## Some experiments in youtube mathematics videos

A couple years ago I was curious how easy it would be to use a graphics tablet as a virtual chalkboard, and produced a handful of very rough YouTube videos to get a feel for the basics of streaming and video editing (much of which I’ve now forgotten how to do). These were the videos in chronological order:

• Introduction to Geometric (Clifford) Algebra.Introduction to Geometric (Clifford) algebra. Interpretation of products of unit vectors, rules for reducing products of unit vectors, and the axioms that justify those rules.
• Geometric Algebra: dot, wedge, cross and vector products.Geometric (Clifford) Algebra introduction, showing the relation between the vector product dot and wedge products, and the cross product.
• Solution of two line intersection using geometric algebra.
• Linear system solution using the wedge product.. This video provides a standalone introduction to the wedge product, the geometry of the wedge product and some properties, and linear system solution as a sample application. In this video the wedge product is introduced independently of any geometric (Clifford) algebra, as an antisymmetric and associative operator. You’ll see that we get Cramer’s rule for free from this solution technique.
• Exponential form of vector products in geometric algebra.In this video, I discussed the exponential form of the product of two vectors.

I showed an example of how two unit vectors, each rotations of zcap orthonormal $$\mathbb{R}^3$$ planes, produce a “complex” exponential in the plane that spans these two vectors.

• Velocity and acceleration in cylindrical coordinates using geometric algebra.I derived the cylindrical coordinate representations of the velocity and acceleration vectors, showing the radial and azimuthal components of each vector.

I also showed how these are related to the dot and wedge product with the radial unit vector.

• Duality transformations in geometric algebra.Duality transformations (pseudoscalar multiplication) will be demonstrated in $$\mathbb{R}^2$$ and $$\mathbb{R}^3$$.

A polar parameterized vector in $$\mathbb{R}^2$$, written in complex exponential form, is multiplied by a unit pseudoscalar for the x-y plane. We see that the result is a vector normal to that vector, with the direction of the normal dependent on the order of multiplication, and the orientation of the pseudoscalar used.

In $$\mathbb{R}^3$$ we see that a vector multiplied by a pseudoscalar yields the bivector that represents the plane that is normal to that vector. The sign of that bivector (or its cyclic orientation) depends on the orientation of the pseudoscalar. The order of multiplication was not mentioned in this case since the $$\mathbb{R}^3$$ pseudoscalar commutes with any grade object (assumed, not proved). An example of a vector with two components in a plane, multiplied by a pseudoscalar was also given, which allowed for a visualization of the bivector that is normal to the original vector.

• Math bait and switch: Fractional integer exponents.When I was a kid, my dad asked me to explain fractional exponents, and perhaps any non-positive integer exponents, to him. He objected to the idea of multiplying something by itself $$1/2$$ times.

I failed to answer the question to his satisfaction. My own son is now reviewing the rules of exponentiation, and it occurred to me (30 years later) why my explanation to Dad failed.

Essentially, there’s a small bait and switch required, and my dad didn’t fall for it.

The meaning that my dad gave to exponentiation was that $$x^n$$ equals $$x$$ times itself $$n$$ times.

Using this rule, it is easy to demonstrate that $$x^a x^b = x^{a + b}$$, and this can be used to justify expressions like $$x^{1/2}$$. However, doing this really means that we’ve switched the definition of exponential, defining an exponential as any number that satisfies the relationship:

$$x^a x^b = x^{a+b}$$,

where $$x^1 = x$$. This slight of hand is required to give meaning to $$x^{1/2}$$ or other exponentials where the exponential argument is any non-positive integer.

Of these videos I just relistened to the wedge product episode, as I had a new lone comment on it, and I couldn’t even remember what I had said. It wasn’t completely horrible, despite the low tech. I was, however, very surprised how soft and gentle my voice was. When I am talking math in person, I get very animated, but attempting to manage the tech was distracting and all the excitement that I’d normally have was obliterated.

I’d love to attempt a manim based presentation of some of this material, but suspect if I do something completely scripted like that, I may not be a very good narrator.

## Curvilinear coordinates and gradient in spacetime, and reciprocal frames.

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

## Motivation.

I started pondering some aspects of spacetime integration theory, and found that there were some aspects of the concepts of reciprocal frames that were not clear to me. In the process of sorting those ideas out for myself, I wrote up the following notes.

In the notes below, I will introduce the many of the prerequisite ideas that are needed to express and apply the fundamental theorem of geometric calculus in a 4D relativistic context. The focus will be the Dirac’s algebra of special relativity, known as STA (Space Time Algebra) in geometric algebra parlance. If desired, it should be clear how to apply these ideas to lower or higher dimensional spaces, and to plain old Euclidean metrics.

### On notation.

In Euclidean space we use bold face reciprocal frame vectors $$\Bx^i \cdot \Bx_j = {\delta^i}_j$$, which nicely distinguishes them from the generalized coordinates $$x_i, x^j$$ associated with the basis or the reciprocal frame, that is
\label{eqn:reciprocalblog:640}
\Bx = x^i \Bx_i = x_j \Bx^j.

On the other hand, it is conventional to use non-bold face for both the four-vectors and their coordinates in STA, such as the following standard basis decomposition
\label{eqn:reciprocalblog:660}
x = x^\mu \gamma_\mu = x_\mu \gamma^\mu.

If we use non-bold face $$x^\mu, x_\nu$$ for the coordinates with respect to a specified frame, then we cannot also use non-bold face for the curvilinear basis vectors.

To resolve this notational ambiguity, I’ve chosen to use bold face $$\Bx^\mu, \Bx_\nu$$ symbols as the curvilinear basis elements in this relativistic context, as we do for Euclidean spaces.

## Definition 1.1: Standard Dirac basis.

The Dirac basis elements are $$\setlr{ \gamma_0, \gamma_1, \gamma_2, \gamma_3 }$$, satisfying
\label{eqn:reciprocalblog:1940}
\gamma_0^2 = 1 = -\gamma_k^2, \quad \forall k = 1,2,3,

and
\label{eqn:reciprocalblog:740}
\gamma_\mu \cdot \gamma_\nu = 0, \quad \forall \mu \ne \nu.

A conventional way of summarizing these orthogonality relationships is $$\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu}$$, where $$\eta_{\mu\nu}$$ are the elements of the metric $$G = \text{diag}(+,-,-,-)$$.

## Definition 1.2: Reciprocal basis for the standard Dirac basis.

We define a reciprocal basis $$\setlr{ \gamma^0, \gamma^1, \gamma^2, \gamma^3}$$ satisfying $$\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu, \forall \mu,\nu \in 0,1,2,3$$.

## Theorem 1.1: Reciprocal basis uniqueness.

This reciprocal basis is unique, and for our choice of metric has the values
\label{eqn:reciprocalblog:1960}
\gamma^0 = \gamma_0, \quad \gamma^k = -\gamma_k, \quad \forall k = 1,2,3.

Proof is left to the reader.

## Definition 1.3: Coordinates.

We define the coordinates of a vector with respect to the standard basis as $$x^\mu$$ satisfying
\label{eqn:reciprocalblog:1980}
x = x^\mu \gamma_\mu,

and define the coordinates of a vector with respect to the reciprocal basis as $$x_\mu$$ satisfying
\label{eqn:reciprocalblog:2000}
x = x_\mu \gamma^\mu,

## Theorem 1.2: Coordinates.

Given the definitions above, we may compute the coordinates of a vector, simply by dotting with the basis elements
\label{eqn:reciprocalblog:2020}
x^\mu = x \cdot \gamma^\mu,

and
\label{eqn:reciprocalblog:2040}
x_\mu = x \cdot \gamma_\mu,

### Start proof:

This follows by straightforward computation
\label{eqn:reciprocalblog:840}
\begin{aligned}
x \cdot \gamma^\mu
&=
\lr{ x^\nu \gamma_\nu } \cdot \gamma^\mu \\
&=
x^\nu \lr{ \gamma_\nu \cdot \gamma^\mu } \\
&=
x^\nu {\delta_\nu}^\mu \\
&=
x^\mu,
\end{aligned}

and
\label{eqn:reciprocalblog:860}
\begin{aligned}
x \cdot \gamma_\mu
&=
\lr{ x_\nu \gamma^\nu } \cdot \gamma_\mu \\
&=
x_\nu \lr{ \gamma^\nu \cdot \gamma_\mu } \\
&=
x_\nu {\delta^\nu}_\mu \\
&=
x_\mu.
\end{aligned}

## Derivative operators.

We’d like to determine the form of the (spacetime) gradient operator. The gradient can be defined in terms of coordinates directly, but we choose an implicit definition, in terms of the directional derivative.

## Definition 1.4: Directional derivative and gradient.

Let $$F = F(x)$$ be a four-vector parameterized multivector. The directional derivative of $$F$$ with respect to the (four-vector) direction $$a$$ is denoted
\label{eqn:reciprocalblog:2060}
\lr{ a \cdot \grad } F = \lim_{\epsilon \rightarrow 0} \frac{ F(x + \epsilon a) – F(x) }{ \epsilon },

where $$\grad$$ is called the space time gradient.

The standard basis representation of the gradient is
\label{eqn:reciprocalblog:2080}

where
\label{eqn:reciprocalblog:2100}
\partial_\mu = \PD{x^\mu}{}.

### Start proof:

The Dirac gradient pops naturally out of the coordinate representation of the directional derivative, as we can see by expanding $$F(x + \epsilon a)$$ in Taylor series
\label{eqn:reciprocalblog:900}
\begin{aligned}
F(x + \epsilon a)
&= F(x) + \epsilon \frac{dF(x + \epsilon a)}{d\epsilon} + O(\epsilon^2) \\
&= F(x) + \epsilon \PD{\lr{x^\mu + \epsilon a^\mu}}{F} \PD{\epsilon}{\lr{x^\mu + \epsilon a^\mu}} \\
&= F(x) + \epsilon \PD{\lr{x^\mu + \epsilon a^\mu}}{F} a^\mu.
\end{aligned}

The directional derivative is
\label{eqn:reciprocalblog:920}
\begin{aligned}
\lim_{\epsilon \rightarrow 0}
\frac{F(x + \epsilon a) – F(x)}{\epsilon}
&=
\lim_{\epsilon \rightarrow 0}\,
a^\mu
\PD{\lr{x^\mu + \epsilon a^\mu}}{F} \\
&=
a^\mu
\PD{x^\mu}{F} \\
&=
\lr{a^\nu \gamma_\nu} \cdot \gamma^\mu \PD{x^\mu}{F} \\
&=
a \cdot \lr{ \gamma^\mu \partial_\mu } F.
\end{aligned}

## Curvilinear bases.

Curvilinear bases are the foundation of the fundamental theorem of multivector calculus. This form of integral calculus is defined over parameterized surfaces (called manifolds) that satisfy some specific non-degeneracy and continuity requirements.

A parameterized vector $$x(u,v, \cdots w)$$ can be thought of as tracing out a hypersurface (curve, surface, volume, …), where the dimension of the hypersurface depends on the number of parameters. At each point, a bases can be constructed from the differentials of the parameterized vector. Such a basis is called the tangent space to the surface at the point in question. Our curvilinear bases will be related to these differentials. We will also be interested in a dual basis that is restricted to the span of the tangent space. This dual basis will be called the reciprocal frame, and line the basis of the tangent space itself, also varies from point to point on the surface.

Fig 1a. One parameter curve, with illustration of tangent space along the curve.

Fig 1b. Two parameter surface, with illustration of tangent space along the surface.

One and two parameter spaces are illustrated in fig. 1a, and 1b.  The tangent space basis at a specific point of a two parameter surface, $$x(u^0, u^1)$$, is illustrated in fig. 1. The differential directions that span the tangent space are
\label{eqn:reciprocalblog:1040}
\begin{aligned}
d\Bx_0 &= \PD{u^0}{x} du^0 \\
d\Bx_1 &= \PD{u^1}{x} du^1,
\end{aligned}

and the tangent space itself is $$\mbox{Span}\setlr{ d\Bx_0, d\Bx_1 }$$. We may form an oriented surface area element $$d\Bx_0 \wedge d\Bx_1$$ over this surface.

Fig 2. Two parameter surface.

Tangent spaces associated with 3 or more parameters cannot be easily visualized in three dimensions, but the idea generalizes algebraically without trouble.

## Definition 1.5: Tangent basis and space.

Given a parameterization $$x = x(u^0, \cdots, u^N)$$, where $$N < 4$$, the span of the vectors
\label{eqn:reciprocalblog:2120}
\Bx_\mu = \PD{u^\mu}{x},

is called the tangent space for the hypersurface associated with the parameterization, and it’s basis is
$$\setlr{ \Bx_\mu }$$.

Later we will see that parameterization constraints must be imposed, as not all surfaces generated by a set of parameterizations are useful for integration theory. In particular, degenerate parameterizations for which the wedge products of the tangent space basis vectors are zero, or those wedge products cannot be inverted, are not physically meaningful. Properly behaved surfaces of this sort are called manifolds.

Having introduced curvilinear coordinates associated with a parameterization, we can now determine the form of the gradient with respect to a parameterization of spacetime.

## Theorem 1.4: Gradient, curvilinear representation.

Given a spacetime parameterization $$x = x(u^0, u^1, u^2, u^3)$$, the gradient with respect to the parameters $$u^\mu$$ is
\label{eqn:reciprocalblog:2140}
\PD{u^\mu}{},

where
\label{eqn:reciprocalblog:2160}

The vectors $$\Bx^\mu$$ are called the reciprocal frame vectors, and the ordered set $$\setlr{ \Bx^0, \Bx^1, \Bx^2, \Bx^3 }$$ is called the reciprocal basis.It is convenient to define $$\partial_\mu \equiv \PDi{u^\mu}{}$$, so that the gradient can be expressed in mixed index representation
\label{eqn:reciprocalblog:2180}

This introduces some notational ambiguity, since we used $$\partial_\mu = \PDi{x^\mu}{}$$ for the standard basis derivative operators too, but we will be careful to be explicit when there is any doubt about what is intended.

### Start proof:

The proof follows by application of the chain rule.
\label{eqn:reciprocalblog:960}
\begin{aligned}
&=
\gamma^\alpha \PD{x^\alpha}{F} \\
&=
\gamma^\alpha
\PD{x^\alpha}{u^\mu}
\PD{u^\mu}{F} \\
&=
\lr{ \grad u^\mu } \PD{u^\mu}{F} \\
&=
\Bx^\mu \PD{u^\mu}{F}.
\end{aligned}

## Theorem 1.5: Reciprocal relationship.

The vectors $$\Bx^\mu = \grad u^\mu$$, and $$\Bx_\mu = \PDi{u^\mu}{x}$$ satisfy the reciprocal relationship
\label{eqn:reciprocalblog:2200}
\Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu.

### Start proof:

\label{eqn:reciprocalblog:1020}
\begin{aligned}
\Bx^\mu \cdot \Bx_\nu
&=
\PD{u^\nu}{x} \\
&=
\lr{
\gamma^\alpha \PD{x^\alpha}{u^\mu}
}
\cdot
\lr{
\PD{u^\nu}{x^\beta} \gamma_\beta
} \\
&=
{\delta^\alpha}_\beta \PD{x^\alpha}{u^\mu}
\PD{u^\nu}{x^\beta} \\
&=
\PD{x^\alpha}{u^\mu} \PD{u^\nu}{x^\alpha} \\
&=
\PD{u^\nu}{u^\mu} \\
&=
{\delta^\mu}_\nu
.
\end{aligned}

### End proof.

It is instructive to consider an example. Here is a parameterization that scales the proper time parameter, and uses polar coordinates in the $$x-y$$ plane.

## Problem: Compute the curvilinear and reciprocal basis.

Given
\label{eqn:reciprocalblog:2360}
x(t,\rho,\theta,z) = c t \gamma_0 + \gamma_1 \rho e^{i \theta} + z \gamma_3,

where $$i = \gamma_1 \gamma_2$$, compute the curvilinear frame vectors and their reciprocals.

The frame vectors are all easy to compute
\label{eqn:reciprocalblog:1180}
\begin{aligned}
\Bx_0 &= \PD{t}{x} = c \gamma_0 \\
\Bx_1 &= \PD{\rho}{x} = \gamma_1 e^{i \theta} \\
\Bx_2 &= \PD{\theta}{x} = \rho \gamma_1 \gamma_1 \gamma_2 e^{i \theta} = – \rho \gamma_2 e^{i \theta} \\
\Bx_3 &= \PD{z}{x} = \gamma_3.
\end{aligned}

The $$\Bx_1$$ vector is radial, $$\Bx^2$$ is perpendicular to that tangent to the same unit circle, as plotted in fig 3.

Fig3: Tangent space direction vectors.

All of these particular frame vectors happen to be mutually perpendicular, something that will not generally be true for a more arbitrary parameterization.

To compute the reciprocal frame vectors, we must express our parameters in terms of $$x^\mu$$ coordinates, and use implicit integration techniques to deal with the coupling of the rotational terms. First observe that
\label{eqn:reciprocalblog:1200}
\gamma_1 e^{i\theta}
= \gamma_1 \lr{ \cos\theta + \gamma_1 \gamma_2 \sin\theta }
= \gamma_1 \cos\theta – \gamma_2 \sin\theta,

so
\label{eqn:reciprocalblog:1220}
\begin{aligned}
x^0 &= c t \\
x^1 &= \rho \cos\theta \\
x^2 &= -\rho \sin\theta \\
x^3 &= z.
\end{aligned}

We can easily evaluate the $$t, z$$ gradients
\label{eqn:reciprocalblog:1240}
\begin{aligned}
\grad t &= \frac{\gamma^1 }{c} \\
\end{aligned}

but the $$\rho, \theta$$ gradients are not as easy. First writing
\label{eqn:reciprocalblog:1260}
\rho^2 = \lr{x^1}^2 + \lr{x^2}^2,

we find
\label{eqn:reciprocalblog:1280}
\begin{aligned}
&= 2 \rho \lr{ \cos\theta \gamma^1 – \sin\theta \gamma^2 } \\
&= 2 \rho \gamma^1 \lr{ \cos\theta – \gamma_1 \gamma^2 \sin\theta } \\
&= 2 \rho \gamma^1 e^{i\theta},
\end{aligned}

so
\label{eqn:reciprocalblog:1300}

For the $$\theta$$ gradient, we can write
\label{eqn:reciprocalblog:1320}
\tan\theta = -\frac{x^2}{x^1},

so
\label{eqn:reciprocalblog:1340}
\begin{aligned}
&= -\frac{\gamma^2}{x^1} – x^2 \frac{-\gamma^1}{\lr{x^1}^2} \\
&= \inv{\lr{x^1}^2} \lr{ – \gamma^2 x^1 + \gamma^1 x^2 } \\
&= \frac{\rho}{\rho^2 \cos^2\theta } \lr{ – \gamma^2 \cos\theta – \gamma^1 \sin\theta } \\
&= -\frac{1}{\rho \cos^2\theta } \gamma^2 \lr{ \cos\theta + \gamma_2 \gamma^1 \sin\theta } \\
&= -\frac{\gamma^2 e^{i\theta} }{\rho \cos^2\theta },
\end{aligned}

or
\label{eqn:reciprocalblog:1360}

In summary,
\label{eqn:reciprocalblog:1380}
\begin{aligned}
\Bx^0 &= \frac{\gamma^0}{c} \\
\Bx^1 &= \gamma^1 e^{i\theta} \\
\Bx^2 &= -\inv{\rho} \gamma^2 e^{i\theta} \\
\Bx^3 &= \gamma^3.
\end{aligned}

Despite being a fairly simple parameterization, it was still fairly difficult to solve for the gradients when the parameterization introduced coupling between the coordinates. In this particular case, we could have solved for the parameters in terms of the coordinates (but it was easier not to), but that will not generally be true. We want a less labor intensive strategy to find the reciprocal frame. When we have a full parameterization of spacetime, then we can do this with nothing more than a matrix inversion.

## Theorem 1.6: Reciprocal frame matrix equations.

Given a spacetime basis $$\setlr{\Bx_0, \cdots \Bx_3}$$, let $$[\Bx_\mu]$$ and $$[\Bx^\nu]$$ be column matrices with the coordinates of these vectors and their reciprocals, with respect to the standard basis $$\setlr{\gamma_0, \gamma_1, \gamma_2, \gamma_3 }$$. Let
\label{eqn:reciprocalblog:2220}
A =
\begin{bmatrix}
[\Bx_0] & \cdots & [\Bx_{3}]
\end{bmatrix}
X =
\begin{bmatrix}
[\Bx^0] & \cdots & [\Bx^{3}]
\end{bmatrix}.

The coordinates of the reciprocal frame vectors can be found by solving
\label{eqn:reciprocalblog:2240}
A^\T G X = 1,

where $$G = \text{diag}(1,-1,-1,-1)$$ and the RHS is an $$4 \times 4$$ identity matrix.

### Start proof:

Let $$\Bx_\mu = {a_\mu}^\alpha \gamma_\alpha, \Bx^\nu = b^{\nu\beta} \gamma_\beta$$, so that
\label{eqn:reciprocalblog:140}
A =
\begin{bmatrix}
{a_\nu}^\mu
\end{bmatrix},

and
\label{eqn:reciprocalblog:160}
X =
\begin{bmatrix}
b^{\nu\mu}
\end{bmatrix},

where $$\mu \in [0,3]$$ are the row indexes and $$\nu \in [0,N-1]$$ are the column indexes. The reciprocal frame satisfies $$\Bx_\mu \cdot \Bx^\nu = {\delta_\mu}^\nu$$, which has the coordinate representation of
\label{eqn:reciprocalblog:180}
\begin{aligned}
\Bx_\mu \cdot \Bx^\nu
&=
\lr{
{a_\mu}^\alpha \gamma_\alpha
}
\cdot
\lr{
b^{\nu\beta} \gamma_\beta
} \\
&=
{a_\mu}^\alpha
\eta_{\alpha\beta}
b^{\nu\beta} \\
&=
{[A^\T G B]_\mu}^\nu,
\end{aligned}

where $$\mu$$ is the row index and $$\nu$$ is the column index.

## Problem: Matrix inversion reciprocals.

For the parameterization of \ref{eqn:reciprocalblog:2360}, find the reciprocal frame vectors by matrix inversion.

We expanded $$\Bx_1$$ explicitly in \ref{eqn:reciprocalblog:1200}. Doing the same for $$\Bx_2$$, we have
\label{eqn:reciprocalblog:1201}
\Bx_2 =
-\rho \gamma_2 e^{i\theta}
= -\rho \gamma_2 \lr{ \cos\theta + \gamma_1 \gamma_2 \sin\theta }
= – \rho \lr{ \gamma_2 \cos\theta + \gamma_1 \sin\theta}.

Reading off the coordinates of our frame vectors, we have
\label{eqn:reciprocalblog:1400}
X =
\begin{bmatrix}
c & 0 & 0 & 0 \\
0 & C & -\rho S & 0 \\
0 & -S & -\rho C & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix},

where $$C = \cos\theta$$ and $$S = \sin\theta$$. We want
\label{eqn:reciprocalblog:1420}
Y =
{\begin{bmatrix}
c & 0 & 0 & 0 \\
0 & -C & S & 0 \\
0 & \rho S & \rho C & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}}^{-1}
=
\begin{bmatrix}
\inv{c} & 0 & 0 & 0 \\
0 & -C & \frac{S}{\rho} & 0 \\
0 & S & \frac{C}{\rho} & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}.

We can read off the coordinates of the reciprocal frame vectors
\label{eqn:reciprocalblog:1440}
\begin{aligned}
\Bx^0 &= \inv{c} \gamma_0 \\
\Bx^1 &= -\cos\theta \gamma_1 + \sin\theta \gamma_2 \\
\Bx^2 &= \inv{\rho} \lr{ \sin\theta \gamma_1 + \cos\theta \gamma_2 } \\
\Bx^3 &= -\gamma_3.
\end{aligned}

Factoring out $$\gamma^1$$ from the $$\Bx^1$$ terms, we find
\label{eqn:reciprocalblog:1460}
\begin{aligned}
\Bx^1
&= -\cos\theta \gamma_1 + \sin\theta \gamma_2 \\
&= \gamma^1 \lr{ \cos\theta + \gamma_1 \gamma_2 \sin\theta } \\
&= \gamma^1 e^{i\theta}.
\end{aligned}

Similarly for $$\Bx^2$$,
\label{eqn:reciprocalblog:1480}
\begin{aligned}
\Bx^2
&= \inv{\rho} \lr{ \sin\theta \gamma_1 + \cos\theta \gamma_2 } \\
&= \frac{\gamma^2}{\rho} \lr{ \sin\theta \gamma_2 \gamma_1 – \cos\theta } \\
&= -\frac{\gamma^2}{\rho} e^{i\theta}.
\end{aligned}

This matches \ref{eqn:reciprocalblog:1380}, as expected, but required only algebraic work to compute.

There will be circumstances where we parameterize only a subset of spacetime, and are interested in calculating quantities associated with such a surface. For example, suppose that
\label{eqn:reciprocalblog:1500}
x(\rho,\theta) = \gamma_1 \rho e^{i \theta},

where $$i = \gamma_1 \gamma_2$$ as before. We are now parameterizing only the $$x-y$$ plane. We will still find
\label{eqn:reciprocalblog:1520}
\begin{aligned}
\Bx_1 &= \gamma_1 e^{i \theta} \\
\Bx_2 &= -\gamma_2 \rho e^{i \theta}.
\end{aligned}

We can compute the reciprocals of these vectors using the gradient method. It’s possible to state matrix equations representing the reciprocal relationship of \ref{eqn:reciprocalblog:2200}, which, in this case, is $$X^\T G Y = 1$$, where the RHS is a $$2 \times 2$$ identity matrix, and $$X, Y$$ are $$4\times 2$$ matrices of coordinates, with
\label{eqn:reciprocalblog:1540}
X =
\begin{bmatrix}
0 & 0 \\
C & -\rho S \\
-S & -\rho C \\
0 & 0
\end{bmatrix}.

We no longer have a square matrix problem to solve, and our solution set is multivalued. In particular, this matrix equation has solutions
\label{eqn:reciprocalblog:1560}
\begin{aligned}
\Bx^1 &= \gamma^1 e^{i\theta} + \alpha \gamma^0 + \beta \gamma^3 \\
\Bx^2 &= -\frac{\gamma^2}{\rho} e^{i\theta} + \alpha’ \gamma^0 + \beta’ \gamma^3.
\end{aligned}

where $$\alpha, \alpha’, \beta, \beta’$$ are arbitrary constants. In the example we considered, we saw that our $$\rho, \theta$$ parameters were functions of only $$x^1, x^2$$, so taking gradients could not introduce any $$\gamma^0, \gamma^3$$ dependence in $$\Bx^1, \Bx^2$$. It seems reasonable to assert that we seek an algebraic method of computing a set of vectors that satisfies the reciprocal relationships, where that set of vectors is restricted to the tangent space. We will need to figure out how to prove that this reciprocal construction is identical to the parameter gradients, but let’s start with figuring out what such a tangent space restricted solution looks like.

## Theorem 1.7: Reciprocal frame for two parameter subspace.

Given two vectors, $$\Bx_1, \Bx_2$$, the vectors $$\Bx^1, \Bx^2 \in \mbox{Span}\setlr{ \Bx_1, \Bx_2 }$$ such that $$\Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu$$ are given by
\label{eqn:reciprocalblog:2260}
\begin{aligned}
\Bx^1 &= \Bx_2 \cdot \inv{\Bx_1 \wedge \Bx_2} \\
\Bx^2 &= -\Bx_1 \cdot \inv{\Bx_1 \wedge \Bx_2},
\end{aligned}

provided $$\Bx_1 \wedge \Bx_2 \ne 0$$ and
$$\lr{ \Bx_1 \wedge \Bx_2 }^2 \ne 0$$.

### Start proof:

The most general set of vectors that satisfy the span constraint are
\label{eqn:reciprocalblog:1580}
\begin{aligned}
\Bx^1 &= a \Bx_1 + b \Bx_2 \\
\Bx^2 &= c \Bx_1 + d \Bx_2.
\end{aligned}

We can use wedge products with either $$\Bx_1$$ or $$\Bx_2$$ to eliminate the other from the RHS
\label{eqn:reciprocalblog:1600}
\begin{aligned}
\Bx^1 \wedge \Bx_2 &= a \lr{ \Bx_1 \wedge \Bx_2 } \\
\Bx^1 \wedge \Bx_1 &= – b \lr{ \Bx_1 \wedge \Bx_2 } \\
\Bx^2 \wedge \Bx_2 &= c \lr{ \Bx_1 \wedge \Bx_2 } \\
\Bx^2 \wedge \Bx_1 &= – d \lr{ \Bx_1 \wedge \Bx_2 },
\end{aligned}

and then dot both sides with $$\Bx_1 \wedge \Bx_2$$ to produce four scalar equations
\label{eqn:reciprocalblog:1640}
\begin{aligned}
a \lr{ \Bx_1 \wedge \Bx_2 }^2
&= \lr{ \Bx^1 \wedge \Bx_2 } \cdot \lr{ \Bx_1 \wedge \Bx_2 } \\
&=
\lr{ \Bx_2 \cdot \Bx_1 } \lr{ \Bx^1 \cdot \Bx_2 }

\lr{ \Bx_2 \cdot \Bx_2 } \lr{ \Bx^1 \cdot \Bx_1 } \\
&=
\lr{ \Bx_2 \cdot \Bx_1 } (0)

\lr{ \Bx_2 \cdot \Bx_2 } (1) \\
&= – \Bx_2 \cdot \Bx_2
\end{aligned}

\label{eqn:reciprocalblog:1660}
\begin{aligned}
– b \lr{ \Bx_1 \wedge \Bx_2 }^2
&=
\lr{ \Bx^1 \wedge \Bx_1 } \cdot \lr{ \Bx_1 \wedge \Bx_2 } \\
&=
\lr{ \Bx^1 \cdot \Bx_2 } \lr{ \Bx_1 \cdot \Bx_1 }

\lr{ \Bx^1 \cdot \Bx_1 } \lr{ \Bx_1 \cdot \Bx_2 } \\
&=
(0) \lr{ \Bx_1 \cdot \Bx_1 }

(1) \lr{ \Bx_1 \cdot \Bx_2 } \\
&= – \Bx_1 \cdot \Bx_2
\end{aligned}

\label{eqn:reciprocalblog:1680}
\begin{aligned}
c \lr{ \Bx_1 \wedge \Bx_2 }^2
&= \lr{ \Bx^2 \wedge \Bx_2 } \cdot \lr{ \Bx_1 \wedge \Bx_2 } \\
&=
\lr{ \Bx_2 \cdot \Bx_1 } \lr{ \Bx^2 \cdot \Bx_2 }

\lr{ \Bx_2 \cdot \Bx_2 } \lr{ \Bx^2 \cdot \Bx_1 } \\
&=
\lr{ \Bx_2 \cdot \Bx_1 } (1)

\lr{ \Bx_2 \cdot \Bx_2 } (0) \\
&= \Bx_2 \cdot \Bx_1
\end{aligned}

\label{eqn:reciprocalblog:1700}
\begin{aligned}
– d \lr{ \Bx_1 \wedge \Bx_2 }^2
&= \lr{ \Bx^2 \wedge \Bx_1 } \cdot \lr{ \Bx_1 \wedge \Bx_2 } \\
&=
\lr{ \Bx_1 \cdot \Bx_1 } \lr{ \Bx^2 \cdot \Bx_2 }

\lr{ \Bx_1 \cdot \Bx_2 } \lr{ \Bx^2 \cdot \Bx_1 } \\
&=
\lr{ \Bx_1 \cdot \Bx_1 } (1)

\lr{ \Bx_1 \cdot \Bx_2 } (0) \\
&= \Bx_1 \cdot \Bx_1.
\end{aligned}

Putting the pieces together we have
\label{eqn:reciprocalblog:1740}
\begin{aligned}
\Bx^1
&= \frac{ – \lr{ \Bx_2 \cdot \Bx_2 } \Bx_1 + \lr{ \Bx_1 \cdot \Bx_2 } \Bx_2
}{\lr{\Bx_1 \wedge \Bx_2}^2} \\
&=
\frac{
\Bx_2 \cdot \lr{ \Bx_1 \wedge \Bx_2 }
}{\lr{\Bx_1 \wedge \Bx_2}^2} \\
&=
\Bx_2 \cdot \inv{\Bx_1 \wedge \Bx_2}
\end{aligned}

\label{eqn:reciprocalblog:1760}
\begin{aligned}
\Bx^2
&=
\frac{ \lr{ \Bx_1 \cdot \Bx_2 } \Bx_1 – \lr{ \Bx_1 \cdot \Bx_1 } \Bx_2
}{\lr{\Bx_1 \wedge \Bx_2}^2} \\
&=
\frac{ -\Bx_1 \cdot \lr{ \Bx_1 \wedge \Bx_2 } }
{\lr{\Bx_1 \wedge \Bx_2}^2} \\
&=
-\Bx_1 \cdot \inv{\Bx_1 \wedge \Bx_2}
\end{aligned}

## Lemma 1.1: Distribution identity.

Given k-vectors $$B, C$$ and a vector $$a$$, where the grade of $$C$$ is greater than that of $$B$$, then
\label{eqn:reciprocalblog:2280}
\lr{a \wedge B} \cdot C = a \cdot \lr{ B \cdot C }.

See [1] for a proof.

## Theorem 1.8: Higher order tangent space reciprocals.

Given an $$N$$ parameter tangent space with basis $$\setlr{ \Bx_0, \Bx_1, \cdots \Bx_{N-1} }$$, the reciprocals are given by
\label{eqn:reciprocalblog:2300}
\Bx^\mu = (-1)^\mu
\lr{ \Bx_0 \wedge \cdots \check{\Bx_\mu} \cdots \wedge \Bx_{N-1} } \cdot I_N^{-1},

where the checked term ($$\check{\Bx_\mu}$$) indicates that all terms are included in the wedges except the $$\Bx_\mu$$ term, and $$I_N = \Bx_0 \wedge \cdots \Bx_{N-1}$$ is the pseudoscalar for the tangent space.

### Start proof:

I’ll outline the proof for the three parameter tangent space case, from which the pattern will be clear. The motivation for this proof is a reexamination of the algebraic structure of the two vector solution. Suppose we have a tangent space basis $$\setlr{\Bx_0, \Bx_1}$$, for which we’ve shown that
\label{eqn:reciprocalblog:1860}
\begin{aligned}
\Bx^0
&= \Bx_1 \cdot \inv{\Bx_0 \wedge \Bx_1} \\
&= \frac{\Bx_1 \cdot \lr{\Bx_0 \wedge \Bx_1} }{\lr{ \Bx_0 \wedge \Bx_1}^2 }.
\end{aligned}

If we dot with $$\Bx_0$$ and $$\Bx_1$$ respectively, we find
\label{eqn:reciprocalblog:1800}
\begin{aligned}
\Bx_0 \cdot \Bx^0
&=
\Bx_0 \cdot \frac{ \Bx_1 \cdot \lr{ \Bx_0 \wedge \Bx_1 } }{\lr{ \Bx_0 \wedge \Bx_1}^2 } \\
&=
\lr{ \Bx_0 \wedge \Bx_1 } \cdot \frac{ \Bx_0 \wedge \Bx_1 }{\lr{ \Bx_0 \wedge \Bx_1}^2 }.
\end{aligned}

We end up with unity as expected. Here the
“factored” out vector is reincorporated into the pseudoscalar using the distribution identity \ref{eqn:reciprocalblog:2280}.
Similarly, dotting with $$\Bx_1$$, we find
\label{eqn:reciprocalblog:0810}
\begin{aligned}
\Bx_1 \cdot \Bx^0
&=
\Bx_1 \cdot \frac{ \Bx_1 \cdot \lr{ \Bx_0 \wedge \Bx_1 } }{\lr{ \Bx_0 \wedge \Bx_1}^2 } \\
&=
\lr{ \Bx_1 \wedge \Bx_1 } \cdot \frac{ \Bx_0 \wedge \Bx_1 }{\lr{ \Bx_0 \wedge \Bx_1}^2 }.
\end{aligned}

This is zero, since wedging a vector with itself is zero. We can perform such an operation in reverse, taking the square of the tangent space pseudoscalar, and factoring out one of the basis vectors. After this, division by that squared pseudoscalar will normalize things.

For a three parameter tangent space with basis $$\setlr{ \Bx_0, \Bx_1, \Bx_2 }$$, we can factor out any of the tangent vectors like so
\label{eqn:reciprocalblog:1880}
\begin{aligned}
\lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 }^2
&= \Bx_0 \cdot \lr{ \lr{ \Bx_1 \wedge \Bx_2 } \cdot \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } } \\
&= (-1) \Bx_1 \cdot \lr{ \lr{ \Bx_0 \wedge \Bx_2 } \cdot \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } } \\
&= (-1)^2 \Bx_2 \cdot \lr{ \lr{ \Bx_0 \wedge \Bx_1 } \cdot \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } }.
\end{aligned}

The toggling of sign reflects the number of permutations required to move the vector of interest to the front of the wedge sequence. Having factored out any one of the vectors, we can rearrange to find that vector that is it’s inverse and perpendicular to all the others.
\label{eqn:reciprocalblog:1900}
\begin{aligned}
\Bx^0 &= (-1)^0 \lr{ \Bx_1 \wedge \Bx_2 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } \\
\Bx^1 &= (-1)^1 \lr{ \Bx_0 \wedge \Bx_2 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } \\
\Bx^2 &= (-1)^2 \lr{ \Bx_0 \wedge \Bx_1 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 }.
\end{aligned}

### End proof.

In the fashion above, should we want the reciprocal frame for all of spacetime given dimension 4 tangent space, we can state it trivially
\label{eqn:reciprocalblog:1920}
\begin{aligned}
\Bx^0 &= (-1)^0 \lr{ \Bx_1 \wedge \Bx_2 \wedge \Bx_3 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 \wedge \Bx_3 } \\
\Bx^1 &= (-1)^1 \lr{ \Bx_0 \wedge \Bx_2 \wedge \Bx_3 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 \wedge \Bx_3 } \\
\Bx^2 &= (-1)^2 \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_3 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 \wedge \Bx_3 } \\
\Bx^3 &= (-1)^3 \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 \wedge \Bx_3 }.
\end{aligned}

This is probably not an efficient way to compute all these reciprocals, since we can utilize a single matrix inversion to solve them in one shot. However, there are theoretical advantages to this construction that will be useful when we get to integration theory.

### On degeneracy.

A small mention of degeneracy was mentioned above. Regardless of metric, $$\Bx_0 \wedge \Bx_1 = 0$$ means that this pair of vectors are colinear. A tangent space with such a pseudoscalar is clearly undesirable, and we must construct parameterizations for which the area element is non-zero in all regions of interest.

Things get more interesting in mixed signature spaces where we can have vectors that square to zero (i.e. lightlike). If the tangent space pseudoscalar has a lightlike factor, then that pseudoscalar will not be invertible. Such a degeneracy will will likely lead to many other troubles, and parameterizations of this sort should be avoided.

This following problem illustrates an example of this sort of degenerate parameterization.

## Problem: Degenerate surface parameterization.

Given a spacetime plane parameterization $$x(u,v) = u a + v b$$, where
\label{eqn:reciprocalblog:480}
a = \gamma_0 + \gamma_1 + \gamma_2 + \gamma_3,

\label{eqn:reciprocalblog:500}
b = \gamma_0 – \gamma_1 + \gamma_2 – \gamma_3,

show that this is a degenerate parameterization, and find the bivector that represents the tangent space. Are these vectors lightlike, spacelike, or timelike? Comment on whether this parameterization represents a physically relevant spacetime surface.

To characterize the vectors, we square them
\label{eqn:reciprocalblog:1080}
a^2 = b^2 =
\gamma_0^2 +
\gamma_1^2 +
\gamma_2^2 +
\gamma_3^2
=
1 – 3
= -2,

so $$a, b$$ are both spacelike vectors. The tangent space is clearly just $$\mbox{Span}\setlr{ a, b } = \mbox{Span}\setlr{ e, f }$$ where
\label{eqn:reciprocalblog:1100}
\begin{aligned}
e &= \gamma_0 + \gamma_2 \\
f &= \gamma_1 + \gamma_3.
\end{aligned}

Observe that $$a = e + f, b = e – f$$, and $$e$$ is lightlike ($$e^2 = 0$$), whereas $$f$$ is spacelike ($$f^2 = -2$$), and $$e \cdot f = 0$$, so $$e f = – f e$$. The bivector for the tangent plane is
\label{eqn:reciprocalblog:1120}
a b
}
=
(e + f) (e – f)
}
=
e^2 – f^2 – 2 e f
}
= -2 e f,

where
\label{eqn:reciprocalblog:1140}
e f = \gamma_{01} + \gamma_{21} + \gamma_{23} + \gamma_{03}.

Because $$e$$ is lightlike (zero square), and $$e f = – f e$$,
the bivector $$e f$$ squares to zero
\label{eqn:reciprocalblog:1780}
\lr{ e f }^2
= -e^2 f^2
= 0,

which shows that the parameterization is degenerate.

This parameterization can also be expressed as
\label{eqn:reciprocalblog:1160}
x(u,v)
= u ( e + f ) + v ( e – f )
= (u + v) e + (u – v) f,

a linear combination of a lightlike and spacelike vector. Intuitively, we expect that a physically meaningful spacetime surface involves linear combinations spacelike vectors, or combinations of a timelike vector with spacelike vectors. This beastie is something entirely different.

### Final notes.

There are a few loose ends above. In particular, we haven’t conclusively proven that the set of reciprocal vectors $$\Bx^\mu = \grad u^\mu$$ are exactly those obtained through algebraic means. For a full parameterization of spacetime, they are necessarily the same, since both are unique. So we know that \ref{eqn:reciprocalblog:1920} must equal the reciprocals obtained by evaluating the gradient for a full parameterization (and this must also equal the reciprocals that we can obtain through matrix inversion.) We have also not proved explicitly that the three parameter construction of the reciprocals in \ref{eqn:reciprocalblog:1900} is in the tangent space, but that is a fairly trivial observation, so that can be left as an exercise for the reader dismissal. Some additional thought about this is probably required, but it seems reasonable to put that on the back burner and move on to some applications.

# References

[1] Peeter Joot. Geometric Algebra for Electrical Engineers. Kindle Direct Publishing, 2019.

## Maxwell’s equation Lagrangian (geometric algebra and tensor formalism)

Maxwell’s equation using geometric algebra Lagrangian.

## Motivation.

In my classical mechanics notes, I’ve got computations of Maxwell’s equation (singular in it’s geometric algebra form) from a Lagrangian in various ways (using a tensor, scalar and multivector Lagrangians), but all of these seem more convoluted than they should be.
Here we do this from scratch, starting with the action principle for field variables, covering:

• Derivation of the relativistic form of the Euler-Lagrange field equations from the covariant form of the action,
• Derivation of Maxwell’s equation (in it’s STA form) from the Maxwell Lagrangian,
• Relationship of the STA Maxwell Lagrangian to the tensor equivalent,
• Relationship of the STA form of Maxwell’s equation to it’s tensor equivalents,
• Relationship of the STA Maxwell’s equation to it’s conventional Gibbs form.
• Show that we may use a multivector valued Lagrangian with all of $$F^2$$, not just the scalar part.

It is assumed that the reader is thoroughly familiar with the STA formalism, and if that is not the case, there is no better reference than [1].

## Theorem 1.1: Relativistic Euler-Lagrange field equations.

Let $$\phi \rightarrow \phi + \delta \phi$$ be any variation of the field, such that the variation
$$\delta \phi = 0$$ vanishes at the boundaries of the action integral
\label{eqn:maxwells:2120}
S = \int d^4 x \LL(\phi, \partial_\nu \phi).

The extreme value of the action is found when the Euler-Lagrange equations
\label{eqn:maxwells:2140}
0 = \PD{\phi}{\LL} – \partial_\nu \PD{(\partial_\nu \phi)}{\LL},

are satisfied. For a Lagrangian with multiple field variables, there will be one such equation for each field.

### Start proof:

To ease the visual burden, designate the variation of the field by $$\delta \phi = \epsilon$$, and perform a first order expansion of the varied Lagrangian
\label{eqn:maxwells:20}
\begin{aligned}
\LL
&\rightarrow
\LL(\phi + \epsilon, \partial_\nu (\phi + \epsilon)) \\
&=
\LL(\phi, \partial_\nu \phi)
+
\PD{\phi}{\LL} \epsilon +
\PD{(\partial_\nu \phi)}{\LL} \partial_\nu \epsilon.
\end{aligned}

The variation of the Lagrangian is
\label{eqn:maxwells:40}
\begin{aligned}
\delta \LL
&=
\PD{\phi}{\LL} \epsilon +
\PD{(\partial_\nu \phi)}{\LL} \partial_\nu \epsilon \\
&=
\PD{\phi}{\LL} \epsilon +
\partial_\nu \lr{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }

\epsilon \partial_\nu \PD{(\partial_\nu \phi)}{\LL},
\end{aligned}

which we may plug into the action integral to find
\label{eqn:maxwells:60}
\delta S
=
\int d^4 x \epsilon \lr{
\PD{\phi}{\LL}

\partial_\nu \PD{(\partial_\nu \phi)}{\LL}
}
+
\int d^4 x
\partial_\nu \lr{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }.

The last integral can be evaluated along the $$dx^\nu$$ direction, leaving
\label{eqn:maxwells:80}
\int d^3 x
\evalbar{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }{\Delta x^\nu},

where $$d^3 x = dx^\alpha dx^\beta dx^\gamma$$ is the product of differentials that does not include $$dx^\nu$$. By construction, $$\epsilon$$ vanishes on the boundary of the action integral so \ref{eqn:maxwells:80} is zero. The action takes its extreme value when
\label{eqn:maxwells:100}
0 = \delta S
=
\int d^4 x \epsilon \lr{
\PD{\phi}{\LL}

\partial_\nu \PD{(\partial_\nu \phi)}{\LL}
}.

The proof is complete after noting that this must hold for all variations of the field $$\epsilon$$, which means that we must have
\label{eqn:maxwells:120}
0 =
\PD{\phi}{\LL}

\partial_\nu \PD{(\partial_\nu \phi)}{\LL}.

### End proof.

Armed with the Euler-Lagrange equations, we can apply them to the Maxwell’s equation Lagrangian, which we will claim has the following form.

## Theorem 1.2: Maxwell’s equation Lagrangian.

Application of the Euler-Lagrange equations to the Lagrangian
\label{eqn:maxwells:2160}
\LL = – \frac{\epsilon_0 c}{2} F \cdot F + J \cdot A,

where $$F = \grad \wedge A$$, yields the vector portion of Maxwell’s equation
\label{eqn:maxwells:2180}
\grad \cdot F = \inv{\epsilon_0 c} J,

which implies
\label{eqn:maxwells:2200}
\grad F = \inv{\epsilon_0 c} J.

This is Maxwell’s equation.

### Start proof:

We wish to apply all of the Euler-Lagrange equations simultaneously (i.e. once for each of the four $$A_\mu$$ components of the potential), and cast it into four-vector form
\label{eqn:maxwells:140}
0 = \gamma_\nu \lr{ \PD{A_\nu}{} – \partial_\mu \PD{(\partial_\mu A_\nu)}{} } \LL.

Since our Lagrangian splits nicely into kinetic and interaction terms, this gives us
\label{eqn:maxwells:160}
0 = \gamma_\nu \lr{ \PD{A_\nu}{(A \cdot J)} + \frac{\epsilon_0 c}{2} \partial_\mu \PD{(\partial_\mu A_\nu)}{ (F \cdot F)} }.

The interaction term above is just
\label{eqn:maxwells:180}
\gamma_\nu \PD{A_\nu}{(A \cdot J)}
=
\gamma_\nu \PD{A_\nu}{(A_\mu J^\mu)}
=
\gamma_\nu J^\nu
=
J,

but the kinetic term takes a bit more work. Let’s start with evaluating
\label{eqn:maxwells:200}
\begin{aligned}
\PD{(\partial_\mu A_\nu)}{ (F \cdot F)}
&=
\PD{(\partial_\mu A_\nu)}{ F } \cdot F
+
F \cdot \PD{(\partial_\mu A_\nu)}{ F } \\
&=
2 \PD{(\partial_\mu A_\nu)}{ F } \cdot F \\
&=
2 \PD{(\partial_\mu A_\nu)}{ (\partial_\alpha A_\beta) } \lr{ \gamma^\alpha \wedge \gamma^\beta } \cdot F \\
&=
2 \lr{ \gamma^\mu \wedge \gamma^\nu } \cdot F.
\end{aligned}

We hit this with the $$\mu$$-partial and expand as a scalar selection to find
\label{eqn:maxwells:220}
\begin{aligned}
\partial_\mu \PD{(\partial_\mu A_\nu)}{ (F \cdot F)}
&=
2 \lr{ \partial_\mu \gamma^\mu \wedge \gamma^\nu } \cdot F \\
&=
– 2 (\gamma^\nu \wedge \grad) \cdot F \\
&=
&=
&=
– 2 \gamma^\nu \cdot \lr{ \grad \cdot F }.
\end{aligned}

Putting all the pieces together yields
\label{eqn:maxwells:240}
0
= J – \epsilon_0 c \gamma_\nu \lr{ \gamma^\nu \cdot \lr{ \grad \cdot F } }
= J – \epsilon_0 c \lr{ \grad \cdot F },

but
\label{eqn:maxwells:260}
\begin{aligned}
&=
&=
&=
\end{aligned}

so the multivector field equations for this Lagrangian are
\label{eqn:maxwells:280}
\grad F = \inv{\epsilon_0 c} J,

as claimed.

## Problem: Correspondence with tensor formalism.

Cast the Lagrangian of \ref{eqn:maxwells:2160} into the conventional tensor form
\label{eqn:maxwells:300}
\LL = \frac{\epsilon_0 c}{4} F_{\mu\nu} F^{\mu\nu} + A^\mu J_\mu.

Also show that the four-vector component of Maxwell’s equation $$\grad \cdot F = J/(\epsilon_0 c)$$ is equivalent to the conventional tensor form of the Gauss-Ampere law
\label{eqn:maxwells:320}
\partial_\mu F^{\mu\nu} = \inv{\epsilon_0 c} J^\nu,

where $$F^{\mu\nu} = \partial^\mu A^\nu – \partial^\nu A^\mu$$ as usual. Also show that the trivector component of Maxwell’s equation $$\grad \wedge F = 0$$ is equivalent to the tensor form of the Gauss-Faraday law
\label{eqn:maxwells:340}
\partial_\alpha \lr{ \epsilon^{\alpha \beta \mu \nu} F_{\mu\nu} } = 0.

To show the Lagrangian correspondence we must expand $$F \cdot F$$ in coordinates
\label{eqn:maxwells:360}
\begin{aligned}
F \cdot F
&=
( \grad \wedge A ) \cdot
( \grad \wedge A ) \\
&=
\lr{ (\gamma^\mu \partial_\mu) \wedge (\gamma^\nu A_\nu) }
\cdot
\lr{ (\gamma^\alpha \partial_\alpha) \wedge (\gamma^\beta A_\beta) } \\
&=
\lr{ \gamma^\mu \wedge \gamma^\nu } \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta }
(\partial_\mu A_\nu )
(\partial^\alpha A^\beta ) \\
&=
\lr{
{\delta^\mu}_\beta
{\delta^\nu}_\alpha

{\delta^\mu}_\alpha
{\delta^\nu}_\beta
}
(\partial_\mu A_\nu )
(\partial^\alpha A^\beta ) \\
&=
– \partial_\mu A_\nu \lr{
\partial^\mu A^\nu

\partial^\nu A^\mu
} \\
&=
– \partial_\mu A_\nu F^{\mu\nu} \\
&=
– \inv{2} \lr{
\partial_\mu A_\nu F^{\mu\nu}
+
\partial_\nu A_\mu F^{\nu\mu}
} \\
&=
– \inv{2} \lr{
\partial_\mu A_\nu

\partial_\nu A_\mu
}
F^{\mu\nu} \\
&=

\inv{2}
F_{\mu\nu}
F^{\mu\nu}.
\end{aligned}

With a substitution of this and $$A \cdot J = A_\mu J^\mu$$ back into the Lagrangian, we recover the tensor form of the Lagrangian.

To recover the tensor form of Maxwell’s equation, we first split it into vector and trivector parts
\label{eqn:maxwells:1580}

Now the vector component may be expanded in coordinates by dotting both sides with $$\gamma^\nu$$ to find
\label{eqn:maxwells:1600}
\inv{\epsilon_0 c} \gamma^\nu \cdot J = J^\nu,

and
\label{eqn:maxwells:1620}
\begin{aligned}
\gamma^\nu \cdot
&=
\partial_\mu \gamma^\nu \cdot \lr{ \gamma^\mu \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta } \partial^\alpha A^\beta } \\
&=
\lr{
{\delta^\mu}_\alpha
{\delta^\nu}_\beta

{\delta^\nu}_\alpha
{\delta^\mu}_\beta
}
\partial_\mu
\partial^\alpha A^\beta \\
&=
\partial_\mu
\lr{
\partial^\mu A^\nu

\partial^\nu A^\mu
} \\
&=
\partial_\mu F^{\mu\nu}.
\end{aligned}

Equating \ref{eqn:maxwells:1600} and \ref{eqn:maxwells:1620} finishes the first part of the job. For the trivector component, we have
\label{eqn:maxwells:1640}
0
= (\gamma^\mu \partial_\mu) \wedge \lr{ \gamma^\alpha \wedge \gamma^\beta } \partial_\alpha A_\beta
= \inv{2} (\gamma^\mu \partial_\mu) \wedge \lr{ \gamma^\alpha \wedge \gamma^\beta } F_{\alpha \beta}.

Wedging with $$\gamma^\tau$$ and then multiplying by $$-2 I$$ we find
\label{eqn:maxwells:1660}
0 = – \lr{ \gamma^\mu \wedge \gamma^\alpha \wedge \gamma^\beta \wedge \gamma^\tau } I \partial_\mu F_{\alpha \beta},

but
\label{eqn:maxwells:1680}
\gamma^\mu \wedge \gamma^\alpha \wedge \gamma^\beta \wedge \gamma^\tau = -I \epsilon^{\mu \alpha \beta \tau},

which leaves us with
\label{eqn:maxwells:1700}
\epsilon^{\mu \alpha \beta \tau} \partial_\mu F_{\alpha \beta} = 0,

as expected.

## Problem: Correspondence of tensor and Gibbs forms of Maxwell’s equations.

Given the identifications

\label{eqn:lorentzForceCovariant:1500}
F^{k0} = E^k,

and
\label{eqn:lorentzForceCovariant:1520}
F^{rs} = -\epsilon^{rst} B^t,

and
\label{eqn:maxwells:1560}
J^\mu = \lr{ c \rho, \BJ },

the reader should satisfy themselves that the traditional Gibbs form of Maxwell’s equations can be recovered from \ref{eqn:maxwells:320}.

The reader is referred to Exercise 3.4 “Electrodynamics, variational principle.” from [2].

## Problem: Correspondence with grad and curl form of Maxwell’s equations.

With $$J = c \rho \gamma_0 + J^k \gamma_k$$ and $$F = \BE + I c \BB$$ show that Maxwell’s equation, as stated in \ref{eqn:maxwells:2200} expand to the conventional div and curl expressions for Maxwell’s equations.

To obtain Maxwell’s equations in their traditional vector forms, we pre-multiply both sides with $$\gamma_0$$
\label{eqn:maxwells:1720}
\gamma_0 \grad F = \inv{\epsilon_0 c} \gamma_0 J,

and then select each grade separately. First observe that the RHS above has scalar and bivector components, as
\label{eqn:maxwells:1740}
\gamma_0 J
=
c \rho + J^k \gamma_0 \gamma_k.

In terms of the spatial bivector basis $$\Be_k = \gamma_k \gamma_0$$, the RHS of \ref{eqn:maxwells:1720} is
\label{eqn:maxwells:1760}
\gamma_0 \frac{J}{\epsilon_0 c} = \frac{\rho}{\epsilon_0} – \mu_0 c \BJ.

For the LHS, first note that
\label{eqn:maxwells:1780}
\begin{aligned}
&=
\gamma_0
\lr{
\gamma_0 \partial^0 +
\gamma_k \partial^k
} \\
&=
\partial_0 – \gamma_0 \gamma_k \partial_k \\
&=
\end{aligned}

We can express all the the LHS of \ref{eqn:maxwells:1720} in the bivector spatial basis, so that Maxwell’s equation in multivector form is
\label{eqn:maxwells:1800}
\lr{ \inv{c} \PD{t}{} + \spacegrad } \lr{ \BE + I c \BB } = \frac{\rho}{\epsilon_0} – \mu_0 c \BJ.

Selecting the scalar, vector, bivector, and trivector grades of both sides (in the spatial basis) gives the following set of respective equations
\label{eqn:maxwells:1840}

\label{eqn:maxwells:1860}
\inv{c} \partial_t \BE + I c \spacegrad \wedge \BB = – \mu_0 c \BJ

\label{eqn:maxwells:1880}
\spacegrad \wedge \BE + I \partial_t \BB = 0

\label{eqn:maxwells:1900}
I c \spacegrad \cdot B = 0,

which we can rewrite after some duality transformations (and noting that $$\mu_0 \epsilon_0 c^2 = 1$$), we have
\label{eqn:maxwells:1940}

\label{eqn:maxwells:1960}
\spacegrad \cross \BB – \mu_0 \epsilon_0 \PD{t}{\BE} = \mu_0 \BJ

\label{eqn:maxwells:1980}
\spacegrad \cross \BE + \PD{t}{\BB} = 0

\label{eqn:maxwells:2000}

which are Maxwell’s equations in their traditional form.

## Problem: Alternative multivector Lagrangian.

Show that a scalar+pseudoscalar Lagrangian of the following form
\label{eqn:maxwells:2220}
\LL = – \frac{\epsilon_0 c}{2} F^2 + J \cdot A,

which omits the scalar selection of the Lagrangian in \ref{eqn:maxwells:2160}, also represents Maxwell’s equation. Discuss the scalar and pseudoscalar components of $$F^2$$, and show why the pseudoscalar inclusion is irrelevant.

The quantity $$F^2 = F \cdot F + F \wedge F$$ has both scalar and pseudoscalar
components. Note that unlike vectors, a bivector wedge in 4D with itself need not be zero (example: $$\gamma_0 \gamma_1 + \gamma_2 \gamma_3$$ wedged with itself).
We can see this multivector nature nicely by expansion in terms of the electric and magnetic fields
\label{eqn:maxwells:2020}
\begin{aligned}
F^2
&= \lr{ \BE + I c \BB }^2 \\
&= \BE^2 – c^2 \BB^2 + I c \lr{ \BE \BB + \BB \BE } \\
&= \BE^2 – c^2 \BB^2 + 2 I c \BE \cdot \BB.
\end{aligned}

Both the scalar and pseudoscalar parts of $$F^2$$ are Lorentz invariant, a requirement of our Lagrangian, but most Maxwell equation Lagrangians only include the scalar $$\BE^2 – c^2 \BB^2$$ component of the field square. If we allow the Lagrangian to be multivector valued, and evaluate the Euler-Lagrange equations, we quickly find the same results
\label{eqn:maxwells:2040}
\begin{aligned}
0
&= \gamma_\nu \lr{ \PD{A_\nu}{} – \partial_\mu \PD{(\partial_\mu A_\nu)}{} } \LL \\
&= \gamma_\nu \lr{ J^\nu + \frac{\epsilon_0 c}{2} \partial_\mu
\lr{
(\gamma^\mu \wedge \gamma^\nu) F
+
F (\gamma^\mu \wedge \gamma^\nu)
}
}.
\end{aligned}

Here some steps are skipped, building on our previous scalar Euler-Lagrange evaluation experience. We have a symmetric product of two bivectors, which we can express as a 0,4 grade selection, since
\label{eqn:maxwells:2060}
\gpgrade{ X F }{0,4} = \inv{2} \lr{ X F + F X },

for any two bivectors $$X, F$$. This leaves
\label{eqn:maxwells:2080}
\begin{aligned}
0
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ (\grad \wedge \gamma^\nu) F }{0,4} \\
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ -\gamma^\nu \grad F + (\gamma^\nu \cdot \grad) F }{0,4} \\
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ -\gamma^\nu \grad F }{0,4} \\
&= J – \epsilon_0 c \gamma_\nu
\lr{
\gamma^\nu \cdot \lr{ \grad \cdot F } + \gamma^\nu \wedge \grad \wedge F
}.
\end{aligned}

However, since $$\grad \wedge F = \grad \wedge \grad \wedge A = 0$$, we see that there is no contribution from the $$F \wedge F$$ pseudoscalar component of the Lagrangian, and we are left with
\label{eqn:maxwells:2100}
\begin{aligned}
0
&= J – \epsilon_0 c (\grad \cdot F) \\
&= J – \epsilon_0 c \grad F,
\end{aligned}

which is Maxwell’s equation, as before.

# References

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

[2] Peeter Joot. Quantum field theory. Kindle Direct Publishing, 2018.

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

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

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

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

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

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

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

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

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

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

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

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

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

or
\label{eqn:staticPotentials:200}

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

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

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

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

### Integral form of the gauge transformation.

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

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

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

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

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

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

# References

## 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

## Motivation

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

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

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

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

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

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

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

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

\label{eqn:chapter3Notes:60}

\label{eqn:chapter3Notes:80}

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

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

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

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

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

### Ideas explored in these notes

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

### No magnetic sources

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

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

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

\label{eqn:potentialMethods:40}

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

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

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

The final two Maxwell equations yield

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

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

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

### No electric sources

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

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

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

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

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

The remaining Maxwell equations provide the required constraints on the potentials

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

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

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

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

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

### Potential operator structure

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

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

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

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

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

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

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

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

or

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

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

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

## Potentials in the STA formalism.

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

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

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

### No magnetic sources

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

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

\label{eqn:potentialMethods:400}

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

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

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

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

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

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

or

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

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

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

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

### No electric sources

Without electric sources, Maxwell’s equation now splits into

\label{eqn:potentialMethods:640}

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

Here the dual of an STA curl yields a solution

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

Substituting this gives

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

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

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

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

So the general solution with both electric and magnetic sources is

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

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

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

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

\label{eqn:potentialMethods:880}

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

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

### Potential operator structure

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

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

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

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

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

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

where

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

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

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

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

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

## Potentials in the 3D Euclidean formalism

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

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

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

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

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

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

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

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

Expanding \ref{eqn:potentialMethods:1320} gives

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

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

The conclusion is that the potential representation of the field is

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

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

## Gauge transformations

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

\label{eqn:potentialMethods:1380}

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

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

The electric field with this changed potential is

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

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

the electric field will also be unaltered by this transformation.

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

\label{eqn:potentialMethods:1460}

and for electric or magnetic sources

\label{eqn:potentialMethods:1480}

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

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

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

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

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

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

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

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

## Lorenz gauge

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

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

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

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

that is

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

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

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

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

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

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

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

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

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

# References

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

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

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

## Corollaries to Stokes and Divergence theorems

In [1] a few problems are set to prove some variations of Stokes theorem. He gives some cool tricks to prove each one using just the classic 3D Stokes and divergence theorems. We can also do them directly from the more general Stokes theorem $$\int d^k \Bx \cdot (\spacegrad \wedge F) = \oint d^{k-1} \Bx \cdot F$$.

## Question: Stokes theorem on scalar function. ([1] pr. 1.60a)

Prove
\label{eqn:stokesCorollariesGriffiths:20}
\int \spacegrad T dV = \oint T d\Ba.

The direct way to prove this is to apply Stokes theorem

\label{eqn:stokesCorollariesGriffiths:80}
\int d^3 \Bx \cdot (\spacegrad \wedge T) = \oint d^2 \Bx \cdot T

Here $$d^3 \Bx = d\Bx_1 \wedge d\Bx_2 \wedge d\Bx_3$$, a pseudoscalar (trivector) volume element, and the wedge and dot products take their most general meanings. For $$k$$-blade $$F$$, and $$k’$$-blade $$F’$$, that is

\label{eqn:stokesCorollariesGriffiths:100}
\begin{aligned}
F \wedge F’ &= \gpgrade{F F’}{k+k’} \\
F \cdot F’ &= \gpgrade{F F’}{\Abs{k-k’}}
\end{aligned}

With $$d^3\Bx = I dV$$, and $$d^2 \Bx = I \ncap dA = I d\Ba$$, we have

\label{eqn:stokesCorollariesGriffiths:120}
\int I dV \spacegrad T = \oint I d\Ba T.

Cancelling the factors of $$I$$ proves the result.

Griffith’s trick to do this was to let $$\Bv = \Bc T$$, where $$\Bc$$ is a constant. For this, the divergence theorem integral is

\label{eqn:stokesCorollariesGriffiths:160}
\begin{aligned}
\int dV \spacegrad \cdot (\Bc T)
&=
\int dV \Bc \cdot \spacegrad T \\
&=
\Bc \cdot \int dV \spacegrad T \\
&=
\oint d\Ba \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Ba T.
\end{aligned}

This is true for any constant $$\Bc$$, so is also true for the unit vectors. This allows for summing projections in each of the unit directions

\label{eqn:stokesCorollariesGriffiths:180}
\begin{aligned}
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad T } \\
&=
\sum \Be_k \lr{ \Be_k \cdot \oint d\Ba T } \\
&=
\oint d\Ba T.
\end{aligned}

## Question: ([1] pr. 1.60b)

Prove
\label{eqn:stokesCorollariesGriffiths:40}
\int \spacegrad \cross \Bv dV = -\oint \Bv \cross d\Ba.

This also follows directly from the general Stokes theorem

\label{eqn:stokesCorollariesGriffiths:200}
\int d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv } = \oint d^2 \Bx \cdot \Bv

The volume integrand is

\label{eqn:stokesCorollariesGriffiths:220}
\begin{aligned}
d^3 \Bx \cdot \lr{ \spacegrad \wedge \Bv }
&=
&=
\end{aligned}

and the surface integrand is
\label{eqn:stokesCorollariesGriffiths:240}
\begin{aligned}
d^2 \Bx \cdot \Bv
&=
\gpgradeone{ I d\Ba \Bv } \\
&=
\gpgradeone{ I (d\Ba \wedge \Bv) } \\
&=
I^2 (d\Ba \cross \Bv) \\
&=
-d\Ba \cross \Bv \\
&=
\Bv \cross d\Ba.
\end{aligned}

Plugging these into \ref{eqn:stokesCorollariesGriffiths:200} proves the result.

Griffiths trick for the same is to apply the divergence theorem to $$\Bv \cross \Bc$$. Such a volume integral is

\label{eqn:stokesCorollariesGriffiths:260}
\begin{aligned}
\int dV \spacegrad \cdot (\Bv \cross \Bc)
&=
\int dV \Bc \cdot (\spacegrad \cross \Bv) \\
&=
\Bc \cdot \int dV \spacegrad \cross \Bv.
\end{aligned}

This must equal
\label{eqn:stokesCorollariesGriffiths:280}
\begin{aligned}
\oint d\Ba \cdot (\Bv \cross \Bc)
&=
\Bc \cdot \oint d\Ba \cross \Bv \\
&=
-\Bc \cdot \oint \Bv \cross d\Ba
\end{aligned}

Again, assembling projections, we have
\label{eqn:stokesCorollariesGriffiths:300}
\begin{aligned}
&=
\sum \Be_k \lr{ \Be_k \cdot \int dV \spacegrad \cross \Bv } \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint \Bv \cross d\Ba } \\
&=
-\oint \Bv \cross d\Ba.
\end{aligned}

## Question: ([1] pr. 1.60e)

Prove
\label{eqn:stokesCorollariesGriffiths:60}
\int \spacegrad T \cross d\Ba = -\oint T d\Bl.

This one follows from
\label{eqn:stokesCorollariesGriffiths:320}
\int d^2 \Bx \cdot \lr{ \spacegrad \wedge T } = \oint d^1 \Bx \cdot T.

The surface integrand can be written
\label{eqn:stokesCorollariesGriffiths:340}
\begin{aligned}
d^2 \Bx \cdot \lr{ \spacegrad \wedge T }
&=
&=
I (d\Ba \wedge \spacegrad T ) \\
&=
I^2 ( d\Ba \cross \spacegrad T ) \\
&=
\end{aligned}

The line integrand is

\label{eqn:stokesCorollariesGriffiths:360}
d^1 \Bx \cdot T = d^1 \Bx T.

Given a two parameter representation of the surface area element $$d^2 \Bx = d\Bx_1 \wedge d\Bx_2$$, the line element representation is
\label{eqn:stokesCorollariesGriffiths:380}
\begin{aligned}
d^1 \Bx
&= (\Bx_1 \wedge d\Bx_2) \cdot \Bx^1 + (d\Bx_1 \wedge \Bx_2) \cdot \Bx^2 \\
&= -d\Bx_2 + d\Bx_1,
\end{aligned}

giving

\label{eqn:stokesCorollariesGriffiths:400}
\begin{aligned}
&=
\int
-\evalbar{\lr{ \PD{u_2}{\Bx} T }}{\Delta u_1} du_2
+\evalbar{\lr{ \PD{u_1}{\Bx} T }}{\Delta u_2} du_1 \\
&=
-\oint d\Bl T,
\end{aligned}

or
\label{eqn:stokesCorollariesGriffiths:420}
=
-\oint d\Bl T.

Griffiths trick for the same is to use $$\Bv = \Bc T$$ for constant $$\Bc$$ in (the usual 3D) Stokes’ theorem. That is

\label{eqn:stokesCorollariesGriffiths:440}
\begin{aligned}
\int d\Ba \cdot (\spacegrad \cross (\Bc T))
&=
\Bc \cdot \int d\Ba \cross \spacegrad T \\
&=
-\Bc \cdot \int \spacegrad T \cross d\Ba \\
&=
\oint d\Bl \cdot (\Bc T) \\
&=
\Bc \cdot \oint d\Bl T.
\end{aligned}

Again assembling projections we have
\label{eqn:stokesCorollariesGriffiths:460}
\begin{aligned}
&=
\sum \Be_k \lr{ \Be_k \cdot \int \spacegrad T \cross d\Ba} \\
&=
-\sum \Be_k \lr{ \Be_k \cdot \oint d\Bl T } \\
&=
-\oint d\Bl T.
\end{aligned}

# References

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

## Motivation

I initially thought that I might submit a problem set solution for ece1228 using Geometric Algebra. In order to justify this, I needed to add an appendix to that problem set that outlined enough of the ideas that such a solution might make sense to the grader.

I ended up changing my mind and reworked the problem entirely, removing any use of GA. Here’s the tutorial I initially considered submitting with that problem.

## Geometric Algebra in a nutshell.

Geometric Algebra defines a non-commutative, associative vector product

\label{eqn:gaTutorial:20}
\begin{aligned}
\Ba \Bb \Bc
&=
(\Ba \Bb) \Bc \\
&=
\Ba (\Bb \Bc),
\end{aligned}

where the square of a vector equals the squared vector magnitude

\label{eqn:gaTutorial:40}
\Ba^2 = \Abs{\Ba}^2,

In Euclidean spaces such a squared vector is always positive, but that is not necessarily the case in the mixed signature spaces used in special relativity.

There are a number of consequences of these two simple vector multiplication rules.

• Squared unit vectors have a unit magnitude (up to a sign). In a Euclidean space such a product is always positive

\label{eqn:gaTutorial:60}
(\Be_1)^2 = 1.

• Products of perpendicular vectors anticommute.

\label{eqn:gaTutorial:80}
\begin{aligned}
2
&=
(\Be_1 + \Be_2)^2 \\
&= (\Be_1 + \Be_2)(\Be_1 + \Be_2) \\
&= \Be_1^2 + \Be_2 \Be_1 + \Be_1 \Be_2 + \Be_2^2 \\
&= 2 + \Be_2 \Be_1 + \Be_1 \Be_2.
\end{aligned}

A product of two perpendicular vectors is called a bivector, and can be used to represent an oriented plane. The last line above shows an example of a scalar and bivector sum, called a multivector. In general Geometric Algebra allows sums of scalars, vectors, bivectors, and higher degree analogues (grades) be summed.

Comparison of the RHS and LHS of \ref{eqn:gaTutorial:80} shows that we must have

\label{eqn:gaTutorial:100}
\Be_2 \Be_1 = -\Be_1 \Be_2.

It is true in general that the product of two perpendicular vectors anticommutes. When, as above, such a product is a product of
two orthonormal vectors, it behaves like a non-commutative imaginary quantity, as it has an imaginary square in Euclidean spaces

\label{eqn:gaTutorial:120}
\begin{aligned}
(\Be_1 \Be_2)^2
&=
(\Be_1 \Be_2)
(\Be_1 \Be_2) \\
&=
\Be_1 (\Be_2
\Be_1) \Be_2 \\
&=
-\Be_1 (\Be_1
\Be_2) \Be_2 \\
&=
-(\Be_1 \Be_1)
(\Be_2 \Be_2) \\
&=-1.
\end{aligned}

Such “imaginary” (unit bivectors) have important applications describing rotations in Euclidean spaces, and boosts in Minkowski spaces.

• The product of three perpendicular vectors, such as

\label{eqn:gaTutorial:140}
I = \Be_1 \Be_2 \Be_3,

is called a trivector. In \R{3}, the product of three orthonormal vectors is called a pseudoscalar for the space, and can represent an oriented volume element. The quantity $$I$$ above is the typical orientation picked for the \R{3} unit pseudoscalar. This quantity also has characteristics of an imaginary number

\label{eqn:gaTutorial:160}
\begin{aligned}
I^2
&=
(\Be_1 \Be_2 \Be_3)
(\Be_1 \Be_2 \Be_3) \\
&=
\Be_1 \Be_2 (\Be_3
\Be_1) \Be_2 \Be_3 \\
&=
-\Be_1 \Be_2 \Be_1
\Be_3 \Be_2 \Be_3 \\
&=
-\Be_1 (\Be_2 \Be_1)
(\Be_3 \Be_2) \Be_3 \\
&=
-\Be_1 (\Be_1 \Be_2)
(\Be_2 \Be_3) \Be_3 \\
&=

\Be_1^2
\Be_2^2
\Be_3^2 \\
&=
-1.
\end{aligned}

• The product of two vectors in \R{3} can be expressed as the sum of a symmetric scalar product and antisymmetric bivector product

\label{eqn:gaTutorial:480}
\begin{aligned}
\Ba \Bb
&=
\sum_{i,j = 1}^n \Be_i \Be_j a_i b_j \\
&=
\sum_{i = 1}^n \Be_i^2 a_i b_i
+
\sum_{0 < i \ne j \le n} \Be_i \Be_j a_i b_j \\ &= \sum_{i = 1}^n a_i b_i + \sum_{0 < i < j \le n} \Be_i \Be_j (a_i b_j - a_j b_i). \end{aligned} The first (symmetric) term is clearly the dot product. The antisymmetric term is designated the wedge product. In general these are written $$\label{eqn:gaTutorial:500} \Ba \Bb = \Ba \cdot \Bb + \Ba \wedge \Bb,$$ where \label{eqn:gaTutorial:520} \begin{aligned} \Ba \cdot \Bb &\equiv \inv{2} \lr{ \Ba \Bb + \Bb \Ba } \\ \Ba \wedge \Bb &\equiv \inv{2} \lr{ \Ba \Bb - \Bb \Ba }, \end{aligned} The coordinate expansion of both can be seen above, but in \R{3} the wedge can also be written $$\label{eqn:gaTutorial:540} \Ba \wedge \Bb = \Be_1 \Be_2 \Be_3 (\Ba \cross \Bb) = I (\Ba \cross \Bb).$$ This allows for an handy dot plus cross product expansion of the vector product $$\label{eqn:gaTutorial:180} \Ba \Bb = \Ba \cdot \Bb + I (\Ba \cross \Bb).$$ This result should be familiar to the student of quantum spin states where one writes $$\label{eqn:gaTutorial:200} (\Bsigma \cdot \Ba) (\Bsigma \cdot \Bb) = (\Ba \cdot \Bb) + i (\Ba \cross \Bb) \cdot \Bsigma.$$ This correspondence is because the Pauli spin basis is a specific matrix representation of a Geometric Algebra, satisfying the same commutator and anticommutator relationships. A number of other algebra structures, such as complex numbers, and quaterions can also be modelled as Geometric Algebra elements.

• It is often useful to utilize the grade selection operator
$$\gpgrade{M}{n}$$ and scalar grade selection operator $$\gpgradezero{M} = \gpgrade{M}{0}$$
to select the scalar, vector, bivector, trivector, or higher grade algebraic elements. For example, operating on vectors $$\Ba, \Bb, \Bc$$, we have

\label{eqn:gaTutorial:580}
\begin{aligned}
&= \Ba \cdot \Bb \\
&=
\Ba (\Bb \cdot \Bc)
+
\Ba \cdot (\Bb \wedge \Bc) \\
&=
\Ba (\Bb \cdot \Bc)
+
(\Ba \cdot \Bb) \Bc

(\Ba \cdot \Bc) \Bb \\
\Ba \wedge \Bb \\
\Ba \wedge \Bb \wedge \Bc.
\end{aligned}

Note that the wedge product of any number of vectors such as $$\Ba \wedge \Bb \wedge \Bc$$ is associative and can be expressed in terms of the complete antisymmetrization of the product of those vectors. A consequence of that is the fact a wedge product that includes any colinear vectors in the product is zero.

## Example: Helmholz equations.

As an example of the power of \ref{eqn:gaTutorial:180}, consider the following Helmholtz equation derivation (wave equations for the electric and magnetic fields in the frequency domain.)

Application of \ref{eqn:gaTutorial:180} to
Maxwell equations in the frequency domain for source free simple media gives

\label{eqn:emtProblemSet1Problem6:340}
\label{eqn:emtProblemSet1Problem6:360}
\spacegrad \BE = -j \omega I \BB

\label{eqn:emtProblemSet1Problem6:380}
\spacegrad I \BB = -j \omega \mu \epsilon \BE.

These equations use the engineering (not physics) sign convention for the phasors where the time domain fields are of the form $$\boldsymbol{\mathcal{E}}(\Br, t) = \textrm{Re}( \BE e^{j\omega t}$$.

Operation with the gradient from the left produces the Helmholtz equation for each of the fields using nothing more than multiplication and simple substitution

\label{eqn:emtProblemSet1Problem6:400}
\label{eqn:emtProblemSet1Problem6:420}
\spacegrad^2 \BE = – \mu \epsilon \omega^2 \BE

\label{eqn:emtProblemSet1Problem6:440}
\spacegrad^2 I \BB = – \mu \epsilon \omega^2 I \BB.

There was no reason to go through the headache of looking up or deriving the expansion of $$\spacegrad \cross (\spacegrad \cross \BA )$$ as is required with the traditional vector algebra demonstration of these identities.

Observe that the usual Helmholtz equation for $$\BB$$ doesn’t have a pseudoscalar factor. That result can be obtained by just cancelling the factors $$I$$ since the \R{3} Euclidean pseudoscalar commutes with all grades (this isn’t the case in \R{2} nor in Minkowski spaces.)

## Example: Factoring the Laplacian.

There are various ways to demonstrate the identity

\label{eqn:gaTutorial:660}

such as the use of (somewhat obscure) tensor contraction techniques. We can also do this with Geometric Algebra (using a different set of obscure techniques) by factoring the Laplacian action on a vector

\label{eqn:gaTutorial:700}
\begin{aligned}
&=
&=
&=
+
%+
&=
+
\end{aligned}

Should we wish to express the last term using cross products, a grade one selection operation can be used
\label{eqn:gaTutorial:680}
\begin{aligned}
&=
&=
&=
&=
&=
\end{aligned}

Here coordinate expansion was not required in any step.

# References

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

[2] L. Dorst, D. Fontijne, and S. Mann. Geometric Algebra for Computer Science. Morgan Kaufmann, San Francisco, 2007.

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

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

## 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.