## New version of classical mechanics notes

I’ve posted a new version of my classical mechanics notes compilation.  This version is not yet live on amazon, but you shouldn’t buy a copy of this “book” anyways, as it is horribly rough (if you want a copy, grab the free PDF instead.)  [I am going to buy a copy so that I can continue to edit a paper copy of it, but nobody else should.]

This version includes additional background material on Space Time Algebra (STA), i.e. the geometric algebra name for the Dirac/Clifford-algebra in 3+1 dimensions.  In particular, I’ve added material on reciprocal frames, the gradient and vector derivatives, line and surface integrals and the fundamental theorem for both.  Some of the integration theory content might make sense to move to a different book, but I’ll keep it with the rest of these STA notes for now.

## 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.

## A couple more reciprocal frame examples.

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

This post logically follows both of the following:

The PDF linked above above contains all the content from this post plus (1.) above [to be edited later into a more logical sequence.]

## More examples.

Here are a few additional examples of reciprocal frame calculations.

## Problem: Unidirectional arbitrary functional dependence.

Let
\label{eqn:reciprocal:2540}
x = a f(u),

where $$a$$ is a constant vector and $$f(u)$$ is some arbitrary differentiable function with a non-zero derivative in the region of interest.

Here we have just a single tangent space direction (a line in spacetime) with tangent vector
\label{eqn:reciprocal:2400}
\Bx_u = a \PD{u}{f} = a f_u,

so we see that the tangent space vectors are just rescaled values of the direction vector $$a$$.
This is a simple enough parameterization that we can compute the reciprocal frame vector explicitly using the gradient. We expect that $$\Bx^u = 1/\Bx_u$$, and find
\label{eqn:reciprocal:2420}
\inv{a} \cdot x = f(u),

but for constant $$a$$, we know that $$\grad a \cdot x = a$$, so taking gradients of both sides we find
\label{eqn:reciprocal:2440}

so the reciprocal vector is
\label{eqn:reciprocal:2460}
\Bx^u = \grad u = \inv{a f_u},

as expected.

## Problem: Linear two variable parameterization.

Let $$x = a u + b v$$, where $$x \wedge a \wedge b = 0$$ represents spacetime plane (also the tangent space.) Find the curvilinear coordinates and their reciprocals.

The frame vectors are easy to compute, as they are just
\label{eqn:reciprocal:1960}
\begin{aligned}
\Bx_u &= \PD{u}{x} = a \\
\Bx_v &= \PD{v}{x} = b.
\end{aligned}

This is an example of a parametric equation that we can easily invert, as we have
\label{eqn:reciprocal:1980}
\begin{aligned}
x \wedge a &= – v \lr{ a \wedge b } \\
x \wedge b &= u \lr{ a \wedge b },
\end{aligned}

so
\label{eqn:reciprocal:2000}
\begin{aligned}
u
&= \inv{ a \wedge b } \cdot \lr{ x \wedge b } \\
&= \inv{ \lr{a \wedge b}^2 } \lr{ a \wedge b } \cdot \lr{ x \wedge b } \\
&=
\frac{
\lr{b \cdot x} \lr{ a \cdot b }

\lr{a \cdot x} \lr{ b \cdot b }
}{ \lr{a \wedge b}^2 }
\end{aligned}

\label{eqn:reciprocal:2020}
\begin{aligned}
v &= -\inv{ a \wedge b } \cdot \lr{ x \wedge a } \\
&= -\inv{ \lr{a \wedge b}^2 } \lr{ a \wedge b } \cdot \lr{ x \wedge a } \\
&=
-\frac{
\lr{b \cdot x} \lr{ a \cdot a }

\lr{a \cdot x} \lr{ a \cdot b }
}{ \lr{a \wedge b}^2 }
\end{aligned}

Recall that $$\grad \lr{ a \cdot x} = a$$, if $$a$$ is a constant, so our gradients are just
\label{eqn:reciprocal:2040}
\begin{aligned}
&=
\frac{
b \lr{ a \cdot b }

a
\lr{ b \cdot b }
}{ \lr{a \wedge b}^2 } \\
&=
b \cdot \inv{ a \wedge b },
\end{aligned}

and
\label{eqn:reciprocal:2060}
\begin{aligned}
&=
-\frac{
b \lr{ a \cdot a }

a \lr{ a \cdot b }
}{ \lr{a \wedge b}^2 } \\
&=
-a \cdot \inv{ a \wedge b }.
\end{aligned}

Expressed in terms of the frame vectors, this is just
\label{eqn:reciprocal:2080}
\begin{aligned}
\Bx^u &= \Bx_v \cdot \inv{ \Bx_u \wedge \Bx_v } \\
\Bx^v &= -\Bx_u \cdot \inv{ \Bx_u \wedge \Bx_v },
\end{aligned}

so we were able to show, for this special two parameter linear case, that the explicit evaluation of the gradients has the exact structure that we intuited that the reciprocals must have, provided they are constrained to the spacetime plane $$a \wedge b$$. It is interesting to observe how this structure falls out of the linear system solution so directly. Also note that these reciprocals are not defined at the origin of the $$(u,v)$$ parameter space.

## Problem: Quadratic two variable parameterization.

Now consider a variation of the previous problem, with $$x = a u^2 + b v^2$$. Find the curvilinear coordinates and their reciprocals.

\label{eqn:reciprocal:2100}
\begin{aligned}
\Bx_u &= \PD{u}{x} = 2 u a \\
\Bx_v &= \PD{v}{x} = 2 v b.
\end{aligned}

Our tangent space is still the $$a \wedge b$$ plane (as is the surface itself), but the spacing of the cells starts getting wider in proportion to $$u, v$$.
Utilizing the work from the previous problem, we have
\label{eqn:reciprocal:2120}
\begin{aligned}
b \cdot \inv{ a \wedge b } \\
-a \cdot \inv{ a \wedge b }.
\end{aligned}

A bit of rearrangement can show that this is equivalent to the reciprocal frame identities. This is a second demonstration that the gradient and the algebraic formulations for the reciprocals match, at least for these special cases of linear non-coupled parameterizations.

## Problem: Reciprocal frame for generalized cylindrical parameterization.

Let the vector parameterization be $$x(\rho,\theta) = \rho e^{-i\theta/2} x(\rho_0, \theta_0) e^{i \theta}$$, where $$i^2 = \pm 1$$ is a unit bivector ($$+1$$ for a boost, and $$-1$$ for a rotation), and where $$\theta, \rho$$ are scalars. Find the tangent space vectors and their reciprocals.

fig. 1. “Cylindrical” boost parameterization.

Note that this is cylindrical parameterization for the rotation case, and traces out hyperbolic regions for the boost case. The boost case is illustrated in fig. 1 where hyperbolas in the light cone are found for boosts of $$\gamma_0$$ with various values of $$\rho$$, and the spacelike hyperbolas are boosts of $$\gamma_1$$, again for various values of $$\rho$$.

The tangent space vectors are
\label{eqn:reciprocal:2480}
\Bx_\rho = \frac{x}{\rho},

and

\label{eqn:reciprocal:2500}
\begin{aligned}
\Bx_\theta
&= -\frac{i}{2} x + x \frac{i}{2} \\
&= x \cdot i.
\end{aligned}

Recall that $$x \cdot i$$ lies perpendicular to $$x$$ (in the plane $$i$$), as illustrated in fig. 2. This means that $$\Bx_\rho$$ and $$\Bx_\theta$$ are orthogonal, so we can find the reciprocal vectors by just inverting them
\label{eqn:reciprocal:2520}
\begin{aligned}
\Bx^\rho &= \frac{\rho}{x} \\
\Bx^\theta &= \frac{1}{x \cdot i}.
\end{aligned}

fig. 2. Projection and rejection geometry.

## Parameterization of a general linear transformation.

Given $$N$$ parameters $$u^0, u^1, \cdots u^{N-1}$$, a general linear transformation from the parameter space to the vector space has the form
\label{eqn:reciprocal:2160}
x =
{a^\alpha}_\beta \gamma_\alpha u^\beta,

where $$\beta \in [0, \cdots, N-1]$$ and $$\alpha \in [0,3]$$.
For such a general transformation, observe that the curvilinear basis vectors are
\label{eqn:reciprocal:2180}
\begin{aligned}
\Bx_\mu
&= \PD{u^\mu}{x} \\
&= \PD{u^\mu}{}
{a^\alpha}_\beta \gamma_\alpha u^\beta \\
&=
{a^\alpha}_\mu \gamma_\alpha.
\end{aligned}

We find an interpretation of $${a^\alpha}_\mu$$ by dotting $$\Bx_\mu$$ with the reciprocal frame vectors of the standard basis
\label{eqn:reciprocal:2200}
\begin{aligned}
\Bx_\mu \cdot \gamma^\nu
&=
{a^\alpha}_\mu \lr{ \gamma_\alpha \cdot \gamma^\nu } \\
&=
{a^\nu}_\mu,
\end{aligned}

so
\label{eqn:reciprocal:2220}
x = \Bx_\mu u^\mu.

We are able to reinterpret \ref{eqn:reciprocal:2160} as a contraction of the tangent space vectors with the parameters, scaling and summing these direction vectors to characterize all the points in the tangent plane.

## Theorem 1.1: Projecting onto the tangent space.

Let $$T$$ represent the tangent space. The projection of a vector onto the tangent space has the form
\label{eqn:reciprocal:2560}
\textrm{Proj}_{\textrm{T}} y = \lr{ y \cdot \Bx^\mu } \Bx_\mu = \lr{ y \cdot \Bx_\mu } \Bx^\mu.

### Start proof:

Let’s designate $$a$$ as the portion of the vector $$y$$ that lies outside of the tangent space
\label{eqn:reciprocal:2260}
y = y^\mu \Bx_\mu + a.

If we knew the coordinates $$y^\mu$$, we would have a recipe for the projection.
Algebraically, requiring that $$a$$ lies outside of the tangent space, is equivalent to stating $$a \cdot \Bx_\mu = a \cdot \Bx^\mu = 0$$. We use that fact, and then take dot products
\label{eqn:reciprocal:2280}
\begin{aligned}
y \cdot \Bx^\nu
&= \lr{ y^\mu \Bx_\mu + a } \cdot \Bx^\nu \\
&= y^\nu,
\end{aligned}

so
\label{eqn:reciprocal:2300}
y = \lr{ y \cdot \Bx^\mu } \Bx_\mu + a.

Similarly, the tangent space projection can be expressed as a linear combination of reciprocal basis elements
\label{eqn:reciprocal:2320}
y = y_\mu \Bx^\mu + a.

Dotting with $$\Bx_\mu$$, we have
\label{eqn:reciprocal:2340}
\begin{aligned}
y \cdot \Bx^\mu
&= \lr{ y_\alpha \Bx^\alpha + a } \cdot \Bx_\mu \\
&= y_\mu,
\end{aligned}

so
\label{eqn:reciprocal:2360}
y = \lr{ y \cdot \Bx^\mu } \Bx_\mu + a.

We find the two stated ways of computing the projection.

Observe that, for the special case that all of $$\setlr{ \Bx_\mu }$$ are orthogonal, the equivalence of these two projection methods follows directly, since
\label{eqn:reciprocal:2380}
\begin{aligned}
\lr{ y \cdot \Bx^\mu } \Bx_\mu
&=
\lr{ y \cdot \inv{\Bx_\mu} } \inv{\Bx^\mu} \\
&=
\lr{ y \cdot \frac{\Bx_\mu}{\lr{\Bx_\mu}^2 } } \frac{\Bx^\mu}{\lr{\Bx^\mu}^2} \\
&=
\lr{ y \cdot \Bx_\mu } \Bx^\mu.
\end{aligned}

## 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.

## Gauge transformation in the Lorentz force Lagrangian.

November 2, 2020 Uncategorized No comments , , ,

## Problem: Lorentz force gauge transformation.

Show that the gauge transformation $$A \rightarrow A + \grad \psi$$ applied to the Lorentz force Lagrangian
\label{eqn:gaugeLorentzSTA:20}
L = \inv{2} m v^2 + q A \cdot v/c,

does not change the equations of motion.

The gauge transformed Lagrangian is
\label{eqn:gaugeLorentzSTA:40}
L = \inv{2} m v^2 + q A \cdot v/c + \frac{q v}{c} \cdot \grad \phi.

We know that the Lorentz force equations are obtained from the first two terms, so need only consider the effects of the new $$\phi$$ dependent term on the action. First observe that
\label{eqn:gaugeLorentzSTA:60}
=
\frac{dx^\mu}{d\tau} \PD{x^\mu}{\phi}
=
\frac{d \phi}{d\tau}.

This means that the action is transformed to
\label{eqn:gaugeLorentzSTA:80}
S
\rightarrow S + \frac{q}{c} \int d\tau \frac{d\phi}{d\tau}
= S + \frac{q}{c} \evalbar{\phi}{\Delta \tau}.

As the action is evaluated over a fixed interval, the gauge transformation only changes the action by a constant, so the equations of motion are unchanged.

# References

## Some nice positive feedback for my book.

Here’s a fun congratulatory email that I received today for my Geometric Algebra for Electrical Engineers book

Peeter ..
I had to email to congratulate you on your geometric algebra book. Like yourself, when I came across it, I was totally blown away and your book, being written from the position of a discoverer rather than an expert, answers most of the questions I was confronted by when reading Doran and Lasenby’s book.
You’re a C++ programmer and from my perspective, when using natural world math, you are constructing a representation of a problem (like code does) except many physicists do not recognize this. They’re doing physics with COBOL (or C with classes!).
congratulations
I couldn’t resist pointing out the irony of his COBOL comment, as my work at LzLabs is now heavily focused on COBOL (and PL/I) compilers and compiler runtimes.  You could say that my work, at work or at play, is all an attempt to transition people away from the evils of legacy COBOL.
For reference the Doran and Lasenby book is phenomenal work, but it is really hard material.  To attempt to read this, you’ll need a thorough understanding of electromagnetism, relativity, tensor algebra, quantum mechanics, advanced classical mechanics, and field theory.  I’m still working on this book, and it’s probably been 12 years since I bought it.  I managed to teach myself some of this material as I went, but also took most of the 4th year UofT undergrad physics courses (and some grad courses) to fill in some of the gaps.
When I titled my book, I included “for Electrical Engineers” in the title.  That titling choice was somewhat derivative, as there were already geometric algebra books “for physicists”,  and “for computer science“.  However, I thought it was also good shorthand for the prerequisites required for the book as “for Electrical Engineers” seemed to be good shorthand for “for a student that has seen electromagnetism in its div, grad, curl form, and doesn’t know special relativity, field theory, differential forms, tensor algebra, or other topics from more advanced physics.”
The relativistic presentation of electromagnetism in Doran and Lasenby, using the Dirac algebra (aka Space Time Algebra (STA)), is much more beautiful than the form that I have used in my book.  However, I was hoping to present the subject in a way that was accessible, and provided a stepping stone for the STA approach when the reader was ready to tackle a next interval of the “learning curve.”

## Motivation.

In my old classical mechanics notes it appears that I did covariant derivations of the Lorentz force equations a number of times, using different trial Lagrangians (relativistic and non-relativistic), and using both geometric algebra and tensor methods. However, none of these appear to have been done concisely, and a number not even coherently.

The following document has been drafted as replacement text for those incoherent classical mechanics notes. I’ll attempt to cover

• a lighting review of the geometric algebra STA (Space Time Algebra),
• relations between Dirac matrix algebra and STA,
• derivation of the relativistic form of the Euler-Lagrange equations from the covariant form of the action,
• relationship of the STA form of the Euler-Lagrange equations to their tensor equivalents,
• derivation of the Lorentz force equation from the STA Lorentz force Lagrangian,
• relationship of the STA Lorentz force equation to its equivalent in the tensor formalism,
• relationship of the STA Lorentz force equation to the traditional vector form.

Note that some of the prerequisite ideas and auxiliary details are presented as problems with solutions. If the reader has sufficient background to attempt those problems themselves, they are encouraged to do so.

The STA and geometric algebra ideas used here are not complete to learn from in isolation. The reader is referred to [1] for a more complete exposition of both STA and geometric algebra.

## Definition 1.1: Index conventions.

Latin indexes $$i, j, k, r, s, t, \cdots$$ are used to designate values in the range $$\setlr{ 1,2,3 }$$. Greek indexes are $$\alpha, \beta, \mu, \nu, \cdots$$ are used for indexes of spacetime quantities $$\setlr{0,1,2,3}$$.
The Einstein convention of implied summation for mixed upper and lower Greek indexes will be used, for example
\begin{equation*}
x^\alpha x_\alpha \equiv \sum_{\alpha = 0}^3 x^\alpha x_\alpha.
\end{equation*}

## Space Time Algebra (STA.)

In the geometric algebra literature, the Dirac algebra of quantum field theory has been rebranded Space Time Algebra (STA). The differences between STA and the Dirac theory that uses matrices ($$\gamma_\mu$$) are as follows

• STA completely omits any representation of the Dirac basis vectors $$\gamma_\mu$$. In particular, any possible matrix representation is irrelevant.
• STA provides a rich set of fundamental operations (grade selection, generalized dot and wedge products for multivector elements, rotation and reflection operations, …)
• Matrix trace, and commutator and anticommutator operations are nowhere to be found in STA, as geometrically grounded equivalents are available instead.
• The “slashed” quantities from Dirac theory, such as $$\gamma_\mu p^\mu$$ are nothing more than vectors in their entirety in STA (where the basis is no longer implicit, as is the case for coordinates.)

Our basis vectors have the following properties.

## Definition 1.2: Standard basis.

Let the four-vector standard basis be designated $$\setlr{\gamma_0, \gamma_1, \gamma_2, \gamma_3 }$$, where the basis vectors satisfy
\label{eqn:lorentzForceCovariant:1540}
\begin{aligned}
\gamma_0^2 &= -\gamma_i^2 = 1 \\
\gamma_\alpha \cdot \gamma_\beta &= 0, \forall \alpha \ne \beta.
\end{aligned}

## Problem: Commutator properties of the STA basis.

In Dirac theory, the commutator properties of the Dirac matrices is considered fundamental, namely
\begin{equation*}
\symmetric{\gamma_\mu}{\gamma_\nu} = 2 \eta_{\mu\nu}.
\end{equation*}

Show that this follows from the axiomatic assumptions of geometric algebra, and describe how the dot and wedge products are related to the anticommutator and commutator products of Dirac theory.

The anticommutator is defined as symmetric sum of products
\label{eqn:lorentzForceCovariant:1040}
\symmetric{\gamma_\mu}{\gamma_\nu}
\equiv
\gamma_\mu \gamma_\nu
+
\gamma_\nu \gamma_\mu,

but this is just twice the dot product in its geometric algebra form $$a b = (a b + ba)/2$$. Observe that the properties of the basis vectors defined in \ref{eqn:lorentzForceCovariant:1540} may be summarized as
\label{eqn:lorentzForceCovariant:1060}
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu},

where $$\eta_{\mu\nu} = \text{diag}(+,-,-,-) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}$$ is the conventional metric tensor. This means
\label{eqn:lorentzForceCovariant:1080}
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu} = 2 \symmetric{\gamma_\mu}{\gamma_\nu},

as claimed.

Similarly, observe that the commutator, defined as the antisymmetric sum of products
\label{eqn:lorentzForceCovariant:1100}
\antisymmetric{\gamma_\mu}{\gamma_\nu} \equiv
\gamma_\mu \gamma_\nu

\gamma_\nu \gamma_\mu,

is twice the wedge product $$a \wedge b = (a b – b a)/2$$. This provides geometric identifications for the respective anti-commutator and commutator products respectively
\label{eqn:lorentzForceCovariant:1120}
\begin{aligned}
\symmetric{\gamma_\mu}{\gamma_\nu} &= 2 \gamma_\mu \cdot \gamma_\nu \\
\antisymmetric{\gamma_\mu}{\gamma_\nu} &= 2 \gamma_\mu \wedge \gamma_\nu,
\end{aligned}

## Definition 1.3: Pseudoscalar.

The pseudoscalar for the space is denoted $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$.

## Problem: Pseudoscalar.

Show that the STA pseudoscalar $$I$$ defined by \ref{eqn:lorentzForceCovariant:1540} satisfies
\begin{equation*}
\tilde{I} = I,
\end{equation*}
where the tilde operator designates reversion. Also show that $$I$$ has the properties of an imaginary number
\begin{equation*}
I^2 = -1.
\end{equation*}
Finally, show that, unlike the spatial pseudoscalar that commutes with all grades, $$I$$ anticommutes with any vector or trivector, and commutes with any bivector.

Since $$\gamma_\alpha \gamma_\beta = -\gamma_\beta \gamma_\alpha$$ for any $$\alpha \ne \beta$$, any permutation of the factors of $$I$$ changes the sign once. In particular
\label{eqn:lorentzForceCovariant:680}
\begin{aligned}
I &=
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&=

\gamma_1
\gamma_2
\gamma_3
\gamma_0 \\
&=

\gamma_2
\gamma_3
\gamma_1
\gamma_0 \\
&=
+
\gamma_3
\gamma_2
\gamma_1
\gamma_0
= \tilde{I}.
\end{aligned}

Using this, we have
\label{eqn:lorentzForceCovariant:700}
\begin{aligned}
I^2
&= I \tilde{I} \\
&=
(
\gamma_0
\gamma_1
\gamma_2
\gamma_3
)(
\gamma_3
\gamma_2
\gamma_1
\gamma_0
) \\
&=
\lr{\gamma_0}^2
\lr{\gamma_1}^2
\lr{\gamma_2}^2
\lr{\gamma_3}^2 \\
&=
(+1)
(-1)
(-1)
(-1) \\
&= -1.
\end{aligned}

To illustrate the anticommutation property with any vector basis element, consider the following two examples:
\label{eqn:lorentzForceCovariant:720}
\begin{aligned}
I \gamma_0 &=
\gamma_0
\gamma_1
\gamma_2
\gamma_3
\gamma_0 \\
&=

\gamma_0
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&=

\gamma_0 I,
\end{aligned}

\label{eqn:lorentzForceCovariant:740}
\begin{aligned}
I \gamma_2
&=
\gamma_0
\gamma_1
\gamma_2
\gamma_3
\gamma_2 \\
&=

\gamma_0
\gamma_1
\gamma_2
\gamma_2
\gamma_3 \\
&=

\gamma_2
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&= -\gamma_2 I.
\end{aligned}

A total of three sign swaps is required to “percolate” any given $$\gamma_\alpha$$ through the factors of $$I$$, resulting in an overall sign change of $$-1$$.

For any bivector basis element $$\alpha \ne \beta$$
\label{eqn:lorentzForceCovariant:760}
\begin{aligned}
I \gamma_\alpha \gamma_\beta
&=
-\gamma_\alpha I \gamma_\beta \\
&=
+\gamma_\alpha \gamma_\beta I.
\end{aligned}

Similarly for any trivector basis element $$\alpha \ne \beta \ne \sigma$$
\label{eqn:lorentzForceCovariant:780}
\begin{aligned}
I \gamma_\alpha \gamma_\beta \gamma_\sigma
&=
-\gamma_\alpha I \gamma_\beta \gamma_\sigma \\
&=
+\gamma_\alpha \gamma_\beta I \gamma_\sigma \\
&=
-\gamma_\alpha \gamma_\beta \gamma_\sigma I.
\end{aligned}

## Definition 1.4: Reciprocal basis.

The reciprocal basis $$\setlr{ \gamma^0, \gamma^1, \gamma^2, \gamma^3 }$$ is defined , such that the property $$\gamma^\alpha \cdot \gamma_\beta = {\delta^\alpha}_\beta$$ holds.

Observe that, $$\gamma^0 = \gamma_0$$ and $$\gamma^i = -\gamma_i$$.

## Theorem 1.1: Coordinates.

Coordinates are defined in terms of dot products with the standard basis, or reciprocal basis
\begin{equation*}
\begin{aligned}
x^\alpha &= x \cdot \gamma^\alpha \\
x_\alpha &= x \cdot \gamma_\alpha,
\end{aligned}
\end{equation*}

### Start proof:

Suppose that a coordinate representation of the following form is assumed
\label{eqn:lorentzForceCovariant:820}
x = x^\alpha \gamma_\alpha = x_\beta \gamma^\beta.

We wish to determine the representation of the $$x^\alpha$$ or $$x_\beta$$ coordinates in terms of $$x$$ and the basis elements. Taking the dot product with any standard basis element, we find
\label{eqn:lorentzForceCovariant:840}
\begin{aligned}
x \cdot \gamma_\mu
&= (x_\beta \gamma^\beta) \cdot \gamma_\mu \\
&= x_\beta {\delta^\beta}_\mu \\
&= x_\mu,
\end{aligned}

as claimed. Similarly, dotting with a reciprocal frame vector, we find
\label{eqn:lorentzForceCovariant:860}
\begin{aligned}
x \cdot \gamma^\mu
&= (x^\beta \gamma_\beta) \cdot \gamma^\mu \\
&= x^\beta {\delta_\beta}^\mu \\
&= x^\mu.
\end{aligned}

### End proof.

Observe that raising or lowering the index of a spatial index toggles the sign of a coordinate, but timelike indexes are left unchanged.
\label{eqn:lorentzForceCovariant:880}
\begin{aligned}
x^0 &= x_0 \\
x^i &= -x_i \\
\end{aligned}

\begin{equation*}
\grad = \gamma^\mu \partial_\mu = \gamma_\nu \partial^\nu,
\end{equation*}
where
\begin{equation*}
\partial_\mu = \PD{x^\mu}{},
\end{equation*}
and
\begin{equation*}
\partial^\mu = \PD{x_\mu}{}.
\end{equation*}

This definition of gradient is consistent with the Dirac gradient (sometimes denoted as a slashed $$\partial$$).

## Definition 1.6: Timelike and spacelike components of a four-vector.

Given a four vector $$x = \gamma_\mu x^\mu$$, that would be designated $$x^\mu = \setlr{ x^0, \Bx}$$ in conventional special relativity, we write
\begin{equation*}
x^0 = x \cdot \gamma_0,
\end{equation*}
and
\begin{equation*}
\Bx = x \wedge \gamma_0,
\end{equation*}
or
\begin{equation*}
x = (x^0 + \Bx) \gamma_0.
\end{equation*}

The spacetime split of a four-vector $$x$$ is relative to the frame. In the relativistic lingo, one would say that it is “observer dependent”, as the same operations with $${\gamma_0}’$$, the timelike basis vector for a different frame, would yield a different set of coordinates.

While the dot and wedge products above provide an effective mechanism to split a four vector into a set of timelike and spacelike quantities, the spatial component of a vector has a bivector representation in STA. Consider the following coordinate expansion of a spatial vector
\label{eqn:lorentzForceCovariant:1000}
\Bx =
x \wedge \gamma_0
=
\lr{ x^\mu \gamma_\mu } \wedge \gamma_0
=
\sum_{k = 1}^3 x^k \gamma_k \gamma_0.

## Definition 1.7: Spatial basis.

We designate
\label{eqn:lorentzForceCovariant:1560}
\Be_i = \gamma_i \gamma_0,

as the standard basis vectors for $$\mathbb{R}^3$$.

In the literature, this bivector representation of the spatial basis may be designated $$\sigma_i = \gamma_i \gamma_0$$, as these bivectors have the properties of the Pauli matrices $$\sigma_i$$. Because I intend to expand these notes to include purely non-relativistic applications, I won’t use the Pauli notation here.

## Problem: Orthonormality of the spatial basis.

Show that the spatial basis $$\setlr{ \Be_1, \Be_2, \Be_3 }$$, defined by \ref{eqn:lorentzForceCovariant:1560}, is orthonormal.

\label{eqn:lorentzForceCovariant:620}
\begin{aligned}
\Be_i \cdot \Be_j
&= \gpgradezero{ \gamma_i \gamma_0 \gamma_j \gamma_0 } \\
&= -\gpgradezero{ \gamma_i \gamma_j } \\
&= – \gamma_i \cdot \gamma_j.
\end{aligned}

This is zero for all $$i \ne j$$, and unity for any $$i = j$$.

## Problem: Spatial pseudoscalar.

Show that the STA pseudoscalar $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$ equals the spatial pseudoscalar $$I = \Be_1 \Be_2 \Be_3$$.

The spatial pseudoscalar, expanded in terms of the STA basis vectors, is
\label{eqn:lorentzForceCovariant:1020}
\begin{aligned}
I
&= \Be_1 \Be_2 \Be_3 \\
&= \lr{ \gamma_1 \gamma_0 }
\lr{ \gamma_2 \gamma_0 }
\lr{ \gamma_3 \gamma_0 } \\
&= \lr{ \gamma_1 \gamma_0 } \gamma_2 \lr{ \gamma_0 \gamma_3 } \gamma_0 \\
&= \lr{ -\gamma_0 \gamma_1 } \gamma_2 \lr{ -\gamma_3 \gamma_0 } \gamma_0 \\
&= \gamma_0 \gamma_1 \gamma_2 \gamma_3 \lr{ \gamma_0 \gamma_0 } \\
&= \gamma_0 \gamma_1 \gamma_2 \gamma_3,
\end{aligned}

as claimed.

## Problem: Characteristics of the Pauli matrices.

The Pauli matrices obey the following anticommutation relations:
\label{eqn:lorentzForceCovariant:660}
\symmetric{ \sigma_a}{\sigma_b } = 2 \delta_{a b},

and commutation relations:
\label{eqn:lorentzForceCovariant:640}
\antisymmetric{ \sigma_a}{ \sigma_b } = 2 i \epsilon_{a b c}\,\sigma_c,

Show how these relate to the geometric algebra dot and wedge products, and determine the geometric algebra representation of the imaginary $$i$$ above.

## Euler-Lagrange equations.

I’ll start at ground zero, with the derivation of the relativistic form of the Euler-Lagrange equations from the action. A relativistic action for a single particle system has the form
\label{eqn:lorentzForceCovariant:20}
S = \int d\tau L(x, \dot{x}),

where $$x$$ is the spacetime coordinate, $$\dot{x} = dx/d\tau$$ is the four-velocity, and $$\tau$$ is proper time.

## Theorem 1.2: Relativistic Euler-Lagrange equations.

Let $$x \rightarrow x + \delta x$$ be any variation of the Lagrangian four-vector coordinates, where $$\delta x = 0$$ at the boundaries of the action integral. The variation of the action is
\label{eqn:lorentzForceCovariant:1580}
\delta S = \int d\tau \delta x \cdot \delta L(x, \dot{x}),

where
\label{eqn:lorentzForceCovariant:1600}

where $$\grad = \gamma^\mu \partial_\mu$$, and where we construct a similar velocity-gradient with respect to the proper-time derivatives of the coordinates $$\grad_v = \gamma^\mu \partial/\partial \dot{x}^\mu$$.The action is extremized when $$\delta S = 0$$, or when $$\delta L = 0$$. This latter condition is called the Euler-Lagrange equations.

### Start proof:

Let $$\epsilon = \delta x$$, and expand the Lagrangian in Taylor series to first order
\label{eqn:lorentzForceCovariant:60}
\begin{aligned}
S &\rightarrow S + \delta S \\
&= \int d\tau L( x + \epsilon, \dot{x} + \dot{\epsilon})
&=
\int d\tau \lr{
L(x, \dot{x}) + \epsilon \cdot \grad L + \dot{\epsilon} \cdot \grad_v L
}.
\end{aligned}

Subtracting off $$S$$ and integrating by parts, leaves
\label{eqn:lorentzForceCovariant:80}
\delta S =
\int d\tau \epsilon \cdot \lr{
}
+
\int d\tau \frac{d}{d\tau} (\grad_v L ) \cdot \epsilon.

The boundary integral
\label{eqn:lorentzForceCovariant:100}
\int d\tau \frac{d}{d\tau} (\grad_v L ) \cdot \epsilon
=
\evalbar{(\grad_v L ) \cdot \epsilon}{\Delta \tau} = 0,

is zero since the variation $$\epsilon$$ is required to vanish on the boundaries. So, if $$\delta S = 0$$, we must have
\label{eqn:lorentzForceCovariant:120}
0 =
\int d\tau \epsilon \cdot \lr{
},

for all variations $$\epsilon$$. Clearly, this requires that
\label{eqn:lorentzForceCovariant:140}

or
\label{eqn:lorentzForceCovariant:145}

which is the coordinate free statement of the Euler-Lagrange equations.

## Problem: Coordinate form of the Euler-Lagrange equations.

Working in coordinates, use the action argument show that the Euler-Lagrange equations have the form
\begin{equation*}
\PD{x^\mu}{L} = \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}
\end{equation*}
Observe that this is identical to the statement of \ref{eqn:lorentzForceCovariant:1600} after contraction with $$\gamma^\mu$$.

In terms of coordinates, the first order Taylor expansion of the action is
\label{eqn:lorentzForceCovariant:180}
\begin{aligned}
S &\rightarrow S + \delta S \\
&= \int d\tau L( x^\alpha + \epsilon^\alpha, \dot{x}^\alpha + \dot{\epsilon}^\alpha) \\
&=
\int d\tau \lr{
L(x^\alpha, \dot{x}^\alpha) + \epsilon^\mu \PD{x^\mu}{L} + \dot{\epsilon}^\mu \PD{\dot{x}^\mu}{L}
}.
\end{aligned}

As before, we integrate by parts to separate out a pure boundary term
\label{eqn:lorentzForceCovariant:200}
\delta S =
\int d\tau \epsilon^\mu
\lr{
\PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}
}
+
\int d\tau \frac{d}{d\tau} \lr{
\epsilon^\mu \PD{\dot{x}^\mu}{L}
}.

The boundary term is killed since $$\epsilon^\mu = 0$$ at the end points of the action integral. We conclude that extremization of the action ($$\delta S = 0$$, for all $$\epsilon^\mu$$) requires
\label{eqn:lorentzForceCovariant:220}
\PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L} = 0.

## Theorem 1.3: Lorentz force.

The relativistic Lagrangian for a charged particle is
\label{eqn:lorentzForceCovariant:1640}
L = \inv{2} m v^2 + q A \cdot v/c.

Application of the Euler-Lagrange equations to this Lagrangian yields the Lorentz-force equation
\label{eqn:lorentzForceCovariant:1660}
\frac{dp}{d\tau} = q F \cdot v/c,

where $$p = m v$$ is the proper momentum, $$F$$ is the Faraday bivector $$F = \grad \wedge A$$, and $$c$$ is the speed of light.

### Start proof:

To make life easier, let’s take advantage of the linearity of the Lagrangian, and break it into the free particle Lagrangian $$L_0 = (1/2) m v^2$$ and a potential term $$L_1 = q A \cdot v/c$$. For the free particle case we have
\label{eqn:lorentzForceCovariant:240}
\begin{aligned}
\delta L_0
&= – \frac{d}{d\tau} (m v) \\
&= – \frac{dp}{d\tau}.
\end{aligned}

For the potential contribution we have
\label{eqn:lorentzForceCovariant:260}
\begin{aligned}
\delta L_1
&= \frac{q}{c} \lr{ \grad (A \cdot v) – \frac{d}{d\tau} \lr{ \grad_v (A \cdot v)} } \\
&= \frac{q}{c} \lr{ \grad (A \cdot v) – \frac{dA}{d\tau} }.
\end{aligned}

The proper time derivative can be evaluated using the chain rule
\label{eqn:lorentzForceCovariant:280}
\frac{dA}{d\tau}
=
\frac{\partial x^\mu}{\partial \tau} \partial_\mu A

Putting all the pieces back together we have
\label{eqn:lorentzForceCovariant:300}
\begin{aligned}
0
&= \delta L \\
&=
-\frac{dp}{d\tau} + \frac{q}{c} \lr{ \grad (A \cdot v) – (v \cdot \grad) A } \\
&=
-\frac{dp}{d\tau} + \frac{q}{c} \lr{ \grad \wedge A } \cdot v.
\end{aligned}

## Problem: Gradient of a squared position vector.

Show that
\begin{equation*}
\grad (a \cdot x) = a,
\end{equation*}
and
\begin{equation*}
\end{equation*}
It should be clear that the same ideas can be used for the velocity gradient, where we obtain $$\grad_v (v^2) = 2 v$$, and $$\grad_v (A \cdot v) = A$$, as used in the derivation above.

The first identity follows easily by expansion in coordinates
\label{eqn:lorentzForceCovariant:320}
\begin{aligned}
&=
\gamma^\mu \partial_\mu a_\alpha x^\alpha \\
&=
\gamma^\mu a_\alpha \delta_\mu^\alpha \\
&=
\gamma^\mu a_\mu \\
&=
a.
\end{aligned}

The second identity follows by linearity of the gradient
\label{eqn:lorentzForceCovariant:340}
\begin{aligned}
&=
&=
\evalbar{\lr{\grad (x \cdot a)}}{a = x}
+
\evalbar{\lr{\grad (b \cdot x)}}{b = x} \\
&=
\evalbar{a}{a = x}
+
\evalbar{b}{b = x} \\
&=
2x.
\end{aligned}

It is desirable to put this relativistic Lorentz force equation into the usual vector and tensor forms for comparison.

## Theorem 1.4: Tensor form of the Lorentz force equation.

The tensor form of the Lorentz force equation is
\label{eqn:lorentzForceCovariant:1620}
\frac{dp^\mu}{d\tau} = \frac{q}{c} F^{\mu\nu} v_\nu,

where the antisymmetric Faraday tensor is defined as $$F^{\mu\nu} = \partial^\mu A^\nu – \partial^\nu A^\mu$$.

### Start proof:

We have only to dot both sides with $$\gamma^\mu$$. On the left we have
\label{eqn:lorentzForceCovariant:380}
\gamma^\mu \cdot \frac{dp}{d\tau}
=
\frac{dp^\mu}{d\tau}.

On the right, we have
\label{eqn:lorentzForceCovariant:400}
\begin{aligned}
\gamma^\mu \cdot \lr{ \frac{q}{c} F \cdot v }
&=
\frac{q}{c} (( \grad \wedge A ) \cdot v ) \cdot \gamma^\mu \\
&=
\frac{q}{c} ( \grad ( A \cdot v ) – (v \cdot \grad) A ) \cdot \gamma^\mu \\
&=
\frac{q}{c} \lr{ (\partial^\mu A^\nu) v_\nu – v_\nu \partial^\nu A^\mu } \\
&=
\frac{q}{c} F^{\mu\nu} v_\nu.
\end{aligned}

## Problem: Tensor expansion of $$F$$.

An alternate way to demonstrate \ref{eqn:lorentzForceCovariant:1620} is to first expand $$F = \grad \wedge A$$ in terms of coordinates, an expansion that can be expressed in terms of a second rank tensor antisymmetric tensor $$F^{\mu\nu}$$. Find that expansion, and re-evaluate the dot products of \ref{eqn:lorentzForceCovariant:400} using that.

\label{eqn:lorentzForceCovariant:900}
\begin{aligned}
F &=
&=
\lr{ \gamma_\mu \partial^\mu } \wedge \lr{ \gamma_\nu A^\nu } \\
&=
\lr{ \gamma_\mu \wedge \gamma_\nu } \partial^\mu A^\nu.
\end{aligned}

To this we can use the usual tensor trick (add self to self, change indexes, and divide by two), to give
\label{eqn:lorentzForceCovariant:920}
\begin{aligned}
F &=
\inv{2} \lr{
\lr{ \gamma_\mu \wedge \gamma_\nu } \partial^\mu A^\nu
+
\lr{ \gamma_\nu \wedge \gamma_\mu } \partial^\nu A^\mu
} \\
&=
\inv{2}
\lr{ \gamma_\mu \wedge \gamma_\nu } \lr{
\partial^\mu A^\nu

\partial^\nu A^\mu
},
\end{aligned}

which is just
\label{eqn:lorentzForceCovariant:940}
F =
\inv{2} \lr{ \gamma_\mu \wedge \gamma_\nu } F^{\mu\nu}.

Now, let’s expand $$(F \cdot v) \cdot \gamma^\mu$$ to compare to the earlier expansion in terms of $$\grad$$ and $$A$$.
\label{eqn:lorentzForceCovariant:960}
\begin{aligned}
(F \cdot v) \cdot \gamma^\mu
&=
\inv{2}
F^{\alpha\nu}
\lr{ \lr{ \gamma_\alpha \wedge \gamma_\nu } \cdot \lr{ \gamma^\beta v_\beta } } \cdot \gamma^\mu \\
&=
\inv{2}
F^{\alpha\nu} v_\beta
\lr{
{\delta_\nu}^\beta {\gamma_\alpha}^\mu

{\delta_\alpha}^\beta {\gamma_\nu}^\mu
} \\
&=
\inv{2}
\lr{
F^{\mu\beta} v_\beta

F^{\beta\mu} v_\beta
} \\
&=
F^{\mu\nu} v_\nu.
\end{aligned}

This alternate expansion illustrates some of the connectivity between the geometric algebra approach and the traditional tensor formalism.

## Problem: Lorentz force direct tensor derivation.

Instead of using the geometric algebra form of the Lorentz force equation as a stepping stone, we may derive the tensor form from the Lagrangian directly, provided the Lagrangian is put into tensor form
\begin{equation*}
L = \inv{2} m v^\mu v_\mu + q A^\mu v_\mu /c.
\end{equation*}
Evaluate the Euler-Lagrange equations in coordinate form and compare to \ref{eqn:lorentzForceCovariant:1620}.

Let $$\delta_\mu L = \gamma_\mu \cdot \delta L$$, so that we can write the Euler-Lagrange equations as
\label{eqn:lorentzForceCovariant:460}
0 = \delta_\mu L = \PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}.

Operating on the kinetic term of the Lagrangian, we have
\label{eqn:lorentzForceCovariant:480}
\delta_\mu L_0 = – \frac{d}{d\tau} m v_\mu.

For the potential term
\label{eqn:lorentzForceCovariant:500}
\begin{aligned}
\delta_\mu L_1
&=
\frac{q}{c} \lr{
v_\nu \PD{x^\mu}{A^\nu} – \frac{d}{d\tau} A_\mu
} \\
&=
\frac{q}{c} \lr{
v_\nu \PD{x^\mu}{A^\nu} – \frac{dx_\alpha}{d\tau} \PD{x_\alpha}{ A_\mu }
} \\
&=
\frac{q}{c} v^\nu \lr{
\partial_\mu A_\nu – \partial_\nu A_\mu
} \\
&=
\frac{q}{c} v^\nu F_{\mu\nu}.
\end{aligned}

Putting the pieces together gives
\label{eqn:lorentzForceCovariant:520}
\frac{d}{d\tau} (m v_\mu) = \frac{q}{c} v^\nu F_{\mu\nu},

which is identical\footnote{Some minor index raising and lowering gymnastics are required.} to the tensor form that we found by expanding the geometric algebra form of Maxwell’s equation in coordinates.

## Theorem 1.5: Vector Lorentz force equation.

Relative to a fixed observer’s frame, the Lorentz force equation of \ref{eqn:lorentzForceCovariant:1660} splits into a spatial rate of change of momentum, and (timelike component) rate of change of energy, as follows
\label{eqn:lorentzForceCovariant:1680}
\begin{aligned}
\ddt{(\gamma m \Bv)} &= q \lr{ \BE + \Bv \cross \BB } \\
\ddt{(\gamma m c^2)} &= q \Bv \cdot \BE,
\end{aligned}

where $$F = \BE + I c \BB$$, $$\gamma = 1/\sqrt{1 – \Bv^2/c^2 }$$.

### Start proof:

The first step is to eliminate the proper time dependencies in the Lorentz force equation. Consider first the coordinate representation of an arbitrary position four-vector $$x$$
\label{eqn:lorentzForceCovariant:1140}
x = c t \gamma_0 + x^k \gamma_k.

The corresponding four-vector velocity is
\label{eqn:lorentzForceCovariant:1160}
v = \ddtau{x} = c \ddtau{t} \gamma_0 + \ddtau{t} \ddt{x^k} \gamma_k.

By construction, $$v^2 = c^2$$ is a Lorentz invariant quantity (this is one of the relativistic postulates), so the LHS of \ref{eqn:lorentzForceCovariant:1160} must have the same square. That is
\label{eqn:lorentzForceCovariant:1240}
c^2 = \lr{ \ddtau{t} }^2 \lr{ c^2 – \Bv^2 },

where $$\Bv = v \wedge \gamma_0$$. This shows that we may make the identification
\label{eqn:lorentzForceCovariant:1260}
\gamma = \ddtau{t} = \inv{1 – \Bv^2/c^2 },

and
\label{eqn:lorentzForceCovariant:1280}
\ddtau{} = \ddtau{t} \ddt{} = \gamma \ddt{}.

We may now factor the four-velocity $$v$$ into its spacetime split
\label{eqn:lorentzForceCovariant:1300}
v = \gamma \lr{ c + \Bv } \gamma_0.

In particular the LHS of the Lorentz force equation can be rewritten as
\label{eqn:lorentzForceCovariant:1320}
\ddtau{p} = \gamma \ddt{}\lr{ \gamma \lr{ c + \Bv } } \gamma_0,

and the RHS of the Lorentz force equation can be rewritten as
\label{eqn:lorentzForceCovariant:1340}
\frac{q}{c} F \cdot v
=
\frac{\gamma q}{c} F \cdot \lr{ (c + \Bv) \gamma_0 }.

Equating timelike and spacelike components leaves us
\label{eqn:lorentzForceCovariant:1380}
\ddt{ (m \gamma c) } = \frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \cdot \gamma_0,

\label{eqn:lorentzForceCovariant:1400}
\ddt{ (m \gamma \Bv) } = \frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \wedge \gamma_0,

Evaluating these products requires some care, but is an essentially manual process. The reader is encouraged to do so once, but the end result may also be obtained easily using software (see lorentzForce.nb in [2]). One finds
\label{eqn:lorentzForceCovariant:1440}
F = \BE + I c \BB
=
E^1 \gamma_{10} +
+ E^2 \gamma_{20} +
+ E^3 \gamma_{30} +
– c B^1 \gamma_{23} +
– c B^2 \gamma_{31} +
– c B^3 \gamma_{12},

\label{eqn:lorentzForceCovariant:1460}
\frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \cdot \gamma_0
= \frac{q}{c} \BE \cdot \Bv,

\label{eqn:lorentzForceCovariant:1480}
\frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \wedge \gamma_0
= q \lr{ \BE + \Bv \cross \BB }.

## Problem: Algebraic spacetime split of the Lorentz force equation.

Derive the results of \ref{eqn:lorentzForceCovariant:1440} through \ref{eqn:lorentzForceCovariant:1480} algebraically.

## Problem: Spacetime split of the Lorentz force tensor equation.

Show that \ref{eqn:lorentzForceCovariant:1680} also follows from the tensor form of the Lorentz force equation (\ref{eqn:lorentzForceCovariant:1620}) provided we identify
\label{eqn:lorentzForceCovariant:1500}
F^{k0} = E^k,

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

Also verify that the identifications of \ref{eqn:lorentzForceCovariant:1500} and \ref{eqn:lorentzForceCovariant:1520} is consistent with the geometric algebra Faraday bivector $$F = \BE + I c \BB$$, and the associated coordinate expansion of the field $$F = (1/2) (\gamma_\mu \wedge \gamma_\nu) F^{\mu\nu}$$.

# 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. Mathematica modules for Geometric Algebra’s GA(2,0), GA(3,0), and GA(1,3), 2017. URL https://github.com/peeterjoot/gapauli. [Online; accessed 24-Oct-2020].

## The many faces of Maxwell’s equations

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

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

## Goals of the project.

This project had a few goals

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

## The many faces of electromagnetism.

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

### Maxwell’s formulation.

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

### Quaterion representation.

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

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

### Minkowski tensor representation.

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

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

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

### Modern tensor formalism.

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

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

### Differential forms.

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

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

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

### Vector formalism.

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

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

### Space time algebra (geometric algebra).

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

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

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

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

### Paravector formalism (geometric algebra).

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

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

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

## Results.

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

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

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

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

# References

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Motivation.

The notation I prefer for relativistic geometric algebra uses Hestenes’ space time algebra (STA) [2], where the basis is a four dimensional space $$\setlr{ \gamma_\mu }$$, subject to Dirac matrix like relations $$\gamma_\mu \cdot \gamma_\nu = \eta_{\mu \nu}$$.

In this formalism a four vector is just the sum of the products of coordinates and basis vectors, for example, using summation convention

\label{eqn:boostToParavector:160}
x = x^\mu \gamma_\mu.

The invariant for a four-vector in STA is just the square of that vector

\label{eqn:boostToParavector:180}
\begin{aligned}
x^2
&= (x^\mu \gamma_\mu) \cdot (x^\nu \gamma_\nu) \\
&= \sum_\mu (x^\mu)^2 (\gamma_\mu)^2 \\
&= (x^0)^2 – \sum_{k = 1}^3 (x^k)^2 \\
&= (ct)^2 – \Bx^2.
\end{aligned}

Recall that a four-vector is time-like if this squared-length is positive, spacelike if negative, and light-like when zero.

Time-like projections are possible by dotting with the “lab-frame” time like basis vector $$\gamma_0$$

\label{eqn:boostToParavector:200}
ct = x \cdot \gamma_0 = x^0,

and space-like projections are wedges with the same

\label{eqn:boostToParavector:220}
\Bx = x \cdot \gamma_0 = x^k \sigma_k,

where sums over Latin indexes $$k \in \setlr{1,2,3}$$ are implied, and where the elements $$\sigma_k$$

\label{eqn:boostToParavector:80}
\sigma_k = \gamma_k \gamma_0.

which are bivectors in STA, can be viewed as an Euclidean vector basis $$\setlr{ \sigma_k }$$.

Rotations in STA involve exponentials of space like bivectors $$\theta = a_{ij} \gamma_i \wedge \gamma_j$$

\label{eqn:boostToParavector:240}
x’ = e^{ \theta/2 } x e^{ -\theta/2 }.

Boosts, on the other hand, have exactly the same form, but the exponentials are with respect to space-time bivectors arguments, such as $$\theta = a \wedge \gamma_0$$, where $$a$$ is any four-vector.

Observe that both boosts and rotations necessarily conserve the space-time length of a four vector (or any multivector with a scalar square).

\label{eqn:boostToParavector:260}
\begin{aligned}
\lr{x’}^2
&=
\lr{ e^{ \theta/2 } x e^{ -\theta/2 } } \lr{ e^{ \theta/2 } x e^{ -\theta/2 } } \\
&=
e^{ \theta/2 } x \lr{ e^{ -\theta/2 } e^{ \theta/2 } } x e^{ -\theta/2 } \\
&=
e^{ \theta/2 } x^2 e^{ -\theta/2 } \\
&=
x^2 e^{ \theta/2 } e^{ -\theta/2 } \\
&=
x^2.
\end{aligned}

## Paravectors.

Paravectors, as used by Baylis [1], represent four-vectors using a Euclidean multivector basis $$\setlr{ \Be_\mu }$$, where $$\Be_0 = 1$$. The conversion between STA and paravector notation requires only multiplication with the timelike basis vector for the lab frame $$\gamma_0$$

\label{eqn:boostToParavector:40}
\begin{aligned}
X
&= x \gamma_0 \\
&= \lr{ x^0 \gamma_0 + x^k \gamma_k } \gamma_0 \\
&= x^0 + x^k \gamma_k \gamma_0 \\
&= x^0 + \Bx \\
&= c t + \Bx,
\end{aligned}

We need a different structure for the invariant length in paravector form. That invariant length is
\label{eqn:boostToParavector:280}
\begin{aligned}
x^2
&=
\lr{ \lr{ ct + \Bx } \gamma_0 }
\lr{ \lr{ ct + \Bx } \gamma_0 } \\
&=
\lr{ \lr{ ct + \Bx } \gamma_0 }
\lr{ \gamma_0 \lr{ ct – \Bx } } \\
&=
\lr{ ct + \Bx }
\lr{ ct – \Bx }.
\end{aligned}

Baylis introduces an involution operator $$\overline{{M}}$$ which toggles the sign of any vector or bivector grades of a multivector. For example, if $$M = a + \Ba + I \Bb + I c$$, where $$a,c \in \mathbb{R}$$ and $$\Ba, \Bb \in \mathbb{R}^3$$ is a multivector with all grades $$0,1,2,3$$, then the involution of $$M$$ is

\label{eqn:boostToParavector:300}
\overline{{M}} = a – \Ba – I \Bb + I c.

Utilizing this operator, the invariant length for a paravector $$X$$ is $$X \overline{{X}}$$.

Let’s consider how boosts and rotations can be expressed in the paravector form. The half angle operator for a boost along the spacelike $$\Bv = v \vcap$$ direction has the form

\label{eqn:boostToParavector:120}
L = e^{ -\vcap \phi/2 },

\label{eqn:boostToParavector:140}
\begin{aligned}
X’
&=
c t’ + \Bx’ \\
&=
x’ \gamma_0 \\
&=
L x L^\dagger \\
&=
e^{ -\vcap \phi/2 } x^\mu \gamma_\mu
e^{ \vcap \phi/2 } \gamma_0 \\
&=
e^{ -\vcap \phi/2 } x^\mu \gamma_\mu \gamma_0
e^{ -\vcap \phi/2 } \\
&=
e^{ -\vcap \phi/2 } \lr{ x^0 + \Bx } e^{ -\vcap \phi/2 } \\
&=
L X L.
\end{aligned}

Because the involution operator toggles the sign of vector grades, it is easy to see that the required invariance is maintained

\label{eqn:boostToParavector:320}
\begin{aligned}
X’ \overline{{X’}}
&=
L X L
\overline{{ L X L }} \\
&=
L X L
\overline{{ L }} \overline{{ X }} \overline{{ L }} \\
&=
L X \overline{{ X }} \overline{{ L }} \\
&=
X \overline{{ X }} L \overline{{ L }} \\
&=
X \overline{{ X }}.
\end{aligned}

Let’s explicitly expand the transformation of \ref{eqn:boostToParavector:140}, so we can relate the rapidity angle $$\phi$$ to the magnitude of the velocity. This is most easily done by splitting the spacelike component $$\Bx$$ of the four vector into its projective and rejective components

\label{eqn:boostToParavector:340}
\begin{aligned}
\Bx
&= \vcap \vcap \Bx \\
&= \vcap \lr{ \vcap \cdot \Bx + \vcap \wedge \Bx } \\
&= \vcap \lr{ \vcap \cdot \Bx } + \vcap \lr{ \vcap \wedge \Bx } \\
&= \Bx_\parallel + \Bx_\perp.
\end{aligned}

The exponential

\label{eqn:boostToParavector:360}
e^{-\vcap \phi/2}
=
\cosh\lr{ \phi/2 }
– \vcap \sinh\lr{ \phi/2 },

commutes with any scalar grades and with $$\Bx_\parallel$$, but anticommutes with $$\Bx_\perp$$, so

\label{eqn:boostToParavector:380}
\begin{aligned}
X’
&=
\lr{ c t + \Bx_\parallel } e^{ -\vcap \phi/2 } e^{ -\vcap \phi/2 }
+
\Bx_\perp e^{ \vcap \phi/2 } e^{ -\vcap \phi/2 } \\
&=
\lr{ c t + \Bx_\parallel } e^{ -\vcap \phi }
+
\Bx_\perp \\
&=
\lr{ c t + \vcap \lr{ \vcap \cdot \Bx } } \lr{ \cosh \phi – \vcap \sinh \phi }
+
\Bx_\perp \\
&=
\Bx_\perp
+
\lr{ c t \cosh\phi – \lr{ \vcap \cdot \Bx} \sinh \phi }
+
\vcap \lr{ \lr{ \vcap \cdot \Bx } \cosh\phi – c t \sinh \phi } \\
&=
\Bx_\perp
+
\cosh\phi \lr{ c t – \lr{ \vcap \cdot \Bx} \tanh \phi }
+
\vcap \cosh\phi \lr{ \vcap \cdot \Bx – c t \tanh \phi }.
\end{aligned}

Employing the argument from [3],
we want $$\phi$$ defined so that this has structure of a Galilean transformation in the limit where $$\phi \rightarrow 0$$. This means we equate

\label{eqn:boostToParavector:400}
\tanh \phi = \frac{v}{c},

so that for small $$\phi$$

\label{eqn:boostToParavector:420}
\Bx’ = \Bx – \Bv t.

We can solving for $$\sinh^2 \phi$$ and $$\cosh^2 \phi$$ in terms of $$v/c$$ using

\label{eqn:boostToParavector:440}
\tanh^2 \phi
= \frac{v^2}{c^2}
=
\frac{ \sinh^2 \phi }{1 + \sinh^2 \phi}
=
\frac{ \cosh^2 \phi – 1 }{\cosh^2 \phi}.

which after picking the positive root required for Galilean equivalence gives
\label{eqn:boostToParavector:460}
\begin{aligned}
\cosh \phi &= \frac{1}{\sqrt{1 – (\Bv/c)^2}} \equiv \gamma \\
\sinh \phi &= \frac{v/c}{\sqrt{1 – (\Bv/c)^2}} = \gamma v/c.
\end{aligned}

The Lorentz boost, written out in full is

\label{eqn:boostToParavector:480}
ct’ + \Bx’
=
\Bx_\perp
+
\gamma \lr{ c t – \frac{\Bv}{c} \cdot \Bx }
+
\gamma \lr{ \vcap \lr{ \vcap \cdot \Bx } – \Bv t }
.

Authors like Chappelle, et al., that also use paravectors [4], specify the form of the Lorentz transformation for the electromagnetic field, but for that transformation reversion is used instead of involution.
I plan to explore that in a later post, starting from the STA formalism that I already understand, and see if I can make sense
of the underlying rationale.

# References

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

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

[3] L. Landau and E. Lifshitz. The Classical theory of fields. Addison-Wesley, 1951.

[4] James M Chappell, Samuel P Drake, Cameron L Seidel, Lachlan J Gunn, and Derek Abbott. Geometric algebra for electrical and electronic engineers. Proceedings of the IEEE, 102 0(9), 2014.