bivector

Potentials in geometric algebra.

December 2, 2023 math and physics play , , , , , , , , , , , , , , , , , , , ,

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Conventional formulation.

The idea behind introducing the scalar potential \( \phi \) and vector potential \( \BA \) is that we can impose a constraint on the form of our observable fields \( \BE, \BB \), (or \( \BD, \BH \)), that reduces the complexity and coupling of Maxwell’s equations. These potentials are not unique, but the types of allowed variations in those potentials (gauge transformations) do not change the observable fields.

The basic idea is that we are looking for representations of the fields that automatically satisfy the pair of source free Maxwell’s equations
\begin{equation}\label{eqn:gapotentials:40}
\begin{aligned}
\spacegrad \cdot \BB &= 0 \\
c \partial_0 \BB + \spacegrad \cross \BE &= 0,
\end{aligned}
\end{equation}
so that the problem is reduced to solving just the remaining source dependent Maxwell’s equations.

The conventional way of constructing these potentials makes use of the identities
\begin{equation}\label{eqn:gapotentials:60}
\begin{aligned}
\spacegrad \cdot \lr{ \spacegrad \cross \Bf } &= 0 \\
\spacegrad \cross \lr{ \spacegrad \chi } &= 0,
\end{aligned}
\end{equation}
where \( \Bf \) is a vector, and \( \chi \) is a scalar. This approach is straightforward. Instead of replicating it, here are a few well known references where such a treatment can be found

  1. section 18-6 potentials and the wave equation in [2] (available online),
  2. section 10.1 The potential formulation in [3], and
  3. section 6.4 Vector and Scalar Potentials, in [4],

Multivector potentials in geometric algebra.

The multivector form of Maxwell’s equation is
\begin{equation}\label{eqn:gapotentials:820}
\lr{ \spacegrad + \partial_0 } F = J,
\end{equation}
where \( \partial_0 = (1/c)\partial/\partial t \), the electromagnetic field \( F = \BE + I c \BB = \BE + I \eta H \) has grades(1,2), and a multivector charge and current density \( J \). Grades(0,1) of the current are the charge and current densities respectively, and if desired, the grade(2,3) portion of the current has the fictitious magnetic charge and current densities (used in microwave and antenna engineering.)

It’s best to consider the case of electric sources, separately from the case of (fictitious) magnetic sources, and then use superposition to construct a potential representation that includes both.

We require a tool, that generalizes the \(\mathbb{R}^3\) cross product curl identities above.

Lemma 1.1: Curl of curl.

Let \( A \in \bigwedge^k \) be a blade of grade \( k \). Then
\begin{equation*}
\nabla \wedge \nabla \wedge A = 0.
\end{equation*}

Observe that for scalar \( A \), this reduces to
\begin{equation}\label{eqn:gapotentials:1740}
\nabla \wedge \nabla A = 0.
\end{equation}
We’ve recently proved this, so we won’t do it again now.

Now we are ready to figure out the structure of the potentials.

Case I. No (fictitious) magnetic sources.

Without magnetic sources, Maxwell’s equation is
\begin{equation}\label{eqn:gapotentials:840}
\lr{ \spacegrad + \partial_0 } F = \gpgrade{J}{0,1},
\end{equation}
This can be split into two equations, one that has just the sources, and one that is source free
\begin{equation}\label{eqn:gapotentials:860}
\gpgrade{ \lr{ \spacegrad + \partial_0 } F }{0,1} = \gpgrade{J}{0,1},
\end{equation}
\begin{equation}\label{eqn:gapotentials:880}
\gpgrade{ \lr{ \spacegrad + \partial_0 } F }{2,3} = 0.
\end{equation}
If you are clever, or have the benefit of having worked out the answer already, you can look directly at \ref{eqn:gapotentials:880} and guess the multivector form for the potential. Hint: you want something closely related to \( F = \lr{ \spacegrad – \partial_0 } A \), where \( A \) has grades(0,1).

If you aren’t that clever, or don’t have a time machine that let’s you look that clever, you’ll have to work it out systematically like the rest of us. We can start by breaking down \( F \) into it’s constituent observer dependent fields. That means that we want to find values for \( \BE, \BH \) that satisfy
\begin{equation}\label{eqn:gapotentials:900}
\gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{ \BE + I \eta \BH } }{2,3} = 0.
\end{equation}
Expanding the multivector factors gives us
\begin{equation}\label{eqn:gapotentials:920}
\begin{aligned}
\gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{ \BE + I \eta \BH } }{2,3}
&=\gpgradetwo{\spacegrad \BE} + \gpgradethree{I \eta \spacegrad \BH} + I \eta \partial 0 \BH \\
&=
\spacegrad \wedge \BE
+ \spacegrad \wedge \lr{ I \eta \BH }
+ I \eta \partial_0 \BH.
\end{aligned}
\end{equation}
Splitting this into one equation for each grade, leaves us with
\begin{equation}\label{eqn:gapotentials:940}
0 = \spacegrad \wedge \BE + I \eta \partial_0 \BH
\end{equation}
\begin{equation}\label{eqn:gapotentials:960}
0 = \spacegrad \wedge \lr{ I \eta \BH }.
\end{equation}
Observe that we could have also written \ref{eqn:gapotentials:960} as \( 0 = I \eta \lr{ \spacegrad \cdot \BH } \), which is the starting point of the conventional non-GA approach.
It’s clear that we want to write \( I \eta \BH = I c \BB \) as a (bivector) curl, and let
\begin{equation}\label{eqn:gapotentials:980}
I \eta \BH = c \spacegrad \wedge \BA.
\end{equation}
It’s a bit sneaky to toss that factor of \( c \) in here, but that’s done to make the units of \( \BA \) turn out in a way that matches the conventional vector potential. If it makes you feel better, you can think of this as an undetermined constant multiplicative undetermined factor that will be used to adjust the dimensions of \( \BA \) down the line.

Having made that choice, \ref{eqn:gapotentials:960} is automatically satisfied, and \ref{eqn:gapotentials:940} is reduced to
\begin{equation}\label{eqn:gapotentials:1000}
\begin{aligned}
0
&= \spacegrad \wedge \BE + I \eta \partial_0 \BH \\
&= \spacegrad \wedge \BE + \partial_0 \spacegrad \wedge \lr{ c \BA } \\
&= \spacegrad \wedge \lr{ \BE + c \partial_0 \BA }.
\end{aligned}
\end{equation}
We can now let
\begin{equation}\label{eqn:gapotentials:1020}
\BE + \partial_0 c \BA = -\spacegrad \phi.
\end{equation}
Again, we had the option of including an arbitrary multiplicative constant, but this time, we managed to find the right switch for our time machine, and look ahead to see that we want that constant to be \( -1 \) in order to have agreement with the conventional result.

We are left with a potential construction for our individual field components
\begin{equation}\label{eqn:gapotentials:1040}
\begin{aligned}
\BE &= -\spacegrad \phi – c \partial_0 \BA \\
I \eta \BH &= c \spacegrad \wedge \BA,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:gapotentials:1060}
F = -\spacegrad \phi – c \partial_0 \BA + c \spacegrad \wedge \BA.
\end{equation}
This automatically satisfies the grades of Maxwell’s equation that are source free, leaving us to solve just
\begin{equation}\label{eqn:gapotentials:1080}
\gpgrade{ \lr{ \spacegrad + \partial_0 } F }{0,1} = \gpgrade{J}{0,1}.
\end{equation}

Multivector potential.

It’s natural to wonder if there is a more structured form for \( F \) than \ref{eqn:gapotentials:1060}, just as we found a GA structure for Maxwell’s equation that eliminated the crazy mix of divs and curls that we had in the original Gibbs representation. Let’s find that structure. To do so, we can enclose \( F \) in a no-op grade selection operation
\begin{equation}\label{eqn:gapotentials:1100}
\begin{aligned}
F
&= \gpgrade{ -\spacegrad \phi – c \partial_0 \BA + c \spacegrad \wedge \BA }{1,2} \\
&= \gpgrade{ -\spacegrad \phi – c \partial_0 \BA + c \spacegrad \BA }{1,2} \\
&= \gpgrade{ \spacegrad \lr{ -\phi + c \BA } – c \partial_0 \BA + \lr{ \partial_0 \phi – \partial_0 \phi } }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA } }{1,2}.
\end{aligned}
\end{equation}

We can now introduce a multivector potential, and express the remaining non-zero grades of Maxwell’s equation in terms of this potential
\begin{equation}\label{eqn:gapotentials:1120}
\begin{aligned}
A &= -\phi + c \BA \\
F &= \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{1,2} \\
\gpgrade{J}{0,1} &= \gpgrade{ \lr{ \spacegrad + \partial_0 } F }{0,1}.
\end{aligned}
\end{equation}

Lorentz gauge.

The grade selection in our representation of \( F \) is a bit annoying, and can be eliminated if we impose additional constraints on the potential. We can write
\begin{equation}\label{eqn:gapotentials:1140}
F =
\lr{ \spacegrad – \partial_0 } A
–
\gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3},
\end{equation}
and then ask what conditions are required for this grade(0,3) selection to be zero. In terms of our constituent potentials, that is
\begin{equation}\label{eqn:gapotentials:1160}
\begin{aligned}
0 &=
\gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3} \\
&=
\gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA } }{0,3} \\
&=
c \spacegrad \cdot \BA + \partial_0 \phi,
\end{aligned}
\end{equation}
This is the Lorentz gauge condition, recognized a bit more easily if written out in terms of the time partials explicitly
\begin{equation}\label{eqn:gapotentials:1180}
\inv{c^2} \PD{t}{\phi} + \spacegrad \cdot \BA = 0.
\end{equation}

We can now write Maxwell’s equations, in the potential formulation, as
\begin{equation}\label{eqn:gapotentials:1200}
\begin{aligned}
A &= -\phi + c \BA \\
F &= \lr{ \spacegrad – \partial_0 } A \\
0 &= \inv{c} \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3} = \inv{c^2} \PD{t}{\phi} + \spacegrad \cdot \BA \\
\gpgrade{J}{0,1} &= \gpgrade{ \lr{ \spacegrad + \partial_0 } F }{0,1} = \lr{ \spacegrad^2 – \partial_{00} } A.
\end{aligned}
\end{equation}
This is quite nice. We have a one to one decoupled relationship between the potential and the current, and are free to use the well known techniques for solving the wave equation (using convolution and a superposition of advanced and retarded Green’s functions for the wave equation operator.)

Gauge transformation.

There’s one more thing that we should look at before moving on to the magnetic sources case, and that’s the question of gauge freedom. We’ve said that the potentials are not unique, but this non-uniqueness has a very specific form.

Since we’ve constructed \( F \) with a grade selection as
\begin{equation}\label{eqn:gapotentials:1220}
F = \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{1,2},
\end{equation}
so it’s clear that any transformation
\begin{equation}\label{eqn:gapotentials:1240}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \psi_{0,3},
\end{equation}
where \( \psi_{0,3} \) is any multivector with grades(0,3) components, will leave \( F \) invariant. That is
\begin{equation}\label{eqn:gapotentials:1260}
\begin{aligned}
A &= -\phi + c \BA \\
&\rightarrow
-\phi + c \BA + \lr{ \spacegrad + \partial_0 } \psi_{0,3} \\
&=
-\phi + c \BA + \lr{ \spacegrad + \partial_0 } \lr{ c \psi + I \bar{\psi} } \\
&=
\lr{ -\phi + c \partial_0 \psi }
+ c \lr{ \BA + \spacegrad \psi }
+ I \spacegrad \bar{\psi}
+ I \partial_0 \bar{\psi}.
\end{aligned}
\end{equation}
We see that the contributions of \( \bar{\psi} \) result in grade(2,3) terms, which are not of interest, and we find that a paired transformation of the potentials
\begin{equation}\label{eqn:gapotentials:1280}
\begin{aligned}
\phi &\rightarrow \phi – \PD{t}{\psi} \\
\BA &\rightarrow \BA + \spacegrad \psi,
\end{aligned}
\end{equation}
called a gauge transformation, leaves the field \( F \) unchanged. This can be expressed slightly more compactly as
\begin{equation}\label{eqn:gapotentials:1300}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } c \psi,
\end{equation}
where, once again, the multiplicative constant \( c \) is included so for consistency with the conventional expression for potential gauge transformation.

Case II. With (fictitious) magnetic sources.

With magnetic sources, Maxwell’s equation is
\begin{equation}\label{eqn:gapotentials:1500}
\lr{ \spacegrad + \partial_0 } F = \gpgrade{J}{2,3}.
\end{equation}
We put this in dual form
\begin{equation}\label{eqn:gapotentials:1520}
\lr{ \spacegrad + \partial_0 } I F = I \gpgrade{J}{2,3},
\end{equation}
which now has the sources all with grades (0,1) as we just analyzed. The dual vector \( I F \), like \( F \), has only grade(1,2) components.

Expanding the source free Maxwell’s equations in terms of \( \BE, \BH \), we have
\begin{equation}\label{eqn:gapotentials:1340}
\begin{aligned}
0
&= \gpgrade{ \lr{ \spacegrad + \partial_0 } I F}{2,3} \\
&= \gpgrade{ \lr{ \spacegrad + \partial_0 } \lr{I \BE – \eta \BH } }{2,3} \\
&= \gpgrade{ I \spacegrad \BE – \eta \spacegrad \BH + I \partial_0 \BE – \eta \partial_0 \BH }{2,3} \\
&= \spacegrad \wedge \lr{ I \BE } – \eta \spacegrad \wedge \BH + I \partial_0 \BE,
\end{aligned}
\end{equation}
or, by grade
\begin{equation}\label{eqn:gapotentials:1360}
0 = \spacegrad \wedge \lr{ I \BE },
\end{equation}
\begin{equation}\label{eqn:gapotentials:1361}
0 = – \eta \spacegrad \wedge \BH + I \partial_0 \BE.
\end{equation}
We see that the dual electric field needs to be a curl to satisfy \ref{eqn:gapotentials:1360}
\begin{equation}\label{eqn:gapotentials:1400}
I \BE = -\eta \spacegrad \wedge c \BF,
\end{equation}
and after substitution into \ref{eqn:gapotentials:1361} we are left with
\begin{equation}\label{eqn:gapotentials:1540}
\begin{aligned}
0
&= – \eta \spacegrad \wedge \BH + \partial_0 \lr{ – \eta \spacegrad \wedge c \BF } \\
&= \eta \spacegrad \wedge \lr{ -\BH – \partial_0 c \BF } \\
\end{aligned}
\end{equation}
We set
\begin{equation}\label{eqn:gapotentials:1420}
-\BH – \partial_0 c \BF = \spacegrad \phi_m,
\end{equation}
Our fields are
\begin{equation}\label{eqn:gapotentials:1440}
\begin{aligned}
\BE &= – \inv{\epsilon} \spacegrad \cross \BF \\
\BH &= -\spacegrad \phi_m – \PD{t}{\BF}.
\end{aligned}
\end{equation}
This has the structure that matches the potential conventions from antenna theory, for example as stated in [1].

Multivector potential.

As with the electrical sources, we expect that we can write this as something like
\begin{equation}\label{eqn:gapotentials:1460}
F = \gpgrade{ \lr{ \spacegrad – \partial_0 } I A }{1,2}.
\end{equation}
Let’s verify that this is the case.
\begin{equation}\label{eqn:gapotentials:1480}
\begin{aligned}
F
&= I \eta \spacegrad \wedge (c \BF) -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF \\
&= \gpgrade{ I \eta \spacegrad \wedge (c \BF) -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF }{1,2} \\
&= \gpgrade{ I \eta \spacegrad c \BF -I \eta \spacegrad \phi_m – I \eta \partial_0 c \BF }{1,2} \\
&= \gpgrade{ I \eta \lr{ \spacegrad \lr{ – \phi_m + c \BF } – \partial_0 c \BF + \partial_0 \phi_m – \partial_0 \phi_m} }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF } }{1,2}.
\end{aligned}
\end{equation}

Lorentz gauge.

Let’s see what constraints we need to write our field in terms of a potential without a grade selection, that is
\begin{equation}\label{eqn:gapotentials:1560}
F = \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF }.
\end{equation}
We need the grade(0,3) components of this multivector to be zero. Those components are
\begin{equation}\label{eqn:gapotentials:1580}
\begin{aligned}
0 &=
\gpgrade{ \lr{ \spacegrad – \partial_0 } I \eta \lr{ – \phi_m + c \BF }}{0,3} \\
&=
\gpgrade{-\spacegrad I \eta \phi_m+\spacegrad I \eta c \BF+ \partial_0 I \eta \phi_m – \partial_0 I \eta c \BF }{0,3} \\
&=
\gpgradethree{ \spacegrad I \eta c \BF }
+ \partial_0 I \eta \phi_m \\
&=
I \eta \lr{ c \lr{ \spacegrad \cdot \BF} + \partial_0 \phi_m },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:gapotentials:1600}
0 = \inv{c^2} \PD{t}{\phi_m} + \spacegrad \cdot \BF.
\end{equation}
This is the Lorentz gauge condition. With this condition we can we can express Maxwell’s equation with magnetic sources, as a forced wave equation
\begin{equation}\label{eqn:gapotentials:1620}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
F &= \lr{ \spacegrad – \partial_0 } A \\
0 &= \inv{c} \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3} = \inv{c^2} \PD{t}{\phi_m} + \spacegrad \cdot \BF \\
\gpgrade{J}{2,3} &= \gpgrade{ \lr{ \spacegrad + \partial_0 } F }{2,3} = \lr{ \spacegrad^2 – \partial_{00} } A.
\end{aligned}
\end{equation}

Gauge transformation.

Without the Lorentz gauge assumption, our potential representation for the field is
\begin{equation}\label{eqn:gapotentials:1640}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
F &= \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{1,2}.
\end{aligned}
\end{equation}
It’s clear that any transformation of the form
\begin{equation}\label{eqn:gapotentials:1660}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \psi_{0,3},
\end{equation}
leaves the field unchanged.
\begin{equation}\label{eqn:gapotentials:1680}
\begin{aligned}
A &= I \eta \lr{ -\phi_m + c \BF } \\
&\rightarrow
I \eta \lr{ -\phi + c \BF } + \lr{ \spacegrad + \partial_0 } \psi_{0,3} \\
&=
I \eta \lr{ -\phi_m + c \BF } + \lr{ \spacegrad + \partial_0 } \lr{ \psi + I \eta c \bar{\psi} } \\
&=
I \eta \lr{
-\phi_m
+ c \partial_0 \bar{\psi}
+ c \BF
+ c \spacegrad \bar{\psi}
}
+ \lr{ \spacegrad + \partial_0 } \psi.
\end{aligned}
\end{equation}
We can drop the \( \psi \) contributions, since this time we want only grades(2,3) in our potential, and find that the
desired form of the gauge transformation, for scalar \( \bar{\psi} \), is
\begin{equation}\label{eqn:gapotentials:1700}
\begin{aligned}
\phi_m &\rightarrow \phi_m – \PD{t}{\bar{\psi}} \\
\BF &\rightarrow \BF + \spacegrad \bar{\psi}.
\end{aligned}
\end{equation}
The multivector form of this is
\begin{equation}\label{eqn:gapotentials:1720}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } I \eta c \bar{\psi}.
\end{equation}

Superposition.

We can now use superposition to construct a potential representation that works for both conventional electric and fictitious magnetic charges and currents.

Without a Lorentz gauge assumption, that is
\begin{equation}\label{eqn:gapotentials:1760}
\begin{aligned}
A &= -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } \\
F &= \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{1,2} \\
J &= \lr{ \spacegrad + \partial_0 } F,
\end{aligned}
\end{equation}
where, given scalar functions \( \psi, \bar{\psi} \), we are free to make gauge transformations of the multivector potential that satisfy
\begin{equation}\label{eqn:gapotentials:1800}
A \rightarrow A + \lr{ \spacegrad + \partial_0 } \lr{ c \psi + I \eta c \bar{\psi} },
\end{equation}

With a Lorentz gauge constraint, we have a wave equation operator acting on \( A \), with the multivector current as a forcing term.
\begin{equation}\label{eqn:gapotentials:1780}
\begin{aligned}
A &= -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } \\
0 &= \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3} \\
F &= \lr{ \spacegrad – \partial_0 } A \\
J &= \lr{ \spacegrad^2 – \partial_{00} } A.
\end{aligned}
\end{equation}

Check.

It’s worth expansion to verify that we got all the dimensional constants write, and compare the results to Maxwell’s equations in their Gibbs form.

Let’s start with an expansion of \( F \) in terms of the potentials
\begin{equation}\label{eqn:gapotentials:1820}
\begin{aligned}
F &=
\gpgrade{\lr{ \spacegrad – \partial_0 } A }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } }{1,2} \\
&=
\gpgrade{ \spacegrad \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } -\partial_0 \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } }{1,2} \\
&=
\gpgrade{ \spacegrad \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF } } -\partial_0 \lr{ c \BA + I \eta c \BF } }{1,2} \\
&=
-\spacegrad \phi + c \spacegrad \wedge \BA – I \eta \spacegrad \phi_m + I \eta c \spacegrad \wedge \BF
-\partial_0 \lr{ c \BA + I \eta c \BF }.
\end{aligned}
\end{equation}
That is
\begin{equation}\label{eqn:gapotentials:1840}
\begin{aligned}
\BE &= -\spacegrad \phi + I \eta c \spacegrad \wedge \BF -c \partial_0 \BA \\
I \eta \BH &= c \spacegrad \wedge \BA – I \eta \spacegrad \phi_m – I \eta c \partial_0 \BF,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:gapotentials:1860}
\begin{aligned}
\BE &= – \spacegrad \phi -\partial_t \BA – \inv{\epsilon} \spacegrad \cross \BF \\
\BH &= – \spacegrad \phi_m – \partial_t \BF + \inv{\mu} \spacegrad \cross \BA.
\end{aligned}
\end{equation}
All is good. This is exactly the form that we expect.

Let’s expand out Maxwell’s equation in terms of this potential representation and see what we get.

Let’s write the total field without the grade(1,2) selection, by subtracting off any grade(0,3) contributions
\begin{equation}\label{eqn:gapotentials:1880}
F = \lr{ \spacegrad – \partial_0 } A – \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3}.
\end{equation}
That difference term is
\begin{equation}\label{eqn:gapotentials:1900}
\begin{aligned}
– \gpgrade{ \lr{ \spacegrad – \partial_0 } A }{0,3}
&=
– \gpgrade{ \lr{ \spacegrad – \partial_0 } \lr{ -\phi + c \BA – I \eta \phi_m + I \eta c \BF } }{0,3} \\
&=
– c \spacegrad \cdot \BA – I \eta c \spacegrad \cdot \BF – \partial_0 \phi – I \eta \partial_0 \phi_m.
\end{aligned}
\end{equation}
The field is nicely split into a multivector term that depends directly on the full multivector potential \( A \), and a difference term that wipes out any scalar and pseudoscalar terms
\begin{equation}\label{eqn:gapotentials:1920}
F
=
\lr{ \spacegrad – \partial_0 } A
– \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } – I \eta \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF }.
\end{equation}

Maxwell’s equations are now reduced to
\begin{equation}\label{eqn:gapotentials:1940}
\lr{ \spacegrad^2 – \partial_{00} } A
–
\lr{ \spacegrad + \partial_0 }
\lr{ \partial_0 \phi + c \spacegrad \cdot \BA }
–
\lr{ \spacegrad + \partial_0 }
I \eta \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF }
= J.
\end{equation}
This splits nicely into a single equation for each grade of \( A, J \) respectively. We write
\begin{equation}\label{eqn:gapotentials:1960}
J = \eta\lr{ c \rho – \BJ } + I \lr{ c \phi_m – \BM },
\end{equation}
so
\begin{equation}\label{eqn:gapotentials:1980}
\begin{aligned}
\lr{ \spacegrad^2 – \partial_{00} } (-\phi) – \partial_0 \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } &= \eta c \rho \\
\lr{ \spacegrad^2 – \partial_{00} } (c \BA) – \spacegrad \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } &= -\eta \BJ \\
\lr{ \spacegrad^2 – \partial_{00} } (I \eta c \BF) – I \eta \partial_0 \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF } &= -I \BM \\
\lr{ \spacegrad^2 – \partial_{00} } (-I \eta \phi_m) – I \eta \spacegrad \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF } &= I c \rho_m.
\end{aligned}
\end{equation}
If we choose the Lorentz gauge conditions
\begin{equation}\label{eqn:gapotentials:2000}
0 = \lr{ \partial_0 \phi + c \spacegrad \cdot \BA } = \lr{ \partial_0 \phi_m + c \spacegrad \cdot \BF },
\end{equation}
all of these equations decouple nicely, leaving us with 8 (scalar) equations in 8 unknowns
\begin{equation}\label{eqn:gapotentials:2020}
\begin{aligned}
\lr{ \spacegrad^2 – \partial_{00} } \phi &= -\frac{\rho}{\epsilon} \\
\lr{ \spacegrad^2 – \partial_{00} } \BA &= -\mu \BJ \\
\lr{ \spacegrad^2 – \partial_{00} } \BF &= -\epsilon \BM \\
\lr{ \spacegrad^2 – \partial_{00} } \phi_m &= – \frac{\rho_m}{\mu}.
\end{aligned}
\end{equation}

Potentials in STA (space time algebra).

All of this was very convoluted. Maxwell’s equation in STA form is considerably simpler, as is the potential formulation.

STA form of Maxwell’s equation.

We identify
\begin{equation}\label{eqn:gapotentials:2040}
\begin{aligned}
\Be_k &= \gamma_k \gamma_0 \\
I &= \Be_1 \Be_2 \Be_3 = \gamma_0 \gamma_1 \gamma_2 \gamma_3 \\
\gamma^\mu \cdot \gamma_\nu &= {\delta^\mu}_\nu.
\end{aligned}
\end{equation}
Our field multivector
\begin{equation}\label{eqn:gapotentials:2060}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= \gamma_{k0} E^k + \eta \gamma_{0123k0} H^k \\
&= \gamma_{k0} E^k + \eta \gamma_{123k} H^k,
\end{aligned}
\end{equation}
now has a pure bivector representation in STA (since \( k \) will always clobber one of the \( 1,2,3 \) indexes.) To find the STA representation of Maxwell’s equation, we simply multiply both sides of our multivector representation, from the left, by \( \gamma_0 \).
\begin{equation}\label{eqn:gapotentials:2080}
\gamma_0 \lr{ \spacegrad + \partial_0 } F = \gamma_0 \lr{ \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_m – \BM } }.
\end{equation}
The LHS is just the spacetime gradient of \( F \), which we can see by expanding the product
\begin{equation}\label{eqn:gapotentials:2100}
\begin{aligned}
\gamma_0 \lr{ \spacegrad + \partial_0 }
&=
\gamma_0 \lr{ \gamma_{k0} \PD{x^k}{} + \PD{x^0}{} } \\
&=
-\gamma_{k} \PD{x^k}{} + \gamma_0 \PD{x^0}{}.
\end{aligned}
\end{equation}
This is the spacetime gradient
\begin{equation}\label{eqn:gapotentials:2120}
\grad \equiv \gamma^k \PD{x^k}{} + \gamma^0 \PD{x^0}{} = \gamma^\mu \partial_\mu.
\end{equation}
Our RHS is
\begin{equation}\label{eqn:gapotentials:2140}
\begin{aligned}
\gamma_0 \lr{ \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_m – \BM } }
&=
\gamma_0 \frac{\rho}{\epsilon} – \gamma_{0k0} \eta (\BJ \cdot \Be_k)
– I \lr{ c \rho_m \gamma_0 – \gamma_{0k0} (\BM \cdot \Be_k) } \\
&=
\gamma_0 \frac{\rho}{\epsilon} + \gamma_k \eta (\BJ \cdot \Be_k)
– I \lr{ c \rho_m \gamma_0 + \gamma_{k} (\BM \cdot \Be_k) }.
\end{aligned}
\end{equation}
If we let
\begin{equation}\label{eqn:gapotentials:2160}
\begin{aligned}
J_e^0 &= \frac{\rho}{\epsilon} \\
J_e^k &= \eta (\BJ \cdot \Be_k) \\
J_m^0 &= c \rho_m \\
J_m^k &= (\BM \cdot \Be_k) \\
J_e &= J_e^\mu \gamma_\mu \\
J_m &= J_m^\mu \gamma_\mu,
\end{aligned}
\end{equation}
then we are left with
\begin{equation}\label{eqn:gapotentials:2180}
\grad F = J_e – I J_m,
\end{equation}
or just
\begin{equation}\label{eqn:gapotentials:2640}
\grad F = J,
\end{equation}
where we now give a different meaning to \( J \) than we had in the multivector formulation. This \( J \) is now a multivector with grade(1,3) components.

Case I: potential formulation for conventional sources.

Much like we did with to find the potential formulation for the multivector form of Maxwell’s equation, we use superposition, and tackle the conventional sources, and fictitious magnetic sources separately.

With no fictitious sources, Maxwell’s equation is
\begin{equation}\label{eqn:gapotentials:2200}
\grad F = J_e,
\end{equation}
which we may split into vector and trivector components
\begin{equation}\label{eqn:gapotentials:2220}
\begin{aligned}
\grad \cdot F &= J_e \\
\grad \wedge F &= 0.
\end{aligned}
\end{equation}
Clearly, the trivector equation can be satified by setting
\begin{equation}\label{eqn:gapotentials:2240}
F = \grad \wedge A,
\end{equation}
for some vector \( A \). We may also make gauge transformations of \( A \) of the form
\begin{equation}\label{eqn:gapotentials:2260}
A \rightarrow A + \grad \psi,
\end{equation}
without changing \( F \), showing that \( A \) is not uniquely determined. With such a representation, Maxwell’s equation is now reduced to
\begin{equation}\label{eqn:gapotentials:2280}
\grad \cdot F = J_e,
\end{equation}
or
\begin{equation}\label{eqn:gapotentials:2300}
\begin{aligned}
J_e
&=
\grad \cdot \lr{ \grad \wedge A } \\
&=
\grad^2 A – \grad \lr{ \grad \cdot A }.
\end{aligned}
\end{equation}
Clearly the equivalent of the Lorentz gauge condition is now just
\begin{equation}\label{eqn:gapotentials:2320}
\grad \cdot A = 0,
\end{equation}
so the Lorentz gauge potential form of Maxwell’s equation is just
\begin{equation}\label{eqn:gapotentials:n}S
\grad^2 A = J_e.
\end{equation}

Case II: potential formulation for fictitious sources.

If we have only fictious sources, Maxwell’s equation is
\begin{equation}\label{eqn:gapotentials:2340}
\grad F = -I J_m,
\end{equation}
or after left multiplication by \( I \) we have
\begin{equation}\label{eqn:gapotentials:2360}
\grad I F = J_m.
\end{equation}
Let \( G = I F \), for the dual field, which is still a bivector. As before, we can split Maxwell’s equations into vector and trivector compoents
\begin{equation}\label{eqn:gapotentials:2380}
\begin{aligned}
\grad \cdot G &= J_m \\
\grad \wedge G &= 0.
\end{aligned}
\end{equation}
We may set
\begin{equation}\label{eqn:gapotentials:2400}
G = \grad \wedge K,
\end{equation}
for vector \( K \). Maxwell’s equation is now reduced to
\begin{equation}\label{eqn:gapotentials:2420}
\grad \cdot G = J_m,
\end{equation}
or
\begin{equation}\label{eqn:gapotentials:2440}
\begin{aligned}
J_m
&=
\grad \cdot \lr{ \grad \wedge K } \\
&=
\grad^2 K – \grad \lr{ \grad \cdot K }.
\end{aligned}
\end{equation}

As before we may make gauge transformations by adding gradient to our potential
\begin{equation}\label{eqn:gapotentials:2460}
K \rightarrow K + \grad \bar{\psi},
\end{equation}
which will not change \( G \). For such sources, the Lorentz gauge condition is \( \grad \cdot K = 0 \). With the Lorentz gauge, Maxwell’s equation is reduced to
\begin{equation}\label{eqn:gapotentials:2480}
\grad^2 K = J_m.
\end{equation}

Superposition.

For non-fictious sources, we have
\begin{equation}\label{eqn:gapotentials:2500}
F = \grad \wedge A
\end{equation}
and for fictious sources, we have
\begin{equation}\label{eqn:gapotentials:2520}
I F = G = \grad \wedge K,
\end{equation}
or
\begin{equation}\label{eqn:gapotentials:2540}
F = -I G = -I \lr{ \grad \wedge K }.
\end{equation}
Combining these results, we have
\begin{equation}\label{eqn:gapotentials:2560}
\begin{aligned}
F
&= \grad \wedge A -I \lr{ \grad \wedge K } \\
&= \gpgradetwo{ \grad \wedge A -I \lr{ \grad \wedge K } } \\
&= \gpgradetwo{ \grad A -I \lr{ \grad K } } \\
&= \gpgradetwo{ \grad \lr{ A + I K } },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:gapotentials:2580}
F = \grad \lr{ A + I K } – \gpgrade{ \grad \lr{ A + I K } }{0,4}.
\end{equation}
Maxwell’s equation is
\begin{equation}\label{eqn:gapotentials:2600}
\grad^2 \lr{ A + I K } – \grad \gpgrade{ \grad \lr{ A + I K } }{0,4} = J.
\end{equation}
With the Lorentz gauge, this splits nicely into one forced wave equation for each vector potential
\begin{equation}\label{eqn:gapotentials:2620}
\begin{aligned}
\grad^2 A &= J_e \\
\grad^2 K &= -J_m.
\end{aligned}
\end{equation}

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.

[2] R.P. Feynman, R.B. Leighton, and M.L. Sands. Feynman lectures on physics, Volume II.[Lectures on physics], chapter The Maxwell Equations. Addison-Wesley Publishing Company. Reading, Massachusetts, 1963. URL https://www.feynmanlectures.caltech.edu/II_18.html.

[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[4] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

Geometric algebra, exact and least squares solutions of two variable linear system

September 25, 2023 math and physics play , , , , , , , , ,

New video (on Google’s CensorshipTube):

Exact system.

Recall that we can use the wedge product to solve linear systems. For example, assuming that \( \Ba, \Bb \) are not colinear, the system
\begin{equation}\label{eqn:cramersProjection:20}
x \Ba + y \Bb = \Bc,
\end{equation}
if it has a solution, can be solved for \( x \) and \( y \) by wedging with \( \Bb \), and \( \Ba \) respectively.
For example, wedging with \( \Bb \), from the right, gives
\begin{equation}\label{eqn:cramersProjection:40}
x \lr{ \Ba \wedge \Bb } + y \lr{ \Bb \wedge \Bb } = \Bc \wedge \Bb,
\end{equation}
but since \( \Bb \wedge \Bb = 0 \), we are left with
\begin{equation}\label{eqn:cramersProjection:60}
x \lr{ \Ba \wedge \Bb } = \Bc \wedge \Bb,
\end{equation}
and since \( \Ba, \Bb \) are not colinear, which means that \( \Ba \wedge \Bb \ne 0 \), we have
\begin{equation}\label{eqn:cramersProjection:80}
x = \inv{ \Ba \wedge \Bb } \Bc \wedge \Bb.
\end{equation}
Similarly, we can wedge with \( \Ba \) (from the left), to find
\begin{equation}\label{eqn:cramersProjection:100}
y = \inv{ \Ba \wedge \Bb } \Ba \wedge \Bc.
\end{equation}
This works because, if the system has a solution, all the bivectors \( \Ba \wedge \Bb \), \( \Ba \wedge \Bc \), and \( \Bb \wedge \Bc \), are all scalar multiples of each other, so we can just divide the two bivectors, and the results must be scalars.

Cramer’s rule.

Incidentally, observe that for \(\mathbb{R}^2\), this is the “Cramer’s rule” solution to the system, since
\begin{equation}\label{eqn:cramersProjection:180}
\Bx \wedge \By = \begin{vmatrix} \Bx & \By \end{vmatrix} \Be_1 \Be_2,
\end{equation}
where we are treating \( \Bx \) and \( \By \) here as column vectors of the coordinates. This means that, after dividing out the plane pseudoscalar \( \Be_1 \Be_2 \), we have
\begin{equation}\label{eqn:cramersProjection:200}
\begin{aligned}
x
&=
\frac{
\begin{vmatrix}
\Bc & \Bb \\
\end{vmatrix}
}{
\begin{vmatrix}
\Ba & \Bb
\end{vmatrix}
} \\
y
&=
\frac{
\begin{vmatrix}
\Ba & \Bc \\
\end{vmatrix}
}{
\begin{vmatrix}
\Ba & \Bb
\end{vmatrix}
}.
\end{aligned}
\end{equation}
This follows the usual Cramer’s rule proscription, where we form determinants of the coordinates of the spanning vectors, replace either of the original vectors in the numerator with the target vector (depending on which variable we seek), and then take ratios of the two determinants.

Least squares solution, using geometry.

Now, let’s consider the case, where the system \ref{eqn:cramersProjection:20} cannot be solved exactly. Geometrically, the best we can do is to try to solve the related “least squares” problem
\begin{equation}\label{eqn:cramersProjection:120}
x \Ba + y \Bb = \Bc_\parallel,
\end{equation}
where \( \Bc_\parallel \) is the projection of \( \Bc \) onto the plane spanned by \( \Ba, \Bb \). Regardless of the value of \( \Bc \), we can always find a solution to this problem. For example, solving for \( x \), we have
\begin{equation}\label{eqn:cramersProjection:160}
\begin{aligned}
x
&= \inv{ \Ba \wedge \Bb } \Bc_\parallel \wedge \Bb \\
&= \inv{ \Ba \wedge \Bb } \cdot \lr{ \Bc_\parallel \wedge \Bb } \\
&= \inv{ \Ba \wedge \Bb } \cdot \lr{ \Bc \wedge \Bb } – \inv{ \Ba \wedge \Bb } \cdot \lr{ \Bc_\perp \wedge \Bb }.
\end{aligned}
\end{equation}
Let’s look at the second term, which can be written
\begin{equation}\label{eqn:cramersProjection:140}
\begin{aligned}
– \inv{ \Ba \wedge \Bb } \cdot \lr{ \Bc_\perp \wedge \Bb }
&=
– \frac{ \Ba \wedge \Bb }{ \lr{ \Ba \wedge \Bb}^2 } \cdot \lr{ \Bc_\perp \wedge \Bb } \\
&\propto
\lr{ \Ba \wedge \Bb } \cdot \lr{ \Bc_\perp \wedge \Bb } \\
&=
\lr{ \lr{ \Ba \wedge \Bb } \cdot \Bc_\perp } \cdot \Bb \\
&=
\lr{ \Ba \lr{ \Bb \cdot \Bc_\perp} – \Bb \lr{ \Ba \cdot \Bc_\perp} } \cdot \Bb \\
&=
0.
\end{aligned}
\end{equation}
The zero above follows because \( \Bc_\perp \) is perpendicular to both \( \Ba \) and \( \Bb \) by construction. Geometrically, we are trying to dot two perpendicular bivectors, where \( \Bb \) is a common factor of those two bivectors, as illustrated in fig. 1.

fig. 1. Perpendicular bivectors.

We see that our least squares solution, to this two variable linear system problem, is
\begin{equation}\label{eqn:cramersProjection:220}
x = \inv{ \Ba \wedge \Bb } \cdot \lr{ \Bc \wedge \Bb }.
\end{equation}
\begin{equation}\label{eqn:cramersProjection:240}
y = \inv{ \Ba \wedge \Bb } \cdot \lr{ \Ba \wedge \Bc }.
\end{equation}

The interesting thing here is how we have managed to connect the geometric notion of the optimal solution, the equivalent of a least squares solution (which we can compute with the Moore-Penrose inverse, or with an SVD (Singular Value Decomposition)), with the entirely geometric notion of selecting for the portion of the desired solution that lies within the span of the set of input vectors, provided that the spanning vectors for that hyperplane are linearly independent.

Least squares solution, using calculus.

I’ve called the projection solution, a least-squares solution, without full justification. Here’s that justification. We define the usual error function, the squared distance from the target, from our superposition position in the plane
\begin{equation}\label{eqn:cramersProjection:300}
\epsilon = \lr{ \Bc – x \Ba – y \Bb }^2,
\end{equation}
and then take partials with respect to \( x, y \), equating each to zero
\begin{equation}\label{eqn:cramersProjection:320}
\begin{aligned}
0 &= \PD{x}{\epsilon} = 2 \lr{ \Bc – x \Ba – y \Bb } \cdot (-\Ba) \\
0 &= \PD{y}{\epsilon} = 2 \lr{ \Bc – x \Ba – y \Bb } \cdot (-\Bb).
\end{aligned}
\end{equation}
This is a two equation, two unknown system, which can be expressed in matrix form as
\begin{equation}\label{eqn:cramersProjection:340}
\begin{bmatrix}
\Ba^2 & \Ba \cdot \Bb \\
\Ba \cdot \Bb & \Bb^2
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
\Ba \cdot \Bc \\
\Bb \cdot \Bc \\
\end{bmatrix}.
\end{equation}
This has solution
\begin{equation}\label{eqn:cramersProjection:360}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\inv{
\begin{vmatrix}
\Ba^2 & \Ba \cdot \Bb \\
\Ba \cdot \Bb & \Bb^2
\end{vmatrix}
}
\begin{bmatrix}
\Bb^2 & -\Ba \cdot \Bb \\
-\Ba \cdot \Bb & \Ba^2
\end{bmatrix}
\begin{bmatrix}
\Ba \cdot \Bc \\
\Bb \cdot \Bc \\
\end{bmatrix}
=
\frac{
\begin{bmatrix}
\Bb^2 \lr{ \Ba \cdot \Bc } – \lr{ \Ba \cdot \Bb} \lr{ \Bb \cdot \Bc } \\
\Ba^2 \lr{ \Bb \cdot \Bc } – \lr{ \Ba \cdot \Bb} \lr{ \Ba \cdot \Bc } \\
\end{bmatrix}
}{
\Ba^2 \Bb^2 – \lr{ \Ba \cdot \Bb }^2
}.
\end{equation}

All of these differences can be expressed as wedge dot products, using the following expansions in reverse
\begin{equation}\label{eqn:cramersProjection:420}
\begin{aligned}
\lr{ \Ba \wedge \Bb } \cdot \lr{ \Bc \wedge \Bd }
&=
\Ba \cdot \lr{ \Bb \cdot \lr{ \Bc \wedge \Bd } } \\
&=
\Ba \cdot \lr{ \lr{\Bb \cdot \Bc} \Bd – \lr{\Bb \cdot \Bd} \Bc } \\
&=
\lr{ \Ba \cdot \Bd } \lr{\Bb \cdot \Bc} – \lr{ \Ba \cdot \Bc }\lr{\Bb \cdot \Bd}.
\end{aligned}
\end{equation}

We find
\begin{equation}\label{eqn:cramersProjection:380}
\begin{aligned}
x
&= \frac{\Bb^2 \lr{ \Ba \cdot \Bc } – \lr{ \Ba \cdot \Bb} \lr{ \Bb \cdot \Bc }}{-\lr{ \Ba \wedge \Bb }^2 } \\
&= \frac{\lr{ \Ba \wedge \Bb } \cdot \lr{ \Bb \wedge \Bc }}{ -\lr{ \Ba \wedge \Bb }^2 } \\
&= \inv{ \Ba \wedge \Bb } \cdot \lr{ \Bc \wedge \Bb },
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:cramersProjection:400}
\begin{aligned}
y
&= \frac{\Ba^2 \lr{ \Bb \cdot \Bc } – \lr{ \Ba \cdot \Bb} \lr{ \Ba \cdot \Bc } }{-\lr{ \Ba \wedge \Bb }^2 } \\
&= \frac{- \lr{ \Ba \wedge \Bb } \cdot \lr{ \Ba \wedge \Bc } }{ -\lr{ \Ba \wedge \Bb }^2 } \\
&= \inv{ \Ba \wedge \Bb } \cdot \lr{ \Ba \wedge \Bc }.
\end{aligned}
\end{equation}
Sure enough, we find what was dubbed our least squares solution, which we now know can be written out as a ratio of (dotted) wedge products.
From \ref{eqn:cramersProjection:340}, it wasn’t obvious that the least squares solution would have a structure that was almost Cramer’s rule like, but having solved this problem using geometry alone, we knew to expect that. It was therefore natural to write the results in terms of wedge products factors, and find the simplest statement of the end result. That end result reduces to Cramer’s rule for the \(\mathbb{R}^2\) special case where the system has an exact solution.

Static load with two forces in a plane, solved a few different ways.

February 12, 2023 math and physics play , , , , , , , , , , , ,

[Click here for a PDF version of this post]

There’s a class of simple statics problems that are pervasive in high school physics and first year engineering classes (for me that CIV102.)  These problems are illustrated in the figures below. Here we have a static load under gravity, and two supporting members (rigid beams or wire lines), which can be under compression, or tension, depending on the geometry.

The problem, given the geometry, is to find the magnitudes of the forces in the two members. The equation to solve is of the form
\begin{equation}\label{eqn:twoForceStaticsProblem:20}
\BF_s + \BF_r + m \Bg = 0.
\end{equation}
The usual way to solve such a problem is to resolve the forces into components. We will do that first here as a review, but then also solve the system using GA techniques, which are arguably simpler or more direct.

Solving as a conventional vector equation.

If we were back in high school we could have written our forces out in vector form
\begin{equation}\label{eqn:twoForceStaticsProblem:160}
\begin{aligned}
\BF_r &= f_r \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } \\
\BF_s &= f_s \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } \\
\Bg &= g \Be_1.
\end{aligned}
\end{equation}
Here the gravitational direction has been pointed along the x-axis.

Our equation to solve is now
\begin{equation}\label{eqn:twoForceStaticsProblem:180}
f_r \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } + f_s \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } + m g \Be_1 = 0.
\end{equation}
This we can solve as a set of scalar equations, one for each of the \( \Be_1 \) and \( \Be_2 \) directions
\begin{equation}\label{eqn:twoForceStaticsProblem:200}
\begin{aligned}
f_r \cos\alpha + f_s \cos\beta + m g &= 0 \\
f_r \sin\alpha + f_s \sin\beta &= 0.
\end{aligned}
\end{equation}
Our solution is
\begin{equation}\label{eqn:twoForceStaticsProblem:220}
\begin{aligned}
\begin{bmatrix}
f_r \\
f_s
\end{bmatrix}
&=
{\begin{bmatrix}
\cos\alpha & \cos\beta \\
\sin\alpha & \sin\beta
\end{bmatrix}}^{-1}
\begin{bmatrix}
– m g \\
0
\end{bmatrix} \\
&=
\inv{
\cos\alpha \sin\beta – \cos\beta \sin\alpha
}
\begin{bmatrix}
\sin\beta & -\cos\beta \\
-\sin\alpha & \cos\alpha
\end{bmatrix}
\begin{bmatrix}
– m g \\
0
\end{bmatrix} \\
&=
\frac{ m g }{ \cos\alpha \sin\beta – \cos\beta \sin\alpha }
\begin{bmatrix}
-\sin\beta \\
\sin\alpha
\end{bmatrix} \\
&=
\frac{ m g }{ \sin\lr{ \beta – \alpha } }
\begin{bmatrix}
-\sin\beta \\
\sin\alpha
\end{bmatrix}.
\end{aligned}
\end{equation}
We have to haul out some trig identities to make a final simplification, but find a solution to the system.

Another approach, is to take cross products with the unit force direction.  First note that
\begin{equation}\label{eqn:twoForceStaticsProblem:240}
\begin{aligned}
\lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } \cross \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta }
&=
\Be_3 \lr{
\cos\alpha \sin\beta – \sin\alpha \cos\beta
} \\
&=
\Be_3 \sin\lr{ \beta – \alpha }.
\end{aligned}
\end{equation}

If we take cross products with each of the unit vectors, we find
\begin{equation}\label{eqn:twoForceStaticsProblem:260}
\begin{aligned}
f_r \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } \cross \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } + m g \Be_1 \cross \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } &= 0 \\
f_s \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } \cross \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } + m g \Be_1 \cross \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } &= 0,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:twoForceStaticsProblem:280}
\begin{aligned}
\Be_3 f_r \sin\lr{ \beta – \alpha } + m g \Be_3 \sin\beta &= 0 \\
-\Be_3 f_s \sin\lr{ \beta – \alpha } + m g \Be_3 \sin\alpha &= 0.
\end{aligned}
\end{equation}
After cancelling the \( \Be_3 \)’s, we find the same result as we did solving the scalar system. This was a fairly direct way to solve the system, but the intermediate cross products were a bit messy. We will try this cross product using the wedge product. Switching from the cross to the wedge, by itself, will not make things any simpler or more complicated, but we can use the complex exponential form of the unit vectors for the forces, and that will make things simpler.

Geometric algebra setup and solution.

As usual for planar problems, let’s write \( i = \Be_1 \Be_2 \) for the plane pseudoscalar, which allows us to write the forces in polar form
\begin{equation}\label{eqn:twoForceStaticsProblem:40}
\begin{aligned}
\BF_r &= f_r \Be_1 e^{i\alpha} \\
\BF_s &= f_s \Be_1 e^{i\beta} \\
\Bg &= g \Be_1,
\end{aligned}
\end{equation}
Our equation to solve is now
\begin{equation}\label{eqn:twoForceStaticsProblem:60}
f_r \Be_1 e^{i\alpha} + f_s \Be_1 e^{i\beta} + m g \Be_1 = 0.
\end{equation}
The solution for either \( f_r \) or \( f_s \) is now trivial, as we only have to take wedge products with the force direction vectors to solve for the magnitudes.  That is
\begin{equation}\label{eqn:twoForceStaticsProblem:80}
\begin{aligned}
f_r \lr{ \Be_1 e^{i\alpha} +} \wedge \lr{ \Be_1 e^{i\beta} } + m g \Be_1 \wedge \lr{ \Be_1 e^{i\beta} } &= 0 \\
f_s \lr{ \Be_1 e^{i\beta} +} \wedge \lr{ \Be_1 e^{i\alpha} } + m g \Be_1 \wedge \lr{ \Be_1 e^{i\alpha} } &= 0.
\end{aligned}
\end{equation}
Writing the wedges as grade two selections, and noting that \( e^{i\theta} \Be_1 = \Be_1 e^{-i\theta } \), we have
\begin{equation}\label{eqn:twoForceStaticsProblem:100}
\begin{aligned}
f_r &= – m g \frac{ \gpgradetwo{\Be_1^2 e^{i\beta}} }{ \gpgradetwo{ \Be_1^2 e^{-i\alpha} e^{i\beta} } } = – m g \frac{ \sin\beta }{ \sin\lr{ \beta – \alpha } } \\
f_s &= – m g \frac{ \gpgradetwo{\Be_1^2 e^{i\alpha}} }{ \gpgradetwo{ \Be_1^2 e^{-i\beta} e^{i\alpha} } } = m g \frac{ \sin\alpha }{ \sin\lr{ \beta – \alpha } }.
\end{aligned}
\end{equation}
The grade selection a unit pseudoscalar factor in both the denominator and numerator, which cancelled out to give the final scalar result.

As a complex variable problem.

Observe that we could have reframed the problem as a multivector problem by left multiplying \ref{eqn:twoForceStaticsProblem:60} by \( \Be_1 \) to find
\begin{equation}\label{eqn:twoForceStaticsProblem:120}
f_r e^{i\alpha} + f_s e^{i\beta} + m g = 0.
\end{equation}
Alternatively, we could have written the equations this way directly as a complex variable problem.

We can now solve for \( f_r \) or \( f_s \) by multiplying by the conjugate of one of the complex exponentials. That is
\begin{equation}\label{eqn:twoForceStaticsProblem:140}
\begin{aligned}
f_r + f_s e^{i\beta} e^{-i\alpha} + m g e^{-i\alpha} &= 0 \\
f_r e^{i\alpha} e^{-i\beta} + f_s + m g e^{-i\beta} &= 0.
\end{aligned}
\end{equation}
Selecting the bivector part of these equations (if interpreted as a multivector equation), or selecting the imaginary (if interpreting as a complex variables equation), will eliminate one of the force magnitudes from each equation, after which we find the same result.

This last approach, treating the problem as either a complex number problem (selecting imaginaries), or multivector problem (selecting bivectors), seems the simplest. We have no messing cross products, nor do we have to haul out the trig identities (the sine difference in the denominator comes practically for free, as it did with the wedge product method.)

Angular momentum bivector in cylindrical and spherical bases.

September 15, 2022 math and physics play , , , , , , ,

[Click here for a PDF version of this post]

Motivation

In a discord thread on the bivector group (a geometric algebra group chat), MoneyKills posts about trouble he has calculating the correct expression for the angular momentum bivector or it’s dual.

This blog post is a more long winded answer than my bivector response and includes this calculation using both cylindrical and spherical coordinates.

Cylindrical coordinates.

The position vector for any point on a plane can be expressed as
\begin{equation}\label{eqn:amomentum:20}
\Br = r \rcap,
\end{equation}
where \( \rcap = \rcap(\phi) \) encodes all the angular dependence of the position vector, and \( r \) is the length along that direction to our point, as illustrated in fig. 1.

fig. 1. Cylindrical coordinates position vector.

The radial unit vector has a compact GA representation
\begin{equation}\label{eqn:amomentum:40}
\rcap = \Be_1 e^{i\phi},
\end{equation}
where \( i = \Be_1 \Be_2 \).

The velocity (or momentum) will have both \( \rcap \) and \( \phicap \) dependence. By chain rule, that velocity is
\begin{equation}\label{eqn:amomentum:60}
\Bv = \dot{r} \rcap + r \dot{\rcap},
\end{equation}
where
\begin{equation}\label{eqn:amomentum:80}
\begin{aligned}
\dot{\rcap}
&= \Be_1 i e^{i\phi} \dot{\phi} \\
&= \Be_2 e^{i\phi} \dot{\phi} \\
&= \phicap \dot{\phi}.
\end{aligned}
\end{equation}
It is left to the reader to show that the vector designated \( \phicap \), is a unit vector and perpendicular to \( \rcap \) (Hint: compute the grade-0 selection of the product of the two to show that they are perpendicular.)

We can now compute the momentum, which is
\begin{equation}\label{eqn:amomentum:100}
\Bp = m \Bv = m \lr{ \dot{r} \rcap + r \dot{\phi} \phicap },
\end{equation}
and the angular momentum bivector
\begin{equation}\label{eqn:amomentum:120}
\begin{aligned}
L
&= \Br \wedge \Bp \\
&= m \lr{ r \rcap } \wedge \lr{ \dot{r} \rcap + r \dot{\phi} \phicap } \\
&= m r^2 \dot{\phi} \rcap \phicap.
\end{aligned}
\end{equation}

This has the \( m r^2 \dot{\phi} \) magnitude that the OP was seeking.

Spherical coordinates.

In spherical coordinates, our position vector is
\begin{equation}\label{eqn:amomentum:140}
\Br = r \lr{ \Be_1 \sin\theta \cos\phi + \Be_2 \sin\theta \sin\phi + \Be_3 \cos\theta },
\end{equation}
as sketched in fig. 2.

fig. 2. Spherical coordinates.

We can factor this into a more compact representation
\begin{equation}\label{eqn:amomentum:160}
\begin{aligned}
\Br
&= r \lr{ \sin\theta \Be_1 (\cos\phi + \Be_{12} \sin\phi ) + \Be_3 \cos\theta } \\
&= r \lr{ \sin\theta \Be_1 e^{\Be_{12} \phi } + \Be_3 \cos\theta } \\
&= r \Be_3 \lr{ \cos\theta + \sin\theta \Be_3 \Be_1 e^{\Be_{12} \phi } }.
\end{aligned}
\end{equation}

It is useful to name two of the bivector terms above, first, we write \( i \) for the azimuthal plane bivector sketched in fig. 3.

Spherical coordinates, azimuthal plane.

\begin{equation}\label{eqn:amomentum:180}
i = \Be_{12},
\end{equation}
and introduce a bivector \( j \) that encodes the \( \Be_3, \rcap \) plane as sketched in fig. 4.

Spherical coordinates, “j-plane”.

\begin{equation}\label{eqn:amomentum:200}
j = \Be_{31} e^{i \phi}.
\end{equation}

Having done so, we now have a compact representation for our position vector
\begin{equation}\label{eqn:amomentum:220}
\begin{aligned}
\Br
&= r \Be_3 \lr{ \cos\theta + j \sin\theta } \\
&= r \Be_3 e^{j \theta}.
\end{aligned}
\end{equation}

This provides us with a nice compact representation of the radial unit vector
\begin{equation}\label{eqn:amomentum:240}
\rcap = \Be_3 e^{j \theta}.
\end{equation}

Just as was the case in cylindrical coordinates, our azimuthal plane unit vector is
\begin{equation}\label{eqn:amomentum:280}
\phicap = \Be_2 e^{i\phi}.
\end{equation}

Now we want to compute the velocity vector. As was the case in cylindrical coordinates, we have
\begin{equation}\label{eqn:amomentum:300}
\Bv = \dot{r} \rcap + r \dot{\rcap},
\end{equation}
but now we need the spherical representation for the \( \rcap \) derivative, which is
\begin{equation}\label{eqn:amomentum:320}
\begin{aligned}
\dot{\rcap}
&=
\PD{\theta}{\rcap} \dot{\theta} + \PD{\phi}{\rcap} \dot{\phi} \\
&=
\Be_3 e^{j\theta} j \dot{\theta} + \Be_3 \sin\theta \PD{\phi}{j} \dot{\phi} \\
&=
\rcap j \dot{\theta} + \Be_3 \sin\theta j i \dot{\phi}.
\end{aligned}
\end{equation}
We can reduce the second multivector term without too much work
\begin{equation}\label{eqn:amomentum:340}
\begin{aligned}
\Be_3 j i
&=
\Be_3 \Be_{31} e^{i\phi} i \\
&=
\Be_3 \Be_{31} i e^{i\phi} \\
&=
\Be_{33112} e^{i\phi} \\
&=
\Be_{2} e^{i\phi} \\
&= \phicap,
\end{aligned}
\end{equation}
so we have
\begin{equation}\label{eqn:amomentum:360}
\dot{\rcap}
=
\rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi}.
\end{equation}

The velocity is
\begin{equation}\label{eqn:amomentum:380}
\Bv = \dot{r} \rcap + r \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} }.
\end{equation}

Now we can finally compute the angular momentum bivector, which is
\begin{equation}\label{eqn:amomentum:400}
\begin{aligned}
L &=
\Br \wedge \Bp \\
&=
m r \rcap \wedge \lr{ \dot{r} \rcap + r \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } } \\
&=
m r^2 \rcap \wedge \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } \\
&=
m r^2 \gpgradetwo{ \rcap \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } },
\end{aligned}
\end{equation}
which is just
\begin{equation}\label{eqn:amomentum:420}
L =
m r^2 \lr{ j \dot{\theta} + \sin\theta \rcap \phicap \dot{\phi} }.
\end{equation}

I was slightly surprised by this result, as I naively expected the cylindrical coordinate result. We have a \( m r^2 \rcap \phicap \dot{\phi} \) term, as was the case in cylindrical coordinates, but scaled down with a \( \sin\theta \) factor. However, this result does make sense. Consider for example, some fixed circular motion with \( \theta = \mathrm{constant} \), as sketched in fig. 5.

fig. 5. Circular motion for constant theta

The radius of this circle is actually \( r \sin\theta \), so the total angular momentum for that motion is scaled down to \( m r^2 \sin\theta \dot{\phi} \), smaller than the maximum circular angular momentum of \( m r^2 \dot{\phi} \) which occurs in the \( \theta = \pi/2 \) azimuthal plane. Similarly, if we have circular motion in the “j-plane”, sketched in fig. 6.

fig. 6. Circular motion for constant phi.

where \( \phi = \mathrm{constant} \), then our angular momentum is \( L = m r^2 j \dot{\theta} \).

A multivector Lagrangian for Maxwell’s equation: A summary of previous exploration.

June 21, 2022 math and physics play , , , , , , , , , , , , , , , , , , , ,

This summarizes the significant parts of the last 8 blog posts.

[Click here for a PDF version of this post]

STA form of Maxwell’s equation.

Maxwell’s equations, with electric and fictional magnetic sources (useful for antenna theory and other engineering applications), are
\begin{equation}\label{eqn:maxwellLagrangian:220}
\begin{aligned}
\spacegrad \cdot \BE &= \frac{\rho}{\epsilon} \\
\spacegrad \cross \BE &= – \BM – \mu \PD{t}{\BH} \\
\spacegrad \cdot \BH &= \frac{\rho_\txtm}{\mu} \\
\spacegrad \cross \BH &= \BJ + \epsilon \PD{t}{\BE}.
\end{aligned}
\end{equation}
We can assemble these into a single geometric algebra equation,
\begin{equation}\label{eqn:maxwellLagrangian:240}
\lr{ \spacegrad + \inv{c} \PD{t}{} } F = \eta \lr{ c \rho – \BJ } + I \lr{ c \rho_{\mathrm{m}} – \BM },
\end{equation}
where \( F = \BE + \eta I \BH = \BE + I c \BB \), \( c = 1/\sqrt{\mu\epsilon}, \eta = \sqrt{(\mu/\epsilon)} \).

By multiplying through by \( \gamma_0 \), making the identification \( \Be_k = \gamma_k \gamma_0 \), and
\begin{equation}\label{eqn:maxwellLagrangian:300}
\begin{aligned}
J^0 &= \frac{\rho}{\epsilon}, \quad J^k = \eta \lr{ \BJ \cdot \Be_k }, \quad J = J^\mu \gamma_\mu \\
M^0 &= c \rho_{\mathrm{m}}, \quad M^k = \BM \cdot \Be_k, \quad M = M^\mu \gamma_\mu \\
\grad &= \gamma^\mu \partial_\mu,
\end{aligned}
\end{equation}
we find the STA form of Maxwell’s equation, including magnetic sources
\begin{equation}\label{eqn:maxwellLagrangian:320}
\grad F = J – I M.
\end{equation}

Decoupling the electric and magnetic fields and sources.

We can utilize two separate four-vector potential fields to split Maxwell’s equation into two parts. Let
\begin{equation}\label{eqn:maxwellLagrangian:1740}
F = F_{\mathrm{e}} + I F_{\mathrm{m}},
\end{equation}
where
\begin{equation}\label{eqn:maxwellLagrangian:1760}
\begin{aligned}
F_{\mathrm{e}} &= \grad \wedge A \\
F_{\mathrm{m}} &= \grad \wedge K,
\end{aligned}
\end{equation}
and \( A, K \) are independent four-vector potential fields. Plugging this into Maxwell’s equation, and employing a duality transformation, gives us two coupled vector grade equations
\begin{equation}\label{eqn:maxwellLagrangian:1780}
\begin{aligned}
\grad \cdot F_{\mathrm{e}} – I \lr{ \grad \wedge F_{\mathrm{m}} } &= J \\
\grad \cdot F_{\mathrm{m}} + I \lr{ \grad \wedge F_{\mathrm{e}} } &= M.
\end{aligned}
\end{equation}
However, since \( \grad \wedge F_{\mathrm{m}} = \grad \wedge F_{\mathrm{e}} = 0 \), by construction, the curls above are killed. We may also add in \( \grad \wedge F_{\mathrm{e}} = 0 \) and \( \grad \wedge F_{\mathrm{m}} = 0 \) respectively, yielding two independent gradient equations
\begin{equation}\label{eqn:maxwellLagrangian:1810}
\begin{aligned}
\grad F_{\mathrm{e}} &= J \\
\grad F_{\mathrm{m}} &= M,
\end{aligned}
\end{equation}
one for each of the electric and magnetic sources and their associated fields.

Tensor formulation.

The electromagnetic field \( F \), is a vector-bivector multivector in the multivector representation of Maxwell’s equation, but is a bivector in the STA representation. The split of \( F \) into it’s electric and magnetic field components is observer dependent, but we may write it without reference to a specific observer frame as
\begin{equation}\label{eqn:maxwellLagrangian:1830}
F = \inv{2} \gamma_\mu \wedge \gamma_\nu F^{\mu\nu},
\end{equation}
where \( F^{\mu\nu} \) is an arbitrary antisymmetric 2nd rank tensor. Maxwell’s equation has a vector and trivector component, which may be split out explicitly using grade selection, to find
\begin{equation}\label{eqn:maxwellLagrangian:360}
\begin{aligned}
\grad \cdot F &= J \\
\grad \wedge F &= -I M.
\end{aligned}
\end{equation}
Further dotting and wedging these equations with \( \gamma^\mu \) allows for extraction of scalar relations
\begin{equation}\label{eqn:maxwellLagrangian:460}
\partial_\nu F^{\nu\mu} = J^{\mu}, \quad \partial_\nu G^{\nu\mu} = M^{\mu},
\end{equation}
where \( G^{\mu\nu} = -(1/2) \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta} \) is also an antisymmetric 2nd rank tensor.

If we treat \( F^{\mu\nu} \) and \( G^{\mu\nu} \) as independent fields, this pair of equations is the coordinate equivalent to \ref{eqn:maxwellLagrangian:1760}, also decoupling the electric and magnetic source contributions to Maxwell’s equation.

Coordinate representation of the Lagrangian.

As observed above, we may choose to express the decoupled fields as curls \( F_{\mathrm{e}} = \grad \wedge A \) or \( F_{\mathrm{m}} = \grad \wedge K \). The coordinate expansion of either field component, given such a representation, is straight forward. For example
\begin{equation}\label{eqn:maxwellLagrangian:1850}
\begin{aligned}
F_{\mathrm{e}}
&= \lr{ \gamma_\mu \partial^\mu } \wedge \lr{ \gamma_\nu A^\nu } \\
&= \inv{2} \lr{ \gamma_\mu \wedge \gamma_\nu } \lr{ \partial^\mu A^\nu – \partial^\nu A^\mu }.
\end{aligned}
\end{equation}

We make the identification \( F^{\mu\nu} = \partial^\mu A^\nu – \partial^\nu A^\mu \), the usual definition of \( F^{\mu\nu} \) in the tensor formalism. In that tensor formalism, the Maxwell Lagrangian is
\begin{equation}\label{eqn:maxwellLagrangian:1870}
\LL = – \inv{4} F_{\mu\nu} F^{\mu\nu} – A_\mu J^\mu.
\end{equation}
We may show this though application of the Euler-Lagrange equations
\begin{equation}\label{eqn:maxwellLagrangian:600}
\PD{A_\mu}{\LL} = \partial_\nu \PD{(\partial_\nu A_\mu)}{\LL}.
\end{equation}
\begin{equation}\label{eqn:maxwellLagrangian:1930}
\begin{aligned}
\PD{(\partial_\nu A_\mu)}{\LL}
&= -\inv{4} (2) \lr{ \PD{(\partial_\nu A_\mu)}{F_{\alpha\beta}} } F^{\alpha\beta} \\
&= -\inv{2} \delta^{[\nu\mu]}_{\alpha\beta} F^{\alpha\beta} \\
&= -\inv{2} \lr{ F^{\nu\mu} – F^{\mu\nu} } \\
&= F^{\mu\nu}.
\end{aligned}
\end{equation}
So \( \partial_\nu F^{\nu\mu} = J^\mu \), the equivalent of \( \grad \cdot F = J \), as expected.

Coordinate-free representation and variation of the Lagrangian.

Because
\begin{equation}\label{eqn:maxwellLagrangian:200}
F^2 =
-\inv{2}
F^{\mu\nu} F_{\mu\nu}
+
\lr{ \gamma_\alpha \wedge \gamma^\beta }
F_{\alpha\mu}
F^{\beta\mu}
+
\frac{I}{4}
\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta},
\end{equation}
we may express the Lagrangian \ref{eqn:maxwellLagrangian:1870} in a coordinate free representation
\begin{equation}\label{eqn:maxwellLagrangian:1890}
\LL = \inv{2} F \cdot F – A \cdot J,
\end{equation}
where \( F = \grad \wedge A \).

We will now show that it is also possible to apply the variational principle to the following multivector Lagrangian
\begin{equation}\label{eqn:maxwellLagrangian:1910}
\LL = \inv{2} F^2 – A \cdot J,
\end{equation}
and recover the geometric algebra form \( \grad F = J \) of Maxwell’s equation in it’s entirety, including both vector and trivector components in one shot.

We will need a few geometric algebra tools to do this.

The first such tool is the notational freedom to let the gradient act bidirectionally on multivectors to the left and right. We will designate such action with over-arrows, sometimes also using braces to limit the scope of the action in question. If \( Q, R \) are multivectors, then the bidirectional action of the gradient in a \( Q, R \) sandwich is
\begin{equation}\label{eqn:maxwellLagrangian:1950}
\begin{aligned}
Q \lrgrad R
&= Q \lgrad R + Q \rgrad R \\
&= \lr{ Q \gamma^\mu \lpartial_\mu } R + Q \lr{ \gamma^\mu \rpartial_\mu R } \\
&= \lr{ \partial_\mu Q } \gamma^\mu R + Q \gamma^\mu \lr{ \partial_\mu R }.
\end{aligned}
\end{equation}
In the final statement, the partials are acting exclusively on \( Q \) and \( R \) respectively, but the \( \gamma^\mu \) factors must remain in place, as they do not necessarily commute with any of the multivector factors.

This bidirectional action is a critical aspect of the Fundamental Theorem of Geometric calculus, another tool that we will require. The specific form of that theorem that we will utilize here is
\begin{equation}\label{eqn:maxwellLagrangian:1970}
\int_V Q d^4 \Bx \lrgrad R = \int_{\partial V} Q d^3 \Bx R,
\end{equation}
where \( d^4 \Bx = I d^4 x \) is the pseudoscalar four-volume element associated with a parameterization of space time. For our purposes, we may assume that parameterization are standard basis coordinates associated with the basis \( \setlr{ \gamma_0, \gamma_1, \gamma_2, \gamma_3 } \). The surface differential form \( d^3 \Bx \) can be given specific meaning, but we do not actually care what that form is here, as all our surface integrals will be zero due to the boundary constraints of the variational principle.

Finally, we will utilize the fact that bivector products can be split into grade \(0,4\) and \( 2 \) components using anticommutator and commutator products, namely, given two bivectors \( F, G \), we have
\begin{equation}\label{eqn:maxwellLagrangian:1990}
\begin{aligned}
\gpgrade{ F G }{0,4} &= \inv{2} \lr{ F G + G F } \\
\gpgrade{ F G }{2} &= \inv{2} \lr{ F G – G F }.
\end{aligned}
\end{equation}

We may now proceed to evaluate the variation of the action for our presumed Lagrangian
\begin{equation}\label{eqn:maxwellLagrangian:2010}
S = \int d^4 x \lr{ \inv{2} F^2 – A \cdot J }.
\end{equation}
We seek solutions of the variational equation \( \delta S = 0 \), that are satisfied for all variations \( \delta A \), where the four-potential variations \( \delta A \) are zero on the boundaries of this action volume (i.e. an infinite spherical surface.)

We may start our variation in terms of \( F \) and \( A \)
\begin{equation}\label{eqn:maxwellLagrangian:1540}
\begin{aligned}
\delta S
&=
\int d^4 x \lr{ \inv{2} \lr{ \delta F } F + F \lr{ \delta F } } – \lr{ \delta A } \cdot J \\
&=
\int d^4 x \gpgrade{ \lr{ \delta F } F – \lr{ \delta A } J }{0,4} \\
&=
\int d^4 x \gpgrade{ \lr{ \grad \wedge \lr{\delta A} } F – \lr{ \delta A } J }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{ \lr{\delta A} \lgrad } F – \lr{ \lr{ \delta A } \cdot \lgrad } F + \lr{ \delta A } J }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{ \lr{\delta A} \lgrad } F + \lr{ \delta A } J }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } J }{0,4},
\end{aligned}
\end{equation}
where we have used arrows, when required, to indicate the directional action of the gradient.

Writing \( d^4 x = -I d^4 \Bx \), we have
\begin{equation}\label{eqn:maxwellLagrangian:1600}
\begin{aligned}
\delta S
&=
-\int_V d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } J }{0,4} \\
&=
-\int_V \gpgrade{ -\lr{\delta A} I d^4 \Bx \lrgrad F – d^4 x \lr{\delta A} \rgrad F + d^4 x \lr{ \delta A } J }{0,4} \\
&=
\int_{\partial V} \gpgrade{ \lr{\delta A} I d^3 \Bx F }{0,4}
+ \int_V d^4 x \gpgrade{ \lr{\delta A} \lr{ \rgrad F – J } }{0,4}.
\end{aligned}
\end{equation}
The first integral is killed since \( \delta A = 0 \) on the boundary. The remaining integrand can be simplified to
\begin{equation}\label{eqn:maxwellLagrangian:1660}
\gpgrade{ \lr{\delta A} \lr{ \rgrad F – J } }{0,4} =
\gpgrade{ \lr{\delta A} \lr{ \grad F – J } }{0},
\end{equation}
where the grade-4 filter has also been discarded since \( \grad F = \grad \cdot F + \grad \wedge F = \grad \cdot F \) since \( \grad \wedge F = \grad \wedge \grad \wedge A = 0 \) by construction, which implies that the only non-zero grades in the multivector \( \grad F – J \) are vector grades. Also, the directional indicator on the gradient has been dropped, since there is no longer any ambiguity. We seek solutions of \( \gpgrade{ \lr{\delta A} \lr{ \grad F – J } }{0} = 0 \) for all variations \( \delta A \), namely
\begin{equation}\label{eqn:maxwellLagrangian:1620}
\boxed{
\grad F = J.
}
\end{equation}
This is Maxwell’s equation in it’s coordinate free STA form, found using the variational principle from a coordinate free multivector Maxwell Lagrangian, without having to resort to a coordinate expansion of that Lagrangian.

Lagrangian for fictitious magnetic sources.

The generalization of the Lagrangian to include magnetic charge and current densities can be as simple as utilizing two independent four-potential fields
\begin{equation}\label{eqn:maxwellLagrangian:n}
\LL = \inv{2} \lr{ \grad \wedge A }^2 – A \cdot J + \alpha \lr{ \inv{2} \lr{ \grad \wedge K }^2 – K \cdot M },
\end{equation}
where \( \alpha \) is an arbitrary multivector constant.

Variation of this Lagrangian provides two independent equations
\begin{equation}\label{eqn:maxwellLagrangian:1840}
\begin{aligned}
\grad \lr{ \grad \wedge A } &= J \\
\grad \lr{ \grad \wedge K } &= M.
\end{aligned}
\end{equation}
We may add these, scaling the second by \( -I \) (recall that \( I, \grad \) anticommute), to find
\begin{equation}\label{eqn:maxwellLagrangian:1860}
\grad \lr{ F_{\mathrm{e}} + I F_{\mathrm{m}} } = J – I M,
\end{equation}
which is \( \grad F = J – I M \), as desired.

It would be interesting to explore whether it is possible find Lagrangian that is dependent on a multivector potential, that would yield \( \grad F = J – I M \) directly, instead of requiring a superposition operation from the two independent solutions. One such possible potential is \( \tilde{A} = A – I K \), for which \( F = \gpgradetwo{ \grad \tilde{A} } = \grad \wedge A + I \lr{ \grad \wedge K } \). The author was not successful constructing such a Lagrangian.