## Solving Maxwell’s equation in freespace: Multivector plane wave representation

The geometric algebra form of Maxwell’s equations in free space (or source free isotopic media with group velocity $$c$$) is the multivector equation
\label{eqn:planewavesMultivector:20}
\lr{ \spacegrad + \inv{c}\PD{t}{} } F(\Bx, t) = 0.

Here $$F = \BE + I c \BB$$ is a multivector with grades 1 and 2 (vector and bivector components). The velocity $$c$$ is called the group velocity since $$F$$, or its components $$\BE, \BH$$ satisfy the wave equation, which can be seen by pre-multiplying with $$\spacegrad – (1/c)\PDi{t}{}$$ to find
\label{eqn:planewavesMultivector:n}
\lr{ \spacegrad^2 – \inv{c^2}\PDSq{t}{} } F(\Bx, t) = 0.

Let’s look at the frequency domain solution of this equation with a presumed phasor representation
\label{eqn:planewavesMultivector:40}
F(\Bx, t) = \textrm{Re} \lr{ F(\Bk) e^{-j \Bk \cdot \Bx + j \omega t} },

where $$j$$ is a scalar imaginary, not necessarily with any geometric interpretation.

Maxwell’s equation reduces to just
\label{eqn:planewavesMultivector:60}
0
=
-j \lr{ \Bk – \frac{\omega}{c} } F(\Bk).

If $$F(\Bk)$$ has a left multivector factor
\label{eqn:planewavesMultivector:80}
F(\Bk) =
\lr{ \Bk + \frac{\omega}{c} } \tilde{F},

where $$\tilde{F}$$ is a multivector to be determined, then
\label{eqn:planewavesMultivector:100}
\begin{aligned}
\lr{ \Bk – \frac{\omega}{c} }
F(\Bk)
&=
\lr{ \Bk – \frac{\omega}{c} }
\lr{ \Bk + \frac{\omega}{c} } \tilde{F} \\
&=
\lr{ \Bk^2 – \lr{\frac{\omega}{c}}^2 } \tilde{F},
\end{aligned}

which is zero if $$\Norm{\Bk} = \ifrac{\omega}{c}$$.

Let $$\kcap = \ifrac{\Bk}{\Norm{\Bk}}$$, and $$\Norm{\Bk} \tilde{F} = F_0 + F_1 + F_2 + F_3$$, where $$F_0, F_1, F_2,$$ and $$F_3$$ are respectively have grades 0,1,2,3. Then
\label{eqn:planewavesMultivector:120}
\begin{aligned}
F(\Bk)
&= \lr{ 1 + \kcap } \lr{ F_0 + F_1 + F_2 + F_3 } \\
&=
F_0 + F_1 + F_2 + F_3
+
\kcap F_0 + \kcap F_1 + \kcap F_2 + \kcap F_3 \\
&=
F_0 + F_1 + F_2 + F_3
+
\kcap F_0 + \kcap \cdot F_1 + \kcap \cdot F_2 + \kcap \cdot F_3
+
\kcap \wedge F_1 + \kcap \wedge F_2 \\
&=
\lr{
F_0 + \kcap \cdot F_1
}
+
\lr{
F_1 + \kcap F_0 + \kcap \cdot F_2
}
+
\lr{
F_2 + \kcap \cdot F_3 + \kcap \wedge F_1
}
+
\lr{
F_3 + \kcap \wedge F_2
}.
\end{aligned}

Since the field $$F$$ has only vector and bivector grades, the grades zero and three components of the expansion above must be zero, or
\label{eqn:planewavesMultivector:140}
\begin{aligned}
F_0 &= – \kcap \cdot F_1 \\
F_3 &= – \kcap \wedge F_2,
\end{aligned}

so
\label{eqn:planewavesMultivector:160}
\begin{aligned}
F(\Bk)
&=
\lr{ 1 + \kcap } \lr{
F_1 – \kcap \cdot F_1 +
F_2 – \kcap \wedge F_2
} \\
&=
\lr{ 1 + \kcap } \lr{
F_1 – \kcap F_1 + \kcap \wedge F_1 +
F_2 – \kcap F_2 + \kcap \cdot F_2
}.
\end{aligned}

The multivector $$1 + \kcap$$ has the projective property of gobbling any leading factors of $$\kcap$$
\label{eqn:planewavesMultivector:180}
\begin{aligned}
(1 + \kcap)\kcap
&= \kcap + 1 \\
&= 1 + \kcap,
\end{aligned}

so for $$F_i \in F_1, F_2$$
\label{eqn:planewavesMultivector:200}
(1 + \kcap) ( F_i – \kcap F_i )
=
(1 + \kcap) ( F_i – F_i )
= 0,

leaving
\label{eqn:planewavesMultivector:220}
F(\Bk)
=
\lr{ 1 + \kcap } \lr{
\kcap \cdot F_2 +
\kcap \wedge F_1
}.

For $$\kcap \cdot F_2$$ to be non-zero $$F_2$$ must be a bivector that lies in a plane containing $$\kcap$$, and $$\kcap \cdot F_2$$ is a vector in that plane that is perpendicular to $$\kcap$$. On the other hand $$\kcap \wedge F_1$$ is non-zero only if $$F_1$$ has a non-zero component that does not lie in along the $$\kcap$$ direction, but $$\kcap \wedge F_1$$, like $$F_2$$ describes a plane that containing $$\kcap$$. This means that having both bivector and vector free variables $$F_2$$ and $$F_1$$ provide more degrees of freedom than required. For example, if $$\BE$$ is any vector, and $$F_2 = \kcap \wedge \BE$$, then
\label{eqn:planewavesMultivector:240}
\begin{aligned}
\lr{ 1 + \kcap }
\kcap \cdot F_2
&=
\lr{ 1 + \kcap }
\kcap \cdot \lr{ \kcap \wedge \BE } \\
&=
\lr{ 1 + \kcap }
\lr{
\BE

\kcap \lr{ \kcap \cdot \BE }
} \\
&=
\lr{ 1 + \kcap }
\kcap \lr{ \kcap \wedge \BE } \\
&=
\lr{ 1 + \kcap }
\kcap \wedge \BE,
\end{aligned}

which has the form $$\lr{ 1 + \kcap } \lr{ \kcap \wedge F_1 }$$, so the solution of the free space Maxwell’s equation can be written
\label{eqn:planewavesMultivector:260}
\boxed{
F(\Bx, t)
=
\textrm{Re} \lr{
\lr{ 1 + \kcap }
\BE\,
e^{-j \Bk \cdot \Bx + j \omega t}
}
,
}

where $$\BE$$ is any vector for which $$\BE \cdot \Bk = 0$$.

## 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.

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

In [1] section 3.2 is a demonstration of the required (curl) form for the magnetic field, and potential form for the electric field.

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

## Maxwell’s equation in GA phasor form.

Maxwell’s equations, omitting magnetic charges and currents, are

\label{eqn:phasorMaxwellsGA:20}

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

\label{eqn:phasorMaxwellsGA:60}

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

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

\label{eqn:phasorMaxwellsGA:140}

\label{eqn:phasorMaxwellsGA:160}

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

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

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

The electric and magnetic field equations, respectively, are

\label{eqn:phasorMaxwellsGA:260}
\spacegrad \BE = \rho/\epsilon_0 -j k c \BB I

\label{eqn:phasorMaxwellsGA:280}
\spacegrad c \BB = \frac{I}{\epsilon_0 c} \BJ + 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:phasorMaxwellsGA:280} by $$I$$), gives Maxwell’s equation in its GA phasor form

\label{eqn:phasorMaxwellsGA:300}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + I c \BB } = \inv{\epsilon_0 c}\lr{c \rho – \BJ}.
}

## 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:phasorMaxwellsGA: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:phasorMaxwellsGA: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:phasorMaxwellsGA:360}
\spacegrad \wedge \BE = – j \omega \BB I

\label{eqn:phasorMaxwellsGA:380}
\spacegrad \cdot \lr{ -\BB I } = \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \BE

\label{eqn:phasorMaxwellsGA:400}

\label{eqn:phasorMaxwellsGA:420}
\spacegrad \wedge (-\BB I) = 0.

## 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 magnetic charges, \ref{eqn:phasorMaxwellsGA:420} shows that the divergence of the dual magnetic field is zero. It it therefore possible to find a potential $$\BA$$ such that

\label{eqn:phasorMaxwellsGA:460}
\BB I = \spacegrad \wedge \BA.

Substituting this into Maxwell-Faraday \ref{eqn:phasorMaxwellsGA:360} gives

\label{eqn:phasorMaxwellsGA:480}
\spacegrad \wedge \lr{ \BE + j \omega \BA } = 0.

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

\label{eqn:phasorMaxwellsGA:500}
\BE + j \omega \BA = -\spacegrad \phi,

for some scalar $$\phi$$. Unlike the $$\BB I = \spacegrad \wedge \BA$$ solution to \ref{eqn:phasorMaxwellsGA:420}, the grade of $$\phi$$ is fixed by the requirement that $$\BE + j \omega \BA$$ is unity (a vector), so a $$\BE + j \omega \BA = \spacegrad \wedge \psi$$, for a higher grade blade $$\psi$$ would not work, despite satisifying the condition $$\spacegrad \wedge \spacegrad \wedge \psi = 0$$.

Substitution of \ref{eqn:phasorMaxwellsGA:500} and \ref{eqn:phasorMaxwellsGA:460} into Ampere’s law \ref{eqn:phasorMaxwellsGA:380} gives

\label{eqn:phasorMaxwellsGA:520}
\begin{aligned}
-\spacegrad \cdot \lr{ \spacegrad \wedge \BA } &= \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \lr{ -\spacegrad \phi -j \omega \BA } \\
\end{aligned}

Rearranging gives

\label{eqn:phasorMaxwellsGA:540}
\spacegrad^2 \BA + k^2 \BA = -\mu_0 \BJ – \spacegrad \lr{ \spacegrad \cdot \BA + j \frac{k}{c} \phi }.

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

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

shows that in \ref{eqn:phasorMaxwellsGA: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:phasorMaxwellsGA:550}
\boxed{
\spacegrad^2 \BA + k^2 \BA = -\mu_0 \BJ.
}

## Maxwell’s equation in Four vector form

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

With

\label{eqn:phasorMaxwellsGA:620}
A = A^\mu \gamma_\mu = \lr{ \phi/c, \BA }

\label{eqn:phasorMaxwellsGA:640}
F = \grad \wedge A = \inv{c} \lr{ \BE + c \BB I }

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

\label{eqn:phasorMaxwellsGA:680}
J = J^\mu \gamma_\mu = \lr{ c \rho, \BJ },

Maxwell’s equation is

\label{eqn:phasorMaxwellsGA:700}
\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:phasorMaxwellsGA: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:phasorMaxwellsGA:660}. The four current portion of the equation comes from

\label{eqn:phasorMaxwellsGA:760}
\begin{aligned}
c \rho – \BJ
&=
\gamma_0 \lr{ \gamma_0 c \rho – \gamma_0 \gamma_a \gamma_0 J^a } \\
&=
\gamma_0 \lr{ \gamma_0 c \rho + \gamma_a J^a } \\
&=
\gamma_0 \lr{ \gamma_\mu J^\mu } \\
&= \gamma_0 J.
\end{aligned}

Taking the curl of the four potential gives

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

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

\label{eqn:phasorMaxwellsGA:800}
\gamma_0 \grad c F = \inv{ \epsilon_0 c } \gamma_0 J,

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

## Helmholtz equation directly from the GA form.

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

\label{eqn:phasorMaxwellsGA:820}
=
=
+
=

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

\label{eqn:phasorMaxwellsGA:840}
\begin{aligned}
&=
&=
\gamma_0 \lr{ \spacegrad + j k }
\gamma_0 \lr{ \spacegrad + j k } A \\
&=
\gamma_0^2 \lr{ -\spacegrad + j k }
\lr{ \spacegrad + j k } A \\
&=
-\lr{ \spacegrad^2 + k^2 } A = \mu_0 J.
\end{aligned}

The timelike component of this gives

\label{eqn:phasorMaxwellsGA:860}
\lr{ \spacegrad^2 + k^2 } \phi = -\rho/\epsilon_0,

and the spacelike components give

\label{eqn:phasorMaxwellsGA:880}
\lr{ \spacegrad^2 + k^2 } \BA = -\mu_0 \BJ,

recovering \ref{eqn:phasorMaxwellsGA: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.