# Peeter Joot's (OLD) Blog.

• ## Recent Comments

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 292,269

# Posts Tagged ‘divergence theorem’

## 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)

Similarly for even grade blades these are

\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.

## Some helpful identities

### 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)

To demonstrate, we start with the expansion of

\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.

## Three dimensional divergence theorem with generally parametrized volume element.

Posted by peeterjoot on April 10, 2011

# Obsolete with potential errors.

This post may be in error.  I wrote this before understanding that the gradient used in Stokes Theorem must be projected onto the tangent space of the parameterized surface, as detailed in Alan MacDonald’s Vector and Geometric Calculus.

See the post ‘stokes theorem in geometric algebra‘ [PDF], where this topic has been revisited with this in mind.

# Original Post:

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

# Motivation.

With the divergence of the energy momentum tensor converted from a volume to a surface integral given by

\begin{aligned}\int_V d^3 \mathbf{x} \partial_\beta T^{\beta \alpha} = \oint_{\partial V} d^2 \sigma^\beta T^{\beta \alpha},\end{aligned} \hspace{\stretch{1}}(1.1)

I got to wondering what a closed form algebraic expression for this curious (and foreign seeming) quantity $d^2 \sigma^\beta$ was. It obviously must be related to the normal to the surface. It seemed to me that a natural way to answer this question was to consider this divergence integral over an arbitrarily parametrized volume. This turns out to be overkill, but a useful seeming digression.

## A generally parametrized parallelepiped volume element.

Suppose we parametrize a volume by specifying that all the points in that volume are covered by the position vector from the origin, given by

\begin{aligned}\mathbf{x} = \mathbf{x}(a_1, a_2, a_3).\end{aligned} \hspace{\stretch{1}}(1.2)

At any point in the volume of interest, we can create a level curve, holding two of the parameters $a_\alpha$ constant, and varying the remaining one. In particular, we can construct three direction vectors along these level curves, one for each parameter not held constant

\begin{aligned}d\mathbf{x}_1 &= da_1 \frac{\partial {\mathbf{x}}}{\partial {a_1}} \\ d\mathbf{x}_2 &= da_2 \frac{\partial {\mathbf{x}}}{\partial {a_2}} \\ d\mathbf{x}_3 &= da_3 \frac{\partial {\mathbf{x}}}{\partial {a_3}}\end{aligned} \hspace{\stretch{1}}(1.3)

The span of these vectors, provided they are non-degenerate, forms a parallelepiped, the volume of which is

\begin{aligned}d^3\mathbf{x} = d\mathbf{x}_3 \cdot (d\mathbf{x}_1 \times d\mathbf{x}_2).\end{aligned} \hspace{\stretch{1}}(1.6)

This volume element can be expanded in a number of ways

\begin{aligned}d^3\mathbf{x} &= \frac{\partial {\mathbf{x}}}{\partial {a_1}} \cdot \left( \frac{\partial {\mathbf{x}}}{\partial {a_2}} \times \frac{\partial {\mathbf{x}}}{\partial {a_3}} \right) \\ &= \frac{\partial {x^\alpha}}{\partial {a_1}} \frac{\partial {x^\beta}}{\partial {a_2}} \frac{\partial {x^\gamma}}{\partial {a_3}} \epsilon_{\alpha \beta \gamma} da_1 da_2 da_3 \\ &= \frac{\partial {x^1}}{\partial {a_\alpha}} \frac{\partial {x^2}}{\partial {a_\beta}} \frac{\partial {x^3}}{\partial {a_\gamma}} \epsilon_{\alpha \beta \gamma} da_1 da_2 da_3 \\ &= \frac{\partial {x^1}}{\partial {a_{[1}}} \frac{\partial {x^2}}{\partial {a_{2}}} \frac{\partial {x^3}}{\partial {a_{3]}}} da_1 da_2 da_2 \\ &= {\left\lvert{ \frac{\partial(x^1, x^2, x^3)}{\partial (a_1, a_2, a_3)}}\right\rvert}da_1 da_2 da_3 \\ \end{aligned}

where the Jacobian determinant is given by

\begin{aligned}{\left\lvert{ \frac{\partial(x^1, x^2, x^3)}{\partial (a_1, a_2, a_3)}}\right\rvert}= \begin{vmatrix} \frac{\partial {x^1}}{\partial {a_1}} & \frac{\partial {x^2}}{\partial {a_1}} & \frac{\partial {x^3}}{\partial {a_1}} \\ \frac{\partial {x^1}}{\partial {a_2}} & \frac{\partial {x^2}}{\partial {a_2}} & \frac{\partial {x^3}}{\partial {a_2}} \\ \frac{\partial {x^1}}{\partial {a_3}} & \frac{\partial {x^2}}{\partial {a_3}} & \frac{\partial {x^3}}{\partial {a_3}}\end{vmatrix}.\end{aligned} \hspace{\stretch{1}}(1.7)

Provided we are interested in a volume for which the sign of this Jacobian determinant does not change sign, our task is to evaluate and reduce the integral

\begin{aligned}\int {\left\lvert{ \frac{\partial(x^1, x^2, x^3)}{\partial (a_1, a_2, a_3)}}\right\rvert}da_1 da_2 da_3 \frac{\partial {T^{\beta \alpha}}}{\partial {x^\beta}}\end{aligned} \hspace{\stretch{1}}(1.8)

to a set (and sum of) two dimensional integrals.

## On the geometry of the surfaces.

Suppose that we integrate over the ranges $[a_{1-}, a_{1+}]$, $[a_{2-}, a_{2+}]$, $[a_{3-}, a_{3+}]$. Observe that the outwards normals along the $a_1 = a_1+$ face is $d\mathbf{n}_{1+} = da_2 da_3 {\partial {\mathbf{x}}}/{\partial {a_2}} \times {\partial {\mathbf{x}}}/{\partial {a_3}}$. This is

\begin{aligned}d\mathbf{n}_{1+} = da_2 da_3 \frac{\partial {\mathbf{x}}}{\partial {a_2}} \times \frac{\partial {\mathbf{x}}}{\partial {a_3}}= da_2 da_3 \frac{\partial {x^\mu}}{\partial {a_2}} \frac{\partial {x^\nu}}{\partial {a_3}} \epsilon_{\mu \nu \gamma} \mathbf{e}_\gamma\end{aligned} \hspace{\stretch{1}}(1.9)

Similarly our normal on the $a_2 = a_{2+}$ face is

\begin{aligned}d\mathbf{n}_{2+} = da_3 da_1 \frac{\partial {\mathbf{x}}}{\partial {a_3}} \times \frac{\partial {\mathbf{x}}}{\partial {a_1}}= da_3 da_1 \frac{\partial {x^\mu}}{\partial {a_3}} \frac{\partial {x^\nu}}{\partial {a_1}} \epsilon_{\mu \nu \gamma} \mathbf{e}_\gamma,\end{aligned} \hspace{\stretch{1}}(1.10)

and on the $a_3 = a_{3+}$ face the outward normal is

\begin{aligned}d\mathbf{n}_{3+} = da_1 da_2 \frac{\partial {\mathbf{x}}}{\partial {a_1}} \times \frac{\partial {\mathbf{x}}}{\partial {a_2}}= da_1 da_2 \frac{\partial {x^\mu}}{\partial {a_1}} \frac{\partial {x^\nu}}{\partial {a_2}} \epsilon_{\mu \nu \gamma} \mathbf{e}_\gamma.\end{aligned} \hspace{\stretch{1}}(1.11)

Along the $a_{\alpha-}$ faces these are just negated. We can summarize these as

\begin{aligned}d\mathbf{n}_{\sigma\pm} = \pm \frac{1}{{2!}} da_\alpha da_\beta \frac{\partial {\mathbf{x}}}{\partial {a_\alpha}} \times \frac{\partial {\mathbf{x}}}{\partial {a_\beta}} \epsilon_{\alpha \beta \sigma}= \pm \frac{1}{{2!}} da_\alpha da_\beta \frac{\partial {x^\mu}}{\partial {a_\alpha}} \frac{\partial {x^\nu}}{\partial {a_\beta}} \epsilon_{\alpha \beta \sigma} \epsilon_{\mu \nu \gamma} \mathbf{e}_\gamma \end{aligned} \hspace{\stretch{1}}(1.12)

## Expansion of the Jacobian determinant

Suppose, to start with, our divergence volume integral 1.8 has just the following term

\begin{aligned}\int d^3 \mathbf{x} \partial_3 M.\end{aligned} \hspace{\stretch{1}}(1.13)

The specifics of how the scalar $M = T^{3 \alpha}$ is indexed will not matter yet, so let’s suppress it. The Jacobian determinant can be expanded along the $\frac{\partial {x^3}}{\partial {a_\alpha}}$ column for

\begin{aligned}\int d^3 \mathbf{x} \partial_3 M&=\int da_1 da_2 da_3{\left\lvert{ \frac{\partial(x^1, x^2, x^3)}{\partial (a_1, a_2, a_3)}}\right\rvert} \frac{\partial {M}}{\partial {x^3}} \\ &=\int da_1 da_2 da_3\left(\frac{\partial {x^1}}{\partial {a_{[1}}} \frac{\partial {x^2}}{\partial {a_{2}}} \frac{\partial {x^3}}{\partial {a_{3]}}} \right)\frac{\partial {M}}{\partial {x^3}} \\ &=\int da_1 da_2 da_3\left(\frac{\partial {x^1}}{\partial {a_{[1}}} \frac{\partial {x^2}}{\partial {a_{2]}}} \frac{\partial {x^3}}{\partial {a_{3}}} +\frac{\partial {x^1}}{\partial {a_{[2}}} \frac{\partial {x^2}}{\partial {a_{3]}}} \frac{\partial {x^3}}{\partial {a_{1}}} +\frac{\partial {x^1}}{\partial {a_{[3}}} \frac{\partial {x^2}}{\partial {a_{1]}}} \frac{\partial {x^3}}{\partial {a_{2}}} \right)\frac{\partial {M}}{\partial {x^3}} \\ &=\int da_1 da_2 da_3\left({\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_1, a_2)}}\right\rvert} \frac{\partial {x^3}}{\partial {a_{3}}} +{\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_2, a_3)}}\right\rvert} \frac{\partial {x^3}}{\partial {a_{1}}} +{\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_3, a_1)}}\right\rvert} \frac{\partial {x^3}}{\partial {a_{2}}} \right)\frac{\partial {M}}{\partial {x^3}} \\ &=\int da_1 da_2 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_1, a_2)}}\right\rvert} \int da_3 \frac{\partial {x^3}}{\partial {a_{3}}} \frac{\partial {M}}{\partial {x^3}} \\ &\qquad +\int da_2 da_3 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_2, a_3)}}\right\rvert} \int da_1 \frac{\partial {x^3}}{\partial {a_{1}}} \frac{\partial {M}}{\partial {x^3}} \\ &\qquad +\int da_3 da_1 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_3, a_1)}}\right\rvert} \int da_2 \frac{\partial {x^3}}{\partial {a_{2}}} \frac{\partial {M}}{\partial {x^3}} \\ &=\int da_1 da_2 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_1, a_2)}}\right\rvert} \int da_3 \frac{\partial {M}}{\partial {a_{3}}} \\ &\qquad +\int da_2 da_3 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_2, a_3)}}\right\rvert} \int da_1 \frac{\partial {M}}{\partial {a_{1}}} \\ &\qquad +\int da_3 da_1 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_3, a_1)}}\right\rvert} \int da_2 \frac{\partial {M}}{\partial {a_{2}}} \\ &=\int da_1 da_2 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_1, a_2)}}\right\rvert} \Bigl( M(a_{3+}) - M(a_{3+}) \Bigr) \\ &\qquad +\int da_2 da_3 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_2, a_3)}}\right\rvert} \Bigl( M(a_{1+}) - M(a_{1+}) \Bigr) \\ &\qquad +\int da_3 da_1 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_3, a_1)}}\right\rvert} \Bigl( M(a_{2+}) - M(a_{2+}) \Bigr)\end{aligned}

Performing the same task (really just performing cyclic permutation of indexes) we can now construct the whole divergence integral

\begin{aligned}\int d^3 \mathbf{x} \partial_\beta T^{\beta \alpha}&=\int da_1 da_2 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_1, a_2)}}\right\rvert} \Bigl( T^{3 \alpha}(a_{3+}) - T^{3 \alpha}(a_{3+}) \Bigr) \\ &\qquad +\int da_2 da_3 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_2, a_3)}}\right\rvert} \Bigl( T^{3 \alpha}(a_{1+}) - T^{3 \alpha}(a_{1+}) \Bigr) \\ &\qquad +\int da_3 da_1 {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_3, a_1)}}\right\rvert} \Bigl( T^{3 \alpha}(a_{2+}) - T^{3 \alpha}(a_{2+}) \Bigr) \\ &+\int da_1 da_2 {\left\lvert{ \frac{\partial(x^2, x^3)}{\partial (a_1, a_2)}}\right\rvert} \Bigl( T^{1 \alpha}(a_{3+}) - T^{1 \alpha}(a_{3+}) \Bigr) \\ &\qquad +\int da_2 da_3 {\left\lvert{ \frac{\partial(x^2, x^3)}{\partial (a_2, a_3)}}\right\rvert} \Bigl( T^{1 \alpha}(a_{1+}) - T^{1 \alpha}(a_{1+}) \Bigr) \\ &\qquad +\int da_3 da_1 {\left\lvert{ \frac{\partial(x^2, x^3)}{\partial (a_3, a_1)}}\right\rvert} \Bigl( T^{1 \alpha}(a_{2+}) - T^{1 \alpha}(a_{2+}) \Bigr) \\ &+\int da_1 da_2 {\left\lvert{ \frac{\partial(x^3, x^1)}{\partial (a_1, a_2)}}\right\rvert} \Bigl( T^{2 \alpha}(a_{3+}) - T^{2 \alpha}(a_{3+}) \Bigr) \\ &\qquad +\int da_2 da_3 {\left\lvert{ \frac{\partial(x^3, x^1)}{\partial (a_2, a_3)}}\right\rvert} \Bigl( T^{2 \alpha}(a_{1+}) - T^{2 \alpha}(a_{1+}) \Bigr) \\ &\qquad +\int da_3 da_1 {\left\lvert{ \frac{\partial(x^3, x^1)}{\partial (a_3, a_1)}}\right\rvert} \Bigl( T^{2 \alpha}(a_{2+}) - T^{2 \alpha}(a_{2+}) \Bigr).\end{aligned}

Regrouping we have

\begin{aligned}\int d^3 \mathbf{x} \partial_\beta T^{\beta \alpha}&=\int da_1 da_2 \left( {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_1, a_2)}}\right\rvert} {\left.{{T^{3 \alpha}}}\right\vert}_{{\Delta a_3}}+{\left\lvert{ \frac{\partial(x^2, x^3)}{\partial (a_1, a_2)}}\right\rvert} {\left.{{T^{1 \alpha}}}\right\vert}_{{\Delta a_3}}+{\left\lvert{ \frac{\partial(x^3, x^1)}{\partial (a_1, a_2)}}\right\rvert} {\left.{{T^{2 \alpha}}}\right\vert}_{{\Delta a_3}}\right) \\ &+\int da_2 da_3 \left( {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_2, a_3)}}\right\rvert} {\left.{{T^{3 \alpha}}}\right\vert}_{{\Delta a_3}}+{\left\lvert{ \frac{\partial(x^2, x^3)}{\partial (a_2, a_3)}}\right\rvert} {\left.{{T^{1 \alpha}}}\right\vert}_{{\Delta a_3}}+{\left\lvert{ \frac{\partial(x^3, x^1)}{\partial (a_2, a_3)}}\right\rvert} {\left.{{T^{2 \alpha}}}\right\vert}_{{\Delta a_3}}\right) \\ &+\int da_3 da_1 \left( {\left\lvert{ \frac{\partial(x^1, x^2)}{\partial (a_3, a_1)}}\right\rvert} {\left.{{T^{3 \alpha}}}\right\vert}_{{\Delta a_3}}+{\left\lvert{ \frac{\partial(x^2, x^3)}{\partial (a_3, a_1)}}\right\rvert} {\left.{{T^{1 \alpha}}}\right\vert}_{{\Delta a_3}}+{\left\lvert{ \frac{\partial(x^3, x^1)}{\partial (a_3, a_1)}}\right\rvert} {\left.{{T^{2 \alpha}}}\right\vert}_{{\Delta a_3}}\right).\end{aligned}

Observe that we can factor these sums utilizing the normals for the parallelepiped volume element

\begin{aligned}\int d^3 \mathbf{x} \partial_\beta T^{\beta \alpha}&=\int da_1 da_2 {\left\lvert{ \frac{\partial(x^\mu, x^\nu)}{\partial (a_1, a_2)}}\right\rvert} \epsilon_{\mu \nu \gamma} \mathbf{e}_\gamma \cdot \mathbf{e}_\beta{\left.{{T^{\beta \alpha}}}\right\vert}_{{\Delta a_3}} \\ &+\int da_2 da_3 {\left\lvert{ \frac{\partial(x^\mu, x^\nu)}{\partial (a_2, a_3)}}\right\rvert} \epsilon_{\mu \nu \gamma} \mathbf{e}_\gamma \cdot \mathbf{e}_\beta{\left.{{T^{\beta \alpha}}}\right\vert}_{{\Delta a_1}} \\ &+\int da_3 da_1 {\left\lvert{ \frac{\partial(x^\mu, x^\nu)}{\partial (a_3, a_1)}}\right\rvert} \epsilon_{\mu \nu \gamma} \mathbf{e}_\gamma \cdot \mathbf{e}_\beta{\left.{{T^{\beta \alpha}}}\right\vert}_{{\Delta a_2}}\end{aligned}

Let’s look at the first of these integrals in more detail. We integrate the values of the $\mathbf{e}_\beta T^{\beta \alpha}$ evaluated on the points of the surface for which $a_3 = a_{3+}$. To perform this integral we dot against the outward normal area element $da_1 da_2 {\partial {x^\mu}}/{\partial {a_1}} {\partial {x^\nu}}/{\partial {a_2}} \epsilon_{\mu\nu\gamma} \mathbf{e}_\gamma$. We do the same, but subtract the integral where $\mathbf{e}_\beta T^{\beta\alpha}$ is evaluated on the surface $a_3 = a_{3-}$, where we dot with the area element that has the inwards normal direction on that surface. This is then done for each of the surfaces of the parallelepiped that we are integrating over.

In terms of the outwards (area scaled) normals $d\mathbf{n}_3, d\mathbf{n}_1, d\mathbf{n}_2$ on the $a_{3+}, a_{1+}$ and $a_{2+}$ surfaces respectively we can write

\begin{aligned}\int d^3 \mathbf{x} \partial_\beta T^{\beta \alpha} = \int d\mathbf{n}_3 \cdot \mathbf{e}_\beta {\left.{{T^{\beta}{\alpha}}}\right\vert}_{{\Delta a_3}}+\int d\mathbf{n}_1 \cdot \mathbf{e}_\beta {\left.{{T^{\beta}{\alpha}}}\right\vert}_{{\Delta a_1}}+\int d\mathbf{n}_2 \cdot \mathbf{e}_\beta {\left.{{T^{\beta}{\alpha}}}\right\vert}_{{\Delta a_2}}.\end{aligned} \hspace{\stretch{1}}(1.14)

This can be written more concisely in index form with

\begin{aligned}d^2 \sigma^\beta = \epsilon_{\mu\nu\beta} \left(\frac{\partial {x^\mu}}{\partial {a_2}}\frac{\partial {x^\nu}}{\partial {a_3}} da_2 da_3+\frac{\partial {x^\mu}}{\partial {a_3}}\frac{\partial {x^\nu}}{\partial {a_1}} da_3 da_1+\frac{\partial {x^\mu}}{\partial {a_1}}\frac{\partial {x^\nu}}{\partial {a_2}} da_1 da_2\right),\end{aligned} \hspace{\stretch{1}}(1.15)

so that the divergence integral is just

\begin{aligned}\int d^3 \mathbf{x} = \int_{\text{over level surfaceslatex a_{1+}, $a_{2+}$, $a_{3+}$}} d^2 \sigma^\beta T^{\beta \alpha}-\int_{\text{over level surfaces $a_{1-}$, $a_{2-}$, $a_{3-}$}} d^2 \sigma^\beta T^{\beta \alpha}\end{aligned} \hspace{\stretch{1}}(1.16)

In each case, for the $a_{\alpha-}$ surfaces, our negated inwards normal form can be redefined so that we integrate over only the outwards normal directions, and we can use the oriented integral notation

\begin{aligned}\int d^3 \mathbf{x} = \oint d^2 \sigma^\beta T^{\beta \alpha},\end{aligned} \hspace{\stretch{1}}(1.17)

To encode (or imply) whether we require a positive or negative sign on the area element tensor of 1.15 for the surface in question.

## A look back, and looking forward.

Now, having performed this long winded calculation, the meaning of $d^2 \sigma^\beta$ becomes clear. What’s also clear is how this could have been arrived at directly utilizing the divergence theorem in its normal vector form. We had only to re-write our equation as a vector equation in terms of the gradient

\begin{aligned}\int_V d^3 \mathbf{x} \frac{\partial {T^{\beta \alpha}}}{\partial {x^\alpha}} = \int_V d^3 \mathbf{x} \boldsymbol{\nabla} \cdot (\mathbf{e}_\beta T^{\beta \alpha}) = \int_{\partial_V} dA \mathbf{n} \cdot \mathbf{e}_\beta T^{\beta \alpha}\end{aligned} \hspace{\stretch{1}}(1.18)

From this we see directly that $d^2 \sigma^\beta = dA \mathbf{n} \cdot \mathbf{e}_\beta$.

Despite there being an easier way to find the form of $d^2 \sigma^\beta$, I still consider this a worthwhile exercise. It hints how one could generalize the arguments to the higher dimensional cases. The main task would be to construct the normals to the hypersurfaces bounding the hypervolume, and how to do this algebraically utilizing determinants may not be too hard (since we want a Jacobian determinant as the hypervolume element in the “volume” integral). We also got more than the normal physics text book proof of the divergence theorem for Cartesian coordinates, and did it here for a general parametrization. This wasn’t a complete argument since we didn’t consider a general surface, broken down into a triangular mesh. We really want volume elements with triangular sides instead of parallelograms.

## PHY450H1S (relativistic electrodynamics) Problem Set 3.

Posted by peeterjoot on March 2, 2011

[Click here for a PDF of this post with nicer formatting]

# Disclaimer.

This problem set is as yet ungraded (although only the second question will be graded).

# Problem 1. Fun with $\epsilon_{\alpha\beta\gamma}$, $\epsilon^{ijkl}$, $F_{ij}$, and the duality of Maxwell’s equations in vacuum.

## 1. Statement. rank 3 spatial antisymmetric tensor identities.

Prove that

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\delta_{\alpha\mu} \delta_{\beta\nu}-\delta_{\alpha\nu} \delta_{\beta\mu}\end{aligned} \hspace{\stretch{1}}(2.1)

and use it to find the familiar relation for

\begin{aligned}(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})\end{aligned} \hspace{\stretch{1}}(2.2)

Also show that

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}=2 \delta_{\alpha\mu}.\end{aligned} \hspace{\stretch{1}}(2.3)

(Einstein summation implied all throughout this problem).

## 1. Solution

We can explicitly expand the (implied) sum over indexes $\gamma$. This is

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\epsilon_{\alpha \beta 1} \epsilon_{\mu \nu 1}+\epsilon_{\alpha \beta 2} \epsilon_{\mu \nu 2}+\epsilon_{\alpha \beta 3} \epsilon_{\mu \nu 3}\end{aligned} \hspace{\stretch{1}}(2.4)

For any $\alpha \ne \beta$ only one term is non-zero. For example with $\alpha,\beta = 2,3$, we have just a contribution from the $\gamma = 1$ part of the sum

\begin{aligned}\epsilon_{2 3 1} \epsilon_{\mu \nu 1}.\end{aligned} \hspace{\stretch{1}}(2.5)

The value of this for $(\mu,\nu) = (\alpha,\beta)$ is

\begin{aligned}(\epsilon_{2 3 1})^2\end{aligned} \hspace{\stretch{1}}(2.6)

whereas for $(\mu,\nu) = (\beta,\alpha)$ we have

\begin{aligned}-(\epsilon_{2 3 1})^2\end{aligned} \hspace{\stretch{1}}(2.7)

Our sum has value one when $(\alpha, \beta)$ matches $(\mu, \nu)$, and value minus one for when $(\mu, \nu)$ are permuted. We can summarize this, by saying that when $\alpha \ne \beta$ we have

\begin{aligned}\boxed{\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\delta_{\alpha\mu} \delta_{\beta\nu}-\delta_{\alpha\nu} \delta_{\beta\mu}.}\end{aligned} \hspace{\stretch{1}}(2.8)

However, observe that when $\alpha = \beta$ the RHS is

\begin{aligned}\delta_{\alpha\mu} \delta_{\alpha\nu}-\delta_{\alpha\nu} \delta_{\alpha\mu} = 0,\end{aligned} \hspace{\stretch{1}}(2.9)

as desired, so this form works in general without any $\alpha \ne \beta$ qualifier, completing this part of the problem.

\begin{aligned}(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})&=(\epsilon_{\alpha \beta \gamma} \mathbf{e}^\alpha A^\beta B^\gamma ) \cdot(\epsilon_{\mu \nu \sigma} \mathbf{e}^\mu C^\nu D^\sigma ) \\ &=\epsilon_{\alpha \beta \gamma} A^\beta B^\gamma\epsilon_{\alpha \nu \sigma} C^\nu D^\sigma \\ &=(\delta_{\beta \nu} \delta_{\gamma\sigma}-\delta_{\beta \sigma} \delta_{\gamma\nu} )A^\beta B^\gammaC^\nu D^\sigma \\ &=A^\nu B^\sigmaC^\nu D^\sigma-A^\sigma B^\nuC^\nu D^\sigma.\end{aligned}

This gives us

\begin{aligned}\boxed{(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})=(\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D})-(\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C}).}\end{aligned} \hspace{\stretch{1}}(2.10)

We have one more identity to deal with.

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}\end{aligned} \hspace{\stretch{1}}(2.11)

We can expand out this (implied) sum slow and dumb as well

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}&=\epsilon_{\alpha 1 2} \epsilon_{\mu 1 2}+\epsilon_{\alpha 2 1} \epsilon_{\mu 2 1} \\ &+\epsilon_{\alpha 1 3} \epsilon_{\mu 1 3}+\epsilon_{\alpha 3 1} \epsilon_{\mu 3 1} \\ &+\epsilon_{\alpha 2 3} \epsilon_{\mu 2 3}+\epsilon_{\alpha 3 2} \epsilon_{\mu 3 2} \\ &=2 \epsilon_{\alpha 1 2} \epsilon_{\mu 1 2}+ 2 \epsilon_{\alpha 1 3} \epsilon_{\mu 1 3}+ 2 \epsilon_{\alpha 2 3} \epsilon_{\mu 2 3}\end{aligned}

Now, observe that for any $\alpha \in (1,2,3)$ only one term of this sum is picked up. For example, with no loss of generality, pick $\alpha = 1$. We are left with only

\begin{aligned}2 \epsilon_{1 2 3} \epsilon_{\mu 2 3}\end{aligned} \hspace{\stretch{1}}(2.12)

This has the value

\begin{aligned}2 (\epsilon_{1 2 3})^2 = 2\end{aligned} \hspace{\stretch{1}}(2.13)

when $\mu = \alpha$ and is zero otherwise. We can therefore summarize the evaluation of this sum as

\begin{aligned}\boxed{\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}= 2\delta_{\alpha\mu},}\end{aligned} \hspace{\stretch{1}}(2.14)

completing this problem.

## 2. Statement. Determinant of three by three matrix.

Prove that for any $3 \times 3$ matrix ${\left\lVert{A_{\alpha\beta}}\right\rVert}$: $\epsilon_{\mu\nu\lambda} A_{\alpha \mu} A_{\beta\nu} A_{\gamma\lambda} = \epsilon_{\alpha \beta \gamma} \text{Det} A$ and that $\epsilon_{\alpha\beta\gamma} \epsilon_{\mu\nu\lambda} A_{\alpha \mu} A_{\beta\nu} A_{\gamma\lambda} = 6 \text{Det} A$.

## 2. Solution

In class Simon showed us how the first identity can be arrived at using the triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \text{Det}(\mathbf{a} \mathbf{b} \mathbf{c})$. It occurred to me later that I’d seen the identity to be proven in the context of Geometric Algebra, but hadn’t recognized it in this tensor form. Basically, a wedge product can be expanded in sums of determinants, and when the dimension of the space is the same as the vector, we have a pseudoscalar times the determinant of the components.

For example, in $\mathbb{R}^{2}$, let’s take the wedge product of a pair of vectors. As preparation for the relativistic $\mathbb{R}^{4}$ case We won’t require an orthonormal basis, but express the vector in terms of a reciprocal frame and the associated components

\begin{aligned}a = a^i e_i = a_j e^j\end{aligned} \hspace{\stretch{1}}(2.15)

where

\begin{aligned}e^i \cdot e_j = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(2.16)

When we get to the relativistic case, we can pick (but don’t have to) the standard basis

\begin{aligned}e_0 &= (1, 0, 0, 0) \\ e_1 &= (0, 1, 0, 0) \\ e_2 &= (0, 0, 1, 0) \\ e_3 &= (0, 0, 0, 1),\end{aligned} \hspace{\stretch{1}}(2.17)

for which our reciprocal frame is implicitly defined by the metric

\begin{aligned}e^0 &= (1, 0, 0, 0) \\ e^1 &= (0, -1, 0, 0) \\ e^2 &= (0, 0, -1, 0) \\ e^3 &= (0, 0, 0, -1).\end{aligned} \hspace{\stretch{1}}(2.21)

Anyways. Back to the problem. Let’s examine the $\mathbb{R}^{2}$ case. Our wedge product in coordinates is

\begin{aligned}a \wedge b=a^i b^j (e_i \wedge e_j)\end{aligned} \hspace{\stretch{1}}(2.25)

Since there are only two basis vectors we have

\begin{aligned}a \wedge b=(a^1 b^2 - a^2 b^1) e_1 \wedge e_2 = \text{Det} {\left\lVert{a^i b^j}\right\rVert} (e_1 \wedge e_2).\end{aligned} \hspace{\stretch{1}}(2.26)

Our wedge product is a product of the determinant of the vector coordinates, times the $\mathbb{R}^{2}$ pseudoscalar $e_1 \wedge e_2$.

This doesn’t look quite like the $\mathbb{R}^{3}$ relation that we want to prove, which had an antisymmetric tensor factor for the determinant. Observe that we get the determinant by picking off the $e_1 \wedge e_2$ component of the bivector result (the only component in this case), and we can do that by dotting with $e^2 \cdot e^1$. To get an antisymmetric tensor times the determinant, we have only to dot with a different pseudoscalar (one that differs by a possible sign due to permutation of the indexes). That is

\begin{aligned}(e^t \wedge e^s) \cdot (a \wedge b)&=a^i b^j (e^t \wedge e^s) \cdot (e_i \wedge e_j) \\ &=a^i b^j\left( {\delta^{s}}_i {\delta^{t}}_j-{\delta^{t}}_i {\delta^{s}}_j \right) \\ &=a^i b^j{\delta^{[t}}_j {\delta^{s]}}_i \\ &=a^i b^j{\delta^{t}}_{[j} {\delta^{s}}_{i]} \\ &=a^{[i} b^{j]}{\delta^{t}}_{j} {\delta^{s}}_{i} \\ &=a^{[s} b^{t]}\end{aligned}

Now, if we write $a^i = A^{1 i}$ and $b^j = A^{2 j}$ we have

\begin{aligned}(e^t \wedge e^s) \cdot (a \wedge b)=A^{1 s} A^{2 t} -A^{1 t} A^{2 s}\end{aligned} \hspace{\stretch{1}}(2.27)

We can write this in two different ways. One of which is

\begin{aligned}A^{1 s} A^{2 t} -A^{1 t} A^{2 s} =\epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert}\end{aligned} \hspace{\stretch{1}}(2.28)

and the other of which is by introducing free indexes for $1$ and $2$, and summing antisymmetrically over these. That is

\begin{aligned}A^{1 s} A^{2 t} -A^{1 t} A^{2 s}=A^{a s} A^{b t} \epsilon_{a b}\end{aligned} \hspace{\stretch{1}}(2.29)

So, we have

\begin{aligned}\boxed{A^{a s} A^{b t} \epsilon_{a b} =A^{1 i} A^{2 j} {\delta^{[t}}_j {\delta^{s]}}_i =\epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert},}\end{aligned} \hspace{\stretch{1}}(2.30)

This result hold regardless of the metric for the space, and does not require that we were using an orthonormal basis. When the metric is Euclidean and we have an orthonormal basis, then all the indexes can be dropped.

The $\mathbb{R}^{3}$ and $\mathbb{R}^{4}$ cases follow in exactly the same way, we just need more vectors in the wedge products.

For the $\mathbb{R}^{3}$ case we have

\begin{aligned}(e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c)&=a^i b^j c^k(e^u \wedge e^t \wedge e^s) \cdot (e_i \wedge e_j \wedge e_k) \\ &=a^i b^j c^k{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i \\ &=a^{[s} b^t c^{u]}\end{aligned}

Again, with $a^i = A^{1 i}$ and $b^j = A^{2 j}$, and $c^k = A^{3 k}$ we have

\begin{aligned}(e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c)=A^{1 i} A^{2 j} A^{3 k}{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i\end{aligned} \hspace{\stretch{1}}(2.31)

and we can choose to write this in either form, resulting in the identity

\begin{aligned}\boxed{\epsilon^{s t u} \text{Det} {\left\lVert{A^{ij}}\right\rVert}=A^{1 i} A^{2 j} A^{3 k}{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i=\epsilon_{a b c} A^{a s} A^{b t} A^{c u}.}\end{aligned} \hspace{\stretch{1}}(2.32)

The $\mathbb{R}^{4}$ case follows exactly the same way, and we have

\begin{aligned}(e^v \wedge e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c \wedge d)&=a^i b^j c^k d^l(e^v \wedge e^u \wedge e^t \wedge e^s) \cdot (e_i \wedge e_j \wedge e_k \wedge e_l) \\ &=a^i b^j c^k d^l{\delta^{[v}}_l{\delta^{u}}_k{\delta^{t}}_j{\delta^{s]}}_i \\ &=a^{[s} b^t c^{u} d^{v]}.\end{aligned}

This time with $a^i = A^{0 i}$ and $b^j = A^{1 j}$, and $c^k = A^{2 k}$, and $d^l = A^{3 l}$ we have

\begin{aligned}\boxed{\epsilon^{s t u v} \text{Det} {\left\lVert{A^{ij}}\right\rVert}=A^{0 i} A^{1 j} A^{2 k} A^{3 l}{\delta^{[v}}_l{\delta^{u}}_k{\delta^{t}}_j{\delta^{s]}}_i=\epsilon_{a b c d} A^{a s} A^{b t} A^{c u} A^{d v}.}\end{aligned} \hspace{\stretch{1}}(2.33)

This one is almost the identity to be established later in problem 1.4. We have only to raise and lower some indexes to get that one. Note that in the Minkowski standard basis above, because $s, t, u, v$ must be a permutation of $0,1,2,3$ for a non-zero result, we must have

\begin{aligned}\epsilon^{s t u v} = (-1)^3 (+1) \epsilon_{s t u v}.\end{aligned} \hspace{\stretch{1}}(2.34)

So raising and lowering the identity above gives us

\begin{aligned}-\epsilon_{s t u v} \text{Det} {\left\lVert{A_{ij}}\right\rVert}=\epsilon^{a b c d} A_{a s} A_{b t} A_{c u} A_{d u}.\end{aligned} \hspace{\stretch{1}}(2.35)

No sign changes were required for the indexes $a, b, c, d$, since they are paired.

Until we did the raising and lowering operations here, there was no specific metric required, so our first result 2.33 is the more general one.

There’s one more part to this problem, doing the antisymmetric sums over the indexes $s, t, \cdots$. For the $\mathbb{R}^{2}$ case we have

\begin{aligned}\epsilon_{s t} \epsilon_{a b} A^{a s} A^{b t}&=\epsilon_{s t} \epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( \epsilon_{1 2} \epsilon^{1 2} +\epsilon_{2 1} \epsilon^{2 1} \right)\text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( 1^2 + (-1)^2\right)\text{Det} {\left\lVert{A^{ij}}\right\rVert}\end{aligned}

We conclude that

\begin{aligned}\boxed{\epsilon_{s t} \epsilon_{a b} A^{a s} A^{b t} = 2! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.36)

For the $\mathbb{R}^{3}$ case we have the same operation

\begin{aligned}\epsilon_{s t u} \epsilon_{a b c} A^{a s} A^{b t} A^{c u}&=\epsilon_{s t u} \epsilon^{s t u} \text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( \epsilon_{1 2 3} \epsilon^{1 2 3} +\epsilon_{1 3 2} \epsilon^{1 3 2} + \cdots\right)\text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=(\pm 1)^2 (3!)\text{Det} {\left\lVert{A^{ij}}\right\rVert}.\end{aligned}

So we conclude

\begin{aligned}\boxed{\epsilon_{s t u} \epsilon_{a b c} A^{a s} A^{b t} A^{c u}= 3! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.37)

It’s clear what the pattern is, and if we evaluate the sum of the antisymmetric tensor squares in $\mathbb{R}^{4}$ we have

\begin{aligned}\epsilon_{s t u v} \epsilon_{s t u v}&=\epsilon_{0 1 2 3} \epsilon_{0 1 2 3}+\epsilon_{0 1 3 2} \epsilon_{0 1 3 2}+\epsilon_{0 2 1 3} \epsilon_{0 2 1 3}+ \cdots \\ &= (\pm 1)^2 (4!),\end{aligned}

So, for our SR case we have

\begin{aligned}\boxed{\epsilon_{s t u v} \epsilon_{a b c d} A^{a s} A^{b t} A^{c u} A^{d v}= 4! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.38)

This was part of question 1.4, albeit in lower index form. Here since all indexes are matched, we have the same result without major change

\begin{aligned}\boxed{\epsilon^{s t u v} \epsilon^{a b c d} A_{a s} A_{b t} A_{c u} A_{d v}= 4! \text{Det} {\left\lVert{A_{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.39)

The main difference is that we are now taking the determinant of a lower index tensor.

## 3. Statement. Rotational invariance of 3D antisymmetric tensor

Use the previous results to show that $\epsilon_{\mu\nu\lambda}$ is invariant under rotations.

## 3. Solution

We apply transformations to coordinates (and thus indexes) of the form

\begin{aligned}x_\mu \rightarrow O_{\mu\nu} x_\nu\end{aligned} \hspace{\stretch{1}}(2.40)

With our tensor transforming as its indexes, we have

\begin{aligned}\epsilon_{\mu\nu\lambda} \rightarrow \epsilon_{\alpha\beta\sigma} O_{\mu\alpha} O_{\nu\beta} O_{\lambda\sigma}.\end{aligned} \hspace{\stretch{1}}(2.41)

We’ve got 2.32, which after dropping indexes, because we are in a Euclidean space, we have

\begin{aligned}\epsilon_{\mu \nu \lambda} \text{Det} {\left\lVert{A_{ij}}\right\rVert} = \epsilon_{\alpha \beta \sigma} A_{\alpha \mu} A_{\beta \nu} A_{\sigma \lambda}.\end{aligned} \hspace{\stretch{1}}(2.42)

Let $A_{i j} = O_{j i}$, which gives us

\begin{aligned}\epsilon_{\mu\nu\lambda} \rightarrow \epsilon_{\mu\nu\lambda} \text{Det} A^\text{T}\end{aligned} \hspace{\stretch{1}}(2.43)

but since $\text{Det} O = \text{Det} O^\text{T}$, we have shown that $\epsilon_{\mu\nu\lambda}$ is invariant under rotation.

## 4. Statement. Rotational invariance of 4D antisymmetric tensor

Use the previous results to show that $\epsilon_{i j k l}$ is invariant under Lorentz transformations.

## 4. Solution

This follows the same way. We assume a transformation of coordinates of the following form

\begin{aligned}(x')^i &= {O^i}_j x^j \\ (x')_i &= {O_i}^j x_j,\end{aligned} \hspace{\stretch{1}}(2.44)

where the determinant of ${O^i}_j = 1$ (sanity check of sign: ${O^i}_j = {\delta^i}_j$).

Our antisymmetric tensor transforms as its coordinates individually

\begin{aligned}\epsilon_{i j k l} &\rightarrow \epsilon_{a b c d} {O_i}^a{O_j}^b{O_k}^c{O_l}^d \\ &= \epsilon^{a b c d} O_{i a}O_{j b}O_{k c}O_{l d} \\ \end{aligned}

Let $P_{ij} = O_{ji}$, and raise and lower all the indexes in 2.46 for

\begin{aligned}-\epsilon_{s t u v} \text{Det} {\left\lVert{P_{ij}}\right\rVert}=\epsilon^{a b c d} P_{a s} P_{b t} P_{c u} P_{d v}.\end{aligned} \hspace{\stretch{1}}(2.46)

We have

\begin{aligned}\epsilon_{i j k l} &= \epsilon^{a b c d} P_{a i}P_{a j}P_{a k}P_{a l} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{P_{ij}}\right\rVert} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{O_{ij}}\right\rVert} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{g_{im} {O^m}_j }\right\rVert} \\ &=-\epsilon_{i j k l} (-1)(1) \\ &=\epsilon_{i j k l}\end{aligned}

Since $\epsilon_{i j k l} = -\epsilon^{i j k l}$ both are therefore invariant under Lorentz transformation.

## 5. Statement. Sum of contracting symmetric and antisymmetric rank 2 tensors

Show that $A^{ij} B_{ij} = 0$ if $A$ is symmetric and $B$ is antisymmetric.

## 5. Solution

We swap indexes in $B$, switch dummy indexes, then swap indexes in $A$

\begin{aligned}A^{i j} B_{i j} &= -A^{i j} B_{j i} \\ &= -A^{j i} B_{i j} \\ &= -A^{i j} B_{i j} \\ \end{aligned}

Our result is the negative of itself, so must be zero.

## 6. Statement. Characteristic equation for the electromagnetic strength tensor

Show that $P(\lambda) = \text{Det} {\left\lVert{F_{i j} - \lambda g_{i j}}\right\rVert}$ is invariant under Lorentz transformations. Consider the polynomial of $P(\lambda)$, also called the characteristic polynomial of the matrix ${\left\lVert{F_{i j}}\right\rVert}$. Find the coefficients of the expansion of $P(\lambda)$ in powers of $\lambda$ in terms of the components of ${\left\lVert{F_{i j}}\right\rVert}$. Use the result to argue that $\mathbf{E} \cdot \mathbf{B}$ and $\mathbf{E}^2 - \mathbf{B}^2$ are Lorentz invariant.

## 6. Solution

### The invariance of the determinant

Let’s consider how any lower index rank 2 tensor transforms. Given a transformation of coordinates

\begin{aligned}(x^i)' &= {O^i}_j x^j \\ (x_i)' &= {O_i}^j x^j ,\end{aligned} \hspace{\stretch{1}}(2.47)

where $\text{Det} {\left\lVert{ {O^i}_j }\right\rVert} = 1$, and ${O_i}^j = {O^m}_n g_{i m} g^{j n}$. Let’s reflect briefly on why this determinant is unit valued. We have

\begin{aligned}(x^i)' (x_i)'= {O_i}^a x^a {O^i}_b x^b = x^b x_b,\end{aligned} \hspace{\stretch{1}}(2.49)

which implies that the transformation product is

\begin{aligned}{O_i}^a {O^i}_b = {\delta^a}_b,\end{aligned} \hspace{\stretch{1}}(2.50)

the identity matrix. The identity matrix has unit determinant, so we must have

\begin{aligned}1 = (\text{Det} \hat{G})^2 (\text{Det} {\left\lVert{ {O^i}_j }\right\rVert})^2.\end{aligned} \hspace{\stretch{1}}(2.51)

Since $\text{Det} \hat{G} = -1$ we have

\begin{aligned}\text{Det} {\left\lVert{ {O^i}_j }\right\rVert} = \pm 1,\end{aligned} \hspace{\stretch{1}}(2.52)

which is all that we can say about the determinant of this class of transformations by considering just invariance. If we restrict the transformations of coordinates to those of the same determinant sign as the identity matrix, we rule out reflections in time or space. This seems to be the essence of the $SO(1,3)$ labeling.

Why dwell on this? Well, I wanted to be clear on the conventions I’d chosen, since parts of the course notes used $\hat{O} = {\left\lVert{O^{i j}}\right\rVert}$, and $X' = \hat{O} X$, and gave that matrix unit determinant. That $O^{i j}$ looks like it is equivalent to my ${O^i}_j$, except that the one in the course notes is loose when it comes to lower and upper indexes since it gives $(x')^i = O^{i j} x^j$.

I’ll write

\begin{aligned}\hat{O} = {\left\lVert{{O^i}_j}\right\rVert},\end{aligned} \hspace{\stretch{1}}(2.53)

and require this (not ${\left\lVert{O^{i j}}\right\rVert}$) to be the matrix with unit determinant. Having cleared the index upper and lower confusion I had trying to reconcile the class notes with the rules for index manipulation, let’s now consider the Lorentz transformation of a lower index rank 2 tensor (not necessarily antisymmetric or symmetric)

We have, transforming in the same fashion as a lower index coordinate four vector (but twice, once for each index)

\begin{aligned}A_{i j} \rightarrow A_{k m} {O_i}^k{O_j}^m.\end{aligned} \hspace{\stretch{1}}(2.54)

The determinant of the transformation tensor ${O_i}^j$ is

\begin{aligned}\text{Det} {\left\lVert{ {O_i}^j }\right\rVert} = \text{Det} {\left\lVert{ g^{i m} {O^m}_n g^{n j} }\right\rVert} = (\text{Det} \hat{G}) (1) (\text{Det} \hat{G} ) = (-1)^2 (1) = 1.\end{aligned} \hspace{\stretch{1}}(2.55)

We see that the determinant of a lower index rank 2 tensor is invariant under Lorentz transformation. This would include our characteristic polynomial $P(\lambda)$.

### Expanding the determinant.

Utilizing 2.39 we can now calculate the characteristic polynomial. This is

\begin{aligned}\text{Det} {\left\lVert{F_{ij} - \lambda g_{ij} }\right\rVert}&= \frac{1}{{4!}}\epsilon^{s t u v} \epsilon^{a b c d} (F_{ a s } - \lambda g_{a s}) (F_{ b t } - \lambda g_{b t}) (F_{ c u } - \lambda g_{c u}) (F_{ d v } - \lambda g_{d v}) \\ &=\frac{1}{{24}}\epsilon^{s t u v} \epsilon_{a b c d} ({F^a}_s - \lambda {g^a}_s) ({F^b}_t - \lambda {g^b}_t) ({F^c}_u - \lambda {g^c}_u) ({F^d}_v - \lambda {g^d}_v) \\ \end{aligned}

However, ${g^a}_b = g_{b c} g^{a c}$, or ${\left\lVert{{g^a}_b}\right\rVert} = \hat{G}^2 = I$. This means we have

\begin{aligned}{g^a}_b = {\delta^a}_b,\end{aligned} \hspace{\stretch{1}}(2.56)

and our determinant is reduced to

\begin{aligned}\begin{aligned}P(\lambda) &=\frac{1}{{24}}\epsilon^{s t u v} \epsilon_{a b c d} \Bigl({F^a}_s {F^b}_t - \lambda( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) + \lambda^2 {\delta^a}_s {\delta^b}_t \Bigr) \\ &\times \qquad \qquad \Bigl({F^c}_u {F^d}_v - \lambda( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) + \lambda^2 {\delta^c}_u {\delta^d}_v \Bigr) \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.57)

If we expand this out we have our powers of $\lambda$ coefficients are

\begin{aligned}\lambda^0 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {F^a}_s {F^b}_t {F^c}_u {F^d}_v \\ \lambda^1 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl(- ({\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) {F^a}_s {F^b}_t - ({\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) {F^c}_u {F^d}_v \Bigr) \\ \lambda^2 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^c}_u {\delta^d}_v {F^a}_s {F^b}_t +( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) ( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) + {\delta^a}_s {\delta^b}_t {F^c}_u {F^d}_v \Bigr) \\ \lambda^3 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl(- ( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) {\delta^c}_u {\delta^d}_v - {\delta^a}_s {\delta^b}_t ( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) \Bigr) \\ \lambda^4 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^a}_s {\delta^b}_t {\delta^c}_u {\delta^d}_v \Bigr) \\ \end{aligned}

By 2.39 the $\lambda^0$ coefficient is just $\text{Det} {\left\lVert{F_{i j}}\right\rVert}$.

The $\lambda^3$ terms can be seen to be zero. For example, the first one is

\begin{aligned}-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {\delta^a}_s {F^b}_t {\delta^c}_u {\delta^d}_v &=-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{s b u v} {F^b}_t \\ &=-\frac{1}{{12}} \delta^{t}_b {F^b}_t \\ &=-\frac{1}{{12}} {F^b}_b \\ &=-\frac{1}{{12}} F^{bu} g_{ub} \\ &= 0,\end{aligned}

where the final equality to zero comes from summing a symmetric and antisymmetric product.

Similarly the $\lambda$ coefficients can be shown to be zero. Again the first as a sample is

\begin{aligned}-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {\delta^c}_u {F^d}_v {F^a}_s {F^b}_t &=-\frac{1}{{24}} \epsilon^{u s t v} \epsilon_{u a b d} {F^d}_v {F^a}_s {F^b}_t \\ &=-\frac{1}{{24}} \delta^{[s}_a\delta^{t}_b\delta^{v]}_d{F^d}_v {F^a}_s {F^b}_t \\ &=-\frac{1}{{24}} {F^a}_{[s}{F^b}_{t}{F^d}_{v]} \\ \end{aligned}

Disregarding the $-1/24$ factor, let’s just expand this antisymmetric sum

\begin{aligned}{F^a}_{[a}{F^b}_{b}{F^d}_{d]}&={F^a}_{a}{F^b}_{b}{F^d}_{d}+{F^a}_{d}{F^b}_{a}{F^d}_{b}+{F^a}_{b}{F^b}_{d}{F^d}_{a}-{F^a}_{a}{F^b}_{d}{F^d}_{b}-{F^a}_{d}{F^b}_{b}{F^d}_{a}-{F^a}_{b}{F^b}_{a}{F^d}_{d} \\ &={F^a}_{d}{F^b}_{a}{F^d}_{b}+{F^a}_{b}{F^b}_{d}{F^d}_{a} \\ \end{aligned}

Of the two terms above that were retained, they are the only ones without a zero ${F^i}_i$ factor. Consider the first part of this remaining part of the sum. Employing the metric tensor, to raise indexes so that the antisymmetry of $F^{ij}$ can be utilized, and then finally relabeling all the dummy indexes we have

\begin{aligned}{F^a}_{d}{F^b}_{a}{F^d}_{b}&=F^{a u}F^{b v}F^{d w}g_{d u}g_{a v}g_{b w} \\ &=(-1)^3F^{u a}F^{v b}F^{w d}g_{d u}g_{a v}g_{b w} \\ &=-(F^{u a}g_{a v})(F^{v b}g_{b w} )(F^{w d}g_{d u})\\ &=-{F^u}_v{F^v}_w{F^w}_u\\ &=-{F^a}_b{F^b}_d{F^d}_a\\ \end{aligned}

This is just the negative of the second term in the sum, leaving us with zero.

Finally, we have for the $\lambda^2$ coefficient ($\times 24$)

\begin{aligned}&\epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^c}_u {\delta^d}_v {F^a}_s {F^b}_t +{\delta^a}_s {F^b}_t {\delta^c}_u {F^d}_v +{\delta^b}_t {F^a}_s {\delta^d}_v {F^c}_u \\ &\qquad +{\delta^b}_t {F^a}_s {\delta^c}_u {F^d}_v +{\delta^a}_s {F^b}_t {\delta^d}_v {F^c}_u + {\delta^a}_s {\delta^b}_t {F^c}_u {F^d}_v \Bigr) \\ &=\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t +\epsilon^{s t u v} \epsilon_{s b u d} {F^b}_t {F^d}_v +\epsilon^{s t u v} \epsilon_{a t c v} {F^a}_s {F^c}_u \\ &\qquad +\epsilon^{s t u v} \epsilon_{a t u d} {F^a}_s {F^d}_v +\epsilon^{s t u v} \epsilon_{s b c v} {F^b}_t {F^c}_u + \epsilon^{s t u v} \epsilon_{s t c d} {F^c}_u {F^d}_v \\ &=\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t +\epsilon^{t v s u } \epsilon_{b d s u} {F^b}_t {F^d}_v +\epsilon^{s u t v} \epsilon_{a c t v} {F^a}_s {F^c}_u \\ &\qquad +\epsilon^{s v t u} \epsilon_{a d t u} {F^a}_s {F^d}_v +\epsilon^{t u s v} \epsilon_{b c s v} {F^b}_t {F^c}_u + \epsilon^{u v s t} \epsilon_{c d s t} {F^c}_u {F^d}_v \\ &=6\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t \\ &=6 (2){\delta^{[s}}_a{\delta^{t]}}_b{F^a}_s {F^b}_t \\ &=12{F^a}_{[a} {F^b}_{b]} \\ &=12( {F^a}_{a} {F^b}_{b} - {F^a}_{b} {F^b}_{a} ) \\ &=-12 {F^a}_{b} {F^b}_{a} \\ &=-12 F^{a b} F_{b a} \\ &=12 F^{a b} F_{a b}\end{aligned}

Therefore, our characteristic polynomial is

\begin{aligned}\boxed{P(\lambda) = \text{Det} {\left\lVert{F_{i j}}\right\rVert} + \frac{\lambda^2}{2} F^{a b} F_{a b} + \lambda^4.}\end{aligned} \hspace{\stretch{1}}(2.58)

Observe that in matrix form our strength tensors are

\begin{aligned}{\left\lVert{ F^{ij} }\right\rVert} &= \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \\ {\left\lVert{ F_{ij} }\right\rVert} &= \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.59)

From these we can compute $F^{a b} F_{a b}$ easily by inspection

\begin{aligned}F^{a b} F_{a b} = 2 (\mathbf{B}^2 - \mathbf{E}^2).\end{aligned} \hspace{\stretch{1}}(2.61)

Computing the determinant is not so easy. The dumb and simple way of expanding by cofactors takes two pages, and yields eventually

\begin{aligned}\text{Det} {\left\lVert{ F^{i j} }\right\rVert} = (\mathbf{E} \cdot \mathbf{B})^2.\end{aligned} \hspace{\stretch{1}}(2.62)

That supplies us with a relation for the characteristic polynomial in $\mathbf{E}$ and $\mathbf{B}$

\begin{aligned}\boxed{P(\lambda) = (\mathbf{E} \cdot \mathbf{B})^2 + \lambda^2 (\mathbf{B}^2 - \mathbf{E}^2) + \lambda^4.}\end{aligned} \hspace{\stretch{1}}(2.63)

Observe that we found this for the special case where $\mathbf{E}$ and $\mathbf{B}$ were perpendicular in homework 2. Observe that when we have that perpendicularity, we can solve for the eigenvalues by inspection

\begin{aligned}\lambda \in \{ 0, 0, \pm \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 } \},\end{aligned} \hspace{\stretch{1}}(2.64)

and were able to diagonalize the matrix ${F^{i}}_j$ to solve the Lorentz force equation in parametric form. When ${\left\lvert{\mathbf{E}}\right\rvert} > {\left\lvert{\mathbf{B}}\right\rvert}$ we had real eigenvalues and an orthogonal diagonalization when $\mathbf{B} = 0$. For the ${\left\lvert{\mathbf{B}}\right\rvert} > {\left\lvert{\mathbf{E}}\right\rvert}$, we had a two purely imaginary eigenvalues, and when $\mathbf{E} = 0$ this was a Hermitian diagonalization. For the general case, when one of $\mathbf{E}$, or $\mathbf{B}$ was zero, things didn’t have the same nice closed form solution.

In general our eigenvalues are

\begin{aligned}\lambda = \pm \frac{1}{{\sqrt{2}}} \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 \pm \sqrt{ (\mathbf{E}^2 - \mathbf{B}^2)^2 - 4 (\mathbf{E} \cdot \mathbf{B})^2 }}.\end{aligned} \hspace{\stretch{1}}(2.65)

For the purposes of this problem we really only wish to show that $\mathbf{E} \cdot \mathbf{B}$ and $\mathbf{E}^2 - \mathbf{B}^2$ are Lorentz invariants. When $\lambda = 0$ we have $P(\lambda) = (\mathbf{E} \cdot \mathbf{B})^2$, a Lorentz invariant. This must mean that $\mathbf{E} \cdot \mathbf{B}$ is itself a Lorentz invariant. Since that is invariant, and we require $P(\lambda)$ to be invariant for any other possible values of $\lambda$, the difference $\mathbf{E}^2 - \mathbf{B}^2$ must also be Lorentz invariant.

## 7. Statement. Show that the pseudoscalar invariant has only boundary effects.

Use integration by parts to show that $\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }$ only depends on the values of $A^i(x)$ at the “boundary” of spacetime (e.g. the “surface” depicted on page 105 of the notes) and hence does not affect the equations of motion for the electromagnetic field.

## 7. Solution

This proceeds in a fairly straightforward fashion

\begin{aligned}\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }&=\int d^4 x \epsilon^{i j k l} (\partial_i A_j - \partial_j A_i) F_{ k l } \\ &=\int d^4 x \epsilon^{i j k l} (\partial_i A_j) F_{ k l } -\epsilon^{j i k l} (\partial_i A_j) F_{ k l } \\ &=2 \int d^4 x \epsilon^{i j k l} (\partial_i A_j) F_{ k l } \\ &=2 \int d^4 x \epsilon^{i j k l} \left( \frac{\partial {}}{\partial {x^i}}(A_j F_{ k l }-A_j \frac{\partial { F_{ k l } }}{\partial {x^i}}\right)\\ \end{aligned}

Now, observe that by the Bianchi identity, this second term is zero

\begin{aligned}\epsilon^{i j k l} \frac{\partial { F_{ k l } }}{\partial {x^i}}=-\epsilon^{j i k l} \partial_i F_{ k l } = 0\end{aligned} \hspace{\stretch{1}}(2.66)

Now we have a set of perfect differentials, and can integrate

\begin{aligned}\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }&= 2 \int d^4 x \epsilon^{i j k l} \frac{\partial {}}{\partial {x^i}}(A_j F_{ k l })\\ &= 2 \int dx^j dx^k dx^l\epsilon^{i j k l} {\left.{{(A_j F_{ k l })}}\right\vert}_{{\Delta x^i}}\\ \end{aligned}

We are left with a only contributions to the integral from the boundary terms on the spacetime hypervolume, three-volume normals bounding the four-volume integration in the original integral.

## 8. Statement. Electromagnetic duality transformations.

Show that the Maxwell equations in vacuum are invariant under the transformation: $F_{i j} \rightarrow \tilde{F}_{i j}$, where $\tilde{F}_{i j} = \frac{1}{{2}} \epsilon_{i j k l} F^{k l}$ is the dual electromagnetic stress tensor. Replacing $F$ with $\tilde{F}$ is known as “electric-magnetic duality”. Explain this name by considering the transformation in terms of $\mathbf{E}$ and $\mathbf{B}$. Are the Maxwell equations with sources invariant under electric-magnetic duality transformations?

## 8. Solution

Let’s first consider the explanation of the name. First recall what the expansions are of $F_{i j}$ and $F^{i j}$ in terms of $\mathbf{E}$ and $\mathbf{E}$. These are

\begin{aligned}F_{0 \alpha} &= \partial_0 A_\alpha - \partial_\alpha A_0 \\ &= -\frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} - \frac{\partial {\phi}}{\partial {x^\alpha}} \\ &= E_\alpha\end{aligned}

with $F^{0 \alpha} = -E^\alpha$, and $E^\alpha = E_\alpha$.

The magnetic field components are

\begin{aligned}F_{\beta \alpha} &= \partial_\beta A_\alpha - \partial_\alpha A_\beta \\ &= -\partial_\beta A^\alpha + \partial_\alpha A^\beta \\ &= \epsilon_{\alpha \beta \sigma} B^\sigma\end{aligned}

with $F^{\beta \alpha} = \epsilon^{\alpha \beta \sigma} B_\sigma$ and $B_\sigma = B^\sigma$.

Now let’s expand the dual tensors. These are

\begin{aligned}\tilde{F}_{0 \alpha} &=\frac{1}{{2}} \epsilon_{0 \alpha i j} F^{i j} \\ &=\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} F^{\beta \sigma} \\ &=\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} \epsilon^{\sigma \beta \mu} B_\mu \\ &=-\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} \epsilon^{\mu \beta \sigma} B_\mu \\ &=-\frac{1}{{2}} (2!) {\delta_\alpha}^\mu B_\mu \\ &=- B_\alpha \\ \end{aligned}

and

\begin{aligned}\tilde{F}_{\beta \alpha} &=\frac{1}{{2}} \epsilon_{\beta \alpha i j} F^{i j} \\ &=\frac{1}{{2}} \left(\epsilon_{\beta \alpha 0 \sigma} F^{0 \sigma} +\epsilon_{\beta \alpha \sigma 0} F^{\sigma 0} \right) \\ &=\epsilon_{0 \beta \alpha \sigma} (-E^\sigma) \\ &=\epsilon_{\alpha \beta \sigma} E^\sigma\end{aligned}

Summarizing we have

\begin{aligned}F_{0 \alpha} &= E^\alpha \\ F^{0 \alpha} &= -E^\alpha \\ F^{\beta \alpha} &= F_{\beta \alpha} = \epsilon_{\alpha \beta \sigma} B^\sigma \\ \tilde{F}_{0 \alpha} &= - B_\alpha \\ \tilde{F}^{0 \alpha} &= B_\alpha \\ \tilde{F}_{\beta \alpha} &= \tilde{F}^{\beta \alpha} = \epsilon_{\alpha \beta \sigma} E^\sigma\end{aligned} \hspace{\stretch{1}}(2.67)

Is there a sign error in the $\tilde{F}_{0 \alpha} = - B_\alpha$ result? Other than that we have the same sort of structure for the tensor with $E$ and $B$ switched around.

Let’s write these in matrix form, to compare

\begin{aligned}\begin{array}{l l l l}{\left\lVert{ \tilde{F}_{i j} }\right\rVert} &= \begin{bmatrix}0 & -B_x & -B_y & -B_z \\ B_x & 0 & -E_z & E_y \\ B_y & E_z & 0 & E_x \\ B_z & -E_y & -E_x & 0 \\ \end{bmatrix} ^{i j} }\right\rVert} &= \begin{bmatrix}0 & B_x & B_y & B_z \\ -B_x & 0 & -E_z & E_y \\ -B_y & E_z & 0 & -E_x \\ -B_z & -E_y & E_x & 0 \\ \end{bmatrix} \\ {\left\lVert{ F^{ij} }\right\rVert} &= \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} }\right\rVert} &= \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0\end{bmatrix}.\end{array}\end{aligned} \hspace{\stretch{1}}(2.73)

From these we can see by inspection that we have

\begin{aligned}\tilde{F}^{i j} F_{ij} = \tilde{F}_{i j} F^{ij} = 4 (\mathbf{E} \cdot \mathbf{B})\end{aligned} \hspace{\stretch{1}}(2.74)

This is consistent with the stated result in [1] (except for a factor of $c$ due to units differences), so it appears the signs above are all kosher.

Now, let’s see if the if the dual tensor satisfies the vacuum equations.

\begin{aligned}\partial_j \tilde{F}^{i j}&=\partial_j \frac{1}{{2}} \epsilon^{i j k l} F_{k l} \\ &=\frac{1}{{2}} \epsilon^{i j k l} \partial_j (\partial_k A_l - \partial_l A_k) \\ &=\frac{1}{{2}} \epsilon^{i j k l} \partial_j \partial_k A_l - \frac{1}{{2}} \epsilon^{i j l k} \partial_k A_l \\ &=\frac{1}{{2}} (\epsilon^{i j k l} - \epsilon^{i j k l} \partial_k A_l \\ &= 0 \qquad\square\end{aligned}

So the first checks out, provided we have no sources. If we have sources, then we see here that Maxwell’s equations do not hold since this would imply that the four current density must be zero.

How about the Bianchi identity? That gives us

\begin{aligned}\epsilon^{i j k l} \partial_j \tilde{F}_{k l} &=\epsilon^{i j k l} \partial_j \frac{1}{{2}} \epsilon_{k l a b} F^{a b} \\ &=\frac{1}{{2}} \epsilon^{k l i j} \epsilon_{k l a b} \partial_j F^{a b} \\ &=\frac{1}{{2}} (2!) {\delta^i}_{[a} {\delta^j}_{b]} \partial_j F^{a b} \\ &=\partial_j (F^{i j} - F^{j i} ) \\ &=2 \partial_j F^{i j} .\end{aligned}

The factor of two is slightly curious. Is there a mistake above? If there is a mistake, it doesn’t change the fact that Maxwell’s equation

\begin{aligned}\partial_k F^{k i} = \frac{4 \pi}{c} j^i\end{aligned} \hspace{\stretch{1}}(2.75)

Gives us zero for the Bianchi identity under source free conditions of $j^i = 0$.

# Problem 2. Transformation properties of $\mathbf{E}$ and $\mathbf{B}$, again.

## 1. Statement

Use the form of $F^{i j}$ from page 82 in the class notes, the transformation law for ${\left\lVert{ F^{i j} }\right\rVert}$ given further down that same page, and the explicit form of the $SO(1,3)$ matrix $\hat{O}$ (say, corresponding to motion in the positive $x_1$ direction with speed $v$) to derive the transformation law of the fields $\mathbf{E}$ and $\mathbf{B}$. Use the transformation law to find the electromagnetic field of a charged particle moving with constant speed $v$ in the positive $x_1$ direction and check that the result agrees with the one that you obtained in Homework 2.

## 1. Solution

Given a transformation of coordinates

\begin{aligned}{x'}^i \rightarrow {O^i}_j x^j\end{aligned} \hspace{\stretch{1}}(3.76)

our rank 2 tensor $F^{i j}$ transforms as

\begin{aligned}F^{i j} \rightarrow {O^i}_aF^{a b}{O^j}_b.\end{aligned} \hspace{\stretch{1}}(3.77)

Introducing matrices

\begin{aligned}\hat{O} &= {\left\lVert{{O^i}_j}\right\rVert} \\ \hat{F} &= {\left\lVert{F^{ij}}\right\rVert} = \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(3.78)

and noting that $\hat{O}^\text{T} = {\left\lVert{{O^j}_i}\right\rVert}$, we can express the electromagnetic strength tensor transformation as

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T}.\end{aligned} \hspace{\stretch{1}}(3.80)

The class notes use ${x'}^i \rightarrow O^{ij} x^j$, which violates our conventions on mixed upper and lower indexes, but the end result 3.80 is the same.

\begin{aligned}{\left\lVert{{O^i}_j}\right\rVert} =\begin{bmatrix}\cosh\alpha & -\sinh\alpha & 0 & 0 \\ -\sinh\alpha & \cosh\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.81)

Writing

\begin{aligned}C &= \cosh\alpha = \gamma \\ S &= -\sinh\alpha = -\gamma \beta,\end{aligned} \hspace{\stretch{1}}(3.82)

we can compute the transformed field strength tensor

\begin{aligned}\hat{F}' &=\begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} \\ &=\begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}- S E_x & -C E_x & -E_y & -E_z \\ C E_x & S E_x & -B_z & B_y \\ C E_y + S B_z & S E_y + C B_z & 0 & -B_x \\ C E_z - S B_y & S E_z - C B_y & B_x & 0 \end{bmatrix} \\ &=\begin{bmatrix}0 & -E_x & -C E_y - S B_z & - C E_z + S B_y \\ E_x & 0 & -S E_y - C B_z & - S E_z + C B_y \\ C E_y + S B_z & S E_y + C B_z & 0 & -B_x \\ C E_z - S B_y & S E_z - C B_y & B_x & 0\end{bmatrix} \\ &=\begin{bmatrix}0 & -E_x & -\gamma(E_y - \beta B_z) & - \gamma(E_z + \beta B_y) \\ E_x & 0 & - \gamma (-\beta E_y + B_z) & \gamma( \beta E_z + B_y) \\ \gamma (E_y - \beta B_z) & \gamma(-\beta E_y + B_z) & 0 & -B_x \\ \gamma (E_z + \beta B_y) & -\gamma(\beta E_z + B_y) & B_x & 0\end{bmatrix}.\end{aligned}

As a check we have the antisymmetry that is expected. There is also a regularity to the end result that is aesthetically pleasing, hinting that things are hopefully error free. In coordinates for $\mathbf{E}$ and $\mathbf{B}$ this is

\begin{aligned}E_x &\rightarrow E_x \\ E_y &\rightarrow \gamma ( E_y - \beta B_z ) \\ E_z &\rightarrow \gamma ( E_z + \beta B_y ) \\ B_z &\rightarrow B_x \\ B_y &\rightarrow \gamma ( B_y + \beta E_z ) \\ B_z &\rightarrow \gamma ( B_z - \beta E_y ) \end{aligned} \hspace{\stretch{1}}(3.84)

Writing $\boldsymbol{\beta} = \mathbf{e}_1 \beta$, we have

\begin{aligned}\boldsymbol{\beta} \times \mathbf{B} = \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \beta & 0 & 0 \\ B_x & B_y & B_z\end{vmatrix} = \mathbf{e}_2 (-\beta B_z) + \mathbf{e}_3( \beta B_y ),\end{aligned} \hspace{\stretch{1}}(3.90)

which puts us enroute to a tidier vector form

\begin{aligned}E_x &\rightarrow E_x \\ E_y &\rightarrow \gamma ( E_y + (\boldsymbol{\beta} \times \mathbf{B})_y ) \\ E_z &\rightarrow \gamma ( E_z + (\boldsymbol{\beta} \times \mathbf{B})_z ) \\ B_z &\rightarrow B_x \\ B_y &\rightarrow \gamma ( B_y - (\boldsymbol{\beta} \times \mathbf{E})_y ) \\ B_z &\rightarrow \gamma ( B_z - (\boldsymbol{\beta} \times \mathbf{E})_z ).\end{aligned} \hspace{\stretch{1}}(3.91)

For a vector $\mathbf{A}$, write $\mathbf{A}_\parallel = (\mathbf{A} \cdot \hat{\mathbf{v}})\hat{\mathbf{v}}$, $\mathbf{A}_\perp = \mathbf{A} - \mathbf{A}_\parallel$, allowing a compact description of the field transformation

\begin{aligned}\mathbf{E} &\rightarrow \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp + \gamma (\boldsymbol{\beta} \times \mathbf{B})_\perp \\ \mathbf{B} &\rightarrow \mathbf{B}_\parallel + \gamma \mathbf{B}_\perp - \gamma (\boldsymbol{\beta} \times \mathbf{E})_\perp.\end{aligned} \hspace{\stretch{1}}(3.97)

Now, we want to consider the field of a moving particle. In the particle’s (unprimed) rest frame the field due to its potential $\phi = q/r$ is

\begin{aligned}\mathbf{E} &= \frac{q}{r^2} \hat{\mathbf{r}} \\ \mathbf{B} &= 0.\end{aligned} \hspace{\stretch{1}}(3.99)

Coordinates for a “stationary” observer, who sees this particle moving along the x-axis at speed $v$ are related by a boost in the $-v$ direction

\begin{aligned}\begin{bmatrix}ct' \\ x' \\ y' \\ z'\end{bmatrix}\begin{bmatrix}\gamma & \gamma (v/c) & 0 & 0 \\ \gamma (v/c) & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}ct \\ x \\ y \\ z\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.101)

Therefore the fields in the observer frame will be

\begin{aligned}\mathbf{E}' &= \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp - \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{B})_\perp = \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp \\ \mathbf{B}' &= \mathbf{B}_\parallel + \gamma \mathbf{B}_\perp + \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{E})_\perp = \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{E})_\perp \end{aligned} \hspace{\stretch{1}}(3.102)

More explicitly with $\mathbf{E} = \frac{q}{r^3}(x, y, z)$ this is

\begin{aligned}\mathbf{E}' &= \frac{q}{r^3}(x, \gamma y, \gamma z) \\ \mathbf{B}' &= \gamma \frac{q v}{c r^3} ( 0, -z, y )\end{aligned} \hspace{\stretch{1}}(3.104)

Comparing to Problem 3 in Problem set 2, I see that this matches the result obtained by separately transforming the gradient, the time partial, and the scalar potential. Actually, if I am being honest, I see that I made a sign error in all the coordinates of $\mathbf{E}'$ when I initially did (this ungraded problem) in problem set 2. That sign error should have been obvious by considering the $v=0$ case which would have mysteriously resulted in inversion of all the coordinates of the observed electric field.

## 2. Statement

A particle is moving with velocity $\mathbf{v}$ in perpendicular $\mathbf{E}$ and $\mathbf{B}$ fields, all given in some particular “stationary” frame of reference.

\begin{enumerate}
\item Show that there exists a frame where the problem of finding the particle trajectory can be reduced to having either only an electric or only a magnetic field.
\item Explain what determines which case takes place.
\item Find the velocity $\mathbf{v}_0$ of that frame relative to the “stationary” frame.
\end{enumerate}

## 2. Solution

\paragraph{Part 1 and 2:} Existence of the transformation.

In the single particle Lorentz trajectory problem we wish to solve

\begin{aligned}m c \frac{du^i}{ds} = \frac{e}{c} F^{i j} u_j,\end{aligned} \hspace{\stretch{1}}(3.106)

which in matrix form we can write as

\begin{aligned}\frac{d U}{ds} = \frac{e}{m c^2} \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.107)

where we write our column vector proper velocity as $U = {\left\lVert{u^i}\right\rVert}$. Under transformation of coordinates ${u'}^i = {O^i}_j x^j$, with $\hat{O} = {\left\lVert{{O^i}_j}\right\rVert}$, this becomes

\begin{aligned}\hat{O} \frac{d U}{ds} = \frac{e}{m c^2} \hat{O} \hat{F} \hat{O}^\text{T} \hat{G} \hat{O} U.\end{aligned} \hspace{\stretch{1}}(3.108)

Suppose we can find eigenvectors for the matrix $\hat{O} \hat{F} \hat{O}^\text{T} \hat{G}$. That is for some eigenvalue $\lambda$, we can find an eigenvector $\Sigma$

\begin{aligned}\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} \Sigma = \lambda \Sigma.\end{aligned} \hspace{\stretch{1}}(3.109)

Rearranging we have

\begin{aligned}(\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I) \Sigma = 0\end{aligned} \hspace{\stretch{1}}(3.110)

and conclude that $\Sigma$ lies in the null space of the matrix $\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I$ and that this difference of matrices must have a zero determinant

\begin{aligned}\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I) = -\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} - \lambda \hat{G}) = 0.\end{aligned} \hspace{\stretch{1}}(3.111)

Since $\hat{G} = \hat{O} \hat{G} \hat{O}^\text{T}$ for any Lorentz transformation $\hat{O}$ in $SO(1,3)$, and $\text{Det} ABC = \text{Det} A \text{Det} B \text{Det} C$ we have

\begin{aligned}\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} - \lambda G)= \text{Det} (\hat{F} - \lambda \hat{G}).\end{aligned} \hspace{\stretch{1}}(3.112)

In problem 1.6, we called this our characteristic equation $P(\lambda) = \text{Det} (\hat{F} - \lambda \hat{G})$. Observe that the characteristic equation is Lorentz invariant for any $\lambda$, which requires that the eigenvalues $\lambda$ are also Lorentz invariants.

In problem 1.6 of this problem set we computed that this characteristic equation expands to

\begin{aligned}P(\lambda) = \text{Det} (\hat{F} - \lambda \hat{G}) = (\mathbf{E} \cdot \mathbf{B})^2 + \lambda^2 (\mathbf{B}^2 - \mathbf{E}^2) + \lambda^4.\end{aligned} \hspace{\stretch{1}}(3.113)

The eigenvalues for the system, also each necessarily Lorentz invariants, are

\begin{aligned}\lambda = \pm \frac{1}{{\sqrt{2}}} \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 \pm \sqrt{ (\mathbf{E}^2 - \mathbf{B}^2)^2 - 4 (\mathbf{E} \cdot \mathbf{B})^2 }}.\end{aligned} \hspace{\stretch{1}}(3.114)

Observe that in the specific case where $\mathbf{E} \cdot \mathbf{B} = 0$, as in this problem, we must have $\mathbf{E}' \cdot \mathbf{B}'$ in all frames, and the two non-zero eigenvalues of our characteristic polynomial are simply

\begin{aligned}\lambda = \pm \sqrt{\mathbf{E}^2 - \mathbf{B}^2}.\end{aligned} \hspace{\stretch{1}}(3.115)

These and $\mathbf{E} \cdot \mathbf{B} = 0$ are the invariants for this system. If we have $\mathbf{E}^2 > \mathbf{B}^2$ in one frame, we must also have ${\mathbf{E}'}^2 > {\mathbf{B}'}^2$ in another frame, still maintaining perpendicular fields. In particular if $\mathbf{B}' = 0$ we maintain real eigenvalues. Similarly if $\mathbf{B}^2 > \mathbf{E}^2$ in some frame, we must always have imaginary eigenvalues, and this is also true in the $\mathbf{E}' = 0$ case.

While the problem can be posed as a pure diagonalization problem (and even solved numerically this way for the general constant fields case), we can also work symbolically, thinking of the trajectories problem as simply seeking a transformation of frames that reduce the scope of the problem to one that is more tractable. That does not have to be the linear transformation that diagonalizes the system. Instead we are free to transform to a frame where one of the two fields $\mathbf{E}'$ or $\mathbf{B}'$ is zero, provided the invariants discussed are maintained.

\paragraph{Part 3:} Finding the boost velocity that wipes out one of the fields.

Let’s now consider a Lorentz boost $\hat{O}$, and seek to solve for the boost velocity that wipes out one of the fields, given the invariants that must be maintained for the system

To make things concrete, suppose that our perpendicular fields are given by $\mathbf{E} = E \mathbf{e}_2$ and $\mathbf{B} = B \mathbf{e}_3$.

Let also assume that we can find the velocity $\mathbf{v}_0$ for which one or more of the transformed fields is zero. Suppose that velocity is

\begin{aligned}\mathbf{v}_0 = v_0 (\alpha_1, \alpha_2, \alpha_3) = v_0 \hat{\mathbf{v}}_0,\end{aligned} \hspace{\stretch{1}}(3.116)

where $\alpha_i$ are the direction cosines of $\mathbf{v}_0$ so that $\sum_i \alpha_i^2 = 1$. We will want to compute the components of $\mathbf{E}$ and $\mathbf{B}$ parallel and perpendicular to this velocity.

Those are

\begin{aligned}\mathbf{E}_\parallel &= E \mathbf{e}_2 \cdot (\alpha_1, \alpha_2, \alpha_3) (\alpha_1, \alpha_2, \alpha_3) \\ &= E \alpha_2 (\alpha_1, \alpha_2, \alpha_3) \\ \end{aligned}

\begin{aligned}\mathbf{E}_\perp &= E \mathbf{e}_2 - \mathbf{E}_\parallel \\ &= E (-\alpha_1 \alpha_2, 1 - \alpha_2^2, -\alpha_2 \alpha_3) \\ &= E (-\alpha_1 \alpha_2, \alpha_1^2 + \alpha_3^2, -\alpha_2 \alpha_3) \\ \end{aligned}

For the magnetic field we have

\begin{aligned}\mathbf{B}_\parallel &= B \alpha_3 (\alpha_1, \alpha_2, \alpha_3),\end{aligned}

and

\begin{aligned}\mathbf{B}_\perp &= B \mathbf{e}_3 - \mathbf{B}_\parallel \\ &= B (-\alpha_1 \alpha_3, -\alpha_2 \alpha_3, \alpha_1^2 + \alpha_2^2) \\ \end{aligned}

Now, observe that $(\boldsymbol{\beta} \times \mathbf{B})_\parallel \propto ((\mathbf{v}_0 \times \mathbf{B}) \cdot \mathbf{v}_0) \mathbf{v}_0$, but this is just zero. So we have $(\boldsymbol{\beta} \times \mathbf{B})_\parallel = \boldsymbol{\beta} \times \mathbf{B}$. So our cross products terms are just

\begin{aligned}\hat{\mathbf{v}}_0 \times \mathbf{B} &= \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & 0 & B \end{vmatrix} = B (\alpha_2, -\alpha_1, 0) \\ \hat{\mathbf{v}}_0 \times \mathbf{E} &= \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & E & 0 \end{vmatrix} = E (-\alpha_3, 0, \alpha_1)\end{aligned}

We can now express how the fields transform, given this arbitrary boost velocity. From 3.97, this is

\begin{aligned}\mathbf{E} &\rightarrow E \alpha_2 (\alpha_1, \alpha_2, \alpha_3) + \gamma E (-\alpha_1 \alpha_2, \alpha_1^2 + \alpha_3^2, -\alpha_2 \alpha_3) + \gamma \frac{v_0^2}{c^2} B (\alpha_2, -\alpha_1, 0) \\ \mathbf{B} &\rightarrowB \alpha_3 (\alpha_1, \alpha_2, \alpha_3)+ \gamma B (-\alpha_1 \alpha_3, -\alpha_2 \alpha_3, \alpha_1^2 + \alpha_2^2) - \gamma \frac{v_0^2}{c^2} E (-\alpha_3, 0, \alpha_1)\end{aligned} \hspace{\stretch{1}}(3.117)

### Zero Electric field case.

Let’s tackle the two cases separately. First when ${\left\lvert{\mathbf{B}}\right\rvert} > {\left\lvert{\mathbf{E}}\right\rvert}$, we can transform to a frame where $\mathbf{E}'=0$. In coordinates from 3.117 this supplies us three sets of equations. These are

\begin{aligned}0 &= E \alpha_2 \alpha_1 (1 - \gamma) + \gamma \frac{v_0^2}{c^2} B \alpha_2 \\ 0 &= E \alpha_2^2 + \gamma E (\alpha_1^2 + \alpha_3^2) - \gamma \frac{v_0^2}{c^2} B \alpha_1 \\ 0 &= E \alpha_2 \alpha_3 (1 - \gamma).\end{aligned} \hspace{\stretch{1}}(3.119)

With an assumed solution the $\mathbf{e}_3$ coordinate equation implies that one of $\alpha_2$ or $\alpha_3$ is zero. Perhaps there are solutions with $\alpha_3 = 0$ too, but inspection shows that $\alpha_2 = 0$ nicely kills off the first equation. Since $\alpha_1^2 + \alpha_2^2 + \alpha_3^2 = 1$, that also implies that we are left with

\begin{aligned}0 = E - \frac{v_0^2}{c^2} B \alpha_1 \end{aligned} \hspace{\stretch{1}}(3.122)

Or

\begin{aligned}\alpha_1 &= \frac{E}{B} \frac{c^2}{v_0^2} \\ \alpha_2 &= 0 \\ \alpha_3 &= \sqrt{1 - \frac{E^2}{B^2} \frac{c^4}{v_0^4} }\end{aligned} \hspace{\stretch{1}}(3.123)

Our velocity was $\mathbf{v}_0 = v_0 (\alpha_1, \alpha_2, \alpha_3)$ solving the problem for the ${\left\lvert{\mathbf{B}}\right\rvert}^2 > {\left\lvert{\mathbf{E}}\right\rvert}^2$ case up to an adjustable constant $v_0$. That constant comes with constraints however, since we must also have our cosine $\alpha_1 \le 1$. Expressed another way, the magnitude of the boost velocity is constrained by the relation

\begin{aligned}\frac{\mathbf{v}_0^2}{c^2} \ge {\left\lvert{\frac{E}{B}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(3.126)

It appears we may also pick the equality case, so one velocity (not unique) that should transform away the electric field is

\begin{aligned}\boxed{\mathbf{v}_0 = c \sqrt{{\left\lvert{\frac{E}{B}}\right\rvert}} \mathbf{e}_1 = \pm c \sqrt{{\left\lvert{\frac{E}{B}}\right\rvert}} \frac{\mathbf{E} \times \mathbf{B}}{{\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}}.}\end{aligned} \hspace{\stretch{1}}(3.127)

This particular boost direction is perpendicular to both fields. Observe that this highlights the invariance condition ${\left\lvert{\frac{E}{B}}\right\rvert} < 1$ since we see this is required for a physically realizable velocity. Boosting in this direction will reduce our problem to one that has only the magnetic field component.

### Zero Magnetic field case.

Now, let’s consider the case where we transform the magnetic field away, the case when our characteristic polynomial has strictly real eigenvalues $\lambda = \pm \sqrt{\mathbf{E}^2 - \mathbf{B}^2}$. In this case, if we write out our equations for the transformed magnetic field and require these to separately equal zero, we have

\begin{aligned}0 &= B \alpha_3 \alpha_1 ( 1 - \gamma ) + \gamma \frac{v_0^2}{c^2} E \alpha_3 \\ 0 &= B \alpha_2 \alpha_3 ( 1 - \gamma ) \\ 0 &= B (\alpha_3^2 + \gamma (\alpha_1^2 + \alpha_2^2)) - \gamma \frac{v_0^2}{c^2} E \alpha_1.\end{aligned} \hspace{\stretch{1}}(3.128)

Similar to before we see that $\alpha_3 = 0$ kills off the first and second equations, leaving just

\begin{aligned}0 = B - \frac{v_0^2}{c^2} E \alpha_1.\end{aligned} \hspace{\stretch{1}}(3.131)

We now have a solution for the family of direction vectors that kill the magnetic field off

\begin{aligned}\alpha_1 &= \frac{B}{E} \frac{c^2}{v_0^2} \\ \alpha_2 &= \sqrt{ 1 - \frac{B^2}{E^2} \frac{c^4}{v_0^4} } \\ \alpha_3 &= 0.\end{aligned} \hspace{\stretch{1}}(3.132)

In addition to the initial constraint that ${\left\lvert{\frac{B}{E}}\right\rvert} < 1$, we have as before, constraints on the allowable values of $v_0$

\begin{aligned}\frac{\mathbf{v}_0^2}{c^2} \ge {\left\lvert{\frac{B}{E}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(3.135)

Like before we can pick the equality $\alpha_1^2 = 1$, yielding a boost direction of

\begin{aligned}\boxed{\mathbf{v}_0 = c \sqrt{{\left\lvert{\frac{B}{E}}\right\rvert}} \mathbf{e}_1 = \pm c \sqrt{{\left\lvert{\frac{B}{E}}\right\rvert}} \frac{\mathbf{E} \times \mathbf{B}}{{\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}}.}\end{aligned} \hspace{\stretch{1}}(3.136)

Again, we see that the invariance condition ${\left\lvert{\mathbf{B}}\right\rvert} < {\left\lvert{\mathbf{E}}\right\rvert}$ is required for a physically realizable velocity if that velocity is entirely perpendicular to the fields.

# Problem 3. Continuity equation for delta function current distributions.

## Statement

Show explicitly that the electromagnetic 4-current $j^i$ for a particle moving with constant velocity (considered in class, p. 100-101 of notes) is conserved $\partial_i j^i = 0$. Give a physical interpretation of this conservation law, for example by integrating $\partial_i j^i$ over some spacetime region and giving an integral form to the conservation law ($\partial_i j^i = 0$ is known as the “continuity equation”).

## Solution

First lets review. Our four current was defined as

\begin{aligned}j^i(x) = \sum_A c e_A \int_{x(\tau)} dx_A^i(\tau) \delta^4(x - x_A(\tau)).\end{aligned} \hspace{\stretch{1}}(4.137)

If each of the trajectories $x_A(\tau)$ represents constant motion we have

\begin{aligned}x_A(\tau) = x_A(0) + \gamma_A \tau ( c, \mathbf{v}_A ).\end{aligned} \hspace{\stretch{1}}(4.138)

The spacetime split of this four vector is

\begin{aligned}x_A^0(\tau) &= x_A^0(0) + \gamma_A \tau c \\ \mathbf{x}_A(\tau) &= \mathbf{x}_A(0) + \gamma_A \tau \mathbf{v},\end{aligned} \hspace{\stretch{1}}(4.139)

with differentials

\begin{aligned}dx_A^0(\tau) &= \gamma_A d\tau c \\ d\mathbf{x}_A(\tau) &= \gamma_A d\tau \mathbf{v}_A.\end{aligned} \hspace{\stretch{1}}(4.141)

Writing out the delta functions explicitly we have

\begin{aligned}\begin{aligned}j^i(x) = \sum_A &c e_A \int_{x(\tau)} dx_A^i(\tau) \delta(x^0 - x_A^0(0) - \gamma_A c \tau) \delta(x^1 - x_A^1(0) - \gamma_A v_A^1 \tau) \\ &\delta(x^2 - x_A^2(0) - \gamma_A v_A^2 \tau) \delta(x^3 - x_A^3(0) - \gamma_A v_A^3 \tau)\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.143)

So our time and space components of the current can be written

\begin{aligned}j^0(x) &= \sum_A c^2 e_A \gamma_A \int_{x(\tau)} d\tau\delta(x^0 - x_A^0(0) - \gamma_A c \tau)\delta^3(\mathbf{x} - \mathbf{x}_A(0) - \gamma_A \mathbf{v}_A \tau) \\ \mathbf{j}(x) &= \sum_A c e_A \mathbf{v}_A \gamma_A \int_{x(\tau)} d\tau\delta(x^0 - x_A^0(0) - \gamma_A c \tau)\delta^3(\mathbf{x} - \mathbf{x}_A(0) - \gamma_A \mathbf{v}_A \tau).\end{aligned} \hspace{\stretch{1}}(4.144)

Each of these integrals can be evaluated with respect to the time coordinate delta function leaving the distribution

\begin{aligned}j^0(x) &= \sum_A c e_A \delta^3(\mathbf{x} - \mathbf{x}_A(0) - \frac{\mathbf{v}_A}{c} (x^0 - x_A^0(0))) \\ \mathbf{j}(x) &= \sum_A e_A \mathbf{v}_A \delta^3(\mathbf{x} - \mathbf{x}_A(0) - \frac{\mathbf{v}_A}{c} (x^0 - x_A^0(0)))\end{aligned} \hspace{\stretch{1}}(4.146)

With this more general expression (multi-particle case) it should be possible to show that the four divergence is zero, however, the problem only asks for one particle. For the one particle case, we can make things really easy by taking the initial point in space and time as the origin, and aligning our velocity with one of the coordinates (say $x$).

Doing so we have the result derived in class

\begin{aligned}j = e \begin{bmatrix}c \\ v \\ 0 \\ 0 \end{bmatrix}\delta(x - v x^0/c)\delta(y)\delta(z).\end{aligned} \hspace{\stretch{1}}(4.148)

Our divergence then has only two portions

\begin{aligned}\frac{\partial {j^0}}{\partial {x^0}} &= e c (-v/c) \delta'(x - v x^0/c) \delta(y) \delta(z) \\ \frac{\partial {j^1}}{\partial {x}} &= e v \delta'(x - v x^0/c) \delta(y) \delta(z).\end{aligned} \hspace{\stretch{1}}(4.149)

and these cancel out when summed. Note that this requires us to be loose with our delta functions, treating them like regular functions that are differentiable.

For the more general multiparticle case, we can treat the sum one particle at a time, and in each case, rotate coordinates so that the four divergence only picks up one term.

As for physical interpretation via integral, we have using the four dimensional divergence theorem

\begin{aligned}\int d^4 x \partial_i j^i = \int j^i dS_i\end{aligned} \hspace{\stretch{1}}(4.151)

where $dS_i$ is the three-volume element perpendicular to a $x^i = \text{constant}$ plane. These volume elements are detailed generally in the text [2], however, they do note that one special case specifically $dS_0 = dx dy dz$, the element of the three-dimensional (spatial) volume “normal” to hyperplanes $ct = \text{constant}$.

Without actually computing the determinants, we have something that is roughly of the form

\begin{aligned}0 = \int j^i dS_i=\int c \rho dx dy dz+\int \mathbf{j} \cdot (\mathbf{n}_x c dt dy dz + \mathbf{n}_y c dt dx dz + \mathbf{n}_z c dt dx dy).\end{aligned} \hspace{\stretch{1}}(4.152)

This is cheating a bit to just write $\mathbf{n}_x, \mathbf{n}_y, \mathbf{n}_z$. Are there specific orientations required by the metric. To be precise we’d have to calculate the determinants detailed in the text, and then do the duality transformations.

Per unit time, we can write instead

\begin{aligned}\frac{\partial {}}{\partial {t}} \int \rho dV= -\int \mathbf{j} \cdot (\mathbf{n}_x dy dz + \mathbf{n}_y dx dz + \mathbf{n}_z dx dy)\end{aligned} \hspace{\stretch{1}}(4.153)

Rather loosely this appears to roughly describe that the rate of change of charge in a volume must be matched with the “flow” of current through the surface within that amount of time.

# References

[1] Wikipedia. Electromagnetic tensor — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 27-February-2011]. http://en.wikipedia.org/w/index.php?title=Electromagnetic_tensor&oldid=414989505.

[2] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

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

Posted by peeterjoot on December 22, 2009

[Click here for a PDF of this post with nicer formatting]

# 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

[Click here for a PDF of this post with nicer formatting]

# 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

[Click here for a PDF of this post with nicer formatting]

# 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

\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(33)

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(34)

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_{k \ne (0,0,0)} {\mathbf{A}_\mathbf{k}}^m 2 \pi i \frac{k_m}{\lambda_m} e^{2\pi i \mathbf{k} \cdot \mathbf{x}} \\ &=2 \pi i \sum_{k \ne (0,0,0)} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{2\pi 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 $\mathbf{k} \ne 0$ 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 $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$ for all $\mathbf{k} \ne 0$ 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(36)

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} \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}} (2 \pi c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right).\end{aligned} \quad\quad\quad(37)

# 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, and any implications of the freedom to pick $\mathbf{A}_0$. I’d also like to construct the Poynting vector $T(\gamma^0) \wedge \gamma_0$, and see what the structure of that is with this Fourier representation.

A general calculation of $T^{\mu\nu}$ for an assumed Fourier solution should be possible too, but working in spatial quantities for the general case is probably torture. A four dimensional Fourier series is likely a superior option for the general case.

# References

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

## Relativistic classical proton electron interaction. (a start).

Posted by peeterjoot on September 15, 2009

[Click here for a PDF of this sequence of posts with nicer formatting]

# Motivation

The problem of a solving for the relativistically correct trajectories of classically interacting proton and electron is one that I’ve wanted to try for a while. Conceptually this is just about the simplest interaction problem in electrodynamics (other than motion of a particle in a field), but it is not obvious to me how to even set up the right equations to solve. I should have the tools now to at least write down the equations to solve, and perhaps solve them too.

Familiarity with Geometric Algebra, and the STA form of the Maxwell and Lorentz force equation will be assumed. Writing $F = \mathbf{E} + c I \mathbf{B}$ for the Faraday bivector, these equations are respectively

\begin{aligned}\nabla F &= J/\epsilon_0 c \\ m\frac{d^2 X}{d\tau} &= \frac{q}{c} F \cdot \frac{dX}{d\tau} \end{aligned} \quad\quad\quad(1)

The possibility of self interaction will also be ignored here. From what I have read this self interaction is more complex than regular two particle interaction.

# With only Coulomb interaction.

With just Coulomb (non-relativistic) interaction setup of the equations of motion for the relative vector difference between the particles is straightforward. Let’s write this out as a reference. Whatever we come up with for the relativistic case should reduce to this at small velocities.

Fixing notation, lets write the proton and electron positions respectively by $\mathbf{r}_p$ and $\mathbf{r}_e$, the proton charge as $Z e$, and the electron charge $-e$. For the forces we have

FIXME: picture

\begin{aligned}\text{Force on electron} &= m_e \frac{d^2 \mathbf{r}_e}{dt^2} = - \frac{1}{{4 \pi \epsilon_0}} Z e^2 \frac{\mathbf{r}_e - \mathbf{r}_p}{{\left\lvert{\mathbf{r}_e - \mathbf{r}_p}\right\rvert}^3} \\ \text{Force on proton} &= m_p \frac{d^2 \mathbf{r}_p}{dt^2} = \frac{1}{{4 \pi \epsilon_0}} Z e^2 \frac{\mathbf{r}_e - \mathbf{r}_p}{{\left\lvert{\mathbf{r}_e - \mathbf{r}_p}\right\rvert}^3} \end{aligned} \quad\quad\quad(3)

Subtracting the two after mass division yields the reduced mass equation for the relative motion

\begin{aligned}\frac{d^2 (\mathbf{r}_e -\mathbf{r}_p)}{dt^2} = - \frac{1}{{4 \pi \epsilon_0}} Z e^2 \left( \frac{1}{{m_e}} + \frac{1}{{m_p}}\right) \frac{\mathbf{r}_e - \mathbf{r}_p}{{\left\lvert{\mathbf{r}_e - \mathbf{r}_p}\right\rvert}^3} \end{aligned} \quad\quad\quad(5)

This is now of the same form as the classical problem of two particle gravitational interaction, with the well known conic solutions.

# Using the divergence equation instead.

While use of the Coulomb force above provides the equation of motion for the relative motion of the charges, how to generalize this to the relativistic case is not entirely clear. For the relativistic case we need to consider all of Maxwell’s equations, and not just the divergence equation. Let’s back up a step and setup the problem using the divergence equation instead of Coulomb’s law. This is a bit closer to the use of all of Maxwell’s equations.

To start off we need a discrete charge expression for the charge density, and can use the delta distribution to express this.

\begin{aligned}0 = \int d^3 x \left( \boldsymbol{\nabla} \cdot \mathbf{E} - \frac{1}{{\epsilon_0}} \left( Z e \delta^3(\mathbf{x} - \mathbf{r}_p) - e \delta^3(\mathbf{x} - \mathbf{r}_e) \right) \right) \end{aligned} \quad\quad\quad(6)

Picking a volume element that only encloses one of the respective charges gives us the Coulomb law for the field produced by those charges as above

\begin{aligned}0 &= \int_{\text{Volume around proton only}} d^3 x \left( \boldsymbol{\nabla} \cdot \mathbf{E}_p - \frac{1}{{\epsilon_0}} Z e \delta^3(\mathbf{x} - \mathbf{r}_p) \right) \\ 0 &= \int_{\text{Volume around electron only}} d^3 x \left( \boldsymbol{\nabla} \cdot \mathbf{E}_e + \frac{1}{{\epsilon_0}} e \delta^3(\mathbf{x} - \mathbf{r}_e) \right) \end{aligned} \quad\quad\quad(7)

Here $\mathbf{E}_p$ and $\mathbf{E}_e$ denote the electric fields due to the proton and electron respectively. Ignoring the possibility of self interaction the Lorentz forces on the particles are

\begin{aligned}\text{Force on proton/electron} = \text{charge of proton/electron times field due to electron/proton} \end{aligned}

In symbols, this is

\begin{aligned}m_p \frac{d^2 \mathbf{r}_p}{dt^2} &= Z e \mathbf{E}_e \\ m_e \frac{d^2 \mathbf{r}_e}{dt^2} &= - e \mathbf{E}_p \end{aligned} \quad\quad\quad(9)

If we were to substitute back into the volume integrals we’d have

\begin{aligned}0 &= \int_{\text{Volume around proton only}} d^3 x \left( -\frac{m_e}{e}\boldsymbol{\nabla} \cdot \frac{d^2 \mathbf{r}_e}{dt^2} - \frac{1}{{\epsilon_0}} Z e \delta^3(\mathbf{x} - \mathbf{r}_p) \right) \\ 0 &= \int_{\text{Volume around electron only}} d^3 x \left( \frac{m_p}{Z e}\boldsymbol{\nabla} \cdot \frac{d^2 \mathbf{r}_p}{dt^2} + \frac{1}{{\epsilon_0}} e \delta^3(\mathbf{x} - \mathbf{r}_e) \right) \end{aligned} \quad\quad\quad(11)

It is tempting to take the differences of these two equations so that we can write this in terms of the relative acceleration $d^2 (\mathbf{r}_e - \mathbf{r}_p)/dt^2$. I did just this initially, and was surprised by a mass term of the form $1/m_e - 1/m_p$ instead of reduced mass, which cannot be right. The key to avoiding this mistake is the proper considerations of the integration volumes. Since the volumes are different and can in fact be entirely disjoint, subtracting these is not possible. For this reason we have to be especially careful if a differential form of the divergence integrals (9) were to be used, as in

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E}_p &= \frac{1}{{\epsilon_0}} Z e \delta^3(\mathbf{x} - \mathbf{r}_p) \\ \boldsymbol{\nabla} \cdot \mathbf{E}_e &= -\frac{1}{{\epsilon_0}} e \delta^3(\mathbf{x} - \mathbf{r}_e) \end{aligned} \quad\quad\quad(13)

The domain of applicability of these equations is no longer explicit, since each has to omit a neighborhood around the other charge. When using a delta distribution to express the point charge density it is probably best to stick with an explicit integral form.

Comparing how far we can get starting with the Gauss’s law instead of the Coulomb force, and looking forward to the relativistic case, it seems likely that solving the field equations due to the respective current densities will be the first required step. Only then can we substitute that field solution back into the Lorentz force equation to complete the search for the particle trajectories.

# Relativistic interaction.

First order of business is an expression for a point charge current density four vector. Following Jackson [1], but switching to vector notation from coordinates, we can apparently employ an arbitrary parametrization for the four-vector particle trajectory $R = R^\mu \gamma_\mu$, as measured in the observer frame, and write

\begin{aligned}J(X) = q c \int d\lambda \frac{dX}{d\lambda} \delta^4 (X - R(\lambda)) \end{aligned} \quad\quad\quad(15)

Here $X = X^\mu \gamma_\mu$ is the four vector event specifying the spacetime position of the current, also as measured in the observer frame. Reparameterizating in terms of time should get us back something more familiar looking

\begin{aligned}J(X) &= q c \int dt \frac{dX}{dt} \delta^4 (X - R(t)) \\ &= q c \int dt \frac{d}{dt} (c t \gamma_0 + \gamma_k X^k)\delta^4 (X - R(t)) \\ &= q c \int dt \frac{d}{dt} (c t + \mathbf{x})\delta^4 (X - R(t)) \gamma_0 \\ &= q c \int dt (c + \mathbf{v})\delta^4 (X - R(t)) \gamma_0 \\ &= q c \int dt' (c + \mathbf{v}(t'))\delta^3 (\mathbf{x} - \mathbf{r}(t')) \delta(c t' - c t) \gamma_0 \\ \end{aligned}

Note that the scaling property of the delta function implies $\delta(c t) = \delta(t)/c$. With the split of the four-volume delta function $\delta^4(X - R(t)) = \delta^3(\mathbf{x} - \mathbf{r}(t)) \delta( {x^0}' - x^0 )$, where $x^0 = c t$, we have an explanation for why Jackson had a factor of $c$ in his representation. I initially thought this factor of $c$ was due to CGS vs SI units! One more Jackson equation decoded. We are left with the following spacetime split for a point charge current density four vector

\begin{aligned}J(X) &= q (c + \mathbf{v}(t))\delta^3 (\mathbf{x} - \mathbf{r}(t)) \gamma_0 \end{aligned} \quad\quad\quad(16)

Comparing to the continuous case where we have $J = \rho ( c + \mathbf{v} ) \gamma_0$, it appears that this works out right. One thing worth noting is that in this time reparameterization I accidentally mixed up $X$, the observation event coordinates of $J(X)$, and $R$, the spacetime trajectory of the particle itself. Despite this, I am saved by the delta function since no contributions to the current can occur on trajectories other than $R$, the worldline of the particle itself. So in the final result it should be correct to interpret