## The many faces of Maxwell’s equations

[Click here for a PDF of this post with nicer formatting (including equation numbering and references)]

The following is a possible introduction for a report for a UofT ECE2500 project associated with writing a small book: “Geometric Algebra for Electrical Engineers”. Given the space constraints for the report I may have to drop much of this, but some of the history of Maxwell’s equations may be of interest, so I thought I’d share before the knife hits the latex.

## Goals of the project.

This project had a few goals

1. Perform a literature review of applications of geometric algebra to the study of electromagnetism. Geometric algebra will be defined precisely later, along with bivector, trivector, multivector and other geometric algebra generalizations of the vector.
2. Identify the subset of the literature that had direct relevance to electrical engineering.
3. Create a complete, and as compact as possible, introduction of the prerequisites required
geometric algebra to problems in electromagnetism.

## The many faces of electromagnetism.

There is a long history of attempts to find more elegant, compact and powerful ways of encoding and working with Maxwell’s equations.

### Maxwell’s formulation.

Maxwell [12] employs some differential operators, including the gradient $$\spacegrad$$ and Laplacian $$\spacegrad^2$$, but the divergence and gradient are always written out in full using coordinates, usually in integral form. Reading the original Treatise highlights how important notation can be, as most modern engineering or physics practitioners would find his original work incomprehensible. A nice translation from Maxwell’s notation to the modern Heaviside-Gibbs notation can be found in [16].

### Quaterion representation.

In his second volume [11] the equations of electromagnetism are stated using quaterions (an extension of complex numbers to three dimensions), but quaternions are not used in the work. The modern form of Maxwell’s equations in quaternion form is
\label{eqn:ece2500report:220}
\begin{aligned}
\inv{2} \antisymmetric{ \frac{d}{dr} }{ \BH } – \inv{2} \symmetric{ \frac{d}{dr} } { c \BD } &= c \rho + \BJ \\
\inv{2} \antisymmetric{ \frac{d}{dr} }{ \BE } + \inv{2} \symmetric{ \frac{d}{dr} }{ c \BB } &= 0,
\end{aligned}

where $$\ifrac{d}{dr} = (1/c) \PDi{t}{} + \Bi \PDi{x}{} + \Bj \PDi{y}{} + \Bk \PDi{z}{}$$ [7] acts bidirectionally, and vectors are expressed in terms of the quaternion basis $$\setlr{ \Bi, \Bj, \Bk }$$, subject to the relations $$\Bi^2 = \Bj^2 = \Bk^2 = -1, \quad \Bi \Bj = \Bk = -\Bj \Bi, \quad \Bj \Bk = \Bi = -\Bk \Bj, \quad \Bk \Bi = \Bj = -\Bi \Bk$$.
There is clearly more structure to these equations than the traditional Heaviside-Gibbs representation that we are used to, which says something for the quaternion model. However, this structure requires notation that is arguably non-intuitive. The fact that the quaterion representation was abandoned long ago by most electromagnetism researchers and engineers supports such an argument.

### Minkowski tensor representation.

Minkowski introduced the concept of a complex time coordinate $$x_4 = i c t$$ for special relativity [3]. Such a four-vector representation can be used for many of the relativistic four-vector pairs of electromagnetism, such as the current $$(c\rho, \BJ)$$, and the energy-momentum Lorentz force relations, and can also be applied to Maxwell’s equations
\label{eqn:ece2500report:140}
\sum_{\mu= 1}^4 \PD{x_\mu}{F_{\mu\nu}} = – 4 \pi j_\nu.
\sum_{\lambda\rho\mu=1}^4
\epsilon_{\mu\nu\lambda\rho}
\PD{x_\mu}{F_{\lambda\rho}} = 0,

where
\label{eqn:ece2500report:160}
F
=
\begin{bmatrix}
0 & B_z & -B_y & -i E_x \\
-B_z & 0 & B_x & -i E_y \\
B_y & -B_x & 0 & -i E_z \\
i E_x & i E_y & i E_z & 0
\end{bmatrix}.

A rank-2 complex (Hermitian) tensor contains all six of the field components. Transformation of coordinates for this representation of the field may be performed exactly like the transformation for any other four-vector. This formalism is described nicely in [13], where the structure used is motivated by transformational requirements. One of the costs of this tensor representation is that we loose the clear separation of the electric and magnetic fields that we are so comfortable with. Another cost is that we loose the distinction between space and time, as separate space and time coordinates have to be projected out of a larger four vector. Both of these costs have theoretical benefits in some applications, particularly for high energy problems where relativity is important, but for the low velocity problems near and dear to electrical engineers who can freely treat space and time independently, the advantages are not clear.

### Modern tensor formalism.

The Minkowski representation fell out of favour in theoretical physics, which settled on a real tensor representation that utilizes an explicit metric tensor $$g_{\mu\nu} = \pm \textrm{diag}(1, -1, -1, -1)$$ to represent the complex inner products of special relativity. In this tensor formalism, Maxwell’s equations are also reduced to a set of two tensor relationships ([10], [8], [5]).
\label{eqn:ece2500report:40}
\begin{aligned}
\partial_\mu F^{\mu \nu} &= \mu_0 J^\nu \\
\epsilon^{\alpha \beta \mu \nu} \partial_\beta F_{\mu \nu} &= 0,
\end{aligned}

where $$F^{\mu\nu}$$ is a \textit{real} rank-2 antisymmetric tensor that contains all six electric and magnetic field components, and $$J^\nu$$ is a four-vector current containing both charge density and current density components. \Cref{eqn:ece2500report:40} provides a unified and simpler theoretical framework for electromagnetism, and is used extensively in physics but not engineering.

### Differential forms.

It has been argued that a differential forms treatment of electromagnetism provides some of the same theoretical advantages as the tensor formalism, without the disadvantages of introducing a hellish mess of index manipulation into the mix. With differential forms it is also possible to express Maxwell’s equations as two equations. The free-space differential forms equivalent [4] to the tensor equations is
\label{eqn:ece2500report:60}
\begin{aligned}
d \alpha &= 0 \\
d *\alpha &= 0,
\end{aligned}

where
\label{eqn:ece2500report:180}
\alpha = \lr{ E_1 dx^1 + E_2 dx^2 + E_3 dx^3 }(c dt) + H_1 dx^2 dx^3 + H_2 dx^3 dx^1 + H_3 dx^1 dx^2.

One of the advantages of this representation is that it is valid even for curvilinear coordinate representations, which are handled naturally in differential forms. However, this formalism also comes with a number of costs. One cost (or benefit), like that of the tensor formalism, is that this is implicitly a relativistic approach subject to non-Euclidean orthonormality conditions $$(dx^i, dx^j) = \delta^{ij}, (dx^i, c dt) = 0, (c dt, c dt) = -1$$. Most grievous of the costs is the requirement to use differentials $$dx^1, dx^2, dx^3, c dt$$, instead of a more familar set of basis vectors, even for non-curvilinear coordinates. This requirement is easily viewed as unnatural, and likely one of the reasons that electromagnetism with differential forms has never become popular.

### Vector formalism.

Euclidean vector algebra, in particular the vector algebra and calculus of $$R^3$$, is the de-facto language of electrical engineering for electromagnetism. Maxwell’s equations in the Heaviside-Gibbs vector formalism are
\label{eqn:ece2500report:20}
\begin{aligned}
\spacegrad \cross \BE &= – \PD{t}{\BB} \\
\spacegrad \cross \BH &= \BJ + \PD{t}{\BD} \\
\spacegrad \cdot \BD &= \rho \\
\end{aligned}

We are all intimately familiar with these equations, with the dot and the cross products, and with gradient, divergence and curl operations that are used to express them.
Given how comfortable we are with this mathematical formalism, there has to be a really good reason to switch to something else.

### Space time algebra (geometric algebra).

An alternative to any of the electrodynamics formalisms described above is STA, the Space Time Algebra. STA is a relativistic geometric algebra that allows Maxwell’s equations to be combined into one equation ([2], [6])
\label{eqn:ece2500report:80}

where
\label{eqn:ece2500report:200}
F = \BE + I c \BB \qquad (= \BE + I \eta \BH)

is a bivector field containing both the electric and magnetic field “vectors”, $$\grad = \gamma^\mu \partial_\mu$$ is the spacetime gradient, $$J$$ is a four vector containing electric charge and current components, and $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$ is the spacetime pseudoscalar, the ordered product of the basis vectors $$\setlr{ \gamma_\mu }$$. The STA representation is explicitly relativistic with a non-Euclidean relationships between the basis vectors $$\gamma_0 \cdot \gamma_0 = 1 = -\gamma_k \cdot \gamma_k, \forall k > 0$$. In this formalism “spatial” vectors $$\Bx = \sum_{k>0} \gamma_k \gamma_0 x^k$$ are represented as spacetime bivectors, requiring a small slight of hand when switching between STA notation and conventional vector representation. Uncoincidentally $$F$$ has exactly the same structure as the 2-form $$\alpha$$ above, provided the differential 1-forms $$dx^\mu$$ are replaced by the basis vectors $$\gamma_\mu$$. However, there is a simple complex structure inherent in the STA form that is not obvious in the 2-form equivalent. The bivector representation of the field $$F$$ directly encodes the antisymmetric nature of $$F^{\mu\nu}$$ from the tensor formalism, and the tensor equivalents of most STA results can be calcualted easily.

Having a single PDE for all of Maxwell’s equations allows for direct Green’s function solution of the field, and has a number of other advantages. There is extensive literature exploring selected applications of STA to electrodynamics. Many theoretical results have been derived using this formalism that require significantly more complex approaches using conventional vector or tensor analysis. Unfortunately, much of the STA literature is inaccessible to the engineering student, practising engineers, or engineering instructors. To even start reading the literature, one must learn geometric algebra, aspects of special relativity and non-Euclidean geometry, generalized integration theory, and even some tensor analysis.

### Paravector formalism (geometric algebra).

In the geometric algebra literature, there are a few authors who have endorsed the use of Euclidean geometric algebras for relativistic applications ([1], [14])
These authors use an Euclidean basis “vector” $$\Be_0 = 1$$ for the timelike direction, along with a standard Euclidean basis $$\setlr{ \Be_i }$$ for the spatial directions. A hybrid scalar plus vector representation of four vectors, called paravectors is employed. Maxwell’s equation is written as a multivector equation
\label{eqn:ece2500report:120}
\lr{ \spacegrad + \inv{c} \PD{t}{} } F = J,

where $$J$$ is a multivector source containing both the electric charge and currents, and $$c$$ is the group velocity for the medium (assumed uniform and isometric). $$J$$ may optionally include the (fictitious) magnetic charge and currents useful in antenna theory. The paravector formalism uses a the hybrid electromagnetic field representation of STA above, however, $$I = \Be_1 \Be_2 \Be_3$$ is interpreted as the $$R^3$$ pseudoscalar, the ordered product of the basis vectors $$\setlr{ \Be_i }$$, and $$F$$ represents a multivector with vector and bivector components. Unlike STA where $$\BE$$ and $$\BB$$ (or $$\BH$$) are interpretted as spacetime bivectors, here they are plain old Euclidian vectors in $$R^3$$, entirely consistent with conventional Heaviyside-Gibbs notation. Like the STA Maxwell’s equation, the paravector form is directly invertible using Green’s function techniques, without requiring the solution of equivalent second order potential problems, nor any requirement to take the derivatives of those potentials to determine the fields.

Lorentz transformation and manipulation of paravectors requires a variety of conjugation, real and imaginary operators, unlike STA where such operations have the same complex exponential structure as any 3D rotation expressed in geometric algebra. The advocates of the paravector representation argue that this provides an effective pedagogical bridge from Euclidean geometry to the Minkowski geometry of special relativity. This author agrees that this form of Maxwell’s equations is the natural choice for an introduction to electromagnetism using geometric algebra, but for relativistic operations, STA is a much more natural and less confusing choice.

## Results.

The end product of this project was a fairly small self contained book, titled “Geometric Algebra for Electrical Engineers”. This book includes an introduction to Euclidean geometric algebra focused on $$R^2$$ and $$R^3$$ (64 pages), an introduction to geometric calculus and multivector Green’s functions (64 pages), and applications to electromagnetism (75 pages). This report summarizes results from this book, omitting most derivations, and attempts to provide an overview that may be used as a road map for the book for further exploration. Many of the fundamental results of electromagnetism are derived directly from the geometric algebra form of Maxwell’s equation in a streamlined and compact fashion. This includes some new results, and many of the existing non-relativistic results from the geometric algebra STA and paravector literature. It will be clear to the reader that it is often simpler to have the electric and magnetic on equal footing, and demonstrates this by deriving most results in terms of the total electromagnetic field $$F$$. Many examples of how to extract the conventional electric and magnetic fields from the geometric algebra results expressed in terms of $$F$$ are given as a bridge between the multivector and vector representations.

The aim of this work was to remove some of the prerequisite conceptual roadblocks that make electromagnetism using geometric algebra inaccessbile. In particular, this project explored non-relativistic applications of geometric algebra to electromagnetism. After derivation from the conventional Heaviside-Gibbs representation of Maxwell’s equations, the paravector representation of Maxwell’s equation is used as the starting point for of all subsequent analysis. However, the paravector literature includes a confusing set of conjugation and real and imaginary selection operations that are tailored for relativisitic applications. These are not neccessary for low velocity applications, and have been avoided completely with the aim of making the subject more accessibility to the engineer.

In the book an attempt has been made to avoid introducing as little new notation as possible. For example, some authors use special notation for the bivector valued magnetic field $$I \BB$$, such as $$\boldsymbol{\mathcal{b}}$$ or $$\Bcap$$. Given the inconsistencies in the literature, $$I \BB$$ (or $$I \BH$$) will be used explicitly for the bivector (magnetic) components of the total electromagnetic field $$F$$. In the geometric algebra literature, there are conflicting conventions for the operator $$\spacegrad + (1/c) \PDi{t}{}$$ which we will call the spacetime gradient after the STA equivalent. For examples of different notations for the spacetime gradient, see [9], [1], and [15]. In the book the spacetime gradient is always written out in full to avoid picking from or explaining some of the subtlties of the competing notations.

Some researchers will find it distasteful that STA and relativity have been avoided completely in this book. Maxwell’s equations are inherently relativistic, and STA expresses the relativistic aspects of electromagnetism in an exceptional and beautiful fashion. However, a student of this book will have learned the geometric algebra and calculus prerequisites of STA. This makes the STA literature much more accessible, especially since most of the results in the book can be trivially translated into STA notation.

# References

[1] William Baylis. Electrodynamics: a modern geometric approach, volume 17. Springer Science \& Business Media, 2004.

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

[3] Albert Einstein. Relativity: The special and the general theory, chapter Minkowski’s Four-Dimensional Space. Princeton University Press, 2015. URL http://www.gutenberg.org/ebooks/5001.

[4] H. Flanders. Differential Forms With Applications to the Physical Sciences. Courier Dover Publications, 1989.

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

[6] David Hestenes. Space-time algebra, volume 1. Springer, 1966.

[7] Peter Michael Jack. Physical space as a quaternion structure, i: Maxwell equations. a brief note. arXiv preprint math-ph/0307038, 2003. URL https://arxiv.org/abs/math-ph/0307038.

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

[9] Bernard Jancewicz. Multivectors and Clifford algebra in electrodynamics. World Scientific, 1988.

[10] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980. ISBN 0750627689.

[11] James Clerk Maxwell. A treatise on electricity and magnetism, volume II. Merchant Books, 1881.

[12] James Clerk Maxwell. A treatise on electricity and magnetism, third edition, volume I. Dover publications, 1891.

[13] M. Schwartz. Principles of Electrodynamics. Dover Publications, 1987.

[14] Chappell et al. A simplified approach to electromagnetism using geometric algebra. arXiv preprint arXiv:1010.4947, 2010.

[15] Chappell et al. Geometric algebra for electrical and electronic engineers. 2014.

[16] Chappell et al. Geometric Algebra for Electrical and Electronic Engineers, 2014

## Spherical gradient, divergence, curl and Laplacian

### Unit vectors

Two of the spherical unit vectors we can immediately write by inspection.

\label{eqn:sphericalLaplacian:20}
\begin{aligned}
\rcap &= \Be_1 \sin\theta \cos\phi + \Be_2 \sin\theta \sin\phi + \Be_3 \cos\theta \\
\phicap &= -\Be_1 \sin\theta + \Be_2 \cos\phi
\end{aligned}

We can compute $$\thetacap$$ by utilizing the right hand triplet property

\label{eqn:sphericalLaplacian:40}
\begin{aligned}
\thetacap
&=
\phicap \cross \rcap \\
&=
\begin{vmatrix}
\Be_1 & \Be_2 & \Be_3 \\
-S_\phi & C_\phi & 0 \\
S_\theta C_\phi & S_\theta S_\phi & C_\theta \\
\end{vmatrix} \\
&=
\Be_1 \lr{ C_\theta C_\phi }
+\Be_2 \lr{ C_\theta S_\phi }
+\Be_3 \lr{ -S_\theta \lr{ S_\phi^2 + C_\phi^2 } } \\
&=
\Be_1 \cos\theta \cos\phi
+\Be_2 \cos\theta \sin\phi
-\Be_3 \sin\theta.
\end{aligned}

Here I’ve used $$C_\theta = \cos\theta, S_\phi = \sin\phi, \cdots$$ as a convenient shorthand. Observe that with $$i = \Be_1 \Be_2$$, these unit vectors admit a small factorization that makes further manipulation easier

\label{eqn:sphericalLaplacian:80}
\boxed{
\begin{aligned}
\rcap &= \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta \\
\thetacap &= \cos\theta \Be_1 e^{i\phi} – \sin\theta \Be_3 \\
\phicap &= \Be_2 e^{i\phi}
\end{aligned}
}

It should also be the case that $$\rcap \thetacap \phicap = I$$, where $$I = \Be_1 \Be_2 \Be_3 = \Be_{123}$$ is the \R{3} pseudoscalar, which is straightforward to check

\label{eqn:sphericalLaplacian:60}
\begin{aligned}
\rcap \thetacap \phicap
&=
\lr{ \Be_1 e^{i\phi} \sin\theta + \Be_3 \cos\theta }
\lr{ \cos\theta \Be_1 e^{i\phi} – \sin\theta \Be_3 }
\Be_2 e^{i\phi} \\
&=
\lr{ \sin\theta \cos\theta – \cos\theta \sin\theta + \Be_{31} e^{i\phi} \lr{ \cos^2\theta + \sin^2\theta } }
\Be_2 e^{i\phi} \\
&=
\Be_{31} \Be_2 e^{-i\phi} e^{i\phi} \\
&=
\Be_{123}.
\end{aligned}

This property could also have been used to compute $$\thetacap$$.

To compute the gradient, note that the coordinate vectors for the spherical parameterization are
\label{eqn:sphericalLaplacian:120}
\begin{aligned}
\Bx_r
&= \PD{r}{\Br} \\
&= \PD{r}{\lr{r \rcap}} \\
&= \rcap + r \PD{r}{\rcap} \\
&= \rcap,
\end{aligned}

\label{eqn:sphericalLaplacian:140}
\begin{aligned}
\Bx_\theta
&= \PD{\theta}{\lr{r \rcap} } \\
&= r \PD{\theta}{} \lr{ S_\theta \Be_1 e^{i\phi} + C_\theta \Be_3 } \\
&= r \PD{\theta}{} \lr{ C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3 } \\
&= r \thetacap,
\end{aligned}

\label{eqn:sphericalLaplacian:160}
\begin{aligned}
\Bx_\phi
&= \PD{\phi}{\lr{r \rcap} } \\
&= r \PD{\phi}{} \lr{ S_\theta \Be_1 e^{i\phi} + C_\theta \Be_3 } \\
&= r S_\theta \Be_2 e^{i\phi} \\
&= r \sin\theta \phicap.
\end{aligned}

Since these are all normal, the dual vectors defined by $$\Bx^j \cdot \Bx_k = \delta^j_k$$, can be obtained by inspection
\label{eqn:sphericalLaplacian:180}
\begin{aligned}
\Bx^r &= \rcap \\
\Bx^\theta &= \inv{r} \thetacap \\
\Bx^\phi &= \inv{r \sin\theta} \phicap.
\end{aligned}

\label{eqn:sphericalLaplacian:200}
\Bx^r \PD{r}{} +
\Bx^\theta \PD{\theta}{} +
\Bx^\phi \PD{\phicap}{},

or
\label{eqn:sphericalLaplacian:240}
\boxed{
=
\rcap \PD{r}{} +
\frac{\thetacap}{r} \PD{\theta}{} +
\frac{\phicap}{r\sin\theta} \PD{\phicap}{}.
}

More information on this general dual-vector technique of computing the gradient in curvilinear coordinate systems can be found in
[2].

### Partials

To compute the divergence, curl and Laplacian, we’ll need the partials of each of the unit vectors $$\PDi{\theta}{\rcap}, \PDi{\phi}{\rcap}, \PDi{\theta}{\thetacap}, \PDi{\phi}{\thetacap}, \PDi{\phi}{\phicap}$$.

The $$\thetacap$$ partials are

\label{eqn:sphericalLaplacian:260}
\begin{aligned}
\PD{\theta}{\thetacap}
&=
\PD{\theta}{} \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \\
&=
-S_\theta \Be_1 e^{i\phi} – C_\theta \Be_3 \\
&=
-\rcap,
\end{aligned}

\label{eqn:sphericalLaplacian:280}
\begin{aligned}
\PD{\phi}{\thetacap}
&=
\PD{\phi}{} \lr{
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \\
&=
C_\theta \Be_2 e^{i\phi} \\
&=
C_\theta \phicap.
\end{aligned}

The $$\phicap$$ partials are

\label{eqn:sphericalLaplacian:300}
\begin{aligned}
\PD{\theta}{\phicap}
&=
\PD{\theta}{} \Be_2 e^{i\phi} \\
&=
0.
\end{aligned}

\label{eqn:sphericalLaplacian:320}
\begin{aligned}
\PD{\phi}{\phicap}
&=
\PD{\phi}{} \Be_2 e^{i \phi} \\
&=
-\Be_1 e^{i \phi} \\
&=
-\rcap \gpgradezero{ \rcap \Be_1 e^{i \phi} }
– \thetacap \gpgradezero{ \thetacap \Be_1 e^{i \phi} }
– \phicap \gpgradezero{ \phicap \Be_1 e^{i \phi} } \\
&=
\Be_1 e^{i\phi} S_\theta + \Be_3 C_\theta
} \Be_1 e^{i \phi} }
C_\theta \Be_1 e^{i\phi} – S_\theta \Be_3
} \Be_1 e^{i \phi} } \\
&=
-\rcap \gpgradezero{ e^{-i\phi} S_\theta e^{i \phi} }
– \thetacap \gpgradezero{ C_\theta e^{-i\phi} e^{i \phi} } \\
&=
-\rcap S_\theta
– \thetacap C_\theta.
\end{aligned}

The $$\rcap$$ partials are were computed as a side effect of evaluating $$\Bx_\theta$$, and $$\Bx_\phi$$, and are

\label{eqn:sphericalLaplacian:340}
\PD{\theta}{\rcap}
=
\thetacap,

\label{eqn:sphericalLaplacian:360}
\PD{\phi}{\rcap}
=
S_\theta \phicap.

In summary
\label{eqn:sphericalLaplacian:380}
\boxed{
\begin{aligned}
\partial_{\theta}{\rcap} &= \thetacap \\
\partial_{\phi}{\rcap} &= S_\theta \phicap \\
\partial_{\theta}{\thetacap} &= -\rcap \\
\partial_{\phi}{\thetacap} &= C_\theta \phicap \\
\partial_{\theta}{\phicap} &= 0 \\
\partial_{\phi}{\phicap} &= -\rcap S_\theta – \thetacap C_\theta.
\end{aligned}
}

### Divergence and curl.

The divergence and curl can be computed from the vector product of the spherical coordinate gradient and the spherical representation of a vector. That is

\label{eqn:sphericalLaplacian:400}

\label{eqn:sphericalLaplacian:420}
\begin{aligned}
&=
\lr{
\rcap \partial_{r}
+ \frac{\thetacap}{r} \partial_{\theta}
+ \frac{\phicap}{rS_\theta} \partial_{\phi}
}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&=
\rcap \partial_{r}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&+ \frac{\thetacap}{r} \partial_{\theta}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&+ \frac{\phicap}{rS_\theta} \partial_{\phicap}
\lr{ \rcap A_r + \thetacap A_\theta + \phicap A_\phi} \\
&=
\lr{ \partial_r A_r + \rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi} \\
&+ \frac{1}{r}
\lr{
\thetacap (\partial_\theta \rcap) A_r + \thetacap (\partial_\theta \thetacap) A_\theta + \thetacap (\partial_\theta \phicap) A_\phi
+\thetacap \rcap \partial_\theta A_r + \partial_\theta A_\theta + \thetacap \phicap \partial_\theta A_\phi
} \\
&+ \frac{1}{rS_\theta}
\lr{
\phicap (\partial_\phi \rcap) A_r + \phicap (\partial_\phi \thetacap) A_\theta + \phicap (\partial_\phi \phicap) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta + \partial_\phi A_\phi
} \\
&=
\lr{ \partial_r A_r + \rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi} \\
&+ \frac{1}{r}
\lr{
\thetacap (\thetacap) A_r + \thetacap (-\rcap) A_\theta + \thetacap (0) A_\phi
+\thetacap \rcap \partial_\theta A_r + \partial_\theta A_\theta + \thetacap \phicap \partial_\theta A_\phi
} \\
&+ \frac{1}{r S_\theta}
\lr{
\phicap (S_\theta \phicap) A_r + \phicap (C_\theta \phicap) A_\theta – \phicap (\rcap S_\theta + \thetacap C_\theta) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta + \partial_\phi A_\phi
}.
\end{aligned}

The scalar component of this is the divergence
\label{eqn:sphericalLaplacian:440}
\begin{aligned}
&=
\partial_r A_r
+ \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
\lr{ S_\theta A_r + C_\theta A_\theta + \partial_\phi A_\phi
} \\
&=
\partial_r A_r
+ 2 \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
C_\theta A_\theta
+ \frac{1}{r S_\theta} \partial_\phi A_\phi \\
&=
\partial_r A_r
+ 2 \frac{A_r}{r}
+ \inv{r} \partial_\theta A_\theta
+ \frac{1}{r S_\theta}
C_\theta A_\theta
+ \frac{1}{r S_\theta} \partial_\phi A_\phi,
\end{aligned}

which can be factored as
\label{eqn:sphericalLaplacian:460}
\boxed{
=
\inv{r^2} \partial_r (r^2 A_r)
+ \inv{r S_\theta} \partial_\theta (S_\theta A_\theta)
+ \frac{1}{r S_\theta} \partial_\phi A_\phi.
}

The bivector grade of $$\spacegrad \BA$$ is the bivector curl
\label{eqn:sphericalLaplacian:480}
\begin{aligned}
&=
\lr{
\rcap \thetacap \partial_r A_\theta + \rcap \phicap \partial_r A_\phi
} \\
\lr{
\thetacap (-\rcap) A_\theta
+\thetacap \rcap \partial_\theta A_r + \thetacap \phicap \partial_\theta A_\phi
} \\
\frac{1}{r S_\theta}
\lr{
-\phicap (\rcap S_\theta + \thetacap C_\theta) A_\phi
+\phicap \rcap \partial_\phi A_r + \phicap \thetacap \partial_\phi A_\theta
} \\
&=
\lr{
\rcap \thetacap \partial_r A_\theta – \phicap \rcap \partial_r A_\phi
} \\
\lr{
\rcap \thetacap A_\theta
-\rcap \thetacap \partial_\theta A_r + \thetacap \phicap \partial_\theta A_\phi
} \\
\frac{1}{r S_\theta}
\lr{
-\phicap \rcap S_\theta A_\phi + \thetacap \phicap C_\theta A_\phi
+\phicap \rcap \partial_\phi A_r – \thetacap \phicap \partial_\phi A_\theta
} \\
&=
\thetacap \phicap \lr{
\inv{r S_\theta} C_\theta A_\phi
+\frac{1}{r} \partial_\theta A_\phi
-\frac{1}{r S_\theta} \partial_\phi A_\theta
} \\
-\partial_r A_\phi
+
\frac{1}{r S_\theta}
\lr{
-S_\theta A_\phi
+ \partial_\phi A_r
}
} \\
\partial_r A_\theta
+ \frac{1}{r} A_\theta
– \inv{r} \partial_\theta A_r
} \\
&=
I
\rcap \lr{
\inv{r S_\theta} \partial_\theta (S_\theta A_\phi)
-\frac{1}{r S_\theta} \partial_\phi A_\theta
}
+ I \thetacap \lr{
\frac{1}{r S_\theta} \partial_\phi A_r
-\inv{r} \partial_r (r A_\phi)
}
+ I \phicap \lr{
\inv{r} \partial_r (r A_\theta)
– \inv{r} \partial_\theta A_r
}
\end{aligned}

This gives
\label{eqn:sphericalLaplacian:500}
\boxed{
=
\rcap \lr{
\inv{r S_\theta} \partial_\theta (S_\theta A_\phi)
-\frac{1}{r S_\theta} \partial_\phi A_\theta
}
+ \thetacap \lr{
\frac{1}{r S_\theta} \partial_\phi A_r
-\inv{r} \partial_r (r A_\phi)
}
+ \phicap \lr{
\inv{r} \partial_r (r A_\theta)
– \inv{r} \partial_\theta A_r
}.
}

This and the divergence result above both check against the back cover of [1].

### Laplacian

Using the divergence and curl it’s possible to compute the Laplacian from those, but we saw in cylindrical coordinates that it was much harder to do it that way than to do it directly.

\label{eqn:sphericalLaplacian:540}
\begin{aligned}
&=
\lr{
\rcap \partial_{r} +
\frac{\thetacap}{r} \partial_{\theta} +
\frac{\phicap}{r S_\theta} \partial_{\phi}
}
\lr{
\rcap \partial_{r} \psi
+ \frac{\thetacap}{r} \partial_{\theta} \psi
+ \frac{\phicap}{r S_\theta} \partial_{\phi} \psi
} \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap}{r} \partial_{\theta} \lr{ \rcap \partial_{r} \psi }
+ \frac{\thetacap}{r^2} \partial_{\theta} \lr{ \thetacap \partial_{\theta} \psi }
+ \frac{\thetacap}{r^2} \partial_{\theta} \lr{ \frac{\phicap}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap}{r S_\theta} \partial_{\phi} \lr{ \rcap \partial_{r} \psi }
+ \frac{\phicap}{r^2 S_\theta} \partial_{\phi} \lr{ \thetacap \partial_{\theta} \psi }
+ \frac{\phicap}{r^2 S_\theta^2} \partial_{\phi} \lr{ \phicap \partial_{\phi} \psi } \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap\rcap}{r} \partial_{\theta} \lr{ \partial_{r} \psi }
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{\thetacap \phicap}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap \rcap}{r S_\theta} \partial_{\phi r} \psi
+ \frac{\phicap\thetacap}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi \\
&
\quad + \frac{\thetacap}{r} (\partial_\theta \rcap) \partial_{r} \psi
+ \frac{\thetacap}{r^2} (\partial_\theta \thetacap) \partial_{\theta} \psi
+ \frac{\thetacap}{r^2} (\partial_\theta \phicap) \frac{\phicap}{S_\theta} \partial_{\phi} \psi \\
&
\quad + \frac{\phicap}{r S_\theta} (\partial_\phi \rcap) \partial_{r} \psi
+ \frac{\phicap}{r^2 S_\theta} (\partial_\phi \thetacap) \partial_{\theta} \psi
+ \frac{\phicap}{r^2 S_\theta^2} (\partial_\phi \phicap) \partial_{\phi} \psi \\
&=
\partial_{rr} \psi
+ \rcap \thetacap \partial_r \lr{ \inv{r} \partial_\theta \psi}
+ \rcap \phicap \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi } \\
&
\quad + \frac{\thetacap\rcap}{r} \partial_{\theta} \lr{ \partial_{r} \psi }
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{\thetacap \phicap}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi } \\
&
\quad + \frac{\phicap \rcap}{r S_\theta} \partial_{\phi r} \psi
+ \frac{\phicap\thetacap}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi \\
&
\quad + \frac{\thetacap}{r} (\thetacap) \partial_{r} \psi
+ \frac{\thetacap}{r^2} (-\rcap) \partial_{\theta} \psi
+ \frac{\thetacap}{r^2} (0) \frac{\phicap}{S_\theta} \partial_{\phi} \psi \\
&
\quad + \frac{\phicap}{r S_\theta} (S_\theta \phicap) \partial_{r} \psi
+ \frac{\phicap}{r^2 S_\theta} (C_\theta \phicap) \partial_{\theta} \psi
+ \frac{\phicap}{r^2 S_\theta^2} (-\rcap S_\theta – \thetacap C_\theta) \partial_{\phi} \psi
\end{aligned}

All the bivector factors are expected to cancel out, but this should be checked. Those with an $$\rcap \thetacap$$ factor are

\label{eqn:sphericalLaplacian:560}
\partial_r \lr{ \inv{r} \partial_\theta \psi}
– \frac{1}{r} \partial_{\theta r} \psi
+ \frac{1}{r^2} \partial_{\theta} \psi
=
-\inv{r^2} \partial_\theta \psi
+\inv{r} \partial_{r \theta} \psi
– \frac{1}{r} \partial_{\theta r} \psi
+ \frac{1}{r^2} \partial_{\theta} \psi
= 0,

and those with a $$\thetacap \phicap$$ factor are
\label{eqn:sphericalLaplacian:580}
\frac{1}{r^2} \partial_{\theta} \lr{ \frac{1}{S_\theta} \partial_{\phi} \psi }
– \frac{1}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} C_\theta \partial_{\phi} \psi
=
– \frac{1}{r^2} \frac{C_\theta}{S_\theta^2} \partial_{\phi} \psi
+ \frac{1}{r^2 S_\theta} \partial_{\theta \phi} \psi
– \frac{1}{r^2 S_\theta} \partial_{\phi\theta} \psi
+ \frac{1}{r^2 S_\theta^2} C_\theta \partial_{\phi} \psi
= 0,

and those with a $$\phicap \rcap$$ factor are
\label{eqn:sphericalLaplacian:600}
– \inv{S_\theta} \partial_r \lr{ \inv{r} \partial_\phi \psi }
+ \frac{1}{r S_\theta} \partial_{\phi r} \psi
– \frac{1}{r^2 S_\theta^2} S_\theta \partial_{\phi} \psi
=
\inv{S_\theta} \frac{1}{r^2} \partial_\phi \psi
– \inv{r S_\theta} \partial_{r \phi} \psi
+ \frac{1}{r S_\theta} \partial_{\phi r} \psi
– \frac{1}{r^2 S_\theta} \partial_{\phi} \psi
= 0.

This leaves
\label{eqn:sphericalLaplacian:620}
=
\partial_{rr} \psi
+ \frac{2}{r} \partial_{r} \psi
+ \frac{1}{r^2} \partial_{\theta \theta} \psi
+ \frac{1}{r^2 S_\theta} C_\theta \partial_{\theta} \psi
+ \frac{1}{r^2 S_\theta^2} \partial_{\phi \phi} \psi.

This factors nicely as

\label{eqn:sphericalLaplacian:640}
\boxed{
=
\inv{r^2} \PD{r}{} \lr{ r^2 \PD{r}{ \psi} }
+ \frac{1}{r^2 \sin\theta} \PD{\theta}{} \lr{ \sin\theta \PD{\theta}{ \psi } }
+ \frac{1}{r^2 \sin\theta^2} \PDSq{\phi}{ \psi}
,
}

which checks against the back cover of Jackson. Here it has been demonstrated explicitly that this operator expression is valid for multivector fields $$\psi$$ as well as scalar fields $$\psi$$.

# References

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

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

## 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 https://arxiv.org/abs/0809.4526.

## 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 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.

## 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.