## Relativistic multivector surface integrals

### Background.

This post is a continuation of:

### Surface integrals.

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

We’ve now covered line integrals and the fundamental theorem for line integrals, so it’s now time to move on to surface integrals.

## Definition 1.1: Surface integral.

Given a two variable parameterization $$x = x(u,v)$$, we write $$d^2\Bx = \Bx_u \wedge \Bx_v du dv$$, and call
\begin{equation*}
\int F d^2\Bx\, G,
\end{equation*}
a surface integral, where $$F,G$$ are arbitrary multivector functions.

Like our multivector line integral, this is intrinsically multivector valued, with a product of $$F$$ with arbitrary grades, a bivector $$d^2 \Bx$$, and $$G$$, also potentially with arbitrary grades. Let’s consider an example.

## Problem: Surface area integral example.

Given the hyperbolic surface parameterization $$x(\rho,\alpha) = \rho \gamma_0 e^{-\vcap \alpha}$$, where $$\vcap = \gamma_{20}$$ evaluate the indefinite integral
\label{eqn:relativisticSurface:40}
\int \gamma_1 e^{\gamma_{21}\alpha} d^2 \Bx\, \gamma_2.

We have $$\Bx_\rho = \gamma_0 e^{-\vcap \alpha}$$ and $$\Bx_\alpha = \rho\gamma_{2} e^{-\vcap \alpha}$$, so
\label{eqn:relativisticSurface:60}
\begin{aligned}
d^2 \Bx
&=
(\Bx_\rho \wedge \Bx_\alpha) d\rho d\alpha \\
&=
\gamma_{0} e^{-\vcap \alpha} \rho\gamma_{2} e^{-\vcap \alpha}
}
d\rho d\alpha \\
&=
\rho \gamma_{02} d\rho d\alpha,
\end{aligned}

so the integral is
\label{eqn:relativisticSurface:80}
\begin{aligned}
\int \rho \gamma_1 e^{\gamma_{21}\alpha} \gamma_{022} d\rho d\alpha
&=
-\inv{2} \rho^2 \int \gamma_1 e^{\gamma_{21}\alpha} \gamma_{0} d\alpha \\
&=
\frac{\gamma_{01}}{2} \rho^2 \int e^{\gamma_{21}\alpha} d\alpha \\
&=
\frac{\gamma_{01}}{2} \rho^2 \gamma^{12} e^{\gamma_{21}\alpha} \\
&=
\frac{\rho^2 \gamma_{20}}{2} e^{\gamma_{21}\alpha}.
\end{aligned}

Because $$F$$ and $$G$$ were both vectors, the resulting integral could only have been a multivector with grades 0,2,4. As it happens, there were no scalar nor pseudoscalar grades in the end result, and we ended up with the spacetime plane between $$\gamma_0$$, and $$\gamma_2 e^{\gamma_{21}\alpha}$$, which are rotations of $$\gamma_2$$ in the x,y plane. This is illustrated in fig. 1 (omitting scale and sign factors.)

fig. 1. Spacetime plane.

## Fundamental theorem for surfaces.

For line integrals we saw that $$d\Bx \cdot \grad = \gpgradezero{ d\Bx \partial }$$, and obtained the fundamental theorem for multivector line integrals by omitting the grade selection and using the multivector operator $$d\Bx \partial$$ in the integrand directly. We have the same situation for surface integrals. In particular, we know that the $$\mathbb{R}^3$$ Stokes theorem can be expressed in terms of $$d^2 \Bx \cdot \spacegrad$$

## Problem: GA form of 3D Stokes’ theorem integrand.

Given an $$\mathbb{R}^3$$ vector field $$\Bf$$, show that
\label{eqn:relativisticSurface:180}
\int dA \ncap \cdot \lr{ \spacegrad \cross \Bf }
=
-\int \lr{d^2\Bx \cdot \spacegrad } \cdot \Bf.

Let $$d^2 \Bx = I \ncap dA$$, implicitly fixing the relative orientation of the bivector area element compared to the chosen surface normal direction.
\label{eqn:relativisticSurface:200}
\begin{aligned}
\int \lr{d^2\Bx \cdot \spacegrad } \cdot \Bf
&=
&=
\int dA \lr{ I \lr{ \ncap \wedge \spacegrad} } \cdot \Bf \\
&=
&=
-\int dA \lr{ \ncap \cross \spacegrad} \cdot \Bf \\
&=
-\int dA \ncap \cdot \lr{ \spacegrad \cross \Bf }.
\end{aligned}

The moral of the story is that the conventional dual form of the $$\mathbb{R}^3$$ Stokes’ theorem can be written directly by projecting the gradient onto the surface area element. Geometrically, this projection operation has a rotational effect as well, since for bivector $$B$$, and vector $$x$$, the bivector-vector dot product $$B \cdot x$$ is the component of $$x$$ that lies in the plane $$B \wedge x = 0$$, but also rotated 90 degrees.

For multivector integration, we do not want an integral operator that includes such dot products. In the line integral case, we were able to achieve the same projective operation by using vector derivative instead of a dot product, and can do the same for the surface integral case. In particular

## Theorem 1.1: Projection of gradient onto the tangent space.

Given a curvilinear representation of the gradient with respect to parameters $$u^0, u^1, u^2, u^3$$
\begin{equation*}
\end{equation*}
the surface projection onto the tangent space associated with any two of those parameters, satisfies
\begin{equation*}
\end{equation*}

### Start proof:

Without loss of generality, we may pick $$u^0, u^1$$ as the parameters associated with the tangent space. The area element for the surface is
\label{eqn:relativisticSurface:100}
d^2 \Bx = \Bx_0 \wedge \Bx_1 \,
du^0 du^1.

Dotting this with the gradient gives
\label{eqn:relativisticSurface:120}
\begin{aligned}
&=
du^0 du^1
\lr{ \Bx_0 \wedge \Bx_1 } \cdot \Bx^\mu \PD{u^\mu}{} \\
&=
du^0 du^1
\lr{
\Bx_0
\lr{\Bx_1 \cdot \Bx^\mu }

\Bx_1
\lr{\Bx_0 \cdot \Bx^\mu }
}
\PD{u^\mu}{} \\
&=
du^0 du^1
\lr{
\Bx_0 \PD{u^1}{}

\Bx_0 \PD{u^1}{}
}.
\end{aligned}

On the other hand, the vector derivative for this surface is
\label{eqn:relativisticSurface:140}
\partial
=
\Bx^0 \PD{u^0}{}
+
\Bx^1 \PD{u^1}{},

so
\label{eqn:relativisticSurface:160}
\begin{aligned}
&=
du^0 du^1\,
\lr{ \Bx_0 \wedge \Bx_1 } \cdot
\lr{
\Bx^0 \PD{u^0}{}
+
\Bx^1 \PD{u^1}{}
} \\
&=
du^0 du^1
\lr{
\Bx_0 \PD{u^1}{}

\Bx_1 \PD{u^0}{}
}.
\end{aligned}

### End proof.

We now want to formulate the geometric algebra form of the fundamental theorem for surface integrals.

## Theorem 1.2: Fundamental theorem for surface integrals.

Given multivector functions $$F, G$$, and surface area element $$d^2 \Bx = \lr{ \Bx_u \wedge \Bx_v }\, du dv$$, associated with a two parameter curve $$x(u,v)$$, then
\begin{equation*}
\int_S F d^2\Bx \lrpartial G = \int_{\partial S} F d^1\Bx G,
\end{equation*}
where $$S$$ is the integration surface, and $$\partial S$$ designates its boundary, and the line integral on the RHS is really short hand for
\begin{equation*}
\int
\evalbar{ \lr{ F (-d\Bx_v) G } }{\Delta u}
+
\int
\evalbar{ \lr{ F (d\Bx_u) G } }{\Delta v},
\end{equation*}
which is a line integral that traverses the boundary of the surface with the opposite orientation to the circulation of the area element.

### Start proof:

The vector derivative for this surface is
\label{eqn:relativisticSurface:220}
\partial =
\Bx^u \PD{u}{}
+
\Bx^v \PD{v}{},

so
\label{eqn:relativisticSurface:240}
F d^2\Bx \lrpartial G
=
\PD{u}{} \lr{ F d^2\Bx\, \Bx^u G }
+
\PD{v}{} \lr{ F d^2\Bx\, \Bx^v G },

where $$d^2\Bx\, \Bx^u$$ is held constant with respect to $$u$$, and $$d^2\Bx\, \Bx^v$$ is held constant with respect to $$v$$ (since the partials of the vector derivative act on $$F, G$$, but not on the area element, nor on the reciprocal vectors of $$\lrpartial$$ itself.) Note that
\label{eqn:relativisticSurface:260}
d^2\Bx \wedge \Bx^u
=
du dv\, \lr{ \Bx_u \wedge \Bx_v } \wedge \Bx^u = 0,

since $$\Bx^u \in sectionpan \setlr{ \Bx_u\, \Bx_v }$$, so
\label{eqn:relativisticSurface:280}
\begin{aligned}
d^2\Bx\, \Bx^u
&=
d^2\Bx \cdot \Bx^u
+
d^2\Bx \wedge \Bx^u \\
&=
d^2\Bx \cdot \Bx^u \\
&=
du dv\, \lr{ \Bx_u \wedge \Bx_v } \cdot \Bx^u \\
&=
-du dv\, \Bx_v.
\end{aligned}

Similarly
\label{eqn:relativisticSurface:300}
\begin{aligned}
d^2\Bx\, \Bx^v
&=
d^2\Bx \cdot \Bx^v \\
&=
du dv\, \lr{ \Bx_u \wedge \Bx_v } \cdot \Bx^v \\
&=
du dv\, \Bx_u.
\end{aligned}

This leaves us with
\label{eqn:relativisticSurface:320}
F d^2\Bx \lrpartial G
=
-du dv\,
\PD{u}{} \lr{ F \Bx_v G }
+
du dv\,
\PD{v}{} \lr{ F \Bx_u G },

where $$\Bx_v, \Bx_u$$ are held constant with respect to $$u,v$$ respectively. Fortuitously, this constant condition can be dropped, since the antisymmetry of the wedge in the area element results in perfect cancellation. If these line elements are not held constant then
\label{eqn:relativisticSurface:340}
\PD{u}{} \lr{ F \Bx_v G }

\PD{v}{} \lr{ F \Bx_u G }
=
F \lr{
\PD{v}{\Bx_u}

\PD{u}{\Bx_v}
} G
+
\lr{
\PD{u}{F} \Bx_v G
+
F \Bx_v \PD{u}{G}
}
+
\lr{
\PD{v}{F} \Bx_u G
+
F \Bx_u \PD{v}{G}
}
,

but the mixed partial contribution is zero
\label{eqn:relativisticSurface:360}
\begin{aligned}
\PD{v}{\Bx_u}

\PD{u}{\Bx_v}
&=
\PD{v}{} \PD{u}{x}

\PD{u}{} \PD{v}{x} \\
&=
0,
\end{aligned}

by equality of mixed partials. We have two perfect differentials, and can evaluate each of these integrals
\label{eqn:relativisticSurface:380}
\begin{aligned}
\int F d^2\Bx \lrpartial G
&=
-\int
du dv\,
\PD{u}{} \lr{ F \Bx_v G }
+
\int
du dv\,
\PD{v}{} \lr{ F \Bx_u G } \\
&=
-\int
dv\,
\evalbar{ \lr{ F \Bx_v G } }{\Delta u}
+
\int
du\,
\evalbar{ \lr{ F \Bx_u G } }{\Delta v} \\
&=
\int
\evalbar{ \lr{ F (-d\Bx_v) G } }{\Delta u}
+
\int
\evalbar{ \lr{ F (d\Bx_u) G } }{\Delta v}.
\end{aligned}

We use the shorthand $$d^1 \Bx = d\Bx_u – d\Bx_v$$ to write
\label{eqn:relativisticSurface:400}
\int_S F d^2\Bx \lrpartial G = \int_{\partial S} F d^1\Bx G,

with the understanding that this is really instructions to evaluate the line integrals in the last step of \ref{eqn:relativisticSurface:380}.

## Problem: Integration in the t,y plane.

Let $$x(t,y) = c t \gamma_0 + y \gamma_2$$. Write out both sides of the fundamental theorem explicitly.

Let’s designate the tangent basis vectors as
\label{eqn:relativisticSurface:420}
\Bx_0 = \PD{t}{x} = c \gamma_0,

and
\label{eqn:relativisticSurface:440}
\Bx_2 = \PD{y}{x} = \gamma_2,

so the vector derivative is
\label{eqn:relativisticSurface:460}
\partial
= \inv{c} \gamma^0 \PD{t}{}
+ \gamma^2 \PD{y}{},

and the area element is
\label{eqn:relativisticSurface:480}
d^2 \Bx = c \gamma_0 \gamma_2.

The fundamental theorem of surface integrals is just a statement that
\label{eqn:relativisticSurface:500}
\int_{t_0}^{t_1} c dt
\int_{y_0}^{y_1} dy
F \gamma_0 \gamma_2 \lr{
\inv{c} \gamma^0 \PD{t}{}
+ \gamma^2 \PD{y}{}
} G
=
\int F \lr{ c \gamma_0 dt – \gamma_2 dy } G,

where the RHS, when stated explicitly, really means
\label{eqn:relativisticSurface:520}
\begin{aligned}
\int &F \lr{ c \gamma_0 dt – \gamma_2 dy } G
=
\int_{t_0}^{t_1} c dt \lr{ F(t,y_1) \gamma_0 G(t, y_1) – F(t,y_0) \gamma_0 G(t, y_0) } \\
\int_{y_0}^{y_1} dy \lr{ F(t_1,y) \gamma_2 G(t_1, y) – F(t_0,y) \gamma_0 G(t_0, y) }.
\end{aligned}

In this particular case, since $$\Bx_0 = c \gamma_0, \Bx_2 = \gamma_2$$ are both constant functions that depend on neither $$t$$ nor $$y$$, it is easy to derive the full expansion of \ref{eqn:relativisticSurface:520} directly from the LHS of \ref{eqn:relativisticSurface:500}.

## Problem: A cylindrical hyperbolic surface.

Generalizing the example surface integral from \ref{eqn:relativisticSurface:40}, let
\label{eqn:relativisticSurface:540}
x(\rho, \alpha) = \rho e^{-\vcap \alpha/2} x(0,1) e^{\vcap \alpha/2},

where $$\rho$$ is a scalar, and $$\vcap = \cos\theta_k\gamma_{k0}$$ is a unit spatial bivector, and $$\cos\theta_k$$ are direction cosines of that vector. This is a composite transformation, where the $$\alpha$$ variation boosts the $$x(0,1)$$ four-vector, and the $$\rho$$ parameter contracts or increases the magnitude of this vector, resulting in $$x$$ spanning a hyperbolic region of spacetime.

Compute the tangent and reciprocal basis, the area element for the surface, and explicitly state both sides of the fundamental theorem.

For the tangent basis vectors we have
\label{eqn:relativisticSurface:560}
\Bx_\rho = \PD{\rho}{x} =
e^{-\vcap \alpha/2} x(0,1) e^{\vcap \alpha/2} = \frac{x}{\rho},

and
\label{eqn:relativisticSurface:580}
\Bx_\alpha = \PD{\alpha}{x} =
\lr{-\vcap/2} x
+
x \lr{ \vcap/2 }
=
x \cdot \vcap.

These vectors $$\Bx_\rho, \Bx_\alpha$$ are orthogonal, as $$x \cdot \vcap$$ is the projection of $$x$$ onto the spacetime plane $$x \wedge \vcap = 0$$, but rotated so that $$x \cdot \lr{ x \cdot \vcap } = 0$$. Because of this orthogonality, the vector derivative for this tangent space is
\label{eqn:relativisticSurface:600}
\partial =
\inv{x \cdot \vcap} \PD{\alpha}{}
+
\frac{\rho}{x}
\PD{\rho}{}
.

The area element is
\label{eqn:relativisticSurface:620}
\begin{aligned}
d^2 \Bx
&=
d\rho d\alpha\,
\frac{x}{\rho} \wedge \lr{ x \cdot \vcap } \\
&=
\inv{\rho} d\rho d\alpha\,
x \lr{ x \cdot \vcap }
.
\end{aligned}

The full statement of the fundamental theorem for this surface is
\label{eqn:relativisticSurface:640}
\int_S
d\rho d\alpha\,
F
\lr{
\inv{\rho} x \lr{ x \cdot \vcap }
}
\lr{
\inv{x \cdot \vcap} \PD{\alpha}{}
+
\frac{\rho}{x}
\PD{\rho}{}
}
G
=
\int_{\partial S}
F \lr{ d\rho \frac{x}{\rho} – d\alpha \lr{ x \cdot \vcap } } G.

As in the previous example, due to the orthogonality of the tangent basis vectors, it’s easy to show find the RHS directly from the LHS.

## Problem: Simple example with non-orthogonal tangent space basis vectors.

Let $$x(u,v) = u a + v b$$, where $$u,v$$ are scalar parameters, and $$a, b$$ are non-null and non-colinear constant four-vectors. Write out the fundamental theorem for surfaces with respect to this parameterization.

The tangent basis vectors are just $$\Bx_u = a, \Bx_v = b$$, with reciprocals
\label{eqn:relativisticSurface:660}
\Bx^u = \Bx_v \cdot \inv{ \Bx_u \wedge \Bx_v } = b \cdot \inv{ a \wedge b },

and
\label{eqn:relativisticSurface:680}
\Bx^v = -\Bx_u \cdot \inv{ \Bx_u \wedge \Bx_v } = -a \cdot \inv{ a \wedge b }.

The fundamental theorem, with respect to this surface, when written out explicitly takes the form
\label{eqn:relativisticSurface:700}
\int F \, du dv\, \lr{ a \wedge b } \inv{ a \wedge b } \cdot \lr{ a \PD{u}{} – b \PD{v}{} } G
=
\int F \lr{ a du – b dv } G.

This is a good example to illustrate the geometry of the line integral circulation.
Suppose that we are integrating over $$u \in [0,1], v \in [0,1]$$. In this case, the line integral really means
\label{eqn:relativisticSurface:720}
\begin{aligned}
\int &F \lr{ a du – b dv } G
=
+
\int F(u,1) (+a du) G(u,1)
+
\int F(u,0) (-a du) G(u,0) \\
\int F(1,v) (-b dv) G(1,v)
+
\int F(0,v) (+b dv) G(0,v),
\end{aligned}

which is a path around the spacetime parallelogram spanned by $$u, v$$, as illustrated in fig. 1, which illustrates the orientation of the bivector area element with the arrows around the exterior of the parallelogram: $$0 \rightarrow a \rightarrow a + b \rightarrow b \rightarrow 0$$.

fig. 2. Line integral orientation.

## Sabine Hossenfelder’s “Lost in Math”

“Lost in Math” is a book that I’ve been curious to read, as I’ve been a subscriber to Sabine’s blog and youtube channel for quite a while.  On her blog and channel, she provides overviews of many topics in physics that are well articulated, as well as what appear to be very well reasoned and researched criticisms of a number of topics (mostly physics related.)  Within the small population of people interested in theoretical physics, I think that she is also very well known for her completely fearlessness, as she appears to have none of the usual social resistance to offending somebody should her statements not be aligned with popular consensus.

This book has a few aspects:

• Interviews with a number of interesting and prominent physicists
• A brutal take on the failures of string theory, supersymmetry, theories of everything, and other research programs that have consumed significant research budgets, but are detached from experimental and observational constraints.
• An argument against the use of beauty, naturalness, and fine tuning avoidance in the constructions of physical theory.  Through the many interviews, we get a glimpse of the specific meanings of these words in the context of modern high level physical theories.
• Some arguments against bigger colliders, given that the current ones have not delivered on their promises of producing new physics.
• A considerable history of modern physics, and background for those wondering what the problems that string theory and supersymmetry have been trying to solve in the first place.
• Some going-forward recommendations.

While there were no equations in this book, it is not a terribly easy read.  I felt that reading this requires considerable physics sophistication.  To level set, while I haven’t studied particle physics or the standard model, I have studied special relativity, electromagnetism, quantum mechanics, and even some introductory QFT, but still found this book fairly difficult (and I admit to nodding off a few times as a result.)  I don’t think this is really a book that aimed at the general public.

If you do have the background to attempt this book, you will probably learn a fair amount, on topics that include, for example: the standard model, general relativity, symmetry breaking, coupling constants, and the cosmological constant.  An example was her nice illustration of symmetry breaking.  We remember touching on this briefly in QFT I, but it was presented in an algebraic and abstract fashion.  At the time I didn’t get the high level view of what this meant (something with higher energy can have symmetries that are impossible at lower energies.)  In this book, this concept is illustrated by a spinning top, which when spinning fast is stable and has rotational symmetry, but once frictional effects start to slow it down, it will start to precess and wobble, and the symmetry that is evident at higher spin rates weakens.  This was a particularly apt justification for the title of the book, as her description of symmetry breaking did not require any mathematics!

Deep in the book, it was pointed out that the equations of the standard model cannot generally be solved, but have to be dealt with using perturbation methods.  In retrospect, this shouldn’t have surprised me, since we generally can’t solve non-harmonic oscillator problems in closed form, and have to resort to numerical methods for most interesting problems.

There were a number of biting statements that triggered laughs while reading this book.  I wish that I’d made notes of more of of those while I read it, but here are two to whet your appetite:

• If you’d been sucking away on a giant jawbreaker for a century, wouldn’t you hope to finally get close to the gum?
• It’s easy enough for us to discard philosophy as useless — because it is useless.

#### On the picture above.

I like reading in the big living room chair behind my desk that our dog Tessa has claimed as her own, so as soon as I get up for coffee (or anything else), she will usually come and plop herself in the chair so that it’s no longer available to me.  If she was lying on the floor, and my wife sits on “her” chair, she will almost always occupy it once Sofia gets up.  Ironically, the picture above was taken just after I had gotten to the section where she was interviewing Chad Orzel, of “How to Teach Quantum Mechanics to your Dog” fame.

## On pet physics theories, Scientology, cosmology, relativity and libertarian tendencies.

In a recent Brian Keating podcast, he asked people to comment if they had their own theory of physics.

I’ve done a lot of exploration of conventional physics, both on my own, and with some in class studies (non-degree undergrad physics courses at UofT, plus grad QM and QFTI), but I don’t have my own personal theories of physics.  It’s enough of a challenge to figure out the existing theories without making up your own$${}^{1}$$.

However, I have had one close encounter with an alternate physics theory, as I had a classmate in undergrad QMI (phy356) that had a personal “Aether” theory of relativity.  He shared that theory with me, but it came in the form of about 50 pages of dense text without equations.  For all intents and purposes, this was a theory that was written in an entirely different language than the rest of physics.  To him, it was all self evident, and he got upset with the suggestion of trying to mathematize it.  A lot of work and thought went into that theory, but it is work that has very little chance of paying off, since it was packaged in a form that made it unpalatable to anybody who is studying conventional physics.  There is also a great deal of work that would be required to “complete” the theory (presuming that could be done), since he would have to show that his theory is not inconsistent with many measurements and experiments, and would not predict nonphysical phenomena.  That was really an impossible task, which he would have found had he attempted to do so.  However, instead of attempting to do that work, he seemed to think that the onus should fall on others to do so.  He had done the work to write what he believed to be a self consistent logical theory that was self evidently true, and shouldn’t have to do anything more.

It is difficult to fully comprehend how he would have arrived at such certainty about his Aether theory, when he did not have the mathematical sophistication to handle the physics found in the theories that he believed his should supplant.  However, in his defence, there are some parts of what I imagine were part of his thought process that I can sympathize with.  The amount of knowledge required to understand the functioning of a even a simple digital watch (not to mention the cell “phone” computers that we now all carry) is absolutely phenomenal.  We are surrounded by so much sophisticated technology that understanding the mechanisms behind it all is practically unknowable.  Much of the world around is us is effectively magic to most people, even those with technical sophistication.  Should there be some sort of catastrophe that wipes out civilization, requiring us to relearn or redevelop everything from first principles, nobody really has the breadth required to reconstruct the world around us.  It is rather humbling to ponder that.

One way of coping with the fact that it is effectively impossible to know how everything works is to not believe in any consensus theories — period.  I think that is the origin of the recent popularization of flat earth models.  I think this was a factor in my classmate’s theory, as he also went on to believe that quantum mechanics was false (or also believed that when I knew him, but never stated it to me.)  People understand that it is impossible to know everything required to build their own satellites, imaging systems, rockets, et-al, (i.e. sophisticated methods of disproving the flat earth theory) and decide to disbelieve everything that they cannot personally prove.  That’s an interesting defence mechanism, but takes things to a rather extreme conclusion.

I have a lot of sympathy for those that do not believe in consensus theories.  Without such disbelief I would not have my current understanding of the world.  It happens that the prevailing consensus theory that I knew growing up was that of Scientology.  Among the many ideas that one finds in Scientology is a statement that relativity is wrong$${}^2$$.  It has been too many years for me to accurately state the reasons that Hubbard stated that relativity was incorrect, but I do seem to recall that one of the arguments had to do with the speed of light being non-constant when bent by a prism $${}^3$$.  I carried some Scientology derived skepticism of relativity into the undergrad “relativistic electrodynamics“$${}^4$$ course that I took back around 2010, but had I not been willing to disregard the consensus beliefs that I had been exposed to up to that point in time, I would not have learned anything from that class.  Throwing away beliefs so that you can form your own conclusions is the key to being able to learn and grow.

I would admit to still carrying around baggage from my early indoctrination, despite not having practised Scientology for 25+ years.  This baggage spans multiple domains.  One example is that I am not subscribed to the religious belief that government and police are my friends.  It is hard to see your father, whom you love and respect, persecuted, and not come away with disrespect for the persecuting institutions.  I now have a rough idea of what Dad back in the Scientology Guardian’s Office did that triggered the ire of the Ontario crown attorneys $${}^5$$.  However, that history definitely colored my views and current attitudes.  In particular, I recognize that back history as a key factor that pushed me so strongly in a libertarian direction.  The libertarian characterization of government as an entity that infringes on personal property and rights seems very reasonable, and aligns perfectly with my experience $${}^6$$.

A second example of indoctrination based disbelief that I recognize that I carry with me is not subscribing to the current popular cosmological models of physics.  The big bang, and the belief that we know to picosecond granularity how the universe was at it’s beginning seems to me very religious.  That belief requires too much extrapolation, and it does not seem any more convincing to me than the Scientology cosmology.  The Scientology cosmology is somewhat ambiguous, and contains both a multiverse story and a steady state but finite model.  In the steady state aspect of that cosmology, the universe that we inhabit is dated with an age of 76 trillion years, but I don’t recall any sort of origin story for the beginning portion of that 76 trillion.  Given the little bits of things that we can actually measure and observe, I have no inclination to sign up for the big bang testiment any more than any other mythical origin story.  Thankfully, I can study almost anything that has practical application in physics or engineering and no amount of belief or disdain in the big bang or other less popular “physics” cosmologies makes any difference.  All of these, whether they be the big bang, cyclic theories, multiverses (of the quantum, thetan-created $${}^7$$, or inflationary varieties), or even the old Scientology 76 trillion years old cosmology of my youth, cannot be measured, proven or disproved.  Just about any cosmology has no impact on anything real.

This throw it all out point of view of cosmology is probably a too cynical and harsh treatment of the subject.  It is certainly not the point of view that most practising physicists would take, but it is imminently practical.  There’s too much that is unknowable, so why waste time on the least knowable aspects of the unknowable when there are so many concrete things that we can learn.

### Footnotes:

[1] The closest that I have come to my own theory of physics is somewhat zealous advocacy for the use of real Clifford algebras in engineering and physics (aka. geometric algebra.)  However, that is not a new theory, it is just a highly effective way to compactly represent many of the entities that we encounter in more clumsy forms in conventional physics.

[2] Hubbard’s sci-fi writing shows that he had knowledge of special relativistic time-dilation, and length-contraction effects.  I seem to recall that Lorentz transformations were mentioned in passing (on either the Student hat course, or in the “PDC” lectures).  I don’t believe that Hubbard had the mathematical sophistication to describe a Lorentz transformation in a quantitative sense.

[3] The traversal of light through matter is a complex affair, considerably different from light in vacuum, where the relativistic constancy applies.  It would be interesting to have a quantitative understanding of the chances of a photon getting through a piece of glass without interacting (absorption and subsequent time delayed spontaneous remission of new photons when the lattice atoms drop back into low energy states.)  There are probably also interactions of photons with the phonons of the lattice itself, and I don’t know how those would be quantified.  However, in short, I bet there is a large chance that most of the light that exits a transparent piece of matter is not the same light that went in, as it is going to come out as photons with different momentum, polarization, frequency, and so forth.  If we measure characteristics of a pulse of light going into and back out of matter, it’s probably somewhat akin to measuring the characteristics of a clementine orange that is thrown at a piece of heavy chicken wire at fastball speeds.  You’ll get some orange peel, seeds, pulp and other constituent components out the other side of the mesh, but shouldn’t really consider the inputs and the outputs to be equivalent entities.

[4] Relativistic electrodynamics was an extremely redundant course title, but was used to distinguish the class from the 3rd year applied electrodynamics course that had none of the tensor, relativistic, Lagrangian, nor four-vector baggage.

[5] Some information about that court case is publicly available, but it would be interesting to see whether I could use the Canadian or Ontario equivalent to the US freedom of information laws to request records from the Ontario crown and the RCMP about the specifics of Dad’s case.  Dad has passed, and was never terribly approachable about the subject when I could have asked him personally.  I did get his spin on the events as well as the media spin, and suspect that the truth is somewhere in between.

[6] This last year will probably push many people towards libertarian-ism (at least the subset of people that are not happy to be conforming sheep, or are too scared not to conform.)  We’ve had countless examples of watching evil bastards in government positions of power impose dictatorial and oppressive covid lockdowns on the poorest and most unfortunate people that they supposedly represent.  Instead, we see the corruption at full scale, with the little guys clobbered, and this covid lockdown scheme essentially being a way to efficiently channel money into the pockets of the rich.  The little guys loose their savings, livelihoods, and get their businesses shut down by fat corrupt bastards that believe they have the authority to decide whether or not you as an individual are essential.  The fat bastards that have the stamp of government authority do not believe that you should have the right to make up your own mind about what levels of risk are acceptable to you or your family.

[7] In Scientology, a sufficiently capable individual is considered capable of creating their own universes, independent of the 76 trillion year old universe that we currently inhabit.  Thetan is the label for the non-corporal form of that individual (i.e. what would be called the spirit or the soul in other religions.)

## Just watched Cloonie’s “Midnight Sky”

I just watched George Clooney’s “Midnight Sky” on netflix.

The movie is visually striking, set on a space ship and on an apocalyptic Earth in +30 years.  Some sort of unspecified radioactive disaster has pretty much wiped out all livable space on Earth.  The movie focuses on the attempt of a sick astronomer to communicate with a space ship that has been off exploring a newly found habitable moon of Jupiter.   They have been out of communication with Earth for a couple years.

I really didn’t understand the foundational premise of the movie.  We have been able to receive communications from satellites that we’ve sent to Jupiter, and a quick google says it’s only ~22 light minutes between Jupiter and Earth.  If that distance is the closest, let’s suppose that it’s a few times that at maximum separation — that’s still only a couple hours separation (guestimating).  Why would the ship have gone completely out of communication with Earth for years while they were on their mission?

There were lots of other holes in the movie, and I wonder if some of those missing pieces were detailed in the book?

Incidentally, the astronomy facility looked really cosy and comfortable for a something located in Antarctica!  There was mention of the poles late in the movie, but early on there was the famous picture of the explorer Scott with his four companions on the wall, which I assumed was meant to give away the location (I recognized that picture from Brian Keating’s book, “Loosing the Nobel Prize”.)

## Brian Keating’s “Losing the Nobel Prize”

I’ve just finished “Loosing the Nobel Prize”, by Brian Keating.  I’d heard the book mentioned in episodes of his “Into the impossible” podcast$${}^1$$.

This is a pretty fun and interesting book, with a few interesting threads woven through it:

• his astronomical and cosmological work,
• a pretty thorough background on a number of astronomical principles and history,
• rationale for a number of the current and past cosmological models,
• how he got close to but missed the Nobel target with his work,
• discussion and criticisms of the Nobel nomination process and rules, and
• DUST!

I had no idea that dust has been the nemesis of astronomers for so many hundreds of years, and will likely continue to be so for hundreds more.  This is not just dust on the lenses, but the dust and other fine matter that pervades the universe and mucks up measurements.  It will be a fitting end for his book to end up dusty on bookshelves around the world once all the purchasers have read it.

The author clearly knows his material well, and presents a thorough background lesson on the history of cosmology, starting way back at the Earth centered model, and moving through the history of competing narratives to the current big bang and inflationary models that seem to have popular consensus.

I’ve never thought much of cosmological ideas, as they go so deep into the territory of extrapolation that they seem worthless to me.  How can you argue that you know what happened $$10^{-17}$$ seconds into the beginning of the universe $${}^2$$, when we can’t solve a three body problem without chaos getting into the mix?  The level of extrapolation that is required for some of these models makes arguments about them seem akin to arguing about how many angels fit on the head of a pin.

What’s kind of sad about cosmological models is how little difference they make.  It doesn’t matter if you subscribe to the current big bang religion, cyclic variations of bang and collapse, steady state, multiverses, or anything else: none of the theories have any practical application to anything that we can see or hear or touch.  I don’t think that my preconceived ideas about the uselessness of cosmology has been changed much by reading this book.  However, I do have a new appreciation for the careful and thorough thought, measurement, and experiment that has gone into building and discarding various models over time.  This book details many of the key experiments and concepts that lie behind some of the models.  It would take a lot of work to fully understand the ideas that were outlined in this book, and that’s not work that I’m inclined to do, but I did enjoy his thorough overview.

Okay, that’s enough of a rant against cosmology.  Don’t let my distaste of that subject dissuade you from reading this book, which is well written, entertaining, informative, and thoughtful.

As a small teaser, here are a couple of selected lines that give a taste for the clever wit that is casually interlaced into the book:

• Trying to interest others in astronomy: “If you can imagine teaching music appreciation to a class filled with tone-deaf students, it was like that, only more disheartening.”
• “It was all worth it, he assured me: because there was only going to be one sunset and one sunrise in the next year at the South Pole, he would take home \$75,000 for a single night’s work!”
• “By the time I arrived at the Pole, it was chilly for summer: -30 C (-25 F).”

#### Footnotes

[1]  I have not worked through all of his back episodes, but his line up of recent guests (Penrose, Susskind, Wilczek, Glashow, …) has been pretty spectacular.

[2] I am probably wrong about the precise levels of granularity that is claimed to be known, but do recall from my teenage reading of Hawking’s Brief History, that he insisted we “know” what happened down to insane levels of precision.

## As your T.A., I have to punish you …

Back in university, I had to implement a reverse polish notation calculator in a software engineering class.  Overall the assignment was pretty stupid, and I entertained myself by generating writing a very compact implementation.  It worked perfectly, but I got a 25/40 (62.5%) grade on it.  That mark was well deserved, although I did not think so at the time.

The grading remarks were actually some of best feedback that I ever received, and also really funny to boot.  I don’t know the name of this old now-nameless TA anymore, but I took his advice to heart, and kept his grading remarks on my wall in my IBM office for years.  That served as an excellent reminder not to write over complicated code.

Today, I found those remarks again, and am posting them for posterity.  Enjoy!

• It is obvious that are a very clever person, but this program is is like a big puzzle, and in understanding it, I appreciated it and enjoyed it, because of your cleverness. However much I enjoyed, it is none the less a very poorly designed program.
• A program should be constructed in the easiest and simplest to understand manner because when you construct very large programs the “complexity” of them will increase greatly.
• A program should not be an intricate puzzle, where you show off how clever you are.
• Your string class is an elephant gun trying to kill a mouse.
• macros Build_binary_op and Binary_op are the worst examples of programming style I have ever seen in my entire life!  Veru c;ever. bit a cardinal sin of programming style.
• Your binary_expr constructor does all the computation.  Not good style.
• Your “expr” class is a baroque mess.
• Although I enjoyed your program, Never write a program like this in your life again.  As your T.A., I have to pushish you so that you do not develop bad habits in the future.  I hate to do it, but I can only give you 25/40 for this “clever puzzle”.

### Reflection.

The only part of this feedback that I would refute was the comment about the string class.  That was a actually a pretty good string implementation.  I didn’t write it because I was a viscous mouse hunter, but because I hit a porting issue with pre-std:: C.  In particular, we had two sets of Solaris machines available to us, and I was using one that had a compiler that included a nice C++ string class.  So, naturally I used it.  For submission, our code had to compile an run on a different Solaris machine, and lo and behold, the string class that all my code was based on was not available.

What I should have done (20/20 hindsight), was throw out my horrendous code, and start over from scratch.  However, I took the more fun approach, and wrote my own string class so that my machine would compile on either machine.

Amusingly, when I worked on IBM LUW, there was a part of the query optimizer code seemed to have learned all it’s tricks from the ugly macros and token pasting that I did in this assignment.  It was truly gross, but there was 10000x more of it than my assignment.  Having been thoroughly punished for my atrocities, I easily recognized this code for the evil it was.  The only way that you could debug that optimizer code, was by running it through the preprocessor, cut and pasting the results, and filtering that cut and paste through something like cindent (these days you would probably use clang-format.)  That code was brutal, and I always wished that it’s authors had had the good luck of having a TA like mine.  That code is probably still part of LUW terrorizing developers.  Apparently the justification for it was that it was originally written by an IBM researcher using templates, but templates couldn’t be used in DB2 code because we didn’t have compiler on all platforms that supported them at the time.

I have used token pasting macros very judiciously and sparingly in the 26 years since I originally used them in this assignment, and I do think that there are a few good uses for that sort of generative code.  However, if you do have to write that sort of code, I think it’s better to write perl (or some other language) code that generates understandable code that can be debugged, instead of relying on token pasting.

## 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.

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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)

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## 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.

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## 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.