## Hardcover physics class notes.

Amazon’s kindle direct publishing invited me to their hardcover trial program, and I’ve now made hardcover versions available of most of my interesting physics notes compilations:

Instead of making hardover versions of my classical mechanics, antenna theory, and electromagnetic theory notes, I have unpublished the paperback versions. These are low quality notes, and I don’t want more people to waste money on them (some have.) The free PDFs of all those notes are still available.

My geometric algebra book is also available in both paperback and hardcover (black and white). I’ve unpublished the color version, as it has a much higher print cost, and I thought it was too confusing to have all the permutations of black-and-white/color and paperback/hardcover.

## A better 3D generalization of the Mandelbrot set.

I’ve been exploring 3D generalizations of the Mandelbrot set:

The iterative equation for the Mandelbrot set can be written in vector form ([1]) as:

\begin{aligned}
\Bz
&\rightarrow
\Bz \Be_1 \Bz + \Bc \\
&=
\Bz \lr{ \Be_1 \cdot \Bz }
+
\Bz \cdot \lr{ \Be_1 \wedge \Bz }
+ \Bc \\
&=
2 \Bz \lr{ \Be_1 \cdot \Bz }

\Bz^2\, \Be_1
+ \Bc
\end{aligned}

Plotting this in 3D was an interesting challenge, but showed that the Mandelbrot set expressed above has rotational symmetry about the x-axis, which is kind of boring.

If all we require for a 3D fractal is to iterate a vector equation that is (presumably) at least quadratic, then we have lots of options. Here’s the first one that comes to mind:

\begin{aligned}
\Bz
&\rightarrow
\gpgradeone{ \Ba \Bz \Bb \Bz \Bc } + \Bd \\
&=
\lr{ \Ba \cdot \Bz } \lr{ \Bz \cross \lr{ \Bc \cross \Bz } }
+
\lr{ \Ba \cross \Bz } \lr{ \Bz \cdot \lr{ \Bc \cross \Bz } }
+ \Bd
.
\end{aligned}

where we iterate starting, as usual with $$\Bz = 0$$ where $$\Bd$$ is the point of interest to test for inclusion in the set. I tried this with
\label{eqn:mandel3:n}
\begin{aligned}
\Ba &= (1,1,1) \\
\Bb &= (1,0,0) \\
\Bc &= (1,-1,0).
\end{aligned}

Here are some slice plots at various values of z

and an animation of the slices with respect to the z-axis:

Here are a couple snapshots from a 3D Paraview rendering of a netCDF dataset of all the escape time values

Data collection and image creation used commit b042acf6ab7a5ba09865490b3f1fedaf0bd6e773 from my Mandelbrot generalization experimentation repository.

# References

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

## Some slices.

As followup to:

here is a bit more experimentation with Paraview slice filtering. This time I saved all the point data with the escape time counts, and rendered it with a few different contours

The default slice filter places the plane in the x-y orientation:

but we can also tilt it in the Paraview render UI

and suppress the contour view to see just the slice

As a very GUI challenged user, I don’t find the interface particularly intuitive, but have at least figured out this one particular slicing task, which is kind of cool.  It’s impressive that the UI can drive interesting computational tasks without having to regenerate or reload any of the raw data itself.  This time I was using the MacOSX Paraview client, which is nicer looking than the Windows version, but has some weird glitches in the file dialogues.

## Analysis.

The graphing play above shows some apparent rotational symmetry our vector equivalent to the Mandelbrot equation

\Bx \rightarrow \Bx \Be_1 \Bx + \Bc.

It was not clear to me if this symmetry existed, as there were artifacts in the plots that made it appear that there was irregularity. However, some thought shows that this irregularity is strictly due to sampling error, and perhaps also due to limitations in the plotting software, as such an uneven surface is probably tricky to deal with.

To see this, here are the first few iterations of the Mandlebrot sequence for an arbitary starting vector $$\Bc$$.

\begin{aligned}
\Bx_0 &= \Bc \\
\Bx_1 &= \Bc \Be_1 \Bc + \Bc \\
\Bx_2 &= \lr{ \Bc \Be_1 \Bc + \Bc } \Be_1 \lr{ \Bc \Be_1 \Bc + \Bc } + \Bc \Be_1 \Bc + \Bc.
\end{aligned}

Now, what happens when we rotate the starting vector $$\Bc$$ in the $$y-z$$ plane. The rotor for such a rotation is

R = \exp\lr{ e_{23} \theta/2 },

where

\Bc \rightarrow R \Bc \tilde{R}.

Observe that if $$\Bc$$ is parallel to the x-axis, then this rotation leaves the starting point invariant, as $$\Be_1$$ commutes with $$R$$. That is

R \Be_1 \tilde{R} =
\Be_1 R \tilde{R} = \Be_1.

Let $$\Bc’ = R \Bc \tilde{R}$$, so that

\Bx_0′ = R \Bc \tilde{R} = R \Bx_0 \tilde{R} .

\begin{aligned}
\Bx_1′
&= R \Bc \tilde{R} \Be_1 R \Bc \tilde{R} + R \Bc \tilde{R} \\
&= R \Bc \Be_1 \Bc \tilde{R} + R \Bc \tilde{R} \\
&= R \lr{ \Bc \Be_1 \Bc R + \Bc } \tilde{R} \\
&= R \Bx_1 \tilde{R}.
\end{aligned}

\label{eqn:m2:n}
\begin{aligned}
\Bx_2′
&= \Bx_1′ \Be_1 \Bx_1′ + \Bc’ \\
&= R \Bx_1 \tilde{R} \Be_1 R \Bx_1 \tilde{R} + R \Bc \tilde{R} \\
&= R \Bx_1 \Be_1 \Bx_1 \tilde{R} + R \Bc \tilde{R} \\
&= R \lr{ \Bx_1 \Be_1 \Bx_1 + \Bc } \tilde{R} \\
&= R \Bx_2 \tilde{R}.
\end{aligned}

The pattern is clear. If we rotate the starting point in the y-z plane, iterating the Mandelbrot sequence results in precisely the same rotation of the x-y plane Mandelbrot sequence. So the apparent rotational symmetry in the 3D iteration of the Mandelbrot vector equation is exactly that. This is an unfortunately boring 3D fractal. All of the interesting fractal nature occurs in the 2D plane, and the rest is just a consequence of rotating that image around the x-axis. We get some interesting fractal artifacts if we slice the rotated Mandelbrot image.

## Some 3D renderings of the Mandelbrot set.

As mentioned previously, using geometric algebra we can convert the iterative equation for the Mandelbrot set from complex number form

z \rightarrow z^2 + c,

to an equivalent vector form

\mathbf{x} \rightarrow \mathbf{x} \mathbf{e} \mathbf{x} + \mathbf{c},

where $$\mathbf{e}$$ represents the x-axis (say). Geometrically, each iteration takes $$\mathbf{e}$$ and reflects it about the direction of $$\mathbf{x}$$, then scales that by $$\mathbf{x}^2$$ and adds $$\mathbf{c}$$.

To get the usual 2D Mandelbrot set, one iterates with vectors that lie only in the x-y plane, but we can treat the Mandelbrot set as a 3D solid if we remove the x-y plane restriction.

Last time I animated slices of the 3D set, but after a whole lot of messing around I managed to save the data for all the interior points of the 3D set in netCDF format, and render the solid using Paraview. Paraview has tons of filters available, and experimenting with them is pretty time consuming, but here are some initial screenshots of the 3D Mandelbrot set:

It’s interesting that much of the characteristic detail of the Mandelbrot set is not visible in the 3D volume, but if we slice that volume, you can then you can see it again.  Here’s a slice taken close to the z=0 plane (but far enough that the “CN tower” portion of the set is not visible)

You can also see some of that detail if the opacity of the rendering is turned way down:

If you look carefully at the images above, you’ll see that the axis labels are wrong.  I think that I’ve screwed up one of the stride related parameters to my putVar call, and I end up with x+z transposed in the axes labels when visualized.

## Notes.

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

## Guts.

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

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

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

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

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

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

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

## Problem: Expansion of 2D area bivector.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

fig. 2. Line integral around rectangular boundary.

## Theorem 1.1: Green’s theorem [1].

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

## Problem: Relationship to Green’s theorem.

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

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

### Start proof:

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

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

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

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

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

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

### End proof.

fig. 3. Path for Green’s theorem.

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

## Theorem 1.2: 2D gradient integrals.

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

### Start proof:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

### End proof.

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

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

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

fig. 4. Outwards oriented normal for Euclidean space.

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

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

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

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

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

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

# References

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

## Some experiments in youtube mathematics videos

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## 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.

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

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

Background for this particular post can be found in

## Motivation.

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

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

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

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

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

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

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

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

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

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

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

## Multivector line integrals.

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

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

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

## Definition 1.1: Line integral.

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

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

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

## Problem: Circular parameterization.

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

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

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

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

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

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

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

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

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

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

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

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

fig. 1. Tangent perpendicularity in mixed metric.

### Perfect differentials.

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

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

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

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

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

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

## Definition 1.2: Vector derivative.

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

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

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

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

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

## Theorem 1.1: Fundamental theorem for line integrals.

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

### Start proof:

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

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

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

## More to come.

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

# References

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

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

## A couple more reciprocal frame examples.

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

This post logically follows both of the following:

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

## More examples.

Here are a few additional examples of reciprocal frame calculations.

## Problem: Unidirectional arbitrary functional dependence.

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

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

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

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

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

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

as expected.

## Problem: Linear two variable parameterization.

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

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

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

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

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

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

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

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

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

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

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

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

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

## Problem: Quadratic two variable parameterization.

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

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

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

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

## Problem: Reciprocal frame for generalized cylindrical parameterization.

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

fig. 1. “Cylindrical” boost parameterization.

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

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

and

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

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

fig. 2. Projection and rejection geometry.

## Parameterization of a general linear transformation.

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

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

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

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

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

## Theorem 1.1: Projecting onto the tangent space.

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

### Start proof:

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

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

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

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

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

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

We find the two stated ways of computing the projection.

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

## Lorentz transformations in Space Time Algebra (STA)

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

## Motivation.

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

## Theorem 1.1: Lorentz transformation.

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

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

### Start proof:

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

### End proof.

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

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

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

fig 1. Projection and rejection geometry.

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

## Theorem 1.2: Boost.

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

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

### Start proof:

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

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

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

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

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

## Theorem 1.3: Spatial rotation.

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

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

### Start proof:

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

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

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

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

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

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

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

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

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

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

## Problem: Verify components relative to boost direction.

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

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

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

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

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

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

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

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

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

as expected.

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

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

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

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

## Problem: Rotation transformation components.

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

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

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

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

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.