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# Posts Tagged ‘geometric algebra’

## new version of Geometric Algebra compilation posted.

Posted by peeterjoot on May 18, 2014

Changelog:

– total rewrite of Stokes theorem content, grouping into new chapter.

– added an Index, and associated \index markup: the new stokes chapter is indexed, as well as \chapter and \section headings.

Other than the above, this version mostly has various bits of accumulated maintainance to the latex code, many of which were scripted:

– Run filter: perl -p -i ~/bin/bracePurgeRef

>> had some older code that used (\eqnref) and (\cite). Strip out the braces (leave the formatting to the ref and cite macros used, and change those globally if desired).

– generate the \myTime command used in FrontBackmatter/Titleback.tex, FrontBackmatter/Titlepage.tex

(changed: make.rules & ~/bin/mkRevInfo)

– remove all Stokes theorem related content, pending a complete rewrite:

\include{calculus/vectorIntegralRelations}
\include{calculus/stokesRevisited}
\include{calculus/stokesNoTensor}

– changes to the implementation of definition, theorem, lemma, and example environments (much prettier boxed backgrounds.)

– Replace many

\begin{align*}
\end{align*}

or

\begin{align}
\end{align}

or



with

\label{eqn:XX:n}
\begin{aligned}

\end{aligned}

or
\label{eqn:XX:n}

– Replace \hbar with \Hbar:

\newcommand{\Hbar}[0]{\,\hbar}

(some package is redefining hbar so that it appears typeset to close)

– use \eqnref instead of \ref

– spelling: indexes -> indices

– move figures to ../../figures/gabook/

(new submodule)

– replace text ‘figure \ref’ with \cref

– replace text ‘equation \ref’ with \eqnref. Equation \ref -> \Eqnref

– Add appendix chapter ‘Mathematica notebooks’

– replace \boxed{} and \myBoxed{} with \boxedEquation{}

– replace \myMathWithDescription -> \mathLabelBox

– prune some old RCS log file content.

– use \prod instead of \Pi

– split make.vars into make.bookvars and cleanup makefile accordingly.

– some introduction of \lr{} insteaad of \left( \right)

– use macros_bm.sty

## Stokes theorem in Geometric algebra

Posted by peeterjoot on May 17, 2014

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

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

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

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

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

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

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

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

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

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

## Theorem 1. Stokes’ Theorem

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

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

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

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

# Basic notation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

# Curvilinear coordinates

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

This is called the vector derivative.

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

# Green’s theorem

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

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

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

## Theorem 2. Green’s Theorem

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

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

Fig 1.1. Infinitesimal loop integral

Over the infinitesimal area, the loop integral decomposes into

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

where the differentials along the curve are

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

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

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

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

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

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

Fig 1.2. Sum of infinitesimal loops

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

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

Special case:

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

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

Here Green’s theorem takes the form

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

# Curl theorem, two volume vector field

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

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

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

that area element is

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

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

Fig 1.3. Oriented area element tiling of a surface

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

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

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

Let’s proceed to expand the inner dot product

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

The complete curl term is thus

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

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

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

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

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

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

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

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

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

Expanded out in full this is

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

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

Fig 1.4. Clockwise loop

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

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

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

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

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

The volume element is an area weighted pseudoscalar for the space

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

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

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

So, the LHS of Stokes theorem takes the coordinate form

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

For the RHS, following fig. 1.5, we have

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

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

Fig 1.5. Euclidean 2D loop

## Example: Cylindrical parameterization

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

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

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

Fig 1.6. Cylindrical polar parameterization

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

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

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

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

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

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

The tangent space unit vectors are

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

and

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

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

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

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

so

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

The tangent space volume element is thus

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

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

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

and

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

Observe that the latter depends on the metric signature.

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

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

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

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

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

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

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

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

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

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

## Example: Composition of boost and rotation

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

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

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

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

and

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

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

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

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

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

and

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

The tangent space vectors are therefore

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

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

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

and

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

By inspection the dual basis for this parameterization is

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

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

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

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

## Example: Dual representation in three dimensions

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

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

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

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

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

Inserting this into an expansion of the curl form we have

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

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

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

leaving

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

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

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

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

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

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

Fig 1.7. Outwards normal

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

and almost have the desired cross product representation

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

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

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

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

# Stokes theorem, three variable volume element parameterization

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

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

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

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

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

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

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

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

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

## Digression

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Fig 1.8. Boundary faces of a spherical parameterization region

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

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

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

This has tangent space basis elements

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

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

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

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

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

The area elements are

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

or

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

The volume element also reduces nicely, and is

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

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

# Stokes theorem, four variable volume element parameterization

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

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

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

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

Our boundary value surface element is therefore

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

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

# Duality and its relation to the pseudoscalar.

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

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

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

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

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

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

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

# Divergence theorem.

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

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

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

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

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

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

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

Stokes theorem takes the form

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

or

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

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

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

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

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

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

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

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

or

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

This allows us to write

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

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

This leaves us with

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

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

## Example: 2D divergence theorem

Let’s start back at

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

On the left our integral can be rewritten as

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

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

For the boundary form we have

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

The duality relations for the tangent space are

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

or

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

Back substitution into the line element gives

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

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

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

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

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

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

Fig 1.9. Normals on area element

We see that

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

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

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

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

## Example: 3D divergence theorem

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

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

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

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

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

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

In the boundary integral our dual two form is

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

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

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

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

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

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

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

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

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

where (sums implied)

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

and

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

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

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

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

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

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

## Example: 4D divergence theorem

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

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

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

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

Here, we’ve written

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

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

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

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

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

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

Here we define the volume scaled normal as

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

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

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

# Volume integral coordinate representations

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

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

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

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

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

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

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

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

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

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

where

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

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

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

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

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

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

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

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

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

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

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

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

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

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

# Final remarks

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

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

# Appendix

## Question: Expand volume elements in coordinates

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

or

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

The final result follows after permuting the indices slightly.

### Lemma 1. Distribution of inner products

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

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

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

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

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

Solving for $B_r b$ in

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

we have

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

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

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

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

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

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

### Lemma 2. Distribution of two bivectors

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

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

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

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

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

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

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

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

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

### Lemma 4. Distribution of two trivectors

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

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

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

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

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

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

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

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

Substituting this back into eq. 1.0.168 proves lemma 4.

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

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

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

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

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

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

### Lemma 6. Duality in a plane

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

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

Satisfying

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

and

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

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

Dotting with $\mathbf{a}$ we have

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

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

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

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

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

This duality relation can be recast with a linear denominator

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

or

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

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

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

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

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

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

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

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

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

### Lemma 8. Pseudoscalar selection

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

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

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

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

or

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

Similarly for odd $k$, we have

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

or

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

Adjusting for the signs completes the proof.

# References

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

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

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

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

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

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

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

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

## Curvilinear coordinates and reciprocal basis

Posted by peeterjoot on March 9, 2014

# Motivation

Here I’d like to explore some ideas from [1] where curvilinear coordinates, manifolds, and the vector derivative are introduced.

# Notation

For simplicity, let’s consider the concrete example of a 2D manifold, 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}}(1.2.1)

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}}(1.2.2)

On the manifold we can calculate a reciprocal basis $\{\mathbf{x}^i\}$, 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}}(1.2.3)

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}}(1.2.4)

(sums over mixed indexes 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}}(1.2.4)

# Examples

Let’s pause for a couple examples that have interesting aspects.

## Example: Circular coordinates on a disk

Consider an infinite disk at height $z_0$, with the origin omitted, parameterized by circular coordinates as in fig. 1.1.

Fig 1.1: Plane with circular coordinates

Points on this surface are

\begin{aligned}\mathbf{x}(r, \theta) = (r \cos\theta, r \sin\theta, z_0).\end{aligned} \hspace{\stretch{1}}(1.3.6)

The manifold basis vectors, defined by eq. 1.2.2 are

\begin{aligned}\begin{aligned}\mathbf{x}_r &= (\cos\theta, \sin\theta, 0) \\ \mathbf{x}_\theta &= r (-\sin\theta, \cos\theta, 0).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.7)

By inspection, the reciprocal basis is

\begin{aligned}\begin{aligned}\mathbf{x}^r &= (\cos\theta, \sin\theta, 0) \\ \mathbf{x}^\theta &= \frac{1}{{r}} (-\sin\theta, \cos\theta, 0).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.3.8)

The first thing to note here is that we cannot reach the points $\mathbf{x}$ of eq. 1.3.6 by linear combination of these basis vectors. Instead these basis vectors only allow us to reach other points on the surface, when already there. For example we cannot actually write

\begin{aligned}\mathbf{x} = x^r \mathbf{x}_r + x^\theta \mathbf{x}_\theta,\end{aligned} \hspace{\stretch{1}}(1.3.9)

unless $z_0 = 0$. This is why eq. 1.2.4 was described as a projective operation (and probably deserves an alternate notation). To recover the original parameterized form of the position vector on the surface, we require

\begin{aligned}\mathbf{x} = x^r \mathbf{x}_r + x^\theta \mathbf{x}_\theta + z_0 \hat{\mathbf{z}}.\end{aligned} \hspace{\stretch{1}}(1.3.10)

The coordinates $x^r, x^\theta$ follow by taking dot products

\begin{aligned}x^r &= \mathbf{x} \cdot \mathbf{x}^r \\ &= (r \cos\theta, r \sin\theta, z_0) \cdot(\cos\theta, \sin\theta, 0) \\ &= r \left( \cos^2 \theta + \sin^2 \theta \right) \\ &= r\end{aligned} \hspace{\stretch{1}}(1.0.11.11)

\begin{aligned}x^\theta &= \mathbf{x} \cdot \mathbf{x}^\theta \\ &= (r \cos\theta, r \sin\theta, z_0) \cdot\frac{1}{{r}} (-\sin\theta, \cos\theta, 0) \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.11.11)

Therefore, a point on the plane, relative to the origin of the plane, in this case, requires just one of the tangent plane basis vectors

\begin{aligned}\mathbf{x} = r \mathbf{x}_r.\end{aligned} \hspace{\stretch{1}}(1.0.11.11)

## Example: Circumference of a circle

Now consider a circular perimeter, as illustrated in fig. 1.2, with the single variable parameterization

Fig 1.2: Circular perimeter

\begin{aligned}\mathbf{x} = r_0 \left( \cos\theta, \sin\theta \right).\end{aligned} \hspace{\stretch{1}}(1.13)

Our tangent space basis is

\begin{aligned}\mathbf{x}_\theta = r_0 \left( -\sin\theta, \cos\theta \right),\end{aligned} \hspace{\stretch{1}}(1.14)

with, by inspection, a reciprocal basis

\begin{aligned}\mathbf{x}^\theta = \frac{1}{{r_0}} \left( -\sin\theta, \cos\theta \right).\end{aligned} \hspace{\stretch{1}}(1.15)

Here we have a curious condition, since the tangent space basis vector is perpendicular to the position vector for the points on the circular surface. So, should we attempt to calculate coordinates using eq. 1.2.4, we just get zero

\begin{aligned}x^\theta &= \mathbf{x} \cdot \mathbf{x}^\theta \\ &= r_0 \left( \cos\theta, \sin\theta \right) \cdot\frac{1}{{r_0}} \left( -\sin\theta, \cos\theta \right) \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.16)

It’s perhaps notable that a coordinate representation using the tangent space basis is possible, but we need to utilize a complex geometry. Assuming

\begin{aligned}\mathbf{x} = x^\theta \mathbf{x}_\theta,\end{aligned} \hspace{\stretch{1}}(1.17)

and writing $i = \mathbf{e}_1 \mathbf{e}_2$ for the pseudoscalar, we can write

\begin{aligned}\begin{aligned}\mathbf{x} &= r_0 \mathbf{e}_1 e^{i\theta} \\ \mathbf{x}_\theta &= r_0 \mathbf{e}_2 e^{i\theta},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.18)

so that, by inversion, the $\theta$ coordinate is

\begin{aligned}x^\theta &= \mathbf{x}\frac{1}{{\mathbf{x}_\theta}} \\ &= \left( r_0 \mathbf{e}_1 e^{i\theta} \right)\left( \frac{ e^{-i\theta } \mathbf{e}_2 }{r_0} \right) \\ &= i,\end{aligned} \hspace{\stretch{1}}(1.19)

or

\begin{aligned}\mathbf{x} = i \mathbf{x}_\theta.\end{aligned} \hspace{\stretch{1}}(1.0.20)

## Example: Surface of a sphere

It is also clear that any parameterization that has radial symmetry will suffer the same issue. For example, for a radial surface in 3D with radius $r_0$ we have

\begin{aligned}\begin{aligned}\mathbf{x} &= r_0 \left( \sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta \right) \\ \mathbf{x}_\theta &= r_0 \left( \cos\theta \cos\phi, \cos\theta \sin\phi, -\sin\theta \right) \\ \mathbf{x}_\phi &= r_0 \left( -\sin\theta \sin\phi, \sin\theta \cos\phi, 0 \right) \\ \mathbf{x}^\theta &= \frac{1}{{r_0}} \left( \cos\theta \cos\phi, \cos\theta \sin\phi, -\sin\theta \right) \\ \mathbf{x}^\phi &= \frac{1}{{r_0 \sin\theta}} \left( -\sin\phi, \cos\phi, 0 \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.21)

The reciprocals here were computed using the mathematica reciprocalFrameSphericalSurface.nb notebook.

Do we have a bivector parameterization of the surface using the tangent space basis? Let’s try

\begin{aligned}\mathbf{x} = x^\theta \mathbf{x}_\theta + x^\phi \mathbf{x}_\phi.\end{aligned} \hspace{\stretch{1}}(1.0.22)

Wedging with $\mathbf{x}_\theta$ and $\mathbf{x}_\phi$, and writing $i = \mathbf{e}_1 \mathbf{e}_2$, respectively yields

\begin{aligned}x^\theta &= \mathbf{x} \wedge \mathbf{x}_\phi \frac{1}{{\mathbf{x}_\theta \wedge \mathbf{x}_\phi}} \\ &= -\mathbf{e}_1 \mathbf{e}_3 \cos \phi - \mathbf{e}_2 \mathbf{e}_3 \sin \phi \\ &= \mathbf{e}_{31} e^{ i \phi}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

\begin{aligned}x^\phi &= -\mathbf{x} \wedge \mathbf{x}_\theta \frac{1}{{\mathbf{x}_\theta \wedge \mathbf{x}_\phi}} \\ &= \mathbf{e}_1 \mathbf{e}_3 \cot \theta \sin \phi + i -\mathbf{e}_2 \mathbf{e}_3 \cot \theta \cos \phi \\ &= \mathbf{e}_2 \mathbf{e}_3 \cot \theta e^{i \phi} + i.\end{aligned} \hspace{\stretch{1}}(1.0.22)

However, substitution back into eq. 1.0.22 shows either pair parameterizes the radial position vector

\begin{aligned}\mathbf{x} = x^\theta \mathbf{x}_\theta = x^\phi \mathbf{x}_\phi.\end{aligned} \hspace{\stretch{1}}(1.0.25)

It is interesting that duality relationships seem to naturally arise attempting to describe points on a surface using the tangent space basis for that surface.

# References

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

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

## Hypersphere volume calculation the easy way

Posted by peeterjoot on January 27, 2013

[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

In problem set three I solved the hypersphere volume problem the hard way. I did it the way that I thought was obvious, starting with the spherical coordinate hypervolume element for an Euclidean space. That build on the volume element calculation I’d previously done for four dimensional Euclidean and Hyperbolic spaces in [1]. This time I avoided any use of Geometric Algebra and wrote out the volume element directly using a Jacobian transformation instead of a wedge product (that leads to the same Jacobian). I then proceeded to integrate the volume element, come up with a recurrence relation for the volume, solve that for even and odd powers, and find a common expression using Gamma functions that was correct for even and odd powers. It was a labourious process, but satisfying since I’d previously tried this calculation and not succeeded.

As and after I did this calculation, which I’ll post later, I figured there had to have been an easier way. In the midterm prep reading of section 5.5 of [2] I found just that method. It’s done there in about six lines, using a trick that seemed really sneaky! I was left with a frustrated feeling, wondering how on earth somebody figured out to do it that way.

After a bit of reflection, I see that the trick is a very sensible approach. I’ll outline that here for 2 and 3 dimensions to show the technique, and the reader can generalize if desired.

# The trick

Switching to spherical or circular coordinates when there is radial symmetry isn’t anything that we could describe as trickery. For example, with $r^2 = x^2 + y^2$ in a two dimensional problem or $r^2 = x^2 + y^2 + z^2$ in a three dimensional problem, it’s clear that we’d do the following respectively if we were evaluating an integral

\begin{aligned}\iint f(r) dx dy = \iint f(r) r dr d\theta = 2 \pi \int f(r) r dr\end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}\iiint f(r) dx dy = \iiint f(r) r^2 dr \sin d\theta d\theta d\phi = 4 \pi \int f(r) r^2 dr\end{aligned} \hspace{\stretch{1}}(1.0.5b)

In fact, for $f(r) = 1$ these give us the area and volume of the circle and sphere respectively.

So, what’s the trick? The first part is the observation that the “area” of a “volume” for the circle and the sphere are found by the derivatives of the “volumes”

\begin{aligned}\frac{d}{dr} \pi r^2 = 2 \pi r\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}\frac{d}{dr} \frac{4 \pi}{3} r^3 = 4 \pi r^2.\end{aligned} \hspace{\stretch{1}}(1.0.2b)

I recall being suprised that this is the case back in high school calculus. When I lost marks for not having the formula for the surface area of a sphere memorized on some test, my calculus teacher told me this title tidbit.

Back then this wasn’t obvious to me, and I complained that it was true only for the perfect symmetry of a sphere. For example, the derivative of an volume of a cube doens’t give the surface area of a cube ($d x^3/dx = 3 x^2 \ne 6 x^2$).

Once we believe or assume that the surface “area” of a hypervolume is the derivative of the volume, we can proceed with the trick. That trick is to express the volume in terms of an unknown constant. For example, for the circle and sphere the generalized “volume”s are respectively

\begin{aligned}V_2 = B r^2\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}V_3 = C r^3\end{aligned} \hspace{\stretch{1}}(1.0.3b)

The perimeter of the circle and surface area of the sphere are then

\begin{aligned}A_2 = 2 B r\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}A_3 = 3 C r^2\end{aligned} \hspace{\stretch{1}}(1.0.4b)

So, if we want to calculate integrals of the form 1.0.1.1 we can write

\begin{aligned}\iint f(r) dx dy = 2 B \int f(r) r dr\end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}\iiint f(r) dx dy = 3 C \int f(r) r^2 dr.\end{aligned} \hspace{\stretch{1}}(1.0.5b)

The essence of the trick is to do such an integral in both Cartesian coordinates to get the left hand sides, and then do the radial right hand side integrals. Comparing these provides the constants $B$ and $C$ and thus completes the “volume” formulas for the circle and sphere.

The function chosen for this integral in the text was a Gaussian exponential $f(r) = e^{-r^2/2}$, something that is strictly radial, and can be integrated over all space. For the 2D case, we’ll integrate

\begin{aligned}\iint e^{-(x^2 +y^2)/2} dx dy = (\sqrt{2 \pi})^2= 2 B \int_0^\infty dr r e^{-r^2/2}= -2 B {\left.{e^{-r^2/2}}\right\vert}_{0}^{\infty} = 2 B.\end{aligned} \hspace{\stretch{1}}(1.0.5b)

We find that $B = \pi$, so the 2D spherical “volume” (the area of the circle) is $V = \pi r^2$.

For the 3D sphere, we have

\begin{aligned}\iiint e^{-(x^2 +y^2 + z^2)/2} dx dy = (\sqrt{2 \pi})^3= 3 C \int_0^\infty dr r^2 e^{-r^2/2}= 3 C \int_0^\infty dr e^{-r^2/2}= 3 C \sqrt{2 \pi}.\end{aligned} \hspace{\stretch{1}}(1.0.5b)

So we have for the volume of the 3D sphere, $V = 4 \pi r^3/3$ as expected. The same idea can be extended to higher dimensional spheres. The text does the even values of $N$. Treating both even and odd values, I’d expect this to yield the result I calculated with the Jacobian volume element method

\begin{aligned}V_{m} = \frac{ \pi^{m/2} R^{m} }{ \Gamma\left( m/2 + 1 \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.5b)

# References

[1] Peeter Joot. Exploring physics with Geometric Algebra, chapter {Spherical and hyperspherical parametrization}. URL http://sites.google.com/site/peeterjoot/math2009/gabook.pdf.

[2] S.K. Ma. Statistical Mechanics. World Scientific, 1985. ISBN 9789971966072. URL http://books.google.ca/books?id=3XyA0SYK4XIC.

## new post of geometric algebra notes, and first post of classicalmechanics notes.

Posted by peeterjoot on January 6, 2013

I’ve posted a new version of my Geometric algebra notes. The changelog (below) for this version is minimal (previous changelog).

There is, however, a significant change in this version of these notes.  I’ve moved a huge chunk out (most of the Lagrangian and Hamiltonian stuff) into a separate pdf on classical mechanics.

Also included in these classical mechanics notes is a few lecture notes from phy354 (UofT Advanced classical mechanics, taught by Prof Erich Poppitz).  I’d audited a few of those lectures, and did the first problem set (and subset of the second).

# gabook changelog

January 04, 2013 Tangent planes and normals in three and four dimensions
Figure out how to express a surface normal in 3d and a “volume” normal in 4d.

Sept 2, 2012 Plane wave solutions in linear isotropic charge free media using Geometric Algebra
Work through the plane wave solution to Maxwell’s equation in linear isotropic charge free media without boundary value constraints. I may have attempted to blunder through this before, but believe this to be more clear than any previous attempts. What’s missing is relating this to polarization states of different types and relationships to Jones vectors and so forth. Also, it’s likely possible to express things in a way that doesn’t require taking any real parts provided one uses the pseudoscalar instead of the scalar complex imaginary appropriately.

Mar 16, 2012 Geometric Algebra. The very quickest introduction.

Jan 27, 2012 Infinitesimal rotations.
Derive the cross product result for infinitesimal rotations with and without GA.

# Classical mechanics changelog

January 06, 2013 Parallel axis theorem
class notes from course audit

January 05, 2013 Problem set 2 (2012)
incomplete attempt at the problem set 2 questions.

December 27, 2012 Dipole Moment induced by a constant electric field

Jul 14, 2012 Some notes on a Landau mechanics problem

Mar 21, 2012 Classical Mechanics Euler Angles

Mar 7, 2012 Rigid body motion.

Feb 29, 2012 Phase Space and Trajectories.

Feb 27, 2012 Potential due to cylindrical distribution.

Feb 24, 2012 Potential for an infinitesimal width infinite plane. Take III

Feb 19, 2012 Attempts at calculating potential distribution for infinite homogeneous plane.

Feb 11, 2012 Runge-Lenz vector conservation
phy354 lecture notes on the Runge-Lenz vector and its use in the Kepler problem.

Jan 24, 2012 PHY354 Advanced Classical Mechanics. Problem set 1.

June 19, 2010 Hoop and spring oscillator problem.
A linear approximation to a hoop and spring problem.

Mar 3, 2010 Notes on Goldstein’s Routh’s procedure.
Puzzle through Routh’s procedure as outlined in Goldstein.

Feb 19, 2010 1D forced harmonic oscillator. Quick solution of non-homogeneous problem.
Solve the one dimensional harmonic oscillator problem using matrix methods.

Jan 1, 2010 Integrating the equation of motion for a one dimensional problem.
Solve for time for an arbitrary one dimensional potential.

Nov 26, 2009 Lagrangian and Euler-Lagrange equation evaluation for the spherical N-pendulum problem

Nov 4, 2009 Spherical polar pendulum for one and multiple masses (Take II)
The constraints required to derive the equations of motion from a bivector parameterized Lagrangian for the multiple spherical pendulum make the problem considerably more complicated than would be the case with a plain scalar parameterization. Take the previous multiple spherical pendulum and rework it with only scalar spherical polar angles. I later rework this once more removing all the geometric algebra, which simplifies it further.

Oct 27, 2009 Spherical polar pendulum for one and multiple masses, and multivector Euler-Lagrange formulation.
Derive the multivector Euler-Lagrange relationships. These were given in Doran/Lasenby but I did not understand it there. Apply these to the multiple spherical pendulum with the Lagrangian expressed in terms of a bivector angle containing all the phi dependence a scalar polar angle.

Sept 26, 2009 Hamiltonian notes.

Sept 22, 2009 Lorentz force from Lagrangian (non-covariant)
Show that the non-covariant Lagrangian from Jackson does produce the Lorentz force law (an exercise for the reader).

Sept 4, 2009 Translation and rotation Noether field currents.
Review Lagrangian field concepts. Derive the field versions of the Euler-Lagrange equations. Calculate the conserved current and conservation law, a divergence, for a Lagrangian with a single parameter symmetry (such as rotation or boost by a scalar angle or rapidity). Next, spacetime symmetries are considered, starting with the question of the symmetry existence, then a calculation of the canonical energy momentum tensor and its associated divergence relation. Next an attempt to use a similar procedure to calculate a conserved current for an incremental spacetime translation. A divergence relation is found, but it is not a conservation relationship having a nonzero difference of energy momentum tensors.

June 17, 2009 Comparison of two covariant Lorentz force Lagrangians
The Lorentz force Lagrangian for a single particle can be expressed in a quadratic fashion much like the classical Kinetic energy based Lagrangian. Compare to the proper time, non quadratic action.

June 5, 2009 Canonical energy momentum tensor and Lagrangian translation
Examine symmetries under translation and spacetime translation and relate to energy and momentum conservation where possible.

April 20, 2009 Tensor derivation of non-dual Maxwell equation from Lagrangian
A tensor only derivation.

April 15, 2009 Lorentz force Lagrangian with conjugate momentum
The Lagrangian can be expressed in a QM like form in terms of a sum of mechanical and electromagnetic momentum, mv + qA/c. The end result is the same and it works out to just be a factorization of the original Lorentz force covariant Lagrangian.

December 02, 2008 Compare some wave equation’s and their Lagrangians
A summary of some wave equation Lagrangians, including wave equations of quantum mechanics.

October 29, 2008 Field form of Noether’s Law

October 22, 2008 Lorentz transform Noether current for interaction Lagrangian

October 19, 2008 Lorentz Invariance of Maxwell Lagrangian

October 13, 2008 Euler Lagrange Equations

October 10, 2008 Derivation of Euler-Lagrange field equations
Derivation of the field form of the Euler Lagrange equations, with applications including Schrodinger’s and Klein-Gordan field equations

October 8, 2008 Revisit Lorentz force from Lagrangian

September 8, 2008 Direct variation of Maxwell equations

Sept 2, 2008 Attempts at solutions for some Goldstein Mechanics problems
Solutions to selected Goldstein Mechanics problems from chapter I and II.

Some of the Goldstein problems in chapter I were also in the Tong problem set. This is some remaining ones and a start at chapter II problems.

Problem 8 from Chapter I was never really completed in my first pass. It looks like I missed the Kinetic term in the Lagrangian too. The question of if angular momentum is conserved in that problem is considered in more detail, and a Noether’s derivation that is specific to the calculation of the conserved “current” for a rotational symmetry is performed. I’d be curious what attack on that question Goldstein was originally thinking of. Although I believe this Noether’s current treatment answers the question in full detail, since it wasn’t covered yet in the text, is there an easier way to get at the result?

September 1, 2008 Vector canonical momentum

August 30, 2008 Short metric tensor explanation
Metric tensor and Lorentz diagonality.

August 25, 2008 Solutions to David Tong’s mf1 Lagrangian problems

August 21, 2008 Covariant Lagrangian, and electrodynamic potential

August 9, 2008 Newton’s Law from Lagrangian

## Tangent planes and normals in three and four dimensions

Posted by peeterjoot on January 4, 2013

# Motivation

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

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

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

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

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

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

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

# 3D surface

We start by considering figure 1:

Figure 1: A portion of a surface in 3D

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

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

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

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

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

The tangent plane is then

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

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

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

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

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

Observe that we can write this as

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

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

# 4D surface

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

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

the differential of this is

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

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

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

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

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

In particular, in each of the directions we have

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

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

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

Our tangent volume in this neighbourhood is

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

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

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

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

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

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

This gives us

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

With the usual 4d gradient definition (sum implied)

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

we have

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

so we can write

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

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

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

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

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

# References

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

## Derivation of Fresnel equations for mixed polarization (using Geometric Algebra)

Posted by peeterjoot on September 25, 2012

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

# Motivation

In [2] we have a derivation of the Fresnel equations for the TE and TM polarization modes. Can we do this for an arbitrary polarization angles?

# Setup

The task at hand is to find evaluate the boundary value constraints. Following the interface plane conventions of [1], and his notation that is

\begin{aligned}\epsilon_1 ( \mathbf{E}_i + \mathbf{E}_r )_z = \epsilon_2 ( \mathbf{E}_t )_z\end{aligned} \hspace{\stretch{1}}(1.2.1a)

\begin{aligned}( \mathbf{B}_i + \mathbf{B}_r )_z = ( \mathbf{B}_t )_z\end{aligned} \hspace{\stretch{1}}(1.2.1b)

\begin{aligned}( \mathbf{E}_i + \mathbf{E}_r )_{x,y} = ( \mathbf{E}_t )_{x,y}\end{aligned} \hspace{\stretch{1}}(1.2.1c)

\begin{aligned}\frac{1}{{\mu_1}} ( \mathbf{B}_i + \mathbf{B}_r )_{x,y} = \frac{1}{{\mu_2}} ( \mathbf{B}_t )_{x,y}\end{aligned} \hspace{\stretch{1}}(1.2.1d)

I’ll work here with a phasor representation directly and not bother with taking real parts, or using tilde notation to mark the vectors as complex.

Our complex magnetic field phasors are related to the electric fields with

\begin{aligned}\mathbf{B} = \frac{1}{{v}} \hat{\mathbf{k}} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.2.2)

Referring to figure (see pdf) shows the geometrical task to tackle, since we’ve got to express all the various unit vectors algebraically. I’ll use Geometric Algebra here to do that for its compact expression of rotations. With

Figure: See pdf: Reflection and transmission of light at an interface

\begin{aligned}j = \mathbf{e}_3 \mathbf{e}_1,\end{aligned} \hspace{\stretch{1}}(1.2.3)

we can express each of the $k$ vector directions by inspection. Those are

\begin{aligned}\hat{\mathbf{k}}_i = \mathbf{e}_3 e^{j \theta_i} = \mathbf{e}_3 \cos\theta_i + \mathbf{e}_1 \sin\theta_i\end{aligned} \hspace{\stretch{1}}(1.2.4a)

\begin{aligned}\hat{\mathbf{k}}_r = -\mathbf{e}_3 e^{-j \theta_r} = -\mathbf{e}_3 \cos\theta_r +\mathbf{e}_1 \sin\theta_r\end{aligned} \hspace{\stretch{1}}(1.2.4b)

\begin{aligned}\hat{\mathbf{k}}_t = \mathbf{e}_3 e^{j \theta_t} = \mathbf{e}_3 \cos\theta_t + \mathbf{e}_1 \sin\theta_t.\end{aligned} \hspace{\stretch{1}}(1.2.4c)

Similarly, the perpendiculars $\hat{\mathbf{m}}_p = \mathbf{e}_2 \times \hat{\mathbf{k}}_p$ are

\begin{aligned}\hat{\mathbf{m}}_i = \mathbf{e}_{1} e^{j \theta_i}= \mathbf{e}_1 \cos\theta_i - \mathbf{e}_3 \sin\theta_i= \mathbf{e}_3 j e^{j \theta_i}\end{aligned} \hspace{\stretch{1}}(1.2.5a)

\begin{aligned}\hat{\mathbf{m}}_r = -\mathbf{e}_{1} e^{-j \theta_r}= -\mathbf{e}_1 \cos\theta_r - \mathbf{e}_3 \sin\theta_r= -\mathbf{e}_3 j e^{-j \theta_r}\end{aligned} \hspace{\stretch{1}}(1.2.5b)

\begin{aligned}\hat{\mathbf{m}}_t = \mathbf{e}_{1 } e^{j \theta_t} = \mathbf{e}_1 \cos\theta_t - \mathbf{e}_3 \sin\theta_t= \mathbf{e}_3 j e^{j \theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.5c)

In [1] problem 9.14 we had to show that the polarization angles for normal incident ($\mathbf{E} \parallel \mathbf{e}_1$) must be the same due to the boundary constraints. Can we also tackle that problem for both this more general angle of incidence and a general polarization? Let’s try so, allowing temporarily for different polarizations of the reflected and transmitted components of the light, calling those polarization angles $\phi_i$, $\phi_r$, and $\phi_t$ respectively. Let’s set the $\phi_i = 0$ polarization aligned such that $\mathbf{E}_i$, $\mathbf{B}_i$ are aligned with the $\mathbf{e}_2$ and $-\hat{\mathbf{m}}_i$ directions respectively, so that the generally polarized phasors are

\begin{aligned}\begin{bmatrix}\mathbf{E}_p \\ \mathbf{B}_p \\ \end{bmatrix}=\begin{bmatrix}\mathbf{e}_2 \\ -\hat{\mathbf{m}}_p \\ \end{bmatrix}e^{ \hat{\mathbf{m}}_p \mathbf{e}_2 \phi_p }\end{aligned} \hspace{\stretch{1}}(1.2.6)

We are now set to at least express our boundary value constraints

\begin{aligned}\epsilon_1 \left( \mathbf{e}_2 E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + \mathbf{e}_2 E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \cdot \mathbf{e}_3 = \epsilon_2 \left( \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7a)

\begin{aligned}\frac{1}{{v_1}} \left( -\hat{\mathbf{m}}_i E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - \hat{\mathbf{m}}_r E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \cdot \mathbf{e}_3 = \frac{1}{{v_2}} \left( -\hat{\mathbf{m}}_t E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7b)

\begin{aligned}\left( \mathbf{e}_2 E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + \mathbf{e}_2 E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \wedge \mathbf{e}_3 = \left( \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( -\hat{\mathbf{m}}_i E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - \hat{\mathbf{m}}_r E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) \wedge \mathbf{e}_3 = \frac{1}{{\mu_2 v_2}} \left( -\hat{\mathbf{m}}_t E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.7d)

Let’s try this in a couple of steps. First with polarization angles set so that one of the fields lies in the plane of the interface (with both variations), and then attempt the general case, first posing the problem in the tranditional way to see what equations fall out, and then using superposition.

Before doing so, let’s introduce a bit of notation to be used throughout. When we wish to refer to all the fields or angles, for example, $\mathbf{E}_i, \mathbf{E}_r, \mathbf{E}_t$ then we’ll write $\mathbf{E}_p$ where $p \in \{i, r, t\}$. Similarily, to refer to just the incident and transmitted components (or angles) we’ll use $\mathbf{E}_q$ where $q \in \{i, t\}$. Following [1] we’ll also write

\begin{aligned}\beta = \frac{\mu_1 v_1} {\mu_2 v_2} \end{aligned} \hspace{\stretch{1}}(1.2.8)

\begin{aligned}\alpha = \frac{\cos\theta_t}{\cos\theta_i},\end{aligned} \hspace{\stretch{1}}(1.2.8)

## Question: Sanity check. Verify for $\mathbf{E}$ parallel to the interface.

For the $\mathbf{E}_p \parallel \mathbf{e}_2$ polarization ($\phi_i = \phi_r = \phi_t$) our phasors are

\begin{aligned}\mathbf{E}_p = \mathbf{e}_2 E_p\end{aligned} \hspace{\stretch{1}}(1.2.9)

\begin{aligned}\mathbf{B}_p = -\frac{1}{{v_p}} \hat{\mathbf{m}}_p E_p\end{aligned} \hspace{\stretch{1}}(1.2.9)

Our boundary value constraints then become

\begin{aligned}\epsilon_1 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) \cdot \mathbf{e}_3 = \epsilon_2 \left( \mathbf{e}_2 E_t \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10a)

\begin{aligned}\frac{1}{{v_1}} \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r \right) \cdot \mathbf{e}_3 = \frac{1}{{v_2}} \left( \hat{\mathbf{m}}_t E_t \right) \cdot \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10b)

\begin{aligned}\left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) \wedge \mathbf{e}_3 = \left( \mathbf{e}_2 E_t \right) \wedge \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(1.2.10c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r \right) \wedge \mathbf{e}_3 = \frac{1}{{\mu_2 v_2}} \left( \hat{\mathbf{m}}_t E_t \right) \wedge \mathbf{e}_3.\end{aligned} \hspace{\stretch{1}}(1.2.10d)

With $\hat{\mathbf{m}}_p$ substitution this is

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.11a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_i} E_i - \mathbf{e}_1 e^{-j \theta_r} E_r \right) }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_t} E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.11b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.11c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_i} E_i -\mathbf{e}_1 e^{-j \theta_r} E_r \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_1 e^{j \theta_t} E_t \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.11d)

Evaluating the grade selections we have a separation into an analogue of real and imaginary parts for

\begin{aligned}0 = 0\end{aligned} \hspace{\stretch{1}}(1.2.12a)

\begin{aligned}\frac{1}{{v_1}} \left( -\sin\theta_i E_i - \sin\theta_r E_r \right) = \frac{1}{{v_2}} \left( -\sin\theta_t E_t \right)\end{aligned} \hspace{\stretch{1}}(1.2.12b)

\begin{aligned}E_i + E_r = E_t\end{aligned} \hspace{\stretch{1}}(1.2.12c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} \left( \cos{\theta_i} E_i - \cos{\theta_r} E_r \right) = \frac{1}{{\mu_2 v_2}} \left( \cos{ \theta_t} E_t \right).\end{aligned} \hspace{\stretch{1}}(1.2.12d)

With $\theta_i = \theta_r$ and $\sin\theta_t/\sin\theta_i = n_1/n_2$ 1.2.12b becomes

\begin{aligned}E_i + E_r = \frac{n_1 v_1}{n_2 v_2} E_t = \frac{v_2 v_1}{v_1 v_2} E_t = E_t,\end{aligned} \hspace{\stretch{1}}(1.2.12d)

so that we find 1.2.12b and 1.2.12c are dependent. We are left with a pair of equations

\begin{aligned}E_i + E_r = E_t\end{aligned} \hspace{\stretch{1}}(1.2.14)

\begin{aligned}E_i - E_r = \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} E_t,\end{aligned} \hspace{\stretch{1}}(1.2.14)

\begin{aligned}2 E_i = \left( 1 + \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} \right) E_t\end{aligned} \hspace{\stretch{1}}(1.2.15)

\begin{aligned}2 E_r = \left( 1 - \frac{\mu_1 v_1}{\mu_2 v_2} \frac{\cos{ \theta_t}}{\cos\theta_i} \right) E_t,\end{aligned} \hspace{\stretch{1}}(1.2.15)

with a final rearrangement to yield

\begin{aligned}\frac{E_t}{E_i}=\frac{2 \mu_2 v_2 \cos\theta_i}{\mu_2 v_2 \cos\theta_i+\mu_1 v_1 \cos\theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.16)

\begin{aligned}\frac{E_r}{E_i}=\frac{\mu_2 v_2 \cos\theta_i-\mu_1 v_1 \cos\theta_t}{\mu_2 v_2 \cos\theta_i+\mu_1 v_1 \cos\theta_t}\end{aligned} \hspace{\stretch{1}}(1.2.16)

Using the $\alpha$ and $\beta$ notation above we have

\begin{aligned}\frac{E_t}{E_i}=\frac{2 }{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.17)

\begin{aligned}\frac{E_r}{E_i}=\frac{1 - \alpha \beta}{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.17)

## Question: Sanity check. Verify for $\mathbf{B}$ parallel to the interface.

As a second sanity check let’s rotate our field polarizations by applying a rotation $e^{\mathbf{e}_2 \hat{\mathbf{m}}_p \pi/2} = \mathbf{e}_2 \hat{\mathbf{m}}_p$ ($\phi_i = \phi_r = \phi_t = -\pi/2$) so that

\begin{aligned}-\hat{\mathbf{m}}_p \rightarrow -\hat{\mathbf{m}}_p \mathbf{e}_2 \hat{\mathbf{m}}_p = \mathbf{e}_2\end{aligned} \hspace{\stretch{1}}(1.2.18)

\begin{aligned}\mathbf{e}_2 \rightarrow \mathbf{e}_2 \mathbf{e}_2 \hat{\mathbf{m}}_p = \hat{\mathbf{m}}_p\end{aligned} \hspace{\stretch{1}}(1.2.18)

This time we have $\mathbf{E}_p \parallel \hat{\mathbf{m}}_p$ and $\mathbf{B}_p \parallel \mathbf{e}_2$. Our boundary value equations become

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_t E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.19a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.19b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_i E_i + \hat{\mathbf{m}}_r E_r \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \hat{\mathbf{m}}_t E_t \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.19c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.19d)

This second equation 1.2.19b is a $0 = 0$ identity, and the remaining after $\hat{\mathbf{m}}_p$ substitution are

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_i} E_i + (-\mathbf{e}_3) j e^{-j \theta_r} E_r \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_t} E_t \right) }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.20a)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_i} E_i + (-\mathbf{e}_3) j e^{-j \theta_r} E_r \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_3 j e^{j \theta_t} E_t \right) }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.20b)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_i + \mathbf{e}_2 E_r \right) }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ \mathbf{e}_3 \left( \mathbf{e}_2 E_t \right) }}\right\rangle}_{2}.\end{aligned} \hspace{\stretch{1}}(1.2.20c)

Simplifying we have

\begin{aligned}\epsilon_1 \left( -\sin \theta_i E_i - \sin{\theta_r} E_r \right) = - \epsilon_2 \sin{\theta_t} E_t \end{aligned} \hspace{\stretch{1}}(1.2.21a)

\begin{aligned} \cos{ \theta_i} E_i - \cos{ \theta_r} E_r = \cos{ \theta_t} E_t\end{aligned} \hspace{\stretch{1}}(1.2.21b)

\begin{aligned}E_i + E_r = \frac{\mu_1 v_1} {\mu_2 v_2} E_t\end{aligned} \hspace{\stretch{1}}(1.2.21c)

We expect an equality

\begin{aligned}\frac{\epsilon_2 \sin\theta_t}{\epsilon_1 \sin\theta_i} = \frac{\mu_1 v_1} {\mu_2 v_2},\end{aligned} \hspace{\stretch{1}}(1.2.21c)

Noting that $\epsilon_p v_p = 1/(v_p \mu_p)$ we find that to be true

\begin{aligned}\frac{\epsilon_2 \sin\theta_t}{\epsilon_1 \sin\theta_i} = \frac{\epsilon_2 n_1}{\epsilon_1 n_2} = \frac{\epsilon_2 v_2}{\epsilon_1 v_1} = \frac{\mu_1 v_1}{\mu_2 v_2} \end{aligned} \hspace{\stretch{1}}(1.2.21c)

we see that 1.2.21a and 1.2.21c are dependent. We are left with the system

\begin{aligned}E_i - E_r = \alpha E_t\end{aligned} \hspace{\stretch{1}}(1.2.24a)

\begin{aligned}E_i + E_r = \beta E_t,\end{aligned} \hspace{\stretch{1}}(1.2.24b)

with solution

\begin{aligned}\frac{E_t}{E_i} = \frac{2 }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.25)

\begin{aligned}\frac{E_r}{E_i} = \frac{\beta - \alpha}{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.25)

## Question: General case. Arbitrary polarization angle.

Determine the set of simulaneous equations that would have to be solved for if the incident polarization angle was allowed to be neither TE nor TM mode.

Substituting our $\hat{\mathbf{m}}_p$ vector expressions into the boundary value constraints we have

\begin{aligned}\epsilon_1 \left\langle{{ \mathbf{e}_3 \mathbf{e}_2 \left( E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) }}\right\rangle = \epsilon_2 \left\langle{{ \mathbf{e}_3 \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle \end{aligned} \hspace{\stretch{1}}(1.2.26a)

\begin{aligned}\frac{1}{{v_1}} \left\langle{{ j e^{j \theta_i} E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - j e^{-j \theta_r} E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } }}\right\rangle = \frac{1}{{v_2}} \left\langle{{ j e^{j \theta_t} E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.26b)

\begin{aligned}{\left\langle{{ \mathbf{e}_3 \mathbf{e}_2 \left( E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } + E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } \right) }}\right\rangle}_{2} = {\left\langle{{ \mathbf{e}_3 \mathbf{e}_2 E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.26c)

\begin{aligned}\frac{1}{{\mu_1 v_1}} {\left\langle{{ j e^{j \theta_i} E_i e^{ \hat{\mathbf{m}}_i \mathbf{e}_2 \phi_i } - j e^{-j \theta_r} E_r e^{ \hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r } }}\right\rangle}_{2} = \frac{1}{{\mu_2 v_2}} {\left\langle{{ j e^{j \theta_t} E_t e^{ \hat{\mathbf{m}}_t \mathbf{e}_2 \phi_t } }}\right\rangle}_{2}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

With $\alpha \in \{i,r\}$ we want to expand some intermediate multivector products

\begin{aligned}\mathbf{e}_{32} e^{\hat{\mathbf{m}}_q \mathbf{e}_2 \phi_q}=\mathbf{e}_{32} \cos \phi_q+\mathbf{e}_{32} \hat{\mathbf{m}}_q \mathbf{e}_2 \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q+\mathbf{e}_{32} \mathbf{e}_3 j e^{j \theta_q} \mathbf{e}_2 \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q-j e^{j \theta_q} \sin{\phi_q}=\mathbf{e}_{32} \cos \phi_q - \mathbf{e}_{31} \cos\theta_q \sin\phi_q+ \sin\theta_q \sin{\phi_q}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}\mathbf{e}_{32} e^{\hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r}=\mathbf{e}_{32} \cos \phi_r+\mathbf{e}_{32} \hat{\mathbf{m}}_r \mathbf{e}_2 \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r+\mathbf{e}_{32} (-\mathbf{e}_3) j e^{-j \theta_r} \mathbf{e}_2 \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r+ j e^{-j \theta_r} \sin{\phi_r}=\mathbf{e}_{32} \cos \phi_r + \mathbf{e}_{31} \cos\theta_r \sin{\phi_r}+ \sin \theta_r \sin{\phi_r}\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}j e^{j \theta_q} e^{\hat{\mathbf{m}}_q \mathbf{e}_2 \phi_q}=j e^{j \theta_q} \left( \cos \phi_q+ \hat{\mathbf{m}}_q \mathbf{e}_2 \sin{\phi_q}\right)=j e^{j \theta_q} \left( \cos \phi_q+ \mathbf{e}_3 j e^{j \theta_q} \mathbf{e}_2 \sin{\phi_q}\right)=j e^{j \theta_q} \left( \cos \phi_q- j e^{-j \theta_q} \mathbf{e}_{32} \sin{\phi_q}\right)=j e^{j \theta_q} \cos \phi_q+ \mathbf{e}_{32} \sin{\phi_q}=\mathbf{e}_{31} \cos {j \theta_q} \cos \phi_q+ \mathbf{e}_{32} \sin{\phi_q}- \sin{ \theta_q} \cos \phi_q\end{aligned} \hspace{\stretch{1}}(1.2.26d)

\begin{aligned}-j e^{-j \theta_r} e^{\hat{\mathbf{m}}_r \mathbf{e}_2 \phi_r}=-j e^{-j \theta_r} \left(\cos\phi_r + \hat{\mathbf{m}}_r \mathbf{e}_2 \sin\phi_r\right)=-j e^{-j \theta_r} \left(\cos\phi_r - \mathbf{e}_3 j e^{-j \theta_r} \mathbf{e}_2 \sin\phi_r\right)=-j e^{-j \theta_r} \left(\cos\phi_r + j e^{j \theta_r} \mathbf{e}_{32} \sin\phi_r\right)=-j e^{-j \theta_r} \cos\phi_r + \mathbf{e}_{32} \sin\phi_r=-\mathbf{e}_{31} \cos{\theta_r} \cos\phi_r + \mathbf{e}_{32} \sin\phi_r- \sin{ \theta_r} \cos\phi_r \end{aligned} \hspace{\stretch{1}}(1.2.26d)

Our boundary value conditions are then

\begin{aligned}\epsilon_1 \left( E_i \sin\theta_i \sin{\phi_i}+ E_r \sin \theta_r \sin{\phi_r}\right) = \epsilon_2 E_t \sin\theta_t \sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{v_1}}\left(E_i \sin{ \theta_i} \cos \phi_i+E_r \sin{ \theta_r} \cos\phi_r \right)=\frac{1}{{v_2}}E_t \sin{ \theta_t} \cos \phi_t\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}E_i \cos \phi_i + E_r \cos \phi_r =E_t \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}-E_i \cos\theta_i \sin\phi_i+ E_r \cos\theta_r \sin{\phi_r}=-E_t \cos\theta_t \sin\phi_t\end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{\mu_1 v_1}}\left(E_i\cos { \theta_i} \cos \phi_i -E_r\cos{\theta_r} \cos\phi_r \right)=\frac{1}{{\mu_2 v_2}}E_t \cos { \theta_t} \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.31)

\begin{aligned}\frac{1}{{\mu_1 v_1}}\left(E_i\sin{\phi_i}+E_r\sin\phi_r\right)=\frac{1}{{\mu_2 v_2}}E_t\sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.31)

Note that the wedge product equations above have been separated into $\mathbf{e}_3 \mathbf{e}_1$ and $\mathbf{e}_3 \mathbf{e}_2$ components, yielding two equations each. Because of 1.2.21c, we see that 1.2.31 and 1.2.31 are dependent. Also, as demonstrated in 1.2.12d we see that 1.2.31 and 1.2.31 are also dependent. We can therefore consider only the last four equations (and still have additional linear dependencies to be discovered.)

Let’s write these as

\begin{aligned}E_i \cos \phi_i + E_r \cos \phi_r =E_t \cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}-E_i \sin\phi_i+ E_r \sin{\phi_r}=-E_t \alpha\sin\phi_t\end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}E_i\cos \phi_i -E_r\cos\phi_r =\alpha \beta E_t\cos \phi_t \end{aligned} \hspace{\stretch{1}}(1.2.32)

\begin{aligned}E_i\sin{\phi_i}+E_r\sin\phi_r=\beta E_t\sin{\phi_t}\end{aligned} \hspace{\stretch{1}}(1.2.32)

Observe that if $\phi_i = \phi_r = \phi_t = 0$ (killing all the sine terms) we recover 1.2.14, and with $\phi_i = \phi_r = \phi_t = \pi/2$ (killing all the cosines) we recover 1.2.24.

Now, if $\phi_i n \pi/2$ we’ve got a different story. Specifically it appears that should we wish to solve for the reflected and transmitted magnitudes, we also have to simulaneously solve for the polarization angles in the reflected and transmitted directions. This is now a problem of solving four simulaneous equations in two linear and two non-linear variables.

Does it make sense that we would have polarization rotation should our initial polarization angle be rotated? I think so. In dicusssing this problem with Prof Thywissen, he strongly suggested treating the problem as a superposition of two light waves. If we consider that, even without attempting to solve the problem, we see that we must have different reflected and transmitted magnitudes associated with the pair of incident waves since we have to calculate each of these with different Fresnel equations. This would have an effect of scaling and rotating the superimposed reflected and transmitted waves.

## Question: General case using using superposition

Using superposition determine the Fresnel equations for an arbitrary incident polarization angle. This should involve solving for both the magnitude and the polarization angle of the reflected and transmitted rays.

For a polarization of $\phi = 0$ and $\phi = \pi/2$ respectively, we have from problems \ref{fresnelAlternatePolarization:pr1-Answer} and \ref{fresnelAlternatePolarization:pr2-Answer}, or from 1.2.32 we have

\begin{aligned}\frac{E_{r \parallel}}{E_{i \parallel}} = \frac{1 - \alpha \beta}{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{t \parallel}}{E_{i \parallel}} = \frac{2 }{1 + \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{r \perp}}{E_{i \perp}} = \frac{ \beta - \alpha }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}\frac{E_{t \perp}}{E_{i \perp}} = \frac{ 2 }{\beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.33)

We can use these results to consider a polarization of $\phi < \pi/2$ as illustrated in figure (see pdf)

Figure: see pdf: Polarization of incident field to be considered

Our incident, reflected, and transmitted fields are

\begin{aligned}\mathbf{E}_i = E_{i} \mathbf{e}_2 e^{\mathbf{e}_2 \hat{\mathbf{m}}_i \phi}\end{aligned} \hspace{\stretch{1}}(1.2.34)

\begin{aligned}\mathbf{E}_r = E_{i \parallel}\frac{1 - \alpha\beta}{1 + \alpha\beta} \mathbf{e}_2 + E_{i \perp}\frac{\beta - \alpha}{\beta + \alpha} \hat{\mathbf{m}}_r\end{aligned} \hspace{\stretch{1}}(1.2.34)

\begin{aligned}\mathbf{E}_t = E_{i \parallel}\frac{2}{1 + \alpha\beta}\mathbf{e}_2+ E_{i \perp}\frac{2}{\beta + \alpha}\hat{\mathbf{m}}_i\end{aligned} \hspace{\stretch{1}}(1.2.34)

However, $E_{i \parallel} = E_i \cos \phi$ and $E_{i \perp} = E_i \sin\phi$ leaving us with

\begin{aligned}\mathbf{E}_i = E_{i} \left( \mathbf{e}_2 \cos\phi + \mathbf{e}_1 e^{j \theta_i} \sin\phi \right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}\mathbf{E}_r = E_i \left(\cos\phi\frac{1 - \alpha\beta}{1 + \alpha\beta} \mathbf{e}_2 - \sin\phi\frac{\beta - \alpha}{\beta + \alpha} \mathbf{e}_1 e^{-j \theta_r}\right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}\mathbf{E}_t = E_i\left(\cos\phi\frac{2}{1 + \alpha\beta}\mathbf{e}_2+ \sin\phi\frac{2}{\beta + \alpha}\mathbf{e}_1 e^{j \theta_t}\right)\end{aligned} \hspace{\stretch{1}}(1.2.35)

We find that the reflected and transmitted polarization angles are respectively

\begin{aligned}\tan \phi_r = \tan \phi\frac{\beta - \alpha}{\beta + \alpha} \frac{1 + \alpha \beta}{1 - \alpha \beta}\end{aligned} \hspace{\stretch{1}}(1.2.36)

\begin{aligned}\tan \phi_t = \tan \phi \frac{ 1 + \alpha \beta}{ \beta + \alpha}\end{aligned} \hspace{\stretch{1}}(1.2.36)

where the associated magnitudes are

\begin{aligned}\frac{E_r}{E_i}= \sqrt{\left(\cos\phi\frac{1 - \alpha\beta}{1 + \alpha\beta} \right)^2+ \left( \sin\phi\frac{\beta - \alpha}{\beta + \alpha} \right)^2}\end{aligned} \hspace{\stretch{1}}(1.2.37)

\begin{aligned}\frac{E_t}{E_i}=\sqrt{\left(\cos\phi\frac{2}{1 + \alpha\beta}\right)^2+\left(\sin\phi\frac{2}{\beta + \alpha}\right)^2}\end{aligned} \hspace{\stretch{1}}(1.2.37)

# References

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

[2] E. Hecht. Optics. 1998.

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

Posted by peeterjoot on September 3, 2012

# Motivation

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

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

# Maxwell’s equations in media

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

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

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

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

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

We merge these using the geometric identity

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

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

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

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

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

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

This can now be added to 1.2.3a for

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

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

# Phasor solutions

We write the electromagnetic field as

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

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

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

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

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

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

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

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

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

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

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

Inserting the exponential form of our assumed solution 1.3.7 we find

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

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

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

Our assumed solution must also satisfy the first order system 1.3.9

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

The constraints on $F_0$ must then be given by

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

With

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

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

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

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

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

This is

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

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

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

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

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

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

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

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

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

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

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

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

# References

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

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

## Putting the stress tensor (and traction vector) into explicit vector form.

Posted by peeterjoot on April 8, 2012

[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.

Exersize 6.1 from [1] is to show that the traction vector can be written in vector form (a rather curious thing to have to say) as

\begin{aligned}\mathbf{t} = -p \hat{\mathbf{n}} + \mu ( 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla})\mathbf{u} + \hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u})).\end{aligned} \hspace{\stretch{1}}(1.1)

Note that the text uses a wedge symbol for the cross product, and I’ve switched to standard notation. I’ve done so because the use of a Geometric-Algebra wedge product also can be used to express this relationship, in which case we would write

\begin{aligned}\mathbf{t} = -p \hat{\mathbf{n}} + \mu ( 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}}).\end{aligned} \hspace{\stretch{1}}(1.2)

In either case we have

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}}=\hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u})=\boldsymbol{\nabla}' (\hat{\mathbf{n}} \cdot \mathbf{u}') - (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u}\end{aligned} \hspace{\stretch{1}}(1.3)

(where the primes indicate the scope of the gradient, showing here that we are operating only on $\mathbf{u}$, and not $\hat{\mathbf{n}}$).

After computing this, lets also compute the stress tensor in cylindrical and spherical coordinates (a portion of that is also problem 6.10), something that this allows us to do fairly easily without having to deal with the second order terms that we encountered doing this by computing the difference of squared displacements.

We’ll work primarily with just the strain tensor portion of the traction vector expressions above, calculating

\begin{aligned}2 {\mathbf{e}}_{\hat{\mathbf{n}}}=2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla})\mathbf{u} + \hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u})=2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla})\mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}}.\end{aligned} \hspace{\stretch{1}}(1.4)

We’ll see that this gives us a nice way to interpret these tensor relationships. The interpretation was less clear when we computed this from the second order difference method, but here we see that we are just looking at the components of the force in each of the respective directions, dependent on which way our normal is specified.

# Verifying the relationship.

\begin{aligned}(\hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u}) + 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u})_i&=n_a (\boldsymbol{\nabla} \times \mathbf{u})_b \epsilon_{a b i} + 2 n_a \partial_a u_i \\ &=n_a \partial_r u_s \epsilon_{r s b} \epsilon_{a b i} + 2 n_a \partial_a u_i \\ &=n_a \partial_r u_s \delta_{ia}^{[rs]} + 2 n_a \partial_a u_i \\ &=n_a ( \partial_i u_a -\partial_a u_i ) + 2 n_a \partial_a u_i \\ &=n_a \partial_i u_a + n_a \partial_a u_i \\ &=n_a (\partial_i u_a + \partial_a u_i) \\ &=\sigma_{i a } n_a\end{aligned}

We can also put the double cross product in wedge product form

\begin{aligned}\hat{\mathbf{n}} \times (\boldsymbol{\nabla} \times \mathbf{u})&=-I \hat{\mathbf{n}} \wedge (\boldsymbol{\nabla} \times \mathbf{u}) \\ &=-\frac{I}{2}\left(\hat{\mathbf{n}} (\boldsymbol{\nabla} \times \mathbf{u})- (\boldsymbol{\nabla} \times \mathbf{u}) \hat{\mathbf{n}}\right) \\ &=-\frac{I}{2}\left(-I \hat{\mathbf{n}} (\boldsymbol{\nabla} \wedge \mathbf{u})+ I (\boldsymbol{\nabla} \wedge \mathbf{u}) \hat{\mathbf{n}}\right) \\ &=-\frac{I^2}{2}\left(- \hat{\mathbf{n}} (\boldsymbol{\nabla} \wedge \mathbf{u})+ (\boldsymbol{\nabla} \wedge \mathbf{u}) \hat{\mathbf{n}}\right) \\ &=(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}}\end{aligned}

Equivalently (and easier) we can just expand the dot product of the wedge and the vector using the relationship

\begin{aligned}\mathbf{a} \cdot (\mathbf{c} \wedge \mathbf{d} \wedge \mathbf{e} \wedge \cdots )=(\mathbf{a} \cdot \mathbf{c}) (\mathbf{d} \wedge \mathbf{e} \wedge \cdots ) - (\mathbf{a} \cdot \mathbf{d}) (\mathbf{c} \wedge \mathbf{e} \wedge \cdots ) +\end{aligned} \hspace{\stretch{1}}(2.5)

so we find

\begin{aligned}((\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{n}} + 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u})_i&=(\boldsymbol{\nabla}' (\mathbf{u}' \cdot \hat{\mathbf{n}})-(\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u}+ 2 (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla}) \mathbf{u})_i \\ &=\partial_i u_a n_a+n_a \partial_a u_i \\ &=\sigma_{ia} n_a.\end{aligned}

# Cylindrical strain tensor.

Let’s now compute the strain tensor (and implicitly the traction vector) in cylindrical coordinates.

Our gradient in cylindrical coordinates is the familiar

\begin{aligned}\boldsymbol{\nabla} = \hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \hat{\boldsymbol{\phi}} \frac{1}{{r }}\frac{\partial {}}{\partial {\phi}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}},\end{aligned} \hspace{\stretch{1}}(3.6)

and our cylindrical velocity is

\begin{aligned}\mathbf{u} = \hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z.\end{aligned} \hspace{\stretch{1}}(3.7)

Our curl is then

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{u}&=\left(\hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \hat{\boldsymbol{\phi}} \frac{1}{{r }}\frac{\partial {}}{\partial {\phi}} + \hat{\mathbf{z}} \frac{\partial {}}{\partial {z}}\right)\wedge\left(\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z\right) \\ &=\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}\left(\partial_r u_\phi -\frac{1}{{r}} \partial_\phi u_r\right)+\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right)+\frac{1}{{r}} \hat{\boldsymbol{\phi}} \wedge \left((\partial_\phi \hat{\mathbf{r}}) u_r+(\partial_\phi \hat{\boldsymbol{\phi}}) u_\phi\right)\end{aligned}

Since $\partial_\phi \hat{\mathbf{r}} = \hat{\boldsymbol{\theta}}$ and $\partial_\phi \hat{\boldsymbol{\phi}} = -\hat{\mathbf{r}}$, we have only one cross term and our curl is

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{u}=\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right).\end{aligned} \hspace{\stretch{1}}(3.8)

We can now move on to compute the directional derivatives and complete the strain calculation in cylindrical coordinates. Let’s consider this computation of the stress for normals in each direction in term.

## With $\hat{\mathbf{n}} = \hat{\mathbf{r}}$.

Our directional derivative component for a $\hat{\mathbf{r}}$ normal direction doesn’t have any cross terms

\begin{aligned}2 (\hat{\mathbf{r}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=2 \partial_r\left(\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z\right) \\ &=2\left(\hat{\mathbf{r}} \partial_r u_r + \hat{\boldsymbol{\phi}} \partial_r u_\phi + \hat{\mathbf{z}} \partial_r u_z\right).\end{aligned}

Projecting our curl bivector onto the $\hat{\mathbf{r}}$ direction we have

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{r}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\mathbf{r}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}) \cdot \hat{\mathbf{r}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+(\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}) \cdot \hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=-\hat{\boldsymbol{\phi}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+\hat{\mathbf{z}}\left(\partial_z u_r - \partial_r u_z\right).\end{aligned}

Putting things together we have

\begin{aligned}2 \mathbf{e}_{\hat{\mathbf{r}}}&=2\left(\hat{\mathbf{r}} \partial_r u_r + \hat{\boldsymbol{\phi}} \partial_r u_\phi + \hat{\mathbf{z}} \partial_r u_z\right)-\hat{\boldsymbol{\phi}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+\hat{\mathbf{z}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=\hat{\mathbf{r}}\left(2 \partial_r u_r\right)+\hat{\boldsymbol{\phi}}\left(2 \partial_r u_\phi-\partial_r u_\phi+\frac{1}{{r}} \partial_\phi u_r- \frac{u_\phi}{r}\right)+\hat{\mathbf{z}}\left(2 \partial_r u_z+\partial_z u_r - \partial_r u_z\right).\end{aligned}

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\mathbf{r}}} = - p \hat{\mathbf{r}} + 2 \mu e_{\hat{\mathbf{r}}},\end{aligned} \hspace{\stretch{1}}(3.9)

we can now read off our components by taking dot products to yield

\begin{subequations}

\begin{aligned}\sigma_{rr}=-p + 2 \mu \frac{\partial {u_r}}{\partial {r}}\end{aligned} \hspace{\stretch{1}}(3.10a)

\begin{aligned}\sigma_{r \phi}=\mu \left( \frac{\partial {u_\phi}}{\partial {r}}+\frac{1}{{r}} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}\right)\end{aligned} \hspace{\stretch{1}}(3.10b)

\begin{aligned}\sigma_{r z}=\mu \left( \frac{\partial {u_z}}{\partial {r}}+\frac{\partial {u_r}}{\partial {z}}\right).\end{aligned} \hspace{\stretch{1}}(3.10c)

\end{subequations}

## With $\hat{\mathbf{n}} = \hat{\boldsymbol{\phi}}$.

Our directional derivative component for a $\hat{\boldsymbol{\phi}}$ normal direction will have some cross terms since both $\hat{\mathbf{r}}$ and $\hat{\boldsymbol{\phi}}$ are functions of $\phi$

\begin{aligned}2 (\hat{\boldsymbol{\phi}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=\frac{2}{r}\partial_\phi\left(\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z\right) \\ &=\frac{2}{r}\left(\hat{\mathbf{r}} \partial_\phi u_r + \hat{\boldsymbol{\phi}} \partial_\phi u_\phi + \hat{\mathbf{z}} \partial_\phi u_z+(\partial_\phi \hat{\mathbf{r}}) u_r + (\partial_\phi \hat{\boldsymbol{\phi}}) u_\phi\right) \\ &=\frac{2}{r}\left(\hat{\mathbf{r}} (\partial_\phi u_r - u_\phi) + \hat{\boldsymbol{\phi}} (\partial_\phi u_\phi + u_r )+ \hat{\mathbf{z}} \partial_\phi u_z\right) \\ \end{aligned}

Projecting our curl bivector onto the $\hat{\boldsymbol{\phi}}$ direction we have

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\phi}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\boldsymbol{\phi}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}) \cdot \hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+(\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}) \cdot \hat{\boldsymbol{\phi}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=\hat{\mathbf{r}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)-\hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)\end{aligned}

Putting things together we have

\begin{aligned}2 \mathbf{e}_{\hat{\boldsymbol{\phi}}}&=\frac{2}{r}\left(\hat{\mathbf{r}} (\partial_\phi u_r - u_\phi) + \hat{\boldsymbol{\phi}} (\partial_\phi u_\phi + u_r )+ \hat{\mathbf{z}} \partial_\phi u_z\right)+\hat{\mathbf{r}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)-\hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right) \\ &=\hat{\mathbf{r}}\left(\frac{1}{r}\partial_\phi u_r-\frac{u_\phi}{r}+\partial_r u_\phi\right)+\frac{2}{r} \hat{\boldsymbol{\phi}}\left(\partial_\phi u_\phi + u_r\right)+\hat{\mathbf{z}}\left(\frac{1}{r} \partial_\phi u_z + \partial_z u_\phi\right).\end{aligned}

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\boldsymbol{\phi}}} = - p \hat{\boldsymbol{\phi}} + 2 \mu e_{\hat{\boldsymbol{\phi}}},\end{aligned} \hspace{\stretch{1}}(3.11)

we can now read off our components by taking dot products to yield

\begin{subequations}

\begin{aligned}\sigma_{\phi \phi}=-p + 2 \mu \left(\frac{1}{{r}}\frac{\partial {u_\phi}}{\partial {\phi}} + \frac{u_r}{r}\right)\end{aligned} \hspace{\stretch{1}}(3.12a)

\begin{aligned}\sigma_{\phi z}=\mu \left(\frac{1}{r} \frac{\partial {u_z}}{\partial {\phi}} + \frac{\partial {u_\phi}}{\partial {z}}\right)\end{aligned} \hspace{\stretch{1}}(3.12b)

\begin{aligned}\sigma_{\phi r}=\mu \left(\frac{1}{r}\frac{\partial {u_r}}{\partial {\phi}}-\frac{u_\phi}{r}+\frac{\partial {u_\phi}}{\partial {r}}\right).\end{aligned} \hspace{\stretch{1}}(3.12c)

\end{subequations}

## With $\hat{\mathbf{n}} = \hat{\mathbf{z}}$.

Like the $\hat{\mathbf{r}}$ normal direction, our directional derivative component for a $\hat{\mathbf{z}}$ normal direction will not have any cross terms

\begin{aligned}2 (\hat{\mathbf{z}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=\partial_z\left(\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\phi}} u_\phi + \hat{\mathbf{z}} u_z\right) \\ &=\hat{\mathbf{r}} \partial_z u_r + \hat{\boldsymbol{\phi}} \partial_z u_\phi + \hat{\mathbf{z}} \partial_z u_z\end{aligned}

Projecting our curl bivector onto the $\hat{\mathbf{z}}$ direction we have

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\phi}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\mathbf{z}}\left(\partial_r u_\phi-\frac{1}{{r}} \partial_\phi u_r+ \frac{u_\phi}{r}\right)+(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{z}}) \cdot \hat{\mathbf{z}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+(\hat{\mathbf{z}} \wedge \hat{\mathbf{r}}) \cdot \hat{\mathbf{z}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=\hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)-\hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right)\end{aligned}

Putting things together we have

\begin{aligned}2 \mathbf{e}_{\hat{\mathbf{z}}}&=2 \hat{\mathbf{r}} \partial_z u_r + 2 \hat{\boldsymbol{\phi}} \partial_z u_\phi + 2 \hat{\mathbf{z}} \partial_z u_z+\hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)-\hat{\mathbf{r}}\left(\partial_z u_r - \partial_r u_z\right) \\ &=\hat{\mathbf{r}}\left(2 \partial_z u_r -\partial_z u_r + \partial_r u_z\right)+\hat{\boldsymbol{\phi}}\left(2 \partial_z u_\phi +\frac{1}{{r}} \partial_\phi u_z- \partial_z u_\phi\right)+\hat{\mathbf{z}}\left(2 \partial_z u_z\right) \\ &=\hat{\mathbf{r}}\left(\partial_z u_r + \partial_r u_z\right)+\hat{\boldsymbol{\phi}}\left(\partial_z u_\phi +\frac{1}{{r}} \partial_\phi u_z\right)+\hat{\mathbf{z}}\left(2 \partial_z u_z\right).\end{aligned}

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\mathbf{z}}} = - p \hat{\mathbf{z}} + 2 \mu e_{\hat{\mathbf{z}}},\end{aligned} \hspace{\stretch{1}}(3.13)

we can now read off our components by taking dot products to yield

\begin{subequations}

\begin{aligned}\sigma_{z z}=-p + 2 \mu \frac{\partial {u_z}}{\partial {z}}\end{aligned} \hspace{\stretch{1}}(3.14a)

\begin{aligned}\sigma_{z r}=\mu \left(\frac{\partial {u_r}}{\partial {z}}+ \frac{\partial {u_z}}{\partial {r}}\right)\end{aligned} \hspace{\stretch{1}}(3.14b)

\begin{aligned}\sigma_{z \phi}=\mu \left(\frac{\partial {u_\phi}}{\partial {z}}+\frac{1}{{r}} \frac{\partial {u_z}}{\partial {\phi}}\right).\end{aligned} \hspace{\stretch{1}}(3.14c)

\end{subequations}

## Summary.

\begin{subequations}

\begin{aligned}\sigma_{rr}=-p + 2 \mu \frac{\partial {u_r}}{\partial {r}}\end{aligned} \hspace{\stretch{1}}(3.15a)

\begin{aligned}\sigma_{\phi \phi}=-p + 2 \mu \left(\frac{1}{{r}}\frac{\partial {u_\phi}}{\partial {\phi}} + \frac{u_r}{r}\right)\end{aligned} \hspace{\stretch{1}}(3.15b)

\begin{aligned}\sigma_{z z}=-p + 2 \mu \frac{\partial {u_z}}{\partial {z}}\end{aligned} \hspace{\stretch{1}}(3.15c)

\begin{aligned}\sigma_{r \phi}=\mu \left( \frac{\partial {u_\phi}}{\partial {r}}+\frac{1}{{r}} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}\right)\end{aligned} \hspace{\stretch{1}}(3.15d)

\begin{aligned}\sigma_{\phi z}=\mu \left(\frac{1}{r} \frac{\partial {u_z}}{\partial {\phi}} + \frac{\partial {u_\phi}}{\partial {z}}\right)\end{aligned} \hspace{\stretch{1}}(3.15e)

\begin{aligned}\sigma_{z r}=\mu \left(\frac{\partial {u_r}}{\partial {z}}+ \frac{\partial {u_z}}{\partial {r}}\right)\end{aligned} \hspace{\stretch{1}}(3.15f)

\end{subequations}

# Spherical strain tensor.

Having done a first order cylindrical derivation of the strain tensor, let’s also do the spherical case for completeness. Would this have much utility in fluids? Perhaps for flow over a spherical barrier?

We need the gradient in spherical coordinates. Recall that our spherical coordinate velocity was

\begin{aligned}\frac{d\mathbf{r}}{dt} = \hat{\mathbf{r}} \dot{r} + \hat{\boldsymbol{\theta}} (r \dot{\theta}) + \hat{\boldsymbol{\phi}} ( r \sin\theta \dot{\phi} ),\end{aligned} \hspace{\stretch{1}}(4.16)

and our gradient mirrors this structure

\begin{aligned}\boldsymbol{\nabla} = \hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \hat{\boldsymbol{\theta}} \frac{1}{{r }}\frac{\partial {}}{\partial {\theta}} + \hat{\boldsymbol{\phi}} \frac{1}{{r \sin\theta}} \frac{\partial {}}{\partial {\phi}}.\end{aligned} \hspace{\stretch{1}}(4.17)

We also previously calculated \inbookref{phy454:continuumL2}{eqn:continuumL2:1010} the unit vector differentials

\begin{subequations}

\begin{aligned}d\hat{\mathbf{r}} = \hat{\boldsymbol{\phi}} \sin\theta d\phi + \hat{\boldsymbol{\theta}} d\theta\end{aligned} \hspace{\stretch{1}}(4.18a)

\begin{aligned}d\hat{\boldsymbol{\theta}} = \hat{\boldsymbol{\phi}} \cos\theta d\phi - \hat{\mathbf{r}} d\theta\end{aligned} \hspace{\stretch{1}}(4.18b)

\begin{aligned}d\hat{\boldsymbol{\phi}} = -(\hat{\mathbf{r}} \sin\theta + \hat{\boldsymbol{\theta}} \cos\theta) d\phi,\end{aligned} \hspace{\stretch{1}}(4.18c)

\end{subequations}

and can use those to read off the partials of all the unit vectors

\begin{aligned}\frac{\partial \hat{\mathbf{r}}}{\partial \{r,\theta, \phi\}} &= \{0, \hat{\boldsymbol{\theta}}, \hat{\boldsymbol{\phi}} \sin\theta \} \\ \frac{\partial \hat{\boldsymbol{\theta}}}{\partial \{r,\theta, \phi\}} &= \{0, -\hat{\mathbf{r}}, \hat{\boldsymbol{\phi}} \cos\theta \} \\ \frac{\partial \hat{\boldsymbol{\phi}}}{\partial \{r,\theta, \phi\}} &= \{0, 0, -\hat{\mathbf{r}} \sin\theta -\hat{\boldsymbol{\theta}} \cos\theta \}.\end{aligned} \hspace{\stretch{1}}(4.19)

Finally, our velocity in spherical coordinates is just

\begin{aligned}\mathbf{u} = \hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi,\end{aligned} \hspace{\stretch{1}}(4.22)

from which we can now compute the curl, and the directional derivative. Starting with the curl we have

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{u}&=\left( \hat{\mathbf{r}} \frac{\partial {}}{\partial {r}} + \hat{\boldsymbol{\theta}} \frac{1}{{r }}\frac{\partial {}}{\partial {\theta}} + \hat{\boldsymbol{\phi}} \frac{1}{{r \sin\theta}} \frac{\partial {}}{\partial {\phi}} \right) \wedge\left( \hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi \right) \\ &=\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r\right)\\ & +\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta\right)\\ & +\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi\right)\\ & +\frac{1}{{r}} \hat{\boldsymbol{\theta}} \wedge \left(u_\theta \underbrace{\partial_\theta \hat{\boldsymbol{\theta}}}_{-\hat{\mathbf{r}}}+u_\phi \underbrace{\partial_\theta \hat{\boldsymbol{\phi}}}_{0}\right)\\ & +\frac{1}{{r \sin\theta}} \hat{\boldsymbol{\phi}} \wedge \left(u_\theta \underbrace{\partial_\phi \hat{\boldsymbol{\theta}}}_{\hat{\boldsymbol{\phi}} \cos\theta}+u_\phi \underbrace{\partial_\phi \hat{\boldsymbol{\phi}}}_{-\hat{\mathbf{r}} \sin\theta - \hat{\boldsymbol{\theta}} \cos\theta}\right).\end{aligned}

So we have

\begin{aligned}\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{u}&=\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ & +\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.23)

## With $\hat{\mathbf{n}} = \hat{\mathbf{r}}$.

The directional derivative portion of our strain is

\begin{aligned}2 (\hat{\mathbf{r}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=2 \partial_r (\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi ) \\ &=2 (\hat{\mathbf{r}} \partial_r u_r + \hat{\boldsymbol{\theta}} \partial_r u_\theta + \hat{\boldsymbol{\phi}} \partial_r u_\phi ).\end{aligned}

The other portion of our strain tensor is

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{r}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}) \cdot \hat{\mathbf{r}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +(\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\mathbf{r}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ & +(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}) \cdot \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right) \\ &=-\hat{\boldsymbol{\theta}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +\hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}

Putting these together we find

\begin{aligned}2 {\mathbf{e}}_{\hat{\mathbf{r}}}&=2 (\hat{\mathbf{r}} \cdot \boldsymbol{\nabla})\mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\mathbf{r}} \\ &=2 (\hat{\mathbf{r}} \partial_r u_r + \hat{\boldsymbol{\theta}} \partial_r u_\theta + \hat{\boldsymbol{\phi}} \partial_r u_\phi )-\hat{\boldsymbol{\theta}}\left(\partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)+\hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right) \\ &=\hat{\mathbf{r}}\left(2 \partial_r u_r\right)+\hat{\boldsymbol{\theta}}\left(2 \partial_r u_\theta-\partial_r u_\theta + \frac{1}{{r}} \partial_\theta u_r - \frac{u_\theta}{r}\right)+\hat{\boldsymbol{\phi}}\left(2 \partial_r u_\phi+ \frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}

Which gives

\begin{aligned}2 {\mathbf{e}}_{\hat{\mathbf{r}}}=\hat{\mathbf{r}}\left(2 \partial_r u_r\right)+\hat{\boldsymbol{\theta}}\left(\partial_r u_\theta+ \frac{1}{{r}} \partial_\theta u_r - \frac{u_\theta}{r}\right)+\hat{\boldsymbol{\phi}}\left(\partial_r u_\phi+ \frac{1}{{r \sin\theta}} \partial_\phi u_r- \frac{u_\phi}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.24)

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\mathbf{r}}} = - p \hat{\mathbf{r}} + 2 \mu e_{\hat{\mathbf{r}}},\end{aligned} \hspace{\stretch{1}}(4.25)

we can now read off our components by taking dot products

\begin{subequations}

\begin{aligned}\sigma_{rr}=-p + 2 \mu \frac{\partial {u_r}}{\partial {r}}\end{aligned} \hspace{\stretch{1}}(4.26a)

\begin{aligned}\sigma_{r \theta}=\mu \left(\frac{\partial {u_\theta}}{\partial {r}}+ \frac{1}{{r}} \frac{\partial {u_r}}{\partial {\theta}} - \frac{u_\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.26b)

\begin{aligned}\sigma_{r \phi}=\mu \left(\frac{\partial {u_\phi}}{\partial {r}}+ \frac{1}{{r \sin\theta}} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}\right).\end{aligned} \hspace{\stretch{1}}(4.26c)

\end{subequations}

This is consistent with (15.20) from [3] (after adjusting for minor notational differences).

## With $\hat{\mathbf{n}} = \hat{\boldsymbol{\theta}}$.

Now let’s do the $\hat{\boldsymbol{\theta}}$ direction. The directional derivative portion of our strain will be a bit more work to compute because we have $\theta$ variation of the unit vectors

\begin{aligned}(\hat{\boldsymbol{\theta}} \cdot \boldsymbol{\nabla}) \mathbf{u} &= \frac{1}{r} \partial_\theta (\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi ) \\ &= \frac{1}{r} \left( \hat{\mathbf{r}} \partial_\theta u_r + \hat{\boldsymbol{\theta}} \partial_\theta u_\theta + \hat{\boldsymbol{\phi}} \partial_\theta u_\phi \right)+\frac{1}{r} \left( (\partial_\theta \hat{\mathbf{r}}) u_r + (\partial_\theta \hat{\boldsymbol{\theta}}) u_\theta + (\partial_\theta \hat{\boldsymbol{\phi}}) u_\phi \right) \\ &= \frac{1}{r}\left(\hat{\mathbf{r}} \partial_\theta u_r + \hat{\boldsymbol{\theta}} \partial_\theta u_\theta + \hat{\boldsymbol{\phi}} \partial_\theta u_\phi \right)+\frac{1}{r} \left( \hat{\boldsymbol{\theta}} u_r - \hat{\mathbf{r}} u_\theta \right).\end{aligned}

So we have

\begin{aligned}2 (\hat{\boldsymbol{\theta}} \cdot \boldsymbol{\nabla}) \mathbf{u}=\frac{2}{r} \hat{\mathbf{r}} (\partial_\theta u_r- u_\theta)+ \frac{2}{r} \hat{\boldsymbol{\theta}} (\partial_\theta u_\theta+ u_r) + \frac{2}{r} \hat{\boldsymbol{\phi}} \partial_\theta u_\phi,\end{aligned} \hspace{\stretch{1}}(4.27)

and can move on to projecting our curl bivector onto the $\hat{\boldsymbol{\theta}}$ direction. That portion of our strain tensor is

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\theta}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}) \cdot \hat{\boldsymbol{\theta}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +(\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\boldsymbol{\theta}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ & +(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}) \cdot \hat{\boldsymbol{\theta}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right) \\ &=\hat{\mathbf{r}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)-\hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right).\end{aligned}

Putting these together we find

\begin{aligned}2 {\mathbf{e}}_{\hat{\boldsymbol{\theta}}}&=2 (\hat{\boldsymbol{\theta}} \cdot \boldsymbol{\nabla})\mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\theta}} \\ &= \frac{2}{r} \hat{\mathbf{r}} (\partial_\theta u_r - u_\theta )+ \frac{2}{r} \hat{\boldsymbol{\theta}} (\partial_\theta u_\theta + u_r )+ \frac{2}{r} \hat{\boldsymbol{\phi}} \partial_\theta u_\phi \\ &+\hat{\mathbf{r}}\left(\partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)-\hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta + \frac{u_\phi \cot\theta}{r}\right).\end{aligned}

Which gives

\begin{aligned}2 {\mathbf{e}}_{\hat{\boldsymbol{\theta}}}=\hat{\mathbf{r}} \left( \frac{1}{r} \partial_\theta u_r + \partial_r u_\theta- \frac{u_\theta}{r}\right)+\hat{\boldsymbol{\theta}} \left( \frac{2}{r} \partial_\theta u_\theta+ \frac{2}{r} u_r\right)+\hat{\boldsymbol{\phi}} \left(\frac{1}{r} \partial_\theta u_\phi+ \frac{1}{{r \sin\theta}} \partial_\phi u_\theta- \frac{u_\phi \cot\theta}{r}\right).\end{aligned} \hspace{\stretch{1}}(4.28)

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\boldsymbol{\theta}}} = - p \hat{\boldsymbol{\theta}} + 2 \mu e_{\hat{\boldsymbol{\theta}}},\end{aligned} \hspace{\stretch{1}}(4.29)

we can now read off our components by taking dot products

\begin{subequations}

\begin{aligned}\sigma_{\theta \theta}=-p+\mu \left( \frac{2}{r} \frac{\partial {u_\theta}}{\partial {\theta}}+ \frac{2}{r} u_r\right)\end{aligned} \hspace{\stretch{1}}(4.30a)

\begin{aligned}\sigma_{\theta \phi}=\mu \left(\frac{1}{r} \frac{\partial {u_\phi}}{\partial {\theta}}+ \frac{1}{{r \sin\theta}} \frac{\partial {u_\theta}}{\partial {\phi}}- \frac{u_\phi \cot\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.30b)

\begin{aligned}\sigma_{\theta r}= \mu \left(\frac{1}{r} \frac{\partial {u_r}}{\partial {\theta}} + \frac{\partial {u_\theta}}{\partial {r}}- \frac{u_\theta}{r}\right).\end{aligned} \hspace{\stretch{1}}(4.30c)

\end{subequations}

This again is consistent with (15.20) from [3].

## With $\hat{\mathbf{n}} = \hat{\boldsymbol{\phi}}$.

Finally, let’s do the $\hat{\boldsymbol{\phi}}$ direction. This directional derivative portion of our strain will also be a bit more work to compute because we have $\hat{\boldsymbol{\phi}}$ variation of the unit vectors

\begin{aligned}(\hat{\boldsymbol{\phi}} \cdot \boldsymbol{\nabla}) \mathbf{u}&=\frac{1}{r \sin\theta} \partial_\phi (\hat{\mathbf{r}} u_r + \hat{\boldsymbol{\theta}} u_\theta + \hat{\boldsymbol{\phi}} u_\phi) \\ &=\frac{1}{r \sin\theta}(\hat{\mathbf{r}} \partial_\phi u_r+\hat{\boldsymbol{\theta}} \partial_\phi u_\theta+\hat{\boldsymbol{\phi}} \partial_\phi u_\phi+(\partial_\phi \hat{\mathbf{r}} )u_r+(\partial_\phi \hat{\boldsymbol{\theta}} )u_\theta+(\partial_\phi \hat{\boldsymbol{\phi}} )u_\phi) \\ &=\frac{1}{r \sin\theta}(\hat{\mathbf{r}} \partial_\phi u_r+\hat{\boldsymbol{\theta}} \partial_\phi u_\theta+\hat{\boldsymbol{\phi}} \partial_\phi u_\phi+\hat{\boldsymbol{\phi}} \sin\thetau_r+\hat{\boldsymbol{\phi}} \cos\thetau_\theta-(\hat{\mathbf{r}} \sin\theta+ \hat{\boldsymbol{\theta}} \cos\theta)u_\phi)\end{aligned}

So we have

\begin{aligned}2 (\hat{\boldsymbol{\phi}} \cdot \boldsymbol{\nabla}) \mathbf{u}=2 \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \frac{u_\phi}{r}\right)+2 \hat{\boldsymbol{\theta}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_\theta-\frac{1}{{r}} \cot\theta u_\phi\right)+2 \hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_\phi+ \frac{1}{{r}} u_r+ \frac{1}{{r}} \cot\theta u_\theta\right),\end{aligned} \hspace{\stretch{1}}(4.31)

and can move on to projecting our curl bivector onto the $\hat{\boldsymbol{\phi}}$ direction. That portion of our strain tensor is

\begin{aligned}(\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\phi}}&=(\hat{\mathbf{r}} \wedge \hat{\boldsymbol{\theta}}) \cdot \hat{\boldsymbol{\phi}}\left( \partial_r u_\theta - \frac{1}{{r}} \partial_\theta u_r + \frac{u_\theta}{r}\right)\\ & +(\hat{\boldsymbol{\theta}} \wedge \hat{\boldsymbol{\phi}}) \cdot \hat{\boldsymbol{\phi}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ & +(\hat{\boldsymbol{\phi}} \wedge \hat{\mathbf{r}}) \cdot \hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right) \\ &=\hat{\boldsymbol{\theta}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)\\ &-\hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}

Putting these together we find

\begin{aligned}2 {\mathbf{e}}_{\hat{\boldsymbol{\theta}}}&=2 (\hat{\boldsymbol{\phi}} \cdot \boldsymbol{\nabla})\mathbf{u} + (\boldsymbol{\nabla} \wedge \mathbf{u}) \cdot \hat{\boldsymbol{\phi}} \\ &=2 \hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \frac{u_\phi}{r}\right)+2 \hat{\boldsymbol{\theta}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_\theta-\frac{1}{{r}} \cot\theta u_\phi\right)+2 \hat{\boldsymbol{\phi}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_\phi+ \frac{1}{{r}} u_r+ \frac{1}{{r}} \cot\theta u_\theta\right) \\ &+\hat{\boldsymbol{\theta}}\left(\frac{1}{{r}} \partial_\theta u_\phi - \frac{1}{{r \sin\theta}} \partial_\phi u_\theta+ \frac{u_\phi \cot\theta}{r}\right)-\hat{\mathbf{r}}\left(\frac{1}{{r \sin\theta}} \partial_\phi u_r - \partial_r u_\phi- \frac{u_\phi}{r}\right).\end{aligned}

Which gives

\begin{aligned}2 {\mathbf{e}}_{\hat{\boldsymbol{\phi}}}=\hat{\mathbf{r}} \left( \frac{ \partial_\phi u_r }{r \sin\theta}- \frac{u_\phi}{r}+ \partial_r u_\phi\right)+\hat{\boldsymbol{\theta}} \left(\frac{\partial_\phi u_\theta}{r \sin\theta}- \frac{u_\phi \cot\theta}{r}+\frac{\partial_\theta u_\phi}{r}\right)+2 \hat{\boldsymbol{\phi}} \left(\frac{\partial_\phi u_\phi}{r \sin\theta}+ \frac{u_r}{r}+ \frac{\cot\theta u_\theta}{r}\right).\end{aligned} \hspace{\stretch{1}}(4.32)

For our stress tensor

\begin{aligned}\boldsymbol{\sigma}_{\hat{\boldsymbol{\phi}}} = - p \hat{\boldsymbol{\phi}} + 2 \mu e_{\hat{\boldsymbol{\phi}}},\end{aligned} \hspace{\stretch{1}}(4.33)

we can now read off our components by taking dot products

\begin{subequations}

\begin{aligned}\sigma_{\phi \phi}=-p+2 \mu \left(\frac{1}{{r \sin\theta}} \frac{\partial {u_\phi}}{\partial {\phi}}+ \frac{u_r}{r}+ \frac{\cot\theta u_\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.34a)

\begin{aligned}\sigma_{\phi r}=\mu \left( \frac{1}{r \sin\theta} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}+ \frac{\partial {u_\phi}}{\partial {r}}\right)\end{aligned} \hspace{\stretch{1}}(4.34b)

\begin{aligned}\sigma_{\phi \theta}= \mu \left(\frac{1}{r \sin\theta} \frac{\partial {u_\theta}}{\partial {\phi}}- \frac{u_\phi \cot\theta}{r}+\frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\theta}}\right).\end{aligned} \hspace{\stretch{1}}(4.34c)

\end{subequations}

This again is consistent with (15.20) from [3].

## Summary

\begin{subequations}

\begin{aligned}\sigma_{rr}=-p + 2 \mu \frac{\partial {u_r}}{\partial {r}}\end{aligned} \hspace{\stretch{1}}(4.35a)

\begin{aligned}\sigma_{\theta \theta}=-p+2 \mu \left( \frac{1}{r} \frac{\partial {u_\theta}}{\partial {\theta}}+ \frac{ u_r }{r}\right)\end{aligned} \hspace{\stretch{1}}(4.35b)

\begin{aligned}\sigma_{\phi \phi}=-p+2 \mu \left(\frac{1}{{r \sin\theta}} \frac{\partial {u_\phi}}{\partial {\phi}}+ \frac{u_r}{r}+ \frac{\cot\theta u_\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.35c)

\begin{aligned}\sigma_{r \theta}=\mu \left(\frac{\partial {u_\theta}}{\partial {r}}+ \frac{1}{{r}} \frac{\partial {u_r}}{\partial {\theta}} - \frac{u_\theta}{r}\right)\end{aligned} \hspace{\stretch{1}}(4.35d)

\begin{aligned}\sigma_{\theta \phi}= \mu \left(\frac{1}{r \sin\theta} \frac{\partial {u_\theta}}{\partial {\phi}}- \frac{u_\phi \cot\theta}{r}+\frac{1}{{r}} \frac{\partial {u_\phi}}{\partial {\theta}}\right).\end{aligned} \hspace{\stretch{1}}(4.35e)

\begin{aligned}\sigma_{\phi r}=\mu \left( \frac{1}{r \sin\theta} \frac{\partial {u_r}}{\partial {\phi}}- \frac{u_\phi}{r}+ \frac{\partial {u_\phi}}{\partial {r}}\right)\end{aligned} \hspace{\stretch{1}}(4.35f)

\end{subequations}

# References

[1] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.

[2] Peeter Joot. Continuum mechanics., chapter {Introduction and strain tensor.} http://sites.google.com/site/peeterjoot2/math2012/phy454.pdf.

[3] L.D. Landau and E.M. Lifshitz. A Course in Theoretical Physics-Fluid Mechanics. Pergamon Press Ltd., 1987.