Geometric Algebra

Inscribed Triangle in circle problem

December 25, 2022 math and physics play , , , , , ,

[Click here for a PDF version of this post]

In the LinkedIn Pre-University Geometric Algebra group, James presents a problem from the MindYourDecisions youtube channel Impossible Viral Problem, as a candidate for solution using geometric algebra.

I tried this out and found a couple ways to solve it. One of those I’ll detail here. I have to admit that part of the reason that I wanted to solve this is that the figure in the beginning of the video really bugged me. The triangle that was inscribed in the circle didn’t have any of the length properties from the problem. I could do much better with a sloppy freehand sketch, but to do a good figure, you have to actually solve for the vertexes of the triangle (once you do that, the area is easy to figure out.)

Formulating the problem.

Having solved the problem, the geometry of the problem is illustrated in fig. 1.

fig. 1. Inscribed triangle in circle.

fig. 1. Inscribed triangle in circle.

I set up the problem so that the \( A,C \) triangle vertices were symmetric with respect to the x-axis, and the \(B \) vertex located elsewhere. I can describe those algebraically as
\begin{equation}\label{eqn:inscribedTriangleProblem:20}
\begin{aligned}
\BA &= r \Be_1 e^{i\theta} \\
\BC &= r \Be_1 e^{-i\theta} \\
\BB &= r \Be_1 e^{i\phi},
\end{aligned}
\end{equation}
where the radius \( r \) and two angles \( \theta, \phi \) are to be determined, and \( i = \Be_1 \Be_1\) the pseudoscalar for the \(x-y\) plane.
The vector pointing to the midpoint of the upper triangular face is given by the average of the \( \BA, \BB \) vectors, which can be seen from
\begin{equation}\label{eqn:inscribedTriangleProblem:40}
\BA + \frac{\BB – \BA}{2} = \frac{\BA + \BB}{2},
\end{equation}
and similarly, the midpoint of the lower face is found at
\begin{equation}\label{eqn:inscribedTriangleProblem:60}
\BC + \frac{\BB – \BC}{2} = \frac{\BB + \BC}{2},
\end{equation}
The problem tells us that the respective lengths of those vectors from the origin are \( r-2, r – 3\) respectively, so
\begin{equation}\label{eqn:inscribedTriangleProblem:80}
\begin{aligned}
r – 2 &= \inv{2} \Abs{ \BA + \BB } \\
r – 3 &= \inv{2} \Abs{ \BB + \BC },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:inscribedTriangleProblem:100}
\begin{aligned}
(r – 2)^2 &= \frac{r^2}{4} \lr{ \Be_1 e^{i\theta} + \Be_1 e^{i\phi} }^2 \\
(r – 3)^2 &= \frac{r^2}{4} \lr{ \Be_1 e^{i\phi} + \Be_1 e^{-i\theta} }^2 \\
\end{aligned}
\end{equation}
Finally, since the midpoint of the right edge is found at \( (r-1)\Be_1 \), it is clear that
\begin{equation}\label{eqn:inscribedTriangleProblem:120}
\frac{r-1}{r} = \cos\theta,
\end{equation}
or
\begin{equation}\label{eqn:inscribedTriangleProblem:140}
r = \inv{1 – \cos\theta}.
\end{equation}
This leaves us with three equations and three unknowns. Unfortunately, these are rather non-linear equations. In the video, a direct method of solving equivalent equations was demonstrated, but I picked the lazy route, and used Mathematica’s NSolve routine, solving for \( r,\theta, \phi\) numerically. Since NSolve has intrinsic complex number support, I made the following substitutions:
\begin{equation}\label{eqn:inscribedTriangleProblem:160}
\begin{aligned}
z &= e^{i\theta} \\
w &= e^{i\phi},
\end{aligned}
\end{equation}
and then plugged those into our relations above, after expanding the squares, to find
\begin{equation}\label{eqn:inscribedTriangleProblem:180}
\begin{aligned}
\lr{ \Be_1 e^{i\theta} + \Be_1 e^{i\phi} }^2
&=
2 + \Be_1 e^{i\theta} \Be_1 e^{i\phi} + \Be_1 e^{i\phi} \Be_1 e^{i\theta} \\
&=
2 + e^{-i\theta} \Be_1^2 e^{i\phi} + e^{-i\phi} \Be_1^2 e^{i\theta} \\
&=
2 + e^{-i\theta} \Be_1^2 e^{i\phi} + e^{-i\phi} \Be_1^2 e^{i\theta} \\
&=
2 + \frac{w}{z} + \frac{z}{w},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:inscribedTriangleProblem:200}
\begin{aligned}
\lr{ \Be_1 e^{i\phi} + \Be_1 e^{-i\theta} }^2
&=
2 + \Be_1 e^{i\phi} \Be_1 e^{-i\theta} + \Be_1 e^{-i\theta} \Be_1 e^{i\phi} \\
&=
2 + e^{-i\phi} e^{-i\theta} + e^{ i\theta} e^{i\phi} \\
&=
2 + w z + \inv{w z}.
\end{aligned}
\end{equation}
This gives us
\begin{equation}\label{eqn:inscribedTriangleProblem:220}
\begin{aligned}
4 \lr{ \frac{r – 2 }{r} }^2 &= 2 + \frac{w}{z} + \frac{z}{w} \\
4 \lr{ \frac{r – 3 }{r} }^2 &= 2 + w z + \inv{w z},
\end{aligned}
\end{equation}
where
\begin{equation}\label{eqn:inscribedTriangleProblem:240}
r = \inv{1 – \inv{2}\lr{ z + \inv{z}}}.
\end{equation}

The NSolve gave me some garbage solutions (like \(\theta = 0 \)) that must have been valid numerically, but did not encode the geometry of the problem, so I added a few additional constraints to the problem, namely
\begin{equation}\label{eqn:inscribedTriangleProblem:260}
\begin{aligned}
z \bar{z} &= 1 \\
w \bar{w} &= 1 \\
\inv{2} \lr{ z + \inv{z} } &\ne 1 \\
1/(1 – (1/2) \textrm{Re}(z + 1/z)) &> 3.
\end{aligned}
\end{equation}
This provided exactly two solutions, but when plotted, they turn out to just be mirror images of each other. After back substitution, the solution illustrated above was given by
\begin{equation}\label{eqn:inscribedTriangleProblem:280}
\begin{aligned}
r &= 3.87939 \\
\theta &= 42.078 \\
\phi &= 164.125,
\end{aligned}
\end{equation}
where these angles are in degrees, not radians.

The triangular area.

There are probably lots of formulas for the area of a triangle (that I have forgotten), but we can compute it easily by doubling the triangle, forming a parallelogram, to find
\begin{equation}\label{eqn:inscribedTriangleProblem:300}
\textrm{Area} = \inv{2} \Abs{ \lr{ \BA – \BC } \wedge {\BC – \BB } },
\end{equation}
or
\begin{equation}\label{eqn:inscribedTriangleProblem:320}
\begin{aligned}
\textrm{Area}^2
&= \frac{-1}{4} \lr{ \lr{ \BA – \BC } \wedge \lr{\BC – \BB } }^2 \\
&= \frac{-1}{4} \lr{ \BA \wedge \BC – \BA \wedge \BB + \BC \wedge \BB }^2 \\
&= \frac{-r^4}{4} \lr{\gpgradetwo{ \Be_1 e^{i\theta} \Be_1 e^{-i\theta} – \Be_1 e^{i\theta} \Be_1 e^{i\phi} + \Be_1 e^{-i\theta} \Be_1 e^{i\phi} }}^2 \\
&= \frac{-r^4}{4} \lr{\gpgradetwo{ e^{-2 i \theta} – e^{i \phi -i\theta} + e^{i\theta + i \phi} }}^2,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:inscribedTriangleProblem:340}
\textrm{Area} = \frac{r^2}{2} \Abs{ -\sin( 2 \theta ) – \sin(\phi- \theta) + \sin(\theta + \phi)}.
\end{equation}
Plugging in \( r, \theta, \phi \), we find
\begin{equation}\label{eqn:inscribedTriangleProblem:360}
\textrm{Area} = 17.1866.
\end{equation}
After computing this value, I then finally watched the original video to compare my answer, and was initially disturbed to find that this wasn’t even one of the possible values. However, that was because the problem itself, as originally stated, didn’t include the correct answer, and my worry that I’d made a mistake was unfounded, as the value I computed matched what was computed in the video (it also looks “about right” visually.)

Interior angles of a regular n-sided polygon: a strange way to find them.

August 8, 2022 math and physics play , , , , , ,

[Click here for a PDF version of this post]

Fig 1. Regular polygon, interior angles.

For reasons that I can’t explain, I woke up this morning dreaming about the interior angles of regular polygons. i.e. the angles \( \pi – \theta \), as illustrated in fig. 1.

The logical way to calculate that angle would be to slice the polygon up into triangles from the center, since each slice would have an interior angle would be \( 2 \pi / N \), and then the problem is just trigonometric. However, in my dream, I was going around the outside, each time rotating by a constant angle, until reaching the original starting point. This was a vector algebra problem, instead of a trigonometric problem, as illustrated in
fig. 2.

fig. 2. Polygon vertex iteration.

I didn’t have the computational power in my dream to solve the problem, and had to write it down when I woke up, to do so. The problem has the structure of a recurrence relation:
\begin{equation}\label{eqn:regularNgon:20}
\Bp_k = \Bp_{k-1} + a \Be_1 \lr{ e^{i\theta} }^{k-1},
\end{equation}
where
\begin{equation}\label{eqn:regularNgon:40}
\Bp_N = \Bp_0.
\end{equation}

We can write these out explicitly for the first few \( k \) to see the pattern
\begin{equation}\label{eqn:regularNgon:60}
\begin{aligned}
\Bp_2
&= \Bp_{1} + a \Be_1 \lr{ e^{i\theta} }^{2-1} \\
&= \Bp_{0} + a \Be_1 \lr{ e^{i\theta} }^{1-1} + a \Be_1 \lr{ e^{i\theta} }^{2-1} \\
&= \Bp_{0} + a \Be_1 \lr{ 1 + \lr{ e^{i\theta} }^{2-1} },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:80}
\Bp_k = \Bp_{0} + a \Be_1 \lr{ 1 + e^{i\theta} + \lr{ e^{i\theta} }^{2-1} + \cdots + \lr{ e^{i\theta} }^{k-1} },
\end{equation}
so the equation to solve (for \(\theta\)) is
\begin{equation}\label{eqn:regularNgon:100}
\Bp_N = \Bp_0 + a \Be_1 \lr{ 1 + \cdots + \lr{ e^{i\theta} }^{N-1} } = \Bp_0,
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:120}
1 + \cdots + \lr{ e^{i\theta} }^{N-1} = 0.
\end{equation}
The LHS is a geometric series of the form
\begin{equation}\label{eqn:regularNgon:140}
S_N = 1 + \alpha + \cdots \alpha^{N-1}.
\end{equation}
Recall that the trick to solve this is noting that
\begin{equation}\label{eqn:regularNgon:160}
\alpha S_N = \alpha + \cdots \alpha^{N-1} + \alpha^N,
\end{equation}
so
\begin{equation}\label{eqn:regularNgon:180}
\alpha S_N – S_N = \alpha^N – 1,
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:200}
S_N = \frac{\alpha^N – 1}{\alpha – 1}.
\end{equation}
For our polygon, we seek a zero numerator, that is
\begin{equation}\label{eqn:regularNgon:220}
e^{N i \theta} = 1,
\end{equation}
and the smallest \( \theta \) solution to this equation is
\begin{equation}\label{eqn:regularNgon:240}
N \theta = 2 \pi,
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:260}
\theta = \frac{2 \pi}{N}.
\end{equation}
The interior angle is the complement of this, since we are going around the outside edge. That is
\begin{equation}\label{eqn:regularNgon:280}
\begin{aligned}
\pi – \theta &= \pi – \frac{2 \pi}{N} \\ &= \frac{ N – 2 }{N} \pi,
\end{aligned}
\end{equation}
and the sum of all the interior angles is
\begin{equation}\label{eqn:regularNgon:300}
N \lr{ \pi – \theta } = \lr{N – 2 } \pi.
\end{equation}

Plugging in some specific values, for \( N = 3, 4, 5, 6 \), we find that the interior angles are \( \pi/3, \pi/2, 3 \pi/5, 4 \pi/6 \), and the respective sums of these interior angles for the entire polygons are \( \pi, 2 \pi, 3 \pi, 4 \pi \).

Like, I said, this isn’t the simplest way to solve this problem. Instead, we could solve for \( 2 \mu \) with respect to interior triangle illustrated in
fig. 3, where

fig. 3. Polygon interior geometry.

\begin{equation}\label{eqn:regularNgon:320}
2 \mu + \frac{ 2\pi}{N} = \pi,
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:340}
2 \mu = \frac{N – 2}{N} \pi,
\end{equation}
as found the hard way. The hard way was kind of fun though.

The toughest problem to solve would be “why on earth was my brain pondering this in the early morning?” I didn’t even go to bed thinking about anything math or geometry related (we finished the night with the brain-dead activity of watching an episode of “Stranger things”.)

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.

Curl of F revisited.

June 20, 2022 math and physics play , , , , , , , , ,

This is the 8th part of a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a multivector Lagrangian representation.

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

There’s an aspect of the previous treatment that has bugged me. We’ve used a Lagrangian
\begin{equation}\label{eqn:fsquared:1440y}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4}, \end{equation}
where \( F = \grad \wedge A \), and found Maxwell’s equation by varying the Lagrangian
\begin{equation}\label{eqn:fsquared:1680}
\grad F = J – I M.
\end{equation}
However, if we decompose this into vector and trivector parts we have
\begin{equation}\label{eqn:fsquared:1700}
\begin{aligned}
\grad \cdot F &= J \\
\grad \wedge F &= -I M,
\end{aligned}
\end{equation}
and then put our original \( F = \grad \wedge A \) back in the magnetic term of this equation, we have a contradiction
\begin{equation}\label{eqn:fsquared:1720}
0 = -I M,
\end{equation}
since
\begin{equation}\label{eqn:fsquared:1880}
\grad \wedge \lr{ \grad \wedge A } = 0,
\end{equation}
provided we have equality of mixed partials for \( A \). The resolution to this contradiction appears to be a requirement to define the field differently. In particular, we can utilize two separate four-vector potential fields to split Maxwell’s equation into two parts. Let
\begin{equation}\label{eqn:fsquared:1740}
F = F_{\mathrm{e}} + I F_{\mathrm{m}},
\end{equation}
where
\begin{equation}\label{eqn:fsquared: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:fsquared: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 \( \grad \wedge F_{\mathrm{m}} = \grad \wedge F_{\mathrm{e}} = 0 \), these decouple trivially, leaving
\begin{equation}\label{eqn:fsquared:1800}
\begin{aligned}
\grad \cdot F_{\mathrm{e}} &= J \\
\grad \cdot F_{\mathrm{m}} &= M.
\end{aligned}
\end{equation}
In fact, again, since \( \grad \wedge F_{\mathrm{m}} = \grad \wedge F_{\mathrm{e}} = 0 \), these are equivalent to two independent gradient equations
\begin{equation}\label{eqn:fsquared: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.

Should we wish to recover these two equations from a Lagrangian, we form a multivector Lagrangian that uses two independent four-vector fields
\begin{equation}\label{eqn:fsquared:1820}
\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:fsquared: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:fsquared:1860}
\grad \lr{ F_{\mathrm{e}} + I F_{\mathrm{m}} } = J – I M,
\end{equation}
which is \( \grad F = J – I M \), as desired. This resolves the eq \ref{eqn:fsquared:1720} conundrum, but the cost is that we essentially have an independent Lagrangian for each of the electric and magnetic sources. I think that is the cost of correctness. Perhaps there is an alternative Lagrangian for the electric+magnetic case that yields all of Maxwell’s equation in one shot. My attempts to formulate one in terms of the total field \( F = F_{\mathrm{e}} + I F_{\mathrm{m}} \) have not been successful.

On the positive side, for non-fictitious electric sources, the case that we care about in physics, we still have the pleasantry of being able to use a simple multivector (coordinate-free) Lagrangian, and vary that in a coordinate free fashion to find Maxwell’s equation. This has an aesthetic quality that is arguably superior to the usual procedure of using the Euler-Lagrange equations and lots of index gymnastics to find the tensor form of Maxwell’s equation (i.e. the vector part of Maxwell’s) and applying the Bianchi identity to fill in the pieces (i.e. the trivector component of Maxwell’s.)

A coordinate free variation of the Maxwell equation multivector Lagrangian.

June 18, 2022 math and physics play , , , , , , , , ,

This is the 7th part of a series on finding Maxwell’s equations (including the fictitious magnetic sources that are useful in engineering) from a multivector Lagrangian representation.

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

For what is now (probably) the final step in this exploration, we now wish to evaluate the variation of the multivector Maxwell Lagrangian
\begin{equation}\label{eqn:fsquared:1440x}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4},
\end{equation}
without resorting to coordinate expansion of any part of \( F = \grad \wedge A \). We’d initially evaluated this, expanding both \( \grad \) and \( A \) in coordinates, and then just \( \grad \), but we can avoid both.
In particular, given a coordinate free Lagrangian, and a coordinate free form of Maxwell’s equation as the final destination, there must be a way to get there directly.

It is clear how to work through the first part of the action variation argument, without resorting to any sort of coordinate expansion
\begin{equation}\label{eqn:fsquared:1540}
\begin{aligned}
\delta S
&=
\int d^4 x \lr{ \inv{2} \lr{ \delta F } F + F \lr{ \delta F } } – \gpgrade{ \lr{ \delta F } \lr{ J – I M } }{0,4} \\
&=
\int d^4 x \gpgrade{ \lr{ \delta F } F – \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
\int d^4 x \gpgrade{ \lr{ \grad \wedge \lr{\delta A} } F – \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{ \lr{\delta A} \grad } F – \lr{ \lr{ \delta A } \cdot \grad } F + \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{ \lr{\delta A} \grad } F + \lr{ \delta A } \lr{ J – I M } }{0,4}.
\end{aligned}
\end{equation}

In the last three lines, it is important to note that \( \grad \) acts bidirectionally, but on \( \delta A \), but not \( F \).
In particular, if \( B, C \) are multivectors, we interpret the bidirectional action of the gradient as
\begin{equation}\label{eqn:fsquared:1560}
\begin{aligned}
B \lrgrad C &=
B \gamma^\mu \lrpartial_\mu C \\
&=
(\partial_\mu B) \gamma^\mu C
+
B \gamma^\mu (\partial_\mu C),
\end{aligned}
\end{equation}
where the partial operators on the first line are bidirectionally acting, and braces have been used in the last line to indicate the scope of the operators in the chain rule expansion.

Let’s also use arrows to clarify the directionality of this first part of the action variation, writing
\begin{equation}\label{eqn:fsquared:1580}
\begin{aligned}
\delta S
&=
-\int d^4 x \gpgrade{ \lr{\delta A} \lgrad F + \lr{ \delta A } \lr{ J – I M } }{0,4} \\
&=
-\int d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } \lr{ J – I M } }{0,4}.
\end{aligned}
\end{equation}
We can cast the first term into an integrand that can be evaluated using the Fundamental Theorem of Geometric Calculus, by introducing a
a parameterization \( x = x(a_\mu) \), for which the tangent space basis vectors are \( \Bx_{a_\mu} = \PDi{a_\mu}{x} \), and the pseudoscalar volume element is
\begin{equation}\label{eqn:fsquared:1640}
d^4 \Bx = \lr{ \Bx_{a_0} \wedge \Bx_{a_1} \wedge \Bx_{a_2} \wedge \Bx_{a_3} } da_0 da_1 da_2 da_3 = I d^4 x.
\end{equation}
Writing \( d^4 x = -I d^4 \Bx \), we have
\begin{equation}\label{eqn:fsquared:1600}
\begin{aligned}
\delta S
&=
-\int_V d^4 x \gpgrade{ \lr{\delta A} \lrgrad F – \lr{\delta A} \rgrad F + \lr{ \delta A } \lr{ J – I M } }{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 } \lr{ J – I M } }{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 + I M } }{0,4}.
\end{aligned}
\end{equation}
The first integral is killed since \( \delta A = 0 \) on the boundary. For the second integral to be zero for all variations \( \delta A \), we must have
\begin{equation}\label{eqn:fsquared:1660}
\gpgrade{ \lr{\delta A} \lr{ \rgrad F – J + I M } }{0,4} = 0,
\end{equation}
but have argued previously that we can drop the grade selection, leaving
\begin{equation}\label{eqn:fsquared:1620}
\boxed{
\grad F = J – I M
},
\end{equation}
where the directional indicator on our gradient has been dropped, since there is no longer any ambiguity. 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.