## Updated notes for ece1229 antenna theory

I’ve now posted a first update of my notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

## Duality transformation

In a discussion of Dirac’s monopoles, [1] introduces a duality transformation, forming electric and magnetic fields by forming a rotation that combines a different pair of electric and magnetic fields. In SI units that transformation becomes

\label{eqn:dualityTransformation:40}
\begin{bmatrix}
\boldsymbol{\mathcal{E}} \\
\eta \boldsymbol{\mathcal{H}}
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\boldsymbol{\mathcal{E}}’ \\
\eta \boldsymbol{\mathcal{H}}’
\end{bmatrix}

\label{eqn:dualityTransformation:60}
\begin{bmatrix}
\boldsymbol{\mathcal{D}} \\
\boldsymbol{\mathcal{B}}/\eta
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\boldsymbol{\mathcal{D}}’ \\
\boldsymbol{\mathcal{B}}’/\eta
\end{bmatrix},

where $$\eta = \sqrt{\mu_0/\epsilon_0}$$. It is left as an exercise to the reader to show that application of these to Maxwell’s equations

\label{eqn:dualityTransformation:100}

\label{eqn:dualityTransformation:120}

\label{eqn:dualityTransformation:140}
-\spacegrad \cross \boldsymbol{\mathcal{E}} – \partial_t \boldsymbol{\mathcal{B}} = \boldsymbol{\mathcal{J}}_{\textrm{m}}

\label{eqn:dualityTransformation:160}
\spacegrad \cross \boldsymbol{\mathcal{H}} – \partial_t \boldsymbol{\mathcal{D}} = \boldsymbol{\mathcal{J}}_{\textrm{e}},

determine a similar relation between the sources. That transformation of Maxwell’s equation is

\label{eqn:dualityTransformation:200}
\spacegrad \cdot \lr{ \cos\theta \boldsymbol{\mathcal{E}}’ + \sin\theta \eta \boldsymbol{\mathcal{H}}’ } = \rho_{\textrm{e}}/\epsilon_0

\label{eqn:dualityTransformation:220}
\spacegrad \cdot \lr{ -\sin\theta \boldsymbol{\mathcal{E}}’/\eta + \cos\theta \boldsymbol{\mathcal{H}}’ } = \rho_{\textrm{m}}/\mu_0

\label{eqn:dualityTransformation:240}
-\spacegrad \cross \lr{ \cos\theta \boldsymbol{\mathcal{E}}’ + \sin\theta \eta \boldsymbol{\mathcal{H}}’ } – \partial_t \lr{ – \sin\theta \eta \boldsymbol{\mathcal{D}}’ + \cos\theta \boldsymbol{\mathcal{B}}’ } = \boldsymbol{\mathcal{J}}_{\textrm{m}}

\label{eqn:dualityTransformation:260}
\spacegrad \cross \lr{ -\sin\theta \boldsymbol{\mathcal{E}}’/\eta + \cos\theta \boldsymbol{\mathcal{H}}’ } – \partial_t \lr{ \cos\theta \boldsymbol{\mathcal{D}}’ + \sin\theta \boldsymbol{\mathcal{B}}’/\eta } = \boldsymbol{\mathcal{J}}_{\textrm{e}}.

A bit of rearranging gives

\label{eqn:dualityTransformation:400}
\begin{bmatrix}
\eta \rho_{\textrm{e}} \\
\rho_{\textrm{m}}
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\eta \rho_{\textrm{e}}’ \\
\rho_{\textrm{m}}’
\end{bmatrix}

\label{eqn:dualityTransformation:420}
\begin{bmatrix}
\eta \boldsymbol{\mathcal{J}}_{\textrm{e}} \\
\boldsymbol{\mathcal{J}}_{\textrm{m}} \\
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\eta \boldsymbol{\mathcal{J}}_{\textrm{e}}’ \\
\boldsymbol{\mathcal{J}}_{\textrm{m}}’ \\
\end{bmatrix}.

For example, with $$\rho_{\textrm{m}} = \boldsymbol{\mathcal{J}}_{\textrm{m}} = 0$$, and $$\theta = \pi/2$$, the transformation of sources is

\label{eqn:dualityTransformation:440}
\begin{aligned}
\rho_{\textrm{e}}’ &= 0 \\
\boldsymbol{\mathcal{J}}_{\textrm{e}}’ &= 0 \\
\rho_{\textrm{m}}’ &= \eta \rho_{\textrm{e}} \\
\boldsymbol{\mathcal{J}}_{\textrm{m}}’ &= \eta \boldsymbol{\mathcal{J}}_{\textrm{e}},
\end{aligned}

and Maxwell’s equations then have only magnetic sources

\label{eqn:dualityTransformation:480}

\label{eqn:dualityTransformation:500}

\label{eqn:dualityTransformation:520}
-\spacegrad \cross \boldsymbol{\mathcal{E}}’ – \partial_t \boldsymbol{\mathcal{B}}’ = \boldsymbol{\mathcal{J}}_{\textrm{m}}’

\label{eqn:dualityTransformation:540}
\spacegrad \cross \boldsymbol{\mathcal{H}}’ – \partial_t \boldsymbol{\mathcal{D}}’ = 0.

Of this relation Jackson points out that “The invariance of the equations of electrodynamics under duality transformations shows that it is a matter of convention to speak of a particle possessing an electric charge, but not magnetic charge.” This is an interesting comment, and worth some additional thought.

# References

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

## Maxwell’s equations with magnetic sources

The form of Maxwell’s equations to be used here are expressed in terms of $$\boldsymbol{\mathcal{E}}$$ and $$\boldsymbol{\mathcal{H}}$$, assume linear media, and do not assume a phasor representation

\label{eqn:energyMomentumWithMagneticSources:120}
\spacegrad \cross \boldsymbol{\mathcal{E}} = – \boldsymbol{\mathcal{M}} – \mu_0 \PD{t}{\boldsymbol{\mathcal{H}}}

\label{eqn:energyMomentumWithMagneticSources:140}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \epsilon_0 \PD{t}{\boldsymbol{\mathcal{E}}}

\label{eqn:energyMomentumWithMagneticSources:160}

\label{eqn:energyMomentumWithMagneticSources:180}

## Energy momentum conservation

With magnetic sources the Poynting and energy conservation relationship has to be adjusted slightly. Let’s derive that result, starting with the divergence of the Poynting vector

\label{eqn:energyMomentumWithMagneticSources:20}
\begin{aligned}
\spacegrad \cdot \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
&=
\boldsymbol{\mathcal{H}} \cdot \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} }
-\boldsymbol{\mathcal{E}} \cdot \lr{ \spacegrad \cross \boldsymbol{\mathcal{H}} } \\
&=
-\boldsymbol{\mathcal{H}} \cdot \lr{ \mu_0 \partial_t \boldsymbol{\mathcal{H}} + \boldsymbol{\mathcal{M}} }
-\boldsymbol{\mathcal{E}} \cdot \lr{ \boldsymbol{\mathcal{J}} + \epsilon_0 \partial_t \boldsymbol{\mathcal{E}} } \\
&=
– \mu_0 \boldsymbol{\mathcal{H}} \cdot \partial_t \boldsymbol{\mathcal{H}} – \boldsymbol{\mathcal{H}} \cdot \boldsymbol{\mathcal{M}}
– \epsilon_0 \boldsymbol{\mathcal{E}} \cdot \partial_t \boldsymbol{\mathcal{E}} – \boldsymbol{\mathcal{E}} \cdot \boldsymbol{\mathcal{J}},
\end{aligned}

or

\label{eqn:energyMomentumWithMagneticSources:40}
\boxed{
\inv{2} \PD{t}{} \lr{ \epsilon_0 \boldsymbol{\mathcal{E}}^2 + \mu_0 \boldsymbol{\mathcal{H}}^2 }
+
\spacegrad \cdot \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
=
– \boldsymbol{\mathcal{H}} \cdot \boldsymbol{\mathcal{M}}
– \boldsymbol{\mathcal{E}} \cdot \boldsymbol{\mathcal{J}}.
}

The usual relationship is only modified by one additional term. Recall from electrodynamics [2] that \ref{eqn:energyMomentumWithMagneticSources:40} (when the magnetic current density $$\boldsymbol{\mathcal{M}}$$ is omitted) is just one of four components of the energy momentum conservation equation

\label{eqn:energyMomentumWithMagneticSources:80}
\partial_\mu T^{\mu \nu} = – \inv{c} F^{\nu \lambda} j_\lambda.

Note that \ref{eqn:energyMomentumWithMagneticSources:80} was likely not in SI units. The next task is to generalize this classical relationship to incorporate the magnetic sources used in antenna theory. With an eye towards the relativistic nature of the energy momentum tensor, it is natural to assume that the remainder of the energy momentum tensor conservation relation can be found by taking the time derivatives of the Poynting vector.

\label{eqn:energyMomentumWithMagneticSources:200}
\PD{t}{} \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
=
\PD{t}{\boldsymbol{\mathcal{E}}} \cross \boldsymbol{\mathcal{H}}
+ \boldsymbol{\mathcal{E}} \cross \PD{t}{\boldsymbol{\mathcal{H}} }
=
\inv{\epsilon_0}
\lr{ \spacegrad \cross \boldsymbol{\mathcal{H}} – \boldsymbol{\mathcal{J}} } \cross \boldsymbol{\mathcal{H}}
+
\inv{\mu_0}
\boldsymbol{\mathcal{E}} \cross
\lr{

\spacegrad \cross \boldsymbol{\mathcal{E}} – \boldsymbol{\mathcal{M}} },

or

\label{eqn:energyMomentumWithMagneticSources:220}
\inv{c^2} \PD{t}{} \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
+
\mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}}
+\epsilon_0
\boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{M}}
=
-\mu_0 \boldsymbol{\mathcal{H}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{H}} }
– \epsilon_0 \boldsymbol{\mathcal{E}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} }.

The $$\mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} \cross \BB$$ is a portion of the Lorentz force equation in its density form. To put \ref{eqn:energyMomentumWithMagneticSources:220} into the desired form, the remainder of the Lorentz force force equation $$\rho \boldsymbol{\mathcal{E}} = \epsilon_0 \boldsymbol{\mathcal{E}} \spacegrad \cdot \boldsymbol{\mathcal{E}}$$ must be added to both sides. To extend the magnetic current term to its full dual (magnetic) Lorentz force structure, the quantity to add to both sides is $$\rho_m \boldsymbol{\mathcal{H}} = \mu_0 \boldsymbol{\mathcal{H}} \spacegrad \cdot \boldsymbol{\mathcal{H}}$$. Performing these manipulations gives

\label{eqn:energyMomentumWithMagneticSources:240}
\inv{c^2} \PD{t}{} \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
+
\rho \BE + \mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}}
+ \rho_m \boldsymbol{\mathcal{H}}
+ \epsilon_0 \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{M}}
=
\mu_0
\lr{
-\boldsymbol{\mathcal{H}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{H}} }
}
+ \epsilon_0
\lr{

\boldsymbol{\mathcal{E}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} }
}.

It seems slightly surprising the sign of the magnetic equivalent of the Lorentz force terms have an alternation of sign. This is, however, consistent with the duality transformations outlined in ([1] table 3.2)

\label{eqn:energyMomentumWithMagneticSources:280}
\rho \rightarrow \rho_m

\label{eqn:energyMomentumWithMagneticSources:300}
\boldsymbol{\mathcal{J}} \rightarrow \boldsymbol{\mathcal{M}}

\label{eqn:energyMomentumWithMagneticSources:320}
\mu_0 \rightarrow \epsilon_0

\label{eqn:energyMomentumWithMagneticSources:340}
\boldsymbol{\mathcal{E}} \rightarrow \boldsymbol{\mathcal{H}}

\label{eqn:energyMomentumWithMagneticSources:360}
\boldsymbol{\mathcal{H}} \rightarrow -\boldsymbol{\mathcal{E}},

for

\label{eqn:energyMomentumWithMagneticSources:380}
\rho \BE + \mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}}
\rightarrow
\rho_m \BH + \epsilon_0 \boldsymbol{\mathcal{M}} \cross \lr{ -\boldsymbol{\mathcal{E}}}
=
\rho_m \BH + \epsilon_0 \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{M}}.

Comfortable that the LHS has the desired structure, the RHS can expressed as a divergence. Just expanding one of the differences of vector products on the RHS does not obviously show that this is possible, for example

\label{eqn:energyMomentumWithMagneticSources:400}
\begin{aligned}
\Be_a \cdot
\lr{

\boldsymbol{\mathcal{E}} \cross \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} }
}
&=
E_a \partial_b E_b

\epsilon_{a b c} E_b \epsilon_{c r s} \partial_r E_s \\
&=
E_a \partial_b E_b

\delta_{a b}^{[r s]} E_b \partial_r E_s \\
&=
E_a \partial_b E_b

E_b \lr{
\partial_a E_b
-\partial_b E_a
} \\
&=
E_a \partial_b E_b
– E_b \partial_a E_b
+ E_b \partial_b E_a.
\end{aligned}

This happens to equal

\label{eqn:energyMomentumWithMagneticSources:420}
\begin{aligned}
\spacegrad \cdot \lr{ \lr{E_a E_b – \inv{2} \delta_{a b} \boldsymbol{\mathcal{E}}^2 } \Be_b }
&=
\partial_b
\lr{E_a E_b – \inv{2} \delta_{a b} \boldsymbol{\mathcal{E}}^2 } \\
&=
E_b \partial_b E_a
+ E_a \partial_b E_b

\inv{2} \delta_{a b} 2 E_c \partial_b E_c \\
i&=
E_b \partial_b E_a
+ E_a \partial_b E_b
– E_b \partial_a E_b.
\end{aligned}

This allows a final formulation of the remaining energy momentum conservation equation in its divergence form. Let

\label{eqn:energyMomentumWithMagneticSources:440}
T^{a b} =
\epsilon_0 \lr{ E_a E_b – \inv{2} \delta_{a b} \boldsymbol{\mathcal{E}}^2 }
+ \mu_0 \lr{ H_a H_b – \inv{2} \delta_{a b} \boldsymbol{\mathcal{H}}^2 },

so that the remaining energy momentum conservation equation, extended to both electric and magnetic sources, is

\label{eqn:energyMomentumWithMagneticSources:460}
\boxed{
\inv{c^2} \PD{t}{} \lr{ \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}} }
+
\rho \BE + \mu_0 \boldsymbol{\mathcal{J}} \cross \boldsymbol{\mathcal{H}}
+ \rho_m \boldsymbol{\mathcal{H}}
+ \epsilon_0 \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{M}}
=
\Be_a \spacegrad \cdot \lr{ T^{a b} \Be_b }.
}

On the LHS we have the rate of change of momentum density, the electric Lorentz force density terms, the dual (magnetic) Lorentz force density terms, and on the RHS the the momentum flux terms.

# References

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

[2] Peeter Joot. Relativistic Electrodynamics., chapter {Energy Momentum Tensor.} peeterjoot.com, 2011. URL http://peeterjoot.com/archives/math2011/phy450.pdf. [Online; accessed 18-February-2015].

## Notes for ece1229 antenna theory

I’ve now posted a first set of notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book.)

• Reading notes for chapter 2 (Fundamental Parameters of Antennas) and chapter 3 (Radiation Integrals and Auxiliary Potential Functions) of the class text.
• Geometric Algebra musings.  How to do formulate Maxwell’s equations when magnetic sources are also included (those modeling magnetic dipoles).
• Some problems for chapter 2 content.

## Dual-Maxwell’s (phasor) equations in Geometric Algebra

These notes repeat (mostly word for word) the previous notes Maxwell’s (phasor) equations in Geometric Algebra. Electric charges and currents have been replaced with magnetic charges and currents, and the appropriate relations modified accordingly.

In [1] section 3.3, treating magnetic charges and currents, and no electric charges and currents, is a demonstration of the required (curl) form for the electric field, and potential form for the electric field. Not knowing what to name this, I’ll call the associated equations the dual-Maxwell’s equations.

I was wondering how this derivation would proceed using the Geometric Algebra (GA) formalism.

## Dual-Maxwell’s equation in GA phasor form.

The dual-Maxwell’s equations, omitting electric charges and currents, are

\label{eqn:phasorDualMaxwellsGA:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}} -\BM

\label{eqn:phasorDualMaxwellsGA:40}

\label{eqn:phasorDualMaxwellsGA:60}

\label{eqn:phasorDualMaxwellsGA:80}

Assuming linear media $$\boldsymbol{\mathcal{B}} = \mu_0 \boldsymbol{\mathcal{H}}$$, $$\boldsymbol{\mathcal{D}} = \epsilon_0 \boldsymbol{\mathcal{E}}$$, and phasor relationships of the form $$\boldsymbol{\mathcal{E}} = \textrm{Re} \lr{ \BE(\Br) e^{j \omega t}}$$ for the fields and the currents, these reduce to

\label{eqn:phasorDualMaxwellsGA:100}
\spacegrad \cross \BE = – j \omega \BB – \BM

\label{eqn:phasorDualMaxwellsGA:120}
\spacegrad \cross \BB = j \omega \epsilon_0 \mu_0 \BE

\label{eqn:phasorDualMaxwellsGA:140}

\label{eqn:phasorDualMaxwellsGA:160}

These four equations can be assembled into a single equation form using the GA identities

\label{eqn:phasorDualMaxwellsGA:200}
\Bf \Bg
= \Bf \cdot \Bg + \Bf \wedge \Bg
= \Bf \cdot \Bg + I \Bf \cross \Bg.

\label{eqn:phasorDualMaxwellsGA:220}
I = \xcap \ycap \zcap.

The electric and magnetic field equations, respectively, are

\label{eqn:phasorDualMaxwellsGA:260}
\spacegrad \BE = – \lr{ \BM + j k c \BB} I

\label{eqn:phasorDualMaxwellsGA:280}
\spacegrad c \BB = c \rho_m + j k \BE I

where $$\omega = k c$$, and $$1 = c^2 \epsilon_0 \mu_0$$ have also been used to eliminate some of the mess of constants.

Summing these (first scaling \ref{eqn:phasorDualMaxwellsGA:280} by $$I$$), gives Maxwell’s equation in its GA phasor form

\label{eqn:phasorDualMaxwellsGA:300}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + I c \BB } = \lr{c \rho – \BM} I.
}

## Preliminaries. Dual magnetic form of Maxwell’s equations.

The arguments of the text showing that a potential representation for the electric and magnetic fields is possible easily translates into GA. To perform this translation, some duality lemmas are required

First consider the cross product of two vectors $$\Bx, \By$$ and the right handed dual $$-\By I$$ of $$\By$$, a bivector, of one of these vectors. Noting that the Euclidean pseudoscalar $$I$$ commutes with all grade multivectors in a Euclidean geometric algebra space, the cross product can be written

\label{eqn:phasorDualMaxwellsGA:320}
\begin{aligned}
\lr{ \Bx \cross \By }
&=
-I \lr{ \Bx \wedge \By } \\
&=
-I \inv{2} \lr{ \Bx \By – \By \Bx } \\
&=
\inv{2} \lr{ \Bx (-\By I) – (-\By I) \Bx } \\
&=
\Bx \cdot \lr{ -\By I }.
\end{aligned}

The last step makes use of the fact that the wedge product of a vector and vector is antisymmetric, whereas the dot product (vector grade selection) of a vector and bivector is antisymmetric. Details on grade selection operators and how to characterize symmetric and antisymmetric products of vectors with blades as either dot or wedge products can be found in [3], [2].

Similarly, the dual of the dot product can be written as

\label{eqn:phasorDualMaxwellsGA:440}
\begin{aligned}
-I \lr{ \Bx \cdot \By }
&=
-I \inv{2} \lr{ \Bx \By + \By \Bx } \\
&=
\inv{2} \lr{ \Bx (-\By I) + (-\By I) \Bx } \\
&=
\Bx \wedge \lr{ -\By I }.
\end{aligned}

These duality transformations are motivated by the observation that in the GA form of Maxwell’s equation the magnetic field shows up in its dual form, a bivector. Spelled out in terms of the dual magnetic field, those equations are

\label{eqn:phasorDualMaxwellsGA:360}
\spacegrad \cdot (-\BE I)= – j \omega \BB – \BM

\label{eqn:phasorDualMaxwellsGA:380}
\spacegrad \wedge \BH = j \omega \epsilon_0 \BE I

\label{eqn:phasorDualMaxwellsGA:400}
\spacegrad \wedge (-\BE I) = 0

\label{eqn:phasorDualMaxwellsGA:420}

## Constructing a potential representation.

The starting point of the argument in the text was the observation that the triple product $$\spacegrad \cdot \lr{ \spacegrad \cross \Bx } = 0$$ for any (sufficiently continuous) vector $$\Bx$$. This triple product is a completely antisymmetric sum, and the equivalent statement in GA is $$\spacegrad \wedge \spacegrad \wedge \Bx = 0$$ for any vector $$\Bx$$. This follows from $$\Ba \wedge \Ba = 0$$, true for any vector $$\Ba$$, including the gradient operator $$\spacegrad$$, provided those gradients are acting on a sufficiently continuous blade.

In the absence of electric charges,
\ref{eqn:phasorDualMaxwellsGA:400} shows that the divergence of the dual electric field is zero. It it therefore possible to find a potential $$\BF$$ such that

\label{eqn:phasorDualMaxwellsGA:460}
-\epsilon_0 \BE I = \spacegrad \wedge \BF.

Substituting this \ref{eqn:phasorDualMaxwellsGA:380} gives

\label{eqn:phasorDualMaxwellsGA:480}
\spacegrad \wedge \lr{ \BH + j \omega \BF } = 0.

This relation is a bivector identity with zero, so will be satisfied if

\label{eqn:phasorDualMaxwellsGA:500}
\BH + j \omega \BF = -\spacegrad \phi_m,

for some scalar $$\phi_m$$. Unlike the $$-\epsilon_0 \BE I = \spacegrad \wedge \BF$$ solution to \ref{eqn:phasorDualMaxwellsGA:400}, the grade of $$\phi_m$$ is fixed by the requirement that $$\BE + j \omega \BF$$ is unity (a vector), so
a $$\BE + j \omega \BF = \spacegrad \wedge \psi$$, for a higher grade blade $$\psi$$ would not work, despite satisfying the condition $$\spacegrad \wedge \spacegrad \wedge \psi = 0$$.

Substitution of \ref{eqn:phasorDualMaxwellsGA:500} and \ref{eqn:phasorDualMaxwellsGA:460} into \ref{eqn:phasorDualMaxwellsGA:380} gives

\label{eqn:phasorDualMaxwellsGA:520}
\begin{aligned}
\spacegrad \cdot \lr{ \spacegrad \wedge \BF } &= -\epsilon_0 \BM – j \omega \epsilon_0 \mu_0 \lr{ -\spacegrad \phi_m -j \omega \BF } \\
\end{aligned}

Rearranging gives

\label{eqn:phasorDualMaxwellsGA:540}
\spacegrad^2 \BF + k^2 \BF = -\epsilon_0 \BM + \spacegrad \lr{ \spacegrad \cdot \BF + j \frac{k}{c} \phi_m }.

The fields $$\BF$$ and $$\phi_m$$ are assumed to be phasors, say $$\boldsymbol{\mathcal{A}} = \textrm{Re} \BF e^{j k c t}$$ and $$\varphi = \textrm{Re} \phi_m e^{j k c t}$$. Grouping the scalar and vector potentials into the standard four vector form
$$F^\mu = \lr{\phi_m/c, \BF}$$, and expanding the Lorentz gauge condition

\label{eqn:phasorDualMaxwellsGA:580}
\begin{aligned}
0
&= \partial_\mu \lr{ F^\mu e^{j k c t}} \\
&= \partial_a \lr{ F^a e^{j k c t}} + \inv{c}\PD{t}{} \lr{ \frac{\phi_m}{c}
e^{j k c t}} \\
&= \spacegrad \cdot \BF e^{j k c t} + \inv{c} j k \phi_m e^{j k c t} \\
&= \lr{ \spacegrad \cdot \BF + j k \phi_m/c } e^{j k c t},
\end{aligned}

shows that in
\ref{eqn:phasorDualMaxwellsGA:540}
the quantity in braces is in fact the Lorentz gauge condition, so in the Lorentz gauge, the vector potential satisfies a non-homogeneous Helmholtz equation.

\label{eqn:phasorDualMaxwellsGA:550}
\boxed{
\spacegrad^2 \BF + k^2 \BF = -\epsilon_0 \BM.
}

## Maxwell’s equation in Four vector form

The four vector form of Maxwell’s equation follows from \ref{eqn:phasorDualMaxwellsGA:300} after pre-multiplying by $$\gamma^0$$.

With

\label{eqn:phasorDualMaxwellsGA:620}
F = F^\mu \gamma_\mu = \lr{ \phi_m/c, \BF }

\label{eqn:phasorDualMaxwellsGA:640}
G = \grad \wedge F = – \epsilon_0 \lr{ \BE + c \BB I } I

\label{eqn:phasorDualMaxwellsGA:660}
\grad = \gamma^\mu \partial_\mu = \gamma^0 \lr{ \spacegrad + j k }

\label{eqn:phasorDualMaxwellsGA:680}
M = M^\mu \gamma_\mu = \lr{ c \rho_m, \BM },

Maxwell’s equation is

\label{eqn:phasorDualMaxwellsGA:720}
\boxed{
}

Here $$\setlr{ \gamma_\mu }$$ is used as the basis of the four vector Minkowski space, with $$\gamma_0^2 = -\gamma_k^2 = 1$$ (i.e. $$\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu$$), and $$\gamma_a \gamma_0 = \sigma_a$$ where $$\setlr{ \sigma_a}$$ is the Pauli basic (i.e. standard basis vectors for \R{3}).

Let’s demonstrate this, one piece at a time. Observe that the action of the spacetime gradient on a phasor, assuming that all time dependence is in the exponential, is

\label{eqn:phasorDualMaxwellsGA:740}
\begin{aligned}
\gamma^\mu \partial_\mu \lr{ \psi e^{j k c t} }
&=
\lr{ \gamma^a \partial_a + \gamma_0 \partial_{c t} } \lr{ \psi e^{j k c t} }
\\
&=
\gamma_0 \lr{ \gamma_0 \gamma^a \partial_a + j k } \lr{ \psi e^{j k c t} } \\
&=
\gamma_0 \lr{ \sigma_a \partial_a + j k } \psi e^{j k c t} \\
&=
\gamma_0 \lr{ \spacegrad + j k } \psi e^{j k c t}
\end{aligned}

This allows the operator identification of \ref{eqn:phasorDualMaxwellsGA:660}. The four current portion of the equation comes from

\label{eqn:phasorDualMaxwellsGA:760}
\begin{aligned}
c \rho_m – \BM
&=
\gamma_0 \lr{ \gamma_0 c \rho_m – \gamma_0 \gamma_a \gamma_0 M^a } \\
&=
\gamma_0 \lr{ \gamma_0 c \rho_m + \gamma_a M^a } \\
&=
\gamma_0 \lr{ \gamma_\mu M^\mu } \\
&= \gamma_0 M.
\end{aligned}

Taking the curl of the four potential gives

\label{eqn:phasorDualMaxwellsGA:780}
\begin{aligned}
&=
\lr{ \gamma^a \partial_a + \gamma_0 j k } \wedge \lr{ \gamma_0 \phi_m/c +
\gamma_b F^b } \\
&=
– \sigma_a \partial_a \phi_m/c + \gamma^a \wedge \gamma_b \partial_a F^b – j k
\sigma_b F^b \\
&=
– \sigma_a \partial_a \phi_m/c + \sigma_a \wedge \sigma_b \partial_a F^b – j k
\sigma_b F^b \\
&= \inv{c} \lr{ – \spacegrad \phi_m – j \omega \BF + c \spacegrad \wedge \BF }
\\
&= \epsilon_0 \lr{ c \BB – \BE I } \\
&= – \epsilon_0 \lr{ \BE + c \BB I } I.
\end{aligned}

Substituting all of these into Maxwell’s \ref{eqn:phasorDualMaxwellsGA:300} gives

\label{eqn:phasorDualMaxwellsGA:800}

which recovers \ref{eqn:phasorDualMaxwellsGA:700} as desired.

## Helmholtz equation directly from the GA form.

It is easier to find \ref{eqn:phasorDualMaxwellsGA:550} from the GA form of Maxwell’s \ref{eqn:phasorDualMaxwellsGA:700} than the traditional curl and divergence equations. Note that

\label{eqn:phasorDualMaxwellsGA:820}
\begin{aligned}
&=
&=
+
&=
\end{aligned}

however, the Lorentz gauge condition $$\partial_\mu F^\mu = \grad \cdot F = 0$$ kills the latter term above. This leaves

\label{eqn:phasorDualMaxwellsGA:840}
\begin{aligned}
&=
&=
\gamma_0 \lr{ \spacegrad + j k }
\gamma_0 \lr{ \spacegrad + j k } F \\
&=
\gamma_0^2 \lr{ -\spacegrad + j k }
\lr{ \spacegrad + j k } F \\
&=
-\lr{ \spacegrad^2 + k^2 } F = -\epsilon_0 M.
\end{aligned}

The timelike component of this gives

\label{eqn:phasorDualMaxwellsGA:860}
\lr{ \spacegrad^2 + k^2 } \phi_m = -\epsilon_0 c \rho_m,

and the spacelike components give

\label{eqn:phasorDualMaxwellsGA:880}
\lr{ \spacegrad^2 + k^2 } \BF = -\epsilon_0 \BM,

recovering \ref{eqn:phasorDualMaxwellsGA:550} as desired.

# 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] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.