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# Posts Tagged ‘bivector’

## Stokes theorem in Geometric algebra

Posted by peeterjoot on May 17, 2014

[Click here for a PDF of this post with nicer formatting  (especially since my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Understanding how to apply Stokes theorem to higher dimensional spaces, non-Euclidean metrics, and with curvilinear coordinates has been a long standing goal.

A traditional answer to these questions can be found in the formalism of differential forms, as covered for example in [2], and [8]. However, both of those texts, despite their small size, are intensely scary. I also found it counter intuitive to have to express all physical quantities as forms, since there are many times when we don’t have any pressing desire to integrate these.

Later I encountered Denker’s straight wire treatment [1], which states that the geometric algebra formulation of Stokes theorem has the form

\begin{aligned}\int_S \nabla \wedge F = \int_{\partial S} F\end{aligned} \hspace{\stretch{1}}(1.0.1)

This is simple enough looking, but there are some important details left out. In particular the grades do not match, so there must be some sort of implied projection or dot product operations too. We also need to understand how to express the hypervolume and hypersurfaces when evaluating these integrals, especially when we want to use curvilinear coordinates.

I’d attempted to puzzle through these details previously. A collection of these attempts, to be removed from my collection of geometric algebra notes, can be found in [4]. I’d recently reviewed all of these and wrote a compact synopsis [5] of all those notes, but in the process of doing so, I realized there was a couple of fundamental problems with the approach I had used.

One detail that was that I failed to understand, was that we have a requirement for treating a infinitesimal region in the proof, then summing over such regions to express the boundary integral. Understanding that the boundary integral form and its dot product are both evaluated only at the end points of the integral region is an important detail that follows from such an argument (as used in proof of Stokes theorem for a 3D Cartesian space in [7].)

I also realized that my previous attempts could only work for the special cases where the dimension of the integration volume also equaled the dimension of the vector space. The key to resolving this issue is the concept of the tangent space, and an understanding of how to express the projection of the gradient onto the tangent space. These concepts are covered thoroughly in [6], which also introduces Stokes theorem as a special case of a more fundamental theorem for integration of geometric algebraic objects. My objective, for now, is still just to understand the generalization of Stokes theorem, and will leave the fundamental theorem of geometric calculus to later.

Now that these details are understood, the purpose of these notes is to detail the Geometric algebra form of Stokes theorem, covering its generalization to higher dimensional spaces and non-Euclidean metrics (i.e. especially those used for special relativity and electromagnetism), and understanding how to properly deal with curvilinear coordinates. This generalization has the form

## Theorem 1. Stokes’ Theorem

For blades $F \in \bigwedge^{s}$, and $m$ volume element $d^k \mathbf{x}, s < k$,

\begin{aligned}\int_V d^k \mathbf{x} \cdot (\boldsymbol{\partial} \wedge F) = \int_{\partial V} d^{k-1} \mathbf{x} \cdot F.\end{aligned}

Here the volume integral is over a $m$ dimensional surface (manifold), $\boldsymbol{\partial}$ is the projection of the gradient onto the tangent space of the manifold, and $\partial V$ indicates integration over the boundary of $V$.

It takes some work to give this more concrete meaning. I will attempt to do so in a gradual fashion, and provide a number of examples that illustrate some of the relevant details.

# Basic notation

A finite vector space, not necessarily Euclidean, with basis $\left\{ {\mathbf{e}_1, \mathbf{e}_2, \cdots} \right\}$ will be assumed to be the generator of the geometric algebra. A dual or reciprocal basis $\left\{ {\mathbf{e}^1, \mathbf{e}^2, \cdots} \right\}$ for this basis can be calculated, defined by the property

\begin{aligned}\mathbf{e}_i \cdot \mathbf{e}^j = {\delta_i}^j.\end{aligned} \hspace{\stretch{1}}(1.1.2)

This is an Euclidean space when $\mathbf{e}_i = \mathbf{e}^i, \forall i$.

To select from a multivector $A$ the grade $k$ portion, say $A_k$ we write

\begin{aligned}A_k = {\left\langle A \right\rangle}_{k}.\end{aligned} \hspace{\stretch{1}}(1.1.3)

The scalar portion of a multivector $A$ will be written as

\begin{aligned}{\left\langle A \right\rangle}_{0} \equiv \left\langle A \right\rangle.\end{aligned} \hspace{\stretch{1}}(1.1.4)

The grade selection operators can be used to define the outer and inner products. For blades $U$, and $V$ of grade $r$ and $s$ respectively, these are

\begin{aligned}{\left\langle U V \right\rangle}_{{\left\lvert {r + s} \right\rvert}} \equiv U \wedge V\end{aligned} \hspace{\stretch{1}}(1.0.5.5)

\begin{aligned}{\left\langle U V \right\rangle}_{{\left\lvert {r - s} \right\rvert}} \equiv U \cdot V.\end{aligned} \hspace{\stretch{1}}(1.0.5.5)

Written out explicitly for odd grade blades $A$ (vector, trivector, …), and vector $\mathbf{a}$ the dot and wedge products are respectively

\begin{aligned}\begin{aligned}\mathbf{a} \wedge A &= \frac{1}{2} (\mathbf{a} A - A \mathbf{a}) \\ \mathbf{a} \cdot A &= \frac{1}{2} (\mathbf{a} A + A \mathbf{a}).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.6)

\begin{aligned}\begin{aligned}\mathbf{a} \wedge A &= \frac{1}{2} (\mathbf{a} A + A \mathbf{a}) \\ \mathbf{a} \cdot A &= \frac{1}{2} (\mathbf{a} A - A \mathbf{a}).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.7)

It will be useful to employ the cyclic scalar reordering identity for the scalar selection operator

\begin{aligned}\left\langle{{\mathbf{a} \mathbf{b} \mathbf{c}}}\right\rangle= \left\langle{{\mathbf{b} \mathbf{c} \mathbf{a}}}\right\rangle= \left\langle{{\mathbf{c} \mathbf{a} \mathbf{b}}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.0.8)

For an $N$ dimensional vector space, a product of $N$ orthonormal (up to a sign) unit vectors is referred to as a pseudoscalar for the space, typically denoted by $I$

\begin{aligned}I = \mathbf{e}_1 \mathbf{e}_2 \cdots \mathbf{e}_N.\end{aligned} \hspace{\stretch{1}}(1.0.9)

The pseudoscalar may commute or anticommute with other blades in the space. We may also form a pseudoscalar for a subspace spanned by vectors $\left\{ {\mathbf{a}, \mathbf{b}, \cdots, \mathbf{c}} \right\}$ by unit scaling the wedge products of those vectors $\mathbf{a} \wedge \mathbf{b} \wedge \cdots \wedge \mathbf{c}$.

# Curvilinear coordinates

For our purposes a manifold can be loosely defined as a parameterized surface. For example, a 2D manifold can be considered a surface in an $n$ dimensional vector space, parameterized by two variables

\begin{aligned}\mathbf{x} = \mathbf{x}(a,b) = \mathbf{x}(u^1, u^2).\end{aligned} \hspace{\stretch{1}}(1.0.10)

Note that the indices here do not represent exponentiation. We can construct a basis for the manifold as

\begin{aligned}\mathbf{x}_i = \frac{\partial {\mathbf{x}}}{\partial {u^i}}.\end{aligned} \hspace{\stretch{1}}(1.0.11)

On the manifold we can calculate a reciprocal basis $\left\{ {\mathbf{x}^i} \right\}$, defined by requiring, at each point on the surface

\begin{aligned}\mathbf{x}^i \cdot \mathbf{x}_j = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(1.0.12)

Associated implicitly with this basis is a curvilinear coordinate representation defined by the projection operation

\begin{aligned}\mathbf{x} = x^i \mathbf{x}_i,\end{aligned} \hspace{\stretch{1}}(1.0.13)

(sums over mixed indices are implied). These coordinates can be calculated by taking dot products with the reciprocal frame vectors

\begin{aligned}\mathbf{x} \cdot \mathbf{x}^i &= x^j \mathbf{x}_j \cdot \mathbf{x}^i \\ &= x^j {\delta_j}^i \\ &= x^i.\end{aligned} \hspace{\stretch{1}}(1.0.13)

In this document all coordinates are with respect to a specific curvilinear basis, and not with respect to the standard basis $\left\{ {\mathbf{e}_i} \right\}$ or its dual basis unless otherwise noted.

Similar to the usual notation for derivatives with respect to the standard basis coordinates we form a lower index partial derivative operator

\begin{aligned}\frac{\partial {}}{\partial {u^i}} \equiv \partial_i,\end{aligned} \hspace{\stretch{1}}(1.0.13)

so that when the complete vector space is spanned by $\left\{ {\mathbf{x}_i} \right\}$ the gradient has the curvilinear representation

\begin{aligned}\boldsymbol{\nabla} = \mathbf{x}^i \frac{\partial {}}{\partial {u^i}}.\end{aligned} \hspace{\stretch{1}}(1.0.13)

This can be motivated by noting that the directional derivative is defined by

\begin{aligned}\mathbf{a} \cdot \boldsymbol{\nabla} f(\mathbf{x}) = \lim_{t \rightarrow 0} \frac{f(\mathbf{x} + t \mathbf{a}) - f(\mathbf{x})}{t}.\end{aligned} \hspace{\stretch{1}}(1.0.17)

When the basis $\left\{ {\mathbf{x}_i} \right\}$ does not span the space, the projection of the gradient onto the tangent space at the point of evaluation

\begin{aligned}\boldsymbol{\partial} = \mathbf{x}^i \partial_i = \sum_i \mathbf{x}_i \frac{\partial {}}{\partial {u^i}}.\end{aligned} \hspace{\stretch{1}}(1.0.18)

This is called the vector derivative.

See [6] for a more complete discussion of the gradient and vector derivatives in curvilinear coordinates.

# Green’s theorem

Given a two parameter ($u,v$) surface parameterization, the curvilinear coordinate representation of a vector $\mathbf{f}$ has the form

\begin{aligned}\mathbf{f} = f_u \mathbf{x}^u + f_v \mathbf{x}^v + f_\perp \mathbf{x}^\perp.\end{aligned} \hspace{\stretch{1}}(1.19)

We assume that the vector space is of dimension two or greater but otherwise unrestricted, and need not have an Euclidean basis. Here $f_\perp \mathbf{x}^\perp$ denotes the rejection of $\mathbf{f}$ from the tangent space at the point of evaluation. Green’s theorem relates the integral around a closed curve to an “area” integral on that surface

## Theorem 2. Green’s Theorem

\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} \mathbf{f} \cdot d\mathbf{l}=\iint \left( {-\frac{\partial {f_u}}{\partial {v}}+\frac{\partial {f_v}}{\partial {u}}} \right)du dv\end{aligned}

Following the arguments used in [7] for Stokes theorem in three dimensions, we first evaluate the loop integral along the differential element of the surface at the point $\mathbf{x}(u_0, v_0)$ evaluated over the range $(du, dv)$, as shown in the infinitesimal loop of fig. 1.1.

Fig 1.1. Infinitesimal loop integral

Over the infinitesimal area, the loop integral decomposes into

\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} \mathbf{f} \cdot d\mathbf{l}=\int \mathbf{f} \cdot d\mathbf{x}_1+\int \mathbf{f} \cdot d\mathbf{x}_2+\int \mathbf{f} \cdot d\mathbf{x}_3+\int \mathbf{f} \cdot d\mathbf{x}_4,\end{aligned} \hspace{\stretch{1}}(1.20)

where the differentials along the curve are

\begin{aligned}\begin{aligned}d\mathbf{x}_1 &= {\left.{{ \frac{\partial {\mathbf{x}}}{\partial {u}} }}\right\vert}_{{v = v_0}} du \\ d\mathbf{x}_2 &= {\left.{{ \frac{\partial {\mathbf{x}}}{\partial {v}} }}\right\vert}_{{u = u_0 + du}} dv \\ d\mathbf{x}_3 &= -{\left.{{ \frac{\partial {\mathbf{x}}}{\partial {u}} }}\right\vert}_{{v = v_0 + dv}} du \\ d\mathbf{x}_4 &= -{\left.{{ \frac{\partial {\mathbf{x}}}{\partial {v}} }}\right\vert}_{{u = u_0}} dv.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.21)

It is assumed that the parameterization change $(du, dv)$ is small enough that this loop integral can be considered planar (regardless of the dimension of the vector space). Making use of the fact that $\mathbf{x}^\perp \cdot \mathbf{x}_\alpha = 0$ for $\alpha \in \left\{ {u,v} \right\}$, the loop integral is

\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} \mathbf{f} \cdot d\mathbf{l}=\int\left( {f_u \mathbf{x}^u + f_v \mathbf{x}^v + f_\perp \mathbf{x}^\perp} \right)\cdot\Bigl(\mathbf{x}_u(u, v_0) du - \mathbf{x}_u(u, v_0 + dv) du+\mathbf{x}_v(u_0 + du, v) dv - \mathbf{x}_v(u_0, v) dv\Bigr)=\int f_u(u, v_0) du - f_u(u, v_0 + dv) du+f_v(u_0 + du, v) dv - f_v(u_0, v) dv\end{aligned} \hspace{\stretch{1}}(1.22)

With the distances being infinitesimal, these differences can be rewritten as partial differentials

\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} \mathbf{f} \cdot d\mathbf{l}=\iint \left( {-\frac{\partial {f_u}}{\partial {v}}+\frac{\partial {f_v}}{\partial {u}}} \right)du dv.\end{aligned} \hspace{\stretch{1}}(1.23)

We can now sum over a larger area as in fig. 1.2

Fig 1.2. Sum of infinitesimal loops

All the opposing oriented loop elements cancel, so the integral around the complete boundary of the surface $\mathbf{x}(u, v)$ is given by the $u,v$ area integral of the partials difference.

We will see that Green’s theorem is a special case of the Curl (Stokes) theorem. This observation will also provide a geometric interpretation of the right hand side area integral of thm. 2, and allow for a coordinate free representation.

Special case:

An important special case of Green’s theorem is for a Euclidean two dimensional space where the vector function is

\begin{aligned}\mathbf{f} = P \mathbf{e}_1 + Q \mathbf{e}_2.\end{aligned} \hspace{\stretch{1}}(1.24)

Here Green’s theorem takes the form

\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} P dx + Q dy=\iint \left( {\frac{\partial {Q}}{\partial {x}}-\frac{\partial {P}}{\partial {y}}} \right)dx dy.\end{aligned} \hspace{\stretch{1}}(1.0.25)

# Curl theorem, two volume vector field

Having examined the right hand side of thm. 1 for the very simplest geometric object $\mathbf{f}$, let’s look at the right hand side, the area integral in more detail. We restrict our attention for now to vectors $\mathbf{f}$ still defined by eq. 1.19.

First we need to assign a meaning to $d^2 \mathbf{x}$. By this, we mean the wedge products of the two differential elements. With

\begin{aligned}d\mathbf{x}_i = du^i \frac{\partial {\mathbf{x}}}{\partial {u^i}} = du^i \mathbf{x}_i,\end{aligned} \hspace{\stretch{1}}(1.26)

that area element is

\begin{aligned}d^2 \mathbf{x}= d\mathbf{x}_1 \wedge d\mathbf{x}_2= du^1 du^2 \mathbf{x}_1 \wedge \mathbf{x}_2.\end{aligned} \hspace{\stretch{1}}(1.0.27)

This is the oriented area element that lies in the tangent plane at the point of evaluation, and has the magnitude of the area of that segment of the surface, as depicted in fig. 1.3.

Fig 1.3. Oriented area element tiling of a surface

Observe that we have no requirement to introduce a normal to the surface to describe the direction of the plane. The wedge product provides the information about the orientation of the place in the space, even when the vector space that our vector lies in has dimension greater than three.

Proceeding with the expansion of the dot product of the area element with the curl, using eq. 1.0.6, eq. 1.0.7, and eq. 1.0.8, and a scalar selection operation, we have

\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \left\langle{{d^2 \mathbf{x} \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)}}\right\rangle \\ &= \left\langle{{d^2 \mathbf{x}\frac{1}{2}\left( { \stackrel{ \rightarrow }{\boldsymbol{\partial}} \mathbf{f} - \mathbf{f} \stackrel{ \leftarrow }{\boldsymbol{\partial}} } \right)}}\right\rangle \\ &= \frac{1}{2}\left\langle{{d^2 \mathbf{x} \left( { \mathbf{x}^i \left( { \partial_i \mathbf{f}} \right) - \left( {\partial_i \mathbf{f}} \right) \mathbf{x}^i } \right)}}\right\rangle \\ &= \frac{1}{2}\left\langle{{\left( { \partial_i \mathbf{f} } \right) d^2 \mathbf{x} \mathbf{x}^i - \left( { \partial_i \mathbf{f} } \right) \mathbf{x}^i d^2 \mathbf{x}}}\right\rangle \\ &= \left\langle{{\left( { \partial_i \mathbf{f} } \right) \left( { d^2 \mathbf{x} \cdot \mathbf{x}^i } \right)}}\right\rangle \\ &= \partial_i \mathbf{f} \cdot\left( { d^2 \mathbf{x} \cdot \mathbf{x}^i } \right).\end{aligned} \hspace{\stretch{1}}(1.28)

Let’s proceed to expand the inner dot product

\begin{aligned}d^2 \mathbf{x} \cdot \mathbf{x}^i &= du^1 du^2\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \mathbf{x}^i \\ &= du^1 du^2\left( {\mathbf{x}_2 \cdot \mathbf{x}^i \mathbf{x}_1-\mathbf{x}_1 \cdot \mathbf{x}^i \mathbf{x}_2} \right) \\ &= du^1 du^2\left( {{\delta_2}^i \mathbf{x}_1-{\delta_1}^i \mathbf{x}_2} \right).\end{aligned} \hspace{\stretch{1}}(1.29)

The complete curl term is thus

\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=du^1 du^2\left( {\frac{\partial {\mathbf{f}}}{\partial {u^2}} \cdot \mathbf{x}_1-\frac{\partial {\mathbf{f}}}{\partial {u^1}} \cdot \mathbf{x}_2} \right)\end{aligned} \hspace{\stretch{1}}(1.30)

This almost has the form of eq. 1.23, although that is not immediately obvious. Working backwards, using the shorthand $u = u^1, v = u^2$, we can show that this coordinate representation can be eliminated

\begin{aligned}-du dv\left( {\frac{\partial {f_v}}{\partial {u}} -\frac{\partial {f_u}}{\partial {v}}} \right) &= du dv\left( {\frac{\partial {}}{\partial {v}}\left( {\mathbf{f} \cdot \mathbf{x}_u} \right)-\frac{\partial {}}{\partial {u}}\left( {\mathbf{f} \cdot \mathbf{x}_v} \right)} \right) \\ &= du dv\left( {\frac{\partial {\mathbf{f}}}{\partial {v}} \cdot \mathbf{x}_u-\frac{\partial {\mathbf{f}}}{\partial {u}} \cdot \mathbf{x}_v+\mathbf{f} \cdot \left( {\frac{\partial {\mathbf{x}_u}}{\partial {v}}-\frac{\partial {\mathbf{x}_v}}{\partial {u}}} \right)} \right) \\ &= du dv \left( {\frac{\partial {\mathbf{f}}}{\partial {v}} \cdot \mathbf{x}_u-\frac{\partial {\mathbf{f}}}{\partial {u}} \cdot \mathbf{x}_v+\mathbf{f} \cdot \left( {\frac{\partial^2 \mathbf{x}}{\partial v \partial u}-\frac{\partial^2 \mathbf{x}}{\partial u \partial v}} \right)} \right) \\ &= du dv \left( {\frac{\partial {\mathbf{f}}}{\partial {v}} \cdot \mathbf{x}_u-\frac{\partial {\mathbf{f}}}{\partial {u}} \cdot \mathbf{x}_v} \right) \\ &= d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right).\end{aligned} \hspace{\stretch{1}}(1.31)

This relates the two parameter surface integral of the curl to the loop integral over its boundary

\begin{aligned}\int d^2 \mathbf{x} \cdot (\boldsymbol{\partial} \wedge \mathbf{f}) = \mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowleft}}\int} \mathbf{f} \cdot d\mathbf{l}.\end{aligned} \hspace{\stretch{1}}(1.0.32)

This is the very simplest special case of Stokes theorem. When written in the general form of Stokes thm. 1

\begin{aligned}\int_A d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f}} \right)=\int_{\partial A} d^1 \mathbf{x} \cdot \mathbf{f}=\int_{\partial A} \left( { d\mathbf{x}_1 - d\mathbf{x}_2 } \right) \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.0.33)

we must remember (the $\partial A$ is to remind us of this) that it is implied that both the vector $\mathbf{f}$ and the differential elements are evaluated on the boundaries of the integration ranges respectively. A more exact statement is

\begin{aligned}\int_{\partial A} d^1 \mathbf{x} \cdot \mathbf{f}=\int {\left.{{\mathbf{f} \cdot d\mathbf{x}_1}}\right\vert}_{{\Delta u^2}}-{\left.{{\mathbf{f} \cdot d\mathbf{x}_2}}\right\vert}_{{\Delta u^1}}=\int {\left.{{f_1}}\right\vert}_{{\Delta u^2}} du^1-{\left.{{f_2}}\right\vert}_{{\Delta u^1}} du^2.\end{aligned} \hspace{\stretch{1}}(1.0.34)

Expanded out in full this is

\begin{aligned}\int {\left.{{\mathbf{f} \cdot d\mathbf{x}_1}}\right\vert}_{{u^2(1)}}-{\left.{{\mathbf{f} \cdot d\mathbf{x}_1}}\right\vert}_{{u^2(0)}}+{\left.{{\mathbf{f} \cdot d\mathbf{x}_2}}\right\vert}_{{u^1(0)}}-{\left.{{\mathbf{f} \cdot d\mathbf{x}_2}}\right\vert}_{{u^1(1)}},\end{aligned} \hspace{\stretch{1}}(1.0.35)

which can be cross checked against fig. 1.4 to demonstrate that this specifies a clockwise orientation. For the surface with oriented area $d\mathbf{x}_1 \wedge d\mathbf{x}_2$, the clockwise loop is designated with line elements (1)-(4), we see that the contributions around this loop (in boxes) match eq. 1.0.35.

Fig 1.4. Clockwise loop

## Example: Green’s theorem, a 2D Cartesian parameterization for a Euclidean space

For a Cartesian 2D Euclidean parameterization of a vector field and the integration space, Stokes theorem should be equivalent to Green’s theorem eq. 1.0.25. Let’s expand both sides of eq. 1.0.32 independently to verify equality. The parameterization is

\begin{aligned}\mathbf{x}(x, y) = x \mathbf{e}_1 + y \mathbf{e}_2.\end{aligned} \hspace{\stretch{1}}(1.36)

Here the dual basis is the basis, and the projection onto the tangent space is just the gradient

\begin{aligned}\boldsymbol{\partial} = \boldsymbol{\nabla}= \mathbf{e}_1 \frac{\partial {}}{\partial {x}}+ \mathbf{e}_2 \frac{\partial {}}{\partial {y}}.\end{aligned} \hspace{\stretch{1}}(1.0.37)

The volume element is an area weighted pseudoscalar for the space

\begin{aligned}d^2 \mathbf{x} = dx dy \frac{\partial {\mathbf{x}}}{\partial {x}} \wedge \frac{\partial {\mathbf{x}}}{\partial {y}} = dx dy \mathbf{e}_1 \mathbf{e}_2,\end{aligned} \hspace{\stretch{1}}(1.0.38)

and the curl of a vector $\mathbf{f} = f_1 \mathbf{e}_1 + f_2 \mathbf{e}_2$ is

\begin{aligned}\boldsymbol{\partial} \wedge \mathbf{f}=\left( {\mathbf{e}_1 \frac{\partial {}}{\partial {x}}+ \mathbf{e}_2 \frac{\partial {}}{\partial {y}}} \right) \wedge\left( {f_1 \mathbf{e}_1 + f_2 \mathbf{e}_2} \right)=\mathbf{e}_1 \mathbf{e}_2\left( {\frac{\partial {f_2}}{\partial {x}}-\frac{\partial {f_1}}{\partial {y}}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.38)

So, the LHS of Stokes theorem takes the coordinate form

\begin{aligned}\int d^2 \mathbf{x} \cdot (\boldsymbol{\partial} \wedge \mathbf{f}) =\iint dx dy\underbrace{\left\langle{{\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_1 \mathbf{e}_2}}\right\rangle}_{=-1}\left( {\frac{\partial {f_2}}{\partial {x}}-\frac{\partial {f_1}}{\partial {y}}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.38)

For the RHS, following fig. 1.5, we have

\begin{aligned}\mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowleft}}\int} \mathbf{f} \cdot d\mathbf{x}=f_2(x_0, y) dy+f_1(x, y_1) dx-f_2(x_1, y) dy-f_1(x, y_0) dx=\int dx \left( {f_1(x, y_1)-f_1(x, y_0)} \right)-\int dy \left( {f_2(x_1, y)-f_2(x_0, y)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.38)

As expected, we can also obtain this by integrating eq. 1.0.38.

Fig 1.5. Euclidean 2D loop

## Example: Cylindrical parameterization

Let’s now consider a cylindrical parameterization of a 4D space with Euclidean metric $++++$ or Minkowski metric $+++-$. For such a space let’s do a brute force expansion of both sides of Stokes theorem to gain some confidence that all is well.

With $\kappa = \mathbf{e}_3 \mathbf{e}_4$, such a space is conveniently parameterized as illustrated in fig. 1.6 as

\begin{aligned}\mathbf{x}(\rho, \theta, h) = x \mathbf{e}_1 + y \mathbf{e}_2 + \rho \mathbf{e}_3 e^{\kappa \theta}.\end{aligned} \hspace{\stretch{1}}(1.42)

Fig 1.6. Cylindrical polar parameterization

Note that the Euclidean case where $\left( {\mathbf{e}_4} \right)^2 = 1$ rejection of the non-axial components of $\mathbf{x}$ expands to

\begin{aligned}\left( { \left( { \mathbf{x} \wedge \mathbf{e}_1 \wedge \mathbf{e}_2} \right) \cdot \mathbf{e}^2 } \right) \cdot \mathbf{e}^1 =\rho \left( { \mathbf{e}_3 \cos\theta + \mathbf{e}_4 \sin \theta } \right),\end{aligned} \hspace{\stretch{1}}(1.43)

whereas for the Minkowski case where $\left( {\mathbf{e}_4} \right)^2 = -1$ we have a hyperbolic expansion

\begin{aligned}\left( { \left( { \mathbf{x} \wedge \mathbf{e}_1 \wedge \mathbf{e}_2} \right) \cdot \mathbf{e}^2 } \right) \cdot \mathbf{e}^1 =\rho \left( { \mathbf{e}_3 \cosh\theta + \mathbf{e}_4 \sinh \theta } \right).\end{aligned} \hspace{\stretch{1}}(1.44)

Within such a space consider the surface along $x = c, y = d$, for which the vectors are parameterized by

\begin{aligned}\mathbf{x}(\rho, \theta) = c \mathbf{e}_1 + d \mathbf{e}_2 + \rho \mathbf{e}_3 e^{\kappa \theta}.\end{aligned} \hspace{\stretch{1}}(1.45)

The tangent space unit vectors are

\begin{aligned}\mathbf{x}_\rho= \frac{\partial {\mathbf{x}}}{\partial {\rho}} = \mathbf{e}_3 e^{\kappa \theta},\end{aligned} \hspace{\stretch{1}}(1.46)

and

\begin{aligned}\mathbf{x}_\theta &= \frac{\partial {\mathbf{x}}}{\partial {\theta}} \\ &= \rho \mathbf{e}_3 \mathbf{e}_3 \mathbf{e}_4 e^{\kappa \theta} \\ &= \rho \mathbf{e}_4 e^{\kappa \theta}.\end{aligned} \hspace{\stretch{1}}(1.47)

Observe that both of these vectors have their origin at the point of evaluation, and aren’t relative to the absolute origin used to parameterize the complete space.

We wish to compute the volume element for the tangent plane. Noting that $\mathbf{e}_3$ and $\mathbf{e}_4$ both anticommute with $\kappa$ we have for $\mathbf{a} \in \text{span} \left\{ {\mathbf{e}_3, \mathbf{e}_4} \right\}$

\begin{aligned}\mathbf{a} e^{\kappa \theta} = e^{-\kappa \theta} \mathbf{a},\end{aligned} \hspace{\stretch{1}}(1.48)

so

\begin{aligned}\mathbf{x}_\theta \wedge \mathbf{x}_\rho &= {\left\langle{{\mathbf{e}_3 e^{\kappa \theta} \rho \mathbf{e}_4 e^{\kappa \theta}}}\right\rangle}_{2} \\ &= \rho {\left\langle{{\mathbf{e}_3 e^{\kappa \theta} e^{-\kappa \theta} \mathbf{e}_4}}\right\rangle}_{2} \\ &= \rho \mathbf{e}_3 \mathbf{e}_4.\end{aligned} \hspace{\stretch{1}}(1.49)

The tangent space volume element is thus

\begin{aligned}d^2 \mathbf{x} = \rho d\rho d\theta \mathbf{e}_3 \mathbf{e}_4.\end{aligned} \hspace{\stretch{1}}(1.50)

With the tangent plane vectors both perpendicular we don’t need the general lemma 6 to compute the reciprocal basis, but can do so by inspection

\begin{aligned}\mathbf{x}^\rho = e^{-\kappa \theta} \mathbf{e}^3,\end{aligned} \hspace{\stretch{1}}(1.0.51)

and

\begin{aligned}\mathbf{x}^\theta = e^{-\kappa \theta} \mathbf{e}^4 \frac{1}{{\rho}}.\end{aligned} \hspace{\stretch{1}}(1.0.52)

Observe that the latter depends on the metric signature.

The vector derivative, the projection of the gradient on the tangent space, is

\begin{aligned}\boldsymbol{\partial} &= \mathbf{x}^\rho \frac{\partial {}}{\partial {\rho}}+\mathbf{x}^\theta \frac{\partial {}}{\partial {\theta}} \\ &= e^{-\kappa \theta} \left( {\mathbf{e}^3 \partial_\rho + \frac{\mathbf{e}^4}{\rho} \partial_\theta } \right).\end{aligned} \hspace{\stretch{1}}(1.0.52)

From this we see that acting with the vector derivative on a scalar radial only dependent function $f(\rho)$ is a vector function that has a radial direction, whereas the action of the vector derivative on an azimuthal only dependent function $g(\theta)$ is a vector function that has only an azimuthal direction. The interpretation of the geometric product action of the vector derivative on a vector function is not as simple since the product will be a multivector.

Expanding the curl in coordinates is messier, but yields in the end when tackled with sufficient care

\begin{aligned}\boldsymbol{\partial} \wedge \mathbf{f} &= {\left\langle{{e^{-\kappa \theta}\left( { e^3 \partial_\rho + \frac{e^4}{\rho} \partial_\theta} \right)\left( { \not{{e_1 x}} + \not{{e_2 y}} + e_3 e^{\kappa \theta } f_\rho + \frac{e^4}{\rho} e^{\kappa \theta } f_\theta} \right)}}\right\rangle}_{2} \\ &= \not{{{\left\langle{{e^{-\kappa \theta} e^3 \partial_\rho \left( { e_3 e^{\kappa \theta } f_\rho} \right)}}\right\rangle}_{2}}}+{\left\langle{{\not{{e^{-\kappa \theta}}} e^3 \partial_\rho \left( { \frac{e^4}{\rho} \not{{e^{\kappa \theta }}} f_\theta} \right)}}\right\rangle}_{2}+{\left\langle{{e^{-\kappa \theta}\frac{e^4}{\rho} \partial_\theta\left( { e_3 e^{\kappa \theta } f_\rho} \right)}}\right\rangle}_{2}+{\left\langle{{e^{-\kappa \theta}\frac{e^4}{\rho} \partial_\theta\left( { \frac{e^4}{\rho} e^{\kappa \theta } f_\theta} \right)}}\right\rangle}_{2} \\ &= \mathbf{e}^3 \mathbf{e}^4 \left( {-\frac{f_\theta}{\rho^2} + \frac{1}{{\rho}} \partial_\rho f_\theta- \frac{1}{{\rho}} \partial_\theta f_\rho} \right)+ \frac{1}{{\rho^2}}{\left\langle{{e^{-\kappa \theta} \left( {\mathbf{e}^4} \right)^2\left( {\mathbf{e}_3 \mathbf{e}_4 f_\theta+ \not{{\partial_\theta f_\theta}}} \right)e^{\kappa \theta}}}\right\rangle}_{2} \\ &= \mathbf{e}^3 \mathbf{e}^4 \left( {-\frac{f_\theta}{\rho^2} + \frac{1}{{\rho}} \partial_\rho f_\theta- \frac{1}{{\rho}} \partial_\theta f_\rho} \right)+ \frac{1}{{\rho^2}}{\left\langle{{\not{{e^{-\kappa \theta} }}\mathbf{e}_3 \mathbf{e}^4 f_\theta\not{{e^{\kappa \theta}}}}}\right\rangle}_{2} \\ &= \frac{\mathbf{e}^3 \mathbf{e}^4 }{\rho}\left( {\partial_\rho f_\theta- \partial_\theta f_\rho} \right).\end{aligned} \hspace{\stretch{1}}(1.0.52)

After all this reduction, we can now state in coordinates the LHS of Stokes theorem explicitly

\begin{aligned}\int d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \int \rho d\rho d\theta \left\langle{{\mathbf{e}_3 \mathbf{e}_4 \mathbf{e}^3 \mathbf{e}^4 }}\right\rangle\frac{1}{{\rho}}\left( {\partial_\rho f_\theta- \partial_\theta f_\rho} \right) \\ &= \int d\rho d\theta\left( {\partial_\theta f_\rho-\partial_\rho f_\theta} \right) \\ &= \int d\rho {\left.{{f_\rho}}\right\vert}_{{\Delta \theta}}- \int d\theta{\left.{{f_\theta}}\right\vert}_{{\Delta \rho}}.\end{aligned} \hspace{\stretch{1}}(1.0.52)

Now compare this to the direct evaluation of the loop integral portion of Stokes theorem. Expressing this using eq. 1.0.34, we have the same result

\begin{aligned}\int d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=\int {\left.{{f_\rho}}\right\vert}_{{\Delta \theta}} d\rho-{\left.{{f_\theta}}\right\vert}_{{\Delta \rho}} d\theta\end{aligned} \hspace{\stretch{1}}(1.0.56)

This example highlights some of the power of Stokes theorem, since the reduction of the volume element differential form was seen to be quite a chore (and easy to make mistakes doing.)

## Example: Composition of boost and rotation

Working in a $\bigwedge^{1,3}$ space with basis $\left\{ {\gamma_0, \gamma_1, \gamma_2, \gamma_3} \right\}$ where $\left( {\gamma_0} \right)^2 = 1$ and $\left( {\gamma_k} \right)^2 = -1, k \in \left\{ {1,2,3} \right\}$, an active composition of boost and rotation has the form

\begin{aligned}\begin{aligned}\mathbf{x}' &= e^{i\alpha/2} \mathbf{x}_0 e^{-i\alpha/2} \\ \mathbf{x}'' &= e^{-j\theta/2} \mathbf{x}' e^{j\theta/2}\end{aligned},\end{aligned} \hspace{\stretch{1}}(1.0.57)

where $i$ is a bivector of a timelike unit vector and perpendicular spacelike unit vector, and $j$ is a bivector of two perpendicular spacelike unit vectors. For example, $i = \gamma_0 \gamma_1$ and $j = \gamma_1 \gamma_2$. For such $i,j$ the respective Lorentz transformation matrices are

\begin{aligned}{\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}}'=\begin{bmatrix}\cosh\alpha & -\sinh\alpha & 0 & 0 \\ -\sinh\alpha & \cosh\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.58)

and

\begin{aligned}{\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}}''=\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & \cos\theta & \sin\theta & 0 \\ 0 & -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}{\begin{bmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}}'.\end{aligned} \hspace{\stretch{1}}(1.0.59)

Let’s calculate the tangent space vectors for this parameterization, assuming that the particle is at an initial spacetime position of $\mathbf{x}_0$. That is

\begin{aligned}\mathbf{x} = e^{-j\theta/2} e^{i\alpha/2} \mathbf{x}_0e^{-i\alpha/2} e^{j\theta/2}.\end{aligned} \hspace{\stretch{1}}(1.0.60)

To calculate the tangent space vectors for this subspace we note that

\begin{aligned}\frac{\partial {\mathbf{x}'}}{\partial {\alpha}} = \frac{i}{2} \mathbf{x}_0 - \mathbf{x}_0 \frac{i}{2} = i \cdot \mathbf{x}_0,\end{aligned} \hspace{\stretch{1}}(1.0.61)

and

\begin{aligned}\frac{\partial {\mathbf{x}''}}{\partial {\theta}} = -\frac{j}{2} \mathbf{x}' + \mathbf{x}' \frac{j}{2} = \mathbf{x}' \cdot j.\end{aligned} \hspace{\stretch{1}}(1.0.62)

The tangent space vectors are therefore

\begin{aligned}\begin{aligned}\mathbf{x}_\alpha &= e^{-j\theta/2} \left( { i \cdot \mathbf{x}_0 } \right)e^{j\theta/2} \\ \mathbf{x}_\theta &= \left( {e^{i\alpha/2} \mathbf{x}_0e^{-i\alpha/2} } \right) \cdot j.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.63)

Continuing a specific example where $i = \gamma_0\gamma_1, j = \gamma_1 \gamma_2$ let’s also pick $\mathbf{x}_0 = \gamma_0$, the spacetime position of a particle at the origin of a frame at that frame’s $c t = 1$. The tangent space vectors for the subspace parameterized by this transformation and this initial position is then reduced to

\begin{aligned}\mathbf{x}_\alpha = -\gamma_1 e^{j \theta} = \gamma_1 \sin\theta + \gamma_2 \cos\theta,\end{aligned} \hspace{\stretch{1}}(1.0.63)

and

\begin{aligned}\mathbf{x}_\theta &= \left( { \gamma_0 e^{-i \alpha} } \right) \cdot j \\ &= \left( { \gamma_0\left( { \cosh\alpha - \gamma_0 \gamma_1 \sinh\alpha } \right)} \right) \cdot \left( { \gamma_1 \gamma_2} \right) \\ &= {\left\langle{{ \left( { \gamma_0 \cosh\alpha - \gamma_1 \sinh\alpha } \right) \gamma_1 \gamma_2 }}\right\rangle}_{1} \\ &= \gamma_2 \sinh\alpha.\end{aligned} \hspace{\stretch{1}}(1.0.63)

By inspection the dual basis for this parameterization is

\begin{aligned}\begin{aligned}\mathbf{x}^\alpha &= \gamma_1 e^{j \theta} \\ \mathbf{x}^\theta &= \frac{\gamma^2}{\sinh\alpha} \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.66)

So, Stokes theorem, applied to a spacetime vector $\mathbf{f}$, for this subspace is

\begin{aligned}\int d\alpha d\theta \sinh\alpha \sin\theta \left( { \gamma_1 \gamma_2 } \right) \cdot \left( {\left( {\gamma_1 e^{j \theta} \partial_\alpha + \frac{\gamma^2}{\sinh\alpha} \partial_\theta} \right)\wedge \mathbf{f}} \right)=\int d\alpha {\left.{{\mathbf{f} \cdot \Bigl( {\gamma^1 e^{j \theta}} \Bigr)}}\right\vert}_{{\theta_0}}^{{\theta_1}}-\int d\theta {\left.{{\mathbf{f} \cdot \Bigl( { \gamma_2 \sinh\alpha } \Bigr)}}\right\vert}_{{\alpha_0}}^{{\alpha_1}}.\end{aligned} \hspace{\stretch{1}}(1.0.67)

Since the point is to avoid the curl integral, we did not actually have to state it explicitly, nor was there any actual need to calculate the dual basis.

## Example: Dual representation in three dimensions

It’s clear that there is a projective nature to the differential form $d^2 \mathbf{x} \cdot \left( {\boldsymbol{\partial} \wedge \mathbf{f}} \right)$. This projective nature allows us, in three dimensions, to re-express Stokes theorem using the gradient instead of the vector derivative, and to utilize the cross product and a normal direction to the plane.

When we parameterize a normal direction to the tangent space, so that for a 2D tangent space spanned by curvilinear coordinates $\mathbf{x}_1$ and $\mathbf{x}_2$ the vector $\mathbf{x}^3$ is normal to both, we can write our vector as

\begin{aligned}\mathbf{f} = f_1 \mathbf{x}^1 + f_2 \mathbf{x}^2 + f_3 \mathbf{x}^3,\end{aligned} \hspace{\stretch{1}}(1.0.68)

and express the orientation of the tangent space area element in terms of a pseudoscalar that includes this normal direction

\begin{aligned}\mathbf{x}_1 \wedge \mathbf{x}_2 =\mathbf{x}^3 \cdot \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) =\mathbf{x}^3 \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right).\end{aligned} \hspace{\stretch{1}}(1.0.69)

Inserting this into an expansion of the curl form we have

\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= du^1 du^2 \left\langle{{\mathbf{x}^3 \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\left( {\left( {\sum_{i=1,2} x^i \partial_i} \right)\wedge\mathbf{f}} \right)}}\right\rangle \\ &= du^1 du^2 \mathbf{x}^3 \cdot \left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\cdot \left( {\boldsymbol{\nabla} \wedge \mathbf{f}} \right)-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\cdot \left( {\mathbf{x}^3 \partial_3 \wedge \mathbf{f}} \right)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.69)

Observe that this last term, the contribution of the component of the gradient perpendicular to the tangent space, has no $\mathbf{x}_3$ components

\begin{aligned}\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\cdot \left( {\mathbf{x}^3 \partial_3 \wedge \mathbf{f}} \right) &= \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right)\cdot \left( {\mathbf{x}^3 \wedge \partial_3 \mathbf{f}} \right) \\ &= \left( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \mathbf{x}^3} \right)\cdot \partial_3 \mathbf{f} \\ &= \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f} \\ &= \mathbf{x}_1 \left( { \mathbf{x}_2 \cdot \partial_3 \mathbf{f} } \right)-\mathbf{x}_2 \left( { \mathbf{x}_1 \cdot \partial_3 \mathbf{f} } \right),\end{aligned} \hspace{\stretch{1}}(1.0.69)

leaving

\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=du^1 du^2 \mathbf{x}^3 \cdot \left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \left( { \boldsymbol{\nabla} \wedge \mathbf{f}} \right)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.69)

Now scale the normal vector and its dual to have unit norm as follows

\begin{aligned}\begin{aligned}\mathbf{x}^3 &= \alpha \hat{\mathbf{x}}^3 \\ \mathbf{x}_3 &= \frac{1}{{\alpha}} \hat{\mathbf{x}}_3,\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.73)

so that for $\beta > 0$, the volume element can be

\begin{aligned}\mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \hat{\mathbf{x}}_3 = \beta I.\end{aligned} \hspace{\stretch{1}}(1.0.73)

This scaling choice is illustrated in fig. 1.7, and represents the “outwards” normal. With such a scaling choice we have

Fig 1.7. Outwards normal

\begin{aligned}\beta du^1 du^2 = dA,\end{aligned} \hspace{\stretch{1}}(1.75)

and almost have the desired cross product representation

\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=dA \hat{\mathbf{x}}^3 \cdot \left( { I \cdot \left( {\boldsymbol{\nabla} \wedge \mathbf{f}} \right) } \right)=dA \hat{\mathbf{x}}^3 \cdot \left( { I \left( {\boldsymbol{\nabla} \wedge \mathbf{f}} \right) } \right).\end{aligned} \hspace{\stretch{1}}(1.76)

With the duality identity $\mathbf{a} \wedge \mathbf{b} = I \left( {\mathbf{a} \times \mathbf{b}} \right)$, we have the traditional 3D representation of Stokes theorem

\begin{aligned}\int d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-\int dA \hat{\mathbf{x}}^3 \cdot \left( {\boldsymbol{\nabla} \times \mathbf{f}} \right) = \mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowleft}}\int} \mathbf{f} \cdot d\mathbf{l}.\end{aligned} \hspace{\stretch{1}}(1.0.77)

Note that the orientation of the loop integral in the traditional statement of the 3D Stokes theorem is counterclockwise instead of clockwise, as written here.

# Stokes theorem, three variable volume element parameterization

We can restate the identity of thm. 1 in an equivalent dot product form.

\begin{aligned}\int_V \left( { d^k \mathbf{x} \cdot \mathbf{x}^i } \right) \cdot \partial_i F = \int_{\partial V} d^{k-1} \mathbf{x} \cdot F.\end{aligned} \hspace{\stretch{1}}(1.0.78)

Here $d^{k-1} \mathbf{x} = \sum_i d^k \mathbf{x} \cdot \mathbf{x}^i$, with the implicit assumption that it and the blade $F$ that it is dotted with, are both evaluated at the end points of integration variable $u^i$ that has been integrated against.

We’ve seen one specific example of this above in the expansions of eq. 1.28, and eq. 1.29, however, the equivalent result of eq. 1.0.78, somewhat magically, applies to any degree blade and volume element provided the degree of the blade is less than that of the volume element (i.e. $s < k$). That magic follows directly from lemma 1.

As an expositional example, consider a three variable volume element parameterization, and a vector blade $\mathbf{f}$

\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \left( { d^3 \mathbf{x} \cdot \mathbf{x}^i } \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \mathbf{x}^i } \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) {\delta_3}^i-\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 } \right) {\delta_2}^i+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) {\delta_1}^i} \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 } \right) \cdot \partial_2 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_1 \mathbf{f}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.78)

It should not be surprising that this has the structure found in the theory of differential forms. Using the differentials for each of the parameterization “directions”, we can write this dot product expansion as

\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=\left( {du^3 \left( { d\mathbf{x}_1 \wedge d\mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f}-du^2 \left( { d\mathbf{x}_1 \wedge d\mathbf{x}_3 } \right) \cdot \partial_2 \mathbf{f}+du^1 \left( { d\mathbf{x}_2 \wedge d\mathbf{x}_3 } \right) \cdot \partial_1 \mathbf{f}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.78)

Observe that the sign changes with each element of $d\mathbf{x}_1 \wedge d\mathbf{x}_2 \wedge d\mathbf{x}_3$ that is skipped. In differential forms, the wedge product composition of 1-forms is an abstract quantity. Here the differentials are just vectors, and their wedge product represents an oriented volume element. This interpretation is likely available in the theory of differential forms too, but is arguably less obvious.

## Digression

As was the case with the loop integral, we expect that the coordinate representation has a representation that can be expressed as a number of antisymmetric terms. A bit of experimentation shows that such a sum, after dropping the parameter space volume element factor, is

\begin{aligned}\mathbf{x}_1 \left( { -\partial_2 f_3 + \partial_3 f_2 } \right)+\mathbf{x}_2 \left( { -\partial_3 f_1 + \partial_1 f_3 } \right)+\mathbf{x}_3 \left( { -\partial_1 f_2 + \partial_2 f_1 } \right) &= \mathbf{x}_1 \left( { -\partial_2 \mathbf{f} \cdot \mathbf{x}_3 + \partial_3 \mathbf{f} \cdot \mathbf{x}_2 } \right)+\mathbf{x}_2 \left( { -\partial_3 \mathbf{f} \cdot \mathbf{x}_1 + \partial_1 \mathbf{f} \cdot \mathbf{x}_3 } \right)+\mathbf{x}_3 \left( { -\partial_1 \mathbf{f} \cdot \mathbf{x}_2 + \partial_2 \mathbf{f} \cdot \mathbf{x}_1 } \right) \\ &= \left( { \mathbf{x}_1 \partial_3 \mathbf{f} \cdot \mathbf{x}_2 -\mathbf{x}_2 \partial_3 \mathbf{f} \cdot \mathbf{x}_1 } \right)+\left( { \mathbf{x}_3 \partial_2 \mathbf{f} \cdot \mathbf{x}_1 -\mathbf{x}_1 \partial_2 \mathbf{f} \cdot \mathbf{x}_3 } \right)+\left( { \mathbf{x}_2 \partial_1 \mathbf{f} \cdot \mathbf{x}_3 -\mathbf{x}_3 \partial_1 \mathbf{f} \cdot \mathbf{x}_2 } \right) \\ &= \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_3 \wedge \mathbf{x}_1 } \right) \cdot \partial_2 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_1 \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.78)

To proceed with the integration, we must again consider an infinitesimal volume element, for which the partial can be evaluated as the difference of the endpoints, with all else held constant. For this three variable parameterization, say, $(u,v,w)$, let’s delimit such an infinitesimal volume element by the parameterization ranges $[u_0,u_0 + du]$, $[v_0,v_0 + dv]$, $[w_0,w_0 + dw]$. The integral is

\begin{aligned}\begin{aligned}\int_{u = u_0}^{u_0 + du}\int_{v = v_0}^{v_0 + dv}\int_{w = w_0}^{w_0 + dw}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)&=\int_{u = u_0}^{u_0 + du}du\int_{v = v_0}^{v_0 + dv}dv{\left.{{ \Bigl( { \left( { \mathbf{x}_u \wedge \mathbf{x}_v } \right) \cdot \mathbf{f} } \Bigr) }}\right\vert}_{{w = w_0}}^{{w_0 + dw}} \\ &-\int_{u = u_0}^{u_0 + du}du\int_{w = w_0}^{w_0 + dw}dw{\left.{{\Bigl( { \left( { \mathbf{x}_u \wedge \mathbf{x}_w } \right) \cdot \mathbf{f} } \Bigr) }}\right\vert}_{{v = v_0}}^{{v_0 + dv}} \\ &+\int_{v = v_0}^{v_0 + dv}dv\int_{w = w_0}^{w_0 + dw}dw{\left.{{\Bigl( { \left( { \mathbf{x}_v \wedge \mathbf{x}_w } \right) \cdot \mathbf{f} } \Bigr) }}\right\vert}_{{u = u_0}}^{{u_0 + du}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.82)

Extending this over the ranges $[u_0,u_0 + \Delta u]$, $[v_0,v_0 + \Delta v]$, $[w_0,w_0 + \Delta w]$, we have proved Stokes thm. 1 for vectors and a three parameter volume element, provided we have a surface element of the form

\begin{aligned}d^2 \mathbf{x} = {\left. \Bigl( {d\mathbf{x}_u \wedge d\mathbf{x}_v } \Bigr) \right\vert}_{w = w_0}^{w_1}-{\left. \Bigl( {d\mathbf{x}_u \wedge d\mathbf{x}_w } \Bigr) \right\vert}_{v = v_0}^{v_1}+{\left. \Bigl( {d\mathbf{x}_v \wedge \mathbf{x}_w } \Bigr) \right\vert}_{ u = u_0 }^{u_1},\end{aligned} \hspace{\stretch{1}}(1.0.82)

where the evaluation of the dot products with $\mathbf{f}$ are also evaluated at the same points.

## Example: Euclidean spherical polar parameterization of 3D subspace

Consider an Euclidean space where a 3D subspace is parameterized using spherical coordinates, as in

\begin{aligned}\mathbf{x}(x, \rho, \theta, \phi) = \mathbf{e}_1 x + \mathbf{e}_4 \rho \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right)=\left( {x, \rho \sin\theta \cos\phi, \rho \sin\theta \sin\phi, \rho \cos\theta} \right).\end{aligned} \hspace{\stretch{1}}(1.0.84)

The tangent space basis for the subspace situated at some fixed $x = x_0$, is easy to calculate, and is found to be

\begin{aligned}\begin{aligned}\mathbf{x}_\rho &= \left( {0, \sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta} \right) =\mathbf{e}_4 \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right) \\ \mathbf{x}_\theta &= \rho \left( {0, \cos\theta \cos\phi, \cos\theta \sin\phi, - \sin\theta} \right) =\rho \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta } \right) \\ \mathbf{x}_\phi &=\rho \left( {0, -\sin\theta \sin\phi, \sin\theta \cos\phi, 0} \right)= \rho \sin\theta \mathbf{e}_3 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.85)

While we can use the general relation of lemma 7 to compute the reciprocal basis. That is

\begin{aligned}\mathbf{a}^{*} = \left( { \mathbf{b} \wedge \mathbf{c} } \right) \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} }}.\end{aligned} \hspace{\stretch{1}}(1.0.86)

However, a naive attempt at applying this without algebraic software is a route that requires a lot of care, and is easy to make mistakes doing. In this case it is really not necessary since the tangent space basis only requires scaling to orthonormalize, satisfying for $i,j \in \left\{ {\rho, \theta, \phi} \right\}$

\begin{aligned}\mathbf{x}_i \cdot \mathbf{x}_j =\begin{bmatrix} 1 & 0 & 0 \\ 0 & \rho^2 & 0 \\ 0 & 0 & \rho^2 \sin^2 \theta \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.0.87)

This allows us to read off the dual basis for the tangent volume by inspection

\begin{aligned}\begin{aligned}\mathbf{x}^\rho &=\mathbf{e}_4 \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right) \\ \mathbf{x}^\theta &= \frac{1}{{\rho}} \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta } \right) \\ \mathbf{x}^\phi &=\frac{1}{{\rho \sin\theta}} \mathbf{e}_3 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.88)

Should we wish to explicitly calculate the curl on the tangent space, we would need these. The area and volume elements are also messy to calculate manually. This expansion can be found in the Mathematica notebook \nbref{sphericalSurfaceAndVolumeElements.nb}, and is

\begin{aligned}\begin{aligned}\mathbf{x}_\theta \wedge \mathbf{x}_\phi &=\rho^2 \sin\theta \left( \mathbf{e}_4 \mathbf{e}_2 \sin\theta \sin\phi + \mathbf{e}_2 \mathbf{e}_3 \cos\theta + \mathbf{e}_3 \mathbf{e}_4 \sin\theta \cos\phi \right) \\ \mathbf{x}_\phi \wedge \mathbf{x}_\rho &=\rho \sin\theta \left(-\mathbf{e}_2 \mathbf{e}_3 \sin\theta -\mathbf{e}_2 \mathbf{e}_4 \cos\theta \sin\phi +\mathbf{e}_3 \mathbf{e}_4\cos\theta \cos\phi \right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\theta &= -\mathbf{e}_4 \rho \left(\mathbf{e}_2\cos\phi +\mathbf{e}_3\sin\phi \right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\theta \wedge \mathbf{x}_\phi &= \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4 \rho^2 \sin\theta \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.89)

Those area elements have a Geometric algebra factorization that are perhaps useful

\begin{aligned}\begin{aligned}\mathbf{x}_\theta \wedge \mathbf{x}_\phi &=-\rho^2 \sin\theta \mathbf{e}_2 \mathbf{e}_3 \exp\left( {-\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right) \\ \mathbf{x}_\phi \wedge \mathbf{x}_\rho &=\rho \sin\theta \mathbf{e}_3 \mathbf{e}_4 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}\exp\left( {\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \theta} \right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\theta &= -\rho \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.0.90)

One of the beauties of Stokes theorem is that we don’t actually have to calculate the dual basis on the tangent space to proceed with the integration. For that calculation above, where we had a normal tangent basis, I still used software was used as an aid, so it is clear that this can generally get pretty messy.

To apply Stokes theorem to a vector field we can use eq. 1.0.82 to write down the integral directly

\begin{aligned}\int_V d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \int_{\partial V} d^2 \mathbf{x} \cdot \mathbf{f} \\ &= \int {\left.{{ \left( { \mathbf{x}_\theta \wedge \mathbf{x}_\phi } \right) \cdot \mathbf{f} }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} d\theta d\phi+\int{\left.{{ \left( { \mathbf{x}_\phi \wedge \mathbf{x}_\rho } \right) \cdot \mathbf{f} }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} d\phi d\rho+\int{\left.{{ \left( { \mathbf{x}_\rho \wedge \mathbf{x}_\theta } \right) \cdot \mathbf{f} }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}} d\rho d\theta.\end{aligned} \hspace{\stretch{1}}(1.0.90)

Observe that eq. 1.0.90 is a vector valued integral that expands to

\begin{aligned}\int {\left.{{ \left( { \mathbf{x}_\theta f_\phi - \mathbf{x}_\phi f_\theta } \right) }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} d\theta d\phi+\int {\left.{{ \left( { \mathbf{x}_\phi f_\rho - \mathbf{x}_\rho f_\phi } \right) }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} d\phi d\rho+\int {\left.{{ \left( { \mathbf{x}_\rho f_\theta - \mathbf{x}_\theta f_\rho } \right) }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}} d\rho d\theta.\end{aligned} \hspace{\stretch{1}}(1.0.92)

This could easily be a difficult integral to evaluate since the vectors $\mathbf{x}_i$ evaluated at the endpoints are still functions of two parameters. An easier integral would result from the application of Stokes theorem to a bivector valued field, say $B$, for which we have

\begin{aligned}\int_V d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right) &= \int_{\partial V} d^2 \mathbf{x} \cdot B \\ &= \int {\left.{{ \left( { \mathbf{x}_\theta \wedge \mathbf{x}_\phi } \right) \cdot B }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} d\theta d\phi+\int{\left.{{ \left( { \mathbf{x}_\phi \wedge \mathbf{x}_\rho } \right) \cdot B }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} d\phi d\rho+\int{\left.{{ \left( { \mathbf{x}_\rho \wedge \mathbf{x}_\theta } \right) \cdot B }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}} d\rho d\theta \\ &= \int {\left.{{ B_{\phi \theta} }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} d\theta d\phi+\int{\left.{{ B_{\rho \phi} }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} d\phi d\rho+\int{\left.{{ B_{\theta \rho} }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}} d\rho d\theta.\end{aligned} \hspace{\stretch{1}}(1.0.92)

There is a geometric interpretation to these oriented area integrals, especially when written out explicitly in terms of the differentials along the parameterization directions. Pulling out a sign explicitly to match the geometry (as we had to also do for the line integrals in the two parameter volume element case), we can write this as

\begin{aligned}\int_{\partial V} d^2 \mathbf{x} \cdot B = -\int {\left.{{ \left( { d\mathbf{x}_\phi \wedge d\mathbf{x}_\theta } \right) \cdot B }}\right\vert}_{{\rho = \rho_0}}^{{\rho_1}} -\int{\left.{{ \left( { d\mathbf{x}_\rho \wedge d\mathbf{x}_\phi } \right) \cdot B }}\right\vert}_{{\theta = \theta_0}}^{{\theta_1}} -\int{\left.{{ \left( { d\mathbf{x}_\theta \wedge d\mathbf{x}_\rho } \right) \cdot B }}\right\vert}_{{\phi = \phi_0}}^{{\phi_1}}.\end{aligned} \hspace{\stretch{1}}(1.0.94)

When written out in this differential form, each of the respective area elements is an oriented area along one of the faces of the parameterization volume, much like the line integral that results from a two parameter volume curl integral. This is visualized in fig. 1.8. In this figure, faces (1) and (3) are “top faces”, those with signs matching the tops of the evaluation ranges eq. 1.0.94, whereas face (2) is a bottom face with a sign that is correspondingly reversed.

Fig 1.8. Boundary faces of a spherical parameterization region

## Example: Minkowski hyperbolic-spherical polar parameterization of 3D subspace

Working with a three parameter volume element in a Minkowski space does not change much. For example in a 4D space with $\left( {\mathbf{e}_4} \right)^2 = -1$, we can employ a hyperbolic-spherical parameterization similar to that used above for the 4D Euclidean space

\begin{aligned}\mathbf{x}(x, \rho, \alpha, \phi)=\left\{ {x, \rho \sinh \alpha \cos\phi, \rho \sinh \alpha \sin\phi, \rho \cosh \alpha} \right\}=\mathbf{e}_1 x + \mathbf{e}_4 \rho \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha } \right).\end{aligned} \hspace{\stretch{1}}(1.0.95)

This has tangent space basis elements

\begin{aligned}\begin{aligned}\mathbf{x}_\rho &= \sinh\alpha \left( { \cos\phi \mathbf{e}_2 + \sin\phi \mathbf{e}_3 } \right) + \cosh\alpha \mathbf{e}_4 = \mathbf{e}_4 \exp\left( {\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\alpha &=\rho \cosh\alpha \left( { \cos\phi \mathbf{e}_2 + \sin\phi \mathbf{e}_3} \right) + \rho \sinh\alpha \mathbf{e}_4=\rho \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( {-\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\phi &=\rho \sinh\alpha \left( { \mathbf{e}_3 \cos\phi - \mathbf{e}_2 \sin\phi} \right) = \rho\sinh\alpha \mathbf{e}_3 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.96)

This is a normal basis, but again not orthonormal. Specifically, for $i,j \in \left\{ {\rho, \theta, \phi} \right\}$ we have

\begin{aligned}\mathbf{x}_i \cdot \mathbf{x}_j =\begin{bmatrix}-1 & 0 & 0 \\ 0 & \rho^2 & 0 \\ 0 & 0 & \rho^2 \sinh^2 \alpha \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(1.0.97)

where we see that the radial vector $\mathbf{x}_\rho$ is timelike. We can form the dual basis again by inspection

\begin{aligned}\begin{aligned}\mathbf{x}_\rho &= -\mathbf{e}_4 \exp\left( {\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\alpha &= \frac{1}{{\rho}} \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( {-\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\phi &= \frac{1}{{\rho\sinh\alpha}} \mathbf{e}_3 e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.98)

The area elements are

\begin{aligned}\begin{aligned}\mathbf{x}_\alpha \wedge \mathbf{x}_\phi &=\rho^2 \sinh\alpha \left(-\mathbf{e}_4 \mathbf{e}_3 \sinh\alpha \cos\phi+\cosh\alpha \mathbf{e}_2 \mathbf{e}_3+\sinh\alpha \sin\phi \mathbf{e}_2 \mathbf{e}_4\right) \\ \mathbf{x}_\phi \wedge \mathbf{x}_\rho &=\rho \sinh\alpha \left(-\mathbf{e}_2 \mathbf{e}_3 \sinh\alpha-\mathbf{e}_2 \mathbf{e}_4 \cosh\alpha \sin\phi+\cosh\alpha \cos\phi \mathbf{e}_3 \mathbf{e}_4\right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\alpha &=-\mathbf{e}_4 \rho \left(\cos\phi \mathbf{e}_2+\sin\phi \mathbf{e}_3\right),\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.99)

or

\begin{aligned}\begin{aligned}\mathbf{x}_\alpha \wedge \mathbf{x}_\phi &=\rho^2 \sinh\alpha \mathbf{e}_2 \mathbf{e}_3 \exp\left( { \mathbf{e}_4 \mathbf{e}_2 e^{-\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha } \right) \\ \mathbf{x}_\phi \wedge \mathbf{x}_\rho &=\rho\sinh\alpha \mathbf{e}_3 \mathbf{e}_4 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \exp\left( {\mathbf{e}_4 \mathbf{e}_2 e^{\mathbf{e}_2 \mathbf{e}_3 \phi} \alpha} \right) \\ \mathbf{x}_\rho \wedge \mathbf{x}_\alpha &=-\mathbf{e}_4 \mathbf{e}_2 \rho e^{\mathbf{e}_2 \mathbf{e}_3 \phi}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.100)

The volume element also reduces nicely, and is

\begin{aligned}\mathbf{x}_\rho \wedge \mathbf{x}_\alpha \wedge \mathbf{x}_\phi = \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4 \rho^2 \sinh\alpha.\end{aligned} \hspace{\stretch{1}}(1.0.101)

The area and volume element reductions were once again messy, done in software using \nbref{sphericalSurfaceAndVolumeElementsMinkowski.nb}. However, we really only need eq. 1.0.96 to perform the Stokes integration.

# Stokes theorem, four variable volume element parameterization

Volume elements for up to four parameters are likely of physical interest, with the four volume elements of interest for relativistic physics in $\bigwedge^{3,1}$ spaces. For example, we may wish to use a parameterization $u^1 = x, u^2 = y, u^3 = z, u^4 = \tau = c t$, with a four volume

\begin{aligned}d^4 \mathbf{x}=d\mathbf{x}_x \wedge d\mathbf{x}_y \wedge d\mathbf{x}_z \wedge d\mathbf{x}_\tau,\end{aligned} \hspace{\stretch{1}}(1.102)

We follow the same procedure to calculate the corresponding boundary surface “area” element (with dimensions of volume in this case). This is

\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= \left( { d^4 \mathbf{x} \cdot \mathbf{x}^i } \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3 du^4\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \mathbf{x}^i } \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3 du_4\left( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) {\delta_4}^i-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) {\delta_3}^i+\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) {\delta_2}^i-\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) {\delta_1}^i} \right) \cdot \partial_i \mathbf{f} \\ &= du^1 du^2 du^3 du^4\left( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_4 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \partial_2 \mathbf{f}-\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \partial_1 \mathbf{f}} \right).\end{aligned} \hspace{\stretch{1}}(1.103)

Our boundary value surface element is therefore

\begin{aligned}d^3 \mathbf{x} = \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3- \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_4+ \mathbf{x}_1 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4- \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4.\end{aligned} \hspace{\stretch{1}}(1.104)

where it is implied that this (and the dot products with $\mathbf{f}$) are evaluated on the boundaries of the integration ranges of the omitted index. This same boundary form can be used for vector, bivector and trivector variations of Stokes theorem.

# Duality and its relation to the pseudoscalar.

Looking to eq. 1.0.181 of lemma 6, and scaling the wedge product $\mathbf{a} \wedge \mathbf{b}$ by its absolute magnitude, we can express duality using that scaled bivector as a pseudoscalar for the plane that spans $\left\{ {\mathbf{a}, \mathbf{b}} \right\}$. Let’s introduce a subscript notation for such scaled blades

\begin{aligned}I_{\mathbf{a}\mathbf{b}} = \frac{\mathbf{a} \wedge \mathbf{b}}{\left\lvert {\mathbf{a} \wedge \mathbf{b}} \right\rvert}.\end{aligned} \hspace{\stretch{1}}(1.105)

This allows us to express the unit vector in the direction of $\mathbf{a}^{*}$ as

\begin{aligned}\widehat{\mathbf{a}^{*}} = \hat{\mathbf{b}} \frac{\left\lvert {\mathbf{a} \wedge \mathbf{b}} \right\rvert}{\mathbf{a} \wedge \mathbf{b}}= \hat{\mathbf{b}} \frac{1}{{I_{\mathbf{a} \mathbf{b}}}}.\end{aligned} \hspace{\stretch{1}}(1.0.106)

Following the pattern of eq. 1.0.181, it is clear how to express the dual vectors for higher dimensional subspaces. For example

or for the unit vector in the direction of $\mathbf{a}^{*}$,

\begin{aligned}\widehat{\mathbf{a}^{*}} = I_{\mathbf{b} \mathbf{c}} \frac{1}{{I_{\mathbf{a} \mathbf{b} \mathbf{c}} }}.\end{aligned}

# Divergence theorem.

When the curl integral is a scalar result we are able to apply duality relationships to obtain the divergence theorem for the corresponding space. We will be able to show that a relationship of the following form holds

\begin{aligned}\int_V dV \boldsymbol{\nabla} \cdot \mathbf{f} = \int_{\partial V} dA_i \hat{\mathbf{n}}^i \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.107)

Here $\mathbf{f}$ is a vector, $\hat{\mathbf{n}}^i$ is normal to the boundary surface, and $dA_i$ is the area of this bounding surface element. We wish to quantify these more precisely, especially because the orientation of the normal vectors are metric dependent. Working a few specific examples will show the pattern nicely, but it is helpful to first consider some aspects of the general case.

First note that, for a scalar Stokes integral we are integrating the vector derivative curl of a blade $F \in \bigwedge^{k-1}$ over a k-parameter volume element. Because the dimension of the space matches the number of parameters, the projection of the gradient onto the tangent space is exactly that gradient

\begin{aligned}\int_V d^k \mathbf{x} \cdot (\boldsymbol{\partial} \wedge F) =\int_V d^k \mathbf{x} \cdot (\boldsymbol{\nabla} \wedge F).\end{aligned} \hspace{\stretch{1}}(1.0.108)

Multiplication of $F$ by the pseudoscalar will always produce a vector. With the introduction of such a dual vector, as in

\begin{aligned}F = I \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.0.108)

Stokes theorem takes the form

\begin{aligned}\int_V d^k \mathbf{x} \cdot {\left\langle{{\boldsymbol{\nabla} I \mathbf{f}}}\right\rangle}_{k}= \int_{\partial V} \left\langle{{ d^{k-1} \mathbf{x} I \mathbf{f}}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(1.0.108)

or

\begin{aligned}\int_V \left\langle{{ d^k \mathbf{x} \boldsymbol{\nabla} I \mathbf{f}}}\right\rangle= \int_{\partial V} \left( { d^{k-1} \mathbf{x} I} \right) \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.0.108)

where we will see that the vector $d^{k-1} \mathbf{x} I$ can roughly be characterized as a normal to the boundary surface. Using primes to indicate the scope of the action of the gradient, cyclic permutation within the scalar selection operator can be used to factor out the pseudoscalar

\begin{aligned}\int_V \left\langle{{ d^k \mathbf{x} \boldsymbol{\nabla} I \mathbf{f}}}\right\rangle &= \int_V \left\langle{{ \mathbf{f}' d^k \mathbf{x} \boldsymbol{\nabla}' I}}\right\rangle \\ &= \int_V {\left\langle{{ \mathbf{f}' d^k \mathbf{x} \boldsymbol{\nabla}'}}\right\rangle}_{k} I \\ &= \int_V(-1)^{k+1} d^k \mathbf{x} \left( { \boldsymbol{\nabla} \cdot \mathbf{f}} \right) I \\ &= (-1)^{k+1} I^2\int_V dV\left( { \boldsymbol{\nabla} \cdot \mathbf{f}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.108)

The second last step uses lemma 8, and the last writes $d^k \mathbf{x} = I^2 \left\lvert {d^k \mathbf{x}} \right\rvert = I^2 dV$, where we have assumed (without loss of generality) that $d^k \mathbf{x}$ has the same orientation as the pseudoscalar for the space. We also assume that the parameterization is non-degenerate over the integration volume (i.e. no $d\mathbf{x}_i = 0$), so the sign of this product cannot change.

Let’s now return to the normal vector $d^{k-1} \mathbf{x} I$. With $d^{k-1} u_i = du^1 du^2 \cdots du^{i-1} du^{i+1} \cdots du^k$ (the $i$ indexed differential omitted), and $I_{ab\cdots c} = (\mathbf{x}_a \wedge \mathbf{x}_b \wedge \cdots \wedge \mathbf{x}_c)/\left\lvert {\mathbf{x}_a \wedge \mathbf{x}_b \wedge \cdots \wedge \mathbf{x}_c} \right\rvert$, we have

\begin{aligned}\begin{aligned}d^{k-1} \mathbf{x} I&=d^{k-1} u_i \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \cdots \wedge \mathbf{x}_k} \right) \cdot \mathbf{x}^i I \\ &= I_{1 2 \cdots (k-1)} I \left\lvert {d\mathbf{x}_1 \wedge d\mathbf{x}_2 \wedge \cdots \wedge d\mathbf{x}_{k-1} } \right\rvert \\ &\quad -I_{1 \cdots (k-2) k} I \left\lvert {d\mathbf{x}_1 \wedge \cdots \wedge d\mathbf{x}_{k-2} \wedge d\mathbf{x}_k} \right\rvert+ \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.113)

We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\mathbf{a}$ with respect to the unit pseudoscalar $I_{\mathbf{b} \cdots \mathbf{c} \mathbf{d}}$ in a subspace spanned by $\left\{ {\mathbf{a}, \cdots \mathbf{c}, \mathbf{d}} \right\}$ is

\begin{aligned}\widehat{\mathbf{a}^{*}} = I_{\mathbf{b} \cdots \mathbf{c} \mathbf{d}} \frac{1}{{ I_{\mathbf{a} \cdots \mathbf{c} \mathbf{d}} }},\end{aligned} \hspace{\stretch{1}}(1.0.114)

or

\begin{aligned}\widehat{\mathbf{a}^{*}} I_{\mathbf{a} \cdots \mathbf{c} \mathbf{d}}^2=I_{\mathbf{b} \cdots \mathbf{c} \mathbf{d}}.\end{aligned} \hspace{\stretch{1}}(1.0.115)

This allows us to write

\begin{aligned}d^{k-1} \mathbf{x} I= I^2 \sum_i \widehat{\mathbf{x}^i} d{A'}_i\end{aligned} \hspace{\stretch{1}}(1.0.116)

where $d{A'}_i = \pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it.

This leaves us with

\begin{aligned}\int_V dV \boldsymbol{\nabla} \cdot \mathbf{f} = (-1)^{k+1} \int_{\partial V} d{A'}_i \widehat{\mathbf{x}^i} \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.117)

To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples.

## Example: 2D divergence theorem

Let’s start back at

\begin{aligned}\int_A \left\langle{{ d^2 \mathbf{x} \boldsymbol{\nabla} I \mathbf{f} }}\right\rangle = \int_{\partial A} \left( { d^1 \mathbf{x} I} \right) \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.118)

On the left our integral can be rewritten as

\begin{aligned}\int_A \left\langle{{ d^2 \mathbf{x} \boldsymbol{\nabla} I \mathbf{f} }}\right\rangle &= -\int_A \left\langle{{ d^2 \mathbf{x} I \boldsymbol{\nabla} \mathbf{f} }}\right\rangle \\ &= -\int_A d^2 \mathbf{x} I \left( { \boldsymbol{\nabla} \cdot \mathbf{f} } \right) \\ &= - I^2 \int_A dA \boldsymbol{\nabla} \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.119)

where $d^2 \mathbf{x} = I dA$ and we pick the pseudoscalar with the same orientation as the volume (area in this case) element $I = (\mathbf{x}_1 \wedge \mathbf{x}_2)/\left\lvert {\mathbf{x}_1 \wedge \mathbf{x}_2} \right\rvert$.

For the boundary form we have

\begin{aligned}d^1 \mathbf{x} = du^2 \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \mathbf{x}^1+ du^1 \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \mathbf{x}^2= -du^2 \mathbf{x}_2 +du^1 \mathbf{x}_1.\end{aligned} \hspace{\stretch{1}}(1.120)

The duality relations for the tangent space are

\begin{aligned}\begin{aligned}\mathbf{x}^2 &= \mathbf{x}_1 \frac{1}{{\mathbf{x}_2 \wedge \mathbf{x}_1}} \\ \mathbf{x}^1 &= \mathbf{x}_2 \frac{1}{{\mathbf{x}_1 \wedge \mathbf{x}_2}}\end{aligned},\end{aligned} \hspace{\stretch{1}}(1.0.121)

or

\begin{aligned}\begin{aligned}\widehat{\mathbf{x}^2} &= -\widehat{\mathbf{x}_1} \frac{1}{I} \\ \widehat{\mathbf{x}^1} &= \widehat{\mathbf{x}_2} \frac{1}{I}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.0.122)

Back substitution into the line element gives

\begin{aligned}d^1 \mathbf{x} = -du^2 \left\lvert {\mathbf{x}_2} \right\rvert \widehat{\mathbf{x}_2}+du^1 \left\lvert {\mathbf{x}_1} \right\rvert \widehat{\mathbf{x}_1}=-du^2 \left\lvert {\mathbf{x}_2} \right\rvert \widehat{\mathbf{x}^1} I-du^1 \left\lvert {\mathbf{x}_1} \right\rvert \widehat{\mathbf{x}^2} I.\end{aligned} \hspace{\stretch{1}}(1.0.122)

Writing (no sum) $du^i \left\lvert {\mathbf{x}_i} \right\rvert = ds_i$, we have

\begin{aligned}d^1 \mathbf{x} I = -\left( { ds_2 \widehat{\mathbf{x}^1} +ds_1 \widehat{\mathbf{x}^2} } \right) I^2.\end{aligned} \hspace{\stretch{1}}(1.0.122)

This provides us a divergence and normal relationship, with $-I^2$ terms on each side that can be canceled. Restoring explicit range evaluation, that is

\begin{aligned}\int_A dA \boldsymbol{\nabla} \cdot \mathbf{f}=\int_{\Delta u^2} {\left.{{ ds_2 \widehat{\mathbf{x}^1} \cdot \mathbf{f}}}\right\vert}_{{\Delta u^1}}+ \int_{\Delta u^1} {\left.{{ ds_1 \widehat{\mathbf{x}^2} \cdot \mathbf{f}}}\right\vert}_{{\Delta u^2}}=\int_{\Delta u^2} {\left.{{ ds_2 \widehat{\mathbf{x}^1} \cdot \mathbf{f}}}\right\vert}_{{u^1(1)}}-\int_{\Delta u^2} {\left.{{ ds_2 \widehat{\mathbf{x}^1} \cdot \mathbf{f}}}\right\vert}_{{u^1(0)}}+ \int_{\Delta u^1} {\left.{{ ds_1 \widehat{\mathbf{x}^2} \cdot \mathbf{f}}}\right\vert}_{{u^2(0)}}- \int_{\Delta u^1} {\left.{{ ds_1 \widehat{\mathbf{x}^2} \cdot \mathbf{f}}}\right\vert}_{{u^2(0)}}.\end{aligned} \hspace{\stretch{1}}(1.0.122)

Let’s consider this graphically for an Euclidean metric as illustrated in fig. 1.9.

Fig 1.9. Normals on area element

We see that

1. along $u^2(0)$ the outwards normal is $-\widehat{\mathbf{x}^2}$,
2. along $u^2(1)$ the outwards normal is $\widehat{\mathbf{x}^2}$,
3. along $u^1(0)$ the outwards normal is $-\widehat{\mathbf{x}^1}$, and
4. along $u^1(1)$ the outwards normal is $\widehat{\mathbf{x}^2}$.

Writing that outwards normal as $\hat{\mathbf{n}}$, we have

\begin{aligned}\int_A dA \boldsymbol{\nabla} \cdot \mathbf{f}= \mathop{\rlap{\ensuremath{\mkern3.5mu\circlearrowright}}\int} ds \hat{\mathbf{n}} \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.126)

Note that we can use the same algebraic notion of outward normal for non-Euclidean spaces, although cannot expect the geometry to look anything like that of the figure.

## Example: 3D divergence theorem

As with the 2D example, let’s start back with

\begin{aligned}\int_V \left\langle{{ d^3 \mathbf{x} \boldsymbol{\nabla} I \mathbf{f} }}\right\rangle = \int_{\partial V} \left( { d^2 \mathbf{x} I} \right) \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.127)

In a 3D space, the pseudoscalar commutes with all grades, so we have

\begin{aligned}\int_V \left\langle{{ d^3 \mathbf{x} \boldsymbol{\nabla} I \mathbf{f} }}\right\rangle=\int_V \left( { d^3 \mathbf{x} I } \right) \boldsymbol{\nabla} \cdot \mathbf{f}=I^2 \int_V dV \boldsymbol{\nabla} \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.128)

where $d^3 \mathbf{x} I = dV I^2$, and we have used a pseudoscalar with the same orientation as the volume element

\begin{aligned}\begin{aligned}I &= \widehat{ \mathbf{x}_{123} } \\ \mathbf{x}_{123} &= \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.129)

In the boundary integral our dual two form is

\begin{aligned}d^2 \mathbf{x} I= du^1 du^2 \mathbf{x}_1 \wedge \mathbf{x}_2+du^3 du^1 \mathbf{x}_3 \wedge \mathbf{x}_1+du^2 du^3 \mathbf{x}_2 \wedge \mathbf{x}_3= \left( { dA_{3} \widehat{ \mathbf{x}_{12} } \frac{1}{I}+dA_{2} \widehat{ \mathbf{x}_{31} } \frac{1}{I}+dA_{1} \widehat{ \mathbf{x}_{23} } \frac{1}{I}} \right) I^2,\end{aligned} \hspace{\stretch{1}}(1.0.129)

where $\mathbf{x}_{ij} = \mathbf{x}_i \wedge \mathbf{x}_j$, and

\begin{aligned}\begin{aligned}dA_1 &= \left\lvert {d\mathbf{x}_2 \wedge d\mathbf{x}_3} \right\rvert \\ dA_2 &= \left\lvert {d\mathbf{x}_3 \wedge d\mathbf{x}_1} \right\rvert \\ dA_3 &= \left\lvert {d\mathbf{x}_1 \wedge d\mathbf{x}_2} \right\rvert.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.131)

Observe that we can do a cyclic permutation of a 3 blade without any change of sign, for example

\begin{aligned}\mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 =-\mathbf{x}_2 \wedge \mathbf{x}_1 \wedge \mathbf{x}_3 =\mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_1.\end{aligned} \hspace{\stretch{1}}(1.0.132)

Because of this we can write the dual two form as we expressed the normals in lemma 7

\begin{aligned}d^2 \mathbf{x} I = \left( { dA_1 \widehat{\mathbf{x}_{23}} \frac{1}{{\widehat{\mathbf{x}_{123}}}} + dA_2 \widehat{\mathbf{x}_{31}} \frac{1}{{\widehat{\mathbf{x}_{231}}}} + dA_3 \widehat{\mathbf{x}_{12}} \frac{1}{{\widehat{\mathbf{x}_{312}}}}} \right) I^2=\left( { dA_1 \widehat{\mathbf{x}^1}+dA_2 \widehat{\mathbf{x}^2}+dA_3 \widehat{\mathbf{x}^3} } \right) I^2.\end{aligned} \hspace{\stretch{1}}(1.0.132)

We can now state the 3D divergence theorem, canceling out the metric and orientation dependent term $I^2$ on both sides

\begin{aligned}\int_V dV \boldsymbol{\nabla} \cdot \mathbf{f}=\int dA \hat{\mathbf{n}} \cdot \mathbf{f},\end{aligned} \hspace{\stretch{1}}(1.0.134)

where (sums implied)

\begin{aligned}dA \hat{\mathbf{n}} = dA_i \widehat{\mathbf{x}^i},\end{aligned} \hspace{\stretch{1}}(1.0.135)

and

\begin{aligned}\begin{aligned}{\left.{{\hat{\mathbf{n}}}}\right\vert}_{{u^i = u^i(1)}} &= \widehat{\mathbf{x}^i} \\ {\left.{{\hat{\mathbf{n}}}}\right\vert}_{{u^i = u^i(0)}} &= -\widehat{\mathbf{x}^i}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.0.136)

The outwards normals at the upper integration ranges of a three parameter surface are depicted in fig. 1.10.

Fig 1.10. Outwards normals on volume at upper integration ranges.

This sign alternation originates with the two form elements $\left( {d\mathbf{x}_i \wedge d\mathbf{x}_j} \right) \cdot F$ from the Stokes boundary integral, which were explicitly evaluated at the endpoints of the integral. That is, for $k \ne i,j$,

\begin{aligned}\int_{\partial V} \left( { d\mathbf{x}_i \wedge d\mathbf{x}_j } \right) \cdot F\equiv\int_{\Delta u^i} \int_{\Delta u^j} {\left.{{\left( { \left( { d\mathbf{x}_i \wedge d\mathbf{x}_j } \right) \cdot F } \right)}}\right\vert}_{{u^k = u^k(1)}}-{\left.{{\left( { \left( { d\mathbf{x}_i \wedge d\mathbf{x}_j } \right) \cdot F } \right)}}\right\vert}_{{u^k = u^k(0)}}\end{aligned} \hspace{\stretch{1}}(1.0.137)

In the context of the divergence theorem, this means that we are implicitly requiring the dot products $\widehat{\mathbf{x}^k} \cdot \mathbf{f}$ to be evaluated specifically at the end points of the integration where $u^k = u^k(1), u^k = u^k(0)$, accounting for the alternation of sign required to describe the normals as uniformly outwards.

## Example: 4D divergence theorem

Applying Stokes theorem to a trivector $T = I \mathbf{f}$ in the 4D case we find

\begin{aligned}-I^2 \int_V d^4 x \boldsymbol{\nabla} \cdot \mathbf{f} = \int_{\partial V} \left( { d^3 \mathbf{x} I} \right) \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.138)

Here the pseudoscalar has been picked to have the same orientation as the hypervolume element $d^4 \mathbf{x} = I d^4 x$. Writing $\mathbf{x}_{ij \cdots k} = \mathbf{x}_i \wedge \mathbf{x}_j \wedge \cdots \mathbf{x}_k$ the dual of the three form is

\begin{aligned}d^3 \mathbf{x} I &= \left( { du^1 du^2 du^3 \mathbf{x}_{123}-du^1 du^2 du^4 \mathbf{x}_{124}+du^1 du^3 du^4 \mathbf{x}_{134}-du^2 du^3 du^4 \mathbf{x}_{234}} \right) I \\ &= \left( { dA^{123} \widehat{ \mathbf{x}_{123} } -dA^{124} \widehat{ \mathbf{x}_{124} } +dA^{134} \widehat{ \mathbf{x}_{134} } -dA^{234} \widehat{ \mathbf{x}_{234} }} \right) I \\ &= \left( { dA^{123} \widehat{ \mathbf{x}_{123} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}} -dA^{124} \widehat{ \mathbf{x}_{124} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}} +dA^{134} \widehat{ \mathbf{x}_{134} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}} -dA^{234} \widehat{ \mathbf{x}_{234} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}}} \right) I^2 \\ &= -\left( { dA^{123} \widehat{ \mathbf{x}_{123} } \frac{1}{{\widehat{\mathbf{x}_{4123} }}} +dA^{124} \widehat{ \mathbf{x}_{124} } \frac{1}{{\widehat{\mathbf{x}_{3412} }}} +dA^{134} \widehat{ \mathbf{x}_{134} } \frac{1}{{\widehat{\mathbf{x}_{2341} }}} +dA^{234} \widehat{ \mathbf{x}_{234} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}}} \right) I^2 \\ &= -\left( { dA^{123} \widehat{ \mathbf{x}_{123} } \frac{1}{{\widehat{\mathbf{x}_{4123} }}} +dA^{124} \widehat{ \mathbf{x}_{412} } \frac{1}{{\widehat{\mathbf{x}_{3412} }}} +dA^{134} \widehat{ \mathbf{x}_{341} } \frac{1}{{\widehat{\mathbf{x}_{2341} }}} +dA^{234} \widehat{ \mathbf{x}_{234} } \frac{1}{{\widehat{\mathbf{x}_{1234} }}}} \right) I^2 \\ &= -\left( { dA^{123} \widehat{ \mathbf{x}^{4} } +dA^{124} \widehat{ \mathbf{x}^{3} } +dA^{134} \widehat{ \mathbf{x}^{2} } +dA^{234} \widehat{ \mathbf{x}^{1} } } \right) I^2\end{aligned} \hspace{\stretch{1}}(1.139)

Here, we’ve written

\begin{aligned}dA^{ijk} = \left\lvert { d\mathbf{x}_i \wedge d\mathbf{x}_j \wedge d\mathbf{x}_k } \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.140)

Observe that the dual representation nicely removes the alternation of sign that we had in the Stokes theorem boundary integral, since each alternation of the wedged vectors in the pseudoscalar changes the sign once.

As before, we define the outwards normals as $\hat{\mathbf{n}} = \pm \widehat{\mathbf{x}^i}$ on the upper and lower integration ranges respectively. The scalar area elements on these faces can be written in a dual form

\begin{aligned}\begin{aligned} dA_4 &= dA^{123} \\ dA_3 &= dA^{124} \\ dA_2 &= dA^{134} \\ dA_1 &= dA^{234} \end{aligned},\end{aligned} \hspace{\stretch{1}}(1.0.141)

so that the 4D divergence theorem looks just like the 2D and 3D cases

\begin{aligned}\int_V d^4 x \boldsymbol{\nabla} \cdot \mathbf{f} = \int_{\partial V} d^3 x \hat{\mathbf{n}} \cdot \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(1.0.142)

Here we define the volume scaled normal as

\begin{aligned}d^3 x \hat{\mathbf{n}} = dA_i \widehat{\mathbf{x}^i}.\end{aligned} \hspace{\stretch{1}}(1.0.143)

As before, we have made use of the implicit fact that the three form (and it’s dot product with $\mathbf{f}$) was evaluated on the boundaries of the integration region, with a toggling of sign on the lower limit of that evaluation that is now reflected in what we have defined as the outwards normal.

We also obtain explicit instructions from this formalism how to compute the “outwards” normal for this surface in a 4D space (unit scaling of the dual basis elements), something that we cannot compute using any sort of geometrical intuition. For free we’ve obtained a result that applies to both Euclidean and Minkowski (or other non-Euclidean) spaces.

# Volume integral coordinate representations

It may be useful to formulate the curl integrals in tensor form. For vectors $\mathbf{f}$, and bivectors $B$, the coordinate representations of those differential forms (\cref{pr:stokesTheoremGeometricAlgebraII:1}) are

\begin{aligned}d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=- d^2 u \epsilon^{ a b } \partial_a f_b\end{aligned} \hspace{\stretch{1}}(1.0.144a)

\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-d^3 u \epsilon^{a b c} \mathbf{x}_a \partial_b f_{c}\end{aligned} \hspace{\stretch{1}}(1.0.144b)

\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-\frac{1}{2} d^4 u \epsilon^{a b c d} \mathbf{x}_a \wedge \mathbf{x}_b \partial_{c} f_{d}\end{aligned} \hspace{\stretch{1}}(1.0.144c)

\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-\frac{1}{2}d^3 u \epsilon^{a b c} \partial_a B_{b c}\end{aligned} \hspace{\stretch{1}}(1.0.144d)

\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-\frac{1}{2} d^4 u \epsilon^{a b c d} \mathbf{x}_a \partial_b B_{cd}\end{aligned} \hspace{\stretch{1}}(1.0.144e)

\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge T } \right)=-d^4 u\left( {\partial_4 T_{123}-\partial_3 T_{124}+\partial_2 T_{134}-\partial_1 T_{234}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.144f)

Here the bivector $B$ and trivector $T$ is expressed in terms of their curvilinear components on the tangent space

\begin{aligned}B = \frac{1}{2} \mathbf{x}^i \wedge \mathbf{x}^j B_{ij} + B_\perp\end{aligned} \hspace{\stretch{1}}(1.0.145a)

\begin{aligned}T = \frac{1}{{3!}} \mathbf{x}^i \wedge \mathbf{x}^j \wedge \mathbf{x}^k T_{ijk} + T_\perp,\end{aligned} \hspace{\stretch{1}}(1.0.145b)

where

\begin{aligned}B_{ij} = \mathbf{x}_j \cdot \left( { \mathbf{x}_i \cdot B } \right) = -B_{ji}.\end{aligned} \hspace{\stretch{1}}(1.0.146a)

\begin{aligned}T_{ijk} = \mathbf{x}_k \cdot \left( { \mathbf{x}_j \cdot \left( { \mathbf{x}_i \cdot B } \right)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.146b)

For the trivector components are also antisymmetric, changing sign with any interchange of indices.

Note that eq. 1.0.144d and eq. 1.0.144f appear much different on the surface, but both have the same structure. This can be seen by writing for former as

\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-d^3 u\left( { \partial_1 B_{2 3} + \partial_2 B_{3 1} + \partial_3 B_{1 2}} \right)=-d^3 u\left( { \partial_3 B_{1 2} - \partial_2 B_{1 3} + \partial_1 B_{2 3}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.146b)

In both of these we have an alternation of sign, where the tensor index skips one of the volume element indices is sequence. We’ve seen in the 4D divergence theorem that this alternation of sign can be related to a duality transformation.

In integral form (no sum over indexes $i$ in $du^i$ terms), these are

\begin{aligned}\int d^2 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=- \epsilon^{ a b } \int {\left.{{du^b f_b}}\right\vert}_{{\Delta u^a}}\end{aligned} \hspace{\stretch{1}}(1.0.148a)

\begin{aligned}\int d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-\epsilon^{a b c} \int du^a du^c{\left.{{\mathbf{x}_a f_{c}}}\right\vert}_{{\Delta u^b}}\end{aligned} \hspace{\stretch{1}}(1.0.148b)

\begin{aligned}\int d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)=-\frac{1}{2} \epsilon^{a b c d} \int du^a du^b du^d{\left.{{\mathbf{x}_a \wedge \mathbf{x}_b f_{d}}}\right\vert}_{{\Delta u^c}}\end{aligned} \hspace{\stretch{1}}(1.0.148c)

\begin{aligned}\int d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-\frac{1}{2}\epsilon^{a b c} \int du^b du^c{\left.{{B_{b c}}}\right\vert}_{{\Delta u^a}}\end{aligned} \hspace{\stretch{1}}(1.0.148d)

\begin{aligned}\int d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)=-\frac{1}{2} \epsilon^{a b c d} \int du^a du^c du^d{\left.{{\mathbf{x}_a B_{cd}}}\right\vert}_{{\Delta u^b}}\end{aligned} \hspace{\stretch{1}}(1.0.148e)

\begin{aligned}\int d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge T } \right)=-\int \left( {du^1 du^2 du^3 {\left.{{T_{123}}}\right\vert}_{{\Delta u^4}}-du^1 du^2 du^4 {\left.{{T_{124}}}\right\vert}_{{\Delta u^3}}+du^1 du^3 du^4 {\left.{{T_{134}}}\right\vert}_{{\Delta u^2}}-du^2 du^3 du^4 {\left.{{T_{234}}}\right\vert}_{{\Delta u^1}}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.148f)

Of these, I suspect that only eq. 1.0.148a and eq. 1.0.148d are of use.

# Final remarks

Because we have used curvilinear coordinates from the get go, we have arrived naturally at a formulation that works for both Euclidean and non-Euclidean geometries, and have demonstrated that Stokes (and the divergence theorem) holds regardless of the geometry or the parameterization. We also know explicitly how to formulate both theorems for any parameterization that we choose, something much more valuable than knowledge that this is possible.

For the divergence theorem we have introduced the concept of outwards normal (for example in 3D, eq. 1.0.136), which still holds for non-Euclidean geometries. We may not be able to form intuitive geometrical interpretations for these normals, but do have an algebraic description of them.

# Appendix

## Question: Expand volume elements in coordinates

Show that the coordinate representation for the volume element dotted with the curl can be represented as a sum of antisymmetric terms. That is

• (a)Prove eq. 1.0.144a
• (b)Prove eq. 1.0.144b
• (c)Prove eq. 1.0.144c
• (d)Prove eq. 1.0.144d
• (e)Prove eq. 1.0.144e
• (f)Prove eq. 1.0.144f

### (a) Two parameter volume, curl of vector

\begin{aligned}d^2 \mathbf{x} \cdot \left( \boldsymbol{\partial} \wedge \mathbf{f} \right) &= d^2 u\Bigl( { \left( \mathbf{x}_1 \wedge \mathbf{x}_2 \right) \cdot \mathbf{x}^i } \Bigr) \cdot \partial_i \mathbf{f} \\ &= d^2 u \left( \mathbf{x}_1 \cdot \partial_2 \mathbf{f}-\mathbf{x}_2 \cdot \partial_1 \mathbf{f} \right) \\ &= d^2 u\left( \partial_2 f_1-\partial_1 f_2 \right) \\ &= - d^2 u \epsilon^{ab} \partial_{a} f_{b}. \qquad\square\end{aligned} \hspace{\stretch{1}}(1.149)

### (b) Three parameter volume, curl of vector

\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right) &= d^3 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \mathbf{x}^i } \Bigr) \cdot \partial_i \mathbf{f} \\ &= d^3 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_3 \wedge \mathbf{x}_1 } \right) \cdot \partial_2 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_1 \mathbf{f}} \Bigr) \\ &= d^3 u\Bigl( {\left( { \mathbf{x}_1 \partial_3 \mathbf{f} \cdot \mathbf{x}_2 -\mathbf{x}_2 \partial_3 \mathbf{f} \cdot \mathbf{x}_1 } \right)+\left( { \mathbf{x}_3 \partial_2 \mathbf{f} \cdot \mathbf{x}_1 -\mathbf{x}_1 \partial_2 \mathbf{f} \cdot \mathbf{x}_3 } \right)+\left( { \mathbf{x}_2 \partial_1 \mathbf{f} \cdot \mathbf{x}_3 -\mathbf{x}_3 \partial_1 \mathbf{f} \cdot \mathbf{x}_2 } \right)} \Bigr) \\ &= d^3 u\Bigl( {\mathbf{x}_1 \left( { -\partial_2 \mathbf{f} \cdot \mathbf{x}_3 + \partial_3 \mathbf{f} \cdot \mathbf{x}_2 } \right)+\mathbf{x}_2 \left( { -\partial_3 \mathbf{f} \cdot \mathbf{x}_1 + \partial_1 \mathbf{f} \cdot \mathbf{x}_3 } \right)+\mathbf{x}_3 \left( { -\partial_1 \mathbf{f} \cdot \mathbf{x}_2 + \partial_2 \mathbf{f} \cdot \mathbf{x}_1 } \right)} \Bigr) \\ &= d^3 u\Bigl( {\mathbf{x}_1 \left( { -\partial_2 f_3 + \partial_3 f_2 } \right)+\mathbf{x}_2 \left( { -\partial_3 f_1 + \partial_1 f_3 } \right)+\mathbf{x}_3 \left( { -\partial_1 f_2 + \partial_2 f_1 } \right)} \Bigr) \\ &= - d^3 u \epsilon^{abc} \partial_b f_c. \qquad\square\end{aligned} \hspace{\stretch{1}}(1.150)

### (c) Four parameter volume, curl of vector

\begin{aligned}\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge \mathbf{f} } \right)&=d^4 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \mathbf{x}^i } \Bigr) \cdot \partial_i \mathbf{f} \\ &=d^4 u\Bigl( {\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_4 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \partial_2 \mathbf{f}-\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \cdot \partial_1 \mathbf{f}} \Bigr) \\ &=d^4 u\Bigl( { \\ &\quad\quad \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \mathbf{x}_3 \cdot \partial_4 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_3 } \right) \mathbf{x}_2 \cdot \partial_4 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \mathbf{x}_1 \cdot \partial_4 \mathbf{f} \\ &\quad-\left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \mathbf{x}_4 \cdot \partial_3 \mathbf{f}+\left( { \mathbf{x}_1 \wedge \mathbf{x}_4 } \right) \mathbf{x}_2 \cdot \partial_3 \mathbf{f}-\left( { \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) \mathbf{x}_1 \cdot \partial_3 \mathbf{f} \\ &\quad+ \left( { \mathbf{x}_1 \wedge \mathbf{x}_3 } \right) \mathbf{x}_4 \cdot \partial_2 \mathbf{f}-\left( { \mathbf{x}_1 \wedge \mathbf{x}_4 } \right) \mathbf{x}_3 \cdot \partial_2 \mathbf{f}+\left( { \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \mathbf{x}_1 \cdot \partial_2 \mathbf{f} \\ &\quad-\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \mathbf{x}_4 \cdot \partial_1 \mathbf{f}+\left( { \mathbf{x}_2 \wedge \mathbf{x}_4 } \right) \mathbf{x}_3 \cdot \partial_1 \mathbf{f}-\left( { \mathbf{x}_3 \wedge \mathbf{x}_4 } \right) \mathbf{x}_2 \cdot \partial_1 \mathbf{f} \\ &\qquad} \Bigr) \\ &=d^4 u\Bigl( {\mathbf{x}_1 \wedge \mathbf{x}_2 \partial_{[4} f_{3]}+\mathbf{x}_1 \wedge \mathbf{x}_3 \partial_{[2} f_{4]}+\mathbf{x}_1 \wedge \mathbf{x}_4 \partial_{[3} f_{2]}+\mathbf{x}_2 \wedge \mathbf{x}_3 \partial_{[4} f_{1]}+\mathbf{x}_2 \wedge \mathbf{x}_4 \partial_{[1} f_{3]}+\mathbf{x}_3 \wedge \mathbf{x}_4 \partial_{[2} f_{1]}} \Bigr) \\ &=- \frac{1}{2} d^4 u \epsilon^{abcd} \mathbf{x}_a \wedge \mathbf{x}_b \partial_{c} f_{d}. \qquad\square\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.151)

### (d) Three parameter volume, curl of bivector

\begin{aligned}\begin{aligned}d^3 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge B } \right)&=d^3 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \mathbf{x}^i } \Bigr) \cdot \partial_i B \\ &=d^3 u\Bigl( { \left( { \mathbf{x}_1 \wedge \mathbf{x}_2 } \right) \cdot \partial_3 B+\left( { \mathbf{x}_3 \wedge \mathbf{x}_1 } \right) \cdot \partial_2 B+\left( { \mathbf{x}_2 \wedge \mathbf{x}_3 } \right) \cdot \partial_1 B} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \partial_3 B } \right) -\mathbf{x}_2 \cdot \left( { \mathbf{x}_1 \cdot \partial_3 B } \right) \\ &\qquad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_1 \cdot \partial_2 B } \right) -\mathbf{x}_1 \cdot \left( { \mathbf{x}_3 \cdot \partial_2 B } \right) \\ &\qquad +\mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \partial_1 B } \right) -\mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot \partial_1 B } \right)} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \partial_3 B - \mathbf{x}_3 \cdot \partial_2 B } \right) \\ &\qquad +\mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \partial_1 B - \mathbf{x}_1 \cdot \partial_3 B } \right) \\ &\qquad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_1 \cdot \partial_2 B - \mathbf{x}_2 \cdot \partial_1 B } \right)} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( {\mathbf{x}_1 \cdot \left( { \partial_3 \left( { \mathbf{x}_2 \cdot B} \right) - \partial_2 \left( { \mathbf{x}_3 \cdot B} \right) } \right) \\ &\qquad +\mathbf{x}_2 \cdot \left( { \partial_1 \left( { \mathbf{x}_3 \cdot B} \right) - \partial_3 \left( { \mathbf{x}_1 \cdot B} \right) } \right) \\ &\qquad +\mathbf{x}_3 \cdot \left( { \partial_2 \left( { \mathbf{x}_1 \cdot B} \right) - \partial_1 \left( { \mathbf{x}_2 \cdot B} \right) } \right)} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( {\partial_2 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_1 \cdot B} \right) } \right) - \partial_3 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_1 \cdot B} \right) } \right) \\ &\qquad+ \partial_3 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot B} \right) } \right) - \partial_1 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot B} \right) } \right) \\ &\qquad+ \partial_1 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot B} \right) } \right) - \partial_2 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_3 \cdot B} \right) } \right)} \Bigr) \\ &=\frac{1}{2} d^3 u\Bigl( {\partial_2 B_{13} - \partial_3 B_{12}+\partial_3 B_{21} - \partial_1 B_{23}+\partial_1 B_{32} - \partial_2 B_{31}} \Bigr) \\ &=d^3 u\Bigl( {\partial_2 B_{13}+\partial_3 B_{21}+\partial_1 B_{32}} \Bigr) \\ &= - \frac{1}{2} d^3 u \epsilon^{abc} \partial_a B_{bc}. \qquad\square\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.152)

### (e) Four parameter volume, curl of bivector

To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\mathbf{x}_{abc}$ for $\mathbf{x}_a \wedge \mathbf{x}_b \wedge \mathbf{x}_c$. The expansion is

\begin{aligned}\begin{aligned}d^4 \mathbf{x} \cdot \left( \boldsymbol{\partial} \wedge B \right) &= d^4 u \left( \mathbf{x}_{1234} \cdot \mathbf{x}^i \right) \cdot \partial_i B \\ &= d^4 u\left( \mathbf{x}_{123} \cdot \partial_4 B - \mathbf{x}_{124} \cdot \partial_3 B + \mathbf{x}_{134} \cdot \partial_2 B - \mathbf{x}_{234} \cdot \partial_1 B \right) \\ &= d^4 u \Bigl( \mathbf{x}_1 \left( \mathbf{x}_{23} \cdot \partial_4 B \right) + \mathbf{x}_2 \left( \mathbf{x}_{32} \cdot \partial_4 B \right) + \mathbf{x}_3 \left( \mathbf{x}_{12} \cdot \partial_4 B \right) \\ &\qquad - \mathbf{x}_1 \left( \mathbf{x}_{24} \cdot \partial_3 B \right) - \mathbf{x}_2 \left( \mathbf{x}_{41} \cdot \partial_3 B \right) - \mathbf{x}_4 \left( \mathbf{x}_{12} \cdot \partial_3 B \right) \\ &\qquad + \mathbf{x}_1 \left( \mathbf{x}_{34} \cdot \partial_2 B \right) + \mathbf{x}_3 \left( \mathbf{x}_{41} \cdot \partial_2 B \right) + \mathbf{x}_4 \left( \mathbf{x}_{13} \cdot \partial_2 B \right) \\ &\qquad - \mathbf{x}_2 \left( \mathbf{x}_{34} \cdot \partial_1 B \right) - \mathbf{x}_3 \left( \mathbf{x}_{42} \cdot \partial_1 B \right) - \mathbf{x}_4 \left( \mathbf{x}_{23} \cdot \partial_1 B \right)} \Bigr) \\ &= d^4 u \Bigl( \mathbf{x}_1 \left( \mathbf{x}_{23} \cdot \partial_4 B + \mathbf{x}_{42} \cdot \partial_3 B + \mathbf{x}_{34} \cdot \partial_2 B \right) \\ &\qquad + \mathbf{x}_2 \left( \mathbf{x}_{32} \cdot \partial_4 B + \mathbf{x}_{14} \cdot \partial_3 B + \mathbf{x}_{43} \cdot \partial_1 B \right) \\ &\qquad + \mathbf{x}_3 \left( \mathbf{x}_{12} \cdot \partial_4 B + \mathbf{x}_{41} \cdot \partial_2 B + \mathbf{x}_{24} \cdot \partial_1 B \right) \\ &\qquad + \mathbf{x}_4 \left( \mathbf{x}_{21} \cdot \partial_3 B + \mathbf{x}_{13} \cdot \partial_2 B + \mathbf{x}_{32} \cdot \partial_1 B \right)} \Bigr) \\ &= - \frac{1}{2} d^4 u \epsilon^{a b c d} \mathbf{x}_a \partial_b B_{c d}. \qquad\square\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.153)

This last step uses an intermediate result from the eq. 1.0.152 expansion above, since each of the four terms has the same structure we have previously observed.

### (f) Four parameter volume, curl of trivector

Using the $\mathbf{x}_{ijk}$ shorthand again, the initial expansion gives

\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge T } \right)=d^4 u\left( {\mathbf{x}_{123} \cdot \partial_4 T - \mathbf{x}_{124} \cdot \partial_3 T + \mathbf{x}_{134} \cdot \partial_2 T - \mathbf{x}_{234} \cdot \partial_1 T} \right).\end{aligned} \hspace{\stretch{1}}(1.0.153)

Applying lemma 4 to expand the inner products within the braces we have

\begin{aligned}\begin{aligned}\mathbf{x}_{123} \cdot \partial_4 T-&\mathbf{x}_{124} \cdot \partial_3 T+\mathbf{x}_{134} \cdot \partial_2 T-\mathbf{x}_{234} \cdot \partial_1 T \\ &=\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \partial_4 T } \right) } \right)-\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_4 \cdot \partial_3 T } \right) } \right) \\ &\quad +\underbrace{\mathbf{x}_1 \cdot \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \partial_2 T } \right) } \right)-\mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \partial_1 T } \right) } \right)}_{\text{Apply cyclic permutations}}\\ &=\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot \partial_4 T } \right) } \right)-\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_4 \cdot \partial_3 T } \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot \partial_2 T } \right) } \right)-\mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_2 \cdot \partial_1 T } \right) } \right) \\ &=\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot\left( {\mathbf{x}_3 \cdot \partial_4 T-\mathbf{x}_4 \cdot \partial_3 T} \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( {\mathbf{x}_1 \cdot \partial_2 T-\mathbf{x}_2 \cdot \partial_1 T} \right) } \right) \\ &=\mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot\left( {\partial_4 \left( { \mathbf{x}_3 \cdot T } \right)-\partial_3 \left( { \mathbf{x}_4 \cdot T } \right)} \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( {\partial_2 \left( { \mathbf{x}_1 \cdot T } \right)-\partial_1 \left( { \mathbf{x}_2 \cdot T } \right)} \right) } \right) \\ &=\mathbf{x}_1 \cdot \partial_4 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot T } \right) } \right)+\mathbf{x}_2 \cdot \partial_3 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_4 \cdot T } \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \partial_2 \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot T } \right) } \right)+\mathbf{x}_4 \cdot \partial_1 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot T } \right) } \right) \\ &-\mathbf{x}_1 \cdot \left( { \left( { \partial_4 \mathbf{x}_2} \right) \cdot \left( { \mathbf{x}_3 \cdot T } \right) } \right)-\mathbf{x}_2 \cdot \left( { \left( { \partial_3 \mathbf{x}_1} \right) \cdot \left( { \mathbf{x}_4 \cdot T } \right) } \right) \\ &\quad -\mathbf{x}_3 \cdot \left( { \left( { \partial_2 \mathbf{x}_4} \right) \cdot \left( { \mathbf{x}_1 \cdot T } \right) } \right)-\mathbf{x}_4 \cdot \left( { \left( { \partial_1 \mathbf{x}_3} \right) \cdot \left( { \mathbf{x}_2 \cdot T } \right) } \right) \\ &=\mathbf{x}_1 \cdot \partial_4 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot T } \right) } \right)+\mathbf{x}_2 \cdot \partial_3 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_4 \cdot T } \right) } \right) \\ &\quad +\mathbf{x}_3 \cdot \partial_2 \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot T } \right) } \right)+\mathbf{x}_4 \cdot \partial_1 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot T } \right) } \right) \\ &+\frac{\partial^2 \mathbf{x}}{\partial u^4 \partial u^2}\cdot\not{{\left( {\mathbf{x}_1 \cdot \left( { \mathbf{x}_3 \cdot T } \right)+\mathbf{x}_3 \cdot \left( { \mathbf{x}_1 \cdot T } \right)} \right)}} \\ &\quad +\frac{\partial^2 \mathbf{x}}{\partial u^1 \partial u^3}\cdot\not{{\left( {\mathbf{x}_2 \cdot \left( { \mathbf{x}_4 \cdot T } \right)+\mathbf{x}_4 \cdot \left( { \mathbf{x}_2 \cdot T } \right)} \right)}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.155)

We can cancel those last terms using lemma 5. Using the same reverse chain rule expansion once more we have

\begin{aligned}\begin{aligned}\mathbf{x}_{123} \cdot \partial_4 T-&\mathbf{x}_{124} \cdot \partial_3 T+\mathbf{x}_{134} \cdot \partial_2 T-\mathbf{x}_{234} \cdot \partial_1 T \\ &=\partial_4 \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot T } \right) } \right) } \right)+\partial_3 \left( { \mathbf{x}_2 \cdot \left( { \mathbf{x}_1 \cdot \left( { \mathbf{x}_4 \cdot T } \right) } \right) } \right)+\partial_2 \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot T } \right) } \right) } \right)+\partial_1 \left( { \mathbf{x}_4 \cdot \left( { \mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot T } \right) } \right) } \right) \\ &-\left( { \partial_4 \mathbf{x}_1} \right)\cdot\not{{\left( {\mathbf{x}_2 \cdot \left( { \mathbf{x}_3 \cdot T } \right)+\mathbf{x}_3 \cdot \left( { \mathbf{x}_2 \cdot T } \right)} \right)}}-\left( { \partial_3 \mathbf{x}_2} \right) \cdot\not{{\left( {\mathbf{x}_1 \cdot \left( { \mathbf{x}_4 \cdot T } \right)\mathbf{x}_4 \cdot \left( { \mathbf{x}_1 \cdot T } \right)} \right)}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.156)

or

\begin{aligned}d^4 \mathbf{x} \cdot \left( { \boldsymbol{\partial} \wedge T } \right)=d^4 u\Bigl( {\partial_4 T_{3 2 1}+\partial_3 T_{4 1 2}+\partial_2 T_{1 4 3}+\partial_1 T_{2 3 4}} \Bigr).\end{aligned} \hspace{\stretch{1}}(1.0.156)

The final result follows after permuting the indices slightly.

### Lemma 1. Distribution of inner products

Given two blades $A_s, B_r$ with grades subject to $s > r > 0$, and a vector $b$, the inner product distributes according to

\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right) = \left( { A_s \cdot b } \right) \cdot B_r.\end{aligned}

This will allow us, for example, to expand a general inner product of the form $d^k \mathbf{x} \cdot (\boldsymbol{\partial} \wedge F)$.

The proof is straightforward, but also mechanical. Start by expanding the wedge and dot products within a grade selection operator

\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right)={\left\langle{{A_s (b \wedge B_r)}}\right\rangle}_{{s - (r + 1)}}=\frac{1}{2} {\left\langle{{A_s \left( {b B_r + (-1)^{r} B_r b} \right) }}\right\rangle}_{{s - (r + 1)}}\end{aligned} \hspace{\stretch{1}}(1.158)

Solving for $B_r b$ in

\begin{aligned}2 b \cdot B_r = b B_r - (-1)^{r} B_r b,\end{aligned} \hspace{\stretch{1}}(1.159)

we have

\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right)=\frac{1}{2} {\left\langle{{ A_s b B_r + A_s \left( { b B_r - 2 b \cdot B_r } \right) }}\right\rangle}_{{s - (r + 1)}}={\left\langle{{ A_s b B_r }}\right\rangle}_{{s - (r + 1)}}-\not{{{\left\langle{{ A_s \left( { b \cdot B_r } \right) }}\right\rangle}_{{s - (r + 1)}}}}.\end{aligned} \hspace{\stretch{1}}(1.160)

The last term above is zero since we are selecting the $s - r - 1$ grade element of a multivector with grades $s - r + 1$ and $s + r - 1$, which has no terms for $r > 0$. Now we can expand the $A_s b$ multivector product, for

\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right)={\left\langle{{ \left( { A_s \cdot b + A_s \wedge b} \right) B_r }}\right\rangle}_{{s - (r + 1)}}.\end{aligned} \hspace{\stretch{1}}(1.161)

The latter multivector (with the wedge product factor) above has grades $s + 1 - r$ and $s + 1 + r$, so this selection operator finds nothing. This leaves

\begin{aligned}A_s \cdot \left( { b \wedge B_r } \right)={\left\langle{{\left( { A_s \cdot b } \right) \cdot B_r+ \left( { A_s \cdot b } \right) \wedge B_r}}\right\rangle}_{{s - (r + 1)}}.\end{aligned} \hspace{\stretch{1}}(1.162)

The first dot products term has grade $s - 1 - r$ and is selected, whereas the wedge term has grade $s - 1 + r \ne s - r - 1$ (for $r > 0$). $\qquad\square$

### Lemma 2. Distribution of two bivectors

For vectors $\mathbf{a}$, $\mathbf{b}$, and bivector $B$, we have

\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} } \right) \cdot B = \frac{1}{2} \left( {\mathbf{a} \cdot \left( { \mathbf{b} \cdot B } \right)-\mathbf{b} \cdot \left( { \mathbf{a} \cdot B } \right)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.163)

Proof follows by applying the scalar selection operator, expanding the wedge product within it, and eliminating any of the terms that cannot contribute grade zero values

\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} } \right) \cdot B &= \left\langle{{\frac{1}{2} \Bigl( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \Bigr) B}}\right\rangle \\ &= \frac{1}{2}\left\langle{{\mathbf{a} \left( { \mathbf{b} \cdot B + \not{{ \mathbf{b} \wedge B }} } \right)-\mathbf{b} \left( { \mathbf{a} \cdot B + \not{{ \mathbf{a} \wedge B }} } \right)}}\right\rangle \\ &= \frac{1}{2}\left\langle{{\mathbf{a} \cdot \left( { \mathbf{b} \cdot B } \right)+\not{{\mathbf{a} \wedge \left( { \mathbf{b} \cdot B } \right)}}-\mathbf{b} \cdot \left( { \mathbf{a} \cdot B } \right)-\not{{\mathbf{b} \wedge \left( { \mathbf{a} \cdot B } \right)}}}}\right\rangle \\ &= \frac{1}{2}\Bigl( {\mathbf{a} \cdot \left( { \mathbf{b} \cdot B } \right)-\mathbf{b} \cdot \left( { \mathbf{a} \cdot B } \right)} \Bigr)\qquad\square\end{aligned} \hspace{\stretch{1}}(1.0.163)

### Lemma 3. Inner product of trivector with bivector

Given a bivector $B$, and trivector $\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}$ where $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are vectors, the inner product is

\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} } \right) \cdot B=\mathbf{a} \Bigl( { \left( { \mathbf{b} \wedge \mathbf{c} } \right) \cdot B } \Bigr)+\mathbf{b} \Bigl( { \left( { \mathbf{c} \wedge \mathbf{a} } \right) \cdot B } \Bigr)+\mathbf{c} \Bigl( { \left( { \mathbf{a} \wedge \mathbf{b} } \right) \cdot B } \Bigr).\end{aligned} \hspace{\stretch{1}}(1.165)

This is also problem 1.1(c) from Exercises 2.1 in [3], and submits to a dumb expansion in successive dot products with a final regrouping. With $B = \mathbf{u} \wedge \mathbf{v}$

\begin{aligned}\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right)\cdot B&={\left\langle{{\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \left( \mathbf{u} \wedge \mathbf{v} \right) }}\right\rangle}_{1} \\ &={\left\langle{{\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right)\left(\mathbf{u} \mathbf{v}- \mathbf{u} \cdot \mathbf{v}\right) }}\right\rangle}_{1} \\ &=\left(\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot \mathbf{u} \right) \cdot \mathbf{v} \\ &=\left( \mathbf{a} \wedge \mathbf{b} \right) \cdot \mathbf{v} \left( \mathbf{c} \cdot \mathbf{u} \right)+\left( \mathbf{c} \wedge \mathbf{a} \right) \cdot \mathbf{v} \left( \mathbf{b} \cdot \mathbf{u} \right)+\left( \mathbf{b} \wedge \mathbf{c} \right) \cdot \mathbf{v} \left( \mathbf{a} \cdot \mathbf{u} \right) \\ &=\mathbf{a}\left( \mathbf{b} \cdot \mathbf{v} \right)\left( \mathbf{c} \cdot \mathbf{u} \right)-\mathbf{b}\left( \mathbf{a} \cdot \mathbf{v} \right)\left( \mathbf{c} \cdot \mathbf{u} \right) \\ &\quad +\mathbf{c}\left( \mathbf{a} \cdot \mathbf{v} \right)\left( \mathbf{b} \cdot \mathbf{u} \right)-\mathbf{a}\left( \mathbf{c} \cdot \mathbf{v} \right)\left( \mathbf{b} \cdot \mathbf{u} \right) \\ &\quad +\mathbf{b}\left( \mathbf{c} \cdot \mathbf{v} \right)\left( \mathbf{a} \cdot \mathbf{u} \right)-\mathbf{c}\left( \mathbf{b} \cdot \mathbf{v} \right)\left( \mathbf{a} \cdot \mathbf{u} \right) \\ &=\mathbf{a}\left( \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) - \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{b} \cdot \mathbf{u} \right) \right)\\ &\quad +\mathbf{b}\left( \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) - \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) \right)\\ &\quad +\mathbf{c}\left( \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{b} \cdot \mathbf{u} \right) - \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) \right) \\ &=\mathbf{a}\left( \mathbf{b} \wedge \mathbf{c} \right)\cdot\left( \mathbf{u} \wedge \mathbf{v} \right)\\ &\quad +\mathbf{b}\left( \mathbf{c} \wedge \mathbf{a} \right)\cdot\left( \mathbf{u} \wedge \mathbf{v} \right)\\ &\quad +\mathbf{c}\left( \mathbf{a} \wedge \mathbf{b} \right) \cdot\left( \mathbf{u} \wedge \mathbf{v} \right)\\ &=\mathbf{a}\left( \mathbf{b} \wedge \mathbf{c} \right)\cdot B+\mathbf{b}\left( \mathbf{c} \wedge \mathbf{a} \right) \cdot B+\mathbf{c}\left( \mathbf{a} \wedge \mathbf{b} \right)\cdot B. \qquad\square\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.166)

### Lemma 4. Distribution of two trivectors

Given a trivector $T$ and three vectors $\mathbf{a}, \mathbf{b}$, and $\mathbf{c}$, the entire inner product can be expanded in terms of any successive set inner products, subject to change of sign with interchange of any two adjacent vectors within the dot product sequence

\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} } \right) \cdot T &= \mathbf{a} \cdot \left( { \mathbf{b} \cdot \left( { \mathbf{c} \cdot T } \right) } \right) \\ &= -\mathbf{a} \cdot \left( { \mathbf{c} \cdot \left( { \mathbf{b} \cdot T } \right) } \right) \\ &= \mathbf{b} \cdot \left( { \mathbf{c} \cdot \left( { \mathbf{a} \cdot T } \right) } \right) \\ &= - \mathbf{b} \cdot \left( { \mathbf{a} \cdot \left( { \mathbf{c} \cdot T } \right) } \right) \\ &= \mathbf{c} \cdot \left( { \mathbf{a} \cdot \left( { \mathbf{b} \cdot T } \right) } \right) \\ &= - \mathbf{c} \cdot \left( { \mathbf{b} \cdot \left( { \mathbf{a} \cdot T } \right) } \right).\end{aligned} \hspace{\stretch{1}}(1.167)

To show this, we first expand within a scalar selection operator

\begin{aligned}\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} } \right) \cdot T&=\left\langle{{\left( { \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} } \right) T}}\right\rangle \\ &=\frac{1}{6}\left\langle{{ \mathbf{a} \mathbf{b} \mathbf{c} T- \mathbf{a} \mathbf{c} \mathbf{b} T+ \mathbf{b} \mathbf{c} \mathbf{a} T- \mathbf{b} \mathbf{a} \mathbf{b} T+ \mathbf{c} \mathbf{a} \mathbf{b} T- \mathbf{c} \mathbf{b} \mathbf{a} T}}\right\rangle \\ \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.168)

Now consider any single term from the scalar selection, such as the first. This can be reordered using the vector dot product identity

\begin{aligned}\left\langle{{ \mathbf{a} \mathbf{b} \mathbf{c} T}}\right\rangle=\left\langle{{ \mathbf{a} \left( { -\mathbf{c} \mathbf{b} + 2 \mathbf{b} \cdot \mathbf{c} } \right) T}}\right\rangle=-\left\langle{{ \mathbf{a} \mathbf{c} \mathbf{b} T}}\right\rangle+2 \mathbf{b} \cdot \mathbf{c} \not{{\left\langle{{ \mathbf{a} T}}\right\rangle}}.\end{aligned} \hspace{\stretch{1}}(1.0.168)

The vector-trivector product in the latter grade selection operation above contributes only bivector and quadvector terms, thus contributing nothing. This can be repeated, showing that

\begin{aligned} \left\langle{{ \mathbf{a} \mathbf{b} \mathbf{c} T }}\right\rangle &= - \left\langle{{ \mathbf{a} \mathbf{c} \mathbf{b} T }}\right\rangle \\ &= + \left\langle{{ \mathbf{b} \mathbf{c} \mathbf{a} T }}\right\rangle \\ &= - \left\langle{{ \mathbf{b} \mathbf{a} \mathbf{c} T }}\right\rangle \\ &= + \left\langle{{ \mathbf{c} \mathbf{a} \mathbf{b} T }}\right\rangle \\ &= - \left\langle{{ \mathbf{c} \mathbf{b} \mathbf{a} T }}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.0.168)

Substituting this back into eq. 1.0.168 proves lemma 4.

### Lemma 5. Permutation of two successive dot products with trivector

Given a trivector $T$ and two vectors $\mathbf{a}$ and $\mathbf{b}$, alternating the order of the dot products changes the sign

\begin{aligned}\mathbf{a} \cdot \left( { \mathbf{b} \cdot T } \right)=-\mathbf{b} \cdot \left( { \mathbf{a} \cdot T } \right).\end{aligned} \hspace{\stretch{1}}(1.171)

This and lemma 4 are clearly examples of a more general identity, but I’ll not try to prove that here. To show this one, we have

\begin{aligned}\mathbf{a} \cdot \left( { \mathbf{b} \cdot T } \right) &= {\left\langle{{ \mathbf{a} \left( { \mathbf{b} \cdot T } \right) }}\right\rangle}_{1} \\ &= \frac{1}{2}{\left\langle{{ \mathbf{a} \mathbf{b} T + \mathbf{a} T \mathbf{b} }}\right\rangle}_{1} \\ &= \frac{1}{2}{\left\langle{{ \left( { -\mathbf{b} \mathbf{a} + \not{{2 \mathbf{a} \cdot \mathbf{b}}}} \right) T + \left( { \mathbf{a} \cdot T} \right) \mathbf{b} + \not{{ \mathbf{a} \wedge T}} \mathbf{b} }}\right\rangle}_{1} \\ &= \frac{1}{2}\left( {-\mathbf{b} \cdot \left( { \mathbf{a} \cdot T } \right)+\left( { \mathbf{a} \cdot T } \right) \cdot \mathbf{b}} \right) \\ &= -\mathbf{b} \cdot \left( { \mathbf{a} \cdot T } \right). \qquad\square\end{aligned} \hspace{\stretch{1}}(1.172)

Cancellation of terms above was because they could not contribute to a grade one selection. We also employed the relation $\mathbf{x} \cdot B = - B \cdot \mathbf{x}$ for bivector $B$ and vector $\mathbf{x}$.

### Lemma 6. Duality in a plane

For a vector $\mathbf{a}$, and a plane containing $\mathbf{a}$ and $\mathbf{b}$, the dual $\mathbf{a}^{*}$ of this vector with respect to this plane is

\begin{aligned}\mathbf{a}^{*} = \frac{\mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right)}{\left( {\mathbf{a} \wedge \mathbf{b}} \right)^2},\end{aligned} \hspace{\stretch{1}}(1.173)

Satisfying

\begin{aligned}\mathbf{a}^{*} \cdot \mathbf{a} = 1,\end{aligned} \hspace{\stretch{1}}(1.174)

and

\begin{aligned}\mathbf{a}^{*} \cdot \mathbf{b} = 0.\end{aligned} \hspace{\stretch{1}}(1.175)

\begin{aligned}\mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right)=\left( { \mathbf{b} \cdot \mathbf{a} } \right) \mathbf{b}-\mathbf{b}^2 \mathbf{a}.\end{aligned} \hspace{\stretch{1}}(1.176)

Dotting with $\mathbf{a}$ we have

\begin{aligned}\mathbf{a} \cdot \left( { \mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right) } \right)=\mathbf{a} \cdot \left( {\left( { \mathbf{b} \cdot \mathbf{a} } \right) \mathbf{b}-\mathbf{b}^2 \mathbf{a}} \right)=\left( { \mathbf{b} \cdot \mathbf{a} } \right)^2 - \mathbf{b}^2 \mathbf{a}^2,\end{aligned} \hspace{\stretch{1}}(1.177)

but dotting with $\mathbf{b}$ yields zero

\begin{aligned}\mathbf{b} \cdot \left( { \mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right) } \right) &= \mathbf{b} \cdot \left( {\left( { \mathbf{b} \cdot \mathbf{a} } \right) \mathbf{b}-\mathbf{b}^2 \mathbf{a}} \right) \\ &= \left( { \mathbf{b} \cdot \mathbf{a} } \right) \mathbf{b}^2 - \mathbf{b}^2 \left( { \mathbf{a} \cdot \mathbf{b} } \right) \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.178)

To complete the proof, we note that the product in eq. 1.177 is just the wedge squared

\begin{aligned}\left( { \mathbf{a} \wedge \mathbf{b}} \right)^2 &= \left\langle{{\left( { \mathbf{a} \wedge \mathbf{b} } \right)^2}}\right\rangle \\ &= \left\langle{{\left( { \mathbf{a} \mathbf{b} - \mathbf{a} \cdot \mathbf{b} } \right)\left( { \mathbf{a} \mathbf{b} - \mathbf{a} \cdot \mathbf{b} } \right)}}\right\rangle \\ &= \left\langle{{\mathbf{a} \mathbf{b} \mathbf{a} \mathbf{b} - 2 \left( {\mathbf{a} \cdot \mathbf{b}} \right) \mathbf{a} \mathbf{b}}}\right\rangle+\left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 \\ &= \left\langle{{\mathbf{a} \mathbf{b} \left( { -\mathbf{b} \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} } \right)}}\right\rangle-\left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 \\ &= \left( { \mathbf{a} \cdot \mathbf{b} } \right)^2-\mathbf{a}^2 \mathbf{b}^2.\end{aligned} \hspace{\stretch{1}}(1.179)

This duality relation can be recast with a linear denominator

\begin{aligned}\mathbf{a}^{*} &= \frac{\mathbf{b} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right)}{\left( {\mathbf{a} \wedge \mathbf{b}} \right)^2} \\ &= \mathbf{b} \frac{\mathbf{a} \wedge \mathbf{b} }{\left( {\mathbf{a} \wedge \mathbf{b}} \right)^2} \\ &= \mathbf{b} \frac{\mathbf{a} \wedge \mathbf{b} }{\left\lvert {\mathbf{a} \wedge \mathbf{b} } \right\rvert} \frac{\left\lvert {\mathbf{a} \wedge \mathbf{b}} \right\rvert}{\mathbf{a} \wedge \mathbf{b} }\frac{1}{{\left( {\mathbf{a} \wedge \mathbf{b}} \right)}},\end{aligned} \hspace{\stretch{1}}(1.180)

or

\begin{aligned}\mathbf{a}^{*} = \mathbf{b} \frac{1}{{\left( {\mathbf{a} \wedge \mathbf{b}} \right)}}.\end{aligned} \hspace{\stretch{1}}(1.0.181)

We can use this form after scaling it appropriately to express duality in terms of the pseudoscalar.

### Lemma 7. Dual vector in a three vector subspace

In the subspace spanned by $\left\{ {\mathbf{a}, \mathbf{b}, \mathbf{c}} \right\}$, the dual of $\mathbf{a}$ is

\begin{aligned}\mathbf{a}^{*} = \mathbf{b} \wedge \mathbf{c} \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}},\end{aligned}

Consider the dot product of $\hat{\mathbf{a}}^{*}$ with $\mathbf{u} \in \left\{ {\mathbf{a}, \mathbf{b}, \mathbf{c}} \right\}$.

\begin{aligned}\mathbf{u} \cdot \mathbf{a}^{*} &= \left\langle{{ \mathbf{u} \mathbf{b} \wedge \mathbf{c} \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle \\ &= \left\langle{{ \mathbf{u} \cdot \left( { \mathbf{b} \wedge \mathbf{c}} \right) \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle+\left\langle{{ \mathbf{u} \wedge \mathbf{b} \wedge \mathbf{c} \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle \\ &= \not{{\left\langle{{ \left( { \left( { \mathbf{u} \cdot \mathbf{b}} \right) \mathbf{c}-\left( {\mathbf{u} \cdot \mathbf{c}} \right) \mathbf{b}} \right)\frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle}}+\left\langle{{ \mathbf{u} \wedge \mathbf{b} \wedge \mathbf{c} \frac{1}{{\mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c}}} }}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.182)

The canceled term is eliminated since it is the product of a vector and trivector producing no scalar term. Substituting $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and noting that $\mathbf{u} \wedge \mathbf{u} = 0$, we have

\begin{aligned}\begin{aligned}\mathbf{a} \cdot \mathbf{a}^{*} &= 1 \\ \mathbf{b} \cdot \mathbf{a}^{*} &= 0 \\ \mathbf{c} \cdot \mathbf{a}^{*} &= 0.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.183)

### Lemma 8. Pseudoscalar selection

For grade $k$ blade $K \in \bigwedge^k$ (i.e. a pseudoscalar), and vectors $\mathbf{a}, \mathbf{b}$, the grade $k$ selection of this blade sandwiched between the vectors is

\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} = (-1)^{k+1} {\left\langle{{K a b}}\right\rangle}_{k} = (-1)^{k+1} K \left( { \mathbf{a} \cdot \mathbf{b}} \right).\end{aligned}

To show this, we have to consider even and odd grades separately. First for even $k$ we have

\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} &= {\left\langle{{ \left( { \mathbf{a} \cdot K + \not{{\mathbf{a} \wedge K}}} \right) \mathbf{b} }}\right\rangle}_{k} \\ &= \frac{1}{2} {\left\langle{{ \left( { \mathbf{a} K - K \mathbf{a} } \right) \mathbf{b} }}\right\rangle}_{k} \\ &= \frac{1}{2} {\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k}-\frac{1}{2} {\left\langle{{ K \mathbf{a} \mathbf{b} }}\right\rangle}_{k},\end{aligned} \hspace{\stretch{1}}(1.184)

or

\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} = -{\left\langle{{ K \mathbf{a} \mathbf{b} }}\right\rangle}_{k} = -K \left( { \mathbf{a} \cdot \mathbf{b}} \right).\end{aligned} \hspace{\stretch{1}}(1.185)

Similarly for odd $k$, we have

\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} &= {\left\langle{{ \left( { \mathbf{a} \cdot K + \not{{\mathbf{a} \wedge K}}} \right) \mathbf{b} }}\right\rangle}_{k} \\ &= \frac{1}{2} {\left\langle{{ \left( { \mathbf{a} K + K \mathbf{a} } \right) \mathbf{b} }}\right\rangle}_{k} \\ &= \frac{1}{2} {\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k}+\frac{1}{2} {\left\langle{{ K \mathbf{a} \mathbf{b} }}\right\rangle}_{k},\end{aligned} \hspace{\stretch{1}}(1.186)

or

\begin{aligned}{\left\langle{{ \mathbf{a} K \mathbf{b} }}\right\rangle}_{k} = {\left\langle{{ K \mathbf{a} \mathbf{b} }}\right\rangle}_{k} = K \left( { \mathbf{a} \cdot \mathbf{b}} \right).\end{aligned} \hspace{\stretch{1}}(1.187)

Adjusting for the signs completes the proof.

# References

[1] John Denker. Magnetic field for a straight wire., 2014. URL http://www.av8n.com/physics/straight-wire.pdf. [Online; accessed 11-May-2014].

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

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

[4] Peeter Joot. Collection of old notes on Stokes theorem in Geometric algebra, 2014. URL https://sites.google.com/site/peeterjoot3/math2014/bigCollectionOfPartiallyIncorrectStokesTheoremMusings.pdf.

[5] Peeter Joot. Synposis of old notes on Stokes theorem in Geometric algebra, 2014. URL https://sites.google.com/site/peeterjoot3/math2014/synopsisOfBigCollectionOfPartiallyIncorrectStokesTheoremMusings.pdf.

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

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

[8] Michael Spivak. Calculus on manifolds, volume 1. Benjamin New York, 1965.

## Tangent planes and normals in three and four dimensions

Posted by peeterjoot on January 4, 2013

# Motivation

I was reviewing the method of Lagrange in my old first year calculus book [1] and found that I needed a review of some of the geometry ideas associated with the gradient (that it is normal to the surface). The approach in the text used 3D level surfaces $f(x, y, z) = c$, which is general but not the most intuitive.

If we define a surface in the simpler explicit form $z = f(x, y)$, then how would you show this normal property? Here we explore this in 3D and 4D, using geometric and wedge products to express the tangent planes and tangent volumes respectively.

In the 4D approach, with a vector $x$ defined by coordinates $x^\mu$ and basis $\{\gamma_\mu\}$ so that

\begin{aligned}x = \gamma_\mu x^\mu,\end{aligned} \hspace{\stretch{1}}(1.1.1)

the reciprocal basis ${\gamma^\mu}$ is defined implicitly by the dot product relations

\begin{aligned}\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu.\end{aligned} \hspace{\stretch{1}}(1.1.2)

Assuming such a basis makes the result general enough that the 4D (or a trivial generalization to N dimensions) holds for both Euclidean spaces as well as mixed metric (i.e. Minkowski) spaces, and avoids having to detail the specific metric in question.

# 3D surface

We start by considering figure 1:

Figure 1: A portion of a surface in 3D

We wish to determine the bivector for the tangent plane in the neighbourhood of the point $\mathbf{p}$

\begin{aligned}\mathbf{p} = ( x, y, f(x, y) ),\end{aligned} \hspace{\stretch{1}}(1.2.3)

then using duality determine the normal vector to that plane at this point. Holding either of the two free parameters constant, we find the tangent vectors on that surface to be

\begin{aligned}\mathbf{p}_1 = \left( dx, 0, \frac{\partial {f}}{\partial {x}} dx \right) \propto \left( 1, 0, \frac{\partial {f}}{\partial {x}} \right) \end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\mathbf{p}_2 = \left( 0, dy, \frac{\partial {f}}{\partial {y}} dy \right) \propto \left( 0, 1, \frac{\partial {f}}{\partial {y}} \right) \end{aligned} \hspace{\stretch{1}}(1.0.4b)

The tangent plane is then

\begin{aligned}\mathbf{p}_1 \wedge \mathbf{p}_2 &= \left( 1, 0, \frac{\partial {f}}{\partial {x}} \right) \wedge\left( 0, 1, \frac{\partial {f}}{\partial {y}} \right) \\ &= \left( \mathbf{e}_1 + \mathbf{e}_3 \frac{\partial {f}}{\partial {x}} \right) \wedge\left( \mathbf{e}_2 + \mathbf{e}_3 \frac{\partial {f}}{\partial {y}} \right) \\ &= \mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_1 \mathbf{e}_3 \frac{\partial {f}}{\partial {y}} + \mathbf{e}_3 \mathbf{e}_2 \frac{\partial {f}}{\partial {x}}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

We can factor out the pseudoscalar 3D volume element $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$, assuming a Euclidean space for which $\mathbf{e}_k^2 = 1$. That is

\begin{aligned}\mathbf{p}_1 \wedge \mathbf{p}_2 = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \left(\mathbf{e}_3- \mathbf{e}_2 \frac{\partial {f}}{\partial {y}} - \mathbf{e}_1 \frac{\partial {f}}{\partial {x}}\right)\end{aligned} \hspace{\stretch{1}}(1.0.6)

Multiplying through by the volume element $I$ we find that the normal to the surface at this point is

\begin{aligned}\mathbf{n} \propto -I(\mathbf{p}_1 \wedge \mathbf{p}_2) = \mathbf{e}_3- \mathbf{e}_1 \frac{\partial {f}}{\partial {x}}- \mathbf{e}_2 \frac{\partial {f}}{\partial {y}}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Observe that we can write this as

\begin{aligned}\boxed{\mathbf{n} = \boldsymbol{\nabla} ( z - f(x, y) ).}\end{aligned} \hspace{\stretch{1}}(1.0.8)

Let’s see how this works in 4D, so that we know how to handle the Minkowski spaces we find in special relativity.

# 4D surface

Now, let’s move up to one additional direction, with

\begin{aligned}x^3 = f(x^0, x^1, x^2).\end{aligned} \hspace{\stretch{1}}(1.0.9)

the differential of this is

\begin{aligned}dx^3 = \sum_{k=0}^2 \frac{\partial {f}}{\partial {x^k}} dx^k = \sum_{k=0}^2 \partial_k f dx^k .\end{aligned} \hspace{\stretch{1}}(1.0.10)

We are going to look at the 3-surface in the neighbourhood of the point

\begin{aligned}p = \left( x^0, x^1, x^2, x^3\right),\end{aligned} \hspace{\stretch{1}}(1.0.11)

so that the tangent vectors in the neighbourhood of this point are in the span of

\begin{aligned}dp = \left( x^0, x^1, x^2, \sum_{k=0}^2 \partial_k dx^k\right).\end{aligned} \hspace{\stretch{1}}(1.0.12)

In particular, in each of the directions we have

\begin{aligned}p_0 \propto ( 1, 0, 0, d_0 f)\end{aligned} \hspace{\stretch{1}}(1.0.13a)

\begin{aligned}p_1 \propto ( 0, 1, 0, d_1 f)\end{aligned} \hspace{\stretch{1}}(1.0.13b)

\begin{aligned}p_2 \propto ( 0, 0, 1, d_2 f)\end{aligned} \hspace{\stretch{1}}(1.0.13c)

Our tangent volume in this neighbourhood is

\begin{aligned}p_0 \wedge p_1 \wedge p_2&=\left( \gamma_0 + \gamma_3 \partial_0 f\right)\wedge\left( \gamma_1 + \gamma_3 \partial_1 f\right)\wedge\left( \gamma_2 + \gamma_3 \partial_2 f\right) \\ &=\left( \gamma_0 \gamma_1 + \gamma_0 \gamma_3 \partial_1 f+ \gamma_3 \gamma_1 \partial_0 f\right)\wedge\left( \gamma_2 + \gamma_3 \partial_2 f\right) \\ &=\gamma_{012} - \gamma_{023} \partial_1 f + \gamma_{123} \partial_0 f + \gamma_{013} \partial_2 f.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Here the shorthand $\gamma_{ijk} = \gamma_i \gamma_j \gamma_k$ has been used. Can we factor out a 4D pseudoscalar from this and end up with a coherent result. We have

\begin{aligned}\gamma_{0123} \gamma^3 = \gamma_{012}\end{aligned} \hspace{\stretch{1}}(1.0.15a)

\begin{aligned}\gamma_{0123} \gamma^1 = \gamma_{023}\end{aligned} \hspace{\stretch{1}}(1.0.15b)

\begin{aligned}\gamma_{0123} \gamma^0 = -\gamma_{123}\end{aligned} \hspace{\stretch{1}}(1.0.15c)

\begin{aligned}\gamma_{0123} \gamma^2 = -\gamma_{013}.\end{aligned} \hspace{\stretch{1}}(1.0.15d)

This gives us

\begin{aligned}d^3 p=p_0 \wedge p_1 \wedge p_2=\gamma_{0123} \left(\gamma^3 - \gamma^1 \partial_1 f- \gamma^0 \partial_0 f- \gamma^2 \partial_2 f\right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

With the usual 4d gradient definition (sum implied)

\begin{aligned}\nabla = \gamma^\mu \partial_\mu,\end{aligned} \hspace{\stretch{1}}(1.0.17)

we have

\begin{aligned}\nabla x^3 = \gamma^\mu \partial_\mu x^3 = \gamma^\mu {\delta_{\mu}}^3= \gamma^3,\end{aligned} \hspace{\stretch{1}}(1.0.18)

so we can write

\begin{aligned}d^3 p = \gamma_{0123} \nabla \left( x^3 - f(x^0, x^1, x^2) \right),\end{aligned} \hspace{\stretch{1}}(1.0.19)

so, finally, the “normal” to this surface volume element at this point is

\begin{aligned}\boxed{n = \nabla \left( x^3 - f(x^0, x^1, x^2) \right).}\end{aligned} \hspace{\stretch{1}}(1.0.20)

This is just like the 3D Euclidean result, with the exception that we need to look at the dual of a 3-volume “surface” instead of our normal 2d surface.

Also note that this is not a metric free result. The metric choice is built into the definition of the gradient 1.0.17 and its associated reciprocal basis. For example with a $1,3$ metric where $\gamma_0^2 = 1, \gamma_k^2 = -1$, we have $\gamma^0 = \gamma_0$ and $\gamma^k = -\gamma_k$.

# References

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

## Plane wave solutions of Maxwell’s equation using Geometric Algebra

Posted by peeterjoot on September 3, 2012

# Motivation

Study of reflection and transmission of radiation in isotropic, charge and current free, linear matter utilizes the plane wave solutions to Maxwell’s equations. These have the structure of phasor equations, with some specific constraints on the components and the exponents.

These constraints are usually derived starting with the plain old vector form of Maxwell’s equations, and it is natural to wonder how this is done directly using Geometric Algebra. [1] provides one such derivation, using the covariant form of Maxwell’s equations. Here’s a slightly more pedestrian way of doing the same.

# Maxwell’s equations in media

We start with Maxwell’s equations for linear matter as found in [2]

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 0\end{aligned} \hspace{\stretch{1}}(1.2.1a)

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = -\frac{\partial {\mathbf{B}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(1.2.1b)

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} = 0\end{aligned} \hspace{\stretch{1}}(1.2.1c)

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{B} = \mu\epsilon \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.1d)

We merge these using the geometric identity

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{a} + I \boldsymbol{\nabla} \times \mathbf{a} = \boldsymbol{\nabla} \mathbf{a},\end{aligned} \hspace{\stretch{1}}(1.2.2)

where $I$ is the 3D pseudoscalar $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$, to find

\begin{aligned}\boldsymbol{\nabla} \mathbf{E} = -I \frac{\partial {\mathbf{B}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(1.2.3a)

\begin{aligned}\boldsymbol{\nabla} \mathbf{B} = I \mu\epsilon \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.3b)

We want dimensions of $1/L$ for the derivative operator on the RHS of 1.2.3b, so we divide through by $\sqrt{\mu\epsilon} I$ for

\begin{aligned}-I \frac{1}{{\sqrt{\mu\epsilon}}} \boldsymbol{\nabla} \mathbf{B} = \sqrt{\mu\epsilon} \frac{\partial {\mathbf{E}}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(1.2.4)

This can now be added to 1.2.3a for

\begin{aligned}\left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \left( \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \right)= 0.\end{aligned} \hspace{\stretch{1}}(1.2.5)

This is Maxwell’s equation in linear isotropic charge and current free matter in Geometric Algebra form.

# Phasor solutions

We write the electromagnetic field as

\begin{aligned}F = \left( \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \right),\end{aligned} \hspace{\stretch{1}}(1.3.6)

so that for vacuum where $1/\sqrt{\mu \epsilon} = c$ we have the usual $F = \mathbf{E} + I c \mathbf{B}$. Assuming a phasor solution of

\begin{aligned}\tilde{F} = F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}\end{aligned} \hspace{\stretch{1}}(1.3.7)

where $F_0$ is allowed to be complex, and the actual field is obtained by taking the real part

\begin{aligned}F = \text{Real} \tilde{F} = \text{Real}(F_0) \cos(\mathbf{k} \cdot \mathbf{x} - \omega t)-\text{Imag}(F_0) \sin(\mathbf{k} \cdot \mathbf{x} - \omega t).\end{aligned} \hspace{\stretch{1}}(1.3.8)

Note carefully that we are using a scalar imaginary $i$, as well as the multivector (pseudoscalar) $I$, despite the fact that both have the square to scalar minus one property.

We now seek the constraints on $\mathbf{k}$, $\omega$, and $F_0$ that allow this to be a solution to 1.2.5

\begin{aligned}0 = \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F}.\end{aligned} \hspace{\stretch{1}}(1.3.9)

As usual in the non-geometric algebra treatment, we observe that any such solution $F$ to Maxwell’s equation is also a wave equation solution. In GA we can do so by right multiplying an operator that has a conjugate form,

\begin{aligned}\begin{aligned}0 &= \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F} \\ &= \left(\boldsymbol{\nabla} - \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) \tilde{F} \\ &=\left( \boldsymbol{\nabla}^2 - \mu\epsilon \frac{\partial^2}{\partial t^2} \right) \tilde{F} \\ &=\left( \boldsymbol{\nabla}^2 - \frac{1}{{v^2}} \frac{\partial^2}{\partial t^2} \right) \tilde{F},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.10)

where $v = 1/\sqrt{\mu\epsilon}$ is the speed of the wave described by this solution.

Inserting the exponential form of our assumed solution 1.3.7 we find

\begin{aligned}0 = -(\mathbf{k}^2 - \omega^2/v^2) F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)},\end{aligned} \hspace{\stretch{1}}(1.3.11)

which implies that the wave number vector $\mathbf{k}$ and the angular frequency $\omega$ are related by

\begin{aligned}v^2 \mathbf{k}^2 = \omega^2.\end{aligned} \hspace{\stretch{1}}(1.3.12)

Our assumed solution must also satisfy the first order system 1.3.9

\begin{aligned}\begin{aligned}0 &= \left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \frac{\partial {}}{\partial {t}} \right) F_0e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)} \\ &=i\left(\mathbf{e}_m k_m - \frac{\omega}{v}\right) F_0e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)} \\ &=i k ( \hat{\mathbf{k}} - 1 ) F_0 e^{i (\mathbf{k} \cdot \mathbf{x} - \omega t)}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.13)

The constraints on $F_0$ must then be given by

\begin{aligned}0 = ( \hat{\mathbf{k}} - 1 ) F_0.\end{aligned} \hspace{\stretch{1}}(1.3.14)

With

\begin{aligned}F_0 = \mathbf{E}_0 + I v \mathbf{B}_0,\end{aligned} \hspace{\stretch{1}}(1.3.15)

we must then have all grades of the multivector equation equal to zero

\begin{aligned}0 = ( \hat{\mathbf{k}} - 1 ) \left(\mathbf{E}_0 + I v \mathbf{B}_0\right).\end{aligned} \hspace{\stretch{1}}(1.3.16)

Writing out all the geometric products, noting that $I$ commutes with all of $\hat{\mathbf{k}}$, $\mathbf{E}_0$, and $\mathbf{B}_0$ and employing the identity $\mathbf{a} \mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \wedge \mathbf{b}$ we have

\begin{aligned}\begin{array}{l l l l l}0 &= \hat{\mathbf{k}} \cdot \mathbf{E}_0 & - \mathbf{E}_0 & + \hat{\mathbf{k}} \wedge \mathbf{E}_0 & I v \hat{\mathbf{k}} \cdot \mathbf{B}_0 \\ & & + I v \hat{\mathbf{k}} \wedge \mathbf{B}_0 & + I v \mathbf{B}_0 &\end{array}\end{aligned} \hspace{\stretch{1}}(1.3.17)

This is

\begin{aligned}0 = \hat{\mathbf{k}} \cdot \mathbf{E}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18a)

\begin{aligned}\mathbf{E}_0 =- \hat{\mathbf{k}} \times v \mathbf{B}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18b)

\begin{aligned}v \mathbf{B}_0 = \hat{\mathbf{k}} \times \mathbf{E}_0 \end{aligned} \hspace{\stretch{1}}(1.3.18c)

\begin{aligned}0 = \hat{\mathbf{k}} \cdot \mathbf{B}_0.\end{aligned} \hspace{\stretch{1}}(1.3.18d)

This and 1.3.12 describe all the constraints on our phasor that are required for it to be a solution. Note that only one of the two cross product equations in are required because the two are not independent. This can be shown by crossing $\hat{\mathbf{k}}$ with 1.3.18b and using the identity

\begin{aligned}\mathbf{a} \times (\mathbf{a} \times \mathbf{b}) = - \mathbf{a}^2 \mathbf{b} + a (\mathbf{a} \cdot \mathbf{b}).\end{aligned} \hspace{\stretch{1}}(1.3.19)

One can find easily that 1.3.18b and 1.3.18c provide the same relationship between the $\mathbf{E}_0$ and $\mathbf{B}_0$ components of $F_0$. Writing out the complete expression for $F_0$ we have

\begin{aligned}\begin{aligned}F_0 &= \mathbf{E}_0 + I v \mathbf{B}_0 \\ &=\mathbf{E}_0 + I \hat{\mathbf{k}} \times \mathbf{E}_0 \\ &=\mathbf{E}_0 + \hat{\mathbf{k}} \wedge \mathbf{E}_0.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.20)

Since $\hat{\mathbf{k}} \cdot \mathbf{E}_0 = 0$, this is

\begin{aligned}F_0 = (1 + \hat{\mathbf{k}}) \mathbf{E}_0.\end{aligned} \hspace{\stretch{1}}(1.3.21)

Had we been clever enough this could have been deduced directly from the 1.3.14 directly, since we require a product that is killed by left multiplication with $\hat{\mathbf{k}} - 1$. Our complete plane wave solution to Maxwell’s equation is therefore given by

\begin{aligned}\begin{aligned}F &= \text{Real}(\tilde{F}) = \mathbf{E} + \frac{I}{\sqrt{\mu\epsilon}} \mathbf{B} \\ \tilde{F} &= (1 \pm \hat{\mathbf{k}}) \mathbf{E}_0 e^{i (\mathbf{k} \cdot \mathbf{x} \mp \omega t)} \\ 0 &= \hat{\mathbf{k}} \cdot \mathbf{E}_0 \\ \mathbf{k}^2 &= \omega^2 \mu \epsilon.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.22)

# References

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

[2] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

## Strain tensor in spherical coordinates

Posted by peeterjoot on January 23, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

## Spherical tensor.

To perform the derivation in spherical coordinates we have some setup to do first, since we need explicit representations of all three unit vectors. The radial vector we can get easily by geometry and find the usual

\begin{aligned}\hat{\mathbf{r}} =\begin{bmatrix}\sin\theta \cos\phi \\ \sin\theta \sin\phi \\ \cos\theta\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.61)

We can get $\hat{\boldsymbol{\phi}}$ by geometrical intuition since it the plane unit vector at angle $\phi$ rotated by $\pi/2$. That is

\begin{aligned}\hat{\boldsymbol{\phi}} =\begin{bmatrix}-\sin\phi \\ \cos\phi \\ 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.62)

We can get $\hat{\boldsymbol{\theta}}$ by utilizing the right handedness of the coordinates since

\begin{aligned}\hat{\boldsymbol{\phi}} \times \hat{\mathbf{r}} = \hat{\boldsymbol{\theta}}\end{aligned} \hspace{\stretch{1}}(3.63)

and find

\begin{aligned}\hat{\boldsymbol{\theta}} =\begin{bmatrix}\cos\theta \cos\phi \\ \cos\theta \sin\phi \\ -\sin\theta\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.64)

\begin{aligned}\begin{aligned}&d\mathbf{l}'^2 - d\mathbf{x}^2 \\ &=2 (dr)^2 \biggl(\frac{\partial u_r}{\partial r}+ \frac{1}{{2}}\frac{\partial u_m}{\partial r} \frac{\partial u_m}{\partial r}\biggr) \\ & + 2 r^2 (d\theta )^2 \biggl(\frac{1}{{r}} u_r + \frac{1}{{2r^2}}(u_r^2 + u_{\theta }^2) - \frac{1}{{r^2}} u_{\theta } \frac{\partial u_r}{\partial \theta }+ \left(\frac{1}{{r}} + \frac{1}{{r^2}}u_r\right) \frac{\partial u_{\theta }}{\partial \theta }+ \frac{1}{{2 r^2}} \frac{\partial u_m}{\partial \theta } \frac{\partial u_m}{\partial \theta }\biggr) \\ &+ 2 r^2 \sin^2\theta (d\phi )^2 \biggl( \frac{1}{{2 r^2 \sin^2\theta}} u_\phi^2+ \frac{1}{{2 r^2 }} u_{\theta }^2 \cot^2\theta+ \frac{1}{{r}} u_r+ \frac{1}{{2 r^2}} u_r^2+ \left(\frac{1}{{r}} + \frac{1}{{r^2}}u_r\right) u_{\theta } \cot\theta \\ &\qquad- \frac{1}{{r^2 \sin\theta}} u_{\phi } \frac{\partial u_r}{\partial \phi }- \frac{1}{{r^2 }} u_{\phi } \frac{\cos\theta}{\sin^2\theta} \frac{\partial u_{\theta }}{\partial \phi }+ \frac{1}{{r^2 }} \frac{\partial u_{\phi }}{\partial \phi } \left(u_{\theta } \frac{\cos\theta}{\sin^2\theta} + \left(r + u_r\right) \frac{1}{{\sin\theta}} \right)+ \frac{1}{{2 r^2 \sin^2\theta}} \frac{\partial u_m}{\partial \phi } \frac{\partial u_m}{\partial \phi }\biggr) \\ & + 2 dr r d\theta \biggl(- \frac{1}{{r}} u_{\theta }+ \frac{1}{{r}} \frac{\partial u_r}{\partial \theta }- \frac{1}{{r}} u_{\theta } \frac{\partial u_r}{\partial r}+ \frac{\partial u_{\theta }}{\partial r} \left(1 + \frac{u_r}{r} \right)+ \frac{1}{{r}} \frac{\partial u_m}{\partial r} \frac{\partial u_m}{\partial \theta }\biggr) \\ & + 2 r^2 \sin\theta d\theta d\phi \biggl(\frac{1}{{r^2 }} u_{\theta } u_{\phi }- \frac{1}{{r^2 \sin\theta}} u_{\theta } \frac{\partial u_r}{\partial \phi }- \frac{1}{{r^2 }} u_{\phi } \frac{\partial u_r}{\partial \theta }- \frac{1}{{r^2 }} u_{\phi } \cot\theta \left(r + u_r + \frac{\partial u_{\theta }}{\partial \theta }\right) \\ &\qquad+ \frac{1}{{r^2 \sin\theta}} \left(r + u_r \right) \frac{\partial u_{\theta }}{\partial \phi }+ \frac{\partial u_{\phi }}{\partial \theta } \left(\frac{u_{\theta }}{r^2} \cot\theta + \frac{1}{{r}} + \frac{u_r}{r^2} \right)+ \frac{1}{{r^2 \sin\theta}} \frac{\partial u_m}{\partial \theta } \frac{\partial u_m}{\partial \phi }\biggr) \\ & + 2 r \sin\theta d\phi dr \biggl(- \frac{1}{{r }} u_{\phi }+ \frac{1}{{r \sin\theta}} \frac{\partial u_r}{\partial \phi }- u_{\phi } \frac{1}{{r }} \frac{\partial u_r}{\partial r}- u_{\phi } \cot\theta \frac{1}{{r }} \frac{\partial u_{\theta }}{\partial r}+ \frac{1}{{r }} \frac{\partial u_{\phi }}{\partial r} \left( u_{\theta } \cot\theta + r + u_r \right)+ \frac{1}{{r \sin\theta}} \frac{\partial u_m}{\partial \phi } \frac{\partial u_m}{\partial r}\biggr)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.65)

### A manual derivation.

Doing the calculation pretty much completely with Mathematica is rather unsatisfying. To set up for it let’s first compute the unit vectors from scratch. I’ll use geometric algebra to do this calculation. Consider figure (\ref{fig:qmTwoExamReflection:continuumL2fig5})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL2fig5}
\caption{Composite rotations for spherical polar unit vectors.}
\end{figure}

We have two sets of rotations, the first is a rotation about the $z$ axis by $\phi$. Writing $i = \mathbf{e}_1 \mathbf{e}_2$ for the unit bivector in the $x,y$ plane, we rotate

\begin{aligned}\mathbf{e}_1' &= \mathbf{e}_1 e^{i\phi} = \mathbf{e}_1 \cos\phi + \mathbf{e}_2 \sin\phi \\ \mathbf{e}_2' &= \mathbf{e}_2 e^{i\phi} = \mathbf{e}_2 \cos\phi - \mathbf{e}_1 \sin\phi \\ \mathbf{e}_3' &= \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(3.66)

Now we rotate in the plane spanned by $\mathbf{e}_3$ and $\mathbf{e}_1'$ by $\theta$. With $j = \mathbf{e}_3 \mathbf{e}_1'$, our vectors in the plane rotate as

\begin{aligned}\mathbf{e}_1'' &= \mathbf{e}_1' e^{j\phi} = \mathbf{e}_1 e^{i\phi} e^{j\theta} \\ \mathbf{e}_3'' &= \mathbf{e}_3' e^{j\theta} = \mathbf{e}_3 e^{j\theta},\end{aligned} \hspace{\stretch{1}}(3.69)

(with $\mathbf{e}_2'' = \mathbf{e}_2$ since $\mathbf{e}_2 \cdot j = 0$).

\begin{aligned}\hat{\boldsymbol{\theta}} = \mathbf{e}_1''&= \mathbf{e}_1 e^{i\phi} e^{j\theta} \\ &= \mathbf{e}_1 e^{i\phi} (\cos\theta + \mathbf{e}_3 \mathbf{e}_1 e^{i\phi} \sin\theta) \\ &= \mathbf{e}_1 e^{i\phi} \cos\theta -\mathbf{e}_3 \sin\theta \\ &= (\mathbf{e}_1 \cos\phi + \mathbf{e}_2 \sin\phi) \cos\theta -\mathbf{e}_3 \sin\theta \\ \end{aligned}

\begin{aligned}\hat{\mathbf{r}} = \mathbf{e}_3''&= \mathbf{e}_3 e^{j\theta} \\ &= \mathbf{e}_3 (\cos\theta + \mathbf{e}_3 \mathbf{e}_1 e^{i\phi} \sin\theta) \\ &= \mathbf{e}_3 \cos\theta + \mathbf{e}_1 e^{i\phi} \sin\theta \\ &= \mathbf{e}_3 \cos\theta + (\mathbf{e}_1 \cos\phi + \mathbf{e}_2 \sin\phi) \sin\theta \\ \end{aligned}

Now, these are all the same relations that we could find with coordinate algebra

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_1 \cos\phi \sin\theta +\mathbf{e}_2 \sin\phi \sin\theta +\mathbf{e}_3 \cos\theta \\ \hat{\boldsymbol{\theta}} &= \mathbf{e}_1 \cos\phi \cos\theta +\mathbf{e}_2 \sin\phi \cos\theta -\mathbf{e}_3 \sin\theta \\ \hat{\boldsymbol{\phi}} &= -\mathbf{e}_1 \sin\phi + \mathbf{e}_2 \cos\phi\end{aligned} \hspace{\stretch{1}}(3.71)

There’s nothing special in this approach if that is as far as we go, but we can put things in a nice tidy form for computation of the differentials of the unit vectors. Introducing the unit pseudoscalar $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$ we can write these in a compact exponential form.

\begin{aligned}\hat{\mathbf{r}}&= (\mathbf{e}_1 \cos\phi +\mathbf{e}_2 \sin\phi ) \sin\theta +\mathbf{e}_3 \cos\theta \\ &= \mathbf{e}_1 e^{i\phi} \sin\theta +\mathbf{e}_3 \cos\theta \\ &= \mathbf{e}_3 ( \cos\theta + \mathbf{e}_3 \mathbf{e}_1 e^{i\phi} \sin\theta ) \\ &= \mathbf{e}_3 ( \cos\theta + \mathbf{e}_3 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_2 e^{i\phi} \sin\theta ) \\ &= \mathbf{e}_3 ( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta ) \\ &= \mathbf{e}_3 e^{ I \hat{\boldsymbol{\phi}} \theta }\end{aligned}

\begin{aligned}\hat{\boldsymbol{\theta}}&=\mathbf{e}_1 \cos\phi \cos\theta +\mathbf{e}_2 \sin\phi \cos\theta -\mathbf{e}_3 \sin\theta \\ &=(\mathbf{e}_1 \cos\phi +\mathbf{e}_2 \sin\phi ) \cos\theta -\mathbf{e}_3 \sin\theta \\ &=\mathbf{e}_1 e^{i\phi} \cos\theta -\mathbf{e}_3 \sin\theta \\ &=\mathbf{e}_1 e^{i\phi} ( \cos\theta - e^{-i\phi} \mathbf{e}_1 \mathbf{e}_3 \sin\theta ) \\ &=\mathbf{e}_1 e^{i\phi} ( \cos\theta - \mathbf{e}_1 \mathbf{e}_3 e^{i\phi} \sin\theta ) \\ &=\mathbf{e}_1 e^{i\phi} ( \cos\theta - \mathbf{e}_1 \mathbf{e}_3 \mathbf{e}_2 \mathbf{e}_2 e^{i\phi} \sin\theta ) \\ &=\mathbf{e}_1 e^{i\phi} ( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta ) \\ &=\mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_2 e^{i\phi} ( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta ) \\ &=i \hat{\boldsymbol{\phi}} e^{I \hat{\boldsymbol{\phi}} \theta}.\end{aligned}

To summarize we have

\begin{aligned}\hat{\boldsymbol{\phi}} &= \mathbf{e}_2 e^{i\phi} \\ \hat{\mathbf{r}} &= \mathbf{e}_3 e^{I\hat{\boldsymbol{\phi}} \theta} \\ \hat{\boldsymbol{\theta}} &= i \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta}.\end{aligned} \hspace{\stretch{1}}(3.74)

Taking differentials we find first

\begin{aligned}d\hat{\boldsymbol{\phi}} = \mathbf{e}_2 e^{i\phi} i d\phi = \hat{\boldsymbol{\phi}} i d\phi\end{aligned}

\begin{aligned}d\hat{\boldsymbol{\theta}}&= d \left( i \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} \right) \\ &= i d \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} + i \hat{\boldsymbol{\phi}} d \left( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta \right) \\ &= i d \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta}+ i \hat{\boldsymbol{\phi}} I (d \hat{\boldsymbol{\phi}}) \sin\theta+ i \hat{\boldsymbol{\phi}} I \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &= i \hat{\boldsymbol{\phi}} i e^{I\hat{\boldsymbol{\phi}} \theta} d\phi+ i \hat{\boldsymbol{\phi}} I \hat{\boldsymbol{\phi}} i \sin\theta d\phi+ i \hat{\boldsymbol{\phi}} I \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &= \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\phi- I \sin\theta d\phi- \mathbf{e}_3 e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &= \hat{\boldsymbol{\phi}} (\cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta) d\phi- I \sin\theta d\phi- \mathbf{e}_3 e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &= \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta\end{aligned}

\begin{aligned}d \hat{\mathbf{r}}&=\mathbf{e}_3 d \left( e^{I\hat{\boldsymbol{\phi}} \theta} \right) \\ &=\mathbf{e}_3 d \left( \cos\theta + I \hat{\boldsymbol{\phi}} \sin\theta \right) \\ &=\mathbf{e}_3 \left( I (d \hat{\boldsymbol{\phi}}) \sin\theta + I \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \right) \\ &=\mathbf{e}_3 \left( I \hat{\boldsymbol{\phi}} i \sin\theta d\phi + I \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \right) \\ &=i \hat{\boldsymbol{\phi}} i \sin\theta d\phi + i \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} d\theta \\ &=\hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta\end{aligned}

Summarizing these differentials we have

\begin{aligned}d\hat{\mathbf{r}} &= \hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta \\ d\hat{\boldsymbol{\theta}} &= \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta \\ d\hat{\boldsymbol{\phi}} &= \hat{\boldsymbol{\phi}} i d\phi\end{aligned} \hspace{\stretch{1}}(3.77)

A final cleanup is required. While $\hat{\boldsymbol{\phi}} i$ is a vector and has a nicely compact form, we need to decompose this into components in the $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\theta}}$ and $\hat{\boldsymbol{\phi}}$ directions. Taking scalar products we have

\begin{aligned}\hat{\boldsymbol{\phi}} \cdot (\hat{\boldsymbol{\phi}} i) = 0\end{aligned}

\begin{aligned}\hat{\mathbf{r}} \cdot (\hat{\boldsymbol{\phi}} i)&=\left\langle{{ \hat{\mathbf{r}} \hat{\boldsymbol{\phi}} i}}\right\rangle \\ &=\left\langle{{ \mathbf{e}_3 e^{I\hat{\boldsymbol{\phi}} \theta} \mathbf{e}_2 e^{i\phi} i}}\right\rangle \\ &=\left\langle{{ \mathbf{e}_3 (\cos\theta + I \mathbf{e}_2 e^{i\phi} \sin\theta) \mathbf{e}_2 e^{i\phi} i}}\right\rangle \\ &=\left\langle{{ I (\cos\theta e^{-i\phi} + I \mathbf{e}_2 \sin\theta) \mathbf{e}_2 }}\right\rangle \\ &=-\sin\theta\end{aligned}

\begin{aligned}\hat{\boldsymbol{\theta}} \cdot (\hat{\boldsymbol{\phi}} i)&=\left\langle{{ \hat{\boldsymbol{\theta}} \hat{\boldsymbol{\phi}} i }}\right\rangle \\ &=\left\langle{{ i \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} \hat{\boldsymbol{\phi}} i }}\right\rangle \\ &=-\left\langle{{ \hat{\boldsymbol{\phi}} e^{I\hat{\boldsymbol{\phi}} \theta} \hat{\boldsymbol{\phi}} }}\right\rangle \\ &=-\left\langle{{ e^{I\hat{\boldsymbol{\phi}} \theta} }}\right\rangle \\ &=- \cos\theta.\end{aligned}

Summarizing once again, but this time in terms of $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\theta}}$ and $\hat{\boldsymbol{\phi}}$ we have

\begin{aligned}d\hat{\mathbf{r}} &= \hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta \\ d\hat{\boldsymbol{\theta}} &= \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta \\ d\hat{\boldsymbol{\phi}} &= -(\hat{\mathbf{r}} \sin\theta + \hat{\boldsymbol{\theta}} \cos\theta) d\phi\end{aligned} \hspace{\stretch{1}}(3.80)

Now we are set to take differentials. With

\begin{aligned}\mathbf{x} = r \hat{\mathbf{r}},\end{aligned} \hspace{\stretch{1}}(3.83)

we have

\begin{aligned}d\mathbf{x} =dr \hat{\mathbf{r}}+ r d\hat{\mathbf{r}}=dr \hat{\mathbf{r}} + \hat{\boldsymbol{\phi}} r \sin\theta d\phi + r \hat{\boldsymbol{\theta}} d\theta.\end{aligned} \hspace{\stretch{1}}(3.84)

Squaring this we get the usual spherical polar line scalar line element

\begin{aligned}d\mathbf{x}^2 = dr^2 + r^2 \sin^2\theta d\phi^2 + r^2 d\theta^2.\end{aligned} \hspace{\stretch{1}}(3.85)

With

\begin{aligned}\mathbf{u} = u_r \hat{\mathbf{r}} + u_\theta \hat{\boldsymbol{\theta}} + u_\phi \hat{\boldsymbol{\phi}},\end{aligned} \hspace{\stretch{1}}(3.86)

our differential is

\begin{aligned}d\mathbf{u}&=du_r \hat{\mathbf{r}} + du_\theta \hat{\boldsymbol{\theta}} + du_\phi \hat{\boldsymbol{\phi}}+ u_r d\hat{\mathbf{r}} + u_\theta d\hat{\boldsymbol{\theta}} + u_\phi d \hat{\boldsymbol{\phi}} \\ &=du_r \hat{\mathbf{r}} + du_\theta \hat{\boldsymbol{\theta}} + du_\phi \hat{\boldsymbol{\phi}}+ u_r \left(\hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta \right)+ u_\theta \left( \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta \right)- u_\phi (\hat{\mathbf{r}} \sin\theta + \hat{\boldsymbol{\theta}} \cos\theta) d\phi\\ &=\hat{\mathbf{r}} \left( du_r - u_\theta d\theta - u_\phi \sin\theta d\phi \right) \\ &+\hat{\boldsymbol{\theta}} \left( du_\theta + u_r d\theta - u_\phi \cos\theta d\phi \right) \\ &+\hat{\boldsymbol{\phi}} \left( du_\phi + u_r \sin\theta d\phi + u_\theta \cos\theta d\phi \right).\end{aligned}

We can add $d\mathbf{x}$ to this and take differences

\begin{aligned}\begin{aligned}(d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2&=\left( du_r - u_\theta d\theta - u_\phi \sin\theta d\phi + dr \right)^2 \\ &+\left( du_\theta + u_r d\theta - u_\phi \cos\theta d\phi + r d\theta \right)^2 \\ &+\left( du_\phi + u_r \sin\theta d\phi + u_\theta \cos\theta d\phi + r \sin\theta d\phi \right)^2\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.87)

For each $m = r,\theta,\phi$ we have

\begin{aligned}du_m=\frac{\partial {u_m}}{\partial {r}} dr +\frac{\partial {u_m}}{\partial {\theta}} d\theta +\frac{\partial {u_m}}{\partial {\phi}} d\phi,\end{aligned} \hspace{\stretch{1}}(3.88)

and plugging through that calculation is really all it takes to derive the textbook result. To do this to first order in $u_m$, we find

\begin{aligned}\frac{1}{{2}} \left((d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2\right)&=du_r dr- u_\theta d\theta dr- u_\phi \sin\theta d\phi dr \\ &+ du_\theta r d\theta+ u_r r d\theta^2- u_\phi r \cos\theta d\phi d\theta \\ &+ r \sin\theta du_\phi d\phi+ r \sin^2\theta u_r d\phi^2+ r \sin\theta \cos\theta u_\theta d\phi^2 \\ &=\left( \frac{\partial {u_r}}{\partial {r}} dr + \frac{\partial {u_r}}{\partial {\theta}} d\theta + \frac{\partial {u_r}}{\partial {\phi}} d\phi \right)dr- u_\theta d\theta dr- u_\phi \sin\theta d\phi dr \\ &+\left( \frac{\partial {u_\theta}}{\partial {r}} dr + \frac{\partial {u_\theta}}{\partial {\theta}} d\theta + \frac{\partial {u_\theta}}{\partial {\phi}} d\phi \right) r d\theta+ u_r r d\theta^2- u_\phi r \cos\theta d\phi d\theta \\ &+\left( \frac{\partial {u_\phi}}{\partial {r}} dr + \frac{\partial {u_\phi}}{\partial {\theta}} d\theta + \frac{\partial {u_\phi}}{\partial {\phi}} d\phi \right)r \sin\theta d\phi+ r \sin^2\theta u_r d\phi^2+ r \sin\theta \cos\theta u_\theta d\phi^2\end{aligned}

Collecting terms we have the result of the text in the braces

\begin{aligned}\begin{aligned}\left((d\mathbf{u} + d\mathbf{x})^2 - d\mathbf{x}^2\right)&=2 dr^2 \left(\frac{\partial {u_r}}{\partial {r}}\right) \\ &+2 r^2 d\theta^2 \left(\frac{1}{{r}} \frac{\partial {u_\theta}}{\partial {\theta}} + u_r \frac{1}{{r}}\right) \\ &+2 r^2 \sin^2\theta d\phi^2 \left(\frac{\partial {u_\phi}}{\partial {\phi}} \frac{1}{{r \sin\theta}} + \frac{1}{{r}} u_r + \frac{1}{{r}} \cot\theta u_\theta\right) \\ &+2 dr r d\theta \left(\frac{1}{{r}} \frac{\partial {u_r}}{\partial {\theta}} - \frac{1}{{r}} u_\theta +\frac{\partial {u_\theta}}{\partial {r}}\right) \\ &+2 r^2 \sin\theta d\theta d\phi \left(\frac{\partial {u_\theta}}{\partial {\phi}} \frac{1}{{r \sin\theta}} - \frac{1}{{r}} u_\phi \cot\theta +\frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\theta}}\right) \\ &+2 r \sin\theta d\phi dr \left(\frac{1}{{r \sin\theta}} \frac{\partial {u_r}}{\partial {\phi}} - \frac{1}{{r}} u_\phi + \frac{\partial {u_\phi}}{\partial {r}}\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.89)

It should be possible to do the calculation to second order too, but to include all the quadratic terms in $u_m$ is again really messy. Trying that with mathematica gives the same results as above using the strictly coordinate algebra approach.

# References

[1] L.D. Landau, EM Lifshitz, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.

## Multivector commutators and Lorentz boosts.

Posted by peeterjoot on October 31, 2010

# Motivation.

In some reading there I found that the electrodynamic field components transform in a reversed sense to that of vectors, where instead of the perpendicular to the boost direction remaining unaffected, those are the parts that are altered.

To explore this, look at the Lorentz boost action on a multivector, utilizing symmetric and antisymmetric products to split that vector into portions effected and unaffected by the boost. For the bivector (electrodynamic case) and the four vector case, examine how these map to dot and wedge (or cross) products.

The underlying motivator for this boost consideration is an attempt to see where equation (6.70) of [1] comes from. We get to this by the very end.

# Guts.

## Structure of the bivector boost.

Recall that we can write our Lorentz boost in exponential form with

\begin{aligned}L &= e^{\alpha \boldsymbol{\sigma}/2} \\ X' &= L^\dagger X L,\end{aligned} \hspace{\stretch{1}}(2.1)

where $\boldsymbol{\sigma}$ is a spatial vector. This works for our bivector field too, assuming the composite transformation is an outermorphism of the transformed four vectors. Applying the boost to both the gradient and the potential our transformed field is then

\begin{aligned}F' &= \nabla' \wedge A' \\ &= (L^\dagger \nabla L) \wedge (L^\dagger A L) \\ &= \frac{1}{{2}} \left((L^\dagger \stackrel{ \rightarrow }{\nabla} L) (L^\dagger A L) -(L^\dagger A L) (L^\dagger \stackrel{ \leftarrow }{\nabla} L)\right) \\ &= \frac{1}{{2}} L^\dagger \left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) L \\ &= L^\dagger (\nabla \wedge A) L.\end{aligned}

Note that arrows were used briefly to indicate that the partials of the gradient are still acting on $A$ despite their vector components being to one side. We are left with the very simple transformation rule

\begin{aligned}F' = L^\dagger F L,\end{aligned} \hspace{\stretch{1}}(2.3)

which has exactly the same structure as the four vector boost.

## Employing the commutator and anticommutator to find the parallel and perpendicular components.

If we apply the boost to a four vector, those components of the four vector that commute with the spatial direction $\boldsymbol{\sigma}$ are unaffected. As an example, which also serves to ensure we have the sign of the rapidity angle $\alpha$ correct, consider $\boldsymbol{\sigma} = \boldsymbol{\sigma}_1$. We have

\begin{aligned}X' = e^{-\alpha \boldsymbol{\sigma}/2} ( x^0 \gamma_0 + x^1 \gamma_1 + x^2 \gamma_2 + x^3 \gamma_3 ) (\cosh \alpha/2 + \gamma_1 \gamma_0 \sinh \alpha/2 )\end{aligned} \hspace{\stretch{1}}(2.4)

We observe that the scalar and $\boldsymbol{\sigma}_1 = \gamma_1 \gamma_0$ components of the exponential commute with $\gamma_2$ and $\gamma_3$ since there is no vector in common, but that $\boldsymbol{\sigma}_1$ anticommutes with $\gamma_0$ and $\gamma_1$. We can therefore write

\begin{aligned}X' &= x^2 \gamma_2 + x^3 \gamma_3 +( x^0 \gamma_0 + x^1 \gamma_1 + ) (\cosh \alpha + \gamma_1 \gamma_0 \sinh \alpha ) \\ &= x^2 \gamma_2 + x^3 \gamma_3 +\gamma_0 ( x^0 \cosh\alpha - x^1 \sinh \alpha )+ \gamma_1 ( x^1 \cosh\alpha - x^0 \sinh \alpha )\end{aligned}

reproducing the familiar matrix result should we choose to write it out. How can we express the commutation property without resorting to components. We could write the four vector as a spatial and timelike component, as in

\begin{aligned}X = x^0 \gamma_0 + \mathbf{x} \gamma_0,\end{aligned} \hspace{\stretch{1}}(2.5)

and further separate that into components parallel and perpendicular to the spatial unit vector $\boldsymbol{\sigma}$ as

\begin{aligned}X = x^0 \gamma_0 + (\mathbf{x} \cdot \boldsymbol{\sigma}) \boldsymbol{\sigma} \gamma_0 + (\mathbf{x} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma} \gamma_0.\end{aligned} \hspace{\stretch{1}}(2.6)

However, it would be nicer to group the first two terms together, since they are ones that are affected by the transformation. It would also be nice to not have to resort to spatial dot and wedge products, since we get into trouble too easily if we try to mix dot and wedge products of four vector and spatial vector components.

What we can do is employ symmetric and antisymmetric products (the anticommutator and commutator respectively). Recall that we can write any multivector product this way, and in particular

\begin{aligned}M \boldsymbol{\sigma} = \frac{1}{{2}} (M \boldsymbol{\sigma} + \boldsymbol{\sigma} M) + \frac{1}{{2}} (M \boldsymbol{\sigma} - \boldsymbol{\sigma} M).\end{aligned} \hspace{\stretch{1}}(2.7)

Left multiplying by the unit spatial vector $\boldsymbol{\sigma}$ we have

\begin{aligned}M = \frac{1}{{2}} (M + \boldsymbol{\sigma} M \boldsymbol{\sigma}) + \frac{1}{{2}} (M - \boldsymbol{\sigma} M \boldsymbol{\sigma}) = \frac{1}{{2}} \left\{{M},{\boldsymbol{\sigma}}\right\} \boldsymbol{\sigma} + \frac{1}{{2}} \left[{M},{\boldsymbol{\sigma}}\right] \boldsymbol{\sigma}.\end{aligned} \hspace{\stretch{1}}(2.8)

When $M = \mathbf{a}$ is a spatial vector this is our familiar split into parallel and perpendicular components with the respective projection and rejection operators

\begin{aligned}\mathbf{a} = \frac{1}{{2}} \left\{\mathbf{a},{\boldsymbol{\sigma}}\right\} \boldsymbol{\sigma} + \frac{1}{{2}} \left[{\mathbf{a}},{\boldsymbol{\sigma}}\right] \boldsymbol{\sigma} = (\mathbf{a} \cdot \boldsymbol{\sigma}) \boldsymbol{\sigma} + (\mathbf{a} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma}.\end{aligned} \hspace{\stretch{1}}(2.9)

However, the more general split employing symmetric and antisymmetric products in 2.8, is something we can use for our four vector and bivector objects too.

Observe that we have the commutation and anti-commutation relationships

\begin{aligned}\left( \frac{1}{{2}} \left\{{M},{\boldsymbol{\sigma}}\right\} \boldsymbol{\sigma} \right) \boldsymbol{\sigma} &= \boldsymbol{\sigma} \left( \frac{1}{{2}} \left\{{M},{\boldsymbol{\sigma}}\right\} \boldsymbol{\sigma} \right) \\ \left( \frac{1}{{2}} \left[{M},{\boldsymbol{\sigma}}\right] \boldsymbol{\sigma} \right) \boldsymbol{\sigma} &= -\boldsymbol{\sigma} \left( \frac{1}{{2}} \left[{M},{\boldsymbol{\sigma}}\right] \boldsymbol{\sigma} \right).\end{aligned} \hspace{\stretch{1}}(2.10)

This split therefore serves to separate the multivector object in question nicely into the portions that are acted on by the Lorentz boost, or left unaffected.

## Application of the symmetric and antisymmetric split to the bivector field.

Let’s apply 2.8 to the spacetime event $X$ again with an x-axis boost $\sigma = \sigma_1$. The anticommutator portion of X in this boost direction is

\begin{aligned}\frac{1}{{2}} \left\{{X},{\boldsymbol{\sigma}_1}\right\} \boldsymbol{\sigma}_1&=\frac{1}{{2}} \left(\left( x^0 \gamma_0 + x^1 \gamma_1 + x^2 \gamma_2 + x^3 \gamma_3 \right)+\gamma_1 \gamma_0\left( x^0 \gamma_0 + x^1 \gamma_1 + x^2 \gamma_2 + x^3 \gamma_3 \right) \gamma_1 \gamma_0\right) \\ &=x^2 \gamma_2 + x^3 \gamma_3,\end{aligned}

whereas the commutator portion gives us

\begin{aligned}\frac{1}{{2}} \left[{X},{\boldsymbol{\sigma}_1}\right] \boldsymbol{\sigma}_1&=\frac{1}{{2}} \left(\left( x^0 \gamma_0 + x^1 \gamma_1 + x^2 \gamma_2 + x^3 \gamma_3 \right)-\gamma_1 \gamma_0\left( x^0 \gamma_0 + x^1 \gamma_1 + x^2 \gamma_2 + x^3 \gamma_3 \right) \gamma_1 \gamma_0\right) \\ &=x^0 \gamma_0 + x^1 \gamma_1.\end{aligned}

We’ve seen that only these commutator portions are acted on by the boost. We have therefore found the desired logical grouping of the four vector $X$ into portions that are left unchanged by the boost and those that are affected. That is

\begin{aligned}\frac{1}{{2}} \left[{X},{\boldsymbol{\sigma}}\right] \boldsymbol{\sigma} &= x^0 \gamma_0 + (\mathbf{x} \cdot \boldsymbol{\sigma}) \boldsymbol{\sigma} \gamma_0 \\ \frac{1}{{2}} \left\{{X},{\boldsymbol{\sigma}}\right\} \boldsymbol{\sigma} &= (\mathbf{x} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma} \gamma_0 \end{aligned} \hspace{\stretch{1}}(2.12)

Let’s now return to the bivector field $F = \nabla \wedge A = \mathbf{E} + I c \mathbf{B}$, and split that multivector into boostable and unboostable portions with the commutator and anticommutator respectively.

Observing that our pseudoscalar $I$ commutes with all spatial vectors we have for the anticommutator parts that will not be affected by the boost

\begin{aligned}\frac{1}{{2}} \left\{{\mathbf{E} + I c \mathbf{B}},{\boldsymbol{\sigma}}\right\} \boldsymbol{\sigma} &= (\mathbf{E} \cdot \boldsymbol{\sigma}) \boldsymbol{\sigma} + I c (\mathbf{B} \cdot \boldsymbol{\sigma}) \boldsymbol{\sigma},\end{aligned} \hspace{\stretch{1}}(2.14)

and for the components that will be boosted we have

\begin{aligned}\frac{1}{{2}} \left[{\mathbf{E} + I c \mathbf{B}},{\boldsymbol{\sigma}}\right] \boldsymbol{\sigma} &= (\mathbf{E} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma} + I c (\mathbf{B} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma}.\end{aligned} \hspace{\stretch{1}}(2.15)

For the four vector case we saw that the components that lay “perpendicular” to the boost direction, were unaffected by the boost. For the field we see the opposite, and the components of the individual electric and magnetic fields that are parallel to the boost direction are unaffected.

Our boosted field is therefore

\begin{aligned}F' = (\mathbf{E} \cdot \boldsymbol{\sigma}) \boldsymbol{\sigma} + I c (\mathbf{B} \cdot \boldsymbol{\sigma}) \boldsymbol{\sigma}+ \left( (\mathbf{E} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma} + I c (\mathbf{B} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma}\right) \left( \cosh \alpha + \boldsymbol{\sigma} \sinh \alpha \right)\end{aligned} \hspace{\stretch{1}}(2.16)

Focusing on just the non-parallel terms we have

\begin{aligned}\left( (\mathbf{E} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma} + I c (\mathbf{B} \wedge \boldsymbol{\sigma}) \boldsymbol{\sigma}\right) \left( \cosh \alpha + \boldsymbol{\sigma} \sinh \alpha \right)&=(\mathbf{E}_\perp + I c \mathbf{B}_\perp ) \cosh\alpha+(I \mathbf{E} \times \boldsymbol{\sigma} - c \mathbf{B} \times \boldsymbol{\sigma} ) \sinh\alpha \\ &=\mathbf{E}_\perp \cosh\alpha - c (\mathbf{B} \times \boldsymbol{\sigma} ) \sinh\alpha + I ( c \mathbf{B}_\perp \cosh\alpha + (\mathbf{E} \times \boldsymbol{\sigma}) \sinh\alpha ) \\ &=\gamma \left(\mathbf{E}_\perp - c (\mathbf{B} \times \boldsymbol{\sigma} ) {\left\lvert{\mathbf{v}}\right\rvert}/c+ I ( c \mathbf{B}_\perp + (\mathbf{E} \times \boldsymbol{\sigma}) {\left\lvert{\mathbf{v}}\right\rvert}/c) \right)\end{aligned}

A final regrouping gives us

\begin{aligned}F'&=\mathbf{E}_\parallel + \gamma \left( \mathbf{E}_\perp - \mathbf{B} \times \mathbf{v} \right)+I c \left( \mathbf{B}_\parallel + \gamma \left( \mathbf{B}_\perp + \mathbf{E} \times \mathbf{v}/c^2 \right) \right)\end{aligned} \hspace{\stretch{1}}(2.17)

In particular when we consider the proton, electron system as in equation (6.70) of [1] where it is stated that the electron will feel a magnetic field given by

\begin{aligned}\mathbf{B} = - \frac{\mathbf{v}}{c} \times \mathbf{E}\end{aligned} \hspace{\stretch{1}}(2.18)

we can see where this comes from. If $F = \mathbf{E} + I c (0)$ is the field acting on the electron, then application of a $\mathbf{v}$ boost to the electron perpendicular to the field (ie: radial motion), we get

\begin{aligned}F' = I c \gamma \mathbf{E} \times \mathbf{v}/c^2 =-I c \gamma \frac{\mathbf{v}}{c^2} \times \mathbf{E}\end{aligned} \hspace{\stretch{1}}(2.19)

We also have an additional $1/c$ factor in our result, but that’s a consequence of the choice of units where the dimensions of $\mathbf{E}$ match $c \mathbf{B}$, whereas in the text we have $\mathbf{E}$ and $\mathbf{B}$ in the same units. We also have an additional $\gamma$ factor, so we must presume that ${\left\lvert{\mathbf{v}}\right\rvert} << c$ in this portion of the text. That is actually a requirement here, for if the electron was already in motion, we'd have to boost a field that also included a magnetic component. A consequence of this is that the final interaction Hamiltonian of (6.75) is necessarily non-relativistic.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

## Derivation of the spherical polar Laplacian

Posted by peeterjoot on October 9, 2010

# Motivation.

In [1] was a Geometric Algebra derivation of the 2D polar Laplacian by squaring the quadient. In [2] was a factorization of the spherical polar unit vectors in a tidy compact form. Here both these ideas are utilized to derive the spherical polar form for the Laplacian, an operation that is strictly algebraic (squaring the gradient) provided we operate on the unit vectors correctly.

# Our rotation multivector.

Our starting point is a pair of rotations. We rotate first in the $x,y$ plane by $\phi$

\begin{aligned}\mathbf{x} &\rightarrow \mathbf{x}' = \tilde{R_\phi} \mathbf{x} R_\phi \\ i &\equiv \mathbf{e}_1 \mathbf{e}_2 \\ R_\phi &= e^{i \phi/2}\end{aligned} \hspace{\stretch{1}}(2.1)

Then apply a rotation in the $\mathbf{e}_3 \wedge (\tilde{R_\phi} \mathbf{e}_1 R_\phi) = \tilde{R_\phi} \mathbf{e}_3 \mathbf{e}_1 R_\phi$ plane

\begin{aligned}\mathbf{x}' &\rightarrow \mathbf{x}'' = \tilde{R_\theta} \mathbf{x}' R_\theta \\ R_\theta &= e^{ \tilde{R_\phi} \mathbf{e}_3 \mathbf{e}_1 R_\phi \theta/2 } = \tilde{R_\phi} e^{ \mathbf{e}_3 \mathbf{e}_1 \theta/2 } R_\phi\end{aligned} \hspace{\stretch{1}}(2.4)

The composition of rotations now gives us

\begin{aligned}\mathbf{x}&\rightarrow \mathbf{x}'' = \tilde{R_\theta} \tilde{R_\phi} \mathbf{x} R_\phi R_\theta = \tilde{R} \mathbf{x} R \\ R &= R_\phi R_\theta = e^{ \mathbf{e}_3 \mathbf{e}_1 \theta/2 } e^{ \mathbf{e}_1 \mathbf{e}_2 \phi/2 }.\end{aligned}

# Expressions for the unit vectors.

The unit vectors in the rotated frame can now be calculated. With $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$ we can calculate

\begin{aligned}\hat{\boldsymbol{\phi}} &= \tilde{R} \mathbf{e}_2 R \\ \hat{\mathbf{r}} &= \tilde{R} \mathbf{e}_3 R \\ \hat{\boldsymbol{\theta}} &= \tilde{R} \mathbf{e}_1 R\end{aligned} \hspace{\stretch{1}}(3.6)

Performing these we get

\begin{aligned}\hat{\boldsymbol{\phi}}&= e^{ -\mathbf{e}_1 \mathbf{e}_2 \phi/2 } e^{ -\mathbf{e}_3 \mathbf{e}_1 \theta/2 } \mathbf{e}_2 e^{ \mathbf{e}_3 \mathbf{e}_1 \theta/2 } e^{ \mathbf{e}_1 \mathbf{e}_2 \phi/2 } \\ &= \mathbf{e}_2 e^{ i \phi },\end{aligned}

and

\begin{aligned}\hat{\mathbf{r}}&= e^{ -\mathbf{e}_1 \mathbf{e}_2 \phi/2 } e^{ -\mathbf{e}_3 \mathbf{e}_1 \theta/2 } \mathbf{e}_3 e^{ \mathbf{e}_3 \mathbf{e}_1 \theta/2 } e^{ \mathbf{e}_1 \mathbf{e}_2 \phi/2 } \\ &= e^{ -\mathbf{e}_1 \mathbf{e}_2 \phi/2 } (\mathbf{e}_3 \cos\theta + \mathbf{e}_1 \sin\theta ) e^{ \mathbf{e}_1 \mathbf{e}_2 \phi/2 } \\ &= \mathbf{e}_3 \cos\theta +\mathbf{e}_1 \sin\theta e^{ \mathbf{e}_1 \mathbf{e}_2 \phi } \\ &= \mathbf{e}_3 (\cos\theta + \mathbf{e}_3 \mathbf{e}_1 \sin\theta e^{ \mathbf{e}_1 \mathbf{e}_2 \phi } ) \\ &= \mathbf{e}_3 e^{I \hat{\boldsymbol{\phi}} \theta},\end{aligned}

and

\begin{aligned}\hat{\boldsymbol{\theta}}&= e^{ -\mathbf{e}_1 \mathbf{e}_2 \phi/2 } e^{ -\mathbf{e}_3 \mathbf{e}_1 \theta/2 } \mathbf{e}_1 e^{ \mathbf{e}_3 \mathbf{e}_1 \theta/2 } e^{ \mathbf{e}_1 \mathbf{e}_2 \phi/2 } \\ &= e^{ -\mathbf{e}_1 \mathbf{e}_2 \phi/2 } ( \mathbf{e}_1 \cos\theta - \mathbf{e}_3 \sin\theta ) e^{ \mathbf{e}_1 \mathbf{e}_2 \phi/2 } \\ &= \mathbf{e}_1 \cos\theta e^{ \mathbf{e}_1 \mathbf{e}_2 \phi/2 } - \mathbf{e}_3 \sin\theta \\ &= i \hat{\boldsymbol{\phi}} \cos\theta - \mathbf{e}_3 \sin\theta \\ &= i \hat{\boldsymbol{\phi}} (\cos\theta + \hat{\boldsymbol{\phi}} i \mathbf{e}_3 \sin\theta ) \\ &= i \hat{\boldsymbol{\phi}} e^{I \hat{\boldsymbol{\phi}} \theta}.\end{aligned}

Summarizing these are

\begin{aligned}\hat{\boldsymbol{\phi}} &= \mathbf{e}_2 e^{ i \phi } \\ \hat{\mathbf{r}} &= \mathbf{e}_3 e^{I \hat{\boldsymbol{\phi}} \theta} \\ \hat{\boldsymbol{\theta}} &= i \hat{\boldsymbol{\phi}} e^{I \hat{\boldsymbol{\phi}} \theta}.\end{aligned} \hspace{\stretch{1}}(3.9)

# Derivatives of the unit vectors.

We’ll need the partials. Most of these can be computed from 3.9 by inspection, and are

\begin{aligned}\partial_r \hat{\boldsymbol{\phi}} &= 0 \\ \partial_r \hat{\mathbf{r}} &= 0 \\ \partial_r \hat{\boldsymbol{\theta}} &= 0 \\ \partial_\theta \hat{\boldsymbol{\phi}} &= 0 \\ \partial_\theta \hat{\mathbf{r}} &= \hat{\mathbf{r}} I \hat{\boldsymbol{\phi}} \\ \partial_\theta \hat{\boldsymbol{\theta}} &= \hat{\boldsymbol{\theta}} I \hat{\boldsymbol{\phi}} \\ \partial_\phi \hat{\boldsymbol{\phi}} &= \hat{\boldsymbol{\phi}} i \\ \partial_\phi \hat{\mathbf{r}} &= \hat{\boldsymbol{\phi}} \sin\theta \\ \partial_\phi \hat{\boldsymbol{\theta}} &= \hat{\boldsymbol{\phi}} \cos\theta\end{aligned} \hspace{\stretch{1}}(4.12)

# Expanding the Laplacian.

We note that the line element is $ds = dr + r d\theta + r\sin\theta d\phi$, so our gradient in spherical coordinates is

\begin{aligned}\boldsymbol{\nabla} &= \hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta + \frac{\hat{\boldsymbol{\phi}}}{r\sin\theta} \partial_\phi.\end{aligned} \hspace{\stretch{1}}(5.21)

We can now evaluate the Laplacian

\begin{aligned}\boldsymbol{\nabla}^2 &=\left( \hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta + \frac{\hat{\boldsymbol{\phi}}}{r\sin\theta} \partial_\phi \right) \cdot\left( \hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta + \frac{\hat{\boldsymbol{\phi}}}{r\sin\theta} \partial_\phi \right).\end{aligned} \hspace{\stretch{1}}(5.22)

Evaluating these one set at a time we have

\begin{aligned}\hat{\mathbf{r}} \partial_r \cdot \left( \hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta + \frac{\hat{\boldsymbol{\phi}}}{r\sin\theta} \partial_\phi \right) &= \partial_{rr},\end{aligned}

and

\begin{aligned}\frac{1}{{r}} \hat{\boldsymbol{\theta}} \partial_\theta \cdot \left( \hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta + \frac{\hat{\boldsymbol{\phi}}}{r\sin\theta} \partial_\phi \right)&=\frac{1}{{r}} \left\langle{{\hat{\boldsymbol{\theta}} \left(\hat{\mathbf{r}} I \hat{\boldsymbol{\phi}} \partial_r + \hat{\mathbf{r}} \partial_{\theta r}+ \frac{\hat{\boldsymbol{\theta}}}{r} \partial_{\theta\theta} + \frac{1}{{r}} \hat{\boldsymbol{\theta}} I \hat{\boldsymbol{\phi}} \partial_\theta+ \hat{\boldsymbol{\phi}} \partial_\theta \frac{1}{{r\sin\theta}} \partial_\phi\right)}}\right\rangle \\ &= \frac{1}{{r}} \partial_r+\frac{1}{{r^2}} \partial_{\theta\theta},\end{aligned}

and

\begin{aligned}\frac{\hat{\boldsymbol{\phi}}}{r\sin\theta} \partial_\phi &\cdot\left( \hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta + \frac{\hat{\boldsymbol{\phi}}}{r\sin\theta} \partial_\phi \right) \\ &=\frac{1}{r\sin\theta} \left\langle{{\hat{\boldsymbol{\phi}}\left(\hat{\boldsymbol{\phi}} \sin\theta \partial_r + \hat{\mathbf{r}} \partial_{\phi r} + \hat{\boldsymbol{\phi}} \cos\theta \frac{1}{r} \partial_\theta + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_{\phi \theta }+ \hat{\boldsymbol{\phi}} i \frac{1}{r\sin\theta} \partial_\phi + \hat{\boldsymbol{\phi}} \frac{1}{r\sin\theta} \partial_{\phi \phi }\right)}}\right\rangle \\ &=\frac{1}{{r}} \partial_r+ \frac{\cot\theta}{r^2}\partial_\theta+ \frac{1}{{r^2 \sin^2\theta}} \partial_{\phi\phi}\end{aligned}

Summing these we have

\begin{aligned}\boldsymbol{\nabla}^2 &=\partial_{rr}+ \frac{2}{r} \partial_r+\frac{1}{{r^2}} \partial_{\theta\theta}+ \frac{\cot\theta}{r^2}\partial_\theta+ \frac{1}{{r^2 \sin^2\theta}} \partial_{\phi\phi}\end{aligned} \hspace{\stretch{1}}(5.23)

This is often written with a chain rule trick to considate the $r$ and $\theta$ partials

\begin{aligned}\boldsymbol{\nabla}^2 \Psi &=\frac{1}{{r}} \partial_{rr} (r \Psi)+ \frac{1}{{r^2 \sin\theta}} \partial_\theta \left( \sin\theta \partial_\theta \Psi \right)+ \frac{1}{{r^2 \sin^2\theta}} \partial_{\psi\psi} \Psi\end{aligned} \hspace{\stretch{1}}(5.24)

It’s simple to verify that this is identical to 5.23.

# References

[2] Peeter Joot. Spherical Polar unit vectors in exponential form. [online]. http://sites.google.com/site/peeterjoot/math2009/sphericalPolarUnit.pdf .

## Fourier transform solutions and associated energy and momentum for the homogeneous Maxwell equation. (rework once more)

Posted by peeterjoot on December 29, 2009

[Click here for a PDF of this post with nicer formatting]. Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.

These notes build on and replace those formerly posted in Energy and momentum for assumed Fourier transform solutions to the homogeneous Maxwell equation.

# Motivation and notation.

In Electrodynamic field energy for vacuum (reworked) [1], building on Energy and momentum for Complex electric and magnetic field phasors [2], a derivation for the energy and momentum density was derived for an assumed Fourier series solution to the homogeneous Maxwell’s equation. Here we move to the continuous case examining Fourier transform solutions and the associated energy and momentum density.

A complex (phasor) representation is implied, so taking real parts when all is said and done is required of the fields. For the energy momentum tensor the Geometric Algebra form, modified for complex fields, is used

\begin{aligned}T(a) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \hspace{\stretch{1}}(1.1)

The assumed four vector potential will be written

\begin{aligned}A(\mathbf{x}, t) = A^\mu(\mathbf{x}, t) \gamma_\mu = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(1.2)

Subject to the requirement that $A$ is a solution of Maxwell’s equation

\begin{aligned}\nabla (\nabla \wedge A) = 0.\end{aligned} \hspace{\stretch{1}}(1.3)

To avoid latex hell, no special notation will be used for the Fourier coefficients,

\begin{aligned}A(\mathbf{k}, t) = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{x}.\end{aligned} \hspace{\stretch{1}}(1.4)

When convenient and unambiguous, this $(\mathbf{k},t)$ dependence will be implied.

Having picked a time and space representation for the field, it will be natural to express both the four potential and the gradient as scalar plus spatial vector, instead of using the Dirac basis. For the gradient this is

\begin{aligned}\nabla &= \gamma^\mu \partial_\mu = (\partial_0 - \boldsymbol{\nabla}) \gamma_0 = \gamma_0 (\partial_0 + \boldsymbol{\nabla}),\end{aligned} \hspace{\stretch{1}}(1.5)

and for the four potential (or the Fourier transform functions), this is

\begin{aligned}A &= \gamma_\mu A^\mu = (\phi + \mathbf{A}) \gamma_0 = \gamma_0 (\phi - \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(1.6)

# Setup

The field bivector $F = \nabla \wedge A$ is required for the energy momentum tensor. This is

\begin{aligned}\nabla \wedge A&= \frac{1}{{2}}\left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \\ &= \frac{1}{{2}}\left( (\stackrel{ \rightarrow }{\partial}_0 - \stackrel{ \rightarrow }{\boldsymbol{\nabla}}) \gamma_0 \gamma_0 (\phi - \mathbf{A})-(\phi + \mathbf{A}) \gamma_0 \gamma_0 (\stackrel{ \leftarrow }{\partial}_0 + \stackrel{ \leftarrow }{\boldsymbol{\nabla}})\right) \\ &= -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \frac{1}{{2}}(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}})\end{aligned}

This last term is a spatial curl and the field is then

\begin{aligned}F = -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}\end{aligned} \hspace{\stretch{1}}(2.7)

Applied to the Fourier representation this is

\begin{aligned}F =\frac{1}{{(\sqrt{2 \pi})^3}} \int\left(- \frac{1}{c} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(2.8)

It is only the real parts of this that we are actually interested in, unless physical meaning can be assigned to the complete complex vector field.

# Constraints supplied by Maxwell’s equation.

A Fourier transform solution of Maxwell’s vacuum equation $\nabla F = 0$ has been assumed. Having expressed the Faraday bivector in terms of spatial vector quantities, it is more convenient to do this back substitution into after pre-multiplying Maxwell’s equation by $\gamma_0$, namely

\begin{aligned}0&= \gamma_0 \nabla F \\ &= (\partial_0 + \boldsymbol{\nabla}) F.\end{aligned} \hspace{\stretch{1}}(3.9)

Applied to the spatially decomposed field as specified in (2.7), this is

\begin{aligned}0&=-\partial_0 \boldsymbol{\nabla} \phi-\partial_{00} \mathbf{A}+ \partial_0 \boldsymbol{\nabla} \wedge \mathbf{A}-\boldsymbol{\nabla}^2 \phi- \boldsymbol{\nabla} \partial_0 \mathbf{A}+ \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &=- \partial_0 \boldsymbol{\nabla} \phi - \boldsymbol{\nabla}^2 \phi- \partial_{00} \mathbf{A}- \boldsymbol{\nabla} \cdot \partial_0 \mathbf{A}+ \boldsymbol{\nabla}^2 \mathbf{A} - \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ) \\ \end{aligned}

All grades of this equation must simultaneously equal zero, and the bivector grades have canceled (assuming commuting space and time partials), leaving two equations of constraint for the system

\begin{aligned}0 &=\boldsymbol{\nabla}^2 \phi + \boldsymbol{\nabla} \cdot \partial_0 \mathbf{A}\end{aligned} \hspace{\stretch{1}}(3.11)

\begin{aligned}0 &=\partial_{00} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A}+ \boldsymbol{\nabla} \partial_0 \phi + \boldsymbol{\nabla} ( \boldsymbol{\nabla} \cdot \mathbf{A} )\end{aligned} \hspace{\stretch{1}}(3.12)

It is immediately evident that a gauge transformation could be immediately helpful to simplify things. In [3] the gauge choice $\boldsymbol{\nabla} \cdot \mathbf{A} = 0$ is used. From (3.11) this implies that $\boldsymbol{\nabla}^2 \phi = 0$. Bohm argues that for this current and charge free case this implies $\phi = 0$, but he also has a periodicity constraint. Without a periodicity constraint it is easy to manufacture non-zero counterexamples. One is a linear function in the space and time coordinates

\begin{aligned}\phi = p x + q y + r z + s t\end{aligned} \hspace{\stretch{1}}(3.13)

This is a valid scalar potential provided that the wave equation for the vector potential is also a solution. We can however, force $\phi = 0$ by making the transformation $A^\mu \rightarrow A^\mu + \partial^\mu \psi$, which in non-covariant notation is

\begin{aligned}\phi &\rightarrow \phi + \frac{1}{c} \partial_t \psi \\ \mathbf{A} &\rightarrow \phi - \boldsymbol{\nabla} \psi\end{aligned} \hspace{\stretch{1}}(3.14)

If the transformed field $\phi' = \phi + \partial_t \psi/c$ can be forced to zero, then the complexity of the associated Maxwell equations are reduced. In particular, antidifferentiation of $\phi = -(1/c) \partial_t \psi$, yields

\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x}, 0) - c \int_{\tau=0}^t \phi(\mathbf{x}, \tau) d\tau.\end{aligned} \hspace{\stretch{1}}(3.16)

Dropping primes, the transformed Maxwell equations now take the form

\begin{aligned}0 &= \partial_t( \boldsymbol{\nabla} \cdot \mathbf{A} )\end{aligned} \hspace{\stretch{1}}(3.17)

\begin{aligned}0 &=\partial_{00} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ).\end{aligned} \hspace{\stretch{1}}(3.18)

There are two classes of solutions that stand out for these equations. If the vector potential is constant in time $\mathbf{A}(\mathbf{x},t) = \mathbf{A}(\mathbf{x})$, Maxwell’s equations are reduced to the single equation

\begin{aligned}0&= - \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ).\end{aligned} \hspace{\stretch{1}}(3.19)

Observe that a gradient can be factored out of this equation

\begin{aligned}- \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} )&=\boldsymbol{\nabla} (-\boldsymbol{\nabla} \mathbf{A} + \boldsymbol{\nabla} \cdot \mathbf{A} ) \\ &=-\boldsymbol{\nabla} (\boldsymbol{\nabla} \wedge \mathbf{A}).\end{aligned}

The solutions are then those $\mathbf{A}$s that satisfy both

\begin{aligned}0 &= \partial_t \mathbf{A} \\ 0 &= \boldsymbol{\nabla} (\boldsymbol{\nabla} \wedge \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(3.20)

In particular any non-time dependent potential $\mathbf{A}$ with constant curl provides a solution to Maxwell’s equations. There may be other solutions to (3.19) too that are more general. Returning to (3.17) a second way to satisfy these equations stands out. Instead of requiring of $\mathbf{A}$ constant curl, constant divergence with respect to the time partial eliminates (3.17). The simplest resulting equations are those for which the divergence is a constant in time and space (such as zero). The solution set are then spanned by the vectors $\mathbf{A}$ for which

\begin{aligned}\text{constant} &= \boldsymbol{\nabla} \cdot \mathbf{A} \end{aligned} \hspace{\stretch{1}}(3.22)

\begin{aligned}0 &= \frac{1}{{c^2}} \partial_{tt} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(3.23)

Any $\mathbf{A}$ that both has constant divergence and satisfies the wave equation will via (2.7) then produce a solution to Maxwell’s equation.

# Maxwell equation constraints applied to the assumed Fourier solutions.

Let’s consider Maxwell’s equations in all three forms, (3.11), (3.20), and (3.22) and apply these constraints to the assumed Fourier solution.

In all cases the starting point is a pair of Fourier transform relationships, where the Fourier transforms are the functions to be determined

\begin{aligned}\phi(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \phi(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.24)

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.25)

## Case I. Constant time vector potential. Scalar potential eliminated by gauge transformation.

From (4.24) we require

\begin{aligned}0 = (2 \pi)^{-3/2} \int \partial_t \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.26)

So the Fourier transform also cannot have any time dependence, and we have

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.27)

What is the curl of this? Temporarily falling back to coordinates is easiest for this calculation

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{A}(\mathbf{k}) e^{i\mathbf{k} \cdot \mathbf{x}}&=\sigma_m \partial_m \wedge \sigma_n A^n(\mathbf{k}) e^{i \mathbf{x} \cdot \mathbf{x}} \\ &=\sigma_m \wedge \sigma_n A^n(\mathbf{k}) i k^m e^{i \mathbf{x} \cdot \mathbf{x}} \\ &=i\mathbf{k} \wedge \mathbf{A}(\mathbf{k}) e^{i \mathbf{x} \cdot \mathbf{x}} \\ \end{aligned}

This gives

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.28)

We want to equate the divergence of this to zero. Neglecting the integral and constant factor this requires

\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \left( i \mathbf{k} \wedge \mathbf{A} e^{i\mathbf{k} \cdot \mathbf{x}} \right) \\ &= {\left\langle{{ \sigma_m \partial_m i (\mathbf{k} \wedge \mathbf{A}) e^{i\mathbf{k} \cdot \mathbf{x}} }}\right\rangle}_{1} \\ &= -{\left\langle{{ \sigma_m (\mathbf{k} \wedge \mathbf{A}) k^m e^{i\mathbf{k} \cdot \mathbf{x}} }}\right\rangle}_{1} \\ &= -\mathbf{k} \cdot (\mathbf{k} \wedge \mathbf{A}) e^{i\mathbf{k} \cdot \mathbf{x}} \\ \end{aligned}

Requiring that the plane spanned by $\mathbf{k}$ and $\mathbf{A}(\mathbf{k})$ be perpendicular to $\mathbf{k}$ implies that $\mathbf{A} \propto \mathbf{k}$. The solution set is then completely described by functions of the form

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{k} \psi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k},\end{aligned} \hspace{\stretch{1}}(4.29)

where $\psi(\mathbf{k})$ is an arbitrary scalar valued function. This is however, an extremely uninteresting solution since the curl is uniformly zero

\begin{aligned}F &= \boldsymbol{\nabla} \wedge \mathbf{A} \\ &= (2 \pi)^{-3/2} \int (i \mathbf{k}) \wedge \mathbf{k} \psi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}

Since $\mathbf{k} \wedge \mathbf{k} = 0$, when all is said and done the $\phi = 0$, $\partial_t \mathbf{A} = 0$ case appears to have no non-trivial (zero) solutions. Moving on, …

## Case II. Constant vector potential divergence. Scalar potential eliminated by gauge transformation.

Next in the order of complexity is consideration of the case (3.22). Here we also have $\phi = 0$, eliminated by gauge transformation, and are looking for solutions with the constraint

\begin{aligned}\text{constant} &= \boldsymbol{\nabla} \cdot \mathbf{A}(\mathbf{x}, t) \\ &= (2 \pi)^{-3/2} \int i \mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}

How can this constraint be enforced? The only obvious way is a requirement for $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t)$ to be zero for all $(\mathbf{k},t)$, meaning that our to be determined Fourier transform coefficients are required to be perpendicular to the wave number vector parameters at all times.

The remainder of Maxwell’s equations, (3.23) impose the addition constraint on the Fourier transform $\mathbf{A}(\mathbf{k},t)$

\begin{aligned}0 &= (2 \pi)^{-3/2} \int \left( \frac{1}{{c^2}} \partial_{tt} \mathbf{A}(\mathbf{k}, t) - i^2 \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.30)

For zero equality for all $\mathbf{x}$ it appears that we require the Fourier transforms $\mathbf{A}(\mathbf{k})$ to be harmonic in time

\begin{aligned}\partial_{tt} \mathbf{A}(\mathbf{k}, t) = - c^2 \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t).\end{aligned} \hspace{\stretch{1}}(4.31)

This has the familiar exponential solutions

\begin{aligned}\mathbf{A}(\mathbf{k}, t) = \mathbf{A}_{\pm}(\mathbf{k}) e^{ \pm i c {\left\lvert{\mathbf{k}}\right\rvert} t },\end{aligned} \hspace{\stretch{1}}(4.32)

also subject to a requirement that $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}) = 0$. Our field, where the $\mathbf{A}_{\pm}(\mathbf{k})$ are to be determined by initial time conditions, is by (2.7) of the form

\begin{aligned}F(\mathbf{x}, t)= \text{Real} \frac{i}{(\sqrt{2\pi})^3} \int \Bigl( -{\left\lvert{\mathbf{k}}\right\rvert} \mathbf{A}_{+}(\mathbf{k}) + \mathbf{k} \wedge \mathbf{A}_{+}(\mathbf{k}) \Bigr) \exp(i \mathbf{k} \cdot \mathbf{x} + i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}+ \text{Real} \frac{i}{(\sqrt{2\pi})^3} \int \Bigl( {\left\lvert{\mathbf{k}}\right\rvert} \mathbf{A}_{-}(\mathbf{k}) + \mathbf{k} \wedge \mathbf{A}_{-}(\mathbf{k}) \Bigr) \exp(i \mathbf{k} \cdot \mathbf{x} - i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.33)

Since $0 = \mathbf{k} \cdot \mathbf{A}_{\pm}(\mathbf{k})$, we have $\mathbf{k} \wedge \mathbf{A}_{\pm}(\mathbf{k}) = \mathbf{k} \mathbf{A}_{\pm}$. This allows for factoring out of ${\left\lvert{\mathbf{k}}\right\rvert}$. The structure of the solution is not changed by incorporating the $i (2\pi)^{-3/2} {\left\lvert{\mathbf{k}}\right\rvert}$ factors into $\mathbf{A}_{\pm}$, leaving the field having the general form

\begin{aligned}F(\mathbf{x}, t)= \text{Real} \int ( \hat{\mathbf{k}} - 1 ) \mathbf{A}_{+}(\mathbf{k}) \exp(i \mathbf{k} \cdot \mathbf{x} + i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}+ \text{Real} \int ( \hat{\mathbf{k}} + 1 ) \mathbf{A}_{-}(\mathbf{k}) \exp(i \mathbf{k} \cdot \mathbf{x} - i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.34)

The original meaning of $\mathbf{A}_{\pm}$ as Fourier transforms of the vector potential is obscured by the tidy up change to absorb ${\left\lvert{\mathbf{k}}\right\rvert}$, but the geometry of the solution is clearer this way.

It is also particularly straightforward to confirm that $\gamma_0 \nabla F = 0$ separately for either half of (4.34).

## Case III. Non-zero scalar potential. No gauge transformation.

Now lets work from (3.11). In particular, a divergence operation can be factored from (3.11), for

\begin{aligned}0 = \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \phi + \partial_0 \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(4.35)

Right off the top, there is a requirement for

\begin{aligned}\text{constant} = \boldsymbol{\nabla} \phi + \partial_0 \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(4.36)

In terms of the Fourier transforms this is

\begin{aligned}\text{constant} = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(i \mathbf{k} \phi(\mathbf{k}, t) + \frac{1}{c} \partial_t \mathbf{A}(\mathbf{k}, t)\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.37)

Are there any ways for this to equal a constant for all $\mathbf{x}$ without requiring that constant to be zero? Assuming no for now, and that this constant must be zero, this implies a coupling between the $\phi$ and $\mathbf{A}$ Fourier transforms of the form

\begin{aligned}\phi(\mathbf{k}, t) = -\frac{1}{{i c \mathbf{k}}} \partial_t \mathbf{A}(\mathbf{k}, t)\end{aligned} \hspace{\stretch{1}}(4.38)

A secondary implication is that $\partial_t \mathbf{A}(\mathbf{k}, t) \propto \mathbf{k}$ or else $\phi(\mathbf{k}, t)$ is not a scalar. We had a transverse solution by requiring via gauge transformation that $\phi = 0$, and here we have instead the vector potential in the propagation direction.

A secondary confirmation that this is a required coupling between the scalar and vector potential can be had by evaluating the divergence equation of (4.35)

\begin{aligned}0 = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(- \mathbf{k}^2 \phi(\mathbf{k}, t) + \frac{i\mathbf{k}}{c} \cdot \partial_t \mathbf{A}(\mathbf{k}, t)\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.39)

Rearranging this also produces (4.38). We want to now substitute this relationship into (3.12).

Starting with just the $\partial_0 \phi - \boldsymbol{\nabla} \cdot \mathbf{A}$ part we have

\begin{aligned}\partial_0 \phi + \boldsymbol{\nabla} \cdot \mathbf{A}&=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{i}{c^2 \mathbf{k}} \partial_{tt} \mathbf{A}(\mathbf{k}, t) + i \mathbf{k} \cdot \mathbf{A}\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.40)

Taking the gradient of this brings down a factor of $i\mathbf{k}$ for

\begin{aligned}\boldsymbol{\nabla} (\partial_0 \phi + \boldsymbol{\nabla} \cdot \mathbf{A})&=-\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{1}{c^2} \partial_{tt} \mathbf{A}(\mathbf{k}, t) + \mathbf{k} (\mathbf{k} \cdot \mathbf{A})\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.41)

(3.12) in its entirety is now

\begin{aligned}0 &=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(- (i\mathbf{k})^2 \mathbf{A}+ \mathbf{k} (\mathbf{k} \cdot \mathbf{A})\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.42)

This isn’t terribly pleasant looking. Perhaps going the other direction. We could write

\begin{aligned}\phi = \frac{i}{c \mathbf{k}} \frac{\partial {\mathbf{A}}}{\partial {t}} = \frac{i}{c} \frac{\partial {\psi}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(4.43)

so that

\begin{aligned}\mathbf{A}(\mathbf{k}, t) = \mathbf{k} \psi(\mathbf{k}, t).\end{aligned} \hspace{\stretch{1}}(4.44)

\begin{aligned}0 &=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{1}{{c^2}} \mathbf{k} \psi_{tt}- \boldsymbol{\nabla}^2 \mathbf{k} \psi + \boldsymbol{\nabla} \frac{i}{c^2} \psi_{tt}+\boldsymbol{\nabla}( \boldsymbol{\nabla} \cdot (\mathbf{k} \psi) )\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \\ \end{aligned}

Note that the gradients here operate on everything to the right, including and especially the exponential. Each application of the gradient brings down an additional $i\mathbf{k}$ factor, and we have

\begin{aligned}\frac{1}{{(\sqrt{2 \pi})^3}} \int \mathbf{k} \Bigl(\frac{1}{{c^2}} \psi_{tt}- i^2 \mathbf{k}^2 \psi + \frac{i^2}{c^2} \psi_{tt}+i^2 \mathbf{k}^2 \psi \Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}

This is identically zero, so we see that this second equation provides no additional information. That is somewhat surprising since there is not a whole lot of constraints supplied by the first equation. The function $\psi(\mathbf{k}, t)$ can be anything. Understanding of this curiosity comes from computation of the Faraday bivector itself. From (2.7), that is

\begin{aligned}F = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(-i \mathbf{k} \frac{i}{c}\psi_t - \frac{1}{c} \mathbf{k} \psi_t + i \mathbf{k} \wedge \mathbf{k} \psi\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.45)

All terms cancel, so we see that a non-zero $\phi$ leads to $F = 0$, as was the case when considering (4.24) (a case that also resulted in $\mathbf{A}(\mathbf{k}) \propto \mathbf{k}$).

Can this Fourier representation lead to a non-transverse solution to Maxwell’s equation? If so, it is not obvious how.

# The energy momentum tensor

The energy momentum tensor is then

\begin{aligned}T(a) &= -\frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(- \frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)+ i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)- i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)a\left(- \frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)- i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.46)

Observing that $\gamma_0$ commutes with spatial bivectors and anticommutes with spatial vectors, and writing $\sigma_\mu = \gamma_\mu \gamma_0$, the tensor splits neatly into scalar and spatial vector components

\begin{aligned}T(\gamma_\mu) \cdot \gamma_0 &= \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)+ i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)}}\right\rangle e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}' \\ T(\gamma_\mu) \wedge \gamma_0 &= \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint{\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)+ i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)}}\right\rangle}_{1}e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.47)

In particular for $\mu = 0$, we have

\begin{aligned}H &\equiv T(\gamma_0) \cdot \gamma_0 = \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)\right)\cdot\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)\right)- (\mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)) \cdot (\mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t))\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}' \\ \mathbf{P} &\equiv T(\gamma_\mu) \wedge \gamma_0 = \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(i\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)\right) \cdot\left(\mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)-i\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)\right)\cdot\left(\mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.49)

Integrating this over all space and identification of the delta function

\begin{aligned}\delta(\mathbf{k}) \equiv \frac{1}{{(2 \pi)^3}} \int e^{i \mathbf{k} \cdot \mathbf{x}} d^3 \mathbf{x},\end{aligned} \hspace{\stretch{1}}(5.51)

reduces the tensor to a single integral in the continuous angular wave number space of $\mathbf{k}$.

\begin{aligned}\int T(a) d^3 \mathbf{x} &= -\frac{\epsilon_0}{2} \text{Real} \int\left(- \frac{1}{c} {{\dot{\mathbf{A}}}}^{*}+ i \mathbf{k} {{\phi}}^{*}- i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)a\left(- \frac{1}{c} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.52)

Or,

\begin{aligned}\int T(\gamma_\mu) \gamma_0 d^3 \mathbf{x} =\frac{\epsilon_0}{2} \text{Real} \int{\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}- i \mathbf{k} {{\phi}}^{*}+ i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}+ i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)}}\right\rangle}_{{0,1}}d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.53)

Multiplying out (5.53) yields for $\int H$

\begin{aligned}\int H d^3 \mathbf{x} &=\frac{\epsilon_0}{2} \int d^3 \mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}}\right\rvert}^2 + \mathbf{k}^2 ({\left\lvert{\phi}\right\rvert}^2 + {\left\lvert{\mathbf{A}}\right\rvert}^2 )- {\left\lvert{\mathbf{k} \cdot \mathbf{A}}\right\rvert}^2+ 2 \frac{\mathbf{k}}{c} \cdot \text{Real}( i {{\phi}}^{*} \dot{\mathbf{A}} )\right)\end{aligned} \hspace{\stretch{1}}(5.54)

Recall that the only non-trivial solution we found for the assumed Fourier transform representation of $F$ was for $\phi = 0$, $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t) = 0$. Thus we have for the energy density integrated over all space, just

\begin{aligned}\int H d^3 \mathbf{x} &=\frac{\epsilon_0}{2} \int d^3 \mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}}\right\rvert}^2 + \mathbf{k}^2 {\left\lvert{\mathbf{A}}\right\rvert}^2 \right).\end{aligned} \hspace{\stretch{1}}(5.55)

Observe that we have the structure of a Harmonic oscillator for the energy of the radiation system. What is the canonical momentum for this system? Will it correspond to the Poynting vector, integrated over all space?

Let’s reduce the vector component of (5.53), after first imposing the $\phi=0$, and $\mathbf{k} \cdot \mathbf{A} = 0$ conditions used to above for our harmonic oscillator form energy relationship. This is

\begin{aligned}\int \mathbf{P} d^3 \mathbf{x} &=\frac{\epsilon_0}{2 c} \text{Real} \int d^3 \mathbf{k} \left( i {\mathbf{A}}^{*}_t \cdot (\mathbf{k} \wedge \mathbf{A})+ i (\mathbf{k} \wedge {\mathbf{A}}^{*}) \cdot \mathbf{A}_t\right) \\ &=\frac{\epsilon_0}{2 c} \text{Real} \int d^3 \mathbf{k} \left( -i ({\mathbf{A}}^{*}_t \cdot \mathbf{A}) \mathbf{k}+ i \mathbf{k} ({\mathbf{A}}^{*} \cdot \mathbf{A}_t)\right)\end{aligned}

This is just

\begin{aligned}\int \mathbf{P} d^3 \mathbf{x} &=\frac{\epsilon_0}{c} \text{Real} i \int \mathbf{k} ({\mathbf{A}}^{*} \cdot \mathbf{A}_t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.56)

Recall that the Fourier transforms for the transverse propagation case had the form $\mathbf{A}(\mathbf{k}, t) = \mathbf{A}_{\pm}(\mathbf{k}) e^{\pm i c {\left\lvert{\mathbf{k}}\right\rvert} t}$, where the minus generated the advanced wave, and the plus the receding wave. With substitution of the vector potential for the advanced wave into the energy and momentum results of (5.55) and (5.56) respectively, we have

\begin{aligned}\int H d^3 \mathbf{x} &= \epsilon_0 \int \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k} \\ \int \mathbf{P} d^3 \mathbf{x} &= \epsilon_0 \int \hat{\mathbf{k}} \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.57)

After a somewhat circuitous route, this has the relativistic symmetry that is expected. In particular the for the complete $\mu=0$ tensor we have after integration over all space

\begin{aligned}\int T(\gamma_0) \gamma_0 d^3 \mathbf{x} = \epsilon_0 \int (1 + \hat{\mathbf{k}}) \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.59)

The receding wave solution would give the same result, but directed as $1 - \hat{\mathbf{k}}$ instead.

Observe that we also have the four divergence conservation statement that is expected

\begin{aligned}\frac{\partial {}}{\partial {t}} \int H d^3 \mathbf{x} + \boldsymbol{\nabla} \cdot \int c \mathbf{P} d^3 \mathbf{x} &= 0.\end{aligned} \hspace{\stretch{1}}(5.60)

This follows trivially since both the derivatives are zero. If the integration region was to be more specific instead of a $0 + 0 = 0$ relationship, we’d have the power flux ${\partial {H}}/{\partial {t}}$ equal in magnitude to the momentum change through a bounding surface. For a more general surface the time and spatial dependencies shouldn’t necessarily vanish, but we should still have this radiation energy momentum conservation.

# References

[1] Peeter Joot. Electrodynamic field energy for vacuum. [online]. http://sites.google.com/site/peeterjoot/math2009/fourierMaxVac.pdf.

[2] Peeter Joot. {Energy and momentum for Complex electric and magnetic field phasors.} [online]. http://sites.google.com/site/peeterjoot/math2009/complexFieldEnergy.pdf.

[3] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

## Energy and momentum for assumed Fourier transform solutions to the homogeneous Maxwell equation.

Posted by peeterjoot on December 22, 2009

# Motivation and notation.

In Electrodynamic field energy for vacuum (reworked) [1], building on Energy and momentum for Complex electric and magnetic field phasors [2] a derivation for the energy and momentum density was derived for an assumed Fourier series solution to the homogeneous Maxwell’s equation. Here we move to the continuous case examining Fourier transform solutions and the associated energy and momentum density.

A complex (phasor) representation is implied, so taking real parts when all is said and done is required of the fields. For the energy momentum tensor the Geometric Algebra form, modified for complex fields, is used

\begin{aligned}T(a) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \hspace{\stretch{1}}(1.1)

The assumed four vector potential will be written

\begin{aligned}A(\mathbf{x}, t) = A^\mu(\mathbf{x}, t) \gamma_\mu = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(1.2)

Subject to the requirement that $A$ is a solution of Maxwell’s equation

\begin{aligned}\nabla (\nabla \wedge A) = 0.\end{aligned} \hspace{\stretch{1}}(1.3)

To avoid latex hell, no special notation will be used for the Fourier coefficients,

\begin{aligned}A(\mathbf{k}, t) = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{x}.\end{aligned} \hspace{\stretch{1}}(1.4)

When convenient and unambiguous, this $(\mathbf{k},t)$ dependence will be implied.

Having picked a time and space representation for the field, it will be natural to express both the four potential and the gradient as scalar plus spatial vector, instead of using the Dirac basis. For the gradient this is

\begin{aligned}\nabla &= \gamma^\mu \partial_\mu = (\partial_0 - \boldsymbol{\nabla}) \gamma_0 = \gamma_0 (\partial_0 + \boldsymbol{\nabla}),\end{aligned} \hspace{\stretch{1}}(1.5)

and for the four potential (or the Fourier transform functions), this is

\begin{aligned}A &= \gamma_\mu A^\mu = (\phi + \mathbf{A}) \gamma_0 = \gamma_0 (\phi - \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(1.6)

# Setup

The field bivector $F = \nabla \wedge A$ is required for the energy momentum tensor. This is

\begin{aligned}\nabla \wedge A&= \frac{1}{{2}}\left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \\ &= \frac{1}{{2}}\left( (\stackrel{ \rightarrow }{\partial}_0 - \stackrel{ \rightarrow }{\boldsymbol{\nabla}}) \gamma_0 \gamma_0 (\phi - \mathbf{A})- (\phi + \mathbf{A}) \gamma_0 \gamma_0 (\stackrel{ \leftarrow }{\partial}_0 + \stackrel{ \leftarrow }{\boldsymbol{\nabla}})\right) \\ &= -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \frac{1}{{2}}(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}}) \end{aligned}

This last term is a spatial curl and the field is then

\begin{aligned}F = -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} \end{aligned} \hspace{\stretch{1}}(2.7)

Applied to the Fourier representation this is

\begin{aligned}F = \frac{1}{{(\sqrt{2 \pi})^3}} \int \left( - \frac{1}{{c}} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(2.8)

The energy momentum tensor is then

\begin{aligned}T(a) &= -\frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint \left( - \frac{1}{{c}} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)+ i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)- i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)a\left( - \frac{1}{{c}} \dot{\mathbf{A}}(\mathbf{k}, t)- i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(2.9)

# The tensor integrated over all space. Energy and momentum?

Integrating this over all space and identification of the delta function

\begin{aligned}\delta(\mathbf{k}) \equiv \frac{1}{{(2 \pi)^3}} \int e^{i \mathbf{k} \cdot \mathbf{x}} d^3 \mathbf{x},\end{aligned} \hspace{\stretch{1}}(3.10)

reduces the tensor to a single integral in the continuous angular wave number space of $\mathbf{k}$.

\begin{aligned}\int T(a) d^3 \mathbf{x} &= -\frac{\epsilon_0}{2} \text{Real} \int \left( - \frac{1}{{c}} {{\dot{\mathbf{A}}}}^{*}+ i \mathbf{k} {{\phi}}^{*}- i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)a\left( - \frac{1}{{c}} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(3.11)

Observing that $\gamma_0$ commutes with spatial bivectors and anticommutes with spatial vectors, and writing $\sigma_\mu = \gamma_\mu \gamma_0$, one has

\begin{aligned}\int T(\gamma_\mu) \gamma_0 d^3 \mathbf{x} = \frac{\epsilon_0}{2} \text{Real} \int {\left\langle{{\left( \frac{1}{{c}} {{\dot{\mathbf{A}}}}^{*}- i \mathbf{k} {{\phi}}^{*}+ i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)\sigma_\mu\left( \frac{1}{{c}} \dot{\mathbf{A}}+ i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)}}\right\rangle}_{{0,1}}d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(3.12)

The scalar and spatial vector grade selection operator has been added for convenience and does not change the result since those are necessarily the only grades anyhow. The post multiplication by the observer frame time basis vector $\gamma_0$ serves to separate the energy and momentum like components of the tensor nicely into scalar and vector aspects. In particular for $T(\gamma^0)$, one could write

\begin{aligned}\int T(\gamma^0) d^3 \mathbf{x} = (H + \mathbf{P}) \gamma_0,\end{aligned} \hspace{\stretch{1}}(3.13)

If these are correctly identified with energy and momentum then it also ought to be true that we have the conservation relationship

\begin{aligned}\frac{\partial {H}}{\partial {t}} + \boldsymbol{\nabla} \cdot (c \mathbf{P}) = 0.\end{aligned} \hspace{\stretch{1}}(3.14)

However, multiplying out (3.12) yields for $H$

\begin{aligned}H &= \frac{\epsilon_0}{2} \int d^3 \mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}}\right\rvert}^2 + \mathbf{k}^2 ({\left\lvert{\phi}\right\rvert}^2 + {\left\lvert{\mathbf{A}}\right\rvert}^2 )- {\left\lvert{\mathbf{k} \cdot \mathbf{A}}\right\rvert}^2 + 2 \frac{\mathbf{k}}{c} \cdot \text{Real}( i {{\phi}}^{*} \dot{\mathbf{A}} )\right)\end{aligned} \hspace{\stretch{1}}(3.15)

The vector component takes a bit more work to reduce

\begin{aligned}\mathbf{P} &= \frac{\epsilon_0}{2} \int d^3 \mathbf{k} \text{Real} \left(\frac{i}{c} ({{\dot{\mathbf{A}}}}^{*} \cdot (\mathbf{k} \wedge \mathbf{A})+ {{\phi}}^{*} \mathbf{k} \cdot (\mathbf{k} \wedge \mathbf{A})+ \frac{i}{c} (\mathbf{k} \wedge {\mathbf{A}}^{*}) \cdot \dot{\mathbf{A}}- \phi (\mathbf{k} \wedge {\mathbf{A}}^{*}) \cdot \mathbf{k}\right) \\ &=\frac{\epsilon_0}{2} \int d^3 \mathbf{k} \text{Real} \left(\frac{i}{c} \left( ({{\dot{\mathbf{A}}}}^{*} \cdot \mathbf{k}) \mathbf{A} -({{\dot{\mathbf{A}}}}^{*} \cdot \mathbf{A}) \mathbf{k} \right)+ {{\phi}}^{*} \left( \mathbf{k}^2 \mathbf{A} - (\mathbf{k} \cdot \mathbf{A}) \mathbf{k} \right)+ \frac{i}{c} \left( ({\mathbf{A}}^{*} \cdot \dot{\mathbf{A}}) \mathbf{k} - (\mathbf{k} \cdot \dot{\mathbf{A}}) {\mathbf{A}}^{*} \right)+ \phi \left( \mathbf{k}^2 {\mathbf{A}}^{*} -({\mathbf{A}}^{*} \cdot \mathbf{k}) \mathbf{k} \right) \right).\end{aligned}

Canceling and regrouping leaves

\begin{aligned}\mathbf{P}&=\epsilon_0 \int d^3 \mathbf{k} \text{Real} \left(\mathbf{A} \left( \mathbf{k}^2 {{\phi}}^{*} + \mathbf{k} \cdot {{\dot{\mathbf{A}}}}^{*} \right)+ \mathbf{k} \left( -{{\phi}}^{*} (\mathbf{k} \cdot \mathbf{A}) + \frac{i}{c} ({\mathbf{A}}^{*} \cdot \dot{\mathbf{A}})\right)\right).\end{aligned} \hspace{\stretch{1}}(3.16)

This has no explicit $\mathbf{x}$ dependence, so the conservation relation (3.14) is violated unless ${\partial {H}}/{\partial {t}} = 0$. There is no reason to assume that will be the case. In the discrete Fourier series treatment, a gauge transformation allowed for elimination of $\phi$, and this implied $\mathbf{k} \cdot \mathbf{A}_\mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k}$ constant. We will probably have a similar result here, eliminating most of the terms in (3.15) and (3.16). Except for the constant $\mathbf{A}_\mathbf{k}$ solution of the field equations there is no obvious way that such a simplified energy expression will have zero derivative.

A more reasonable conclusion is that this approach is flawed. We ought to be looking at the divergence relation as a starting point, and instead of integrating over all space, instead employing Gauss’s theorem to convert the divergence integral into a surface integral. Without math, the conservation relationship probably ought to be expressed as energy change in a volume is matched by the momentum change through the surface. However, without an integral over all space, we do not get the nice delta function cancellation observed above. How to proceed is not immediately clear. Stepping back to review applications of Gauss’s theorem is probably a good first step.

# References

[1] Peeter Joot. Electrodynamic field energy for vacuum. [online]. http://sites.google.com/site/peeterjoot/math2009/fourierMaxVac.pdf.

[2] Peeter Joot. {Energy and momentum for Complex electric and magnetic field phasors.} [online]. http://sites.google.com/site/peeterjoot/math2009/complexFieldEnergy.pdf.

## Electrodynamic field energy for vacuum (reworked)

Posted by peeterjoot on December 21, 2009

# Previous version.

Reducing the products in the Dirac basis makes life more complicated then it needs to be (became obvious when attempting to derive an expression for the Poynting integral).

# Motivation.

From Energy and momentum for Complex electric and magnetic field phasors [PDF] how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism is now understood. To recap, for the field $F = \mathbf{E} + I c \mathbf{B}$, where $\mathbf{E}$ and $\mathbf{B}$ may be complex vectors we have for Maxwell’s equation

\begin{aligned}\nabla F = J/\epsilon_0 c.\end{aligned} \quad\quad\quad(1)

This is a doubly complex representation, with the four vector pseudoscalar $I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$ acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution $F$ of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary $i$, as in $\mathbf{E} = \text{Real}(\mathbf{E}) + i \text{Imag}(\mathbf{E})$, is a commuting imaginary, commuting with all the multivector elements in the algebra.

The real valued, four vector, energy momentum tensor $T(a)$ was found to be

\begin{aligned}T(a) = \frac{\epsilon_0}{4} \Bigl( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} \Bigr) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \quad\quad\quad(2)

To supply some context that gives meaning to this tensor the associated conservation relationship was found to be

\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(3)

and in particular for $a = \gamma^0$, this four vector divergence takes the form

\begin{aligned}\frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ \boldsymbol{\nabla} \cdot \frac{1}{{\mu_0}} \text{Real} (\mathbf{E} \times {\mathbf{B}}^{*} )+ \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) = 0,\end{aligned} \quad\quad\quad(4)

relating the energy term $T^{00} = T(\gamma^0) \cdot \gamma^0$ and the Poynting spatial vector $T(\gamma^0) \wedge \gamma^0$ with the current density and electric field product that constitutes the energy portion of the Lorentz force density.

Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [2]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.

# Setup

Let’s assume a Fourier representation for the four vector potential $A$ for the field $F = \nabla \wedge A$. That is

\begin{aligned}A = \sum_{\mathbf{k}} A_\mathbf{k}(t) e^{i \mathbf{k} \cdot \mathbf{x}},\end{aligned} \quad\quad\quad(5)

where summation is over all angular wave number triplets $\mathbf{k} = 2 \pi (k_1/\lambda_1, k_2/\lambda_2, k_3/\lambda_3)$. The Fourier coefficients $A_\mathbf{k} = {A_\mathbf{k}}^\mu \gamma_\mu$ are allowed to be complex valued, as is the resulting four vector $A$, and the associated bivector field $F$.

Fourier inversion, with $V = \lambda_1 \lambda_2 \lambda_3$, follows from

\begin{aligned}\delta_{\mathbf{k}', \mathbf{k}} =\frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} e^{ i \mathbf{k}' \cdot \mathbf{x}} e^{-i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(6)

but only this orthogonality relationship and not the Fourier coefficients themselves

\begin{aligned}A_\mathbf{k} = \frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} A(\mathbf{x}, t) e^{- i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(7)

will be of interest here. Evaluating the curl for this potential yields

\begin{aligned}F = \nabla \wedge A= \sum_{\mathbf{k}} \left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^m \wedge A_\mathbf{k} \frac{2 \pi i k_m}{\lambda_m} \right) e^{i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(8)

Since the four vector potential has been expressed using an explicit split into time and space components it will be natural to re express the bivector field in terms of scalar and (spatial) vector potentials, with the Fourier coefficients. Writing $\sigma_m = \gamma_m \gamma_0$ for the spatial basis vectors, ${A_\mathbf{k}}^0 = \phi_\mathbf{k}$, and $\mathbf{A} = A^k \sigma_k$, this is

\begin{aligned}A_\mathbf{k} = (\phi_\mathbf{k} + \mathbf{A}_\mathbf{k}) \gamma_0.\end{aligned} \quad\quad\quad(9)

The Faraday bivector field $F$ is then

\begin{aligned}F = \sum_\mathbf{k} \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) e^{i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(10)

This is now enough to express the energy momentum tensor $T(\gamma^\mu)$

\begin{aligned}T(\gamma^\mu) &= -\frac{\epsilon_0}{2} \sum_{\mathbf{k},\mathbf{k}'}\text{Real} \left(\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}'})}}^{*} + i \mathbf{k}' {{\phi_{\mathbf{k}'}}}^{*} - i \mathbf{k}' \wedge {{\mathbf{A}_{\mathbf{k}'}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x}}\right).\end{aligned} \quad\quad\quad(11)

It will be more convenient to work with a scalar plus bivector (spatial vector) form of this tensor, and right multiplication by $\gamma_0$ produces such a split

\begin{aligned}T(\gamma^\mu) \gamma_0 = \left\langle{{T(\gamma^\mu) \gamma_0}}\right\rangle + \sigma_a \left\langle{{ \sigma_a T(\gamma^\mu) \gamma_0 }}\right\rangle\end{aligned} \quad\quad\quad(12)

The primary object of this treatment will be consideration of the $\mu = 0$ components of the tensor, which provide a split into energy density $T(\gamma^0) \cdot \gamma_0$, and Poynting vector (momentum density) $T(\gamma^0) \wedge \gamma_0$.

Our first step is to integrate (12) over the volume $V$. This integration and the orthogonality relationship (6), removes the exponentials, leaving

\begin{aligned}\int T(\gamma^\mu) \cdot \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) \gamma_0 }}\right\rangle \\ \int T(\gamma^\mu) \wedge \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) \gamma_0}}\right\rangle \end{aligned} \quad\quad\quad(13)

Because $\gamma_0$ commutes with the spatial bivectors, and anticommutes with the spatial vectors, the remainder of the Dirac basis vectors in these expressions can be eliminated

\begin{aligned}\int T(\gamma^0) \cdot \gamma_0&= -\frac{\epsilon_0 V }{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(15)

\begin{aligned}\int T(\gamma^0) \wedge \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(16)

\begin{aligned}\int T(\gamma^m) \cdot \gamma_0&= \frac{\epsilon_0 V }{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \sigma_m\left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(17)

\begin{aligned}\int T(\gamma^m) \wedge \gamma_0&= \frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \sigma_m\left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle.\end{aligned} \quad\quad\quad(18)

# Expanding the energy momentum tensor components.

## Energy

In (15) only the bivector-bivector and vector-vector products produce any scalar grades. Except for the bivector product this can be done by inspection. For that part we utilize the identity

\begin{aligned}\left\langle{{ (\mathbf{k} \wedge \mathbf{a}) (\mathbf{k} \wedge \mathbf{b}) }}\right\rangle= (\mathbf{a} \cdot \mathbf{k}) (\mathbf{b} \cdot \mathbf{k}) - \mathbf{k}^2 (\mathbf{a} \cdot \mathbf{b}).\end{aligned} \quad\quad\quad(19)

This leaves for the energy $H = \int T(\gamma^0) \cdot \gamma_0$ in the volume

\begin{aligned}H = \frac{\epsilon_0 V}{2} \sum_\mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2 +\mathbf{k}^2 \left( {\left\lvert{\phi_\mathbf{k}}\right\rvert}^2 + {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right) - {\left\lvert{\mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2+ \frac{2}{c} \text{Real} \left( i {{\phi_\mathbf{k}}}^{*} \cdot \dot{\mathbf{A}}_\mathbf{k} \right)\right)\end{aligned} \quad\quad\quad(20)

We are left with a completely real expression, and one without any explicit Geometric Algebra. This does not look like the Harmonic oscillator Hamiltonian that was expected. A gauge transformation to eliminate $\phi_\mathbf{k}$ and an observation about when $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum.

## Momentum

Now move on to (16). For the factors other than $\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form

\begin{aligned}\sigma_a \left\langle{{ \sigma_a \mathbf{a} (\mathbf{k} \wedge \mathbf{c}) }}\right\rangle&=\sigma_a (\mathbf{c} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{k}) - \sigma_a (\mathbf{k} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{c}) \\ &=\mathbf{c} (\mathbf{a} \cdot \mathbf{k}) - \mathbf{k} (\mathbf{a} \cdot \mathbf{c}),\end{aligned}

and the other

\begin{aligned}\sigma_a \left\langle{{ \sigma_a (\mathbf{k} \wedge \mathbf{c}) \mathbf{a} }}\right\rangle&=\sigma_a (\mathbf{k} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{c}) - \sigma_a (\mathbf{c} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{k}) \\ &=\mathbf{k} (\mathbf{a} \cdot \mathbf{c}) - \mathbf{c} (\mathbf{a} \cdot \mathbf{k}).\end{aligned}

The momentum $\mathbf{P} = \int T(\gamma^0) \wedge \gamma_0$ in this volume follows by computation of

\begin{aligned}&\sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \\ &= i \mathbf{A}_\mathbf{k} \left( \left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} \right) \cdot \mathbf{k} \right) - i \mathbf{k} \left( \left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} \right) \cdot \mathbf{A}_\mathbf{k} \right) \\ &- i \mathbf{k} \left( \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} \right) \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right) + i {{\mathbf{A}_{\mathbf{k}}}}^{*} \left( \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} \right) \cdot \mathbf{k} \right)\end{aligned}

All the products are paired in nice conjugates, taking real parts, and premultiplication with $-\epsilon_0 V/2$ gives the desired result. Observe that two of these terms cancel, and another two have no real part. Those last are

\begin{aligned}-\frac{\epsilon_0 V \mathbf{k}}{2 c} \text{Real} \left( i {{(\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{A}_\mathbf{k}+\dot{\mathbf{A}}_\mathbf{k} \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right)&=-\frac{\epsilon_0 V \mathbf{k}}{2 c} \text{Real} \left( i \frac{d}{dt} \mathbf{A}_\mathbf{k} \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right)\end{aligned}

Taking the real part of this pure imaginary $i {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2$ is zero, leaving just

\begin{aligned}\mathbf{P} &= \epsilon_0 V \sum_{\mathbf{k}}\text{Real} \left(i \mathbf{A}_\mathbf{k} \left( \frac{1}{{c}} {{\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{k} \right)+ \mathbf{k}^2 \phi_\mathbf{k} {{ \mathbf{A}_\mathbf{k} }}^{*}- \mathbf{k} {{\phi_\mathbf{k}}}^{*} (\mathbf{k} \cdot \mathbf{A}_\mathbf{k})\right)\end{aligned} \quad\quad\quad(21)

I am not sure why exactly, but I actually expected a term with ${\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2$, quadratic in the vector potential. Is there a mistake above?

## Gauge transformation to simplify the Hamiltonian.

In (20) something that looked like the Harmonic oscillator was expected. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.

If we are to change our four vector potential $A \rightarrow A + \nabla \psi$, then Maxwell’s equation takes the form

\begin{aligned}J/\epsilon_0 c = \nabla (\nabla \wedge (A + \nabla \psi) = \nabla (\nabla \wedge A) + \nabla (\underbrace{\nabla \wedge \nabla \psi}_{=0}),\end{aligned} \quad\quad\quad(22)

which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form

\begin{aligned}A^\mu \rightarrow A^\mu + \partial_\mu \psi,\end{aligned} \quad\quad\quad(23)

and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a $\psi$ such that $A^0 + \partial_0 \psi = 0$. That is

\begin{aligned}\sum_\mathbf{k} \phi_\mathbf{k}(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \frac{1}{{c}} \partial_t \psi = 0.\end{aligned} \quad\quad\quad(24)

Or,

\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x},0) -\frac{1}{{c}} \sum_\mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x}} \int_{\tau=0}^t \phi_\mathbf{k}(\tau).\end{aligned} \quad\quad\quad(25)

With such a transformation, the $\phi_\mathbf{k}$ and $\dot{\mathbf{A}}_\mathbf{k}$ cross term in the Hamiltonian (20) vanishes, as does the $\phi_\mathbf{k}$ term in the four vector square of the last term, leaving just

\begin{aligned}H = \frac{\epsilon_0}{c^2} V \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} \Bigl((c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 + {\left\lvert{ ( c \mathbf{k}) \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2+ {\left\lvert{ c \mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(26)

Additionally, wedging (5) with $\gamma_0$ now does not loose any information so our potential Fourier series is reduced to just

\begin{aligned}\mathbf{A} &= \sum_{\mathbf{k}} \mathbf{A}_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \\ \mathbf{A}_\mathbf{k} &= \frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} \mathbf{A}(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3.\end{aligned} \quad\quad\quad(27)

The desired harmonic oscillator form would be had in (26) if it were not for the $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While $\mathbf{A} = A \wedge \gamma_0$, the lack of an $A^0$ component means that this can be inverted as

\begin{aligned}A = \mathbf{A} \gamma_0 = -\gamma_0 \mathbf{A}.\end{aligned} \quad\quad\quad(29)

The gradient can also be factored scalar and spatial vector components

\begin{aligned}\nabla = \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) = ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0.\end{aligned} \quad\quad\quad(30)

So, with this $A^0 = 0$ gauge choice the bivector field $F$ is

\begin{aligned}F = \nabla \wedge A = \frac{1}{{2}} \left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \end{aligned} \quad\quad\quad(31)

From the left the gradient action on $A$ is

\begin{aligned}\stackrel{ \rightarrow }{\nabla} A &= ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0 (-\gamma_0 \mathbf{A}) \\ &= ( -\partial_0 + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} ) \mathbf{A},\end{aligned}

and from the right

\begin{aligned}A \stackrel{ \leftarrow }{\nabla}&= \mathbf{A} \gamma_0 \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \mathbf{A} ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \partial_0 \mathbf{A} + \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \end{aligned}

Taking the difference we have

\begin{aligned}F &= \frac{1}{{2}} \Bigl( -\partial_0 \mathbf{A} + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \partial_0 \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \Bigr).\end{aligned}

Which is just

\begin{aligned}F = -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}.\end{aligned} \quad\quad\quad(32)

For this vacuum case, premultiplication of Maxwell’s equation by $\gamma_0$ gives

\begin{aligned}0 &= \gamma_0 \nabla ( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= (\partial_0 + \boldsymbol{\nabla})( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} - \partial_0 \boldsymbol{\nabla} \cdot \mathbf{A} - \partial_0 \boldsymbol{\nabla} \wedge \mathbf{A} + \partial_0 ( \boldsymbol{\nabla} \wedge \mathbf{A} ) + \underbrace{\boldsymbol{\nabla} \cdot ( \boldsymbol{\nabla} \wedge \mathbf{A} ) }_{\boldsymbol{\nabla}^2 \mathbf{A} - \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A})}+ \underbrace{\boldsymbol{\nabla} \wedge ( \boldsymbol{\nabla} \wedge \mathbf{A} )}_{=0} \\ \end{aligned}

The spatial bivector and trivector grades are all zero. Equating the remaining scalar and vector components to zero separately yields a pair of equations in $\mathbf{A}$

\begin{aligned}0 &= \partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) \\ 0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) \end{aligned} \quad\quad\quad(33)

If the divergence of the vector potential is constant we have just a wave equation. Let’s see what that divergence is with the assumed Fourier representation

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A} &=\sum_{\mathbf{k} \ne (0,0,0)} {\mathbf{A}_\mathbf{k}}^m 2 \pi i \frac{k_m}{\lambda_m} e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=i \sum_{\mathbf{k} \ne (0,0,0)} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=i \sum_\mathbf{k} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} \end{aligned}

Since $\mathbf{A}_\mathbf{k} = \mathbf{A}_\mathbf{k}(t)$, there are two ways for $\partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) = 0$. For each $\mathbf{k}$ there must be a requirement for either $\mathbf{A}_\mathbf{k} \cdot \mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k} = \text{constant}$. The constant $\mathbf{A}_\mathbf{k}$ solution to the first equation appears to represent a standing spatial wave with no time dependence. Is that of any interest?

The more interesting seeming case is where we have some non-static time varying state. In this case, if $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$, the second of these Maxwell’s equations is just the vector potential wave equation, since the divergence is zero. That is

\begin{aligned}0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} \end{aligned} \quad\quad\quad(35)

Solving this isn’t really what is of interest, since the objective was just to determine if the divergence could be assumed to be zero. This shows then, that if the transverse solution to Maxwell’s equation is picked, the Hamiltonian for this field, with this gauge choice, becomes

\begin{aligned}H = \frac{\epsilon_0}{c^2} V \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} (c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right).\end{aligned} \quad\quad\quad(36)

How does the gauge choice alter the Poynting vector? From (21), all the $\phi_\mathbf{k}$ dependence in that integrated momentum density is lost

\begin{aligned}\mathbf{P} &= \epsilon_0 V \sum_{\mathbf{k}}\text{Real} \left(i \mathbf{A}_\mathbf{k} \left( \frac{1}{{c}} {{\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{k} \right)\right).\end{aligned} \quad\quad\quad(37)

The $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$ solutions to Maxwell’s equation are seen to result in zero momentum for this infinite periodic field. My expectation was something of the form $c \mathbf{P} = H \hat{\mathbf{k}}$, so intuition is either failing me, or my math is failing me, or this contrived periodic field solution leads to trouble.

# Conclusions and followup.

The objective was met, a reproduction of Bohm’s Harmonic oscillator result using a complex exponential Fourier series instead of separate sine and cosines.

The reason for Bohm’s choice to fix zero divergence as the gauge choice upfront is now clear. That automatically cuts complexity from the results. Figuring out how to work this problem with complex valued potentials and also using the Geometric Algebra formulation probably also made the work a bit more difficult since blundering through both simultaneously was required instead of just one at a time.

This was an interesting exercise though, since doing it this way I am able to understand all the intermediate steps. Bohm employed some subtler argumentation to eliminate the scalar potential $\phi$ upfront, and I have to admit I did not follow his logic, whereas blindly following where the math leads me all makes sense.

As a bit of followup, I’d like to consider the constant $\mathbf{A}_\mathbf{k}$ case in more detail, and any implications of the freedom to pick $\mathbf{A}_0$.

The general calculation of $T^{\mu\nu}$ for the assumed Fourier solution should be possible too, but was not attempted. Doing that general calculation with a four dimensional Fourier series is likely tidier than working with scalar and spatial variables as done here.

Now that the math is out of the way (except possibly for the momentum which doesn’t seem right), some discussion of implications and applications is also in order. My preference is to let the math sink-in a bit first and mull over the momentum issues at leisure.

# References

[2] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

## Electrodynamic field energy for vacuum.

Posted by peeterjoot on December 19, 2009

# Motivation.

We now know how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism. To recap, for the field $F = \mathbf{E} + I c \mathbf{B}$, where $\mathbf{E}$ and $\mathbf{B}$ may be complex vectors we have for Maxwell’s equation

\begin{aligned}\nabla F = J/\epsilon_0 c.\end{aligned} \quad\quad\quad(1)

This is a doubly complex representation, with the four vector pseudoscalar $I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$ acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution $F$ of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary $i$, as in $\mathbf{E} = \text{Real}(\mathbf{E}) + i \text{Imag}(\mathbf{E})$, is a commuting imaginary, commuting with all the multivector elements in the algebra.

The real valued, four vector, energy momentum tensor $T(a)$ was found to be

\begin{aligned}T(a) = \frac{\epsilon_0}{4} \Bigl( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} \Bigr) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \quad\quad\quad(2)

To supply some context that gives meaning to this tensor the associated conservation relationship was found to be

\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(3)

and in particular for $a = \gamma^0$, this four vector divergence takes the form

\begin{aligned}\frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ \boldsymbol{\nabla} \cdot \frac{1}{{\mu_0}} \text{Real} (\mathbf{E} \times {\mathbf{B}}^{*} )+ \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) = 0,\end{aligned} \quad\quad\quad(4)

relating the energy term $T^{00} = T(\gamma^0) \cdot \gamma^0$ and the Poynting spatial vector $T(\gamma^0) \wedge \gamma^0$ with the current density and electric field product that constitutes the energy portion of the Lorentz force density.

Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [1]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.

# Setup

Let’s assume a Fourier representation for the four vector potential $A$ for the field $F = \nabla \wedge A$. That is

\begin{aligned}A = \sum_{\mathbf{k}} A_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}},\end{aligned} \quad\quad\quad(5)

where summation is over all wave number triplets $\mathbf{k} = (p/\lambda_1,q/\lambda_2,r/\lambda_3)$. The Fourier coefficients $A_\mathbf{k} = {A_\mathbf{k}}^\mu \gamma_\mu$ are allowed to be complex valued, as is the resulting four vector $A$, and the associated bivector field $F$.

Fourier inversion follows from

\begin{aligned}\delta_{\mathbf{k}', \mathbf{k}} =\frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} e^{2 \pi i \mathbf{k}' \cdot \mathbf{x}} e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(6)

but only this orthogonality relationship and not the Fourier coefficients themselves

\begin{aligned}A_\mathbf{k} = \frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} A(\mathbf{x}, t) e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(7)

will be of interest here. Evaluating the curl for this potential yields

\begin{aligned}F = \nabla \wedge A= \sum_{\mathbf{k}} \left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \sum_{m=1}^3 \gamma^m \wedge A_\mathbf{k} \frac{2 \pi i k_m}{\lambda_m} \right) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(8)

We can now form the energy density

\begin{aligned}U = T(\gamma^0) \cdot \gamma^0=-\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} \gamma^0 F \gamma^0 \Bigr).\end{aligned} \quad\quad\quad(9)

With implied summation over all repeated integer indexes (even without matching uppers and lowers), this is

\begin{aligned}U =-\frac{\epsilon_0}{2} \sum_{\mathbf{k}', \mathbf{k}} \text{Real} \left\langle{{\left( \frac{1}{{c}} \gamma^0 \wedge {{\dot{A}_{\mathbf{k}'}}}^{*} - \gamma^m \wedge {{A_{\mathbf{k}'}}}^{*} \frac{2 \pi i k_m'}{\lambda_m} \right) e^{-2 \pi i \mathbf{k}' \cdot \mathbf{x}}\gamma^0\left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^n \wedge A_\mathbf{k} \frac{2 \pi i k_n}{\lambda_n} \right) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}\gamma^0}}\right\rangle.\end{aligned} \quad\quad\quad(10)

The grade selection used here doesn’t change the result since we already have a scalar, but will just make it convenient to filter out any higher order products that will cancel anyways. Integrating over the volume element and taking advantage of the orthogonality relationship (6), the exponentials are removed, leaving the energy contained in the volume

\begin{aligned}H = -\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2}\sum_{\mathbf{k}} \text{Real} \left\langle{{\left( \frac{1}{{c}} \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} - \gamma^m \wedge {{A_{\mathbf{k}}}}^{*} \frac{2 \pi i k_m}{\lambda_m} \right) \gamma^0\left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^n \wedge A_\mathbf{k} \frac{2 \pi i k_n}{\lambda_n} \right) \gamma^0}}\right\rangle.\end{aligned} \quad\quad\quad(11)

# First reduction of the Hamiltonian.

Let’s take the products involved in sequence one at a time, and evaluate, later adding and taking real parts if required all of

\begin{aligned}\frac{1}{{c^2}}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) \gamma^0 (\gamma^0 \wedge \dot{A}_\mathbf{k}) \gamma^0 }}\right\rangle &=-\frac{1}{{c^2}}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) (\gamma^0 \wedge \dot{A}_\mathbf{k}) }}\right\rangle \end{aligned} \quad\quad\quad(12)

\begin{aligned}- \frac{2 \pi i k_m}{c \lambda_m} \left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) \gamma^0 ( \gamma^0 \wedge \dot{A}_\mathbf{k} ) \gamma^0}}\right\rangle &=\frac{2 \pi i k_m}{c \lambda_m} \left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) ( \gamma^0 \wedge \dot{A}_\mathbf{k} ) }}\right\rangle \end{aligned} \quad\quad\quad(13)

\begin{aligned}\frac{2 \pi i k_n}{c \lambda_n} \left\langle{{ ( \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) \gamma^0 ( \gamma^n \wedge A_\mathbf{k} ) \gamma^0}}\right\rangle &=-\frac{2 \pi i k_n}{c \lambda_n} \left\langle{{ ( \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) ( \gamma^n \wedge A_\mathbf{k} ) }}\right\rangle \end{aligned} \quad\quad\quad(14)

\begin{aligned}-\frac{4 \pi^2 k_m k_n}{\lambda_m \lambda_n}\left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) \gamma^0(\gamma^n \wedge A_\mathbf{k} ) \gamma^0}}\right\rangle. &\end{aligned} \quad\quad\quad(15)

The expectation is to obtain a Hamiltonian for the field that has the structure of harmonic oscillators, where the middle two products would have to be zero or sum to zero or have real parts that sum to zero. The first is expected to contain only products of ${\left\lvert{{\dot{A}_\mathbf{k}}^m}\right\rvert}^2$, and the last only products of ${\left\lvert{{A_\mathbf{k}}^m}\right\rvert}^2$.

While initially guessing that (13) and (14) may cancel, this isn’t so obviously the case. The use of cyclic permutation of multivectors within the scalar grade selection operator $\left\langle{{A B}}\right\rangle = \left\langle{{B A}}\right\rangle$ plus a change of dummy summation indexes in one of the two shows that this sum is of the form $Z + {{Z}}^{*}$. This sum is intrinsically real, so we can neglect one of the two doubling the other, but we will still be required to show that the real part of either is zero.

Lets reduce these one at a time starting with (12), and write $\dot{A}_\mathbf{k} = \kappa$ temporarily

\begin{aligned}\left\langle{{ (\gamma^0 \wedge {{\kappa}}^{*} ) (\gamma^0 \wedge \kappa }}\right\rangle &={\kappa^m}^{{*}} \kappa^{m'}\left\langle{{ \gamma^0 \gamma_m \gamma^0 \gamma_{m'} }}\right\rangle \\ &=-{\kappa^m}^{{*}} \kappa^{m'}\left\langle{{ \gamma_m \gamma_{m'} }}\right\rangle \\ &={\kappa^m}^{{*}} \kappa^{m'}\delta_{m m'}.\end{aligned}

So the first of our Hamiltonian terms is

\begin{aligned}\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_\mathbf{k}}}^{*} ) (\gamma^0 \wedge \dot{A}_\mathbf{k} }}\right\rangle &=\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2}{\left\lvert{{{\dot{A}}_{\mathbf{k}}}^m}\right\rvert}^2.\end{aligned} \quad\quad\quad(16)

Note that summation over $m$ is still implied here, so we’d be better off with a spatial vector representation of the Fourier coefficients $\mathbf{A}_\mathbf{k} = A_\mathbf{k} \wedge \gamma_0$. With such a notation, this contribution to the Hamiltonian is

\begin{aligned}\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2} \dot{\mathbf{A}}_\mathbf{k} \cdot {{\dot{\mathbf{A}}_\mathbf{k}}}^{*}.\end{aligned} \quad\quad\quad(17)

To reduce (13) and (13), this time writing $\kappa = A_\mathbf{k}$, we can start with just the scalar selection

\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) ( \gamma^0 \wedge \dot{\kappa} ) }}\right\rangle &=\Bigl( \gamma^m {{(\kappa^0)}}^{*} - {{\kappa}}^{*} \underbrace{(\gamma^m \cdot \gamma^0)}_{=0} \Bigr) \cdot \dot{\kappa} \\ &={{(\kappa^0)}}^{*} \dot{\kappa}^m\end{aligned}

Thus the contribution to the Hamiltonian from (13) and (13) is

\begin{aligned}\frac{2 \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \pi k_m}{c \lambda_m} \text{Real} \Bigl( i {{(A_\mathbf{k}^0)}}^{*} \dot{A_\mathbf{k}}^m \Bigl)=\frac{2 \pi \epsilon_0 \lambda_1 \lambda_2 \lambda_3}{c} \text{Real} \Bigl( i {{(A_\mathbf{k}^0)}}^{*} \mathbf{k} \cdot \dot{\mathbf{A}}_\mathbf{k} \Bigl).\end{aligned} \quad\quad\quad(18)

Most definitively not zero in general. Our final expansion (15) is the messiest. Again with $A_\mathbf{k} = \kappa$ for short, the grade selection of this term in coordinates is

\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) \gamma^0 (\gamma^n \wedge \kappa ) \gamma^0 }}\right\rangle&=- {{\kappa_\mu}}^{*} \kappa^\nu \left\langle{{ (\gamma^m \wedge \gamma^\mu) \gamma^0 (\gamma_n \wedge \gamma_\nu) \gamma^0 }}\right\rangle\end{aligned} \quad\quad\quad(19)

Expanding this out yields

\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) \gamma^0 (\gamma^n \wedge \kappa ) \gamma^0 }}\right\rangle&=- ( {\left\lvert{\kappa_0}\right\rvert}^2 - {\left\lvert{A^a}\right\rvert}^2 ) \delta_{m n} + {{A^n}}^{*} A^m.\end{aligned} \quad\quad\quad(20)

The contribution to the Hamiltonian from this, with $\phi_\mathbf{k} = A^0_\mathbf{k}$, is then

\begin{aligned}2 \pi^2 \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \Bigl(-\mathbf{k}^2 {{\phi_\mathbf{k}}}^{*} \phi_\mathbf{k} + \mathbf{k}^2 ({{\mathbf{A}_\mathbf{k}}}^{*} \cdot \mathbf{A}_\mathbf{k})+ (\mathbf{k} \cdot {{\mathbf{A}_k}}^{*}) (\mathbf{k} \cdot \mathbf{A}_k)\Bigr).\end{aligned} \quad\quad\quad(21)

A final reassembly of the Hamiltonian from the parts (17) and (18) and (21) is then

\begin{aligned}H = \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2 c^2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{2 \pi}{c} \text{Real} \Bigl( i {{ \phi_\mathbf{k} }}^{*} (\mathbf{k} \cdot \dot{\mathbf{A}}_\mathbf{k}) \Bigl)+2 \pi^2 \Bigl(\mathbf{k}^2 ( -{\left\lvert{\phi_\mathbf{k}}\right\rvert}^2 + {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 ) + {\left\lvert{\mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(22)

This is finally reduced to a completely real expression, and one without any explicit Geometric Algebra. All the four vector Fourier vector potentials written out explicitly in terms of the spacetime split $A_\mathbf{k} = (\phi_\mathbf{k}, \mathbf{A}_\mathbf{k})$, which is natural since an explicit time and space split was the starting point.

# Gauge transformation to simplify the Hamiltonian.

While (22) has considerably simpler form than (11), what was expected, was something that looked like the Harmonic oscillator. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.

If we are to change our four vector potential $A \rightarrow A + \nabla \psi$, then Maxwell’s equation takes the form

\begin{aligned}J/\epsilon_0 c = \nabla (\nabla \wedge (A + \nabla \psi) = \nabla (\nabla \wedge A) + \nabla (\underbrace{\nabla \wedge \nabla \psi}_{=0}),\end{aligned} \quad\quad\quad(23)

which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form

\begin{aligned}A^\mu \rightarrow A^\mu + \partial_\mu \psi,\end{aligned} \quad\quad\quad(24)

and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a $\psi$ such that $A^0 + \partial_0 \psi = 0$. That is

\begin{aligned}\sum_\mathbf{k} \phi_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} + \frac{1}{{c}} \partial_t \psi = 0.\end{aligned} \quad\quad\quad(25)

Or,

\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x},0) -\frac{1}{{c}} \sum_\mathbf{k} e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \int_{\tau=0}^t \phi_\mathbf{k}(\tau).\end{aligned} \quad\quad\quad(26)

With such a transformation, the $\phi_\mathbf{k}$ and $\dot{\mathbf{A}}_\mathbf{k}$ cross term in the Hamiltonian (22) vanishes, as does the $\phi_\mathbf{k}$ term in the four vector square of the last term, leaving just

\begin{aligned}H = \frac{\epsilon_0}{c^2} \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} \Bigl((2 \pi c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 + {\left\lvert{ ( 2 \pi c \mathbf{k}) \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(27)

Additionally, wedging (5) with $\gamma_0$ now does not loose any information so our potential Fourier series is reduced to just

\begin{aligned}\mathbf{A} &= \sum_{\mathbf{k}} \mathbf{A}_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \\ \mathbf{A}_\mathbf{k} &= \frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} \mathbf{A}(\mathbf{x}, t) e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3.\end{aligned} \quad\quad\quad(28)

The desired harmonic oscillator form would be had in (27) if it were not for the $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While $\mathbf{A} = A \wedge \gamma_0$, the lack of an $A^0$ component means that this can be inverted as

\begin{aligned}A = \mathbf{A} \gamma_0 = -\gamma_0 \mathbf{A}.\end{aligned} \quad\quad\quad(30)

The gradient can also be factored scalar and spatial vector components

\begin{aligned}\nabla = \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) = ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0.\end{aligned} \quad\quad\quad(31)

So, with this $A^0 = 0$ gauge choice the bivector field $F$ is

\begin{aligned}F = \nabla \wedge A = \frac{1}{{2}} \left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \end{aligned} \quad\quad\quad(32)

From the left the gradient action on $A$ is

\begin{aligned}\stackrel{ \rightarrow }{\nabla} A &= ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0 (-\gamma_0 \mathbf{A}) \\ &= ( -\partial_0 + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} ) \mathbf{A},\end{aligned}

and from the right

\begin{aligned}A \stackrel{ \leftarrow }{\nabla}&= \mathbf{A} \gamma_0 \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \mathbf{A} ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \partial_0 \mathbf{A} + \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \end{aligned}

Taking the difference we have