## 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.

## New version of classical mechanics notes

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

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

## 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.

## Generalizing Ampere’s law using geometric algebra.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

# References

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

## Magnetic moment for a localized magnetostatic current

### Motivation.

I was once again reading my Jackson [2]. This time I found that his presentation of magnetic moment didn’t really make sense to me. Here’s my own pass through it, filling in a number of details. As I did last time, I’ll also translate into SI units as I go.

### Vector potential.

The Biot-Savart expression for the magnetic field can be factored into a curl expression using the usual tricks

\label{eqn:magneticMomentJackson:20}
\begin{aligned}
\BB
&= \frac{\mu_0}{4\pi} \int \frac{\BJ(\Bx’) \cross (\Bx – \Bx’)}{\Abs{\Bx – \Bx’}^3} d^3 x’ \\
&= -\frac{\mu_0}{4\pi} \int \BJ(\Bx’) \cross \spacegrad \inv{\Abs{\Bx – \Bx’}} d^3 x’ \\
&= \frac{\mu_0}{4\pi} \spacegrad \cross \int \frac{\BJ(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’,
\end{aligned}

so the vector potential, through its curl, defines the magnetic field $$\BB = \spacegrad \cross \BA$$ is given by

\label{eqn:magneticMomentJackson:40}
\BA(\Bx) = \frac{\mu_0}{4 \pi} \int \frac{J(\Bx’)}{\Abs{\Bx – \Bx’}} d^3 x’.

If the current source is localized (zero outside of some finite region), then there will always be a region for which $$\Abs{\Bx} \gg \Abs{\Bx’}$$, so the denominator yields to Taylor expansion

\label{eqn:magneticMomentJackson:60}
\begin{aligned}
\inv{\Abs{\Bx – \Bx’}}
&=
\inv{\Abs{\Bx}} \lr{1 + \frac{\Abs{\Bx’}^2}{\Abs{\Bx}^2} – 2 \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^2} }^{-1/2} \\
&\approx
\inv{\Abs{\Bx}} \lr{ 1 + \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^2} } \\
&=
\inv{\Abs{\Bx}} + \frac{\Bx \cdot \Bx’}{\Abs{\Bx}^3}.
\end{aligned}

so the vector potential, far enough away from the current source is
\label{eqn:magneticMomentJackson:80}
\BA(\Bx)
=
\frac{\mu_0}{4 \pi} \int \frac{J(\Bx’)}{\Abs{\Bx}} d^3 x’
+\frac{\mu_0}{4 \pi} \int \frac{(\Bx \cdot \Bx’)J(\Bx’)}{\Abs{\Bx}^3} d^3 x’.

Jackson uses a sneaky trick to show that the first integral is killed for a localized source. That trick appears to be based on evaluating the following divergence

\label{eqn:magneticMomentJackson:100}
\begin{aligned}
&=
+
&=
(\Be_k \partial_k x_i) \cdot\BJ \\
&=
\delta_{ki} J_k \\
&=
J_i.
\end{aligned}

Note that this made use of the fact that $$\spacegrad \cdot \BJ = 0$$ for magnetostatics. This provides a way to rewrite the current density as a divergence

\label{eqn:magneticMomentJackson:120}
\begin{aligned}
\int \frac{J(\Bx’)}{\Abs{\Bx}} d^3 x’
&=
\Be_i \int \frac{\spacegrad’ \cdot (x_i’ \BJ(\Bx’))}{\Abs{\Bx}} d^3 x’ \\
&=
\frac{\Be_i}{\Abs{\Bx}} \int \spacegrad’ \cdot (x_i’ \BJ(\Bx’)) d^3 x’ \\
&=
\frac{1}{\Abs{\Bx}} \oint \Bx’ (d\Ba \cdot \BJ(\Bx’)).
\end{aligned}

When $$\BJ$$ is localized, this is zero provided we pick the integration surface for the volume outside of that localization region.

It is now desired to rewrite $$\int \Bx \cdot \Bx’ \BJ$$ as a triple cross product since the dot product of such a triple cross product has exactly this term in it

\label{eqn:magneticMomentJackson:140}
\begin{aligned}
– \Bx \cross \int \Bx’ \cross \BJ
&=
\int (\Bx \cdot \Bx’) \BJ

\int (\Bx \cdot \BJ) \Bx’ \\
&=
\int (\Bx \cdot \Bx’) \BJ

\Be_k x_i \int J_i x_k’,
\end{aligned}

so
\label{eqn:magneticMomentJackson:160}
\int (\Bx \cdot \Bx’) \BJ
=
– \Bx \cross \int \Bx’ \cross \BJ
+
\Be_k x_i \int J_i x_k’.

To get of this second term, the next sneaky trick is to consider the following divergence

\label{eqn:magneticMomentJackson:180}
\begin{aligned}
\oint d\Ba’ \cdot (\BJ(\Bx’) x_i’ x_j’)
&=
\int dV’ \spacegrad’ \cdot (\BJ(\Bx’) x_i’ x_j’) \\
&=
+
\int dV’ \BJ \cdot \spacegrad’ (x_i’ x_j’) \\
&=
\int dV’ J_k \cdot \lr{ x_i’ \partial_k x_j’ + x_j’ \partial_k x_i’ } \\
&=
\int dV’ \lr{J_k x_i’ \delta_{kj} + J_k x_j’ \delta_{ki}} \\
&=
\int dV’ \lr{J_j x_i’ + J_i x_j’}.
\end{aligned}

The surface integral is once again zero, which means that we have an antisymmetric relationship in integrals of the form

\label{eqn:magneticMomentJackson:200}
\int J_j x_i’ = -\int J_i x_j’.

Now we can use the tensor algebra trick of writing $$y = (y + y)/2$$,

\label{eqn:magneticMomentJackson:220}
\begin{aligned}
\int (\Bx \cdot \Bx’) \BJ
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\Be_k x_i \int J_i x_k’ \\
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\inv{2} \Be_k x_i \int \lr{ J_i x_k’ + J_i x_k’ } \\
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\inv{2} \Be_k x_i \int \lr{ J_i x_k’ – J_k x_i’ } \\
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\inv{2} \Be_k x_i \int (\BJ \cross \Bx’)_j \epsilon_{ikj} \\
&=
– \Bx \cross \int \Bx’ \cross \BJ

\inv{2} \epsilon_{kij} \Be_k x_i \int (\BJ \cross \Bx’)_j \\
&=
– \Bx \cross \int \Bx’ \cross \BJ

\inv{2} \Bx \cross \int \BJ \cross \Bx’ \\
&=
– \Bx \cross \int \Bx’ \cross \BJ
+
\inv{2} \Bx \cross \int \Bx’ \cross \BJ \\
&=
-\inv{2} \Bx \cross \int \Bx’ \cross \BJ,
\end{aligned}

so

\label{eqn:magneticMomentJackson:240}
\BA(\Bx) \approx \frac{\mu_0}{4 \pi \Abs{\Bx}^3} \lr{ -\frac{\Bx}{2} } \int \Bx’ \cross \BJ(\Bx’) d^3 x’.

Letting

\label{eqn:magneticMomentJackson:260}
\boxed{
\Bm = \inv{2} \int \Bx’ \cross \BJ(\Bx’) d^3 x’,
}

the far field approximation of the vector potential is
\label{eqn:magneticMomentJackson:280}
\boxed{
\BA(\Bx) = \frac{\mu_0}{4 \pi} \frac{\Bm \cross \Bx}{\Abs{\Bx}^3}.
}

Note that when the current is restricted to an infintisimally thin loop, the magnetic moment reduces to

\label{eqn:magneticMomentJackson:300}
\Bm(\Bx) = \frac{I}{2} \int \Bx \cross d\Bl’.

Refering to [1] (pr. 1.60), this can be seen to be $$I$$ times the “vector-area” integral.

# References

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

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

## Jackson’s electrostatic self energy analysis

### Motivation

I was reading my Jackson [1], which characteristically had the statement “the […] integral can easily be shown to have the value $$4 \pi$$”, in a discussion of electrostatic energy and self energy. After a few attempts and a couple of pages of calculations, I figured out how this can be easily shown.

### Context

Let me walk through the context that leads to the “easy” integral, and then the evaluation of that integral. Unlike my older copy of Jackson, I’ll do this in SI units.

The starting point is a statement that the work done (potential energy) of one charge $$q_i$$ in a set of $$n$$ charges, where that charge is brought to its position $$\Bx_i$$ from infinity, is

\label{eqn:electrostaticJacksonSelfEnergy:20}
W_i = q_i \Phi(\Bx_i),

where the potential energy due to the rest of the charge configuration is

\label{eqn:electrostaticJacksonSelfEnergy:40}
\Phi(\Bx_i) = \inv{4 \pi \epsilon} \sum_{i \ne j} \frac{q_j}{\Abs{\Bx_i – \Bx_j}}.

This means that the total potential energy, making sure not to double count, to move all the charges in from infinity is

\label{eqn:electrostaticJacksonSelfEnergy:60}
W = \inv{4 \pi \epsilon} \sum_{1 \le i < j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}. This sum over all unique pairs is somewhat unwieldy, so it can be adjusted by explicitly double counting with a corresponding divide by two $$\label{eqn:electrostaticJacksonSelfEnergy:80} W = \inv{2} \inv{4 \pi \epsilon} \sum_{1 \le i \ne j \le n} \frac{q_i q_j}{\Abs{\Bx_i - \Bx_j}}.$$ The point that causes the trouble later is the continuum equivalent to this relationship, which is $$\label{eqn:electrostaticJacksonSelfEnergy:100} W = \inv{8 \pi \epsilon} \int \frac{\rho(\Bx) \rho(\Bx')}{\Abs{\Bx - \Bx'}} d^3 \Bx d^3 \Bx',$$ or $$\label{eqn:electrostaticJacksonSelfEnergy:120} W = \inv{2} \int \rho(\Bx) \Phi(\Bx) d^3 \Bx.$$ There's a subtlety here that is often passed over. When the charge densities represent point charges $$\rho(\Bx) = q \delta^3(\Bx - \Bx')$$ are located at, notice that this integral equivalent is evaluated over all space, including the spaces that the charges that the charges are located at. Ignoring that subtlety, this potential energy can be expressed in terms of the electric field, and then integrated by parts \label{eqn:electrostaticJacksonSelfEnergy:140} \begin{aligned} W &= \inv{2 } \int (\spacegrad \cdot (\epsilon \BE)) \Phi(\Bx) d^3 \Bx \\ &= \frac{\epsilon}{2 } \int \lr{ \spacegrad \cdot (\BE \Phi) - (\spacegrad \Phi) \cdot \BE } d^3 \Bx \\ &= \frac{\epsilon}{2 } \oint dA \ncap \cdot (\BE \Phi) + \frac{\epsilon}{2 } \int \BE \cdot \BE d^3 \Bx. \end{aligned} The presumption is that $$\BE \Phi$$ falls off as the bounds of the integration volume tends to infinity. That leaves us with an energy density proportional to the square of the field $$\label{eqn:electrostaticJacksonSelfEnergy:160} w = \frac{\epsilon}{2 } \BE^2.$$

### Inconsistency

It’s here that Jackson points out the inconsistency between \ref{eqn:electrostaticJacksonSelfEnergy:160} and the original
discrete analogue \ref{eqn:electrostaticJacksonSelfEnergy:80} that this was based on. The energy density is positive definite, whereas the discrete potential energy can be negative if there is a difference in the sign of the charges.

Here Jackson uses a two particle charge distribution to help resolve this conundrum. For a superposition $$\BE = \BE_1 + \BE_2$$, we have

\label{eqn:electrostaticJacksonSelfEnergy:180}
\BE
=
\inv{4 \pi \epsilon} \frac{q_1 (\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3}
+ \inv{4 \pi \epsilon} \frac{q_2 (\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3},

so the energy density is
\label{eqn:electrostaticJacksonSelfEnergy:200}
w =
\frac{1}{32 \pi^2 \epsilon} \frac{q_1^2}{\Abs{\Bx – \Bx_1}^4 }
+
\frac{1}{32 \pi^2 \epsilon} \frac{q_2^2}{\Abs{\Bx – \Bx_2}^4 }
+
2 \frac{q_1 q_2}{32 \pi^2 \epsilon}
\frac{(\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3} \cdot
\frac{(\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3}.

The discrete potential had only an interaction energy, whereas the potential from this squared field has an interaction energy plus two self energy terms. Those two strictly positive self energy terms are what forces this field energy positive, independent of the sign of the interaction energy density. Jackson makes a change of variables of the form

\label{eqn:electrostaticJacksonSelfEnergy:220}
\begin{aligned}
\Brho &= (\Bx – \Bx_1)/R \\
R &= \Abs{\Bx_1 – \Bx_2} \\
\ncap &= (\Bx_1 – \Bx_2)/R,
\end{aligned}

for which we find

\label{eqn:electrostaticJacksonSelfEnergy:240}
\Bx = \Bx_1 + R \Brho,

so
\label{eqn:electrostaticJacksonSelfEnergy:260}
\Bx – \Bx_2 =
\Bx_1 – \Bx_2 + R \Brho
R (\ncap + \Brho),

and
\label{eqn:electrostaticJacksonSelfEnergy:280}
d^3 \Bx = R^3 d^3 \Brho,

so the total interaction energy is
\label{eqn:electrostaticJacksonSelfEnergy:300}
\begin{aligned}
W_{\textrm{int}}
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon}
\int d^3 \Bx
\frac{(\Bx – \Bx_1)}{\Abs{\Bx – \Bx_1}^3} \cdot
\frac{(\Bx – \Bx_2)}{\Abs{\Bx – \Bx_2}^3} \\
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon}
\int R^3 d^3 \Brho
\frac{ R \Brho }{ R^3 \Abs{\Brho}^3 } \cdot
\frac{R (\ncap + \Brho)}{R^3 \Abs{\ncap + \Brho}^3} \\
&=
\frac{q_1 q_2}{16 \pi^2 \epsilon R}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}.
\end{aligned}

Evaluating this integral is what Jackson calls easy. The technique required is to express the integrand in terms of gradients in the $$\Brho$$ coordinate system

\label{eqn:electrostaticJacksonSelfEnergy:320}
\begin{aligned}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
&=
\int d^3 \Brho
\cdot
\lr{ – \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} } \\
&=
\int d^3 \Brho
\cdot
\lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} }.
\end{aligned}

I found it somewhat non-trivial to find the exact form of the chain rule that is required to simplify this integral, but after some trial and error, figured it out by working backwards from
\label{eqn:electrostaticJacksonSelfEnergy:340}
\spacegrad_\Brho^2 \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
=
+

In integral form this is
\label{eqn:electrostaticJacksonSelfEnergy:360}
\begin{aligned}
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
&=
\int d^3 \Brho’
+
\int d^3 \Brho
\spacegrad_\Brho \cdot \lr{ \inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho \inv{ \Abs{\Brho} } } \\
&=
\int d^3 \Brho’
\lr{ \spacegrad_{\Brho’} \inv{\Abs{\Brho’ – \ncap} } \cdot \spacegrad_{\Brho’} \inv{ \Abs{\Brho’} } }
+
\int d^3 \Brho’
\inv{\Abs{\Brho’ – \ncap}} \spacegrad_{\Brho’}^2 \inv{ \Abs{\Brho’} } \\
&+
\int d^3 \Brho
+
\int d^3 \Brho
\inv{\Abs{\ncap + \Brho}} \spacegrad_\Brho^2 \inv{ \Abs{\Brho} } \\
&=
2 \int d^3 \Brho
&- 4 \pi
\int d^3 \Brho’
\inv{\Abs{\Brho’ – \ncap}} \delta^3(\Brho’)
– 4 \pi
\int d^3 \Brho
\inv{\Abs{\Brho + \ncap}} \delta^3(\Brho) \\
&=
2 \int d^3 \Brho
– 8 \pi.
\end{aligned}

This used the Laplacian representation of the delta function $$\delta^3(\Bx) = -(1/4\pi) \spacegrad^2 (1/\Abs{\Bx})$$. Back-substitution gives

\label{eqn:electrostaticJacksonSelfEnergy:380}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
=
4 \pi
+
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}.

We can argue that this last integral tends to zero, since

\label{eqn:electrostaticJacksonSelfEnergy:400}
\begin{aligned}
\oint dA’ \ncap’ \cdot \spacegrad_\Brho \inv{ \Abs{\Brho} \Abs{\ncap + \Brho}}
&=
\oint dA’ \ncap’ \cdot \lr{
\lr{ \spacegrad_\Brho \inv{ \Abs{\Brho}} } \inv{\Abs{\ncap + \Brho}}
+
\inv{ \Abs{\Brho}} \lr{ \spacegrad_\Brho \inv{\Abs{\ncap + \Brho}} }
} \\
&=
-\oint dA’ \ncap’ \cdot \lr{
\frac{ \Brho } {\inv{ \Abs{\Brho}}^3 } \inv{\Abs{\ncap + \Brho}}
+
\inv{ \Abs{\Brho}} \frac{ (\Brho + \ncap) }{ \Abs{\ncap + \Brho}^3 }
} \\
&=
-\oint dA’ \inv{\Abs{\Brho} \Abs{\Brho + \ncap}}
\lr{
\frac{ \ncap’ \cdot \Brho }{
{\Abs{\Brho}}^2 }
+\frac{ \ncap’ \cdot (\Brho + \ncap) }{
{\Abs{\Brho + \ncap}}^2 }
}.
\end{aligned}

The integrand in this surface integral is of $$O(1/\rho^3)$$ so tends to zero on an infinite surface in the $$\Brho$$ coordinate system. This completes the “easy” integral, leaving

\label{eqn:electrostaticJacksonSelfEnergy:420}
\int d^3 \Brho
\frac{ \Brho }{ \Abs{\Brho}^3 } \cdot
\frac{(\ncap + \Brho)}{ \Abs{\ncap + \Brho}^3}
=
4 \pi.

The total field energy can now be expressed as a sum of the self energies and the interaction energy
\label{eqn:electrostaticJacksonSelfEnergy:440}
W =
\frac{1}{32 \pi^2 \epsilon} \int d^3 \Bx \frac{q_1^2}{\Abs{\Bx – \Bx_1}^4 }
+
\frac{1}{32 \pi^2 \epsilon} \int d^3 \Bx \frac{q_2^2}{\Abs{\Bx – \Bx_2}^4 }
+ \inv{ 4 \pi \epsilon}
\frac{q_1 q_2}{\Abs{\Bx_1 – \Bx_2} }.

The interaction energy is exactly the potential energies for the two particles, the this total energy in the field is biased in the positive direction by the pair of self energies. It is interesting that the energy obtained from integrating the field energy density contains such self energy terms, but I don’t know exactly what to make of them at this point in time.

# References

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

## Helmholtz theorem

This is a problem from ece1228. I attempted solutions in a number of ways. One using Geometric Algebra, one devoid of that algebra, and then this method, which combined aspects of both. Of the three methods I tried to obtain this result, this is the most compact and elegant. It does however, require a fair bit of Geometric Algebra knowledge, including the Fundamental Theorem of Geometric Calculus, as detailed in [1], [3] and [2].

## Question: Helmholtz theorem

Prove the first Helmholtz’s theorem, i.e. if vector $$\BM$$ is defined by its divergence

\label{eqn:helmholtzDerviationMultivector:20}

and its curl
\label{eqn:helmholtzDerviationMultivector:40}

within a region and its normal component $$\BM_{\textrm{n}}$$ over the boundary, then $$\BM$$ is
uniquely specified.

The gradient of the vector $$\BM$$ can be written as a single even grade multivector

\label{eqn:helmholtzDerviationMultivector:60}
= s + I \BC.

We will use this to attempt to discover the relation between the vector $$\BM$$ and its divergence and curl. We can express $$\BM$$ at the point of interest as a convolution with the delta function at all other points in space

\label{eqn:helmholtzDerviationMultivector:80}
\BM(\Bx) = \int_V dV’ \delta(\Bx – \Bx’) \BM(\Bx’).

The Laplacian representation of the delta function in \R{3} is

\label{eqn:helmholtzDerviationMultivector:100}
\delta(\Bx – \Bx’) = -\inv{4\pi} \spacegrad^2 \inv{\Abs{\Bx – \Bx’}},

so $$\BM$$ can be represented as the following convolution

\label{eqn:helmholtzDerviationMultivector:120}
\BM(\Bx) = -\inv{4\pi} \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’).

Using this relation and proceeding with a few applications of the chain rule, plus the fact that $$\spacegrad 1/\Abs{\Bx – \Bx’} = -\spacegrad’ 1/\Abs{\Bx – \Bx’}$$, we find

\label{eqn:helmholtzDerviationMultivector:720}
\begin{aligned}
-4 \pi \BM(\Bx)
&= \int_V dV’ \spacegrad^2 \inv{\Abs{\Bx – \Bx’}} \BM(\Bx’) \\
} } \\
&=
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
\frac{s(\Bx’) + I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}
} \\
&=
\ncap \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}
}
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
\frac{I\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}

By inserting a no-op grade selection operation in the second step, the trivector terms that would show up in subsequent steps are automatically filtered out. This leaves us with a boundary term dependent on the surface and the normal and tangential components of $$\BM$$. Added to that is a pair of volume integrals that provide the unique dependence of $$\BM$$ on its divergence and curl. When the surface is taken to infinity, which requires $$\Abs{\BM}/\Abs{\Bx – \Bx’} \rightarrow 0$$, then the dependence of $$\BM$$ on its divergence and curl is unique.

In order to express final result in traditional vector algebra form, a couple transformations are required. The first is that

\label{eqn:helmholtzDerviationMultivector:800}
\gpgradeone{ \Ba I \Bb } = I^2 \Ba \cross \Bb = -\Ba \cross \Bb.

For the grade selection in the boundary integral, note that

\label{eqn:helmholtzDerviationMultivector:740}
\begin{aligned}
&=
+
&=
+
&=

\end{aligned}

These give

\label{eqn:helmholtzDerviationMultivector:721}
\boxed{
\begin{aligned}
\BM(\Bx)
&=
\spacegrad \inv{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}}

\spacegrad \cross \inv{4\pi} \int_{\partial V} dA’ \ncap \cross \frac{\BM(\Bx’)}{\Abs{\Bx – \Bx’}} \\
\frac{s(\Bx’)}{\Abs{\Bx – \Bx’}}
\frac{\BC(\Bx’)}{\Abs{\Bx – \Bx’}}.
\end{aligned}
}

# References

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

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

[3] Garret Sobczyk and Omar Le’on S’anchez. Fundamental theorem of calculus. Advances in Applied Clifford Algebras, 21:221–231, 2011. URL https://arxiv.org/abs/0809.4526.

## Does the divergence and curl uniquely determine the vector?

A problem posed in the ece1228 problem set was the following

### Helmholtz theorem.

Prove the first Helmholtz’s theorem, i.e. if vector $$\BM$$ is defined by its divergence

\label{eqn:emtProblemSet1Problem5:20}

and its curl
\label{eqn:emtProblemSet1Problem5:40}

within a region and its normal component $$\BM_{\textrm{n}}$$ over the boundary, then $$\BM$$ is uniquely specified.

### Solution.

This problem screams for an attempt using Geometric Algebra techniques, since
the gradient of this vector can be written as a single even grade multivector

\label{eqn:emtProblemSet1Problem5AppendixGA:60}
\begin{aligned}
&= s + I \BC.
\end{aligned}

Observe that the Laplacian of $$\BM$$ is vector valued

\label{eqn:emtProblemSet1Problem5AppendixGA:400}

This means that $$\spacegrad \BC$$ must be a bivector $$\spacegrad \BC = \spacegrad \wedge \BC$$, or that $$\BC$$ has zero divergence

\label{eqn:emtProblemSet1Problem5AppendixGA:420}

This required constraint on $$\BC$$ will show up in subsequent analysis. An equivalent problem to the one posed
is to show that the even grade multivector equation $$\spacegrad \BM = s + I \BC$$ has an inverse given the constraint
specified by \ref{eqn:emtProblemSet1Problem5AppendixGA:420}.

The Green’s function for the gradient can be found in [1], where it is used to generalize the Cauchy integral equations to higher dimensions.

\label{eqn:emtProblemSet1Problem5AppendixGA:80}
\begin{aligned}
G(\Bx ; \Bx’) &= \inv{4 \pi} \frac{ \Bx – \Bx’ }{\Abs{\Bx – \Bx’}^3} \\
\end{aligned}

The inversion equation is an application of the Fundamental Theorem of (Geometric) Calculus, with the gradient operating bidirectionally on the Green’s function and the vector function

\label{eqn:emtProblemSet1Problem5AppendixGA:100}
\begin{aligned}
\oint_{\partial V} G(\Bx, \Bx’) d^2 \Bx’ \BM(\Bx’)
&=
\int_V G(\Bx, \Bx’) d^3 \Bx \lrspacegrad’ \BM(\Bx’) \\
&=
\int_V d^3 \Bx (G(\Bx, \Bx’) \lspacegrad’) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) (\spacegrad’ \BM(\Bx’)) \\
&=
-\int_V d^3 \Bx \delta(\Bx – \By) \BM(\Bx’)
+
\int_V d^3 \Bx G(\Bx, \Bx’) \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-I \BM(\Bx)
+
\inv{4 \pi} \int_V d^3 \Bx \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}

The integrals are in terms of the primed coordinates so that the end result is a function of $$\Bx$$. To rearrange for $$\BM$$, let $$d^3 \Bx’ = I dV’$$, and $$d^2 \Bx’ \ncap(\Bx’) = I dA’$$, then right multiply with the pseudoscalar $$I$$, noting that in \R{3} the pseudoscalar commutes with any grades

\label{eqn:emtProblemSet1Problem5AppendixGA:440}
\begin{aligned}
\BM(\Bx)
&=
I \oint_{\partial V} G(\Bx, \Bx’) I dA’ \ncap \BM(\Bx’)

I \inv{4 \pi} \int_V I dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) } \\
&=
-\oint_{\partial V} dA’ G(\Bx, \Bx’) \ncap \BM(\Bx’)
+
\inv{4 \pi} \int_V dV’ \frac{ \Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \lr{ s(\Bx’) + I \BC(\Bx’) }.
\end{aligned}

This can be decomposed into a vector and a trivector equation. Let $$\Br = \Bx – \Bx’ = r \rcap$$, and note that

\label{eqn:emtProblemSet1Problem5AppendixGA:500}
\begin{aligned}
&=
\gpgradeone{ I \rcap \BC } \\
&=
I \rcap \wedge \BC \\
&=
-\rcap \cross \BC,
\end{aligned}

so this pair of equations can be written as

\label{eqn:emtProblemSet1Problem5AppendixGA:520}
\begin{aligned}
\BM(\Bx)
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\gpgradeone{ \rcap \ncap \BM(\Bx’) }}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) } \\
0
&=
-\inv{4 \pi} \oint_{\partial V} dA’ \frac{\rcap}{r^2} \wedge \ncap \wedge \BM(\Bx’)
+
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}.
\end{aligned}

Consider the last integral in the pseudoscalar equation above. Since we expect no pseudoscalar components, this must be zero, or cancel perfectly. It’s not obvious that this is the case, but a transformation to a surface integral shows the constraints required for that to be the case. To do so note

\label{eqn:emtProblemSet1Problem5AppendixGA:540}
\begin{aligned}
&= -\spacegrad’ \inv{\Bx – \Bx’} \\
&=
-\frac{\Bx – \Bx’}{\Abs{\Bx – \Bx’}^3} \\
&= -\frac{\rcap}{r^2}.
\end{aligned}

Using this and the chain rule we have

\label{eqn:emtProblemSet1Problem5AppendixGA:560}
\begin{aligned}
\frac{I}{4 \pi} \int_V dV’ \frac{ \rcap \cdot \BC(\Bx’) }{r^2}
&=
\frac{I}{4 \pi} \int_V dV’ \lr{ \spacegrad’ \inv{ r } } \cdot \BC(\Bx’) \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r}

\frac{I}{4 \pi} \int_V dV’ \frac{ \spacegrad’ \cdot \BC(\Bx’) }{r} \\
&=
\frac{I}{4 \pi} \int_V dV’ \spacegrad’ \cdot \frac{\BC(\Bx’)}{r} \\
&=
\frac{I}{4 \pi} \int_{\partial V} dA’ \ncap(\Bx’) \cdot \frac{\BC(\Bx’)}{r}.
\end{aligned}

The divergence of $$\BC$$ above was killed by recalling the constraint \ref{eqn:emtProblemSet1Problem5AppendixGA:420}. This means that we can rewrite entirely as surface integral and eventually reduced to a single triple product

\label{eqn:emtProblemSet1Problem5AppendixGA:580}
\begin{aligned}
0
&=
-\frac{I}{4 \pi} \oint_{\partial V} dA’ \lr{
\frac{\rcap}{r^2} \cdot (\ncap \cross \BM(\Bx’))
-\ncap \cdot \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
\frac{\rcap}{r^2} \cross \BM(\Bx’)
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
+ \frac{\BC(\Bx’)}{r}
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’ \ncap \cdot \lr{
} \\
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\frac{\BM(\Bx’) \cross \ncap}{r}
&=
\frac{I}{4 \pi} \oint_{\partial V} dA’
\frac{\BM(\Bx’) \cross \ncap}{r}.
\end{aligned}

### Final results.

Assembling things back into a single multivector equation, the complete inversion integral for $$\BM$$ is

\label{eqn:emtProblemSet1Problem5AppendixGA:600}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\partial V} dA’
\lr{
\frac{\BM(\Bx’) \wedge \ncap}{r}
}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) –
\frac{\rcap}{r^2} \cross \BC(\Bx’) }.

This shows that vector $$\BM$$ can be recovered uniquely from $$s, \BC$$ when $$\Abs{\BM}/r^2$$ vanishes on an infinite surface. If we restrict attention to a finite surface, we have to add to the fixed solution a specific solution that depends on the value of $$\BM$$ on that surface. The vector portion of that surface integrand contains

\label{eqn:emtProblemSet1Problem5AppendixGA:640}
\begin{aligned}
&=
\rcap (\ncap \cdot \BM )
+
\rcap \cdot (\ncap \wedge \BM ) \\
&=
\rcap (\ncap \cdot \BM )
+
(\rcap \cdot \ncap) \BM

(\rcap \cdot \BM ) \ncap.
\end{aligned}

The constraints required by a zero triple product $$\spacegrad’ \cdot (\BM(\Bx’) \cross \ncap(\Bx’))$$ are complicated on a such a general finite surface. Consider instead, for simplicity, the case of a spherical surface, which can be analyzed more easily. In that case the outward normal of the surface centred on the test charge point $$\Bx$$ is $$\ncap = -\rcap$$. The pseudoscalar integrand is not generally killed unless the divergence of its tangential component on this surface is zero. One way that this can occur is for $$\BM \cross \ncap = 0$$, so that $$-\gpgradeone{ \rcap \ncap \BM } = \BM = (\BM \cdot \ncap) \ncap = \BM_{\textrm{n}}$$.

This gives

\label{eqn:emtProblemSet1Problem5AppendixGA:620}
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
+
\inv{4 \pi} \int_V dV’ \lr{
\frac{\rcap}{r^2} s(\Bx’) +
\BC(\Bx’) \cross \frac{\rcap}{r^2} },

or, in terms of potential functions, which is arguably tidier

\label{eqn:emtProblemSet1Problem5AppendixGA:300}
\boxed{
\BM(\Bx)
=
\inv{4 \pi} \oint_{\Abs{\Bx – \Bx’} = r} dA’ \frac{\BM_{\textrm{n}}(\Bx’)}{r^2}
-\spacegrad \int_V dV’ \frac{ s(\Bx’)}{ 4 \pi r }
+\spacegrad \cross \int_V dV’ \frac{ \BC(\Bx’) }{ 4 \pi r }.
}

### Commentary

I attempted this problem in three different ways. My first approach (above) assembled the divergence and curl relations above into a single (Geometric Algebra) multivector gradient equation and applied the vector valued Green’s function for the gradient to invert that equation. That approach logically led from the differential equation for $$\BM$$ to the solution for $$\BM$$ in terms of $$s$$ and $$\BC$$. However, this strategy introduced some complexities that make me doubt the correctness of the associated boundary analysis.

Even if the details of the boundary handling in my multivector approach is not correct, I thought that approach was interesting enough to share.

# References

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

## Motivation

fig 1. Two surfaces normal to the interface.

Most electrodynamics textbooks either start with or contain a treatment of boundary value conditions. These typically involve evaluating Maxwell’s equations over areas or volumes of decreasing height, such as those illustrated in fig. 1, and fig. 2. These represent surfaces and volumes where the height is allowed to decrease to infinitesimal levels, and are traditionally used to find the boundary value constraints of the normal and tangential components of the electric and magnetic fields.

fig 2. A pillbox volume encompassing the interface.

More advanced topics, such as evaluation of the Fresnel reflection and transmission equations, also rely on similar consideration of boundary value constraints. I’ve wondered for a long time how the Fresnel equations could be attacked by looking at the boundary conditions for the combined field $$F = \BE + I c \BB$$, instead of the considering them separately.

## A unified approach.

The Geometric Algebra (and relativistic tensor) formulations of Maxwell’s equations put the electric and magnetic fields on equal footings. It is in fact possible to specify the boundary value constraints on the fields without first separating Maxwell’s equations into their traditional forms. The starting point in Geometric Algebra is Maxwell’s equation, premultiplied by a stationary observer’s timelike basis vector

\label{eqn:maxwellBoundaryConditions:20}
\gamma_0 \grad F = \inv{\epsilon_0 c} \gamma_0 J,

or

\label{eqn:maxwellBoundaryConditions:40}
\lr{ \partial_0 + \spacegrad} F = \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0}.

The electrodynamic field $$F = \BE + I c \BB$$ is a multivector in this spatial domain (whereas it is a bivector in the spacetime algebra domain), and has vector and bivector components. The product of the spatial gradient and the field can still be split into dot and curl components $$\spacegrad M = \spacegrad \cdot M + \spacegrad \wedge M$$. If $$M = \sum M_i$$, where $$M_i$$ is an grade $$i$$ blade, then we give this the Hestenes’ [1] definitions

\label{eqn:maxwellBoundaryConditions:60}
\begin{aligned}
\end{aligned}

With that said, Maxwell’s equation can be rearranged into a pair of multivector equations

\label{eqn:maxwellBoundaryConditions:80}
\begin{aligned}
\spacegrad \cdot F &= \gpgrade{-\partial_0 F + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c}}{0,1} \\
\end{aligned}

The latter equation can be integrated with Stokes theorem, but we need to apply a duality transformation to the latter in order to apply Stokes to it

\label{eqn:maxwellBoundaryConditions:120}
\begin{aligned}
&=
&=
&=
&=
\end{aligned}

so

\label{eqn:maxwellBoundaryConditions:100}
\begin{aligned}
\spacegrad \wedge (I F) &= I \lr{ -\inv{c} \partial_t \BE + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c} } \\
\spacegrad \wedge F &= -I \partial_t \BB.
\end{aligned}

Integrating each of these over the pillbox volume gives

\label{eqn:maxwellBoundaryConditions:140}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F)
&=
\int_{V} d^3 \Bx \cdot \lr{ I \lr{ -\inv{c} \partial_t \BE + \frac{\rho}{\epsilon_0} – \frac{\BJ}{\epsilon_0 c} } } \\
\oint_{\partial V} d^2 \Bx \cdot F
&=
– \partial_t \int_{V} d^3 \Bx \cdot \lr{ I \BB }.
\end{aligned}

In the absence of charges and currents on the surface, and if the height of the volume is reduced to zero, the volume integrals vanish, and only the upper surfaces of the pillbox contribute to the surface integrals.

\label{eqn:maxwellBoundaryConditions:200}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F) &= 0 \\
\oint_{\partial V} d^2 \Bx \cdot F &= 0.
\end{aligned}

With a multivector $$F$$ in the mix, the geometric meaning of these integrals is not terribly clear. They do describe the boundary conditions, but to see exactly what those are, we can now resort to the split of $$F$$ into its electric and magnetic fields. Let’s look at the non-dual integral to start with

\label{eqn:maxwellBoundaryConditions:160}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot F
&=
\oint_{\partial V} d^2 \Bx \cdot \lr{ \BE + I c \BB } \\
&=
\oint_{\partial V} d^2 \Bx \cdot \BE + I c d^2 \Bx \wedge \BB \\
&=
0.
\end{aligned}

No component of $$\BE$$ that is normal to the surface contributes to $$d^2 \Bx \cdot \BE$$, whereas only components of $$\BB$$ that are normal contribute to $$d^2 \Bx \wedge \BB$$. That means that we must have tangential components of $$\BE$$ and the normal components of $$\BB$$ matching on the surfaces

\label{eqn:maxwellBoundaryConditions:180}
\begin{aligned}
\lr{\BE_2 \wedge \ncap} \ncap – \lr{\BE_1 \wedge (-\ncap)} (-\ncap) &= 0 \\
\lr{\BB_2 \cdot \ncap} \ncap – \lr{\BB_1 \cdot (-\ncap)} (-\ncap) &= 0 .
\end{aligned}

Similarly, for the dot product of the dual field, this is

\label{eqn:maxwellBoundaryConditions:220}
\begin{aligned}
\oint_{\partial V} d^2 \Bx \cdot (I F)
&=
\oint_{\partial V} d^2 \Bx \cdot (I \BE – c \BB) \\
&=
\oint_{\partial V} I d^2 \Bx \wedge \BE – c d^2 \Bx \cdot \BB.
\end{aligned}

For this integral, only the normal components of $$\BE$$ contribute, and only the tangential components of $$\BB$$ contribute. This means that

\label{eqn:maxwellBoundaryConditions:240}
\begin{aligned}
\lr{\BE_2 \cdot \ncap} \ncap – \lr{\BE_1 \cdot (-\ncap)} (-\ncap) &= 0 \\
\lr{\BB_2 \wedge \ncap} \ncap – \lr{\BB_1 \wedge (-\ncap)} (-\ncap) &= 0.
\end{aligned}

This is why we end up with a seemingly strange mix of tangential and normal components of the electric and magnetic fields. These constraints can be summarized as

\label{eqn:maxwellBoundaryConditions:260}
\begin{aligned}
( \BE_2 – \BE_1 ) \cdot \ncap &= 0 \\
( \BE_2 – \BE_1 ) \wedge \ncap &= 0 \\
( \BB_2 – \BB_1 ) \cdot \ncap &= 0 \\
( \BB_2 – \BB_1 ) \wedge \ncap &= 0
\end{aligned}

These relationships are usually expressed in terms of all of $$\BE, \BD, \BB$$ and $$\BH$$. Because I’d started with Maxwell’s equations for free space, I don’t have the $$\epsilon$$ and $$\mu$$ factors that produce those more general relationships. Those more general boundary value relationships are usually the starting point for the Fresnel interface analysis. It is also possible to further generalize these relationships to include charges and currents on the surface.

# References

[1] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.