Fundamental theorem of geometric calculus for line integrals (relativistic.)

December 16, 2020 math and physics play area element, bivector, boost, cross product, dot product, Fundamental Theorem of Calculus, Fundamental Theorem of Geometric Calculus, fundamental theorem of integral calculus, Geometric Calculus, line element, line integral, Lorentz invariance, multivector, multivector integral, normal, parameterization, pseudoscalar, reciprocal frame, reciprocal frame vector, scalar selection, space time algebra, spacetime, STA, tangent space, tangent vector, vector integral, volume element, wedge product

[This post is best viewed in PDF form, due to latex elements that I could not format with wordpress mathjax.]

Background for this particular post can be found in

Motivation.

I’ve been slowly working my way towards a statement of the fundamental theorem of integral calculus, where the functions being integrated are elements of the Dirac algebra (space time multivectors in the geometric algebra parlance.)

This is interesting because we want to be able to do line, surface, 3-volume and 4-volume space time integrals. We have many \(\mathbb{R}^3\) integral theorems
\begin{equation}\label{eqn:fundamentalTheoremOfGC:40a}
\int_A^B d\Bl \cdot \spacegrad f = f(B) – f(A),
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:60a}
\int_S dA\, \ncap \cross \spacegrad f = \int_{\partial S} d\Bx\, f,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:80a}
\int_S dA\, \ncap \cdot \lr{ \spacegrad \cross \Bf} = \int_{\partial S} d\Bx \cdot \Bf,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:100a}
\int_S dx dy \lr{ \PD{y}{P} – \PD{x}{Q} }
=
\int_{\partial S} P dx + Q dy,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:120a}
\int_V dV\, \spacegrad f = \int_{\partial V} dA\, \ncap f,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:140a}
\int_V dV\, \spacegrad \cross \Bf = \int_{\partial V} dA\, \ncap \cross \Bf,
\end{equation}
\begin{equation}\label{eqn:fundamentalTheoremOfGC:160a}
\int_V dV\, \spacegrad \cdot \Bf = \int_{\partial V} dA\, \ncap \cdot \Bf,
\end{equation}
and want to know how to generalize these to four dimensions and also make sure that we are handling the relativistic mixed signature correctly. If our starting point was the mess of equations above, we’d be in trouble, since it is not obvious how these generalize. All the theorems with unit normals have to be handled completely differently in four dimensions since we don’t have a unique normal to any given spacetime plane.
What comes to our rescue is the Fundamental Theorem of Geometric Calculus (FTGC), which has the form
\begin{equation}\label{eqn:fundamentalTheoremOfGC:40}
\int F d^n \Bx\, \lrpartial G = \int F d^{n-1} \Bx\, G,
\end{equation}
where \(F,G\) are multivectors functions (i.e. sums of products of vectors.) We’ve seen ([2], [1]) that all the identities above are special cases of the fundamental theorem.

Do we need any special care to state the FTGC correctly for our relativistic case? It turns out that the answer is no! Tangent and reciprocal frame vectors do all the heavy lifting, and we can use the fundamental theorem as is, even in our mixed signature space. The only real change that we need to make is use spacetime gradient and vector derivative operators instead of their spatial equivalents. We will see how this works below. Note that instead of starting with \ref{eqn:fundamentalTheoremOfGC:40} directly, I will attempt to build up to that point in a progressive fashion that is hopefully does not require the reader to make too many unjustified mental leaps.

Multivector line integrals.

We want to define multivector line integrals to start with. Recall that in \(\mathbb{R}^3\) we would say that for scalar functions \( f\), the integral
\begin{equation}\label{eqn:fundamentalTheoremOfGC:180b}
\int d\Bx\, f = \int f d\Bx,
\end{equation}
is a line integral. Also, for vector functions \( \Bf \) we call
\begin{equation}\label{eqn:fundamentalTheoremOfGC:200}
\int d\Bx \cdot \Bf = \inv{2} \int d\Bx\, \Bf + \Bf d\Bx.
\end{equation}
a line integral. In order to generalize line integrals to multivector functions, we will allow our multivector functions to be placed on either or both sides of the differential.

Definition 1.1: Line integral.

Given a single variable parameterization \( x = x(u) \), we write \( d^1\Bx = \Bx_u du \), and call
\begin{equation}\label{eqn:fundamentalTheoremOfGC:220a}
\int F d^1\Bx\, G,
\end{equation}
a line integral, where \( F,G \) are arbitrary multivector functions.

We must be careful not to reorder any of the factors in the integrand, since the differential may not commute with either \( F \) or \( G \). Here is a simple example where the integrand has a product of a vector and differential.

Problem: Circular parameterization.

Given a circular parameterization \( x(\theta) = \gamma_1 e^{-i\theta} \), where \( i = \gamma_1 \gamma_2 \), the unit bivector for the \(x,y\) plane. Compute the line integral
\begin{equation}\label{eqn:fundamentalTheoremOfGC:100}
\int_0^{\pi/4} F(\theta)\, d^1 \Bx\, G(\theta),
\end{equation}
where \( F(\theta) = \Bx^\theta + \gamma_3 + \gamma_1 \gamma_0 \) is a multivector valued function, and \( G(\theta) = \gamma_0 \) is vector valued.

Answer

The tangent vector for the curve is
\begin{equation}\label{eqn:fundamentalTheoremOfGC:60}
\Bx_\theta
= -\gamma_1 \gamma_1 \gamma_2 e^{-i\theta}
= \gamma_2 e^{-i\theta},
\end{equation}
with reciprocal vector \( \Bx^\theta = e^{i \theta} \gamma^2 \). The differential element is \( d^1 \Bx = \gamma_2 e^{-i\theta} d\theta \), so the integrand is
\begin{equation}\label{eqn:fundamentalTheoremOfGC:80}
\begin{aligned}
\int_0^{\pi/4} \lr{ \Bx^\theta + \gamma_3 + \gamma_1 \gamma_0 } d^1 \Bx\, \gamma_0
&=
\int_0^{\pi/4} \lr{ e^{i\theta} \gamma^2 + \gamma_3 + \gamma_1 \gamma_0 } \gamma_2 e^{-i\theta} d\theta\, \gamma_0 \\
&=
\frac{\pi}{4} \gamma_0 + \lr{ \gamma_{32} + \gamma_{102} } \inv{-i} \lr{ e^{-i\pi/4} – 1 } \gamma_0 \\
&=
\frac{\pi}{4} \gamma_0 + \inv{\sqrt{2}} \lr{ \gamma_{32} + \gamma_{102} } \gamma_{120} \lr{ 1 – \gamma_{12} } \\
&=
\frac{\pi}{4} \gamma_0 + \inv{\sqrt{2}} \lr{ \gamma_{310} + 1 } \lr{ 1 – \gamma_{12} }.
\end{aligned}
\end{equation}
Observe how care is required not to reorder any terms. This particular end result is a multivector with scalar, vector, bivector, and trivector grades, but no pseudoscalar component. The grades in the end result depend on both the function in the integrand and on the path. For example, had we integrated all the way around the circle, the end result would have been the vector \( 2 \pi \gamma_0 \) (i.e. a \( \gamma_0 \) weighted unit circle circumference), as all the other grades would have been killed by the complex exponential integrated over a full period.

Problem: Line integral for boosted time direction vector.

Let \( x = e^{\vcap \alpha/2} \gamma_0 e^{-\vcap \alpha/2} \) represent the spacetime curve of all the boosts of \( \gamma_0 \) along a specific velocity direction vector, where \( \vcap = (v \wedge \gamma_0)/\Norm{v \wedge \gamma_0} \) is a unit spatial bivector for any constant vector \( v \). Compute the line integral
\begin{equation}\label{eqn:fundamentalTheoremOfGC:240}
\int x\, d^1 \Bx.
\end{equation}

Answer

Observe that \( \vcap \) and \( \gamma_0 \) anticommute, so we may write our boost as a one sided exponential
\begin{equation}\label{eqn:fundamentalTheoremOfGC:260}
x(\alpha) = \gamma_0 e^{-\vcap \alpha} = e^{\vcap \alpha} \gamma_0 = \lr{ \cosh\alpha + \vcap \sinh\alpha } \gamma_0.
\end{equation}
The tangent vector is just
\begin{equation}\label{eqn:fundamentalTheoremOfGC:280}
\Bx_\alpha = \PD{\alpha}{x} = e^{\vcap\alpha} \vcap \gamma_0.
\end{equation}
Let’s get a bit of intuition about the nature of this vector. It’s square is
\begin{equation}\label{eqn:fundamentalTheoremOfGC:300}
\begin{aligned}
\Bx_\alpha^2
&=
e^{\vcap\alpha} \vcap \gamma_0
e^{\vcap\alpha} \vcap \gamma_0 \\
&=
-e^{\vcap\alpha} \vcap e^{-\vcap\alpha} \vcap (\gamma_0)^2 \\
&=
-1,
\end{aligned}
\end{equation}
so we see that the tangent vector is a spacelike unit vector. As the vector representing points on the curve is necessarily timelike (due to Lorentz invariance), these two must be orthogonal at all points. Let’s confirm this algebraically
\begin{equation}\label{eqn:fundamentalTheoremOfGC:320}
\begin{aligned}
x \cdot \Bx_\alpha
&=
\gpgradezero{ e^{\vcap \alpha} \gamma_0 e^{\vcap \alpha} \vcap \gamma_0 } \\
&=
\gpgradezero{ e^{-\vcap \alpha} e^{\vcap \alpha} \vcap (\gamma_0)^2 } \\
&=
\gpgradezero{ \vcap } \\
&= 0.
\end{aligned}
\end{equation}
Here we used \( e^{\vcap \alpha} \gamma_0 = \gamma_0 e^{-\vcap \alpha} \), and \( \gpgradezero{A B} = \gpgradezero{B A} \). Geometrically, we have the curious fact that the direction vectors to points on the curve are perpendicular (with respect to our relativistic dot product) to the tangent vectors on the curve, as illustrated in fig. 1.

fig. 1. Tangent perpendicularity in mixed metric.

Perfect differentials.

Having seen a couple examples of multivector line integrals, let’s now move on to figure out the structure of a line integral that has a “perfect” differential integrand. We can take a hint from the \(\mathbb{R}^3\) vector result that we already know, namely
\begin{equation}\label{eqn:fundamentalTheoremOfGC:120}
\int_A^B d\Bl \cdot \spacegrad f = f(B) – f(A).
\end{equation}
It seems reasonable to guess that the relativistic generalization of this is
\begin{equation}\label{eqn:fundamentalTheoremOfGC:140}
\int_A^B dx \cdot \grad f = f(B) – f(A).
\end{equation}
Let’s check that, by expanding in coordinates
\begin{equation}\label{eqn:fundamentalTheoremOfGC:160}
\begin{aligned}
\int_A^B dx \cdot \grad f
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \partial_\mu f \\
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \PD{x^\mu}{f} \\
&=
\int_A^B d\tau \frac{df}{d\tau} \\
&=
f(B) – f(A).
\end{aligned}
\end{equation}
If we drop the dot product, will we have such a nice result? Let’s see:
\begin{equation}\label{eqn:fundamentalTheoremOfGC:180}
\begin{aligned}
\int_A^B dx \grad f
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \gamma_\mu \gamma^\nu \partial_\nu f \\
&=
\int_A^B d\tau \frac{dx^\mu}{d\tau} \PD{x^\mu}{f}
+
\int_A^B
d\tau
\sum_{\mu \ne \nu} \gamma_\mu \gamma^\nu
\frac{dx^\mu}{d\tau} \PD{x^\nu}{f}.
\end{aligned}
\end{equation}
This scalar component of this integrand is a perfect differential, but the bivector part of the integrand is a complete mess, that we have no hope of generally integrating. It happens that if we consider one of the simplest parameterization examples, we can get a strong hint of how to generalize the differential operator to one that ends up providing a perfect differential. In particular, let’s integrate over a linear constant path, such as \( x(\tau) = \tau \gamma_0 \). For this path, we have
\begin{equation}\label{eqn:fundamentalTheoremOfGC:200a}
\begin{aligned}
\int_A^B dx \grad f
&=
\int_A^B \gamma_0 d\tau \lr{
\gamma^0 \partial_0 +
\gamma^1 \partial_1 +
\gamma^2 \partial_2 +
\gamma^3 \partial_3 } f \\
&=
\int_A^B d\tau \lr{
\PD{\tau}{f} +
\gamma_0 \gamma^1 \PD{x^1}{f} +
\gamma_0 \gamma^2 \PD{x^2}{f} +
\gamma_0 \gamma^3 \PD{x^3}{f}
}.
\end{aligned}
\end{equation}
Just because the path does not have any \( x^1, x^2, x^3 \) component dependencies does not mean that these last three partials are neccessarily zero. For example \( f = f(x(\tau)) = \lr{ x^0 }^2 \gamma_0 + x^1 \gamma_1 \) will have a non-zero contribution from the \( \partial_1 \) operator. In that particular case, we can easily integrate \( f \), but we have to know the specifics of the function to do the integral. However, if we had a differential operator that did not include any component off the integration path, we would ahve a perfect differential. That is, if we were to replace the gradient with the projection of the gradient onto the tangent space, we would have a perfect differential. We see that the function of the dot product in \ref{eqn:fundamentalTheoremOfGC:140} has the same effect, as it rejects any component of the gradient that does not lie on the tangent space.

Definition 1.2: Vector derivative.

Given a spacetime manifold parameterized by \( x = x(u^0, \cdots u^{N-1}) \), with tangent vectors \( \Bx_\mu = \PDi{u^\mu}{x} \), and reciprocal vectors \( \Bx^\mu \in \textrm{Span}\setlr{\Bx_\nu} \), such that \( \Bx^\mu \cdot \Bx_\nu = {\delta^\mu}_\nu \), the vector derivative is defined as
\begin{equation}\label{eqn:fundamentalTheoremOfGC:240a}
\partial = \sum_{\mu = 0}^{N-1} \Bx^\mu \PD{u^\mu}{}.
\end{equation}
Observe that if this is a full parameterization of the space (\(N = 4\)), then the vector derivative is identical to the gradient. The vector derivative is the projection of the gradient onto the tangent space at the point of evaluation.Furthermore, we designate \( \lrpartial \) as the vector derivative allowed to act bidirectionally, as follows
\begin{equation}\label{eqn:fundamentalTheoremOfGC:260a}
R \lrpartial S
=
R \Bx^\mu \PD{u^\mu}{S}
+
\PD{u^\mu}{R} \Bx^\mu S,
\end{equation}
where \( R, S \) are multivectors, and summation convention is implied. In this bidirectional action,
the vector factors of the vector derivative must stay in place (as they do not neccessarily commute with \( R,S\)), but the derivative operators apply in a chain rule like fashion to both functions.

Noting that \( \Bx_u \cdot \grad = \Bx_u \cdot \partial \), we may rewrite the scalar line integral identity \ref{eqn:fundamentalTheoremOfGC:140} as
\begin{equation}\label{eqn:fundamentalTheoremOfGC:220}
\int_A^B dx \cdot \partial f = f(B) – f(A).
\end{equation}
However, as our example hinted at, the fundamental theorem for line integrals has a multivector generalization that does not rely on a dot product to do the tangent space filtering, and is more powerful. That generalization has the following form.

Theorem 1.1: Fundamental theorem for line integrals.

Given multivector functions \( F, G \), and a single parameter curve \( x(u) \) with line element \( d^1 \Bx = \Bx_u du \), then
\begin{equation}\label{eqn:fundamentalTheoremOfGC:280a}
\int_A^B F d^1\Bx \lrpartial G = F(B) G(B) – F(A) G(A).
\end{equation}

Start proof:

Writing out the integrand explicitly, we find
\begin{equation}\label{eqn:fundamentalTheoremOfGC:340}
\int_A^B F d^1\Bx \lrpartial G
=
\int_A^B \lr{
\PD{\alpha}{F} d\alpha\, \Bx_\alpha \Bx^\alpha G
+
F d\alpha\, \Bx_\alpha \Bx^\alpha \PD{\alpha}{G }
}
\end{equation}
However for a single parameter curve, we have \( \Bx^\alpha = 1/\Bx_\alpha \), so we are left with
\begin{equation}\label{eqn:fundamentalTheoremOfGC:360}
\begin{aligned}
\int_A^B F d^1\Bx \lrpartial G
&=
\int_A^B d\alpha\, \PD{\alpha}{(F G)} \\
&=
\evalbar{F G}{B}
–
\evalbar{F G}{A}.
\end{aligned}
\end{equation}

End proof.

More to come.

In the next installment we will explore surface integrals in spacetime, and the generalization of the fundamental theorem to multivector space time integrals.

References

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

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

Lorentz transformations in Space Time Algebra (STA)

December 12, 2020 math and physics play bivector, boost, Dirac algebra, dot product, exponential, four vector, Geometric Algebra, grade selection, Lorentz transform, projection, pseudoscalar, rejection, rotation, space time algebra, spacetime plane, spatial bivector, spatial plane, STA, wedge product

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

Motivation.

One of the remarkable features of geometric algebra are the complex exponential sandwiches that can be used to encode rotations in any dimension, or rotation like operations like Lorentz transformations in Minkowski spaces. In this post, we show some examples that unpack the geometric algebra expressions for Lorentz transformations operations of this sort. In particular, we will look at the exponential sandwich operations for spatial rotations and Lorentz boosts in the Dirac algebra, known as Space Time Algebra (STA) in geometric algebra circles, and demonstrate that these sandwiches do have the desired effects.

Lorentz transformations.

Theorem 1.1: Lorentz transformation.

The transformation
\begin{equation}\label{eqn:lorentzTransform:580}
x \rightarrow e^{B} x e^{-B} = x’,
\end{equation}
where \( B = a \wedge b \), is an STA 2-blade for any two linearly independent four-vectors \( a, b \), is a norm preserving, that is
\begin{equation}\label{eqn:lorentzTransform:600}
x^2 = {x’}^2.
\end{equation}

Start proof:

The proof is disturbingly trivial in this geometric algebra form
\begin{equation}\label{eqn:lorentzTransform:40}
\begin{aligned}
{x’}^2
&=
e^{B} x e^{-B} e^{B} x e^{-B} \\
&=
e^{B} x x e^{-B} \\
&=
x^2 e^{B} e^{-B} \\
&=
x^2.
\end{aligned}
\end{equation}

End proof.

In particular, observe that we did not need to construct the usual infinitesimal representations of rotation and boost transformation matrices or tensors in order to demonstrate that we have spacetime invariance for the transformations. The rough idea of such a transformation is that the exponential commutes with components of the four-vector that lie off the spacetime plane specified by the bivector \( B \), and anticommutes with components of the four-vector that lie in the plane. The end result is that the sandwich operation simplifies to
\begin{equation}\label{eqn:lorentzTransform:60}
x’ = x_\parallel e^{-B} + x_\perp,
\end{equation}
where \( x = x_\perp + x_\parallel \) and \( x_\perp \cdot B = 0 \), and \( x_\parallel \wedge B = 0 \). In particular, using \( x = x B B^{-1} = \lr{ x \cdot B + x \wedge B } B^{-1} \), we find that
\begin{equation}\label{eqn:lorentzTransform:80}
\begin{aligned}
x_\parallel &= \lr{ x \cdot B } B^{-1} \\
x_\perp &= \lr{ x \wedge B } B^{-1}.
\end{aligned}
\end{equation}
When \( B \) is a spacetime plane \( B = b \wedge \gamma_0 \), then this exponential has a hyperbolic nature, and we end up with a Lorentz boost. When \( B \) is a spatial bivector, we end up with a single complex exponential, encoding our plane old 3D rotation. More general \( B \)’s that encode composite boosts and rotations are also possible, but \( B \) must be invertible (it should have no lightlike factors.) The rough geometry of these projections is illustrated in fig 1, where the spacetime plane is represented by \( B \).

fig 1. Projection and rejection geometry.

What is not so obvious is how to pick \( B \)’s that correspond to specific rotation axes or boost directions. Let’s consider each of those cases in turn.

Theorem 1.2: Boost.

The boost along a direction vector \( \vcap \) and rapidity \( \alpha \) is given by
\begin{equation}\label{eqn:lorentzTransform:620}
x’ = e^{-\vcap \alpha/2} x e^{\vcap \alpha/2},
\end{equation}
where \( \vcap = \gamma_{k0} \cos\theta^k \) is an STA bivector representing a spatial direction with direction cosines \( \cos\theta^k \).

Start proof:

We want to demonstrate that this is equivalent to the usual boost formulation. We can start with decomposition of the four-vector \( x \) into components that lie in and off of the spacetime plane \( \vcap \).
\begin{equation}\label{eqn:lorentzTransform:100}
\begin{aligned}
x
&= \lr{ x^0 + \Bx } \gamma_0 \\
&= \lr{ x^0 + \Bx \vcap^2 } \gamma_0 \\
&= \lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap + \lr{ \Bx \wedge \vcap} \vcap } \gamma_0,
\end{aligned}
\end{equation}
where \( \Bx = x \wedge \gamma_0 \). The first two components lie in the boost plane, whereas the last is the spatial component of the vector that lies perpendicular to the boost plane. Observe that \( \vcap \) anticommutes with the dot product term and commutes with he wedge product term, so we have
\begin{equation}\label{eqn:lorentzTransform:120}
\begin{aligned}
x’
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap } \vcap } \gamma_0
e^{\vcap \alpha/2 }
e^{\vcap \alpha/2 }
+
\lr{ \Bx \wedge \vcap } \vcap \gamma_0
e^{-\vcap \alpha/2 }
e^{\vcap \alpha/2 } \\
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap } \vcap } \gamma_0
e^{\vcap \alpha }
+
\lr{ \Bx \wedge \vcap } \vcap \gamma_0.
\end{aligned}
\end{equation}
Noting that \( \vcap^2 = 1 \), we may expand the exponential in hyperbolic functions, and find that the boosted portion of the vector expands as
\begin{equation}\label{eqn:lorentzTransform:260}
\begin{aligned}
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0 e^{\vcap \alpha}
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0 \lr{ \cosh\alpha + \vcap \sinh \alpha} \\
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \lr{ \cosh\alpha – \vcap \sinh \alpha} \gamma_0 \\
&=
\lr{ x^0 \cosh\alpha – \lr{ \Bx \cdot \vcap} \sinh \alpha} \gamma_0
+
\lr{ -x^0 \sinh \alpha + \lr{ \Bx \cdot \vcap} \cosh \alpha } \vcap \gamma_0.
\end{aligned}
\end{equation}
We are left with
\begin{equation}\label{eqn:lorentzTransform:320}
\begin{aligned}
x’
&=
\lr{ x^0 \cosh\alpha – \lr{ \Bx \cdot \vcap} \sinh \alpha} \gamma_0
+
\lr{ \lr{ \Bx \cdot \vcap} \cosh \alpha -x^0 \sinh \alpha } \vcap \gamma_0
+
\lr{ \Bx \wedge \vcap} \vcap \gamma_0 \\
&=
\begin{bmatrix}
\gamma_0 & \vcap \gamma_0
\end{bmatrix}
\begin{bmatrix}
\cosh\alpha & – \sinh\alpha \\
-\sinh\alpha & \cosh\alpha
\end{bmatrix}
\begin{bmatrix}
x^0 \\
\Bx \cdot \vcap
\end{bmatrix}
+
\lr{ \Bx \wedge \vcap} \vcap \gamma_0,
\end{aligned}
\end{equation}
which has the desired Lorentz boost structure. Of course, this is usually seen with \( \vcap = \gamma_{10} \) so that the components in the coordinate column vector are \( (ct, x) \).

End proof.

Theorem 1.3: Spatial rotation.

Given two linearly independent spatial bivectors \( \Ba = a^k \gamma_{k0}, \Bb = b^k \gamma_{k0} \), a rotation of \(\theta\) radians in the plane of \( \Ba, \Bb \) from \( \Ba \) towards \( \Bb \), is given by
\begin{equation}\label{eqn:lorentzTransform:640}
x’ = e^{-i\theta} x e^{i\theta},
\end{equation}
where \( i = (\Ba \wedge \Bb)/\Abs{\Ba \wedge \Bb} \), is a unit (spatial) bivector.

Start proof:

Without loss of generality, we may pick \( i = \acap \bcap \), where \( \acap^2 = \bcap^2 = 1 \), and \( \acap \cdot \bcap = 0 \). With such an orthonormal basis for the plane, we can decompose our four vector into portions that lie in and off the plane
\begin{equation}\label{eqn:lorentzTransform:400}
\begin{aligned}
x
&= \lr{ x^0 + \Bx } \gamma_0 \\
&= \lr{ x^0 + \Bx i i^{-1} } \gamma_0 \\
&= \lr{ x^0 + \lr{ \Bx \cdot i } i^{-1} + \lr{ \Bx \wedge i } i^{-1} } \gamma_0.
\end{aligned}
\end{equation}
The projective term lies in the plane of rotation, whereas the timelike and spatial rejection term are perpendicular. That is
\begin{equation}\label{eqn:lorentzTransform:420}
\begin{aligned}
x_\parallel &= \lr{ \Bx \cdot i } i^{-1} \gamma_0 \\
x_\perp &= \lr{ x^0 + \lr{ \Bx \wedge i } i^{-1} } \gamma_0,
\end{aligned}
\end{equation}
where \( x_\parallel \wedge i = 0 \), and \( x_\perp \cdot i = 0 \). The plane pseudoscalar \( i \) anticommutes with \( x_\parallel \), and commutes with \( x_\perp \), so
\begin{equation}\label{eqn:lorentzTransform:440}
\begin{aligned}
x’
&= e^{-i\theta/2} \lr{ x_\parallel + x_\perp } e^{i\theta/2} \\
&= x_\parallel e^{i\theta} + x_\perp.
\end{aligned}
\end{equation}
However
\begin{equation}\label{eqn:lorentzTransform:460}
\begin{aligned}
\lr{ \Bx \cdot i } i^{-1}
&=
\lr{ \Bx \cdot \lr{ \acap \wedge \bcap } } \bcap \acap \\
&=
\lr{\Bx \cdot \acap} \bcap \bcap \acap
-\lr{\Bx \cdot \bcap} \acap \bcap \acap \\
&=
\lr{\Bx \cdot \acap} \acap
+\lr{\Bx \cdot \bcap} \bcap,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:lorentzTransform:480}
\begin{aligned}
x_\parallel e^{i\theta}
&=
\lr{
\lr{\Bx \cdot \acap} \acap
+
\lr{\Bx \cdot \bcap} \bcap
}
\gamma_0
\lr{
\cos\theta + \acap \bcap \sin\theta
} \\
&=
\acap \lr{
\lr{\Bx \cdot \acap} \cos\theta
–
\lr{\Bx \cdot \bcap} \sin\theta
}
\gamma_0
+
\bcap \lr{
\lr{\Bx \cdot \acap} \sin\theta
+
\lr{\Bx \cdot \bcap} \cos\theta
}
\gamma_0,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:lorentzTransform:500}
x’
=
\begin{bmatrix}
\acap & \bcap
\end{bmatrix}
\begin{bmatrix}
\cos\theta & – \sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\Bx \cdot \acap \\
\Bx \cdot \bcap \\
\end{bmatrix}
\gamma_0
+
\lr{ x \wedge i} i^{-1} \gamma_0.
\end{equation}
Observe that this rejection term can be explicitly expanded to
\begin{equation}\label{eqn:lorentzTransform:520}
\lr{ \Bx \wedge i} i^{-1} \gamma_0 =
x –
\lr{ \Bx \cdot \acap } \acap \gamma_0
–
\lr{ \Bx \cdot \acap } \acap \gamma_0.
\end{equation}
This is the timelike component of the vector, plus the spatial component that is normal to the plane. This exponential sandwich transformation rotates only the portion of the vector that lies in the plane, and leaves the rest (timelike and normal) untouched.

End proof.

Problems.

Problem: Verify components relative to boost direction.

In the proof of thm. 1.2, the vector \( x \) was expanded in terms of the spacetime split. An alternate approach, is to expand as
\begin{equation}\label{eqn:lorentzTransform:340}
\begin{aligned}
x
&= x \vcap^2 \\
&= \lr{ x \cdot \vcap + x \wedge \vcap } \vcap \\
&= \lr{ x \cdot \vcap } \vcap + \lr{ x \wedge \vcap } \vcap.
\end{aligned}
\end{equation}
Show that
\begin{equation}\label{eqn:lorentzTransform:360}
\lr{ x \cdot \vcap } \vcap
=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0,
\end{equation}
and
\begin{equation}\label{eqn:lorentzTransform:380}
\lr{ x \wedge \vcap } \vcap
=
\lr{ \Bx \wedge \vcap} \vcap \gamma_0.
\end{equation}

Answer

Let \( x = x^\mu \gamma_\mu \), so that
\begin{equation}\label{eqn:lorentzTransform:160}
\begin{aligned}
x \cdot \vcap
&=
\gpgradeone{ x^\mu \gamma_\mu \cos\theta^b \gamma_{b 0} } \\
&=
x^\mu \cos\theta^b \gpgradeone{ \gamma_\mu \gamma_{b 0} }
.
\end{aligned}
\end{equation}
The \( \mu = 0 \) component of this grade selection is
\begin{equation}\label{eqn:lorentzTransform:180}
\gpgradeone{ \gamma_0 \gamma_{b 0} }
=
-\gamma_b,
\end{equation}
and for \( \mu = a \ne 0 \), we have
\begin{equation}\label{eqn:lorentzTransform:200}
\gpgradeone{ \gamma_a \gamma_{b 0} }
=
-\delta_{a b} \gamma_0,
\end{equation}
so we have
\begin{equation}\label{eqn:lorentzTransform:220}
\begin{aligned}
x \cdot \vcap
&=
x^0 \cos\theta^b (-\gamma_b)
+
x^a \cos\theta^b (-\delta_{ab} \gamma_0 ) \\
&=
-x^0 \vcap \gamma_0
–
x^b \cos\theta^b \gamma_0 \\
&=
– \lr{ x^0 \vcap + \Bx \cdot \vcap } \gamma_0,
\end{aligned}
\end{equation}
where \( \Bx = x \wedge \gamma_0 \) is the spatial portion of the four vector \( x \) relative to the stationary observer frame. Since \( \vcap \) anticommutes with \( \gamma_0 \), the component of \( x \) in the spacetime plane \( \vcap \) is
\begin{equation}\label{eqn:lorentzTransform:240}
\lr{ x \cdot \vcap } \vcap =
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0,
\end{equation}
as expected.

For the rejection term, we have
\begin{equation}\label{eqn:lorentzTransform:280}
x \wedge \vcap
=
x^\mu \cos\theta^s \gpgradethree{ \gamma_\mu \gamma_{s 0} }.
\end{equation}
The \( \mu = 0 \) term clearly contributes nothing, leaving us with:
\begin{equation}\label{eqn:lorentzTransform:300}
\begin{aligned}
\lr{ x \wedge \vcap } \vcap
&=
\lr{ x \wedge \vcap } \cdot \vcap \\
&=
x^r \cos\theta^s \cos\theta^t \lr{ \lr{ \gamma_r \wedge \gamma_{s}} \gamma_0 } \cdot \lr{ \gamma_{t0} } \\
&=
x^r \cos\theta^s \cos\theta^t \gpgradeone{
\lr{ \gamma_r \wedge \gamma_{s} } \gamma_0 \gamma_{t0}
} \\
&=
-x^r \cos\theta^s \cos\theta^t \lr{ \gamma_r \wedge \gamma_{s}} \cdot \gamma_t \\
&=
-x^r \cos\theta^s \cos\theta^t \lr{ -\gamma_r \delta_{st} + \gamma_s \delta_{rt} } \\
&=
x^r \cos\theta^t \cos\theta^t \gamma_r
–
x^t \cos\theta^s \cos\theta^t \gamma_s \\
&=
\Bx \gamma_0
– (\Bx \cdot \vcap) \vcap \gamma_0 \\
&=
\lr{ \Bx \wedge \vcap} \vcap \gamma_0,
\end{aligned}
\end{equation}
as expected. Is there a clever way to demonstrate this without resorting to coordinates?

Problem: Rotation transformation components.

Given a unit spatial bivector \( i = \acap \bcap \), where \( \acap \cdot \bcap = 0 \) and \( i^2 = -1 \), show that
\begin{equation}\label{eqn:lorentzTransform:540}
\lr{ x \cdot i } i^{-1}
=
\lr{ \Bx \cdot i } i^{-1} \gamma_0
=
\lr{\Bx \cdot \acap } \acap \gamma_0
+
\lr{\Bx \cdot \bcap } \bcap \gamma_0,
\end{equation}
and
\begin{equation}\label{eqn:lorentzTransform:560}
\lr{ x \wedge i } i^{-1}
=
\lr{ \Bx \wedge i } i^{-1} \gamma_0
=
x –
\lr{\Bx \cdot \acap } \acap \gamma_0
–
\lr{\Bx \cdot \bcap } \bcap \gamma_0.
\end{equation}
Also show that \( i \) anticommutes with \( \lr{ x \cdot i } i^{-1} \) and commutes with \( \lr{ x \wedge i } i^{-1} \).

Answer

This problem is left for the reader, as I don’t feel like typing out my solution.

The first part of this problem can be done in the tedious coordinate approach used above, but hopefully there is a better way.

For the last (commutation) part of the problem, here is a hint. Let \( x \wedge i = n i \), where \( n \cdot i = 0 \). The result then follows easily.

Maxwell’s equation Lagrangian (geometric algebra and tensor formalism)

November 1, 2020 math and physics play action, action principle, bivector, current density, dot product, Euler-Lagrange equations, four potential, four vector, Gauss-Ampere law, Gauss-Faraday law, interaction term, kinetic term, Lagrangian density, Lorentz invariant, Maxwell's equation, multivector, pseudoscalar, pseudoscalar selection, scalar selection, space time algebra, trivector, variation, wedge product

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Maxwell’s equation using geometric algebra Lagrangian.

Motivation.

In my classical mechanics notes, I’ve got computations of Maxwell’s equation (singular in it’s geometric algebra form) from a Lagrangian in various ways (using a tensor, scalar and multivector Lagrangians), but all of these seem more convoluted than they should be.
Here we do this from scratch, starting with the action principle for field variables, covering:

Derivation of the relativistic form of the Euler-Lagrange field equations from the covariant form of the action,
Derivation of Maxwell’s equation (in it’s STA form) from the Maxwell Lagrangian,
Relationship of the STA Maxwell Lagrangian to the tensor equivalent,
Relationship of the STA form of Maxwell’s equation to it’s tensor equivalents,
Relationship of the STA Maxwell’s equation to it’s conventional Gibbs form.
Show that we may use a multivector valued Lagrangian with all of \( F^2 \), not just the scalar part.

It is assumed that the reader is thoroughly familiar with the STA formalism, and if that is not the case, there is no better reference than [1].

Field action.

Theorem 1.1: Relativistic Euler-Lagrange field equations.

Let \( \phi \rightarrow \phi + \delta \phi \) be any variation of the field, such that the variation
\( \delta \phi = 0 \) vanishes at the boundaries of the action integral
\begin{equation}\label{eqn:maxwells:2120}
S = \int d^4 x \LL(\phi, \partial_\nu \phi).
\end{equation}
The extreme value of the action is found when the Euler-Lagrange equations
\begin{equation}\label{eqn:maxwells:2140}
0 = \PD{\phi}{\LL} – \partial_\nu \PD{(\partial_\nu \phi)}{\LL},
\end{equation}
are satisfied. For a Lagrangian with multiple field variables, there will be one such equation for each field.

Start proof:

To ease the visual burden, designate the variation of the field by \( \delta \phi = \epsilon \), and perform a first order expansion of the varied Lagrangian
\begin{equation}\label{eqn:maxwells:20}
\begin{aligned}
\LL
&\rightarrow
\LL(\phi + \epsilon, \partial_\nu (\phi + \epsilon)) \\
&=
\LL(\phi, \partial_\nu \phi)
+
\PD{\phi}{\LL} \epsilon +
\PD{(\partial_\nu \phi)}{\LL} \partial_\nu \epsilon.
\end{aligned}
\end{equation}
The variation of the Lagrangian is
\begin{equation}\label{eqn:maxwells:40}
\begin{aligned}
\delta \LL
&=
\PD{\phi}{\LL} \epsilon +
\PD{(\partial_\nu \phi)}{\LL} \partial_\nu \epsilon \\
&=
\PD{\phi}{\LL} \epsilon +
\partial_\nu \lr{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }
–
\epsilon \partial_\nu \PD{(\partial_\nu \phi)}{\LL},
\end{aligned}
\end{equation}
which we may plug into the action integral to find
\begin{equation}\label{eqn:maxwells:60}
\delta S
=
\int d^4 x \epsilon \lr{
\PD{\phi}{\LL}
–
\partial_\nu \PD{(\partial_\nu \phi)}{\LL}
}
+
\int d^4 x
\partial_\nu \lr{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }.
\end{equation}
The last integral can be evaluated along the \( dx^\nu \) direction, leaving
\begin{equation}\label{eqn:maxwells:80}
\int d^3 x
\evalbar{ \PD{(\partial_\nu \phi)}{\LL} \epsilon }{\Delta x^\nu},
\end{equation}
where \( d^3 x = dx^\alpha dx^\beta dx^\gamma \) is the product of differentials that does not include \( dx^\nu \). By construction, \( \epsilon \) vanishes on the boundary of the action integral so \ref{eqn:maxwells:80} is zero. The action takes its extreme value when
\begin{equation}\label{eqn:maxwells:100}
0 = \delta S
=
\int d^4 x \epsilon \lr{
\PD{\phi}{\LL}
–
\partial_\nu \PD{(\partial_\nu \phi)}{\LL}
}.
\end{equation}
The proof is complete after noting that this must hold for all variations of the field \( \epsilon \), which means that we must have
\begin{equation}\label{eqn:maxwells:120}
0 =
\PD{\phi}{\LL}
–
\partial_\nu \PD{(\partial_\nu \phi)}{\LL}.
\end{equation}

End proof.

Armed with the Euler-Lagrange equations, we can apply them to the Maxwell’s equation Lagrangian, which we will claim has the following form.

Theorem 1.2: Maxwell’s equation Lagrangian.

Application of the Euler-Lagrange equations to the Lagrangian
\begin{equation}\label{eqn:maxwells:2160}
\LL = – \frac{\epsilon_0 c}{2} F \cdot F + J \cdot A,
\end{equation}
where \( F = \grad \wedge A \), yields the vector portion of Maxwell’s equation
\begin{equation}\label{eqn:maxwells:2180}
\grad \cdot F = \inv{\epsilon_0 c} J,
\end{equation}
which implies
\begin{equation}\label{eqn:maxwells:2200}
\grad F = \inv{\epsilon_0 c} J.
\end{equation}
This is Maxwell’s equation.

Start proof:

We wish to apply all of the Euler-Lagrange equations simultaneously (i.e. once for each of the four \(A_\mu\) components of the potential), and cast it into four-vector form
\begin{equation}\label{eqn:maxwells:140}
0 = \gamma_\nu \lr{ \PD{A_\nu}{} – \partial_\mu \PD{(\partial_\mu A_\nu)}{} } \LL.
\end{equation}
Since our Lagrangian splits nicely into kinetic and interaction terms, this gives us
\begin{equation}\label{eqn:maxwells:160}
0 = \gamma_\nu \lr{ \PD{A_\nu}{(A \cdot J)} + \frac{\epsilon_0 c}{2} \partial_\mu \PD{(\partial_\mu A_\nu)}{ (F \cdot F)} }.
\end{equation}
The interaction term above is just
\begin{equation}\label{eqn:maxwells:180}
\gamma_\nu \PD{A_\nu}{(A \cdot J)}
=
\gamma_\nu \PD{A_\nu}{(A_\mu J^\mu)}
=
\gamma_\nu J^\nu
=
J,
\end{equation}
but the kinetic term takes a bit more work. Let’s start with evaluating
\begin{equation}\label{eqn:maxwells:200}
\begin{aligned}
\PD{(\partial_\mu A_\nu)}{ (F \cdot F)}
&=
\PD{(\partial_\mu A_\nu)}{ F } \cdot F
+
F \cdot \PD{(\partial_\mu A_\nu)}{ F } \\
&=
2 \PD{(\partial_\mu A_\nu)}{ F } \cdot F \\
&=
2 \PD{(\partial_\mu A_\nu)}{ (\partial_\alpha A_\beta) } \lr{ \gamma^\alpha \wedge \gamma^\beta } \cdot F \\
&=
2 \lr{ \gamma^\mu \wedge \gamma^\nu } \cdot F.
\end{aligned}
\end{equation}
We hit this with the \(\mu\)-partial and expand as a scalar selection to find
\begin{equation}\label{eqn:maxwells:220}
\begin{aligned}
\partial_\mu \PD{(\partial_\mu A_\nu)}{ (F \cdot F)}
&=
2 \lr{ \partial_\mu \gamma^\mu \wedge \gamma^\nu } \cdot F \\
&=
– 2 (\gamma^\nu \wedge \grad) \cdot F \\
&=
– 2 \gpgradezero{ (\gamma^\nu \wedge \grad) F } \\
&=
– 2 \gpgradezero{ \gamma^\nu \grad F – \gamma^\nu \cdot \grad F } \\
&=
– 2 \gamma^\nu \cdot \lr{ \grad \cdot F }.
\end{aligned}
\end{equation}
Putting all the pieces together yields
\begin{equation}\label{eqn:maxwells:240}
0
= J – \epsilon_0 c \gamma_\nu \lr{ \gamma^\nu \cdot \lr{ \grad \cdot F } }
= J – \epsilon_0 c \lr{ \grad \cdot F },
\end{equation}
but
\begin{equation}\label{eqn:maxwells:260}
\begin{aligned}
\grad \cdot F
&=
\grad F – \grad \wedge F \\
&=
\grad F – \grad \wedge (\grad \wedge A) \\
&=
\grad F,
\end{aligned}
\end{equation}
so the multivector field equations for this Lagrangian are
\begin{equation}\label{eqn:maxwells:280}
\grad F = \inv{\epsilon_0 c} J,
\end{equation}
as claimed.

End proof.

Problem: Correspondence with tensor formalism.

Cast the Lagrangian of \ref{eqn:maxwells:2160} into the conventional tensor form
\begin{equation}\label{eqn:maxwells:300}
\LL = \frac{\epsilon_0 c}{4} F_{\mu\nu} F^{\mu\nu} + A^\mu J_\mu.
\end{equation}
Also show that the four-vector component of Maxwell’s equation \( \grad \cdot F = J/(\epsilon_0 c) \) is equivalent to the conventional tensor form of the Gauss-Ampere law
\begin{equation}\label{eqn:maxwells:320}
\partial_\mu F^{\mu\nu} = \inv{\epsilon_0 c} J^\nu,
\end{equation}
where \( F^{\mu\nu} = \partial^\mu A^\nu – \partial^\nu A^\mu \) as usual. Also show that the trivector component of Maxwell’s equation \( \grad \wedge F = 0 \) is equivalent to the tensor form of the Gauss-Faraday law
\begin{equation}\label{eqn:maxwells:340}
\partial_\alpha \lr{ \epsilon^{\alpha \beta \mu \nu} F_{\mu\nu} } = 0.
\end{equation}

Answer

To show the Lagrangian correspondence we must expand \( F \cdot F \) in coordinates
\begin{equation}\label{eqn:maxwells:360}
\begin{aligned}
F \cdot F
&=
( \grad \wedge A ) \cdot
( \grad \wedge A ) \\
&=
\lr{ (\gamma^\mu \partial_\mu) \wedge (\gamma^\nu A_\nu) }
\cdot
\lr{ (\gamma^\alpha \partial_\alpha) \wedge (\gamma^\beta A_\beta) } \\
&=
\lr{ \gamma^\mu \wedge \gamma^\nu } \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta }
(\partial_\mu A_\nu )
(\partial^\alpha A^\beta ) \\
&=
\lr{
{\delta^\mu}_\beta
{\delta^\nu}_\alpha
–
{\delta^\mu}_\alpha
{\delta^\nu}_\beta
}
(\partial_\mu A_\nu )
(\partial^\alpha A^\beta ) \\
&=
– \partial_\mu A_\nu \lr{
\partial^\mu A^\nu
–
\partial^\nu A^\mu
} \\
&=
– \partial_\mu A_\nu F^{\mu\nu} \\
&=
– \inv{2} \lr{
\partial_\mu A_\nu F^{\mu\nu}
+
\partial_\nu A_\mu F^{\nu\mu}
} \\
&=
– \inv{2} \lr{
\partial_\mu A_\nu
–
\partial_\nu A_\mu
}
F^{\mu\nu} \\
&=
–
\inv{2}
F_{\mu\nu}
F^{\mu\nu}.
\end{aligned}
\end{equation}
With a substitution of this and \( A \cdot J = A_\mu J^\mu \) back into the Lagrangian, we recover the tensor form of the Lagrangian.

To recover the tensor form of Maxwell’s equation, we first split it into vector and trivector parts
\begin{equation}\label{eqn:maxwells:1580}
\grad \cdot F + \grad \wedge F = \inv{\epsilon_0 c} J.
\end{equation}
Now the vector component may be expanded in coordinates by dotting both sides with \( \gamma^\nu \) to find
\begin{equation}\label{eqn:maxwells:1600}
\inv{\epsilon_0 c} \gamma^\nu \cdot J = J^\nu,
\end{equation}
and
\begin{equation}\label{eqn:maxwells:1620}
\begin{aligned}
\gamma^\nu \cdot
\lr{ \grad \cdot F }
&=
\partial_\mu \gamma^\nu \cdot \lr{ \gamma^\mu \cdot \lr{ \gamma_\alpha \wedge \gamma_\beta } \partial^\alpha A^\beta } \\
&=
\lr{
{\delta^\mu}_\alpha
{\delta^\nu}_\beta
–
{\delta^\nu}_\alpha
{\delta^\mu}_\beta
}
\partial_\mu
\partial^\alpha A^\beta \\
&=
\partial_\mu
\lr{
\partial^\mu A^\nu
–
\partial^\nu A^\mu
} \\
&=
\partial_\mu F^{\mu\nu}.
\end{aligned}
\end{equation}
Equating \ref{eqn:maxwells:1600} and \ref{eqn:maxwells:1620} finishes the first part of the job. For the trivector component, we have
\begin{equation}\label{eqn:maxwells:1640}
0
= \grad \wedge F
= (\gamma^\mu \partial_\mu) \wedge \lr{ \gamma^\alpha \wedge \gamma^\beta } \partial_\alpha A_\beta
= \inv{2} (\gamma^\mu \partial_\mu) \wedge \lr{ \gamma^\alpha \wedge \gamma^\beta } F_{\alpha \beta}.
\end{equation}
Wedging with \( \gamma^\tau \) and then multiplying by \( -2 I \) we find
\begin{equation}\label{eqn:maxwells:1660}
0 = – \lr{ \gamma^\mu \wedge \gamma^\alpha \wedge \gamma^\beta \wedge \gamma^\tau } I \partial_\mu F_{\alpha \beta},
\end{equation}
but
\begin{equation}\label{eqn:maxwells:1680}
\gamma^\mu \wedge \gamma^\alpha \wedge \gamma^\beta \wedge \gamma^\tau = -I \epsilon^{\mu \alpha \beta \tau},
\end{equation}
which leaves us with
\begin{equation}\label{eqn:maxwells:1700}
\epsilon^{\mu \alpha \beta \tau} \partial_\mu F_{\alpha \beta} = 0,
\end{equation}
as expected.

Problem: Correspondence of tensor and Gibbs forms of Maxwell’s equations.

Given the identifications

\begin{equation}\label{eqn:lorentzForceCovariant:1500}
F^{k0} = E^k,
\end{equation}
and
\begin{equation}\label{eqn:lorentzForceCovariant:1520}
F^{rs} = -\epsilon^{rst} B^t,
\end{equation}
and
\begin{equation}\label{eqn:maxwells:1560}
J^\mu = \lr{ c \rho, \BJ },
\end{equation}
the reader should satisfy themselves that the traditional Gibbs form of Maxwell’s equations can be recovered from \ref{eqn:maxwells:320}.

Answer

The reader is referred to Exercise 3.4 “Electrodynamics, variational principle.” from [2].

Problem: Correspondence with grad and curl form of Maxwell’s equations.

With \( J = c \rho \gamma_0 + J^k \gamma_k \) and \( F = \BE + I c \BB \) show that Maxwell’s equation, as stated in \ref{eqn:maxwells:2200} expand to the conventional div and curl expressions for Maxwell’s equations.

Answer

To obtain Maxwell’s equations in their traditional vector forms, we pre-multiply both sides with \( \gamma_0 \)
\begin{equation}\label{eqn:maxwells:1720}
\gamma_0 \grad F = \inv{\epsilon_0 c} \gamma_0 J,
\end{equation}
and then select each grade separately. First observe that the RHS above has scalar and bivector components, as
\begin{equation}\label{eqn:maxwells:1740}
\gamma_0 J
=
c \rho + J^k \gamma_0 \gamma_k.
\end{equation}
In terms of the spatial bivector basis \( \Be_k = \gamma_k \gamma_0 \), the RHS of \ref{eqn:maxwells:1720} is
\begin{equation}\label{eqn:maxwells:1760}
\gamma_0 \frac{J}{\epsilon_0 c} = \frac{\rho}{\epsilon_0} – \mu_0 c \BJ.
\end{equation}
For the LHS, first note that
\begin{equation}\label{eqn:maxwells:1780}
\begin{aligned}
\gamma_0 \grad
&=
\gamma_0
\lr{
\gamma_0 \partial^0 +
\gamma_k \partial^k
} \\
&=
\partial_0 – \gamma_0 \gamma_k \partial_k \\
&=
\inv{c} \PD{t}{} + \spacegrad.
\end{aligned}
\end{equation}
We can express all the the LHS of \ref{eqn:maxwells:1720} in the bivector spatial basis, so that Maxwell’s equation in multivector form is
\begin{equation}\label{eqn:maxwells:1800}
\lr{ \inv{c} \PD{t}{} + \spacegrad } \lr{ \BE + I c \BB } = \frac{\rho}{\epsilon_0} – \mu_0 c \BJ.
\end{equation}
Selecting the scalar, vector, bivector, and trivector grades of both sides (in the spatial basis) gives the following set of respective equations
\begin{equation}\label{eqn:maxwells:1840}
\spacegrad \cdot \BE = \frac{\rho}{\epsilon_0}
\end{equation}
\begin{equation}\label{eqn:maxwells:1860}
\inv{c} \partial_t \BE + I c \spacegrad \wedge \BB = – \mu_0 c \BJ
\end{equation}
\begin{equation}\label{eqn:maxwells:1880}
\spacegrad \wedge \BE + I \partial_t \BB = 0
\end{equation}
\begin{equation}\label{eqn:maxwells:1900}
I c \spacegrad \cdot B = 0,
\end{equation}
which we can rewrite after some duality transformations (and noting that \( \mu_0 \epsilon_0 c^2 = 1 \)), we have
\begin{equation}\label{eqn:maxwells:1940}
\spacegrad \cdot \BE = \frac{\rho}{\epsilon_0}
\end{equation}
\begin{equation}\label{eqn:maxwells:1960}
\spacegrad \cross \BB – \mu_0 \epsilon_0 \PD{t}{\BE} = \mu_0 \BJ
\end{equation}
\begin{equation}\label{eqn:maxwells:1980}
\spacegrad \cross \BE + \PD{t}{\BB} = 0
\end{equation}
\begin{equation}\label{eqn:maxwells:2000}
\spacegrad \cdot B = 0,
\end{equation}
which are Maxwell’s equations in their traditional form.

Problem: Alternative multivector Lagrangian.

Show that a scalar+pseudoscalar Lagrangian of the following form
\begin{equation}\label{eqn:maxwells:2220}
\LL = – \frac{\epsilon_0 c}{2} F^2 + J \cdot A,
\end{equation}
which omits the scalar selection of the Lagrangian in \ref{eqn:maxwells:2160}, also represents Maxwell’s equation. Discuss the scalar and pseudoscalar components of \( F^2 \), and show why the pseudoscalar inclusion is irrelevant.

Answer

The quantity \( F^2 = F \cdot F + F \wedge F \) has both scalar and pseudoscalar
components. Note that unlike vectors, a bivector wedge in 4D with itself need not be zero (example: \( \gamma_0 \gamma_1 + \gamma_2 \gamma_3 \) wedged with itself).
We can see this multivector nature nicely by expansion in terms of the electric and magnetic fields
\begin{equation}\label{eqn:maxwells:2020}
\begin{aligned}
F^2
&= \lr{ \BE + I c \BB }^2 \\
&= \BE^2 – c^2 \BB^2 + I c \lr{ \BE \BB + \BB \BE } \\
&= \BE^2 – c^2 \BB^2 + 2 I c \BE \cdot \BB.
\end{aligned}
\end{equation}
Both the scalar and pseudoscalar parts of \( F^2 \) are Lorentz invariant, a requirement of our Lagrangian, but most Maxwell equation Lagrangians only include the scalar \( \BE^2 – c^2 \BB^2 \) component of the field square. If we allow the Lagrangian to be multivector valued, and evaluate the Euler-Lagrange equations, we quickly find the same results
\begin{equation}\label{eqn:maxwells:2040}
\begin{aligned}
0
&= \gamma_\nu \lr{ \PD{A_\nu}{} – \partial_\mu \PD{(\partial_\mu A_\nu)}{} } \LL \\
&= \gamma_\nu \lr{ J^\nu + \frac{\epsilon_0 c}{2} \partial_\mu
\lr{
(\gamma^\mu \wedge \gamma^\nu) F
+
F (\gamma^\mu \wedge \gamma^\nu)
}
}.
\end{aligned}
\end{equation}
Here some steps are skipped, building on our previous scalar Euler-Lagrange evaluation experience. We have a symmetric product of two bivectors, which we can express as a 0,4 grade selection, since
\begin{equation}\label{eqn:maxwells:2060}
\gpgrade{ X F }{0,4} = \inv{2} \lr{ X F + F X },
\end{equation}
for any two bivectors \( X, F \). This leaves
\begin{equation}\label{eqn:maxwells:2080}
\begin{aligned}
0
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ (\grad \wedge \gamma^\nu) F }{0,4} \\
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ -\gamma^\nu \grad F + (\gamma^\nu \cdot \grad) F }{0,4} \\
&= J + \epsilon_0 c \gamma_\nu \gpgrade{ -\gamma^\nu \grad F }{0,4} \\
&= J – \epsilon_0 c \gamma_\nu
\lr{
\gamma^\nu \cdot \lr{ \grad \cdot F } + \gamma^\nu \wedge \grad \wedge F
}.
\end{aligned}
\end{equation}
However, since \( \grad \wedge F = \grad \wedge \grad \wedge A = 0 \), we see that there is no contribution from the \( F \wedge F \) pseudoscalar component of the Lagrangian, and we are left with
\begin{equation}\label{eqn:maxwells:2100}
\begin{aligned}
0
&= J – \epsilon_0 c (\grad \cdot F) \\
&= J – \epsilon_0 c \grad F,
\end{aligned}
\end{equation}
which is Maxwell’s equation, as before.

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. Quantum field theory. Kindle Direct Publishing, 2018.

Lagrangian for the Lorentz force equation.

October 24, 2020 math and physics play action, anti-commutator, commutator, coordinates, covariant, Dirac algebra, Dirac matrix, dot product, Einstein convention, extremize action, Faraday bivector, Faraday tensor, frame, Geometric Algebra, gradient, Lorentz force, Lorentz force Lagrangian, Lorentz invariant, minimize action, observer frame, Pauli matrix, proper momentum, proper time, proper velocity, pseudoscalar, reciprocal basis, relativity, spacelike, spacetime gradient, STA, standard basis, tensor algebra, timelike, wedge product

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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.

Conventions.

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
\begin{equation}\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}
\end{equation}

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.

Answer

The anticommutator is defined as symmetric sum of products
\begin{equation}\label{eqn:lorentzForceCovariant:1040}
\symmetric{\gamma_\mu}{\gamma_\nu}
\equiv
\gamma_\mu \gamma_\nu
+
\gamma_\nu \gamma_\mu,
\end{equation}
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
\begin{equation}\label{eqn:lorentzForceCovariant:1060}
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu},
\end{equation}
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
\begin{equation}\label{eqn:lorentzForceCovariant:1080}
\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu} = 2 \symmetric{\gamma_\mu}{\gamma_\nu},
\end{equation}
as claimed.

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

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.

Answer

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
\begin{equation}\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}
\end{equation}
Using this, we have
\begin{equation}\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}
\end{equation}
To illustrate the anticommutation property with any vector basis element, consider the following two examples:
\begin{equation}\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}
\end{equation}
\begin{equation}\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}
\end{equation}
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 \( \alpha \ne \beta \)
\begin{equation}\label{eqn:lorentzForceCovariant:760}
\begin{aligned}
I \gamma_\alpha \gamma_\beta
&=
-\gamma_\alpha I \gamma_\beta \\
&=
+\gamma_\alpha \gamma_\beta I.
\end{aligned}
\end{equation}

Similarly for any trivector basis element \( \alpha \ne \beta \ne \sigma \)
\begin{equation}\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}
\end{equation}

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
\begin{equation}\label{eqn:lorentzForceCovariant:820}
x = x^\alpha \gamma_\alpha = x_\beta \gamma^\beta.
\end{equation}
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
\begin{equation}\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}
\end{equation}
as claimed. Similarly, dotting with a reciprocal frame vector, we find
\begin{equation}\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{equation}

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.
\begin{equation}\label{eqn:lorentzForceCovariant:880}
\begin{aligned}
x^0 &= x_0 \\
x^i &= -x_i \\
\end{aligned}
\end{equation}

Definition 1.5: Spacetime gradient.

The spacetime gradient operator is
\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
\begin{equation}\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.
\end{equation}

Definition 1.7: Spatial basis.

We designate
\begin{equation}\label{eqn:lorentzForceCovariant:1560}
\Be_i = \gamma_i \gamma_0,
\end{equation}
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.

Answer

\begin{equation}\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}
\end{equation}
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 \( I = \Be_1 \Be_2 \Be_3 \).

Answer

The spatial pseudoscalar, expanded in terms of the STA basis vectors, is
\begin{equation}\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}
\end{equation}
as claimed.

Problem: Characteristics of the Pauli matrices.

The Pauli matrices obey the following anticommutation relations:
\begin{equation}\label{eqn:lorentzForceCovariant:660}
\symmetric{ \sigma_a}{\sigma_b } = 2 \delta_{a b},
\end{equation}
and commutation relations:
\begin{equation}\label{eqn:lorentzForceCovariant:640}
\antisymmetric{ \sigma_a}{ \sigma_b } = 2 i \epsilon_{a b c}\,\sigma_c,
\end{equation}
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
\begin{equation}\label{eqn:lorentzForceCovariant:20}
S = \int d\tau L(x, \dot{x}),
\end{equation}
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
\begin{equation}\label{eqn:lorentzForceCovariant:1580}
\delta S = \int d\tau \delta x \cdot \delta L(x, \dot{x}),
\end{equation}
where
\begin{equation}\label{eqn:lorentzForceCovariant:1600}
\delta L = \grad L – \frac{d}{d\tau} (\grad_v L),
\end{equation}
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
\begin{equation}\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}
\end{equation}
Subtracting off \( S \) and integrating by parts, leaves
\begin{equation}\label{eqn:lorentzForceCovariant:80}
\delta S =
\int d\tau \epsilon \cdot \lr{
\grad L – \frac{d}{d\tau} \grad_v L
}
+
\int d\tau \frac{d}{d\tau} (\grad_v L ) \cdot \epsilon.
\end{equation}
The boundary integral
\begin{equation}\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,
\end{equation}
is zero since the variation \( \epsilon \) is required to vanish on the boundaries. So, if \( \delta S = 0 \), we must have
\begin{equation}\label{eqn:lorentzForceCovariant:120}
0 =
\int d\tau \epsilon \cdot \lr{
\grad L – \frac{d}{d\tau} \grad_v L
},
\end{equation}
for all variations \( \epsilon \). Clearly, this requires that
\begin{equation}\label{eqn:lorentzForceCovariant:140}
\delta L = \grad L – \frac{d}{d\tau} (\grad_v L) = 0,
\end{equation}
or
\begin{equation}\label{eqn:lorentzForceCovariant:145}
\grad L = \frac{d}{d\tau} (\grad_v L),
\end{equation}
which is the coordinate free statement of the Euler-Lagrange equations.

End proof.

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 \( \gamma^\mu \).

Answer

In terms of coordinates, the first order Taylor expansion of the action is
\begin{equation}\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}
\end{equation}
As before, we integrate by parts to separate out a pure boundary term
\begin{equation}\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}
}.
\end{equation}
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
\begin{equation}\label{eqn:lorentzForceCovariant:220}
\PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L} = 0.
\end{equation}

Lorentz force equation.

Theorem 1.3: Lorentz force.

The relativistic Lagrangian for a charged particle is
\begin{equation}\label{eqn:lorentzForceCovariant:1640}
L = \inv{2} m v^2 + q A \cdot v/c.
\end{equation}
Application of the Euler-Lagrange equations to this Lagrangian yields the Lorentz-force equation
\begin{equation}\label{eqn:lorentzForceCovariant:1660}
\frac{dp}{d\tau} = q F \cdot v/c,
\end{equation}
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
\begin{equation}\label{eqn:lorentzForceCovariant:240}
\begin{aligned}
\delta L_0
&= \grad L_0 – \frac{d}{d\tau} (\grad_v L_0) \\
&= – \frac{d}{d\tau} (m v) \\
&= – \frac{dp}{d\tau}.
\end{aligned}
\end{equation}
For the potential contribution we have
\begin{equation}\label{eqn:lorentzForceCovariant:260}
\begin{aligned}
\delta L_1
&= \grad L_1 – \frac{d}{d\tau} (\grad_v 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}
\end{equation}
The proper time derivative can be evaluated using the chain rule
\begin{equation}\label{eqn:lorentzForceCovariant:280}
\frac{dA}{d\tau}
=
\frac{\partial x^\mu}{\partial \tau} \partial_\mu A
= (v \cdot \grad) A.
\end{equation}
Putting all the pieces back together we have
\begin{equation}\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}
\end{equation}

End proof.

Problem: Gradient of a squared position vector.

Show that
\begin{equation*}
\grad (a \cdot x) = a,
\end{equation*}
and
\begin{equation*}
\grad x^2 = 2 x.
\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.

Answer

The first identity follows easily by expansion in coordinates
\begin{equation}\label{eqn:lorentzForceCovariant:320}
\begin{aligned}
\grad (a \cdot x)
&=
\gamma^\mu \partial_\mu a_\alpha x^\alpha \\
&=
\gamma^\mu a_\alpha \delta_\mu^\alpha \\
&=
\gamma^\mu a_\mu \\
&=
a.
\end{aligned}
\end{equation}
The second identity follows by linearity of the gradient
\begin{equation}\label{eqn:lorentzForceCovariant:340}
\begin{aligned}
\grad x^2
&=
\grad (x \cdot x) \\
&=
\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}
\end{equation}

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
\begin{equation}\label{eqn:lorentzForceCovariant:1620}
\frac{dp^\mu}{d\tau} = \frac{q}{c} F^{\mu\nu} v_\nu,
\end{equation}
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
\begin{equation}\label{eqn:lorentzForceCovariant:380}
\gamma^\mu \cdot \frac{dp}{d\tau}
=
\frac{dp^\mu}{d\tau}.
\end{equation}
On the right, we have
\begin{equation}\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}
\end{equation}

End proof.

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.

Answer

\begin{equation}\label{eqn:lorentzForceCovariant:900}
\begin{aligned}
F &=
\grad \wedge A \\
&=
\lr{ \gamma_\mu \partial^\mu } \wedge \lr{ \gamma_\nu A^\nu } \\
&=
\lr{ \gamma_\mu \wedge \gamma_\nu } \partial^\mu A^\nu.
\end{aligned}
\end{equation}
To this we can use the usual tensor trick (add self to self, change indexes, and divide by two), to give
\begin{equation}\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}
\end{equation}
which is just
\begin{equation}\label{eqn:lorentzForceCovariant:940}
F =
\inv{2} \lr{ \gamma_\mu \wedge \gamma_\nu } F^{\mu\nu}.
\end{equation}
Now, let’s expand \( (F \cdot v) \cdot \gamma^\mu \) to compare to the earlier expansion in terms of \( \grad \) and \( A \).
\begin{equation}\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}
\end{equation}
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}.

Answer

Let \( \delta_\mu L = \gamma_\mu \cdot \delta L \), so that we can write the Euler-Lagrange equations as
\begin{equation}\label{eqn:lorentzForceCovariant:460}
0 = \delta_\mu L = \PD{x^\mu}{L} – \frac{d}{d\tau} \PD{\dot{x}^\mu}{L}.
\end{equation}
Operating on the kinetic term of the Lagrangian, we have
\begin{equation}\label{eqn:lorentzForceCovariant:480}
\delta_\mu L_0 = – \frac{d}{d\tau} m v_\mu.
\end{equation}
For the potential term
\begin{equation}\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}
\end{equation}
Putting the pieces together gives
\begin{equation}\label{eqn:lorentzForceCovariant:520}
\frac{d}{d\tau} (m v_\mu) = \frac{q}{c} v^\nu F_{\mu\nu},
\end{equation}
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
\begin{equation}\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}
\end{equation}
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 \( x \)
\begin{equation}\label{eqn:lorentzForceCovariant:1140}
x = c t \gamma_0 + x^k \gamma_k.
\end{equation}
The corresponding four-vector velocity is
\begin{equation}\label{eqn:lorentzForceCovariant:1160}
v = \ddtau{x} = c \ddtau{t} \gamma_0 + \ddtau{t} \ddt{x^k} \gamma_k.
\end{equation}
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
\begin{equation}\label{eqn:lorentzForceCovariant:1240}
c^2 = \lr{ \ddtau{t} }^2 \lr{ c^2 – \Bv^2 },
\end{equation}
where \( \Bv = v \wedge \gamma_0 \). This shows that we may make the identification
\begin{equation}\label{eqn:lorentzForceCovariant:1260}
\gamma = \ddtau{t} = \inv{1 – \Bv^2/c^2 },
\end{equation}
and
\begin{equation}\label{eqn:lorentzForceCovariant:1280}
\ddtau{} = \ddtau{t} \ddt{} = \gamma \ddt{}.
\end{equation}
We may now factor the four-velocity \( v \) into its spacetime split
\begin{equation}\label{eqn:lorentzForceCovariant:1300}
v = \gamma \lr{ c + \Bv } \gamma_0.
\end{equation}
In particular the LHS of the Lorentz force equation can be rewritten as
\begin{equation}\label{eqn:lorentzForceCovariant:1320}
\ddtau{p} = \gamma \ddt{}\lr{ \gamma \lr{ c + \Bv } } \gamma_0,
\end{equation}
and the RHS of the Lorentz force equation can be rewritten as
\begin{equation}\label{eqn:lorentzForceCovariant:1340}
\frac{q}{c} F \cdot v
=
\frac{\gamma q}{c} F \cdot \lr{ (c + \Bv) \gamma_0 }.
\end{equation}
Equating timelike and spacelike components leaves us
\begin{equation}\label{eqn:lorentzForceCovariant:1380}
\ddt{ (m \gamma c) } = \frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \cdot \gamma_0,
\end{equation}
\begin{equation}\label{eqn:lorentzForceCovariant:1400}
\ddt{ (m \gamma \Bv) } = \frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \wedge \gamma_0,
\end{equation}
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
\begin{equation}\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},
\end{equation}
\begin{equation}\label{eqn:lorentzForceCovariant:1460}
\frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \cdot \gamma_0
= \frac{q}{c} \BE \cdot \Bv,
\end{equation}
\begin{equation}\label{eqn:lorentzForceCovariant:1480}
\frac{q}{c} \lr{ F \cdot \lr{ (c + \Bv) \gamma_0 } } \wedge \gamma_0
= q \lr{ \BE + \Bv \cross \BB }.
\end{equation}

End proof.

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
\begin{equation}\label{eqn:lorentzForceCovariant:1500}
F^{k0} = E^k,
\end{equation}
and
\begin{equation}\label{eqn:lorentzForceCovariant:1520}
F^{rs} = -\epsilon^{rst} B^t.
\end{equation}

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].

[Series intro] An introduction to geometric algebra.

July 25, 2020 Geometric Algebra for Electrical Engineers bases, change of basis, complex numbers, coordinates, dot product, dot product space, Euclidean spaces, Geometric Algebra, linear transformation, matrices, metric, multivector space, non-Euclidean spaces, real numbers, vector product, vector representations, vector space, vectors

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
\begin{equation}\label{eqn:multivector:20}
\Bx \By = \Bx \cdot \By + \Bx \wedge \By.
\end{equation}
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
\begin{equation}\label{eqn:multivector:40}
\Bx \wedge \By =
\inv{2} \lr{
\Bx \By – \By \Bx
}.
\end{equation}
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.

dot product

Fundamental theorem of geometric calculus for line integrals (relativistic.)

Motivation.

Multivector line integrals.

Definition 1.1: Line integral.

Problem: Circular parameterization.

Answer

Problem: Line integral for boosted time direction vector.

Answer

Perfect differentials.

Definition 1.2: Vector derivative.

Theorem 1.1: Fundamental theorem for line integrals.

Start proof:

End proof.

More to come.

References

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Lorentz transformations in Space Time Algebra (STA)

Motivation.

Lorentz transformations.

Theorem 1.1: Lorentz transformation.

Start proof:

End proof.

Theorem 1.2: Boost.

Start proof:

End proof.

Theorem 1.3: Spatial rotation.

Start proof:

End proof.

Problems.

Problem: Verify components relative to boost direction.

Answer

Problem: Rotation transformation components.

Answer

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Maxwell’s equation Lagrangian (geometric algebra and tensor formalism)

Motivation.

Field action.

Theorem 1.1: Relativistic Euler-Lagrange field equations.

Start proof:

End proof.

Theorem 1.2: Maxwell’s equation Lagrangian.

Start proof:

End proof.

Problem: Correspondence with tensor formalism.

Answer

Problem: Correspondence of tensor and Gibbs forms of Maxwell’s equations.

Answer

Problem: Correspondence with grad and curl form of Maxwell’s equations.

Answer

Problem: Alternative multivector Lagrangian.

Answer

References

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Lagrangian for the Lorentz force equation.

Motivation.

Conventions.

Definition 1.1: Index conventions.

Space Time Algebra (STA.)

Definition 1.2: Standard basis.

Problem: Commutator properties of the STA basis.

Answer

Definition 1.3: Pseudoscalar.

Problem: Pseudoscalar.

Answer

Definition 1.4: Reciprocal basis.

Theorem 1.1: Coordinates.

Start proof:

End proof.

Definition 1.5: Spacetime gradient.

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

Definition 1.7: Spatial basis.

Problem: Orthonormality of the spatial basis.

Answer

Problem: Spatial pseudoscalar.

Answer

Problem: Characteristics of the Pauli matrices.