math and physics play

Hardcover physics class notes.

March 13, 2021 math and physics play , , , , , , , , , , , , , ,

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Instead of making hardover versions of my classical mechanics, antenna theory, and electromagnetic theory notes, I have unpublished the paperback versions.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 I’ve unpublished the color version, as it has a much higher print cost, and I thought it was too confusing to have all the permutations of black-and-white/color and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.

February 9, 2021 math and physics play , , , , , , ,

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{equation}
\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}
\end{equation}
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{equation}
\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}
\end{equation}
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
\begin{equation}\label{eqn:mandel3:n}
\begin{aligned}
\Ba &= (1,1,1) \\
\Bb &= (1,0,0) \\
\Bc &= (1,-1,0).
\end{aligned}
\end{equation}
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.

Slicing of the 3D Mandelbrot set, and analysis.

February 8, 2021 math and physics play , , , , , , ,

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
\begin{equation}
\Bx \rightarrow \Bx \Be_1 \Bx + \Bc.
\end{equation}
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{equation}
\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}
\end{equation}

Now, what happens when we rotate the starting vector \( \Bc \) in the \( y-z \) plane. The rotor for such a rotation is
\begin{equation}
R = \exp\lr{ e_{23} \theta/2 },
\end{equation}
where
\begin{equation}
\Bc \rightarrow R \Bc \tilde{R}.
\end{equation}
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
\begin{equation}
R \Be_1 \tilde{R} =
\Be_1 R \tilde{R} = \Be_1.
\end{equation}
Let \( \Bc’ = R \Bc \tilde{R} \), so that
\begin{equation}
\Bx_0′ = R \Bc \tilde{R} = R \Bx_0 \tilde{R} .
\end{equation}
\begin{equation}
\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}
\end{equation}
\begin{equation}\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}
\end{equation}

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.

February 7, 2021 math and physics play , , , , ,

As mentioned previously, using geometric algebra we can convert the iterative equation for the Mandelbrot set from complex number form
\begin{equation}
z \rightarrow z^2 + c,
\end{equation}
to an equivalent vector form
\begin{equation}
\mathbf{x} \rightarrow \mathbf{x} \mathbf{e} \mathbf{x} + \mathbf{c},
\end{equation}
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.

Unpacking the fundamental theorem of multivector calculus in two dimensions

January 18, 2021 math and physics play , , , , , , , , , , , , , , , , , , ,

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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:20}
\int_S F d\Bx \lrpartial G = \evalbar{F G}{\Delta S},
\end{equation}
and
\begin{equation}\label{eqn:unpackingFundamentalTheorem:40}
\int_S F d^2\Bx \lrpartial G = \oint_{\partial S} F d\Bx G.
\end{equation}
The first case is trivial. Given a parameterizated curve \( x = x(u) \), it just states
\begin{equation}\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)),
\end{equation}
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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:80}
d^2 \Bx = d\Bx_u \wedge d \Bx_v.
\end{equation}
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.

Answer

With parameterization \( x = x(u,v) = x^\alpha e_\alpha = x_\alpha e^\alpha \), we have
\begin{equation}\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,
\end{equation}
or
\begin{equation}\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.
\end{equation}
The upper and lower index pseudoscalars are related by
\begin{equation}\label{eqn:unpackingFundamentalTheorem:180}
e^1 e^2 e_1 e_2 =
-e^1 e^2 e_2 e_1 =
-1,
\end{equation}
so with \( I = e_1 e_2 \),
\begin{equation}\label{eqn:unpackingFundamentalTheorem:200}
e^1 e^2 = -I^{-1},
\end{equation}
leaving us with
\begin{equation}\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}.
\end{equation}
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
\begin{equation}\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,
\end{equation}
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.)

Answer

From \ref{eqn:unpackingFundamentalTheorem:140} we see that the fundamental theorem takes the form
\begin{equation}\label{eqn:unpackingFundamentalTheorem:220}
\int_S dx^1 dx^2\, F I \lrgrad G = \oint_{\partial S} F d\Bx G.
\end{equation}
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:
\begin{equation}\label{eqn:unpackingFundamentalTheorem:240}
I \grad
=
e_1 e_2 \lr{ e^1 \partial_1 + e^2 \partial_2 }
=
-e_2 \partial_1 + e_1 \partial_2.
\end{equation}

  • \( F = 1 \) and \( G \in \bigwedge^0 \) or \( G \in \bigwedge^2 \). For \( F = 1 \) and scalar or bivector \( G \) we have
    \begin{equation}\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,
    \end{equation}
    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
    \begin{equation}\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) }.
    \end{equation}
    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
    \begin{equation}\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,
    \end{equation}
    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
    \begin{equation}\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,
    \end{equation}
    and bivector relations
    \begin{equation}\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.
    \end{equation}
    To expand \ref{eqn:unpackingFundamentalTheorem:320}, let
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:360}
    G = g_1 e^1 + g_2 e^2,
    \end{equation}
    for which
    \begin{equation}\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,
    \end{equation}
    and
    \begin{equation}\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,
    \end{equation}
    so \ref{eqn:unpackingFundamentalTheorem:320} expands to
    \begin{equation}\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}.
    \end{equation}
    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
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:560}
    dA (I \grad) \cdot G
    =
    dA \gpgradezero{ I \grad G }
    =
    dA I \lr{ \grad \wedge G },
    \end{equation}
    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
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:440}
    G = g^1 e_1 + g^2 e_2,
    \end{equation}
    so
    \begin{equation}\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 },
    \end{equation}
    and
    \begin{equation}\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 }.
    \end{equation}
    So \ref{eqn:unpackingFundamentalTheorem:340}, after multiplication of both sides by \( I^{-1} \), is
    \begin{equation}\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}.
    \end{equation}

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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:580}
dA (I \grad) \wedge G
=
dA \gpgradetwo{ I \grad G }
=
dA I \lr{ \grad \cdot G },
\end{equation}
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 \( Q \).

Answer

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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:600}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }.
\end{equation}
On the RHS we have
\begin{equation}\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) }.
\end{equation}
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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:640}
\int dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }
=
\oint dx P + dy Q,
\end{equation}
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
\begin{equation}\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)},
\end{equation}
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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:940}
\int I^{-1} d^2 \Bx \rgrad G = \oint I^{-1} d\Bx G,
\end{equation}
and
\begin{equation}\label{eqn:unpackingFundamentalTheorem:960}
\int F d^2 \Bx \lgrad I^{-1} = \oint F d\Bx I^{-1}.
\end{equation}
Starting with \ref{eqn:unpackingFundamentalTheorem:940} we find
\begin{equation}\label{eqn:unpackingFundamentalTheorem:700}
\int I^{-1} J du dv I \rgrad G = \oint d\Bx G,
\end{equation}
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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:720}
I = \frac{\Bx_u \wedge \Bx_v}{J},
\end{equation}
so
\begin{equation}\label{eqn:unpackingFundamentalTheorem:740}
I^{-1} = \frac{J}{\Bx_u \wedge \Bx_v}.
\end{equation}
Also note that \( \lr{\Bx_u \wedge \Bx_v}^{-1} = \Bx^v \wedge \Bx^u \), so
\begin{equation}\label{eqn:unpackingFundamentalTheorem:760}
I^{-1} = J \lr{ \Bx^v \wedge \Bx^u },
\end{equation}
and
\begin{equation}\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 },
\end{equation}
so the right acting gradient integral is
\begin{equation}\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},
\end{equation}
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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:820}
\int J du dv F I \lgrad I^{-1} = \oint F d\Bx I^{-1}.
\end{equation}
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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:840}
\int J du dv F I \lgrad I^{-1} = \oint F d\Bx I^{-1},
\end{equation}
so
\begin{equation}\label{eqn:unpackingFundamentalTheorem:850}
-\int J du dv F I I^{-1} \lgrad = -\oint F I^{-1} d\Bx.
\end{equation}
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
\begin{equation}\label{eqn:unpackingFundamentalTheorem:860}
\lr{ I a } \cdot a
=
\gpgradezero{ I a a }
=
a^2 \gpgradezero{ I }
= 0,
\end{equation}
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
    \begin{equation}\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,
    \end{equation}
    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
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:900}
    \begin{aligned}
    \int dt dx \grad G
    &=
    \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}
    \end{equation}
    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
    \begin{equation}\label{eqn:unpackingFundamentalTheorem:920}
    \begin{aligned}
    \int dx dy \grad G
    &=
    \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}
    \end{equation}
    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.