## Notes.

Due to limitations in the MathJax-Latex package, all the oriented integrals in this blog post should be interpreted as having a clockwise orientation. [See the PDF version of this post for more sophisticated formatting.]

## Guts.

Given a two dimensional generating vector space, there are two instances of the fundamental theorem for multivector integration
\label{eqn:unpackingFundamentalTheorem:20}
\int_S F d\Bx \lrpartial G = \evalbar{F G}{\Delta S},

and
\label{eqn:unpackingFundamentalTheorem:40}
\int_S F d^2\Bx \lrpartial G = \oint_{\partial S} F d\Bx G.

The first case is trivial. Given a parameterizated curve $$x = x(u)$$, it just states
\label{eqn:unpackingFundamentalTheorem:60}
\int_{u(0)}^{u(1)} du \PD{u}{}\lr{FG} = F(u(1))G(u(1)) – F(u(0))G(u(0)),

for all multivectors $$F, G$$, regardless of the signature of the underlying space.

The surface integral is more interesting. Let’s first look at the area element for this surface integral, which is
\label{eqn:unpackingFundamentalTheorem:80}
d^2 \Bx = d\Bx_u \wedge d \Bx_v.

Geometrically, this has the area of the parallelogram spanned by $$d\Bx_u$$ and $$d\Bx_v$$, but weighted by the pseudoscalar of the space. This is explored algebraically in the following problem and illustrated in fig. 1.

fig. 1. 2D vector space and area element.

## Problem: Expansion of 2D area bivector.

Let $$\setlr{e_1, e_2}$$ be an orthonormal basis for a two dimensional space, with reciprocal frame $$\setlr{e^1, e^2}$$. Expand the area bivector $$d^2 \Bx$$ in coordinates relating the bivector to the Jacobian and the pseudoscalar.

With parameterization $$x = x(u,v) = x^\alpha e_\alpha = x_\alpha e^\alpha$$, we have
\label{eqn:unpackingFundamentalTheorem:120}
\Bx_u \wedge \Bx_v
=
\lr{ \PD{u}{x^\alpha} e_\alpha } \wedge
\lr{ \PD{v}{x^\beta} e_\beta }
=
\PD{u}{x^\alpha}
\PD{v}{x^\beta}
e_\alpha
e_\beta
=
\PD{(u,v)}{(x^1,x^2)} e_1 e_2,

or
\label{eqn:unpackingFundamentalTheorem:160}
\Bx_u \wedge \Bx_v
=
\lr{ \PD{u}{x_\alpha} e^\alpha } \wedge
\lr{ \PD{v}{x_\beta} e^\beta }
=
\PD{u}{x_\alpha}
\PD{v}{x_\beta}
e^\alpha
e^\beta
=
\PD{(u,v)}{(x_1,x_2)} e^1 e^2.

The upper and lower index pseudoscalars are related by
\label{eqn:unpackingFundamentalTheorem:180}
e^1 e^2 e_1 e_2 =
-e^1 e^2 e_2 e_1 =
-1,

so with $$I = e_1 e_2$$,
\label{eqn:unpackingFundamentalTheorem:200}
e^1 e^2 = -I^{-1},

leaving us with
\label{eqn:unpackingFundamentalTheorem:140}
d^2 \Bx
= \PD{(u,v)}{(x^1,x^2)} du dv\, I
= -\PD{(u,v)}{(x_1,x_2)} du dv\, I^{-1}.

We see that the area bivector is proportional to either the upper or lower index Jacobian and to the pseudoscalar for the space.

We may write the fundamental theorem for a 2D space as
\label{eqn:unpackingFundamentalTheorem:680}
\int_S du dv \, \PD{(u,v)}{(x^1,x^2)} F I \lrgrad G = \oint_{\partial S} F d\Bx G,

where we have dispensed with the vector derivative and use the gradient instead, since they are identical in a two parameter two dimensional space. Of course, unless we are using $$x^1, x^2$$ as our parameterization, we still want the curvilinear representation of the gradient $$\grad = \Bx^u \PDi{u}{} + \Bx^v \PDi{v}{}$$.

## Problem: Standard basis expansion of fundamental surface relation.

For a parameterization $$x = x^1 e_1 + x^2 e_2$$, where $$\setlr{ e_1, e_2 }$$ is a standard (orthogonal) basis, expand the fundamental theorem for surface integrals for the single sided $$F = 1$$ case. Consider functions $$G$$ of each grade (scalar, vector, bivector.)

From \ref{eqn:unpackingFundamentalTheorem:140} we see that the fundamental theorem takes the form
\label{eqn:unpackingFundamentalTheorem:220}
\int_S dx^1 dx^2\, F I \lrgrad G = \oint_{\partial S} F d\Bx G.

In a Euclidean space, the operator $$I \lrgrad$$, is a $$\pi/2$$ rotation of the gradient, but has a rotated like structure in all metrics:
\label{eqn:unpackingFundamentalTheorem:240}
=
e_1 e_2 \lr{ e^1 \partial_1 + e^2 \partial_2 }
=
-e_2 \partial_1 + e_1 \partial_2.

• $$F = 1$$ and $$G \in \bigwedge^0$$ or $$G \in \bigwedge^2$$. For $$F = 1$$ and scalar or bivector $$G$$ we have
\label{eqn:unpackingFundamentalTheorem:260}
\int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } G = \oint_{\partial S} d\Bx G,

where, for $$x^1 \in [x^1(0),x^1(1)]$$ and $$x^2 \in [x^2(0),x^2(1)]$$, the RHS written explicitly is
\label{eqn:unpackingFundamentalTheorem:280}
\oint_{\partial S} d\Bx G
=
\int dx^1 e_1
\lr{ G(x^1, x^2(1)) – G(x^1, x^2(0)) }
– dx^2 e_2
\lr{ G(x^1(1),x^2) – G(x^1(0), x^2) }.

This is sketched in fig. 2. Since a 2D bivector $$G$$ can be written as $$G = I g$$, where $$g$$ is a scalar, we may write the pseudoscalar case as
\label{eqn:unpackingFundamentalTheorem:300}
\int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } g = \oint_{\partial S} d\Bx g,

after right multiplying both sides with $$I^{-1}$$. Algebraically the scalar and pseudoscalar cases can be thought of as identical scalar relationships.
• $$F = 1, G \in \bigwedge^1$$. For $$F = 1$$ and vector $$G$$ the 2D fundamental theorem for surfaces can be split into scalar
\label{eqn:unpackingFundamentalTheorem:320}
\int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot G = \oint_{\partial S} d\Bx \cdot G,

and bivector relations
\label{eqn:unpackingFundamentalTheorem:340}
\int_S dx^1 dx^2\, \lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G = \oint_{\partial S} d\Bx \wedge G.

To expand \ref{eqn:unpackingFundamentalTheorem:320}, let
\label{eqn:unpackingFundamentalTheorem:360}
G = g_1 e^1 + g_2 e^2,

for which
\label{eqn:unpackingFundamentalTheorem:380}
\lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot G
=
\lr{ -e_2 \partial_1 + e_1 \partial_2 } \cdot
\lr{ g_1 e^1 + g_2 e^2 }
=
\partial_2 g_1 – \partial_1 g_2,

and
\label{eqn:unpackingFundamentalTheorem:400}
d\Bx \cdot G
=
\lr{ dx^1 e_1 – dx^2 e_2 } \cdot \lr{ g_1 e^1 + g_2 e^2 }
=
dx^1 g_1 – dx^2 g_2,

so \ref{eqn:unpackingFundamentalTheorem:320} expands to
\label{eqn:unpackingFundamentalTheorem:500}
\int_S dx^1 dx^2\, \lr{ \partial_2 g_1 – \partial_1 g_2 }
=
\int
\evalbar{dx^1 g_1}{\Delta x^2} – \evalbar{ dx^2 g_2 }{\Delta x^1}.

This coordinate expansion illustrates how the pseudoscalar nature of the area element results in a duality transformation, as we end up with a curl like operation on the LHS, despite the dot product nature of the decomposition that we used. That can also be seen directly for vector $$G$$, since
\label{eqn:unpackingFundamentalTheorem:560}
=
=
dA I \lr{ \grad \wedge G },

since the scalar selection of $$I \lr{ \grad \cdot G }$$ is zero.In the grade-2 relation \ref{eqn:unpackingFundamentalTheorem:340}, we expect a pseudoscalar cancellation on both sides, leaving a scalar (divergence-like) relationship. This time, we use upper index coordinates for the vector $$G$$, letting
\label{eqn:unpackingFundamentalTheorem:440}
G = g^1 e_1 + g^2 e_2,

so
\label{eqn:unpackingFundamentalTheorem:460}
\lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G
=
\lr{ -e_2 \partial_1 + e_1 \partial_2 } \wedge G
\lr{ g^1 e_1 + g^2 e_2 }
=
e_1 e_2 \lr{ \partial_1 g^1 + \partial_2 g^2 },

and
\label{eqn:unpackingFundamentalTheorem:480}
d\Bx \wedge G
=
\lr{ dx^1 e_1 – dx^2 e_2 } \wedge
\lr{ g^1 e_1 + g^2 e_2 }
=
e_1 e_2 \lr{ dx^1 g^2 + dx^2 g^1 }.

So \ref{eqn:unpackingFundamentalTheorem:340}, after multiplication of both sides by $$I^{-1}$$, is
\label{eqn:unpackingFundamentalTheorem:520}
\int_S dx^1 dx^2\,
\lr{ \partial_1 g^1 + \partial_2 g^2 }
=
\int
\evalbar{dx^1 g^2}{\Delta x^2} + \evalbar{dx^2 g^1 }{\Delta x^1}.

As before, we’ve implicitly performed a duality transformation, and end up with a divergence operation. That can be seen directly without coordinate expansion, by rewriting the wedge as a grade two selection, and expanding the gradient action on the vector $$G$$, as follows
\label{eqn:unpackingFundamentalTheorem:580}
=
=
dA I \lr{ \grad \cdot G },

since $$I \lr{ \grad \wedge G }$$ has only a scalar component.

fig. 2. Line integral around rectangular boundary.

## Theorem 1.1: Green’s theorem [1].

Let $$S$$ be a Jordan region with a piecewise-smooth boundary $$C$$. If $$P, Q$$ are continuously differentiable on an open set that contains $$S$$, then
\begin{equation*}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} } = \oint P dx + Q dy.
\end{equation*}

## Problem: Relationship to Green’s theorem.

If the space is Euclidean, show that \ref{eqn:unpackingFundamentalTheorem:500} and \ref{eqn:unpackingFundamentalTheorem:520} are both instances of Green’s theorem with suitable choices of $$P$$ and $$Q$$.

I will omit the subtleties related to general regions and consider just the case of an infinitesimal square region.

### Start proof:

Let’s start with \ref{eqn:unpackingFundamentalTheorem:500}, with $$g_1 = P$$ and $$g_2 = Q$$, and $$x^1 = x, x^2 = y$$, the RHS is
\label{eqn:unpackingFundamentalTheorem:600}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }.

On the RHS we have
\label{eqn:unpackingFundamentalTheorem:620}
\int \evalbar{dx P}{\Delta y} – \evalbar{ dy Q }{\Delta x}
=
\int dx \lr{ P(x, y_1) – P(x, y_0) } – \int dy \lr{ Q(x_1, y) – Q(x_0, y) }.

This pair of integrals is plotted in fig. 3, from which we see that \ref{eqn:unpackingFundamentalTheorem:620} can be expressed as the line integral, leaving us with
\label{eqn:unpackingFundamentalTheorem:640}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }
=
\oint dx P + dy Q,

which is Green’s theorem over the infinitesimal square integration region.

For the equivalence of \ref{eqn:unpackingFundamentalTheorem:520} to Green’s theorem, let $$g^2 = P$$, and $$g^1 = -Q$$. Plugging into the LHS, we find the Green’s theorem integrand. On the RHS, the integrand expands to
\label{eqn:unpackingFundamentalTheorem:660}
\evalbar{dx g^2}{\Delta y} + \evalbar{dy g^1 }{\Delta x}
=
dx \lr{ P(x,y_1) – P(x, y_0)}
+
dy \lr{ -Q(x_1, y) + Q(x_0, y)},

which is exactly what we found in \ref{eqn:unpackingFundamentalTheorem:620}.

### End proof.

fig. 3. Path for Green’s theorem.

We may also relate multivector gradient integrals in 2D to the normal integral around the boundary of the bounding curve. That relationship is as follows.

## Theorem 1.2: 2D gradient integrals.

\begin{equation*}
\begin{aligned}
\int J du dv \rgrad G &= \oint I^{-1} d\Bx G = \int J \lr{ \Bx^v du + \Bx^u dv } G \\
\int J du dv F \lgrad &= \oint F I^{-1} d\Bx = \int J F \lr{ \Bx^v du + \Bx^u dv },
\end{aligned}
\end{equation*}
where $$J = \partial(x^1, x^2)/\partial(u,v)$$ is the Jacobian of the parameterization $$x = x(u,v)$$. In terms of the coordinates $$x^1, x^2$$, this reduces to
\begin{equation*}
\begin{aligned}
\int dx^1 dx^2 \rgrad G &= \oint I^{-1} d\Bx G = \int \lr{ e^2 dx^1 + e^1 dx^2 } G \\
\int dx^1 dx^2 F \lgrad &= \oint G I^{-1} d\Bx = \int F \lr{ e^2 dx^1 + e^1 dx^2 }.
\end{aligned}
\end{equation*}
The vector $$I^{-1} d\Bx$$ is orthogonal to the tangent vector along the boundary, and for Euclidean spaces it can be identified as the outwards normal.

### Start proof:

Respectively setting $$F = 1$$, and $$G = 1$$ in \ref{eqn:unpackingFundamentalTheorem:680}, we have
\label{eqn:unpackingFundamentalTheorem:940}
\int I^{-1} d^2 \Bx \rgrad G = \oint I^{-1} d\Bx G,

and
\label{eqn:unpackingFundamentalTheorem:960}
\int F d^2 \Bx \lgrad I^{-1} = \oint F d\Bx I^{-1}.

Starting with \ref{eqn:unpackingFundamentalTheorem:940} we find
\label{eqn:unpackingFundamentalTheorem:700}
\int I^{-1} J du dv I \rgrad G = \oint d\Bx G,

to find $$\int dx^1 dx^2 \rgrad G = \oint I^{-1} d\Bx G$$, as desireed. In terms of a parameterization $$x = x(u,v)$$, the pseudoscalar for the space is
\label{eqn:unpackingFundamentalTheorem:720}
I = \frac{\Bx_u \wedge \Bx_v}{J},

so
\label{eqn:unpackingFundamentalTheorem:740}
I^{-1} = \frac{J}{\Bx_u \wedge \Bx_v}.

Also note that $$\lr{\Bx_u \wedge \Bx_v}^{-1} = \Bx^v \wedge \Bx^u$$, so
\label{eqn:unpackingFundamentalTheorem:760}
I^{-1} = J \lr{ \Bx^v \wedge \Bx^u },

and
\label{eqn:unpackingFundamentalTheorem:780}
I^{-1} d\Bx
= I^{-1} \cdot d\Bx
= J \lr{ \Bx^v \wedge \Bx^u } \cdot \lr{ \Bx_u du – \Bx_v dv }
= J \lr{ \Bx^v du + \Bx^u dv },

so the right acting gradient integral is
\label{eqn:unpackingFundamentalTheorem:800}
\int J du dv \grad G =
\int
\evalbar{J \Bx^v G}{\Delta v} du + \evalbar{J \Bx^u G dv}{\Delta u},

which we write in abbreviated form as $$\int J \lr{ \Bx^v du + \Bx^u dv} G$$.

For the $$G = 1$$ case, from \ref{eqn:unpackingFundamentalTheorem:960} we find
\label{eqn:unpackingFundamentalTheorem:820}
\int J du dv F I \lgrad I^{-1} = \oint F d\Bx I^{-1}.

However, in a 2D space, regardless of metric, we have $$I a = – a I$$ for any vector $$a$$ (i.e. $$\grad$$ or $$d\Bx$$), so we may commute the outer pseudoscalars in
\label{eqn:unpackingFundamentalTheorem:840}
\int J du dv F I \lgrad I^{-1} = \oint F d\Bx I^{-1},

so
\label{eqn:unpackingFundamentalTheorem:850}
-\int J du dv F I I^{-1} \lgrad = -\oint F I^{-1} d\Bx.

After cancelling the negative sign on both sides, we have the claimed result.

To see that $$I a$$, for any vector $$a$$ is normal to $$a$$, we can compute the dot product
\label{eqn:unpackingFundamentalTheorem:860}
\lr{ I a } \cdot a
=
=
= 0,

since the scalar selection of a bivector is zero. Since $$I^{-1} = \pm I$$, the same argument shows that $$I^{-1} d\Bx$$ must be orthogonal to $$d\Bx$$.

### End proof.

Let’s look at the geometry of the normal $$I^{-1} \Bx$$ in a couple 2D vector spaces. We use an integration volume of a unit square to simplify the boundary term expressions.

• Euclidean: With a parameterization $$x(u,v) = u\Be_1 + v \Be_2$$, and Euclidean basis vectors $$(\Be_1)^2 = (\Be_2)^2 = 1$$, the fundamental theorem integrated over the rectangle $$[x_0,x_1] \times [y_0,y_1]$$ is
\label{eqn:unpackingFundamentalTheorem:880}
\int dx dy \grad G =
\int
\Be_2 \lr{ G(x,y_1) – G(x,y_0) } dx +
\Be_1 \lr{ G(x_1,y) – G(x_0,y) } dy,

Each of the terms in the integrand above are illustrated in fig. 4, and we see that this is a path integral weighted by the outwards normal.

fig. 4. Outwards oriented normal for Euclidean space.

• Spacetime: Let $$x(u,v) = u \gamma_0 + v \gamma_1$$, where $$(\gamma_0)^2 = -(\gamma_1)^2 = 1$$. With $$u = t, v = x$$, the gradient integral over a $$[t_0,t_1] \times [x_0,x_1]$$ of spacetime is
\label{eqn:unpackingFundamentalTheorem:900}
\begin{aligned}
&=
\int
\gamma^1 dt \lr{ G(t, x_1) – G(t, x_0) }
+
\gamma^0 dx \lr{ G(t_1, x) – G(t_1, x) } \\
&=
\int
\gamma_1 dt \lr{ -G(t, x_1) + G(t, x_0) }
+
\gamma_0 dx \lr{ G(t_1, x) – G(t_1, x) }
.
\end{aligned}

With $$t$$ plotted along the horizontal axis, and $$x$$ along the vertical, each of the terms of this integrand is illustrated graphically in fig. 5. For this mixed signature space, there is no longer any good geometrical characterization of the normal.

fig. 5. Orientation of the boundary normal for a spacetime basis.

• Spacelike:
Let $$x(u,v) = u \gamma_1 + v \gamma_2$$, where $$(\gamma_1)^2 = (\gamma_2)^2 = -1$$. With $$u = x, v = y$$, the gradient integral over a $$[x_0,x_1] \times [y_0,y_1]$$ of this space is
\label{eqn:unpackingFundamentalTheorem:920}
\begin{aligned}
&=
\int
\gamma^2 dx \lr{ G(x, y_1) – G(x, y_0) }
+
\gamma^1 dy \lr{ G(x_1, y) – G(x_1, y) } \\
&=
\int
\gamma_2 dx \lr{ -G(x, y_1) + G(x, y_0) }
+
\gamma_1 dy \lr{ -G(x_1, y) + G(x_1, y) }
.
\end{aligned}

Referring to fig. 6. where the elements of the integrand are illustrated, we see that the normal $$I^{-1} d\Bx$$ for the boundary of this region can be characterized as inwards.

fig. 6. Inwards oriented normal for a Dirac spacelike basis.

# References

[1] S.L. Salas and E. Hille. Calculus: one and several variables. Wiley New York, 1990.

## 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.

## Fundamental theorem of geometric calculus for line integrals (relativistic.)

[This post is best viewed in PDF form, due to latex elements that I could not format with wordpress mathjax.]

Background for this particular post can be found in

## Motivation.

I’ve been slowly working my way towards a statement of the fundamental theorem of integral calculus, where the functions being integrated are elements of the Dirac algebra (space time multivectors in the geometric algebra parlance.)

This is interesting because we want to be able to do line, surface, 3-volume and 4-volume space time integrals. We have many $$\mathbb{R}^3$$ integral theorems
\label{eqn:fundamentalTheoremOfGC:40a}
\int_A^B d\Bl \cdot \spacegrad f = f(B) – f(A),

\label{eqn:fundamentalTheoremOfGC:60a}
\int_S dA\, \ncap \cross \spacegrad f = \int_{\partial S} d\Bx\, f,

\label{eqn:fundamentalTheoremOfGC:80a}
\int_S dA\, \ncap \cdot \lr{ \spacegrad \cross \Bf} = \int_{\partial S} d\Bx \cdot \Bf,

\label{eqn:fundamentalTheoremOfGC:100a}
\int_S dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }
=
\int_{\partial S} P dx + Q dy,

\label{eqn:fundamentalTheoremOfGC:120a}
\int_V dV\, \spacegrad f = \int_{\partial V} dA\, \ncap f,

\label{eqn:fundamentalTheoremOfGC:140a}
\int_V dV\, \spacegrad \cross \Bf = \int_{\partial V} dA\, \ncap \cross \Bf,

\label{eqn:fundamentalTheoremOfGC:160a}
\int_V dV\, \spacegrad \cdot \Bf = \int_{\partial V} dA\, \ncap \cdot \Bf,

and want to know how to generalize these to four dimensions and also make sure that we are handling the relativistic mixed signature correctly. If our starting point was the mess of equations above, we’d be in trouble, since it is not obvious how these generalize. All the theorems with unit normals have to be handled completely differently in four dimensions since we don’t have a unique normal to any given spacetime plane.
What comes to our rescue is the Fundamental Theorem of Geometric Calculus (FTGC), which has the form
\label{eqn:fundamentalTheoremOfGC:40}
\int F d^n \Bx\, \lrpartial G = \int F d^{n-1} \Bx\, G,

where $$F,G$$ are multivectors functions (i.e. sums of products of vectors.) We’ve seen ([2], [1]) that all the identities above are special cases of the fundamental theorem.

Do we need any special care to state the FTGC correctly for our relativistic case? It turns out that the answer is no! Tangent and reciprocal frame vectors do all the heavy lifting, and we can use the fundamental theorem as is, even in our mixed signature space. The only real change that we need to make is use spacetime gradient and vector derivative operators instead of their spatial equivalents. We will see how this works below. Note that instead of starting with \ref{eqn:fundamentalTheoremOfGC:40} directly, I will attempt to build up to that point in a progressive fashion that is hopefully does not require the reader to make too many unjustified mental leaps.

## Multivector line integrals.

We want to define multivector line integrals to start with. Recall that in $$\mathbb{R}^3$$ we would say that for scalar functions $$f$$, the integral
\label{eqn:fundamentalTheoremOfGC:180b}
\int d\Bx\, f = \int f d\Bx,

is a line integral. Also, for vector functions $$\Bf$$ we call
\label{eqn:fundamentalTheoremOfGC:200}
\int d\Bx \cdot \Bf = \inv{2} \int d\Bx\, \Bf + \Bf d\Bx.

a line integral. In order to generalize line integrals to multivector functions, we will allow our multivector functions to be placed on either or both sides of the differential.

## Definition 1.1: Line integral.

Given a single variable parameterization $$x = x(u)$$, we write $$d^1\Bx = \Bx_u du$$, and call
\label{eqn:fundamentalTheoremOfGC:220a}
\int F d^1\Bx\, G,

a line integral, where $$F,G$$ are arbitrary multivector functions.

We must be careful not to reorder any of the factors in the integrand, since the differential may not commute with either $$F$$ or $$G$$. Here is a simple example where the integrand has a product of a vector and differential.

## Problem: Circular parameterization.

Given a circular parameterization $$x(\theta) = \gamma_1 e^{-i\theta}$$, where $$i = \gamma_1 \gamma_2$$, the unit bivector for the $$x,y$$ plane. Compute the line integral
\label{eqn:fundamentalTheoremOfGC:100}
\int_0^{\pi/4} F(\theta)\, d^1 \Bx\, G(\theta),

where $$F(\theta) = \Bx^\theta + \gamma_3 + \gamma_1 \gamma_0$$ is a multivector valued function, and $$G(\theta) = \gamma_0$$ is vector valued.

The tangent vector for the curve is
\label{eqn:fundamentalTheoremOfGC:60}
\Bx_\theta
= -\gamma_1 \gamma_1 \gamma_2 e^{-i\theta}
= \gamma_2 e^{-i\theta},

with reciprocal vector $$\Bx^\theta = e^{i \theta} \gamma^2$$. The differential element is $$d^1 \Bx = \gamma_2 e^{-i\theta} d\theta$$, so the integrand is
\label{eqn:fundamentalTheoremOfGC:80}
\begin{aligned}
\int_0^{\pi/4} \lr{ \Bx^\theta + \gamma_3 + \gamma_1 \gamma_0 } d^1 \Bx\, \gamma_0
&=
\int_0^{\pi/4} \lr{ e^{i\theta} \gamma^2 + \gamma_3 + \gamma_1 \gamma_0 } \gamma_2 e^{-i\theta} d\theta\, \gamma_0 \\
&=
\frac{\pi}{4} \gamma_0 + \lr{ \gamma_{32} + \gamma_{102} } \inv{-i} \lr{ e^{-i\pi/4} – 1 } \gamma_0 \\
&=
\frac{\pi}{4} \gamma_0 + \inv{\sqrt{2}} \lr{ \gamma_{32} + \gamma_{102} } \gamma_{120} \lr{ 1 – \gamma_{12} } \\
&=
\frac{\pi}{4} \gamma_0 + \inv{\sqrt{2}} \lr{ \gamma_{310} + 1 } \lr{ 1 – \gamma_{12} }.
\end{aligned}

Observe how care is required not to reorder any terms. This particular end result is a multivector with scalar, vector, bivector, and trivector grades, but no pseudoscalar component. The grades in the end result depend on both the function in the integrand and on the path. For example, had we integrated all the way around the circle, the end result would have been the vector $$2 \pi \gamma_0$$ (i.e. a $$\gamma_0$$ weighted unit circle circumference), as all the other grades would have been killed by the complex exponential integrated over a full period.

## Problem: Line integral for boosted time direction vector.

Let $$x = e^{\vcap \alpha/2} \gamma_0 e^{-\vcap \alpha/2}$$ represent the spacetime curve of all the boosts of $$\gamma_0$$ along a specific velocity direction vector, where $$\vcap = (v \wedge \gamma_0)/\Norm{v \wedge \gamma_0}$$ is a unit spatial bivector for any constant vector $$v$$. Compute the line integral
\label{eqn:fundamentalTheoremOfGC:240}
\int x\, d^1 \Bx.

Observe that $$\vcap$$ and $$\gamma_0$$ anticommute, so we may write our boost as a one sided exponential
\label{eqn:fundamentalTheoremOfGC:260}
x(\alpha) = \gamma_0 e^{-\vcap \alpha} = e^{\vcap \alpha} \gamma_0 = \lr{ \cosh\alpha + \vcap \sinh\alpha } \gamma_0.

The tangent vector is just
\label{eqn:fundamentalTheoremOfGC:280}
\Bx_\alpha = \PD{\alpha}{x} = e^{\vcap\alpha} \vcap \gamma_0.

Let’s get a bit of intuition about the nature of this vector. It’s square is
\label{eqn:fundamentalTheoremOfGC:300}
\begin{aligned}
\Bx_\alpha^2
&=
e^{\vcap\alpha} \vcap \gamma_0
e^{\vcap\alpha} \vcap \gamma_0 \\
&=
-e^{\vcap\alpha} \vcap e^{-\vcap\alpha} \vcap (\gamma_0)^2 \\
&=
-1,
\end{aligned}

so we see that the tangent vector is a spacelike unit vector. As the vector representing points on the curve is necessarily timelike (due to Lorentz invariance), these two must be orthogonal at all points. Let’s confirm this algebraically
\label{eqn:fundamentalTheoremOfGC:320}
\begin{aligned}
x \cdot \Bx_\alpha
&=
\gpgradezero{ e^{\vcap \alpha} \gamma_0 e^{\vcap \alpha} \vcap \gamma_0 } \\
&=
\gpgradezero{ e^{-\vcap \alpha} e^{\vcap \alpha} \vcap (\gamma_0)^2 } \\
&=
&= 0.
\end{aligned}

Here we used $$e^{\vcap \alpha} \gamma_0 = \gamma_0 e^{-\vcap \alpha}$$, and $$\gpgradezero{A B} = \gpgradezero{B A}$$. Geometrically, we have the curious fact that the direction vectors to points on the curve are perpendicular (with respect to our relativistic dot product) to the tangent vectors on the curve, as illustrated in fig. 1.

fig. 1. Tangent perpendicularity in mixed metric.

### Perfect differentials.

Having seen a couple examples of multivector line integrals, let’s now move on to figure out the structure of a line integral that has a “perfect” differential integrand. We can take a hint from the $$\mathbb{R}^3$$ vector result that we already know, namely
\label{eqn:fundamentalTheoremOfGC:120}
\int_A^B d\Bl \cdot \spacegrad f = f(B) – f(A).

It seems reasonable to guess that the relativistic generalization of this is
\label{eqn:fundamentalTheoremOfGC:140}
\int_A^B dx \cdot \grad f = f(B) – f(A).

Let’s check that, by expanding in coordinates
\label{eqn:fundamentalTheoremOfGC:160}
\begin{aligned}
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \partial_\mu f \\
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \PD{x^\mu}{f} \\
&=
\int_A^B d\tau \frac{df}{d\tau} \\
&=
f(B) – f(A).
\end{aligned}

If we drop the dot product, will we have such a nice result? Let’s see:
\label{eqn:fundamentalTheoremOfGC:180}
\begin{aligned}
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \gamma_\mu \gamma^\nu \partial_\nu f \\
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \PD{x^\mu}{f}
+
\int_A^B
d\tau
\sum_{\mu \ne \nu} \gamma_\mu \gamma^\nu
\frac{dx^\mu}{d\tau} \PD{x^\nu}{f}.
\end{aligned}

This scalar component of this integrand is a perfect differential, but the bivector part of the integrand is a complete mess, that we have no hope of generally integrating. It happens that if we consider one of the simplest parameterization examples, we can get a strong hint of how to generalize the differential operator to one that ends up providing a perfect differential. In particular, let’s integrate over a linear constant path, such as $$x(\tau) = \tau \gamma_0$$. For this path, we have
\label{eqn:fundamentalTheoremOfGC:200a}
\begin{aligned}
&=
\int_A^B \gamma_0 d\tau \lr{
\gamma^0 \partial_0 +
\gamma^1 \partial_1 +
\gamma^2 \partial_2 +
\gamma^3 \partial_3 } f \\
&=
\int_A^B d\tau \lr{
\PD{\tau}{f} +
\gamma_0 \gamma^1 \PD{x^1}{f} +
\gamma_0 \gamma^2 \PD{x^2}{f} +
\gamma_0 \gamma^3 \PD{x^3}{f}
}.
\end{aligned}

Just because the path does not have any $$x^1, x^2, x^3$$ component dependencies does not mean that these last three partials are neccessarily zero. For example $$f = f(x(\tau)) = \lr{ x^0 }^2 \gamma_0 + x^1 \gamma_1$$ will have a non-zero contribution from the $$\partial_1$$ operator. In that particular case, we can easily integrate $$f$$, but we have to know the specifics of the function to do the integral. However, if we had a differential operator that did not include any component off the integration path, we would ahve a perfect differential. That is, if we were to replace the gradient with the projection of the gradient onto the tangent space, we would have a perfect differential. We see that the function of the dot product in \ref{eqn:fundamentalTheoremOfGC:140} has the same effect, as it rejects any component of the gradient that does not lie on the tangent space.

## Definition 1.2: Vector derivative.

Given a spacetime manifold parameterized by $$x = x(u^0, \cdots u^{N-1})$$, with tangent vectors $$\Bx_\mu = \PDi{u^\mu}{x}$$, and reciprocal vectors $$\Bx^\mu \in \textrm{Span}\setlr{\Bx_\nu}$$, such that $$\Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu$$, the vector derivative is defined as
\label{eqn:fundamentalTheoremOfGC:240a}
\partial = \sum_{\mu = 0}^{N-1} \Bx^\mu \PD{u^\mu}{}.

Observe that if this is a full parameterization of the space ($$N = 4$$), then the vector derivative is identical to the gradient. The vector derivative is the projection of the gradient onto the tangent space at the point of evaluation.Furthermore, we designate $$\lrpartial$$ as the vector derivative allowed to act bidirectionally, as follows
\label{eqn:fundamentalTheoremOfGC:260a}
R \lrpartial S
=
R \Bx^\mu \PD{u^\mu}{S}
+
\PD{u^\mu}{R} \Bx^\mu S,

where $$R, S$$ are multivectors, and summation convention is implied. In this bidirectional action,
the vector factors of the vector derivative must stay in place (as they do not neccessarily commute with $$R,S$$), but the derivative operators apply in a chain rule like fashion to both functions.

Noting that $$\Bx_u \cdot \grad = \Bx_u \cdot \partial$$, we may rewrite the scalar line integral identity \ref{eqn:fundamentalTheoremOfGC:140} as
\label{eqn:fundamentalTheoremOfGC:220}
\int_A^B dx \cdot \partial f = f(B) – f(A).

However, as our example hinted at, the fundamental theorem for line integrals has a multivector generalization that does not rely on a dot product to do the tangent space filtering, and is more powerful. That generalization has the following form.

## Theorem 1.1: Fundamental theorem for line integrals.

Given multivector functions $$F, G$$, and a single parameter curve $$x(u)$$ with line element $$d^1 \Bx = \Bx_u du$$, then
\label{eqn:fundamentalTheoremOfGC:280a}
\int_A^B F d^1\Bx \lrpartial G = F(B) G(B) – F(A) G(A).

### Start proof:

Writing out the integrand explicitly, we find
\label{eqn:fundamentalTheoremOfGC:340}
\int_A^B F d^1\Bx \lrpartial G
=
\int_A^B \lr{
\PD{\alpha}{F} d\alpha\, \Bx_\alpha \Bx^\alpha G
+
F d\alpha\, \Bx_\alpha \Bx^\alpha \PD{\alpha}{G }
}

However for a single parameter curve, we have $$\Bx^\alpha = 1/\Bx_\alpha$$, so we are left with
\label{eqn:fundamentalTheoremOfGC:360}
\begin{aligned}
\int_A^B F d^1\Bx \lrpartial G
&=
\int_A^B d\alpha\, \PD{\alpha}{(F G)} \\
&=
\evalbar{F G}{B}

\evalbar{F G}{A}.
\end{aligned}

## More to come.

In the next installment we will explore surface integrals in spacetime, and the generalization of the fundamental theorem to multivector space time integrals.

# References

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

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

## Lorentz transformations in Space Time Algebra (STA)

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

## Motivation.

One of the remarkable features of geometric algebra are the complex exponential sandwiches that can be used to encode rotations in any dimension, or rotation like operations like Lorentz transformations in Minkowski spaces. In this post, we show some examples that unpack the geometric algebra expressions for Lorentz transformations operations of this sort. In particular, we will look at the exponential sandwich operations for spatial rotations and Lorentz boosts in the Dirac algebra, known as Space Time Algebra (STA) in geometric algebra circles, and demonstrate that these sandwiches do have the desired effects.

## Theorem 1.1: Lorentz transformation.

The transformation
\label{eqn:lorentzTransform:580}
x \rightarrow e^{B} x e^{-B} = x’,

where $$B = a \wedge b$$, is an STA 2-blade for any two linearly independent four-vectors $$a, b$$, is a norm preserving, that is
\label{eqn:lorentzTransform:600}
x^2 = {x’}^2.

### Start proof:

The proof is disturbingly trivial in this geometric algebra form
\label{eqn:lorentzTransform:40}
\begin{aligned}
{x’}^2
&=
e^{B} x e^{-B} e^{B} x e^{-B} \\
&=
e^{B} x x e^{-B} \\
&=
x^2 e^{B} e^{-B} \\
&=
x^2.
\end{aligned}

### End proof.

In particular, observe that we did not need to construct the usual infinitesimal representations of rotation and boost transformation matrices or tensors in order to demonstrate that we have spacetime invariance for the transformations. The rough idea of such a transformation is that the exponential commutes with components of the four-vector that lie off the spacetime plane specified by the bivector $$B$$, and anticommutes with components of the four-vector that lie in the plane. The end result is that the sandwich operation simplifies to
\label{eqn:lorentzTransform:60}
x’ = x_\parallel e^{-B} + x_\perp,

where $$x = x_\perp + x_\parallel$$ and $$x_\perp \cdot B = 0$$, and $$x_\parallel \wedge B = 0$$. In particular, using $$x = x B B^{-1} = \lr{ x \cdot B + x \wedge B } B^{-1}$$, we find that
\label{eqn:lorentzTransform:80}
\begin{aligned}
x_\parallel &= \lr{ x \cdot B } B^{-1} \\
x_\perp &= \lr{ x \wedge B } B^{-1}.
\end{aligned}

When $$B$$ is a spacetime plane $$B = b \wedge \gamma_0$$, then this exponential has a hyperbolic nature, and we end up with a Lorentz boost. When $$B$$ is a spatial bivector, we end up with a single complex exponential, encoding our plane old 3D rotation. More general $$B$$’s that encode composite boosts and rotations are also possible, but $$B$$ must be invertible (it should have no lightlike factors.) The rough geometry of these projections is illustrated in fig 1, where the spacetime plane is represented by $$B$$.

fig 1. Projection and rejection geometry.

What is not so obvious is how to pick $$B$$’s that correspond to specific rotation axes or boost directions. Let’s consider each of those cases in turn.

## Theorem 1.2: Boost.

The boost along a direction vector $$\vcap$$ and rapidity $$\alpha$$ is given by
\label{eqn:lorentzTransform:620}
x’ = e^{-\vcap \alpha/2} x e^{\vcap \alpha/2},

where $$\vcap = \gamma_{k0} \cos\theta^k$$ is an STA bivector representing a spatial direction with direction cosines $$\cos\theta^k$$.

### Start proof:

We want to demonstrate that this is equivalent to the usual boost formulation. We can start with decomposition of the four-vector $$x$$ into components that lie in and off of the spacetime plane $$\vcap$$.
\label{eqn:lorentzTransform:100}
\begin{aligned}
x
&= \lr{ x^0 + \Bx } \gamma_0 \\
&= \lr{ x^0 + \Bx \vcap^2 } \gamma_0 \\
&= \lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap + \lr{ \Bx \wedge \vcap} \vcap } \gamma_0,
\end{aligned}

where $$\Bx = x \wedge \gamma_0$$. The first two components lie in the boost plane, whereas the last is the spatial component of the vector that lies perpendicular to the boost plane. Observe that $$\vcap$$ anticommutes with the dot product term and commutes with he wedge product term, so we have
\label{eqn:lorentzTransform:120}
\begin{aligned}
x’
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap } \vcap } \gamma_0
e^{\vcap \alpha/2 }
e^{\vcap \alpha/2 }
+
\lr{ \Bx \wedge \vcap } \vcap \gamma_0
e^{-\vcap \alpha/2 }
e^{\vcap \alpha/2 } \\
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap } \vcap } \gamma_0
e^{\vcap \alpha }
+
\lr{ \Bx \wedge \vcap } \vcap \gamma_0.
\end{aligned}

Noting that $$\vcap^2 = 1$$, we may expand the exponential in hyperbolic functions, and find that the boosted portion of the vector expands as
\label{eqn:lorentzTransform:260}
\begin{aligned}
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0 e^{\vcap \alpha}
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0 \lr{ \cosh\alpha + \vcap \sinh \alpha} \\
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \lr{ \cosh\alpha – \vcap \sinh \alpha} \gamma_0 \\
&=
\lr{ x^0 \cosh\alpha – \lr{ \Bx \cdot \vcap} \sinh \alpha} \gamma_0
+
\lr{ -x^0 \sinh \alpha + \lr{ \Bx \cdot \vcap} \cosh \alpha } \vcap \gamma_0.
\end{aligned}

We are left with
\label{eqn:lorentzTransform:320}
\begin{aligned}
x’
&=
\lr{ x^0 \cosh\alpha – \lr{ \Bx \cdot \vcap} \sinh \alpha} \gamma_0
+
\lr{ \lr{ \Bx \cdot \vcap} \cosh \alpha -x^0 \sinh \alpha } \vcap \gamma_0
+
\lr{ \Bx \wedge \vcap} \vcap \gamma_0 \\
&=
\begin{bmatrix}
\gamma_0 & \vcap \gamma_0
\end{bmatrix}
\begin{bmatrix}
\cosh\alpha & – \sinh\alpha \\
-\sinh\alpha & \cosh\alpha
\end{bmatrix}
\begin{bmatrix}
x^0 \\
\Bx \cdot \vcap
\end{bmatrix}
+
\lr{ \Bx \wedge \vcap} \vcap \gamma_0,
\end{aligned}

which has the desired Lorentz boost structure. Of course, this is usually seen with $$\vcap = \gamma_{10}$$ so that the components in the coordinate column vector are $$(ct, x)$$.

## Theorem 1.3: Spatial rotation.

Given two linearly independent spatial bivectors $$\Ba = a^k \gamma_{k0}, \Bb = b^k \gamma_{k0}$$, a rotation of $$\theta$$ radians in the plane of $$\Ba, \Bb$$ from $$\Ba$$ towards $$\Bb$$, is given by
\label{eqn:lorentzTransform:640}
x’ = e^{-i\theta} x e^{i\theta},

where $$i = (\Ba \wedge \Bb)/\Abs{\Ba \wedge \Bb}$$, is a unit (spatial) bivector.

### Start proof:

Without loss of generality, we may pick $$i = \acap \bcap$$, where $$\acap^2 = \bcap^2 = 1$$, and $$\acap \cdot \bcap = 0$$. With such an orthonormal basis for the plane, we can decompose our four vector into portions that lie in and off the plane
\label{eqn:lorentzTransform:400}
\begin{aligned}
x
&= \lr{ x^0 + \Bx } \gamma_0 \\
&= \lr{ x^0 + \Bx i i^{-1} } \gamma_0 \\
&= \lr{ x^0 + \lr{ \Bx \cdot i } i^{-1} + \lr{ \Bx \wedge i } i^{-1} } \gamma_0.
\end{aligned}

The projective term lies in the plane of rotation, whereas the timelike and spatial rejection term are perpendicular. That is
\label{eqn:lorentzTransform:420}
\begin{aligned}
x_\parallel &= \lr{ \Bx \cdot i } i^{-1} \gamma_0 \\
x_\perp &= \lr{ x^0 + \lr{ \Bx \wedge i } i^{-1} } \gamma_0,
\end{aligned}

where $$x_\parallel \wedge i = 0$$, and $$x_\perp \cdot i = 0$$. The plane pseudoscalar $$i$$ anticommutes with $$x_\parallel$$, and commutes with $$x_\perp$$, so
\label{eqn:lorentzTransform:440}
\begin{aligned}
x’
&= e^{-i\theta/2} \lr{ x_\parallel + x_\perp } e^{i\theta/2} \\
&= x_\parallel e^{i\theta} + x_\perp.
\end{aligned}

However
\label{eqn:lorentzTransform:460}
\begin{aligned}
\lr{ \Bx \cdot i } i^{-1}
&=
\lr{ \Bx \cdot \lr{ \acap \wedge \bcap } } \bcap \acap \\
&=
\lr{\Bx \cdot \acap} \bcap \bcap \acap
-\lr{\Bx \cdot \bcap} \acap \bcap \acap \\
&=
\lr{\Bx \cdot \acap} \acap
+\lr{\Bx \cdot \bcap} \bcap,
\end{aligned}

so
\label{eqn:lorentzTransform:480}
\begin{aligned}
x_\parallel e^{i\theta}
&=
\lr{
\lr{\Bx \cdot \acap} \acap
+
\lr{\Bx \cdot \bcap} \bcap
}
\gamma_0
\lr{
\cos\theta + \acap \bcap \sin\theta
} \\
&=
\acap \lr{
\lr{\Bx \cdot \acap} \cos\theta

\lr{\Bx \cdot \bcap} \sin\theta
}
\gamma_0
+
\bcap \lr{
\lr{\Bx \cdot \acap} \sin\theta
+
\lr{\Bx \cdot \bcap} \cos\theta
}
\gamma_0,
\end{aligned}

so
\label{eqn:lorentzTransform:500}
x’
=
\begin{bmatrix}
\acap & \bcap
\end{bmatrix}
\begin{bmatrix}
\cos\theta & – \sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\Bx \cdot \acap \\
\Bx \cdot \bcap \\
\end{bmatrix}
\gamma_0
+
\lr{ x \wedge i} i^{-1} \gamma_0.

Observe that this rejection term can be explicitly expanded to
\label{eqn:lorentzTransform:520}
\lr{ \Bx \wedge i} i^{-1} \gamma_0 =
x –
\lr{ \Bx \cdot \acap } \acap \gamma_0

\lr{ \Bx \cdot \acap } \acap \gamma_0.

This is the timelike component of the vector, plus the spatial component that is normal to the plane. This exponential sandwich transformation rotates only the portion of the vector that lies in the plane, and leaves the rest (timelike and normal) untouched.

## Problem: Verify components relative to boost direction.

In the proof of thm. 1.2, the vector $$x$$ was expanded in terms of the spacetime split. An alternate approach, is to expand as
\label{eqn:lorentzTransform:340}
\begin{aligned}
x
&= x \vcap^2 \\
&= \lr{ x \cdot \vcap + x \wedge \vcap } \vcap \\
&= \lr{ x \cdot \vcap } \vcap + \lr{ x \wedge \vcap } \vcap.
\end{aligned}

Show that
\label{eqn:lorentzTransform:360}
\lr{ x \cdot \vcap } \vcap
=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0,

and
\label{eqn:lorentzTransform:380}
\lr{ x \wedge \vcap } \vcap
=
\lr{ \Bx \wedge \vcap} \vcap \gamma_0.

Let $$x = x^\mu \gamma_\mu$$, so that
\label{eqn:lorentzTransform:160}
\begin{aligned}
x \cdot \vcap
&=
\gpgradeone{ x^\mu \gamma_\mu \cos\theta^b \gamma_{b 0} } \\
&=
x^\mu \cos\theta^b \gpgradeone{ \gamma_\mu \gamma_{b 0} }
.
\end{aligned}

The $$\mu = 0$$ component of this grade selection is
\label{eqn:lorentzTransform:180}
=
-\gamma_b,

and for $$\mu = a \ne 0$$, we have
\label{eqn:lorentzTransform:200}
=
-\delta_{a b} \gamma_0,

so we have
\label{eqn:lorentzTransform:220}
\begin{aligned}
x \cdot \vcap
&=
x^0 \cos\theta^b (-\gamma_b)
+
x^a \cos\theta^b (-\delta_{ab} \gamma_0 ) \\
&=
-x^0 \vcap \gamma_0

x^b \cos\theta^b \gamma_0 \\
&=
– \lr{ x^0 \vcap + \Bx \cdot \vcap } \gamma_0,
\end{aligned}

where $$\Bx = x \wedge \gamma_0$$ is the spatial portion of the four vector $$x$$ relative to the stationary observer frame. Since $$\vcap$$ anticommutes with $$\gamma_0$$, the component of $$x$$ in the spacetime plane $$\vcap$$ is
\label{eqn:lorentzTransform:240}
\lr{ x \cdot \vcap } \vcap =
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0,

as expected.

For the rejection term, we have
\label{eqn:lorentzTransform:280}
x \wedge \vcap
=
x^\mu \cos\theta^s \gpgradethree{ \gamma_\mu \gamma_{s 0} }.

The $$\mu = 0$$ term clearly contributes nothing, leaving us with:
\label{eqn:lorentzTransform:300}
\begin{aligned}
\lr{ x \wedge \vcap } \vcap
&=
\lr{ x \wedge \vcap } \cdot \vcap \\
&=
x^r \cos\theta^s \cos\theta^t \lr{ \lr{ \gamma_r \wedge \gamma_{s}} \gamma_0 } \cdot \lr{ \gamma_{t0} } \\
&=
\lr{ \gamma_r \wedge \gamma_{s} } \gamma_0 \gamma_{t0}
} \\
&=
-x^r \cos\theta^s \cos\theta^t \lr{ \gamma_r \wedge \gamma_{s}} \cdot \gamma_t \\
&=
-x^r \cos\theta^s \cos\theta^t \lr{ -\gamma_r \delta_{st} + \gamma_s \delta_{rt} } \\
&=
x^r \cos\theta^t \cos\theta^t \gamma_r

x^t \cos\theta^s \cos\theta^t \gamma_s \\
&=
\Bx \gamma_0
– (\Bx \cdot \vcap) \vcap \gamma_0 \\
&=
\lr{ \Bx \wedge \vcap} \vcap \gamma_0,
\end{aligned}

as expected. Is there a clever way to demonstrate this without resorting to coordinates?

## Problem: Rotation transformation components.

Given a unit spatial bivector $$i = \acap \bcap$$, where $$\acap \cdot \bcap = 0$$ and $$i^2 = -1$$, show that
\label{eqn:lorentzTransform:540}
\lr{ x \cdot i } i^{-1}
=
\lr{ \Bx \cdot i } i^{-1} \gamma_0
=
\lr{\Bx \cdot \acap } \acap \gamma_0
+
\lr{\Bx \cdot \bcap } \bcap \gamma_0,

and
\label{eqn:lorentzTransform:560}
\lr{ x \wedge i } i^{-1}
=
\lr{ \Bx \wedge i } i^{-1} \gamma_0
=
x –
\lr{\Bx \cdot \acap } \acap \gamma_0

\lr{\Bx \cdot \bcap } \bcap \gamma_0.

Also show that $$i$$ anticommutes with $$\lr{ x \cdot i } i^{-1}$$ and commutes with $$\lr{ x \wedge i } i^{-1}$$.

This problem is left for the reader, as I don’t feel like typing out my solution.

The first part of this problem can be done in the tedious coordinate approach used above, but hopefully there is a better way.

For the last (commutation) part of the problem, here is a hint. Let $$x \wedge i = n i$$, where $$n \cdot i = 0$$. The result then follows easily.

## 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.

## Mathematica notebooks updated, and a bivector addition visualization.

February 10, 2019 math and physics play No comments , , ,

This blog now has a copy of all my Mathematica notebooks (as of Feb 10, 2019), complete with a chronological index.  I hadn’t updated that index since 2014, and it was quite stale.

I’ve also added an additional level of per-directory indexing.  For example, you can now look at just the notebooks for my book, Geometric Algebra for Electrical Engineers.  That was possible before, but you would have had to clone the entire git repository to be able to do so easily.

This update includes a new notebook written today, which has a Manipulate visualization of 3D bivector addition that is kind of fun.

Bivector addition, at least in 3D, can be done graphically almost like vector addition.  Instead of trying to add the planes (which can be done, as in the neat illustration in Geometric Algebra for Computer Science), you can do the task more simply by connecting the normals head to tail, where each of the normals are scaled by the area of the bivector (i.e. it’s absolute magnitude).  The resulting bivector has an area equal to the length of that sum of normals, and a “direction” perpendicular to that resulting normal.  This fun little Manipulate lets you interactively visualize this process, by changing the radius of a set of summed bivectors, each oriented in a different direction, and observing the effects of doing so.

Of course, you can interpret this visualization as nothing more than a representation of addition of cross products, if you were to interpret the vector representing a cross product as an oriented area with a normal equal to that cross product (where the normal’s magnitude equals the area, as in this bivector addition visualization.)  This works out nicely because of the duality relationship between the cross and wedge product, and the duality relationship between 3D bivectors and their normals.

## 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

## Generalizing Ampere’s law using geometric algebra.

The question I’d like to explore in this post is how Ampere’s law, the relationship between the line integral of the magnetic field to current (i.e. the enclosed current)
\label{eqn:flux:20}
\oint_{\partial A} d\Bx \cdot \BH = -\int_A \ncap \cdot \BJ,

generalizes to geometric algebra where Maxwell’s equations for a statics configuration (all time derivatives zero) is
\label{eqn:flux:40}

where the multivector fields and currents are
\label{eqn:flux:60}
\begin{aligned}
F &= \BE + I \eta \BH \\
J &= \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_\txtm – \BM }.
\end{aligned}

Here (fictitious) the magnetic charge and current densities that can be useful in antenna theory have been included in the multivector current for generality.

My presumption is that it should be possible to utilize the fundamental theorem of geometric calculus for expressing the integral over an oriented surface to its boundary, but applied directly to Maxwell’s equation. That integral theorem has the form
\label{eqn:flux:80}
\int_A d^2 \Bx \boldpartial F = \oint_{\partial A} d\Bx F,

where $$d^2 \Bx = d\Ba \wedge d\Bb$$ is a two parameter bivector valued surface, and $$\boldpartial$$ is vector derivative, the projection of the gradient onto the tangent space. I won’t try to explain all of geometric calculus here, and refer the interested reader to [1], which is an excellent reference on geometric calculus and integration theory.

The gotcha is that we actually want a surface integral with $$\spacegrad F$$. We can split the gradient into the vector derivative a normal component
\label{eqn:flux:160}

so
\label{eqn:flux:100}
=
\int_A d^2 \Bx \boldpartial F
+
\int_A d^2 \Bx \ncap \lr{ \ncap \cdot \spacegrad } F,

so
\label{eqn:flux:120}
\begin{aligned}
\oint_{\partial A} d\Bx F
&=
\int_A d^2 \Bx \lr{ J – \ncap \lr{ \ncap \cdot \spacegrad } F } \\
&=
\int_A dA \lr{ I \ncap J – \lr{ \ncap \cdot \spacegrad } I F }
\end{aligned}

This is not nearly as nice as the magnetic flux relationship which was nicely split with the current and fields nicely separated. The $$d\Bx F$$ product has all possible grades, as does the $$d^2 \Bx J$$ product (in general). Observe however, that the normal term on the right has only grades 1,2, so we can split our line integral relations into pairs with and without grade 1,2 components
\label{eqn:flux:140}
\begin{aligned}
&=
\int_A dA \gpgrade{ I \ncap J }{0,3} \\
&=
\int_A dA \lr{ \gpgrade{ I \ncap J }{1,2} – \lr{ \ncap \cdot \spacegrad } I F }.
\end{aligned}

Let’s expand these explicitly in terms of the component fields and densities to check against the conventional relationships, and see if things look right. The line integrand expands to
\label{eqn:flux:180}
\begin{aligned}
d\Bx F
&=
d\Bx \lr{ \BE + I \eta \BH }
=
d\Bx \cdot \BE + I \eta d\Bx \cdot \BH
+
d\Bx \wedge \BE + I \eta d\Bx \wedge \BH \\
&=
d\Bx \cdot \BE
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
+ I \eta (d\Bx \cdot \BH),
\end{aligned}

the current integrand expands to
\label{eqn:flux:200}
\begin{aligned}
I \ncap J
&=
I \ncap
\lr{
\frac{\rho}{\epsilon} – \eta \BJ + I \lr{ c \rho_\txtm – \BM }
} \\
&=
\ncap I \frac{\rho}{\epsilon} – \eta \ncap I \BJ – \ncap c \rho_\txtm + \ncap \BM \\
&=
\ncap \cdot \BM
+ \eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
– \eta I (\ncap \cdot \BJ).
\end{aligned}

We are left with
\label{eqn:flux:220}
\begin{aligned}
\oint_{\partial A}
\lr{
d\Bx \cdot \BE + I \eta (d\Bx \cdot \BH)
}
&=
\int_A dA
\lr{
\ncap \cdot \BM – \eta I (\ncap \cdot \BJ)
} \\
\oint_{\partial A}
\lr{
– \eta (d\Bx \cross \BH)
+ I (d\Bx \cross \BE )
}
&=
\int_A dA
\lr{
\eta (\ncap \cross \BJ)
– \ncap c \rho_\txtm
+ I (\ncap \cross \BM)
+ \ncap I \frac{\rho}{\epsilon}
-\PD{n}{} \lr{ I \BE – \eta \BH }
}.
\end{aligned}

This is a crazy mess of dots, crosses, fields and sources. We can split it into one equation for each grade, which will probably look a little more regular. That is
\label{eqn:flux:240}
\begin{aligned}
\oint_{\partial A} d\Bx \cdot \BE &= \int_A dA \ncap \cdot \BM \\
\oint_{\partial A} d\Bx \cross \BH
&=
\int_A dA
\lr{
– \ncap \cross \BJ
+ \frac{ \ncap \rho_\txtm }{\mu}
– \PD{n}{\BH}
} \\
\oint_{\partial A} d\Bx \cross \BE &=
\int_A dA
\lr{
\ncap \cross \BM
+ \frac{\ncap \rho}{\epsilon}
– \PD{n}{\BE}
} \\
\oint_{\partial A} d\Bx \cdot \BH &= -\int_A dA \ncap \cdot \BJ \\
\end{aligned}

The first and last equations could have been obtained much more easily from Maxwell’s equations in their conventional form more easily. The two cross product equations with the normal derivatives are not familiar to me, even without the fictitious magnetic sources. It is somewhat remarkable that so much can be packed into one multivector equation:
\label{eqn:flux:260}
\oint_{\partial A} d\Bx F
=
I \int_A dA \lr{ \ncap J – \PD{n}{F} }.

# References

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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