a-z)^2 }}
}
+
\zcap \lr{
\inv{\sqrt{ \rho^2 + (b-z)^2 }}
-\inv{\sqrt{ \rho^2 + (a-z)^2 }}
}
}.

For has a PV limit of \ref{eqn:lineCharge:40} at $$z = 0$$, and also for the finite case, has the $$\zcap$$ field component that was obtained when the field was obtained by direct integration.

### Conclusions

• We have to evaluate the potential at all points in space, not just on the axis that we evaluate the field on (should we choose to do so).
• In this case, we found that it was not directly meaningful to take the limit of a potential distribution. We can, however, compute the field from a potential for a finite charge distribution,
take the limit of that field, and then calculate the corresponding potential for the infinite distribution.

Is there a more robust mechanism that can be used to directly calculate the potential for an infinite charge distribution, instead of calculating the potential from the field of such an infinite distribution?

I think that were things go wrong is that the integral of \ref{eqn:lineCharge:180} does not apply to charge distributions that are not finite on the infinite range $$z \in [-\infty, \infty]$$. That solution was obtained by utilizing an all-space Green’s function, and the boundary term in that Green’s analysis was assumed to tend to zero. That isn’t the case when the charge distribution is $$\lambda_0 \delta( z )$$.

# References

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