## Progressing towards coordinate free form of the Euler-Lagrange equations for Maxwell’s equation

This is the 6th part of a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a multivector Lagrangian representation.

[Click here for a PDF version of this series of posts, up to and including this one.]  The first, second, third, fourth, and fifth parts are also available here on this blog.

We managed to find Maxwell’s equation in it’s STA form by variation of a multivector Lagrangian, with respect to a four-vector field (the potential). That approach differed from the usual variation with respect to the coordinates of that four-vector, or the use of the Euler-Lagrange equations with respect to those coordinates.

### Euler-Lagrange equations.

Having done so, an immediate question is whether we can express the Euler-Lagrange equations with respect to the four-potential in it’s entirety, instead of the coordinates of that vector. I have some intuition about how to completely avoid that use of coordinates, but first we can get part way there.

Consider a general Lagrangian, dependent on a field $$A$$ and all it’s derivatives $$\partial_\mu A$$
\label{eqn:fsquared:1180}
\LL = \LL( A, \partial_\mu A ).

The variational principle requires
\label{eqn:fsquared:1200}
0 = \delta S = \int d^4 x \delta \LL( A, \partial_\mu A ).

That variation can be expressed as a limiting parametric operation as follows
\label{eqn:fsquared:1220}
\delta S
= \int d^4 x
\lr{
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )
+
\sum_\mu
\lim_{t \rightarrow 0} \ddt{} \LL( \partial_\mu A + t \delta \partial_\mu A )
}

We eventually want a coordinate free expression for the variation, but we’ll use them to get there. We can expand the first derivative by chain rule as
\label{eqn:fsquared:1240}
\begin{aligned}
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )
&=
\lim_{t \rightarrow 0} \PD{(A^\alpha + t \delta A^\alpha)}{\LL} \PD{t}{}(A^\alpha + t \delta A^\alpha) \\
&=
\PD{A^\alpha}{\LL} \delta A^\alpha.
\end{aligned}

This has the structure of a directional derivative $$A$$. In particular, let
\label{eqn:fsquared:1260}

so we have
\label{eqn:fsquared:1280}
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )

Similarly,
\label{eqn:fsquared:1300}
\lim_{t \rightarrow 0} \ddt{} \LL( \partial_\mu A + t \delta \partial_\mu A )
=
\PD{(\partial_\mu A^\alpha)}{\LL} \delta \partial_\mu A^\alpha,

so we can define a gradient with respect to each of the derivatives of $$A$$ as
\label{eqn:fsquared:1320}
\grad_{\partial_\mu A} = \gamma^\alpha \PD{(\partial_\mu A^\alpha)}{}.

Our variation can now be expressed in a somewhat coordinate free form
\label{eqn:fsquared:1340}
\delta S = \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\delta \partial_\mu A} \cdot \grad_{\partial_\mu A} } \LL
}.

We now sum implicitly over pairs of indexes $$\mu$$ (i.e. we are treating $$\grad_{\partial_\mu A}$$ as an upper index entity). We can now proceed with our chain rule expansion
\label{eqn:fsquared:1360}
\begin{aligned}
\delta S
&= \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\delta \partial_\mu A} \cdot \grad_{\partial_\mu A} } \LL
} \\
&= \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\partial_\mu \delta A} \cdot \grad_{\partial_\mu A} } \LL
} \\
&= \int d^4 x \lr{
+ \partial_\mu \lr{ \lr{ \delta A \cdot \grad_{\partial_\mu A} } \LL }
– \lr{\PD{x^\mu}{} \delta A \cdot \grad_{\partial_\mu A} \LL}_{\delta A}
}.
\end{aligned}

As usual, we kill off the boundary term, by insisting that $$\delta A = 0$$ on the boundary, leaving us with a four-vector form of the field Euler-Lagrange equations
\label{eqn:fsquared:1380}
\lr{\delta A \cdot \grad_A} \LL = \lr{\PD{x^\mu}{} \delta A \cdot \grad_{\partial_\mu A} \LL}_{\delta A},

where the RHS derivatives are taken with $$\delta A$$ held fixed. We seek solutions of this equation that hold for all variations $$\delta A$$.

### Application to the Maxwell Lagrangian.

For the Maxwell application we need a few helper calculations. The first, given a multivector $$B$$, is
\label{eqn:fsquared:1400}
\begin{aligned}
\lr{ \delta A \cdot \grad_A } A B
&=
\delta A^\alpha \PD{A^\alpha}{} \gamma_\beta A^\beta B \\
&=
\delta A^\alpha \gamma_\alpha B \\
&=
\lr{ \delta A } B.
\end{aligned}

Now let’s compute, for multivector $$B$$
\label{eqn:fsquared:1420}
\begin{aligned}
\lr{ \delta A \cdot \grad_{\partial_\mu A} } B F
&=
\delta A^\alpha \PD{(\partial_\mu A^\alpha)} B \lr{ \gamma^\beta \wedge \partial_\beta \lr{ \gamma_\pi A^\pi } } \\
&=
\delta A^\alpha B \lr{ \gamma^\mu \wedge \gamma_\alpha } \\
&=
B \lr{ \gamma^\mu \wedge \delta A }.
\end{aligned}

Our Lagrangian is
\label{eqn:fsquared:1440}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4},

so
\label{eqn:fsquared:1460}
=
-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4},

and
\label{eqn:fsquared:1480}
\begin{aligned}
\lr{ \delta A \cdot \grad_{\partial_\mu A} } \inv{2} F^2
&=
\inv{2} \lr{ F \lr{ \gamma^\mu \wedge \delta A } + \lr{ \gamma^\mu \wedge \delta A } F } \\
&=
\lr{ \gamma^\mu \wedge \delta A } F
}{0,4} \\
&=
\lr{ \delta A \wedge \gamma^\mu } F
}{0,4} \\
&=
\delta A \gamma^\mu F

\lr{ \delta A \cdot \gamma^\mu } F
}{0,4} \\
&=
\delta A \gamma^\mu F
}{0,4}.
\end{aligned}

Taking derivatives (holding $$\delta A$$ fixed), we have
\label{eqn:fsquared:1500}
\begin{aligned}
-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4}
&=
\delta A \partial_\mu \gamma^\mu F
}{0,4} \\
&=
}{0,4}.
\end{aligned}

We’ve already seen that the solution can be expressed without grade selection as
\label{eqn:fsquared:1520}
\grad F = \lr{ J – I M },

which is Maxwell’s equation in it’s STA form. It’s not clear that this is really any less work, but it’s a step towards a coordinate free evaluation of the Maxwell Lagrangian (at least not having to use the coordinates $$A^\mu$$ as we have to do in the tensor formalism.)

## A couple more reciprocal frame examples.

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

This post logically follows both of the following:

The PDF linked above above contains all the content from this post plus (1.) above [to be edited later into a more logical sequence.]

## More examples.

Here are a few additional examples of reciprocal frame calculations.

## Problem: Unidirectional arbitrary functional dependence.

Let
\label{eqn:reciprocal:2540}
x = a f(u),

where $$a$$ is a constant vector and $$f(u)$$ is some arbitrary differentiable function with a non-zero derivative in the region of interest.

Here we have just a single tangent space direction (a line in spacetime) with tangent vector
\label{eqn:reciprocal:2400}
\Bx_u = a \PD{u}{f} = a f_u,

so we see that the tangent space vectors are just rescaled values of the direction vector $$a$$.
This is a simple enough parameterization that we can compute the reciprocal frame vector explicitly using the gradient. We expect that $$\Bx^u = 1/\Bx_u$$, and find
\label{eqn:reciprocal:2420}
\inv{a} \cdot x = f(u),

but for constant $$a$$, we know that $$\grad a \cdot x = a$$, so taking gradients of both sides we find
\label{eqn:reciprocal:2440}

so the reciprocal vector is
\label{eqn:reciprocal:2460}
\Bx^u = \grad u = \inv{a f_u},

as expected.

## Problem: Linear two variable parameterization.

Let $$x = a u + b v$$, where $$x \wedge a \wedge b = 0$$ represents spacetime plane (also the tangent space.) Find the curvilinear coordinates and their reciprocals.

The frame vectors are easy to compute, as they are just
\label{eqn:reciprocal:1960}
\begin{aligned}
\Bx_u &= \PD{u}{x} = a \\
\Bx_v &= \PD{v}{x} = b.
\end{aligned}

This is an example of a parametric equation that we can easily invert, as we have
\label{eqn:reciprocal:1980}
\begin{aligned}
x \wedge a &= – v \lr{ a \wedge b } \\
x \wedge b &= u \lr{ a \wedge b },
\end{aligned}

so
\label{eqn:reciprocal:2000}
\begin{aligned}
u
&= \inv{ a \wedge b } \cdot \lr{ x \wedge b } \\
&= \inv{ \lr{a \wedge b}^2 } \lr{ a \wedge b } \cdot \lr{ x \wedge b } \\
&=
\frac{
\lr{b \cdot x} \lr{ a \cdot b }

\lr{a \cdot x} \lr{ b \cdot b }
}{ \lr{a \wedge b}^2 }
\end{aligned}

\label{eqn:reciprocal:2020}
\begin{aligned}
v &= -\inv{ a \wedge b } \cdot \lr{ x \wedge a } \\
&= -\inv{ \lr{a \wedge b}^2 } \lr{ a \wedge b } \cdot \lr{ x \wedge a } \\
&=
-\frac{
\lr{b \cdot x} \lr{ a \cdot a }

\lr{a \cdot x} \lr{ a \cdot b }
}{ \lr{a \wedge b}^2 }
\end{aligned}

Recall that $$\grad \lr{ a \cdot x} = a$$, if $$a$$ is a constant, so our gradients are just
\label{eqn:reciprocal:2040}
\begin{aligned}
&=
\frac{
b \lr{ a \cdot b }

a
\lr{ b \cdot b }
}{ \lr{a \wedge b}^2 } \\
&=
b \cdot \inv{ a \wedge b },
\end{aligned}

and
\label{eqn:reciprocal:2060}
\begin{aligned}
&=
-\frac{
b \lr{ a \cdot a }

a \lr{ a \cdot b }
}{ \lr{a \wedge b}^2 } \\
&=
-a \cdot \inv{ a \wedge b }.
\end{aligned}

Expressed in terms of the frame vectors, this is just
\label{eqn:reciprocal:2080}
\begin{aligned}
\Bx^u &= \Bx_v \cdot \inv{ \Bx_u \wedge \Bx_v } \\
\Bx^v &= -\Bx_u \cdot \inv{ \Bx_u \wedge \Bx_v },
\end{aligned}

so we were able to show, for this special two parameter linear case, that the explicit evaluation of the gradients has the exact structure that we intuited that the reciprocals must have, provided they are constrained to the spacetime plane $$a \wedge b$$. It is interesting to observe how this structure falls out of the linear system solution so directly. Also note that these reciprocals are not defined at the origin of the $$(u,v)$$ parameter space.

## Problem: Quadratic two variable parameterization.

Now consider a variation of the previous problem, with $$x = a u^2 + b v^2$$. Find the curvilinear coordinates and their reciprocals.

\label{eqn:reciprocal:2100}
\begin{aligned}
\Bx_u &= \PD{u}{x} = 2 u a \\
\Bx_v &= \PD{v}{x} = 2 v b.
\end{aligned}

Our tangent space is still the $$a \wedge b$$ plane (as is the surface itself), but the spacing of the cells starts getting wider in proportion to $$u, v$$.
Utilizing the work from the previous problem, we have
\label{eqn:reciprocal:2120}
\begin{aligned}
b \cdot \inv{ a \wedge b } \\
-a \cdot \inv{ a \wedge b }.
\end{aligned}

A bit of rearrangement can show that this is equivalent to the reciprocal frame identities. This is a second demonstration that the gradient and the algebraic formulations for the reciprocals match, at least for these special cases of linear non-coupled parameterizations.

## Problem: Reciprocal frame for generalized cylindrical parameterization.

Let the vector parameterization be $$x(\rho,\theta) = \rho e^{-i\theta/2} x(\rho_0, \theta_0) e^{i \theta}$$, where $$i^2 = \pm 1$$ is a unit bivector ($$+1$$ for a boost, and $$-1$$ for a rotation), and where $$\theta, \rho$$ are scalars. Find the tangent space vectors and their reciprocals.

fig. 1. “Cylindrical” boost parameterization.

Note that this is cylindrical parameterization for the rotation case, and traces out hyperbolic regions for the boost case. The boost case is illustrated in fig. 1 where hyperbolas in the light cone are found for boosts of $$\gamma_0$$ with various values of $$\rho$$, and the spacelike hyperbolas are boosts of $$\gamma_1$$, again for various values of $$\rho$$.

The tangent space vectors are
\label{eqn:reciprocal:2480}
\Bx_\rho = \frac{x}{\rho},

and

\label{eqn:reciprocal:2500}
\begin{aligned}
\Bx_\theta
&= -\frac{i}{2} x + x \frac{i}{2} \\
&= x \cdot i.
\end{aligned}

Recall that $$x \cdot i$$ lies perpendicular to $$x$$ (in the plane $$i$$), as illustrated in fig. 2. This means that $$\Bx_\rho$$ and $$\Bx_\theta$$ are orthogonal, so we can find the reciprocal vectors by just inverting them
\label{eqn:reciprocal:2520}
\begin{aligned}
\Bx^\rho &= \frac{\rho}{x} \\
\Bx^\theta &= \frac{1}{x \cdot i}.
\end{aligned}

fig. 2. Projection and rejection geometry.

## Parameterization of a general linear transformation.

Given $$N$$ parameters $$u^0, u^1, \cdots u^{N-1}$$, a general linear transformation from the parameter space to the vector space has the form
\label{eqn:reciprocal:2160}
x =
{a^\alpha}_\beta \gamma_\alpha u^\beta,

where $$\beta \in [0, \cdots, N-1]$$ and $$\alpha \in [0,3]$$.
For such a general transformation, observe that the curvilinear basis vectors are
\label{eqn:reciprocal:2180}
\begin{aligned}
\Bx_\mu
&= \PD{u^\mu}{x} \\
&= \PD{u^\mu}{}
{a^\alpha}_\beta \gamma_\alpha u^\beta \\
&=
{a^\alpha}_\mu \gamma_\alpha.
\end{aligned}

We find an interpretation of $${a^\alpha}_\mu$$ by dotting $$\Bx_\mu$$ with the reciprocal frame vectors of the standard basis
\label{eqn:reciprocal:2200}
\begin{aligned}
\Bx_\mu \cdot \gamma^\nu
&=
{a^\alpha}_\mu \lr{ \gamma_\alpha \cdot \gamma^\nu } \\
&=
{a^\nu}_\mu,
\end{aligned}

so
\label{eqn:reciprocal:2220}
x = \Bx_\mu u^\mu.

We are able to reinterpret \ref{eqn:reciprocal:2160} as a contraction of the tangent space vectors with the parameters, scaling and summing these direction vectors to characterize all the points in the tangent plane.

## Theorem 1.1: Projecting onto the tangent space.

Let $$T$$ represent the tangent space. The projection of a vector onto the tangent space has the form
\label{eqn:reciprocal:2560}
\textrm{Proj}_{\textrm{T}} y = \lr{ y \cdot \Bx^\mu } \Bx_\mu = \lr{ y \cdot \Bx_\mu } \Bx^\mu.

### Start proof:

Let’s designate $$a$$ as the portion of the vector $$y$$ that lies outside of the tangent space
\label{eqn:reciprocal:2260}
y = y^\mu \Bx_\mu + a.

If we knew the coordinates $$y^\mu$$, we would have a recipe for the projection.
Algebraically, requiring that $$a$$ lies outside of the tangent space, is equivalent to stating $$a \cdot \Bx_\mu = a \cdot \Bx^\mu = 0$$. We use that fact, and then take dot products
\label{eqn:reciprocal:2280}
\begin{aligned}
y \cdot \Bx^\nu
&= \lr{ y^\mu \Bx_\mu + a } \cdot \Bx^\nu \\
&= y^\nu,
\end{aligned}

so
\label{eqn:reciprocal:2300}
y = \lr{ y \cdot \Bx^\mu } \Bx_\mu + a.

Similarly, the tangent space projection can be expressed as a linear combination of reciprocal basis elements
\label{eqn:reciprocal:2320}
y = y_\mu \Bx^\mu + a.

Dotting with $$\Bx_\mu$$, we have
\label{eqn:reciprocal:2340}
\begin{aligned}
y \cdot \Bx^\mu
&= \lr{ y_\alpha \Bx^\alpha + a } \cdot \Bx_\mu \\
&= y_\mu,
\end{aligned}

so
\label{eqn:reciprocal:2360}
y = \lr{ y \cdot \Bx^\mu } \Bx_\mu + a.

We find the two stated ways of computing the projection.

Observe that, for the special case that all of $$\setlr{ \Bx_\mu }$$ are orthogonal, the equivalence of these two projection methods follows directly, since
\label{eqn:reciprocal:2380}
\begin{aligned}
\lr{ y \cdot \Bx^\mu } \Bx_\mu
&=
\lr{ y \cdot \inv{\Bx_\mu} } \inv{\Bx^\mu} \\
&=
\lr{ y \cdot \frac{\Bx_\mu}{\lr{\Bx_\mu}^2 } } \frac{\Bx^\mu}{\lr{\Bx^\mu}^2} \\
&=
\lr{ y \cdot \Bx_\mu } \Bx^\mu.
\end{aligned}

## Curvilinear coordinates and gradient in spacetime, and reciprocal frames.

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

## Motivation.

I started pondering some aspects of spacetime integration theory, and found that there were some aspects of the concepts of reciprocal frames that were not clear to me. In the process of sorting those ideas out for myself, I wrote up the following notes.

In the notes below, I will introduce the many of the prerequisite ideas that are needed to express and apply the fundamental theorem of geometric calculus in a 4D relativistic context. The focus will be the Dirac’s algebra of special relativity, known as STA (Space Time Algebra) in geometric algebra parlance. If desired, it should be clear how to apply these ideas to lower or higher dimensional spaces, and to plain old Euclidean metrics.

### On notation.

In Euclidean space we use bold face reciprocal frame vectors $$\Bx^i \cdot \Bx_j = {\delta^i}_j$$, which nicely distinguishes them from the generalized coordinates $$x_i, x^j$$ associated with the basis or the reciprocal frame, that is
\label{eqn:reciprocalblog:640}
\Bx = x^i \Bx_i = x_j \Bx^j.

On the other hand, it is conventional to use non-bold face for both the four-vectors and their coordinates in STA, such as the following standard basis decomposition
\label{eqn:reciprocalblog:660}
x = x^\mu \gamma_\mu = x_\mu \gamma^\mu.

If we use non-bold face $$x^\mu, x_\nu$$ for the coordinates with respect to a specified frame, then we cannot also use non-bold face for the curvilinear basis vectors.

To resolve this notational ambiguity, I’ve chosen to use bold face $$\Bx^\mu, \Bx_\nu$$ symbols as the curvilinear basis elements in this relativistic context, as we do for Euclidean spaces.

## Definition 1.1: Standard Dirac basis.

The Dirac basis elements are $$\setlr{ \gamma_0, \gamma_1, \gamma_2, \gamma_3 }$$, satisfying
\label{eqn:reciprocalblog:1940}
\gamma_0^2 = 1 = -\gamma_k^2, \quad \forall k = 1,2,3,

and
\label{eqn:reciprocalblog:740}
\gamma_\mu \cdot \gamma_\nu = 0, \quad \forall \mu \ne \nu.

A conventional way of summarizing these orthogonality relationships is $$\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu}$$, where $$\eta_{\mu\nu}$$ are the elements of the metric $$G = \text{diag}(+,-,-,-)$$.

## Definition 1.2: Reciprocal basis for the standard Dirac basis.

We define a reciprocal basis $$\setlr{ \gamma^0, \gamma^1, \gamma^2, \gamma^3}$$ satisfying $$\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu, \forall \mu,\nu \in 0,1,2,3$$.

## Theorem 1.1: Reciprocal basis uniqueness.

This reciprocal basis is unique, and for our choice of metric has the values
\label{eqn:reciprocalblog:1960}
\gamma^0 = \gamma_0, \quad \gamma^k = -\gamma_k, \quad \forall k = 1,2,3.

Proof is left to the reader.

## Definition 1.3: Coordinates.

We define the coordinates of a vector with respect to the standard basis as $$x^\mu$$ satisfying
\label{eqn:reciprocalblog:1980}
x = x^\mu \gamma_\mu,

and define the coordinates of a vector with respect to the reciprocal basis as $$x_\mu$$ satisfying
\label{eqn:reciprocalblog:2000}
x = x_\mu \gamma^\mu,

## Theorem 1.2: Coordinates.

Given the definitions above, we may compute the coordinates of a vector, simply by dotting with the basis elements
\label{eqn:reciprocalblog:2020}
x^\mu = x \cdot \gamma^\mu,

and
\label{eqn:reciprocalblog:2040}
x_\mu = x \cdot \gamma_\mu,

### Start proof:

This follows by straightforward computation
\label{eqn:reciprocalblog:840}
\begin{aligned}
x \cdot \gamma^\mu
&=
\lr{ x^\nu \gamma_\nu } \cdot \gamma^\mu \\
&=
x^\nu \lr{ \gamma_\nu \cdot \gamma^\mu } \\
&=
x^\nu {\delta_\nu}^\mu \\
&=
x^\mu,
\end{aligned}

and
\label{eqn:reciprocalblog:860}
\begin{aligned}
x \cdot \gamma_\mu
&=
\lr{ x_\nu \gamma^\nu } \cdot \gamma_\mu \\
&=
x_\nu \lr{ \gamma^\nu \cdot \gamma_\mu } \\
&=
x_\nu {\delta^\nu}_\mu \\
&=
x_\mu.
\end{aligned}

## Derivative operators.

We’d like to determine the form of the (spacetime) gradient operator. The gradient can be defined in terms of coordinates directly, but we choose an implicit definition, in terms of the directional derivative.

## Definition 1.4: Directional derivative and gradient.

Let $$F = F(x)$$ be a four-vector parameterized multivector. The directional derivative of $$F$$ with respect to the (four-vector) direction $$a$$ is denoted
\label{eqn:reciprocalblog:2060}
\lr{ a \cdot \grad } F = \lim_{\epsilon \rightarrow 0} \frac{ F(x + \epsilon a) – F(x) }{ \epsilon },

where $$\grad$$ is called the space time gradient.

The standard basis representation of the gradient is
\label{eqn:reciprocalblog:2080}

where
\label{eqn:reciprocalblog:2100}
\partial_\mu = \PD{x^\mu}{}.

### Start proof:

The Dirac gradient pops naturally out of the coordinate representation of the directional derivative, as we can see by expanding $$F(x + \epsilon a)$$ in Taylor series
\label{eqn:reciprocalblog:900}
\begin{aligned}
F(x + \epsilon a)
&= F(x) + \epsilon \frac{dF(x + \epsilon a)}{d\epsilon} + O(\epsilon^2) \\
&= F(x) + \epsilon \PD{\lr{x^\mu + \epsilon a^\mu}}{F} \PD{\epsilon}{\lr{x^\mu + \epsilon a^\mu}} \\
&= F(x) + \epsilon \PD{\lr{x^\mu + \epsilon a^\mu}}{F} a^\mu.
\end{aligned}

The directional derivative is
\label{eqn:reciprocalblog:920}
\begin{aligned}
\lim_{\epsilon \rightarrow 0}
\frac{F(x + \epsilon a) – F(x)}{\epsilon}
&=
\lim_{\epsilon \rightarrow 0}\,
a^\mu
\PD{\lr{x^\mu + \epsilon a^\mu}}{F} \\
&=
a^\mu
\PD{x^\mu}{F} \\
&=
\lr{a^\nu \gamma_\nu} \cdot \gamma^\mu \PD{x^\mu}{F} \\
&=
a \cdot \lr{ \gamma^\mu \partial_\mu } F.
\end{aligned}

## Curvilinear bases.

Curvilinear bases are the foundation of the fundamental theorem of multivector calculus. This form of integral calculus is defined over parameterized surfaces (called manifolds) that satisfy some specific non-degeneracy and continuity requirements.

A parameterized vector $$x(u,v, \cdots w)$$ can be thought of as tracing out a hypersurface (curve, surface, volume, …), where the dimension of the hypersurface depends on the number of parameters. At each point, a bases can be constructed from the differentials of the parameterized vector. Such a basis is called the tangent space to the surface at the point in question. Our curvilinear bases will be related to these differentials. We will also be interested in a dual basis that is restricted to the span of the tangent space. This dual basis will be called the reciprocal frame, and line the basis of the tangent space itself, also varies from point to point on the surface.

Fig 1a. One parameter curve, with illustration of tangent space along the curve.

Fig 1b. Two parameter surface, with illustration of tangent space along the surface.

One and two parameter spaces are illustrated in fig. 1a, and 1b.  The tangent space basis at a specific point of a two parameter surface, $$x(u^0, u^1)$$, is illustrated in fig. 1. The differential directions that span the tangent space are
\label{eqn:reciprocalblog:1040}
\begin{aligned}
d\Bx_0 &= \PD{u^0}{x} du^0 \\
d\Bx_1 &= \PD{u^1}{x} du^1,
\end{aligned}

and the tangent space itself is $$\mbox{Span}\setlr{ d\Bx_0, d\Bx_1 }$$. We may form an oriented surface area element $$d\Bx_0 \wedge d\Bx_1$$ over this surface.

Fig 2. Two parameter surface.

Tangent spaces associated with 3 or more parameters cannot be easily visualized in three dimensions, but the idea generalizes algebraically without trouble.

## Definition 1.5: Tangent basis and space.

Given a parameterization $$x = x(u^0, \cdots, u^N)$$, where $$N < 4$$, the span of the vectors
\label{eqn:reciprocalblog:2120}
\Bx_\mu = \PD{u^\mu}{x},

is called the tangent space for the hypersurface associated with the parameterization, and it’s basis is
$$\setlr{ \Bx_\mu }$$.

Later we will see that parameterization constraints must be imposed, as not all surfaces generated by a set of parameterizations are useful for integration theory. In particular, degenerate parameterizations for which the wedge products of the tangent space basis vectors are zero, or those wedge products cannot be inverted, are not physically meaningful. Properly behaved surfaces of this sort are called manifolds.

Having introduced curvilinear coordinates associated with a parameterization, we can now determine the form of the gradient with respect to a parameterization of spacetime.

## Theorem 1.4: Gradient, curvilinear representation.

Given a spacetime parameterization $$x = x(u^0, u^1, u^2, u^3)$$, the gradient with respect to the parameters $$u^\mu$$ is
\label{eqn:reciprocalblog:2140}
\PD{u^\mu}{},

where
\label{eqn:reciprocalblog:2160}

The vectors $$\Bx^\mu$$ are called the reciprocal frame vectors, and the ordered set $$\setlr{ \Bx^0, \Bx^1, \Bx^2, \Bx^3 }$$ is called the reciprocal basis.It is convenient to define $$\partial_\mu \equiv \PDi{u^\mu}{}$$, so that the gradient can be expressed in mixed index representation
\label{eqn:reciprocalblog:2180}

This introduces some notational ambiguity, since we used $$\partial_\mu = \PDi{x^\mu}{}$$ for the standard basis derivative operators too, but we will be careful to be explicit when there is any doubt about what is intended.

### Start proof:

The proof follows by application of the chain rule.
\label{eqn:reciprocalblog:960}
\begin{aligned}
&=
\gamma^\alpha \PD{x^\alpha}{F} \\
&=
\gamma^\alpha
\PD{x^\alpha}{u^\mu}
\PD{u^\mu}{F} \\
&=
\lr{ \grad u^\mu } \PD{u^\mu}{F} \\
&=
\Bx^\mu \PD{u^\mu}{F}.
\end{aligned}

## Theorem 1.5: Reciprocal relationship.

The vectors $$\Bx^\mu = \grad u^\mu$$, and $$\Bx_\mu = \PDi{u^\mu}{x}$$ satisfy the reciprocal relationship
\label{eqn:reciprocalblog:2200}
\Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu.

### Start proof:

\label{eqn:reciprocalblog:1020}
\begin{aligned}
\Bx^\mu \cdot \Bx_\nu
&=
\PD{u^\nu}{x} \\
&=
\lr{
\gamma^\alpha \PD{x^\alpha}{u^\mu}
}
\cdot
\lr{
\PD{u^\nu}{x^\beta} \gamma_\beta
} \\
&=
{\delta^\alpha}_\beta \PD{x^\alpha}{u^\mu}
\PD{u^\nu}{x^\beta} \\
&=
\PD{x^\alpha}{u^\mu} \PD{u^\nu}{x^\alpha} \\
&=
\PD{u^\nu}{u^\mu} \\
&=
{\delta^\mu}_\nu
.
\end{aligned}

### End proof.

It is instructive to consider an example. Here is a parameterization that scales the proper time parameter, and uses polar coordinates in the $$x-y$$ plane.

## Problem: Compute the curvilinear and reciprocal basis.

Given
\label{eqn:reciprocalblog:2360}
x(t,\rho,\theta,z) = c t \gamma_0 + \gamma_1 \rho e^{i \theta} + z \gamma_3,

where $$i = \gamma_1 \gamma_2$$, compute the curvilinear frame vectors and their reciprocals.

The frame vectors are all easy to compute
\label{eqn:reciprocalblog:1180}
\begin{aligned}
\Bx_0 &= \PD{t}{x} = c \gamma_0 \\
\Bx_1 &= \PD{\rho}{x} = \gamma_1 e^{i \theta} \\
\Bx_2 &= \PD{\theta}{x} = \rho \gamma_1 \gamma_1 \gamma_2 e^{i \theta} = – \rho \gamma_2 e^{i \theta} \\
\Bx_3 &= \PD{z}{x} = \gamma_3.
\end{aligned}

The $$\Bx_1$$ vector is radial, $$\Bx^2$$ is perpendicular to that tangent to the same unit circle, as plotted in fig 3.

Fig3: Tangent space direction vectors.

All of these particular frame vectors happen to be mutually perpendicular, something that will not generally be true for a more arbitrary parameterization.

To compute the reciprocal frame vectors, we must express our parameters in terms of $$x^\mu$$ coordinates, and use implicit integration techniques to deal with the coupling of the rotational terms. First observe that
\label{eqn:reciprocalblog:1200}
\gamma_1 e^{i\theta}
= \gamma_1 \lr{ \cos\theta + \gamma_1 \gamma_2 \sin\theta }
= \gamma_1 \cos\theta – \gamma_2 \sin\theta,

so
\label{eqn:reciprocalblog:1220}
\begin{aligned}
x^0 &= c t \\
x^1 &= \rho \cos\theta \\
x^2 &= -\rho \sin\theta \\
x^3 &= z.
\end{aligned}

We can easily evaluate the $$t, z$$ gradients
\label{eqn:reciprocalblog:1240}
\begin{aligned}
\grad t &= \frac{\gamma^1 }{c} \\
\end{aligned}

but the $$\rho, \theta$$ gradients are not as easy. First writing
\label{eqn:reciprocalblog:1260}
\rho^2 = \lr{x^1}^2 + \lr{x^2}^2,

we find
\label{eqn:reciprocalblog:1280}
\begin{aligned}
&= 2 \rho \lr{ \cos\theta \gamma^1 – \sin\theta \gamma^2 } \\
&= 2 \rho \gamma^1 \lr{ \cos\theta – \gamma_1 \gamma^2 \sin\theta } \\
&= 2 \rho \gamma^1 e^{i\theta},
\end{aligned}

so
\label{eqn:reciprocalblog:1300}

For the $$\theta$$ gradient, we can write
\label{eqn:reciprocalblog:1320}
\tan\theta = -\frac{x^2}{x^1},

so
\label{eqn:reciprocalblog:1340}
\begin{aligned}
&= -\frac{\gamma^2}{x^1} – x^2 \frac{-\gamma^1}{\lr{x^1}^2} \\
&= \inv{\lr{x^1}^2} \lr{ – \gamma^2 x^1 + \gamma^1 x^2 } \\
&= \frac{\rho}{\rho^2 \cos^2\theta } \lr{ – \gamma^2 \cos\theta – \gamma^1 \sin\theta } \\
&= -\frac{1}{\rho \cos^2\theta } \gamma^2 \lr{ \cos\theta + \gamma_2 \gamma^1 \sin\theta } \\
&= -\frac{\gamma^2 e^{i\theta} }{\rho \cos^2\theta },
\end{aligned}

or
\label{eqn:reciprocalblog:1360}

In summary,
\label{eqn:reciprocalblog:1380}
\begin{aligned}
\Bx^0 &= \frac{\gamma^0}{c} \\
\Bx^1 &= \gamma^1 e^{i\theta} \\
\Bx^2 &= -\inv{\rho} \gamma^2 e^{i\theta} \\
\Bx^3 &= \gamma^3.
\end{aligned}

Despite being a fairly simple parameterization, it was still fairly difficult to solve for the gradients when the parameterization introduced coupling between the coordinates. In this particular case, we could have solved for the parameters in terms of the coordinates (but it was easier not to), but that will not generally be true. We want a less labor intensive strategy to find the reciprocal frame. When we have a full parameterization of spacetime, then we can do this with nothing more than a matrix inversion.

## Theorem 1.6: Reciprocal frame matrix equations.

Given a spacetime basis $$\setlr{\Bx_0, \cdots \Bx_3}$$, let $$[\Bx_\mu]$$ and $$[\Bx^\nu]$$ be column matrices with the coordinates of these vectors and their reciprocals, with respect to the standard basis $$\setlr{\gamma_0, \gamma_1, \gamma_2, \gamma_3 }$$. Let
\label{eqn:reciprocalblog:2220}
A =
\begin{bmatrix}
[\Bx_0] & \cdots & [\Bx_{3}]
\end{bmatrix}
X =
\begin{bmatrix}
[\Bx^0] & \cdots & [\Bx^{3}]
\end{bmatrix}.

The coordinates of the reciprocal frame vectors can be found by solving
\label{eqn:reciprocalblog:2240}
A^\T G X = 1,

where $$G = \text{diag}(1,-1,-1,-1)$$ and the RHS is an $$4 \times 4$$ identity matrix.

### Start proof:

Let $$\Bx_\mu = {a_\mu}^\alpha \gamma_\alpha, \Bx^\nu = b^{\nu\beta} \gamma_\beta$$, so that
\label{eqn:reciprocalblog:140}
A =
\begin{bmatrix}
{a_\nu}^\mu
\end{bmatrix},

and
\label{eqn:reciprocalblog:160}
X =
\begin{bmatrix}
b^{\nu\mu}
\end{bmatrix},

where $$\mu \in [0,3]$$ are the row indexes and $$\nu \in [0,N-1]$$ are the column indexes. The reciprocal frame satisfies $$\Bx_\mu \cdot \Bx^\nu = {\delta_\mu}^\nu$$, which has the coordinate representation of
\label{eqn:reciprocalblog:180}
\begin{aligned}
\Bx_\mu \cdot \Bx^\nu
&=
\lr{
{a_\mu}^\alpha \gamma_\alpha
}
\cdot
\lr{
b^{\nu\beta} \gamma_\beta
} \\
&=
{a_\mu}^\alpha
\eta_{\alpha\beta}
b^{\nu\beta} \\
&=
{[A^\T G B]_\mu}^\nu,
\end{aligned}

where $$\mu$$ is the row index and $$\nu$$ is the column index.

## Problem: Matrix inversion reciprocals.

For the parameterization of \ref{eqn:reciprocalblog:2360}, find the reciprocal frame vectors by matrix inversion.

We expanded $$\Bx_1$$ explicitly in \ref{eqn:reciprocalblog:1200}. Doing the same for $$\Bx_2$$, we have
\label{eqn:reciprocalblog:1201}
\Bx_2 =
-\rho \gamma_2 e^{i\theta}
= -\rho \gamma_2 \lr{ \cos\theta + \gamma_1 \gamma_2 \sin\theta }
= – \rho \lr{ \gamma_2 \cos\theta + \gamma_1 \sin\theta}.

Reading off the coordinates of our frame vectors, we have
\label{eqn:reciprocalblog:1400}
X =
\begin{bmatrix}
c & 0 & 0 & 0 \\
0 & C & -\rho S & 0 \\
0 & -S & -\rho C & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix},

where $$C = \cos\theta$$ and $$S = \sin\theta$$. We want
\label{eqn:reciprocalblog:1420}
Y =
{\begin{bmatrix}
c & 0 & 0 & 0 \\
0 & -C & S & 0 \\
0 & \rho S & \rho C & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}}^{-1}
=
\begin{bmatrix}
\inv{c} & 0 & 0 & 0 \\
0 & -C & \frac{S}{\rho} & 0 \\
0 & S & \frac{C}{\rho} & 0 \\
0 & 0 & 0 & -1 \\
\end{bmatrix}.

We can read off the coordinates of the reciprocal frame vectors
\label{eqn:reciprocalblog:1440}
\begin{aligned}
\Bx^0 &= \inv{c} \gamma_0 \\
\Bx^1 &= -\cos\theta \gamma_1 + \sin\theta \gamma_2 \\
\Bx^2 &= \inv{\rho} \lr{ \sin\theta \gamma_1 + \cos\theta \gamma_2 } \\
\Bx^3 &= -\gamma_3.
\end{aligned}

Factoring out $$\gamma^1$$ from the $$\Bx^1$$ terms, we find
\label{eqn:reciprocalblog:1460}
\begin{aligned}
\Bx^1
&= -\cos\theta \gamma_1 + \sin\theta \gamma_2 \\
&= \gamma^1 \lr{ \cos\theta + \gamma_1 \gamma_2 \sin\theta } \\
&= \gamma^1 e^{i\theta}.
\end{aligned}

Similarly for $$\Bx^2$$,
\label{eqn:reciprocalblog:1480}
\begin{aligned}
\Bx^2
&= \inv{\rho} \lr{ \sin\theta \gamma_1 + \cos\theta \gamma_2 } \\
&= \frac{\gamma^2}{\rho} \lr{ \sin\theta \gamma_2 \gamma_1 – \cos\theta } \\
&= -\frac{\gamma^2}{\rho} e^{i\theta}.
\end{aligned}

This matches \ref{eqn:reciprocalblog:1380}, as expected, but required only algebraic work to compute.

There will be circumstances where we parameterize only a subset of spacetime, and are interested in calculating quantities associated with such a surface. For example, suppose that
\label{eqn:reciprocalblog:1500}
x(\rho,\theta) = \gamma_1 \rho e^{i \theta},

where $$i = \gamma_1 \gamma_2$$ as before. We are now parameterizing only the $$x-y$$ plane. We will still find
\label{eqn:reciprocalblog:1520}
\begin{aligned}
\Bx_1 &= \gamma_1 e^{i \theta} \\
\Bx_2 &= -\gamma_2 \rho e^{i \theta}.
\end{aligned}

We can compute the reciprocals of these vectors using the gradient method. It’s possible to state matrix equations representing the reciprocal relationship of \ref{eqn:reciprocalblog:2200}, which, in this case, is $$X^\T G Y = 1$$, where the RHS is a $$2 \times 2$$ identity matrix, and $$X, Y$$ are $$4\times 2$$ matrices of coordinates, with
\label{eqn:reciprocalblog:1540}
X =
\begin{bmatrix}
0 & 0 \\
C & -\rho S \\
-S & -\rho C \\
0 & 0
\end{bmatrix}.

We no longer have a square matrix problem to solve, and our solution set is multivalued. In particular, this matrix equation has solutions
\label{eqn:reciprocalblog:1560}
\begin{aligned}
\Bx^1 &= \gamma^1 e^{i\theta} + \alpha \gamma^0 + \beta \gamma^3 \\
\Bx^2 &= -\frac{\gamma^2}{\rho} e^{i\theta} + \alpha’ \gamma^0 + \beta’ \gamma^3.
\end{aligned}

where $$\alpha, \alpha’, \beta, \beta’$$ are arbitrary constants. In the example we considered, we saw that our $$\rho, \theta$$ parameters were functions of only $$x^1, x^2$$, so taking gradients could not introduce any $$\gamma^0, \gamma^3$$ dependence in $$\Bx^1, \Bx^2$$. It seems reasonable to assert that we seek an algebraic method of computing a set of vectors that satisfies the reciprocal relationships, where that set of vectors is restricted to the tangent space. We will need to figure out how to prove that this reciprocal construction is identical to the parameter gradients, but let’s start with figuring out what such a tangent space restricted solution looks like.

## Theorem 1.7: Reciprocal frame for two parameter subspace.

Given two vectors, $$\Bx_1, \Bx_2$$, the vectors $$\Bx^1, \Bx^2 \in \mbox{Span}\setlr{ \Bx_1, \Bx_2 }$$ such that $$\Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu$$ are given by
\label{eqn:reciprocalblog:2260}
\begin{aligned}
\Bx^1 &= \Bx_2 \cdot \inv{\Bx_1 \wedge \Bx_2} \\
\Bx^2 &= -\Bx_1 \cdot \inv{\Bx_1 \wedge \Bx_2},
\end{aligned}

provided $$\Bx_1 \wedge \Bx_2 \ne 0$$ and
$$\lr{ \Bx_1 \wedge \Bx_2 }^2 \ne 0$$.

### Start proof:

The most general set of vectors that satisfy the span constraint are
\label{eqn:reciprocalblog:1580}
\begin{aligned}
\Bx^1 &= a \Bx_1 + b \Bx_2 \\
\Bx^2 &= c \Bx_1 + d \Bx_2.
\end{aligned}

We can use wedge products with either $$\Bx_1$$ or $$\Bx_2$$ to eliminate the other from the RHS
\label{eqn:reciprocalblog:1600}
\begin{aligned}
\Bx^1 \wedge \Bx_2 &= a \lr{ \Bx_1 \wedge \Bx_2 } \\
\Bx^1 \wedge \Bx_1 &= – b \lr{ \Bx_1 \wedge \Bx_2 } \\
\Bx^2 \wedge \Bx_2 &= c \lr{ \Bx_1 \wedge \Bx_2 } \\
\Bx^2 \wedge \Bx_1 &= – d \lr{ \Bx_1 \wedge \Bx_2 },
\end{aligned}

and then dot both sides with $$\Bx_1 \wedge \Bx_2$$ to produce four scalar equations
\label{eqn:reciprocalblog:1640}
\begin{aligned}
a \lr{ \Bx_1 \wedge \Bx_2 }^2
&= \lr{ \Bx^1 \wedge \Bx_2 } \cdot \lr{ \Bx_1 \wedge \Bx_2 } \\
&=
\lr{ \Bx_2 \cdot \Bx_1 } \lr{ \Bx^1 \cdot \Bx_2 }

\lr{ \Bx_2 \cdot \Bx_2 } \lr{ \Bx^1 \cdot \Bx_1 } \\
&=
\lr{ \Bx_2 \cdot \Bx_1 } (0)

\lr{ \Bx_2 \cdot \Bx_2 } (1) \\
&= – \Bx_2 \cdot \Bx_2
\end{aligned}

\label{eqn:reciprocalblog:1660}
\begin{aligned}
– b \lr{ \Bx_1 \wedge \Bx_2 }^2
&=
\lr{ \Bx^1 \wedge \Bx_1 } \cdot \lr{ \Bx_1 \wedge \Bx_2 } \\
&=
\lr{ \Bx^1 \cdot \Bx_2 } \lr{ \Bx_1 \cdot \Bx_1 }

\lr{ \Bx^1 \cdot \Bx_1 } \lr{ \Bx_1 \cdot \Bx_2 } \\
&=
(0) \lr{ \Bx_1 \cdot \Bx_1 }

(1) \lr{ \Bx_1 \cdot \Bx_2 } \\
&= – \Bx_1 \cdot \Bx_2
\end{aligned}

\label{eqn:reciprocalblog:1680}
\begin{aligned}
c \lr{ \Bx_1 \wedge \Bx_2 }^2
&= \lr{ \Bx^2 \wedge \Bx_2 } \cdot \lr{ \Bx_1 \wedge \Bx_2 } \\
&=
\lr{ \Bx_2 \cdot \Bx_1 } \lr{ \Bx^2 \cdot \Bx_2 }

\lr{ \Bx_2 \cdot \Bx_2 } \lr{ \Bx^2 \cdot \Bx_1 } \\
&=
\lr{ \Bx_2 \cdot \Bx_1 } (1)

\lr{ \Bx_2 \cdot \Bx_2 } (0) \\
&= \Bx_2 \cdot \Bx_1
\end{aligned}

\label{eqn:reciprocalblog:1700}
\begin{aligned}
– d \lr{ \Bx_1 \wedge \Bx_2 }^2
&= \lr{ \Bx^2 \wedge \Bx_1 } \cdot \lr{ \Bx_1 \wedge \Bx_2 } \\
&=
\lr{ \Bx_1 \cdot \Bx_1 } \lr{ \Bx^2 \cdot \Bx_2 }

\lr{ \Bx_1 \cdot \Bx_2 } \lr{ \Bx^2 \cdot \Bx_1 } \\
&=
\lr{ \Bx_1 \cdot \Bx_1 } (1)

\lr{ \Bx_1 \cdot \Bx_2 } (0) \\
&= \Bx_1 \cdot \Bx_1.
\end{aligned}

Putting the pieces together we have
\label{eqn:reciprocalblog:1740}
\begin{aligned}
\Bx^1
&= \frac{ – \lr{ \Bx_2 \cdot \Bx_2 } \Bx_1 + \lr{ \Bx_1 \cdot \Bx_2 } \Bx_2
}{\lr{\Bx_1 \wedge \Bx_2}^2} \\
&=
\frac{
\Bx_2 \cdot \lr{ \Bx_1 \wedge \Bx_2 }
}{\lr{\Bx_1 \wedge \Bx_2}^2} \\
&=
\Bx_2 \cdot \inv{\Bx_1 \wedge \Bx_2}
\end{aligned}

\label{eqn:reciprocalblog:1760}
\begin{aligned}
\Bx^2
&=
\frac{ \lr{ \Bx_1 \cdot \Bx_2 } \Bx_1 – \lr{ \Bx_1 \cdot \Bx_1 } \Bx_2
}{\lr{\Bx_1 \wedge \Bx_2}^2} \\
&=
\frac{ -\Bx_1 \cdot \lr{ \Bx_1 \wedge \Bx_2 } }
{\lr{\Bx_1 \wedge \Bx_2}^2} \\
&=
-\Bx_1 \cdot \inv{\Bx_1 \wedge \Bx_2}
\end{aligned}

## Lemma 1.1: Distribution identity.

Given k-vectors $$B, C$$ and a vector $$a$$, where the grade of $$C$$ is greater than that of $$B$$, then
\label{eqn:reciprocalblog:2280}
\lr{a \wedge B} \cdot C = a \cdot \lr{ B \cdot C }.

See [1] for a proof.

## Theorem 1.8: Higher order tangent space reciprocals.

Given an $$N$$ parameter tangent space with basis $$\setlr{ \Bx_0, \Bx_1, \cdots \Bx_{N-1} }$$, the reciprocals are given by
\label{eqn:reciprocalblog:2300}
\Bx^\mu = (-1)^\mu
\lr{ \Bx_0 \wedge \cdots \check{\Bx_\mu} \cdots \wedge \Bx_{N-1} } \cdot I_N^{-1},

where the checked term ($$\check{\Bx_\mu}$$) indicates that all terms are included in the wedges except the $$\Bx_\mu$$ term, and $$I_N = \Bx_0 \wedge \cdots \Bx_{N-1}$$ is the pseudoscalar for the tangent space.

### Start proof:

I’ll outline the proof for the three parameter tangent space case, from which the pattern will be clear. The motivation for this proof is a reexamination of the algebraic structure of the two vector solution. Suppose we have a tangent space basis $$\setlr{\Bx_0, \Bx_1}$$, for which we’ve shown that
\label{eqn:reciprocalblog:1860}
\begin{aligned}
\Bx^0
&= \Bx_1 \cdot \inv{\Bx_0 \wedge \Bx_1} \\
&= \frac{\Bx_1 \cdot \lr{\Bx_0 \wedge \Bx_1} }{\lr{ \Bx_0 \wedge \Bx_1}^2 }.
\end{aligned}

If we dot with $$\Bx_0$$ and $$\Bx_1$$ respectively, we find
\label{eqn:reciprocalblog:1800}
\begin{aligned}
\Bx_0 \cdot \Bx^0
&=
\Bx_0 \cdot \frac{ \Bx_1 \cdot \lr{ \Bx_0 \wedge \Bx_1 } }{\lr{ \Bx_0 \wedge \Bx_1}^2 } \\
&=
\lr{ \Bx_0 \wedge \Bx_1 } \cdot \frac{ \Bx_0 \wedge \Bx_1 }{\lr{ \Bx_0 \wedge \Bx_1}^2 }.
\end{aligned}

We end up with unity as expected. Here the
“factored” out vector is reincorporated into the pseudoscalar using the distribution identity \ref{eqn:reciprocalblog:2280}.
Similarly, dotting with $$\Bx_1$$, we find
\label{eqn:reciprocalblog:0810}
\begin{aligned}
\Bx_1 \cdot \Bx^0
&=
\Bx_1 \cdot \frac{ \Bx_1 \cdot \lr{ \Bx_0 \wedge \Bx_1 } }{\lr{ \Bx_0 \wedge \Bx_1}^2 } \\
&=
\lr{ \Bx_1 \wedge \Bx_1 } \cdot \frac{ \Bx_0 \wedge \Bx_1 }{\lr{ \Bx_0 \wedge \Bx_1}^2 }.
\end{aligned}

This is zero, since wedging a vector with itself is zero. We can perform such an operation in reverse, taking the square of the tangent space pseudoscalar, and factoring out one of the basis vectors. After this, division by that squared pseudoscalar will normalize things.

For a three parameter tangent space with basis $$\setlr{ \Bx_0, \Bx_1, \Bx_2 }$$, we can factor out any of the tangent vectors like so
\label{eqn:reciprocalblog:1880}
\begin{aligned}
\lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 }^2
&= \Bx_0 \cdot \lr{ \lr{ \Bx_1 \wedge \Bx_2 } \cdot \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } } \\
&= (-1) \Bx_1 \cdot \lr{ \lr{ \Bx_0 \wedge \Bx_2 } \cdot \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } } \\
&= (-1)^2 \Bx_2 \cdot \lr{ \lr{ \Bx_0 \wedge \Bx_1 } \cdot \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } }.
\end{aligned}

The toggling of sign reflects the number of permutations required to move the vector of interest to the front of the wedge sequence. Having factored out any one of the vectors, we can rearrange to find that vector that is it’s inverse and perpendicular to all the others.
\label{eqn:reciprocalblog:1900}
\begin{aligned}
\Bx^0 &= (-1)^0 \lr{ \Bx_1 \wedge \Bx_2 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } \\
\Bx^1 &= (-1)^1 \lr{ \Bx_0 \wedge \Bx_2 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } \\
\Bx^2 &= (-1)^2 \lr{ \Bx_0 \wedge \Bx_1 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 }.
\end{aligned}

### End proof.

In the fashion above, should we want the reciprocal frame for all of spacetime given dimension 4 tangent space, we can state it trivially
\label{eqn:reciprocalblog:1920}
\begin{aligned}
\Bx^0 &= (-1)^0 \lr{ \Bx_1 \wedge \Bx_2 \wedge \Bx_3 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 \wedge \Bx_3 } \\
\Bx^1 &= (-1)^1 \lr{ \Bx_0 \wedge \Bx_2 \wedge \Bx_3 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 \wedge \Bx_3 } \\
\Bx^2 &= (-1)^2 \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_3 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 \wedge \Bx_3 } \\
\Bx^3 &= (-1)^3 \lr{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 } \cdot \inv{ \Bx_0 \wedge \Bx_1 \wedge \Bx_2 \wedge \Bx_3 }.
\end{aligned}

This is probably not an efficient way to compute all these reciprocals, since we can utilize a single matrix inversion to solve them in one shot. However, there are theoretical advantages to this construction that will be useful when we get to integration theory.

### On degeneracy.

A small mention of degeneracy was mentioned above. Regardless of metric, $$\Bx_0 \wedge \Bx_1 = 0$$ means that this pair of vectors are colinear. A tangent space with such a pseudoscalar is clearly undesirable, and we must construct parameterizations for which the area element is non-zero in all regions of interest.

Things get more interesting in mixed signature spaces where we can have vectors that square to zero (i.e. lightlike). If the tangent space pseudoscalar has a lightlike factor, then that pseudoscalar will not be invertible. Such a degeneracy will will likely lead to many other troubles, and parameterizations of this sort should be avoided.

This following problem illustrates an example of this sort of degenerate parameterization.

## Problem: Degenerate surface parameterization.

Given a spacetime plane parameterization $$x(u,v) = u a + v b$$, where
\label{eqn:reciprocalblog:480}
a = \gamma_0 + \gamma_1 + \gamma_2 + \gamma_3,

\label{eqn:reciprocalblog:500}
b = \gamma_0 – \gamma_1 + \gamma_2 – \gamma_3,

show that this is a degenerate parameterization, and find the bivector that represents the tangent space. Are these vectors lightlike, spacelike, or timelike? Comment on whether this parameterization represents a physically relevant spacetime surface.

To characterize the vectors, we square them
\label{eqn:reciprocalblog:1080}
a^2 = b^2 =
\gamma_0^2 +
\gamma_1^2 +
\gamma_2^2 +
\gamma_3^2
=
1 – 3
= -2,

so $$a, b$$ are both spacelike vectors. The tangent space is clearly just $$\mbox{Span}\setlr{ a, b } = \mbox{Span}\setlr{ e, f }$$ where
\label{eqn:reciprocalblog:1100}
\begin{aligned}
e &= \gamma_0 + \gamma_2 \\
f &= \gamma_1 + \gamma_3.
\end{aligned}

Observe that $$a = e + f, b = e – f$$, and $$e$$ is lightlike ($$e^2 = 0$$), whereas $$f$$ is spacelike ($$f^2 = -2$$), and $$e \cdot f = 0$$, so $$e f = – f e$$. The bivector for the tangent plane is
\label{eqn:reciprocalblog:1120}
a b
}
=
(e + f) (e – f)
}
=
e^2 – f^2 – 2 e f
}
= -2 e f,

where
\label{eqn:reciprocalblog:1140}
e f = \gamma_{01} + \gamma_{21} + \gamma_{23} + \gamma_{03}.

Because $$e$$ is lightlike (zero square), and $$e f = – f e$$,
the bivector $$e f$$ squares to zero
\label{eqn:reciprocalblog:1780}
\lr{ e f }^2
= -e^2 f^2
= 0,

which shows that the parameterization is degenerate.

This parameterization can also be expressed as
\label{eqn:reciprocalblog:1160}
x(u,v)
= u ( e + f ) + v ( e – f )
= (u + v) e + (u – v) f,

a linear combination of a lightlike and spacelike vector. Intuitively, we expect that a physically meaningful spacetime surface involves linear combinations spacelike vectors, or combinations of a timelike vector with spacelike vectors. This beastie is something entirely different.

### Final notes.

There are a few loose ends above. In particular, we haven’t conclusively proven that the set of reciprocal vectors $$\Bx^\mu = \grad u^\mu$$ are exactly those obtained through algebraic means. For a full parameterization of spacetime, they are necessarily the same, since both are unique. So we know that \ref{eqn:reciprocalblog:1920} must equal the reciprocals obtained by evaluating the gradient for a full parameterization (and this must also equal the reciprocals that we can obtain through matrix inversion.) We have also not proved explicitly that the three parameter construction of the reciprocals in \ref{eqn:reciprocalblog:1900} is in the tangent space, but that is a fairly trivial observation, so that can be left as an exercise for the reader dismissal. Some additional thought about this is probably required, but it seems reasonable to put that on the back burner and move on to some applications.

# References

[1] Peeter Joot. Geometric Algebra for Electrical Engineers. Kindle Direct Publishing, 2019.

## Motivation.

In my old classical mechanics notes it appears that I did covariant derivations of the Lorentz force equations a number of times, using different trial Lagrangians (relativistic and non-relativistic), and using both geometric algebra and tensor methods. However, none of these appear to have been done concisely, and a number not even coherently.

The following document has been drafted as replacement text for those incoherent classical mechanics notes. I’ll attempt to cover

• a lighting review of the geometric algebra STA (Space Time Algebra),
• relations between Dirac matrix algebra and STA,
• derivation of the relativistic form of the Euler-Lagrange equations from the covariant form of the action,
• relationship of the STA form of the Euler-Lagrange equations to their tensor equivalents,
• derivation of the Lorentz force equation from the STA Lorentz force Lagrangian,
• relationship of the STA Lorentz force equation to its equivalent in the tensor formalism,
• relationship of the STA Lorentz force equation to the traditional vector form.

Note that some of the prerequisite ideas and auxiliary details are presented as problems with solutions. If the reader has sufficient background to attempt those problems themselves, they are encouraged to do so.

The STA and geometric algebra ideas used here are not complete to learn from in isolation. The reader is referred to [1] for a more complete exposition of both STA and geometric algebra.

## Definition 1.1: Index conventions.

Latin indexes $$i, j, k, r, s, t, \cdots$$ are used to designate values in the range $$\setlr{ 1,2,3 }$$. Greek indexes are $$\alpha, \beta, \mu, \nu, \cdots$$ are used for indexes of spacetime quantities $$\setlr{0,1,2,3}$$.
The Einstein convention of implied summation for mixed upper and lower Greek indexes will be used, for example
\begin{equation*}
x^\alpha x_\alpha \equiv \sum_{\alpha = 0}^3 x^\alpha x_\alpha.
\end{equation*}

## Space Time Algebra (STA.)

In the geometric algebra literature, the Dirac algebra of quantum field theory has been rebranded Space Time Algebra (STA). The differences between STA and the Dirac theory that uses matrices ($$\gamma_\mu$$) are as follows

• STA completely omits any representation of the Dirac basis vectors $$\gamma_\mu$$. In particular, any possible matrix representation is irrelevant.
• STA provides a rich set of fundamental operations (grade selection, generalized dot and wedge products for multivector elements, rotation and reflection operations, …)
• Matrix trace, and commutator and anticommutator operations are nowhere to be found in STA, as geometrically grounded equivalents are available instead.
• The “slashed” quantities from Dirac theory, such as $$\gamma_\mu p^\mu$$ are nothing more than vectors in their entirety in STA (where the basis is no longer implicit, as is the case for coordinates.)

Our basis vectors have the following properties.

## Definition 1.2: Standard basis.

Let the four-vector standard basis be designated $$\setlr{\gamma_0, \gamma_1, \gamma_2, \gamma_3 }$$, where the basis vectors satisfy
\label{eqn:lorentzForceCovariant:1540}
\begin{aligned}
\gamma_0^2 &= -\gamma_i^2 = 1 \\
\gamma_\alpha \cdot \gamma_\beta &= 0, \forall \alpha \ne \beta.
\end{aligned}

## Problem: Commutator properties of the STA basis.

In Dirac theory, the commutator properties of the Dirac matrices is considered fundamental, namely
\begin{equation*}
\symmetric{\gamma_\mu}{\gamma_\nu} = 2 \eta_{\mu\nu}.
\end{equation*}

Show that this follows from the axiomatic assumptions of geometric algebra, and describe how the dot and wedge products are related to the anticommutator and commutator products of Dirac theory.

The anticommutator is defined as symmetric sum of products
\label{eqn:lorentzForceCovariant:1040}
\symmetric{\gamma_\mu}{\gamma_\nu}
\equiv
\gamma_\mu \gamma_\nu
+
\gamma_\nu \gamma_\mu,

but this is just twice the dot product in its geometric algebra form $$a b = (a b + ba)/2$$. Observe that the properties of the basis vectors defined in \ref{eqn:lorentzForceCovariant:1540} may be summarized as
\label{eqn:lorentzForceCovariant:1060}
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu},

where $$\eta_{\mu\nu} = \text{diag}(+,-,-,-) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}$$ is the conventional metric tensor. This means
\label{eqn:lorentzForceCovariant:1080}
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu} = 2 \symmetric{\gamma_\mu}{\gamma_\nu},

as claimed.

Similarly, observe that the commutator, defined as the antisymmetric sum of products
\label{eqn:lorentzForceCovariant:1100}
\antisymmetric{\gamma_\mu}{\gamma_\nu} \equiv
\gamma_\mu \gamma_\nu

\gamma_\nu \gamma_\mu,

is twice the wedge product $$a \wedge b = (a b – b a)/2$$. This provides geometric identifications for the respective anti-commutator and commutator products respectively
\label{eqn:lorentzForceCovariant:1120}
\begin{aligned}
\symmetric{\gamma_\mu}{\gamma_\nu} &= 2 \gamma_\mu \cdot \gamma_\nu \\
\antisymmetric{\gamma_\mu}{\gamma_\nu} &= 2 \gamma_\mu \wedge \gamma_\nu,
\end{aligned}

## Definition 1.3: Pseudoscalar.

The pseudoscalar for the space is denoted $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$.

## Problem: Pseudoscalar.

Show that the STA pseudoscalar $$I$$ defined by \ref{eqn:lorentzForceCovariant:1540} satisfies
\begin{equation*}
\tilde{I} = I,
\end{equation*}
where the tilde operator designates reversion. Also show that $$I$$ has the properties of an imaginary number
\begin{equation*}
I^2 = -1.
\end{equation*}
Finally, show that, unlike the spatial pseudoscalar that commutes with all grades, $$I$$ anticommutes with any vector or trivector, and commutes with any bivector.

Since $$\gamma_\alpha \gamma_\beta = -\gamma_\beta \gamma_\alpha$$ for any $$\alpha \ne \beta$$, any permutation of the factors of $$I$$ changes the sign once. In particular
\label{eqn:lorentzForceCovariant:680}
\begin{aligned}
I &=
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&=

\gamma_1
\gamma_2
\gamma_3
\gamma_0 \\
&=

\gamma_2
\gamma_3
\gamma_1
\gamma_0 \\
&=
+
\gamma_3
\gamma_2
\gamma_1
\gamma_0
= \tilde{I}.
\end{aligned}

Using this, we have
\label{eqn:lorentzForceCovariant:700}
\begin{aligned}
I^2
&= I \tilde{I} \\
&=
(
\gamma_0
\gamma_1
\gamma_2
\gamma_3
)(
\gamma_3
\gamma_2
\gamma_1
\gamma_0
) \\
&=
\lr{\gamma_0}^2
\lr{\gamma_1}^2
\lr{\gamma_2}^2
\lr{\gamma_3}^2 \\
&=
(+1)
(-1)
(-1)
(-1) \\
&= -1.
\end{aligned}

To illustrate the anticommutation property with any vector basis element, consider the following two examples:
\label{eqn:lorentzForceCovariant:720}
\begin{aligned}
I \gamma_0 &=
\gamma_0
\gamma_1
\gamma_2
\gamma_3
\gamma_0 \\
&=

\gamma_0
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&=

\gamma_0 I,
\end{aligned}

\label{eqn:lorentzForceCovariant:740}
\begin{aligned}
I \gamma_2
&=
\gamma_0
\gamma_1
\gamma_2
\gamma_3
\gamma_2 \\
&=

\gamma_0
\gamma_1
\gamma_2
\gamma_2
\gamma_3 \\
&=

\gamma_2
\gamma_0
\gamma_1
\gamma_2
\gamma_3 \\
&= -\gamma_2 I.
\end{aligned}

A total of three sign swaps is required to “percolate” any given $$\gamma_\alpha$$ through the factors of $$I$$, resulting in an overall sign change of $$-1$$.

For any bivector basis element $$\alpha \ne \beta$$
\label{eqn:lorentzForceCovariant:760}
\begin{aligned}
I \gamma_\alpha \gamma_\beta
&=
-\gamma_\alpha I \gamma_\beta \\
&=
+\gamma_\alpha \gamma_\beta I.
\end{aligned}

Similarly for any trivector basis element $$\alpha \ne \beta \ne \sigma$$
\label{eqn:lorentzForceCovariant:780}
\begin{aligned}
I \gamma_\alpha \gamma_\beta \gamma_\sigma
&=
-\gamma_\alpha I \gamma_\beta \gamma_\sigma \\
&=
+\gamma_\alpha \gamma_\beta I \gamma_\sigma \\
&=
-\gamma_\alpha \gamma_\beta \gamma_\sigma I.
\end{aligned}

## Definition 1.4: Reciprocal basis.

The reciprocal basis $$\setlr{ \gamma^0, \gamma^1, \gamma^2, \gamma^3 }$$ is defined , such that the property $$\gamma^\alpha \cdot \gamma_\beta = {\delta^\alpha}_\beta$$ holds.

Observe that, $$\gamma^0 = \gamma_0$$ and $$\gamma^i = -\gamma_i$$.

## Theorem 1.1: Coordinates.

Coordinates are defined in terms of dot products with the standard basis, or reciprocal basis
\begin{equation*}
\begin{aligned}
x^\alpha &= x \cdot \gamma^\alpha \\
x_\alpha &= x \cdot \gamma_\alpha,
\end{aligned}
\end{equation*}

### Start proof:

Suppose that a coordinate representation of the following form is assumed
\label{eqn:lorentzForceCovariant:820}
x = x^\alpha \gamma_\alpha = x_\beta \gamma^\beta.

We wish to determine the representation of the $$x^\alpha$$ or $$x_\beta$$ coordinates in terms of $$x$$ and the basis elements. Taking the dot product with any standard basis element, we find
\label{eqn:lorentzForceCovariant:840}
\begin{aligned}
x \cdot \gamma_\mu
&= (x_\beta \gamma^\beta) \cdot \gamma_\mu \\
&= x_\beta {\delta^\beta}_\mu \\
&= x_\mu,
\end{aligned}

as claimed. Similarly, dotting with a reciprocal frame vector, we find
\label{eqn:lorentzForceCovariant:860}
\begin{aligned}
x \cdot \gamma^\mu
&= (x^\beta \gamma_\beta) \cdot \gamma^\mu \\
&= x^\beta {\delta_\beta}^\mu \\
&= x^\mu.
\end{aligned}

### End proof.

Observe that raising or lowering the index of a spatial index toggles the sign of a coordinate, but timelike indexes are left unchanged.
\label{eqn:lorentzForceCovariant:880}
\begin{aligned}
x^0 &= x_0 \\
x^i &= -x_i \\
\end{aligned}

\begin{equation*}
\grad = \gamma^\mu \partial_\mu = \gamma_\nu \partial^\nu,
\end{equation*}
where
\begin{equation*}
\partial_\mu = \PD{x^\mu}{},
\end{equation*}
and
\begin{equation*}
\partial^\mu = \PD{x_\mu}{}.
\end{equation*}

This definition of gradient is consistent with the Dirac gradient (sometimes denoted as a slashed $$\partial$$).

## Definition 1.6: Timelike and spacelike components of a four-vector.

Given a four vector $$x = \gamma_\mu x^\mu$$, that would be designated $$x^\mu = \setlr{ x^0, \Bx}$$ in conventional special relativity, we write
\begin{equation*}
x^0 = x \cdot \gamma_0,
\end{equation*}
and
\begin{equation*}
\Bx = x \wedge \gamma_0,
\end{equation*}
or
\begin{equation*}
x = (x^0 + \Bx) \gamma_0.
\end{equation*}

The spacetime split of a four-vector $$x$$ is relative to the frame. In the relativistic lingo, one would say that it is “observer dependent”, as the same operations with $${\gamma_0}’$$, the timelike basis vector for a different frame, would yield a different set of coordinates.

While the dot and wedge products above provide an effective mechanism to split a four vector into a set of timelike and spacelike quantities, the spatial component of a vector has a bivector representation in STA. Consider the following coordinate expansion of a spatial vector
\label{eqn:lorentzForceCovariant:1000}
\Bx =
x \wedge \gamma_0
=
\lr{ x^\mu \gamma_\mu } \wedge \gamma_0
=
\sum_{k = 1}^3 x^k \gamma_k \gamma_0.

## Definition 1.7: Spatial basis.

We designate
\label{eqn:lorentzForceCovariant:1560}
\Be_i = \gamma_i \gamma_0,

as the standard basis vectors for $$\mathbb{R}^3$$.

In the literature, this bivector representation of the spatial basis may be designated $$\sigma_i = \gamma_i \gamma_0$$, as these bivectors have the properties of the Pauli matrices $$\sigma_i$$. Because I intend to expand these notes to include purely non-relativistic applications, I won’t use the Pauli notation here.

## Problem: Orthonormality of the spatial basis.

Show that the spatial basis $$\setlr{ \Be_1, \Be_2, \Be_3 }$$, defined by \ref{eqn:lorentzForceCovariant:1560}, is orthonormal.

\label{eqn:lorentzForceCovariant:620}
\begin{aligned}
\Be_i \cdot \Be_j
&= \gpgradezero{ \gamma_i \gamma_0 \gamma_j \gamma_0 } \\
&= -\gpgradezero{ \gamma_i \gamma_j } \\
&= – \gamma_i \cdot \gamma_j.
\end{aligned}

This is zero for all $$i \ne j$$, and unity for any $$i = j$$.

## Problem: Spatial pseudoscalar.

Show that the STA pseudoscalar $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$ equals the spatial pseudoscalar $$I = \Be_1 \Be_2 \Be_3$$.

The spatial pseudoscalar, expanded in terms of the STA basis vectors, is
\label{eqn:lorentzForceCovariant:1020}
\begin{aligned}
I
&= \Be_1 \Be_2 \Be_3 \\
&= \lr{ \gamma_1 \gamma_0 }
\lr{ \gamma_2 \gamma_0 }
\lr{ \gamma_3 \gamma_0 } \\
&= \lr{ \gamma_1 \gamma_0 } \gamma_2 \lr{ \gamma_0 \gamma_3 } \gamma_0 \\
&= \lr{ -\gamma_0 \gamma_1 } \gamma_2 \lr{ -\gamma_3 \gamma_0 } \gamma_0 \\
&= \gamma_0 \gamma_1 \gamma_2 \gamma_3 \lr{ \gamma_0 \gamma_0 } \\
&= \gamma_0 \gamma_1 \gamma_2 \gamma_3,
\end{aligned}

as claimed.

## Problem: Characteristics of the Pauli matrices.

The Pauli matrices obey the following anticommutation relations:
\label{eqn:lorentzForceCovariant:660}
\symmetric{ \sigma_a}{\sigma_b } = 2 \delta_{a b},

and commutation relations:
\label{eqn:lorentzForceCovariant:640}
\antisymmetric{ \sigma_a}{ \sigma_b } = 2 i \epsilon_{a b c}\,\sigma_c,

Show how these relate to the geometric algebra dot and wedge products, and determine the geometric algebra representation of the imaginary $$i$$ above.

## Euler-Lagrange equations.

I’ll start at ground zero, with the derivation of the relativistic form of the Euler-Lagrange equations from the action. A relativistic action for a single particle system has the form
\label{eqn:lorentzForceCovariant:20}
S = \int d\tau L(x, \dot{x}),

where $$x$$ is the spacetime coordinate, $$\dot{x} = dx/d\tau$$ is the four-velocity, and $$\tau$$ is proper time.

## Theorem 1.2: Relativistic Euler-Lagrange equations.

Let $$x \rightarrow x + \delta x$$ be any variation of the Lagrangian four-vector coordinates, where $$\delta x = 0$$ at the boundaries of the action integral. The variation of the action is
\label{eqn:lorentzForceCovariant:1580}
\delta S = \int d\tau \delta x \cdot \delta L(x, \dot{x}),

where
\label{eqn:lorentzForceCovariant:1600}

where $$\grad = \gamma^\mu \partial_\mu$$, and where we construct a similar velocity-gradient with respect to the proper-time derivatives of the coordinates $$\grad_v = \gamma^\mu \partial/\partial \dot{x}^\mu$$.The action is extremized when $$\delta S = 0$$, or when $$\delta L = 0$$. This latter condition is called the Euler-Lagrange equations.

### Start proof:

Let $$\epsilon = \delta x$$, and expand the Lagrangian in Taylor series to first order
\label{eqn:lorentzForceCovariant:60}
\begin{aligned}
S &\rightarrow S + \delta S \\
&= \int d\tau L( x + \epsilon, \dot{x} + \dot{\epsilon})
&=
\int d\tau \lr{
L(x, \dot{x}) + \epsilon \cdot \grad L + \dot{\epsilon} \cdot \grad_v L
}.
\end{aligned}

Subtracting off $$S$$ and integrating by parts, leaves
\label{eqn:lorentzForceCovariant:80}
\delta S =
\int d\tau \epsilon \cdot \lr{
}
+
\int d\tau \frac{d}{d\tau} (\grad_v L ) \cdot \epsilon.

The boundary integral
\label{eqn:lorentzForceCovariant:100}
\int d\tau \frac{d}{d\tau} (\grad_v L ) \cdot \epsilon
=
\evalbar{(\grad_v L ) \cdot \epsilon}{\Delta \tau} = 0,

is zero since the variation $$\epsilon$$ is required to vanish on the boundaries. So, if $$\delta S = 0$$, we must have
\label{eqn:lorentzForceCovariant:120}
0 =
\int d\tau \epsilon \cdot \lr{
},

for all variations $$\epsilon$$. Clearly, this requires that
\label{eqn:lorentzForceCovariant:140}

or
\label{eqn:lorentzForceCovariant:145}

which is the coordinate free statement of the Euler-Lagrange equations.

## Problem: Coordinate form of the Euler-Lagrange equations.

Working in coordinates, use the action argument show that the Euler-Lagrange equations have the form
\begin{equation*}
\PD{x^\mu}{L} = \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}
\end{equation*}
Observe that this is identical to the statement of \ref{eqn:lorentzForceCovariant:1600} after contraction with $$\gamma^\mu$$.

In terms of coordinates, the first order Taylor expansion of the action is
\label{eqn:lorentzForceCovariant:180}
\begin{aligned}
S &\rightarrow S + \delta S \\
&= \int d\tau L( x^\alpha + \epsilon^\alpha, \dot{x}^\alpha + \dot{\epsilon}^\alpha) \\
&=
\int d\tau \lr{
L(x^\alpha, \dot{x}^\alpha) + \epsilon^\mu \PD{x^\mu}{L} + \dot{\epsilon}^\mu \PD{\dot{x}^\mu}{L}
}.
\end{aligned}

As before, we integrate by parts to separate out a pure boundary term
\label{eqn:lorentzForceCovariant:200}
\delta S =
\int d\tau \epsilon^\mu
\lr{
\PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}
}
+
\int d\tau \frac{d}{d\tau} \lr{
\epsilon^\mu \PD{\dot{x}^\mu}{L}
}.

The boundary term is killed since $$\epsilon^\mu = 0$$ at the end points of the action integral. We conclude that extremization of the action ($$\delta S = 0$$, for all $$\epsilon^\mu$$) requires
\label{eqn:lorentzForceCovariant:220}
\PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L} = 0.

## Theorem 1.3: Lorentz force.

The relativistic Lagrangian for a charged particle is
\label{eqn:lorentzForceCovariant:1640}
L = \inv{2} m v^2 + q A \cdot v/c.

Application of the Euler-Lagrange equations to this Lagrangian yields the Lorentz-force equation
\label{eqn:lorentzForceCovariant:1660}
\frac{dp}{d\tau} = q F \cdot v/c,

where $$p = m v$$ is the proper momentum, $$F$$ is the Faraday bivector $$F = \grad \wedge A$$, and $$c$$ is the speed of light.

### Start proof:

To make life easier, let’s take advantage of the linearity of the Lagrangian, and break it into the free particle Lagrangian $$L_0 = (1/2) m v^2$$ and a potential term $$L_1 = q A \cdot v/c$$. For the free particle case we have
\label{eqn:lorentzForceCovariant:240}
\begin{aligned}
\delta L_0
&= – \frac{d}{d\tau} (m v) \\
&= – \frac{dp}{d\tau}.
\end{aligned}

For the potential contribution we have
\label{eqn:lorentzForceCovariant:260}
\begin{aligned}
\delta L_1
&= \frac{q}{c} \lr{ \grad (A \cdot v) – \frac{d}{d\tau} \lr{ \grad_v (A \cdot v)} } \\
&= \frac{q}{c} \lr{ \grad (A \cdot v) – \frac{dA}{d\tau} }.
\end{aligned}

The proper time derivative can be evaluated using the chain rule
\label{eqn:lorentzForceCovariant:280}
\frac{dA}{d\tau}
=
\frac{\partial x^\mu}{\partial \tau} \partial_\mu A

Putting all the pieces back together we have
\label{eqn:lorentzForceCovariant:300}
\begin{aligned}
0
&= \delta L \\
&=
-\frac{dp}{d\tau} + \frac{q}{c} \lr{ \grad (A \cdot v) – (v \cdot \grad) A } \\
&=
-\frac{dp}{d\tau} + \frac{q}{c} \lr{ \grad \wedge A } \cdot v.
\end{aligned}

## Problem: Gradient of a squared position vector.

Show that
\begin{equation*}
\grad (a \cdot x) = a,
\end{equation*}
and
\begin{equation*}
\end{equation*}
It should be clear that the same ideas can be used for the velocity gradient, where we obtain $$\grad_v (v^2) = 2 v$$, and $$\grad_v (A \cdot v) = A$$, as used in the derivation above.

The first identity follows easily by expansion in coordinates
\label{eqn:lorentzForceCovariant:320}
\begin{aligned}
&=
\gamma^\mu \partial_\mu a_\alpha x^\alpha \\
&=
\gamma^\mu a_\alpha \delta_\mu^\alpha \\
&=
\gamma^\mu a_\mu \\
&=
a.
\end{aligned}

The second identity follows by linearity of the gradient
\label{eqn:lorentzForceCovariant:340}
\begin{aligned}
&=
&=
\evalbar{\lr{\grad (x \cdot a)}}{a = x}
+
\evalbar{\lr{\grad (b \cdot x)}}{b = x} \\
&=
\evalbar{a}{a = x}
+
\evalbar{b}{b = x} \\
&=
2x.
\end{aligned}

It is desirable to put this relativistic Lorentz force equation into the usual vector and tensor forms for comparison.

## Theorem 1.4: Tensor form of the Lorentz force equation.

The tensor form of the Lorentz force equation is
\label{eqn:lorentzForceCovariant:1620}
\frac{dp^\mu}{d\tau} = \frac{q}{c} F^{\mu\nu} v_\nu,

where the antisymmetric Faraday tensor is defined as $$F^{\mu\nu} = \partial^\mu A^\nu – \partial^\nu A^\mu$$.

### Start proof:

We have only to dot both sides with $$\gamma^\mu$$. On the left we have
\label{eqn:lorentzForceCovariant:380}
\gamma^\mu \cdot \frac{dp}{d\tau}
=
\frac{dp^\mu}{d\tau}.

On the right, we have
\label{eqn:lorentzForceCovariant:400}
\begin{aligned}
\gamma^\mu \cdot \lr{ \frac{q}{c} F \cdot v }
&=
\frac{q}{c} (( \grad \wedge A ) \cdot v ) \cdot \gamma^\mu \\
&=
\frac{q}{c} ( \grad ( A \cdot v ) – (v \cdot \grad) A ) \cdot \gamma^\mu \\
&=
\frac{q}{c} \lr{ (\partial^\mu A^\nu) v_\nu – v_\nu \partial^\nu A^\mu } \\
&=
\frac{q}{c} F^{\mu\nu} v_\nu.
\end{aligned}

## Problem: Tensor expansion of $$F$$.

An alternate way to demonstrate \ref{eqn:lorentzForceCovariant:1620} is to first expand $$F = \grad \wedge A$$ in terms of coordinates, an expansion that can be expressed in terms of a second rank tensor antisymmetric tensor $$F^{\mu\nu}$$. Find that expansion, and re-evaluate the dot products of \ref{eqn:lorentzForceCovariant:400} using that.

\label{eqn:lorentzForceCovariant:900}
\begin{aligned}
F &=
&=
\lr{ \gamma_\mu \partial^\mu } \wedge \lr{ \gamma_\nu A^\nu } \\
&=
\lr{ \gamma_\mu \wedge \gamma_\nu } \partial^\mu A^\nu.
\end{aligned}

To this we can use the usual tensor trick (add self to self, change indexes, and divide by two), to give
\label{eqn:lorentzForceCovariant:920}
\begin{aligned}
F &=
\inv{2} \lr{
\lr{ \gamma_\mu \wedge \gamma_\nu } \partial^\mu A^\nu
+
\lr{ \gamma_\nu \wedge \gamma_\mu } \partial^\nu A^\mu
} \\
&=
\inv{2}
\lr{ \gamma_\mu \wedge \gamma_\nu } \lr{
\partial^\mu A^\nu

\partial^\nu A^\mu
},
\end{aligned}

which is just
\label{eqn:lorentzForceCovariant:940}
F =
\inv{2} \lr{ \gamma_\mu \wedge \gamma_\nu } F^{\mu\nu}.

Now, let’s expand $$(F \cdot v) \cdot \gamma^\mu$$ to compare to the earlier expansion in terms of $$\grad$$ and $$A$$.
\label{eqn:lorentzForceCovariant:960}
\begin{aligned}
(F \cdot v) \cdot \gamma^\mu
&=
\inv{2}
F^{\alpha\nu}
\lr{ \lr{ \gamma_\alpha \wedge \gamma_\nu } \cdot \lr{ \gamma^\beta v_\beta } } \cdot \gamma^\mu \\
&=
\inv{2}
F^{\alpha\nu} v_\beta
\lr{
{\delta_\nu}^\beta {\gamma_\alpha}^\mu

{\delta_\alpha}^\beta {\gamma_\nu}^\mu
} \\
&=
\inv{2}
\lr{
F^{\mu\beta} v_\beta

F^{\beta\mu} v_\beta
} \\
&=
F^{\mu\nu} v_\nu.
\end{aligned}

This alternate expansion illustrates some of the connectivity between the geometric algebra approach and the traditional tensor formalism.

## Problem: Lorentz force direct tensor derivation.

Instead of using the geometric algebra form of the Lorentz force equation as a stepping stone, we may derive the tensor form from the Lagrangian directly, provided the Lagrangian is put into tensor form
\begin{equation*}
L = \inv{2} m v^\mu v_\mu + q A^\mu v_\mu /c.
\end{equation*}
Evaluate the Euler-Lagrange equations in coordinate form and compare to \ref{eqn:lorentzForceCovariant:1620}.

Let $$\delta_\mu L = \gamma_\mu \cdot \delta L$$, so that we can write the Euler-Lagrange equations as
\label{eqn:lorentzForceCovariant:460}
0 = \delta_\mu L = \PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}.

Operating on the kinetic term of the Lagrangian, we have
\label{eqn:lorentzForceCovariant:480}
\delta_\mu L_0 = – \frac{d}{d\tau} m v_\mu.

For the potential term
\label{eqn:lorentzForceCovariant:500}
\begin{aligned}
\delta_\mu L_1
&=
\frac{q}{c} \lr{
v_\nu \PD{x^\mu}{A^\nu} – \frac{d}{d\tau} A_\mu
} \\
&=
\frac{q}{c} \lr{
v_\nu \PD{x^\mu}{A^\nu} – \frac{dx_\alpha}{d\tau} \PD{x_\alpha}{ A_\mu }
} \\
&=
\frac{q}{c} v^\nu \lr{
\partial_\mu A_\nu – \partial_\nu A_\mu
} \\
&=
\frac{q}{c} v^\nu F_{\mu\nu}.
\end{aligned}

Putting the pieces together gives
\label{eqn:lorentzForceCovariant:520}
\frac{d}{d\tau} (m v_\mu) = \frac{q}{c} v^\nu F_{\mu\nu},

which is identical\footnote{Some minor index raising and lowering gymnastics are required.} to the tensor form that we found by expanding the geometric algebra form of Maxwell’s equation in coordinates.

## Theorem 1.5: Vector Lorentz force equation.

Relative to a fixed observer’s frame, the Lorentz force equation of \ref{eqn:lorentzForceCovariant:1660} splits into a spatial rate of change of momentum, and (timelike component) rate of change of energy, as follows
\label{eqn:lorentzForceCovariant:1680}
\begin{aligned}
\ddt{(\gamma m \Bv)} &= q \lr{ \BE + \Bv \cross \BB } \\
\ddt{(\gamma m c^2)} &= q \Bv \cdot \BE,
\end{aligned}

where $$F = \BE + I c \BB$$, $$\gamma = 1/\sqrt{1 – \Bv^2/c^2 }$$.

### Start proof:

The first step is to eliminate the proper time dependencies in the Lorentz force equation. Consider first the coordinate representation of an arbitrary position four-vector $$x$$
\label{eqn:lorentzForceCovariant:1140}
x = c t \gamma_0 + x^k \gamma_k.

The corresponding four-vector velocity is
\label{eqn:lorentzForceCovariant:1160}
v = \ddtau{x} = c \ddtau{t} \gamma_0 + \ddtau{t} \ddt{x^k} \gamma_k.

By construction, $$v^2 = c^2$$ is a Lorentz invariant quantity (this is one of the relativistic postulates), so the LHS of \ref{eqn:lorentzForceCovariant:1160} must have the same square. That is
\label{eqn:lorentzForceCovariant:1240}
c^2 = \lr{ \ddtau{t} }^2 \lr{ c^2 – \Bv^2 },

where $$\Bv = v \wedge \gamma_0$$. This shows that we may make the identification
\label{eqn:lorentzForceCovariant:1260}
\gamma = \ddtau{t} = \inv{1 – \Bv^2/c^2 },

and
\label{eqn:lorentzForceCovariant:1280}
\ddtau{} = \ddtau{t} \ddt{} = \gamma \ddt{}.

We may now factor the four-velocity $$v$$ into its spacetime split
\label{eqn:lorentzForceCovariant:1300}
v = \gamma \lr{ c + \Bv } \gamma_0.

In particular the LHS of the Lorentz force equation can be rewritten as
\label{eqn:lorentzForceCovariant:1320}
\ddtau{p} = \gamma \ddt{}\lr{ \gamma \lr{ c + \Bv } } \gamma_0,

and the RHS of the Lorentz force equation can be rewritten as
\label{eqn:lorentzForceCovariant:1340}
\frac{q}{c} F \cdot v
=
\frac{\gamma q}{c} F \cdot \lr{ (c + \Bv) \gamma_0 }.

Equating timelike and spacelike components leaves us
\label{eqn:lorentzForceCovariant:1380}
\ddt{ (m \gamma c) } = \frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \cdot \gamma_0,

\label{eqn:lorentzForceCovariant:1400}
\ddt{ (m \gamma \Bv) } = \frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \wedge \gamma_0,

Evaluating these products requires some care, but is an essentially manual process. The reader is encouraged to do so once, but the end result may also be obtained easily using software (see lorentzForce.nb in [2]). One finds
\label{eqn:lorentzForceCovariant:1440}
F = \BE + I c \BB
=
E^1 \gamma_{10} +
+ E^2 \gamma_{20} +
+ E^3 \gamma_{30} +
– c B^1 \gamma_{23} +
– c B^2 \gamma_{31} +
– c B^3 \gamma_{12},

\label{eqn:lorentzForceCovariant:1460}
\frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \cdot \gamma_0
= \frac{q}{c} \BE \cdot \Bv,

\label{eqn:lorentzForceCovariant:1480}
\frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \wedge \gamma_0
= q \lr{ \BE + \Bv \cross \BB }.

## Problem: Algebraic spacetime split of the Lorentz force equation.

Derive the results of \ref{eqn:lorentzForceCovariant:1440} through \ref{eqn:lorentzForceCovariant:1480} algebraically.

## Problem: Spacetime split of the Lorentz force tensor equation.

Show that \ref{eqn:lorentzForceCovariant:1680} also follows from the tensor form of the Lorentz force equation (\ref{eqn:lorentzForceCovariant:1620}) provided we identify
\label{eqn:lorentzForceCovariant:1500}
F^{k0} = E^k,

and
\label{eqn:lorentzForceCovariant:1520}
F^{rs} = -\epsilon^{rst} B^t.

Also verify that the identifications of \ref{eqn:lorentzForceCovariant:1500} and \ref{eqn:lorentzForceCovariant:1520} is consistent with the geometric algebra Faraday bivector $$F = \BE + I c \BB$$, and the associated coordinate expansion of the field $$F = (1/2) (\gamma_\mu \wedge \gamma_\nu) F^{\mu\nu}$$.

# References

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

[2] Peeter Joot. Mathematica modules for Geometric Algebra’s GA(2,0), GA(3,0), and GA(1,3), 2017. URL https://github.com/peeterjoot/gapauli. [Online; accessed 24-Oct-2020].

## What’s in the pipe.

It’s been a while since I did any math or physics writing. This is the first post in a series where I plan to work my way systematically from an introduction of vectors, to the axioms of geometric algebra.  I plan to start with an introduction of vectors as directed “arrows”, building on that to discuss coordinates, tuples, and column matrix representations, and representation independent ideas. With those basics established, I’ll remind the reader about how generalized vector and dot product spaces are defined and give some examples. Finally, with the foundation of vectors and vector spaces in place, I’ll introduce the concept of a multivector space, and the geometric product, and start unpacking the implications of the axioms that follow naturally from this train of thought.

The applications that I plan to include in this series will be restricted to Euclidean spaces (i.e. where length is given by the Pythagorean law), primarily those of 2 and 3 dimensions.  However, it will be good to also lay the foundations for the non-Euclidean spaces that we encounter in relativistic electromagnetism (there is actually no other kind), and in computer graphics applications of geometric algebra, especially since we can do so nearly for free.  I plan to try to introduce the requisite ideas (i.e. the metric, which allows for a generalized dot product) by discussing Euclidean non-orthonormal bases.  Such bases have applications in condensed matter physics where there are useful for modelling crystal and lattice structure, and provide a hands conceptual bridge to a set of ideas that might otherwise seem abstract and without “real world” application.

## Motivation.

Many introductions to geometric algebra start by first introducing the dot product, then bivectors and the wedge product, and eventually define the product of two vectors as the synthetic sum of the dot and wedge
\label{eqn:multivector:20}
\Bx \By = \Bx \cdot \By + \Bx \wedge \By.

It takes a fair amount of work to do this well. In the seminal work [4] a few pages are taken for each of the dot and wedge products, showing the similarities and building up ideas, before introducing the geometric product in this fashion. In [2] the authors take a phenomenal five chapters to build up the context required to introduce the geometric product.  I am not disparaging the authors for taking that long to build up the ideas, as their introduction of the subject is exceedingly clear and thorough, and they do a lot more than the minumum required to define the geometric product.

The strategy to introduce the geometric product as a sum of dot and wedge can result in considerable confusion, especially since the wedge product is often defined in terms of the geometric product
\label{eqn:multivector:40}
\Bx \wedge \By =
\inv{2} \lr{
\Bx \By – \By \Bx
}.

The whole subject can appear like a chicken and egg problem. I personally found the subject very confusing initially, and had considerable difficulty understanding which of the many identities of geometric algebra were the most fundamental. For this reason, I found the axiomatic approach of [1] very refreshing. The cavaet with that work is that is is exceptionally terse, as they jammed a reformulation of most of physics using geometric algebra into that single book, and it would have been thousands of pages had they tried to make it readable by mere mortals.

When I wrote my own book on geometric algebra, I had the intuition that the way to introduce the subject ought to be like the vector space in abstract linear algebra. The construct of a vector space is a curious and indirect way to define a vector. Vectors are not defined as entities, but simply as members of a vector space, a space that is required to have a set of properties. I thought that the same approach would probably work with multivectors, which could be defined as members of a multivector space, a mathematical construction with a set of properties.

I did try this approach, but was not fully satisfied with what I wrote. I think that dissatisfaction was because I tried to define the multivector first. To define the multivector, I first introduced a whole set of prerequisite ideas (bivector, trivector, blade, k-vector, vector product, …), but that was also problematic, since the vector multiplication idea required for those concepts wasn’t fully defined until the multivector space itself was defined.

My approach shows some mathematical cowardness. Had I taken the approach of the vector space fully to heart, the multivector could have been defined as a member of a multivector space, and all the other ideas follow from that. In this multi-part series, I’m going to play with this approach anew, and see how it works out.  If it does work, I’ll see if I can incorporate this approach into a new version of my book.

## Review and background.

In this series, I’m going to assume a reader interested in geometric algebra, is probably also familiar with a wide variety of concepts, including but not limited to

• vectors,
• coordinates,
• matrices,
• basis,
• change of basis,
• dot product,
• real and complex numbers,
• rotations and translations,
• vector spaces, and
• linear transformations.

Despite those assumptions, as mentioned above, I’m going to attempt to build up the basics of vector representation and vector spaces in a systematic fashion, starting from a very elementary level.

My reasons for doing so are mainly to explore the logical sequencing of the ideas required.  I’ve always found well crafted pedagogical sequences rewarding, and will hopefully construct one here that is appreciated by anybody who chooses to follow along.

## Next time.

As preparation for the next article in this series, the reader is asked to watch a short lesson from Vector, not so supervillain extraordinaire (Despicable Me).

## References

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

[2] L. Dorst, D. Fontijne, and S. Mann. Geometric Algebra for Computer Science. Morgan Kaufmann, San Francisco, 2007.

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