multivector

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

September 25, 2023 math and physics play No comments , , , , , , , , ,

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.)

A multivector Lagrangian for Maxwell’s equation, w/ electric and magnetic current density four-vector sources

June 29, 2022 math and physics play , , , , , , , ,

[Click here for a PDF version of this and previous related posts .]

Initially I had trouble generalizing the multivector Lagrangian to include both the electric and magnetic sources without using two independent potentials. However, this can be done, provided one is careful enough. Recall that we found that a useful formulation for the field in terms of two potentials is
\begin{equation}\label{eqn:maxwellLagrangian:2050}
F = F_{\mathrm{e}} + I F_{\mathrm{m}},
\end{equation}
where
\begin{equation}\label{eqn:maxwellLagrangian:2070}
\begin{aligned}
F_{\mathrm{e}} = \grad \wedge A \\
F_{\mathrm{m}} = \grad \wedge K,
\end{aligned}
\end{equation}
and where \( A, K \) are arbitrary four-vector potentials.
Use of two potentials allowed us to decouple Maxwell’s equations into two separate gradient equations. We don’t want to do that now, but let’s see how we can combine the two fields into a single multivector potential. Letting the gradient act bidirectionally, and introducing a dummy grade-two selection into the mix, we have
\begin{equation}\label{eqn:maxwellLagrangian:2090}
\begin{aligned}
F
&= \rgrad \wedge A + I \lr{ \rgrad \wedge K } \\
&= – A \wedge \lgrad – I \lr{ K \wedge \lgrad } \\
&= -\gpgradetwo{ A \wedge \lgrad + I \lr{ K \wedge \lgrad } } \\
&= -\gpgradetwo{ A \lgrad + I K \lgrad } \\
&= -\gpgradetwo{ \lr{ A + I K } \lgrad }.
\end{aligned}
\end{equation}
Now, we call
\begin{equation}\label{eqn:maxwellLagrangian:2110}
N = A + I K,
\end{equation}
(a 1,3 multivector), the multivector potential, and write the electromagnetic field not in terms of curls explicitly, but using a grade-2 selection filter
\begin{equation}\label{eqn:maxwellLagrangian:2130}
F = -\gpgradetwo{ N \lgrad }.
\end{equation}

We can now form the following multivector Lagrangian
\begin{equation}\label{eqn:maxwellLagrangian:2150}
\LL = \inv{2} F^2 – \gpgrade{ N \lr{ J – I M } }{0,4},
\end{equation}
and vary the action to (eventually) find our multivector Maxwell’s equation, without ever resorting to coordinates. We have
\begin{equation}\label{eqn:maxwellLagrangian:2170}
\begin{aligned}
\delta S
&= \int d^4 x \inv{2} \lr{ \lr{ \delta F } F + F \lr{ \delta F } } – \gpgrade{ \delta N \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ \lr{ \delta F } F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ -\gpgradetwo{ \lr{ \delta N} \lgrad } F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ -\gpgradetwo{ \lr{ \delta N} \lrgrad } F +\gpgradetwo{ \lr{ \delta N} \rgrad } F – \lr{ \delta N } \lr{ J – I M } }{0,4}.
\end{aligned}
\end{equation}
The \( \lrgrad \) term can be evaluated using the fundamential theorem of GC, and will be zero, as \( \delta N = 0 \) on the boundary. Let’s look at the next integrand term a bit more carefully
\begin{equation}\label{eqn:maxwellLagrangian:2190}
\begin{aligned}
\gpgrade{ \gpgradetwo{ \lr{ \delta N} \rgrad } F }{0,4}
&=
\gpgrade{ \gpgradetwo{ \lr{ \lr{ \delta A } + I \lr{ \delta K } } \rgrad } F }{0,4} \\
&=
\gpgrade{ \lr{ \lr{\delta A} \wedge \rgrad + I \lr{ \lr{ \delta K } \wedge \rgrad }} F }{0,4} \\
&=
\gpgrade{ \lr{\delta A} \rgrad F – \lr{ \lr{\delta A} \cdot \rgrad} F + I \lr{ \delta K } \rgrad F – I \lr{ \lr{ \delta K } \cdot \rgrad} F }{0,4} \\
&=
\gpgrade{ \lr{\delta A} \rgrad F + I \lr{ \delta K } \rgrad F }{0,4} \\
&=
\gpgrade{ \lr{ \lr{\delta A} + I \lr{ \delta K} } \rgrad F }{0,4} \\
&=
\gpgrade{ \lr{ \delta N} \rgrad F }{0,4},
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:maxwellLagrangian:2210}
\begin{aligned}
\delta S
&= \int d^4 x \gpgrade{ \lr{ \delta N} \rgrad F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ \lr{ \delta N} \lr{ \rgrad F – \lr{ J – I M } } }{0,4}.
\end{aligned}
\end{equation}
for this to be zero for all variations \( \delta N \) of the 1,3-multivector potential \( N \), we must have
\begin{equation}\label{eqn:maxwellLagrangian:2230}
\grad F = J – I M.
\end{equation}
This is Maxwell’s equation, as desired, including both electric and (if desired) magnetic sources.

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.

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

June 17, 2022 math and physics play , , , , , , , , , ,

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

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

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

Euler-Lagrange equations.

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

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

The variational principle requires
\begin{equation}\label{eqn:fsquared:1200}
0 = \delta S = \int d^4 x \delta \LL( A, \partial_\mu A ).
\end{equation}
That variation can be expressed as a limiting parametric operation as follows
\begin{equation}\label{eqn:fsquared:1220}
\delta S
= \int d^4 x
\lr{
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )
+
\sum_\mu
\lim_{t \rightarrow 0} \ddt{} \LL( \partial_\mu A + t \delta \partial_\mu A )
}
\end{equation}
We eventually want a coordinate free expression for the variation, but we’ll use them to get there. We can expand the first derivative by chain rule as
\begin{equation}\label{eqn:fsquared:1240}
\begin{aligned}
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )
&=
\lim_{t \rightarrow 0} \PD{(A^\alpha + t \delta A^\alpha)}{\LL} \PD{t}{}(A^\alpha + t \delta A^\alpha) \\
&=
\PD{A^\alpha}{\LL} \delta A^\alpha.
\end{aligned}
\end{equation}
This has the structure of a directional derivative \( A \). In particular, let
\begin{equation}\label{eqn:fsquared:1260}
\grad_A = \gamma^\alpha \PD{A^\alpha}{},
\end{equation}
so we have
\begin{equation}\label{eqn:fsquared:1280}
\lim_{t \rightarrow 0} \ddt{} \LL( A + t \delta A )
= \delta A \cdot \grad_A.
\end{equation}
Similarly,
\begin{equation}\label{eqn:fsquared:1300}
\lim_{t \rightarrow 0} \ddt{} \LL( \partial_\mu A + t \delta \partial_\mu A )
=
\PD{(\partial_\mu A^\alpha)}{\LL} \delta \partial_\mu A^\alpha,
\end{equation}
so we can define a gradient with respect to each of the derivatives of \(A \) as
\begin{equation}\label{eqn:fsquared:1320}
\grad_{\partial_\mu A} = \gamma^\alpha \PD{(\partial_\mu A^\alpha)}{}.
\end{equation}
Our variation can now be expressed in a somewhat coordinate free form
\begin{equation}\label{eqn:fsquared:1340}
\delta S = \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\delta \partial_\mu A} \cdot \grad_{\partial_\mu A} } \LL
}.
\end{equation}
We now sum implicitly over pairs of indexes \( \mu \) (i.e. we are treating \( \grad_{\partial_\mu A} \) as an upper index entity). We can now proceed with our chain rule expansion
\begin{equation}\label{eqn:fsquared:1360}
\begin{aligned}
\delta S
&= \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\delta \partial_\mu A} \cdot \grad_{\partial_\mu A} } \LL
} \\
&= \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL + \lr{ \lr{\partial_\mu \delta A} \cdot \grad_{\partial_\mu A} } \LL
} \\
&= \int d^4 x \lr{
\lr{\delta A \cdot \grad_A} \LL
+ \partial_\mu \lr{ \lr{ \delta A \cdot \grad_{\partial_\mu A} } \LL }
– \lr{\PD{x^\mu}{} \delta A \cdot \grad_{\partial_\mu A} \LL}_{\delta A}
}.
\end{aligned}
\end{equation}
As usual, we kill off the boundary term, by insisting that \( \delta A = 0 \) on the boundary, leaving us with a four-vector form of the field Euler-Lagrange equations
\begin{equation}\label{eqn:fsquared:1380}
\lr{\delta A \cdot \grad_A} \LL = \lr{\PD{x^\mu}{} \delta A \cdot \grad_{\partial_\mu A} \LL}_{\delta A},
\end{equation}
where the RHS derivatives are taken with \(\delta A \) held fixed. We seek solutions of this equation that hold for all variations \( \delta A \).

Application to the Maxwell Lagrangian.

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

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

Our Lagrangian is
\begin{equation}\label{eqn:fsquared:1440}
\LL = \inv{2} F^2 – \gpgrade{A \lr{ J – I M } }{0,4},
\end{equation}
so
\begin{equation}\label{eqn:fsquared:1460}
\lr{\delta A \cdot \grad_A} \LL
=
-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4},
\end{equation}
and
\begin{equation}\label{eqn:fsquared:1480}
\begin{aligned}
\lr{ \delta A \cdot \grad_{\partial_\mu A} } \inv{2} F^2
&=
\inv{2} \lr{ F \lr{ \gamma^\mu \wedge \delta A } + \lr{ \gamma^\mu \wedge \delta A } F } \\
&=
\gpgrade{
\lr{ \gamma^\mu \wedge \delta A } F
}{0,4} \\
&=
-\gpgrade{
\lr{ \delta A \wedge \gamma^\mu } F
}{0,4} \\
&=
-\gpgrade{
\delta A \gamma^\mu F
–
\lr{ \delta A \cdot \gamma^\mu } F
}{0,4} \\
&=
-\gpgrade{
\delta A \gamma^\mu F
}{0,4}.
\end{aligned}
\end{equation}
Taking derivatives (holding \( \delta A \) fixed), we have
\begin{equation}\label{eqn:fsquared:1500}
\begin{aligned}
-\gpgrade{ \lr{ \delta A } \lr{ J – I M } }{0,4}
&=
-\gpgrade{
\delta A \partial_\mu \gamma^\mu F
}{0,4} \\
&=
-\gpgrade{
\delta A \grad F
}{0,4}.
\end{aligned}
\end{equation}
We’ve already seen that the solution can be expressed without grade selection as
\begin{equation}\label{eqn:fsquared:1520}
\grad F = \lr{ J – I M },
\end{equation}
which is Maxwell’s equation in it’s STA form. It’s not clear that this is really any less work, but it’s a step towards a coordinate free evaluation of the Maxwell Lagrangian (at least not having to use the coordinates \( A^\mu \) as we have to do in the tensor formalism.)

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