Stokes Theorem

The Fundamental Theorem of (Geometric) Calculus is a generalization of Stokes theorem to multivector integrals. Notationally, it looks like Stokes theorem with all the dot and wedge products removed. It is worth restating Stokes theorem and all the definitions associated with it for reference

Stokes’ Theorem

For blades $$F \in \bigwedge^{s}$$, and $$m$$ volume element $$d^k \Bx, s < k$$, \begin{equation*} \int_V d^k \Bx \cdot (\boldpartial \wedge F) = \oint_{\partial V} d^{k-1} \Bx \cdot F. \end{equation*} This is a loaded and abstract statement, and requires many definitions to make it useful

• The volume integral is over a $$m$$ dimensional surface (manifold).
• Integration over the boundary of the manifold $$V$$ is indicated by $$\partial V$$.
• This manifold is assumed to be spanned by a parameterized vector $$\Bx(u^1, u^2, \cdots, u^k)$$.
• A curvilinear coordinate basis $$\setlr{ \Bx_i }$$ can be defined on the manifold by
\label{eqn:fundamentalTheoremOfCalculus:40}
\Bx_i \equiv \PD{u^i}{\Bx} \equiv \partial_i \Bx.

• A dual basis $$\setlr{\Bx^i}$$ reciprocal to the tangent vector basis $$\Bx_i$$ can be calculated subject to the requirement $$\Bx_i \cdot \Bx^j = \delta_i^j$$.
• The vector derivative $$\boldpartial$$, the projection of the gradient onto the tangent space of the manifold, is defined by
\label{eqn:fundamentalTheoremOfCalculus:100}
\boldpartial = \Bx^i \partial_i = \sum_{i=1}^k \Bx_i \PD{u^i}{}.

• The volume element is defined by
\label{eqn:fundamentalTheoremOfCalculus:60}
d^k \Bx = d\Bx_1 \wedge d\Bx_2 \cdots \wedge d\Bx_k,

where

\label{eqn:fundamentalTheoremOfCalculus:80}
d\Bx_k = \Bx_k du^k,\qquad \text{(no sum)}.

• The volume element is non-zero on the manifold, or $$\Bx_1 \wedge \cdots \wedge \Bx_k \ne 0$$.
• The surface area element $$d^{k-1} \Bx$$, is defined by
\label{eqn:fundamentalTheoremOfCalculus:120}
d^{k-1} \Bx = \sum_{i = 1}^k (-1)^{k-i} d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k,

where $$\widehat{d\Bx_i}$$ indicates the omission of $$d\Bx_i$$.

• My proof for this theorem was restricted to a simple “rectangular” volume parameterized by the ranges
$$[u^1(0), u^1(1) ] \otimes [u^2(0), u^2(1) ] \otimes \cdots \otimes [u^k(0), u^k(1) ]$$

• The precise meaning that should be given to oriented area integral is
\label{eqn:fundamentalTheoremOfCalculus:140}
\oint_{\partial V} d^{k-1} \Bx \cdot F
=
\sum_{i = 1}^k (-1)^{k-i} \int \evalrange{
\lr{ \lr{ d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k } \cdot F }
}{u^i = u^i(0)}{u^i(1)},

where both the a area form and the blade $$F$$ are evaluated at the end points of the parameterization range.

After the work of stating exactly what is meant by this theorem, most of the proof follows from the fact that for $$s < k$$ the volume curl dot product can be expanded as $$\label{eqn:fundamentalTheoremOfCalculus:160} \int_V d^k \Bx \cdot (\boldpartial \wedge F) = \int_V d^k \Bx \cdot (\Bx^i \wedge \partial_i F) = \int_V \lr{ d^k \Bx \cdot \Bx^i } \cdot \partial_i F.$$ Each of the $$du^i$$ integrals can be evaluated directly, since each of the remaining $$d\Bx_j = du^j \PDi{u^j}{}, i \ne j$$ is calculated with $$u^i$$ held fixed. This allows for the integration over a rectangular'' parameterization region, proving the theorem for such a volume parameterization. A more general proof requires a triangulation of the volume and surface, but the basic principle of the theorem is evident, without that additional work.

Fundamental Theorem of Calculus

There is a Geometric Algebra generalization of Stokes theorem that does not have the blade grade restriction of Stokes theorem. In [2] this is stated as

\label{eqn:fundamentalTheoremOfCalculus:180}
\int_V d^k \Bx \boldpartial F = \oint_{\partial V} d^{k-1} \Bx F.

A similar expression is used in [1] where it is also pointed out there is a variant with the vector derivative acting to the left

\label{eqn:fundamentalTheoremOfCalculus:200}
\int_V F d^k \Bx \boldpartial = \oint_{\partial V} F d^{k-1} \Bx.

In [3] it is pointed out that a bidirectional formulation is possible, providing the most general expression of the Fundamental Theorem of (Geometric) Calculus

\label{eqn:fundamentalTheoremOfCalculus:220}
\boxed{
\int_V F d^k \Bx \boldpartial G = \oint_{\partial V} F d^{k-1} \Bx G.
}

Here the vector derivative acts both to the left and right on $$F$$ and $$G$$. The specific action of this operator is
\label{eqn:fundamentalTheoremOfCalculus:240}
\begin{aligned}
F \boldpartial G
&=
(F \boldpartial) G
+
F (\boldpartial G) \\
&=
(\partial_i F) \Bx^i G
+
F \Bx^i (\partial_i G).
\end{aligned}

The fundamental theorem can be demonstrated by direct expansion. With the vector derivative $$\boldpartial$$ and its partials $$\partial_i$$ acting bidirectionally, that is

\label{eqn:fundamentalTheoremOfCalculus:260}
\begin{aligned}
\int_V F d^k \Bx \boldpartial G
&=
\int_V F d^k \Bx \Bx^i \partial_i G \\
&=
\int_V F \lr{ d^k \Bx \cdot \Bx^i + d^k \Bx \wedge \Bx^i } \partial_i G.
\end{aligned}

Both the reciprocal frame vectors and the curvilinear basis span the tangent space of the manifold, since we can write any reciprocal frame vector as a set of projections in the curvilinear basis

\label{eqn:fundamentalTheoremOfCalculus:280}
\Bx^i = \sum_j \lr{ \Bx^i \cdot \Bx^j } \Bx_j,

so $$\Bx^i \in sectionpan \setlr{ \Bx_j, j \in [1,k] }$$.
This means that $$d^k \Bx \wedge \Bx^i = 0$$, and

\label{eqn:fundamentalTheoremOfCalculus:300}
\begin{aligned}
\int_V F d^k \Bx \boldpartial G
&=
\int_V F \lr{ d^k \Bx \cdot \Bx^i } \partial_i G \\
&=
\sum_{i = 1}^{k}
\int_V
du^1 du^2 \cdots \widehat{ du^i} \cdots du^k
F \lr{
(-1)^{k-i}
\Bx_1 \wedge \Bx_2 \cdots \widehat{\Bx_i} \cdots \wedge \Bx_k } \partial_i G du^i \\
&=
\sum_{i = 1}^{k}
(-1)^{k-i}
\int_{u^1}
\int_{u^2}
\cdots
\int_{u^{i-1}}
\int_{u^{i+1}}
\cdots
\int_{u^k}
\evalrange{ \lr{
F d\Bx_1 \wedge d\Bx_2 \cdots \widehat{d\Bx_i} \cdots \wedge d\Bx_k G
}
}{u^i = u^i(0)}{u^i(1)}.
\end{aligned}

Adding in the same notational sugar that we used in Stokes theorem, this proves the Fundamental theorem \ref{eqn:fundamentalTheoremOfCalculus:220} for “rectangular” parameterizations. Note that such a parameterization need not actually be rectangular.

Example: Application to Maxwell’s equation

{example:fundamentalTheoremOfCalculus:1}

Maxwell’s equation is an example of a first order gradient equation

\label{eqn:fundamentalTheoremOfCalculus:320}
\grad F = \inv{\epsilon_0 c} J.

Integrating over a four-volume (where the vector derivative equals the gradient), and applying the Fundamental theorem, we have

\label{eqn:fundamentalTheoremOfCalculus:340}
\inv{\epsilon_0 c} \int d^4 x J = \oint d^3 x F.

Observe that the surface area element product with $$F$$ has both vector and trivector terms. This can be demonstrated by considering some examples

\label{eqn:fundamentalTheoremOfCalculus:360}
\begin{aligned}
\gamma_{012} \gamma_{01} &\propto \gamma_2 \\
\gamma_{012} \gamma_{23} &\propto \gamma_{023}.
\end{aligned}

On the other hand, the four volume integral of $$J$$ has only trivector parts. This means that the integral can be split into a pair of same-grade equations

\label{eqn:fundamentalTheoremOfCalculus:380}
\begin{aligned}
\inv{\epsilon_0 c} \int d^4 x \cdot J &=
\oint \gpgradethree{ d^3 x F} \\
0 &=
\oint d^3 x \cdot F.
\end{aligned}

The first can be put into a slightly tidier form using a duality transformation
\label{eqn:fundamentalTheoremOfCalculus:400}
\begin{aligned}
&=
-\gpgradethree{ d^3 x I^2 F} \\
&=
\gpgradethree{ I d^3 x I F} \\
&=
(I d^3 x) \wedge (I F).
\end{aligned}

Letting $$n \Abs{d^3 x} = I d^3 x$$, this gives

\label{eqn:fundamentalTheoremOfCalculus:420}
\oint \Abs{d^3 x} n \wedge (I F) = \inv{\epsilon_0 c} \int d^4 x \cdot J.

Note that this normal is normal to a three-volume subspace of the spacetime volume. For example, if one component of that spacetime surface area element is $$\gamma_{012} c dt dx dy$$, then the normal to that area component is $$\gamma_3$$.

A second set of duality transformations

\label{eqn:fundamentalTheoremOfCalculus:440}
\begin{aligned}
n \wedge (IF)
&=
&=
&=
-\gpgradethree{ I (n \cdot F)} \\
&=
-I (n \cdot F),
\end{aligned}

and
\label{eqn:fundamentalTheoremOfCalculus:460}
\begin{aligned}
I d^4 x \cdot J
&=
\gpgradeone{ I d^4 x \cdot J } \\
&=
\gpgradeone{ I d^4 x J } \\
&=
\gpgradeone{ (I d^4 x) J } \\
&=
(I d^4 x) J,
\end{aligned}

can further tidy things up, leaving us with

\label{eqn:fundamentalTheoremOfCalculus:500}
\boxed{
\begin{aligned}
\oint \Abs{d^3 x} n \cdot F &= \inv{\epsilon_0 c} \int (I d^4 x) J \\
\oint d^3 x \cdot F &= 0.
\end{aligned}
}

The Fundamental theorem of calculus immediately provides relations between the Faraday bivector $$F$$ and the four-current $$J$$.

References

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

[2] A. Macdonald. Vector and Geometric Calculus. CreateSpace Independent Publishing Platform, 2012.

[3] Garret Sobczyk and Omar Le\’on S\’anchez. Fundamental theorem of calculus. Advances in Applied Clifford Algebras, 21\penalty0 (1):\penalty0 221–231, 2011. URL http://arxiv.org/abs/0809.4526.

Stokes integrals for Maxwell’s equations in Geometric Algebra

Recall that the relativistic form of Maxwell’s equation in Geometric Algebra is

\label{eqn:maxwellStokes:20}
\grad F = \inv{c \epsilon_0} J.

where $$\grad = \gamma^\mu \partial_\mu$$ is the spacetime gradient, and $$J = (c\rho, \BJ) = J^\mu \gamma_\mu$$ is the four (vector) current density. The pseudoscalar for the space is denoted $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$, where the basis elements satisfy $$\gamma_0^2 = 1 = -\gamma_k^2$$, and a dual basis satisfies $$\gamma_\mu \cdot \gamma^\nu = \delta_\mu^\nu$$. The electromagnetic field $$F$$ is a composite multivector $$F = \BE + I c \BB$$. This is actually a bivector because spatial vectors have a bivector representation in the space time algebra of the form $$\BE = E^k \gamma_k \gamma_0$$.

Previously, I wrote out the Stokes integrals for Maxwell’s equation in GA form using some three parameter spacetime manifold volumes. This time I’m going to use two and three parameter spatial volumes, again with the Geometric Algebra form of Stokes theorem.

Multiplication by a timelike unit vector transforms Maxwell’s equation from their relativistic form. When that vector is the standard basis timelike unit vector $$\gamma_0$$, we obtain Maxwell’s equations from the point of view of a stationary observer

\label{eqn:stokesMaxwellSpaceTimeSplit:40}
\lr{\partial_0 + \spacegrad} \lr{ \BE + c I \BB } = \inv{\epsilon_0 c} \lr{ c \rho – \BJ },

Extracting the scalar, vector, bivector, and trivector grades respectively, we have
\label{eqn:stokesMaxwellSpaceTimeSplit:60}
\begin{aligned}
\spacegrad \cdot \BE &= \frac{\rho}{\epsilon_0} \\
c I \spacegrad \wedge \BB &= -\partial_0 \BE – \inv{\epsilon_0 c} \BJ \\
\spacegrad \wedge \BE &= – I c \partial_0 \BB \\
c I \spacegrad \cdot \BB &= 0.
\end{aligned}

Each of these can be written as a curl equation

\label{eqn:stokesMaxwellSpaceTimeSplit:80}
\boxed{
\begin{aligned}
\spacegrad \wedge (I \BE) &= I \frac{\rho}{\epsilon_0} \\
\inv{\mu_0} \spacegrad \wedge \BB &= \epsilon_0 I \partial_t \BE + I \BJ \\
\spacegrad \wedge \BE &= -I \partial_t \BB \\
\spacegrad \wedge (I \BB) &= 0,
\end{aligned}
}

a form that allows for direct application of Stokes integrals. The first and last of these require a three parameter volume element, whereas the two bivector grade equations can be integrated using either two or three parameter volume elements. Suppose that we have can parameterize the space with parameters $$u, v, w$$, for which the gradient has the representation

\label{eqn:stokesMaxwellSpaceTimeSplit:100}
\spacegrad = \Bx^u \partial_u + \Bx^v \partial_v + \Bx^w \partial_w,

but we integrate over a two parameter subset of this space spanned by $$\Bx(u,v)$$, with area element

\label{eqn:stokesMaxwellSpaceTimeSplit:120}
\begin{aligned}
d^2 \Bx
&= d\Bx_u \wedge d\Bx_v \\
&=
\PD{u}{\Bx}
\wedge
\PD{v}{\Bx}
\,du dv \\
&=
\Bx_u
\wedge
\Bx_v
\,du dv,
\end{aligned}

as illustrated in fig. 1.

fig. 1. Two parameter manifold.

Our curvilinear coordinates $$\Bx_u, \Bx_v, \Bx_w$$ are dual to the reciprocal basis $$\Bx^u, \Bx^v, \Bx^w$$, but we won’t actually have to calculate that reciprocal basis. Instead we need only know that it can be calculated and is defined by the relations $$\Bx_a \cdot \Bx^b = \delta_a^b$$. Knowing that we can reduce (say),

\label{eqn:stokesMaxwellSpaceTimeSplit:140}
\begin{aligned}
d^2 \Bx \cdot ( \spacegrad \wedge \BE )
&=
d^2 \Bx \cdot ( \Bx^a \partial_a \wedge \BE ) \\
&=
(\Bx_u \wedge \Bx_v) \cdot ( \Bx^a \wedge \partial_a \BE ) \,du dv \\
&=
(((\Bx_u \wedge \Bx_v) \cdot \Bx^a) \cdot \partial_a \BE \,du dv \\
&=
d\Bx_u \cdot \partial_v \BE \,dv
-d\Bx_v \cdot \partial_u \BE \,du,
\end{aligned}

Because each of the differentials, for example $$d\Bx_u = (\PDi{u}{\Bx}) du$$, is calculated with the other (i.e.$$v$$) held constant, this is directly integrable, leaving

\label{eqn:stokesMaxwellSpaceTimeSplit:160}
\begin{aligned}
\int d^2 \Bx \cdot ( \spacegrad \wedge \BE )
&=
\int \evalrange{\lr{d\Bx_u \cdot \BE}}{v=0}{v=1}
-\int \evalrange{\lr{d\Bx_v \cdot \BE}}{u=0}{u=1} \\
&=
\oint d\Bx \cdot \BE.
\end{aligned}

That direct integration of one of the parameters, while the others are held constant, is the basic idea behind Stokes theorem.

The pseudoscalar grade Maxwell’s equations from \ref{eqn:stokesMaxwellSpaceTimeSplit:80} require a three parameter volume element to apply Stokes theorem to. Again, allowing for curvilinear coordinates such a differential expands as

\label{eqn:stokesMaxwellSpaceTimeSplit:180}
\begin{aligned}
d^3 \Bx \cdot (\spacegrad \wedge (I\BB))
&=
(( \Bx_u \wedge \Bx_v \wedge \Bx_w ) \cdot \Bx^a ) \cdot \partial_a (I\BB) \,du dv dw \\
&=
(d\Bx_u \wedge d\Bx_v) \cdot \partial_w (I\BB) dw
+(d\Bx_v \wedge d\Bx_w) \cdot \partial_u (I\BB) du
+(d\Bx_w \wedge d\Bx_u) \cdot \partial_v (I\BB) dv.
\end{aligned}

Like the two parameter volume, this is directly integrable

\label{eqn:stokesMaxwellSpaceTimeSplit:200}
\int
d^3 \Bx \cdot (\spacegrad \wedge (I\BB))
=
\int \evalbar{(d\Bx_u \wedge d\Bx_v) \cdot (I\BB) }{\Delta w}
+\int \evalbar{(d\Bx_v \wedge d\Bx_w) \cdot (I\BB)}{\Delta u}
+\int \evalbar{(d\Bx_w \wedge d\Bx_u) \cdot (I\BB)}{\Delta v}.

After some thought (or a craft project such as that of fig. 2) is can be observed that this is conceptually an oriented surface integral

fig. 2. Oriented three parameter surface.

Noting that

\label{eqn:stokesMaxwellSpaceTimeSplit:221}
\begin{aligned}
d^2 \Bx \cdot (I\Bf)
&= \gpgradezero{ d^2 \Bx I B } \\
&= I (d^2\Bx \wedge \Bf)
\end{aligned}

we can now write down the results of application of Stokes theorem to each of Maxwell’s equations in their curl forms

\label{eqn:stokesMaxwellSpaceTimeSplit:220}
\boxed{
\begin{aligned}
\oint d\Bx \cdot \BE &= -I \partial_t \int d^2 \Bx \wedge \BB \\
\inv{\mu_0} \oint d\Bx \cdot \BB &= \epsilon_0 I \partial_t \int d^2 \Bx \wedge \BE + I \int d^2 \Bx \wedge \BJ \\
\oint d^2 \Bx \wedge \BE &= \inv{\epsilon_0} \int (d^3 \Bx \cdot I) \rho \\
\oint d^2 \Bx \wedge \BB &= 0.
\end{aligned}
}

In the three parameter surface integrals the specific meaning to apply to $$d^2 \Bx \wedge \Bf$$ is
\label{eqn:stokesMaxwellSpaceTimeSplit:240}
\oint d^2 \Bx \wedge \Bf
=
\int \evalbar{\lr{d\Bx_u \wedge d\Bx_v \wedge \Bf}}{\Delta w}
+\int \evalbar{\lr{d\Bx_v \wedge d\Bx_w \wedge \Bf}}{\Delta u}
+\int \evalbar{\lr{d\Bx_w \wedge d\Bx_u \wedge \Bf}}{\Delta v}.

Note that in each case only the component of the vector $$\Bf$$ that is projected onto the normal to the area element contributes.

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:

Maxwell’s equations in tensor form with magnetic sources

Following the principle that one should always relate new formalisms to things previously learned, I’d like to know what Maxwell’s equations look like in tensor form when magnetic sources are included. As a verification that the previous Geometric Algebra form of Maxwell’s equation that includes magnetic sources is correct, I’ll start with the GA form of Maxwell’s equation, find the tensor form, and then verify that the vector form of Maxwell’s equations can be recovered from the tensor form.

Tensor form

With four-vector potential $$A$$, and bivector electromagnetic field $$F = \grad \wedge A$$, the GA form of Maxwell’s equation is

\label{eqn:gaMagneticSourcesToTensorToVector:20}
\grad F = \frac{J}{\epsilon_0 c} + M I.

The left hand side can be unpacked into vector and trivector terms $$\grad F = \grad \cdot F + \grad \wedge F$$, which happens to also separate the sources nicely as a side effect

\label{eqn:gaMagneticSourcesToTensorToVector:60}
\grad \cdot F = \frac{J}{\epsilon_0 c}

\label{eqn:gaMagneticSourcesToTensorToVector:80}
\grad \wedge F = M I.

The electric source equation can be unpacked into tensor form by dotting with the four vector basis vectors. With the usual definition $$F^{\alpha \beta} = \partial^\alpha A^\beta – \partial^\beta A^\alpha$$, that is

\label{eqn:gaMagneticSourcesToTensorToVector:100}
\begin{aligned}
\gamma^\mu \cdot \lr{ \grad \cdot F }
&=
\gamma^\mu \cdot \lr{ \grad \cdot \lr{ \grad \wedge A } } \\
&=
\gamma^\mu \cdot \lr{ \gamma^\nu \partial_\nu \cdot
\lr{ \gamma_\alpha \partial^\alpha \wedge \gamma_\beta A^\beta } } \\
&=
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta
} } \partial_\nu \partial^\alpha A^\beta \\
&=
\inv{2}
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta } }
\partial_\nu F^{\alpha \beta} \\
&=
\inv{2} \delta^{\nu \mu}_{[\alpha \beta]} \partial_\nu F^{\alpha \beta} \\
&=
\inv{2} \partial_\nu F^{\nu \mu}

\inv{2} \partial_\nu F^{\mu \nu} \\
&=
\partial_\nu F^{\nu \mu}.
\end{aligned}

So the first tensor equation is

\label{eqn:gaMagneticSourcesToTensorToVector:120}
\boxed{
\partial_\nu F^{\nu \mu} = \inv{c \epsilon_0} J^\mu.
}

To unpack the magnetic source portion of Maxwell’s equation, put it first into dual form, so that it has four vectors on each side

\label{eqn:gaMagneticSourcesToTensorToVector:140}
\begin{aligned}
M
&= – \lr{ \grad \wedge F} I \\
&= -\frac{1}{2} \lr{ \grad F I – F I \grad } \\
&= – \grad \cdot \lr{ F I }.
\end{aligned}

Dotting with $$\gamma^\mu$$ gives

\label{eqn:gaMagneticSourcesToTensorToVector:160}
\begin{aligned}
M^\mu
&= \gamma^\mu \cdot \lr{ \grad \cdot \lr{ – F I } } \\
&= \gamma^\mu \cdot \lr{ \gamma^\nu \partial_\nu \cdot \lr{ -\frac{1}{2}
\gamma^\alpha \wedge \gamma^\beta I F_{\alpha \beta} } } \\
&= -\inv{2}
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{ \gamma^\alpha \wedge \gamma^\beta I } }
}
\partial_\nu F_{\alpha \beta}.
\end{aligned}

This scalar grade selection is a complete antisymmetrization of the indexes

\label{eqn:gaMagneticSourcesToTensorToVector:180}
\begin{aligned}
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{ \gamma^\alpha \wedge \gamma^\beta I } }
}
&=
\gamma^\mu \cdot \lr{ \gamma^\nu \cdot \lr{
\gamma^\alpha \gamma^\beta
\gamma_0 \gamma_1 \gamma_2 \gamma_3
} }
} \\
&=
\gamma_0 \gamma_1 \gamma_2 \gamma_3
\gamma^\mu \gamma^\nu \gamma^\alpha \gamma^\beta
} \\
&=
\delta^{\mu \nu \alpha \beta}_{3 2 1 0} \\
&=
\epsilon^{\mu \nu \alpha \beta },
\end{aligned}

so the magnetic source portion of Maxwell’s equation, in tensor form, is

\label{eqn:gaMagneticSourcesToTensorToVector:200}
\boxed{
\inv{2} \epsilon^{\nu \alpha \beta \mu}
\partial_\nu F_{\alpha \beta}
=
M^\mu.
}

Relating the tensor to the fields

The electromagnetic field has been identified with the electric and magnetic fields by

\label{eqn:gaMagneticSourcesToTensorToVector:220}
F = \boldsymbol{\mathcal{E}} + c \mu_0 \boldsymbol{\mathcal{H}} I ,

or in coordinates

\label{eqn:gaMagneticSourcesToTensorToVector:240}
\inv{2} \gamma_\mu \wedge \gamma_\nu F^{\mu \nu}
= E^a \gamma_a \gamma_0 + c \mu_0 H^a \gamma_a \gamma_0 I.

By forming the dot product sequence $$F^{\alpha \beta} = \gamma^\beta \cdot \lr{ \gamma^\alpha \cdot F }$$, the electric and magnetic field components can be related to the tensor components. The electric field components follow by inspection and are

\label{eqn:gaMagneticSourcesToTensorToVector:260}
E^b = \gamma^0 \cdot \lr{ \gamma^b \cdot F } = F^{b 0}.

The magnetic field relation to the tensor components follow from

\label{eqn:gaMagneticSourcesToTensorToVector:280}
\begin{aligned}
F^{r s}
&= F_{r s} \\
&= \gamma_s \cdot \lr{ \gamma_r \cdot \lr{ c \mu_0 H^a \gamma_a \gamma_0 I
} } \\
&=
c \mu_0 H^a \gpgradezero{ \gamma_s \gamma_r \gamma_a \gamma_0 I } \\
&=
c \mu_0 H^a \gpgradezero{ -\gamma^0 \gamma^1 \gamma^2 \gamma^3
\gamma_s \gamma_r \gamma_a \gamma_0 } \\
&=
c \mu_0 H^a \gpgradezero{ -\gamma^1 \gamma^2 \gamma^3
\gamma_s \gamma_r \gamma_a } \\
&=
– c \mu_0 H^a \delta^{[3 2 1]}_{s r a} \\
&=
c \mu_0 H^a \epsilon_{ s r a }.
\end{aligned}

Expanding this for each pair of spacelike coordinates gives

\label{eqn:gaMagneticSourcesToTensorToVector:320}
F^{1 2} = c \mu_0 H^3 \epsilon_{ 2 1 3 } = – c \mu_0 H^3

\label{eqn:gaMagneticSourcesToTensorToVector:340}
F^{2 3} = c \mu_0 H^1 \epsilon_{ 3 2 1 } = – c \mu_0 H^1

\label{eqn:gaMagneticSourcesToTensorToVector:360}
F^{3 1} = c \mu_0 H^2 \epsilon_{ 1 3 2 } = – c \mu_0 H^2,

or

\label{eqn:gaMagneticSourcesToTensorToVector:380}
\boxed{
\begin{aligned}
E^1 &= F^{1 0} \\
E^2 &= F^{2 0} \\
E^3 &= F^{3 0} \\
H^1 &= -\inv{c \mu_0} F^{2 3} \\
H^2 &= -\inv{c \mu_0} F^{3 1} \\
H^3 &= -\inv{c \mu_0} F^{1 2}.
\end{aligned}
}

Recover the vector equations from the tensor equations

Starting with the non-dual Maxwell tensor equation, expanding the timelike index gives

\label{eqn:gaMagneticSourcesToTensorToVector:480}
\begin{aligned}
\inv{c \epsilon_0} J^0
&= \inv{\epsilon_0} \rho \\
&=
\partial_\nu F^{\nu 0} \\
&=
\partial_1 F^{1 0}
+\partial_2 F^{2 0}
+\partial_3 F^{3 0}
\end{aligned}

This is Gauss’s law

\label{eqn:gaMagneticSourcesToTensorToVector:500}
\boxed{
=
\rho/\epsilon_0.
}

For a spacelike index, any one is representive. Expanding index 1 gives

\label{eqn:gaMagneticSourcesToTensorToVector:520}
\begin{aligned}
\inv{c \epsilon_0} J^1
&= \partial_\nu F^{\nu 1} \\
&= \inv{c} \partial_t F^{0 1}
+ \partial_2 F^{2 1}
+ \partial_3 F^{3 1} \\
&= -\inv{c} E^1
+ \partial_2 (c \mu_0 H^3)
+ \partial_3 (-c \mu_0 H^2) \\
&=
\lr{ -\inv{c} \PD{t}{\boldsymbol{\mathcal{E}}} + c \mu_0 \spacegrad \cross \boldsymbol{\mathcal{H}} } \cdot \Be_1.
\end{aligned}

Extending this to the other indexes and multiplying through by $$\epsilon_0 c$$ recovers the Ampere-Maxwell equation (assuming linear media)

\label{eqn:gaMagneticSourcesToTensorToVector:540}
\boxed{
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}.
}

The expansion of the 0th free (timelike) index of the dual Maxwell tensor equation is

\label{eqn:gaMagneticSourcesToTensorToVector:400}
\begin{aligned}
M^0
&=
\inv{2} \epsilon^{\nu \alpha \beta 0}
\partial_\nu F_{\alpha \beta} \\
&=
-\inv{2} \epsilon^{0 \nu \alpha \beta}
\partial_\nu F_{\alpha \beta} \\
&=
-\inv{2}
\lr{
\partial_1 (F_{2 3} – F_{3 2})
+\partial_2 (F_{3 1} – F_{1 3})
+\partial_3 (F_{1 2} – F_{2 1})
} \\
&=

\lr{
\partial_1 F_{2 3}
+\partial_2 F_{3 1}
+\partial_3 F_{1 2}
} \\
&=

\lr{
\partial_1 (- c \mu_0 H^1 ) +
\partial_2 (- c \mu_0 H^2 ) +
\partial_3 (- c \mu_0 H^3 )
},
\end{aligned}

but $$M^0 = c \rho_m$$, giving us Gauss’s law for magnetism (with magnetic charge density included)

\label{eqn:gaMagneticSourcesToTensorToVector:420}
\boxed{
}

For the spacelike indexes of the dual Maxwell equation, only one need be computed (say 1), and cyclic permutation will provide the rest. That is

\label{eqn:gaMagneticSourcesToTensorToVector:440}
\begin{aligned}
M^1
&= \inv{2} \epsilon^{\nu \alpha \beta 1} \partial_\nu F_{\alpha \beta} \\
&=
\inv{2} \lr{ \partial_2 \lr{F_{3 0} – F_{0 3}} }
+\inv{2} \lr{ \partial_3 \lr{F_{0 2} – F_{0 2}} }
+\inv{2} \lr{ \partial_0 \lr{F_{2 3} – F_{3 2}} } \\
&=
– \partial_2 F^{3 0}
+ \partial_3 F^{2 0}
+ \partial_0 F_{2 3} \\
&=
-\partial_2 E^3 + \partial_3 E^2 + \inv{c} \PD{t}{} \lr{ – c \mu_0 H^1 } \\
&= – \lr{ \spacegrad \cross \boldsymbol{\mathcal{E}} + \mu_0 \PD{t}{\boldsymbol{\mathcal{H}}} } \cdot \Be_1.
\end{aligned}

Extending this to the rest of the coordinates gives the Maxwell-Faraday equation (as extended to include magnetic current density sources)

\label{eqn:gaMagneticSourcesToTensorToVector:460}
\boxed{
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\boldsymbol{\mathcal{M}} – \mu_0 \PD{t}{\boldsymbol{\mathcal{H}}}.
}

This takes things full circle, going from the vector differential Maxwell’s equations, to the Geometric Algebra form of Maxwell’s equation, to Maxwell’s equations in tensor form, and back to the vector form. Not only is the tensor form of Maxwell’s equations with magnetic sources now known, the translation from the tensor and vector formalism has also been verified, and miraculously no signs or factors of 2 were lost or gained in the process.

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.

Phasor form of (extended) Maxwell’s equations in Geometric Algebra

Separate examinations of the phasor form of Maxwell’s equation (with electric charges and current densities), and the Dual Maxwell’s equation (i.e. allowing magnetic charges and currents) were just performed. Here the structure of these equations with both electric and magnetic charges and currents will be examined.

The vector curl and divergence form of Maxwell’s equations are

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

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

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:60}

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:80}

In phasor form these are

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:100}
\spacegrad \cross \BE = – j k c \BB -\BM

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:120}
\spacegrad \cross \BH = \BJ + j k c \BD

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:140}

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:160}

Switching to $$\BE = \BD/\epsilon_0, \BB = \mu_0 \BH$$ fields (even though these aren’t the primary fields in engineering), gives

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:180}
\spacegrad \cross \BE = – j k (c \BB) -\BM

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:200}
\spacegrad \cross (c \BB) = \frac{\BJ}{\epsilon_0 c} + j k \BE

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:220}

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:240}
\spacegrad \cdot (c \BB) = c \rho_m.

Finally, using

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:260}
\Bf \Bg = \Bf \cdot \Bg + I \Bf \cross \Bg,

the divergence and curl contributions of each of the fields can be grouped

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:300}
\spacegrad \BE = \rho/\epsilon_0 – \lr{ j k (c \BB) +\BM} I

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:320}
\spacegrad (c \BB I) = c \rho_m I – \lr{ \frac{\BJ}{\epsilon_0 c} + j k \BE },

or

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:340}
\spacegrad \lr{ \BE + c \BB I }
=
\rho/\epsilon_0 – \lr{ j k (c \BB) +\BM} I
+
c \rho_m I – \lr{ \frac{\BJ}{\epsilon_0 c} + j k \BE }.

Regrouping gives Maxwell’s equations including both electric and magnetic sources
\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:360}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + c \BB I }
=
\inv{\epsilon_0 c} \lr{ c \rho – \BJ }
+ \lr{ c \rho_m – \BM } I.
}

It was observed that these can be put into a tidy four vector form by premultiplying by $$\gamma_0$$, where

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:400}
J = \gamma_\mu J^\mu = \lr{ c \rho, \BJ }

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:420}
M = \gamma_\mu M^\mu = \lr{ c \rho_m, \BM }

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:440}
\grad = \gamma_0 \lr{ \spacegrad + j k } = \gamma^k \partial_k + j k \gamma_0,

That gives

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:460}
\boxed{
\grad \lr{ \BE + c \BB I } = \frac{J}{\epsilon_0 c} + M I.
}

When there were only electric sources, it was observed that potential solutions were of the form $$\BE + c \BB I \propto \grad \wedge A$$, whereas when there was only magnetic sources it was observed that potential solutions were of the form $$\BE + c \BB I \propto (\grad \wedge F) I$$. It seems reasonable to attempt a trial solution that contains both such contributions, say

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:480}
\BE + c \BB I = \grad \wedge A_{\textrm{e}} + \grad \wedge A_{\textrm{m}} I.

Without any loss of generality Lorentz gauge conditions can be imposed on the four-vector fields $$A_{\textrm{e}}, A_{\textrm{m}}$$. Those conditions are

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:500}

Since $$\grad X = \grad \cdot X + \grad \wedge X$$, for any four vector $$X$$, the trial solution \ref{eqn:phasorMaxwellsWithElectricAndMagneticCharges:480} is reduced to

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:520}
\BE + c \BB I = \grad A_{\textrm{e}} + \grad A_{\textrm{m}} I.

Maxwell’s equation is now

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:540}
\begin{aligned}
\frac{J}{\epsilon_0 c} + M I
&=
\grad^2 \lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
\gamma_0 \lr{ \spacegrad + j k }
\gamma_0 \lr{ \spacegrad + j k }
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
\lr{ -\spacegrad + j k }
\lr{ \spacegrad + j k }
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I } \\
&=
\lr{ A_{\textrm{e}} + A_{\textrm{m}} I }.
\end{aligned}

Notice how tidily this separates into vector and trivector components. Those are

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:580}
-\lr{ \spacegrad^2 + k^2 } A_{\textrm{e}} = \frac{J}{\epsilon_0 c}

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:600}
-\lr{ \spacegrad^2 + k^2 } A_{\textrm{m}} = M.

The result is a single Helmholtz equation for each of the electric and magnetic four-potentials, and both can be solved completely independently. This was claimed in class, but now the underlying reason is clear.

Because a single frequency phasor relationship was implied the scalar components of each of these four potentials is determined by the Lorentz gauge condition. For example

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:620}
\begin{aligned}
0
&=
\spacegrad \cdot \lr{ A_{\textrm{e}} e^{j k c t} } \\
&=
\lr{ \gamma^0 \inv{c} \PD{t}{} + \gamma^k \PD{x^k}{} } \cdot
\lr{
\gamma_0 A_{\textrm{e}}^0 e^{j k c t}
+ \gamma_m A_{\textrm{e}}^m e^{j k c t}
} \\
&=
\lr{ \gamma^0 j k + \gamma^r \PD{x^r}{} } \cdot
\lr{
\gamma_0 A_{\textrm{e}}^0
+ \gamma_s A_{\textrm{e}}^s
}
e^{j k c t} \\
&=
\lr{
j k
A_{\textrm{e}}^0
+
\BA_{\textrm{e}}
}
e^{j k c t},
\end{aligned}

so

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:640}
A_{\textrm{e}}^0
=\frac{ j} { k }
\BA_{\textrm{e}}.

The same sort of relationship will apply to the magnetic potential too. This means that the Helmholtz equations can be solved in the three vector space as

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:680}
\lr{ \spacegrad^2 + k^2 } \BA_{\textrm{e}} = -\frac{\BJ}{\epsilon_0 c}

\label{eqn:phasorMaxwellsWithElectricAndMagneticCharges:700}
\lr{ \spacegrad^2 + k^2 } \BA_{\textrm{m}} = -\BM.

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.