pseudoscalar

Equation of a hyperplane, and shortest distance between two hyperplanes.

December 13, 2024 math and physics play , , , , , , ,

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Scalar equation for a hyperplane.

In our last post, we found, in a round about way, that

Theorem 1.1:

The equation of a \(\mathbb{R}^N\) hyperplane, with distance \( d \) from the origin, and normal \( \mathbf{\hat{n}} \) is
\begin{equation*}
\Bx \cdot \mathbf{\hat{n}} = d.
\end{equation*}

Start proof:

Let \( \beta = \setlr{ \mathbf{\hat{f}}_1, \cdots \mathbf{\hat{f}}_{N-1} } \) be an orthonormal basis for the hyperplane normal to \( \mathbf{\hat{n}} \), and \( \Bd = d \mathbf{\hat{n}} \) be the vector in that hyperplane, closest to the origin, as illustrated in fig. 1.

 

fig 1. R^3 plane with normal n-cap

The hyperplane \( d \) distant from the origin with normal \( \mathbf{\hat{n}} \) has the parametric representation
\begin{equation}\label{eqn:hyperplaneGeometry:40}
\Bx(a_1, \cdots, a_{N-1}) = d \mathbf{\hat{n}} + \sum_{i = 1}^{N-1} a_i \mathbf{\hat{f}}_i.
\end{equation}
Equivalently, suppressing the parameterization, with \( \Bx = \Bx(a_1, \cdots, a_{N-1}) \), representing any vector in that hyperplane, by dotting with \( \mathbf{\hat{n}} \), we have
\begin{equation}\label{eqn:hyperplaneGeometry:60}
\Bx \cdot \mathbf{\hat{n}} = d \mathbf{\hat{n}} \cdot \mathbf{\hat{n}},
\end{equation}
where all the \( \mathbf{\hat{f}}_i \cdot \mathbf{\hat{n}} \) dot products are zero by construction. Since \( \mathbf{\hat{n}} \cdot \mathbf{\hat{n}} = 0 \), the proof is complete.

End proof.

Incidentally, observe we can also write the hyperplane equation in dual form, as
\begin{equation}\label{eqn:hyperplaneGeometry:220}
\Bx \wedge (\mathbf{\hat{n}} I) = d I,
\end{equation}
where \( I \) is an \(\mathbb{R}^N\) pseudoscalar (such as \( I = \mathbf{\hat{n}} \mathbf{\hat{f}}_1 \cdots \mathbf{\hat{f}}_{N-1} \)).

Our previous parallel plane separation problem.

The standard \(\mathbb{R}^3\) scalar form for an equation of a plane is
\begin{equation}\label{eqn:hyperplaneGeometry:80}
a x + b y + c z = d,
\end{equation}
where \( d \) looses it’s geometrical meaning. If we form \( \Bn = (a,b,c) \), then we can rewrite this as
\begin{equation}\label{eqn:hyperplaneGeometry:100}
\Bx \cdot \Bn = d,
\end{equation}
for this representation of an equation of a plane, we see that \( d/\Norm{\Bn} \) is the shortest distance from the origin to the plane. This means that if we have a pair of parallel plane equations
\begin{equation}\label{eqn:hyperplaneGeometry:120}
\begin{aligned}
\Bx \cdot \Bn &= d_1 \\
\Bx \cdot \Bn &= d_2,
\end{aligned}
\end{equation}
then the distance between those planes, by inspection, is
\begin{equation}\label{eqn:hyperplaneGeometry:140}
\Abs{ \frac{d_2}{\Norm{\Bn}} – \frac{d_1}{\Norm{\Bn}} },
\end{equation}
which reduces to just \( \Abs{d_2 – d_1} \) if \( \Bn \) is a unit normal for the plane. In our previous post, the problem to solve was to find the shortest distance between the parallel planes given by
\begin{equation}\label{eqn:hyperplaneGeometry:160}
\begin{aligned}
x – y + 2 z &= -3 \\
3 x – 3 y + 6 z &= 1.
\end{aligned}
\end{equation}
The more natural geometrical form for these plane equations is
\begin{equation}\label{eqn:hyperplaneGeometry:180}
\begin{aligned}
\Bx \cdot \mathbf{\hat{n}} &= -\frac{3}{\sqrt{6}} \\
\Bx \cdot \mathbf{\hat{n}} &= \inv{3 \sqrt{6}},
\end{aligned}
\end{equation}
where \( \mathbf{\hat{n}} = (1,-1,2)/\sqrt{6} \), as illustrated in fig. 2.

fig. 2. The two planes.

 

Given that representation, we can find the distance between the planes just by taking the absolute difference of the respective distances to the origin
\begin{equation}\label{eqn:hyperplaneGeometry:200}
\begin{aligned}
\Abs{ -\frac{3}{\sqrt{6}} – \inv{3 \sqrt{6}} }
&= \frac{\sqrt{6}}{6} \lr{ 3 + \inv{3} } \\
&= \frac{10}{18} \sqrt{6} \\
&= \frac{5}{9} \sqrt{6}.
\end{aligned}
\end{equation}

Triangle area problem: REVISITED.

March 31, 2024 math and physics play , , , , ,

[Click here for a PDF version of this post]

On LinkedIn, James asked for ideas about how to solve What is the total area of ABC? You should be able to solve this! using geometric algebra.

I found a couple ways, and this last variation is pretty cool.

fig. 1. Triangle with given areas.

To start with I’ve re-sketched the triangle with the areas slightly more to scale in fig. 1, where areas \( A_1 = 40, A_2 = 30, A_3 = 35, A_4 = 84 \) are given. The aim is to find the total area \( \sum A_i \).

If we had the vertex and center locations as vectors, we could easily compute the total area, but we don’t. We also don’t know the locations of the edge intersections, but can calculate those, as they satisfy
\begin{equation}\label{eqn:triangle_area_problem:20}
\begin{aligned}
\BD &= s_1 \BA = \BB + t_1 \lr{ \BC – \BB } \\
\BE &= s_2 \BC = \BA + t_2 \lr{ \BB – \BA } \\
\BF &= s_3 \BB = \BA + t_3 \lr{ \BC – \BA }.
\end{aligned}
\end{equation}
It turns out that the problem is over specified, and we will only need \( \BD, \BE \). To find those, we may eliminate the \( t_i \)’s by wedging appropriately (or equivalently, using Cramer’s rule), to find
\begin{equation}\label{eqn:triangle_area_problem:40}
\begin{aligned}
s_1 \BA \wedge \lr{ \BC – \BB } &= \BB \wedge \lr{ \BC – \BB } \\
s_2 \BC \wedge \lr{ \BB – \BA } &= \BA \wedge \lr{ \BB – \BA },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:triangle_area_problem:60}
\begin{aligned}
s_1 &= \frac{\BB \wedge \BC }{\BA \wedge \lr{ \BC – \BB }} \\
s_2 &= \frac{\BA \wedge \BB }{\BC \wedge \lr{ \BB – \BA }}.
\end{aligned}
\end{equation}
Now let’s introduce some scalar area variables, each pseudoscalar multiples of bivector area elements, with \( i = \Be_{1} \Be_2 \)
\begin{equation}\label{eqn:triangle_area_problem:81}
\begin{aligned}
X &= \lr{ \BA \wedge \BB } i^{-1} = \begin{vmatrix} \BA & \BB \end{vmatrix} \\
Y &= \lr{ \BC \wedge \BB } i^{-1} = \begin{vmatrix} \BC & \BB \end{vmatrix} \\
Z &= \lr{ \BA \wedge \BC } i^{-1} = \begin{vmatrix} \BA & \BC \end{vmatrix},
\end{aligned}
\end{equation}
Note that the orientation of all of these has been picked to have a positive orientation matching the figure, and that the
triangle area that we seek for this problem is \( 1/2 \Abs{ \BA \wedge \BB } = X/2 \).

The intersection parameters, after cancelling pseudoscalar factors, are
\begin{equation}\label{eqn:triangle_area_problem:100}
\begin{aligned}
s_1 &= \frac{\BB \wedge \BC }{\BA \wedge \BC – \BA \wedge \BB } = \frac{-Y}{Z – X} \\
s_2 &= \frac{\BA \wedge \BB }{\BC \wedge \BB – \BC \wedge \BA } = \frac{X}{Y + Z},
\end{aligned}
\end{equation}
so the intersection points are
\begin{equation}\label{eqn:triangle_area_problem:120}
\begin{aligned}
\BD &= \BA \frac{Y}{X – Z} \\
\BE &= \BC \frac{X}{Y + Z}.
\end{aligned}
\end{equation}
Observe that both scalar factors are positive (i.e.: \( X > Z \).)

We may now express all the known areas in terms of our area variables
\begin{equation}\label{eqn:triangle_area_problem:140}
\begin{aligned}
A_1 &= \inv{2} \lr{ \BD \wedge \BC } i^{-1} \\
A_1 + A_2 &= \inv{2} \lr{ \BA \wedge \BC } i^{-1} \\
A_1 + A_2 + A_3 &= \inv{2} \lr{ \BA \wedge \BE } i^{-1} \\
A_2 &= \inv{2} \lr{ \lr{\BA – \BD} \wedge \lr{ \BC – \BD } } i^{-1}\\
A_3 &= \inv{2} \lr{ \lr{\BA – \BC} \wedge \lr{ \BE – \BC } } i^{-1}\\
A_5 &= \inv{2} \lr{ \lr{\BB – \BC} \wedge \lr{ \BF – \BC } } i^{-1}.
\end{aligned}
\end{equation}

As mentioned, the problem is over specified, and we can get away with just the first three of these relations to solve for total area. Eliminating \( \BD, \BE \) from those, gives us
\begin{equation}\label{eqn:triangle_area_problem:180}
A_1 = \inv{2} \frac{Y}{X – Z} \lr{ \BA \wedge \BC } i^{-1} = \frac{Z}{2} \lr{ \frac{Y}{X – Z} },
\end{equation}
\begin{equation}\label{eqn:triangle_area_problem:460}
A_1 + A_2 = \inv{2} \lr{ \BA \wedge \BC } i^{-1} = \frac{Z}{2},
\end{equation}
and
\begin{equation}\label{eqn:triangle_area_problem:400}
\begin{aligned}
A_1 + A_2 + A_3 &= \inv{2} \lr{ \BA \wedge \BE } i^{-1} \\
&= \inv{2} \lr{ \BA \wedge \BC } \frac{X}{Y + Z} \\
&= \frac{Z}{2} \frac{X}{Y + Z}.
\end{aligned}
\end{equation}

Let’s eliminate \( Z \) to start with, leaving
\begin{equation}\label{eqn:triangle_area_problem:420}
\begin{aligned}
A_1 \lr{ X – 2 A_1 – 2 A_2 } &= Y \lr{ A_1 + A_2 } \\
\lr{ A_1 + A_2 + A_3 } \lr{ Y + 2 A_1 + 2 A_2 } &= \lr{ A_1 + A_2 } X.
\end{aligned}
\end{equation}
Solving for \( Y \) yields
\begin{equation}\label{eqn:triangle_area_problem:380}
Y = – 2 A_1 – 2 A_2 + \frac{ \lr{A_1 + A_2} X }{ A_1 + A_2 + A_3 } = \lr{ A_1 + A_2 } \lr{ -2 + \frac{X}{A_1 + A_2 + A_3 } },
\end{equation}
and back substution leaves us with a linear equation in \( X \)
\begin{equation}\label{eqn:triangle_area_problem:480}
\lr{ A_1 + A_2}^2 \lr{ -2 + \frac{X}{A_1 + A_2 + A_3 } } = A_1 \lr{ X – 2 A_1 – 2 A_2 }.
\end{equation}

This is easily solved to find
\begin{equation}\label{eqn:triangle_area_problem:500}
\frac{X}{2} = \frac{ \lr{ A_1 + A_2} A_2 \lr{ A_1 + A_2 + A_3 } }{A_2 \lr{ A_1 + A_2} – A_1 A_3 }.
\end{equation}
Plugging in the numeric values for the problem solves it, giving a total triangular area of \( \inv{2} \lr{\BA \wedge \BB } i^{-1} = X/2 = 315 \).

Now, I’ll have to watch the video and see how he solved it.

Bivector transformation, and reciprocal frame for column vectors of a transformation

January 21, 2024 math and physics play , , , ,

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

The author of a book draft I am reading pointed out if a vector transforms as
\begin{equation}\label{eqn:adjoint:760}
\Bv \rightarrow M \Bv,
\end{equation}
then cross products must transform as
\begin{equation}\label{eqn:adjoint:780}
\Ba \cross \Bb \rightarrow \lr{ \textrm{adj}\, M }^\T \lr{ \Ba \cross \Bb }.
\end{equation}
Bivectors clearly must transform in the same fashion. We also noticed that the adjoint is related to the reciprocal frame vectors of the columns of \( M \), but didn’t examine the reciprocal frame formulation of the adjoint in any detail.

Before we do that, let’s consider a slightly simpler case, the transformation of a pseudoscalar. That is
\begin{equation}\label{eqn:adjoint:800}
\begin{aligned}
M(\Ba) \wedge M(\Bb) \wedge M(\Bc)
&\rightarrow
\sum_{ijk}
\lr{ \Bm_i a_i } \wedge
\lr{ \Bm_j a_j } \wedge
\lr{ \Bm_k a_k } \\
&=
\sum_{ijk}
\lr{ \Bm_i \wedge \Bm_j \wedge \Bm_k } a_i b_j c_k \\
&=
\sum_{ijk}
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \epsilon_{ijk} a_i b_j c_k \\
&=
\Abs{M}
\sum_{ijk} \epsilon_{ijk} a_i b_j c_k \\
&=
\Abs{M} \lr{ \Ba \wedge \Bb \wedge \Bc }.
\end{aligned}
\end{equation}
This is a well known geometric algebra result (called an outermorphism transformation.)

It’s somewhat amusing that an outermorphism transformation with two wedged vectors is a bit more complicated to express than the same for three. Let’s see if we can find a coordinate free form for such a transformation.
\begin{equation}\label{eqn:adjoint:820}
\begin{aligned}
M(\Ba) \wedge M(\Bb)
&=
\sum_{ij} \lr{ \Bm_i a_i } \wedge \lr{ \Bm_j b_j } \\
&=
\sum_{ij} \lr{ \Bm_i \wedge \Bm_j } a_i b_j \\
&=
\sum_{i < j} \lr{ \Bm_i \wedge \Bm_j }
\begin{vmatrix}
a_i & a_j \\
b_i & b_j
\end{vmatrix} \\
&=
\sum_{i < j} \lr{ \Bm_i \wedge \Bm_j } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_j \wedge \Be_i } }.
\end{aligned}
\end{equation}

Recall that the reciprocal frame with respect to the basis \( \setlr{ \Bm_1, \Bm_2, \Bm_3 } \), assuming this is a non-degenerate basis, has elements of the form
\begin{equation}\label{eqn:adjoint:840}
\begin{aligned}
\Bm^1 &= \lr{ \Bm_2 \wedge \Bm_3 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \\
\Bm^2 &= \lr{ \Bm_3 \wedge \Bm_1 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 } \\
\Bm^3 &= \lr{ \Bm_1 \wedge \Bm_2 } \inv{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }.
\end{aligned}
\end{equation}
This can be flipped around as
\begin{equation}\label{eqn:adjoint:860}
\begin{aligned}
\Bm_2 \wedge \Bm_3 &= \Bm^1 \Abs{M} I \\
\Bm_3 \wedge \Bm_1 &= \Bm^2 \Abs{M} I \\
\Bm_1 \wedge \Bm_2 &= \Bm^3 \Abs{M} I \\
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:adjoint:880}
\begin{aligned}
M&(\Ba) \wedge M(\Bb) \\
&=
\lr{ \Bm_1 \wedge \Bm_2 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }
+
\lr{ \Bm_2 \wedge \Bm_3 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_3 \wedge \Be_2 } }
+
\lr{ \Bm_3 \wedge \Bm_1 } \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_1 \wedge \Be_3 } } \\
&=
I \Abs{M} \lr{
\Bm^3 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }
+
\Bm^1 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_3 \wedge \Be_2 } }
+
\Bm^2 \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_1 \wedge \Be_3 } }
}
\end{aligned}
\end{equation}

Let’s see if we can simplify one of these double index quantities
\begin{equation}\label{eqn:adjoint:900}
\begin{aligned}
I \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } }
&=
\gpgradethree{ I \lr{ \lr{ \Ba \wedge \Bb } \cdot \lr{ \Be_2 \wedge \Be_1 } } } \\
&=
\gpgradethree{ I \lr{ \Ba \wedge \Bb } \lr{ \Be_2 \wedge \Be_1 } } \\
&=
\gpgradethree{ \lr{ \Ba \wedge \Bb } \Be_{12321} } \\
&=
\gpgradethree{ \lr{ \Ba \wedge \Bb } \Be_{3} } \\
&=
\Ba \wedge \Bb \wedge \Be_3.
\end{aligned}
\end{equation}
We have
\begin{equation}\label{eqn:adjoint:920}
M(\Ba) \wedge M(\Bb) = \Abs{M} \lr{
\lr{ \Ba \wedge \Bb \wedge \Be_1 } \Bm^1
+
\lr{ \Ba \wedge \Bb \wedge \Be_2 } \Bm^2
+
\lr{ \Ba \wedge \Bb \wedge \Be_3 } \Bm^3
}.
\end{equation}

Using summation convention, we can now express the transformation of a bivector \( B \) as just
\begin{equation}\label{eqn:adjoint:940}
B \rightarrow \Abs{M} \lr{ B \wedge \Be_i } \Bm^i.
\end{equation}
If we are interested in the transformation of a pseudovector \( \Bv \) defined implicitly as the dual of a bivector \( B = I \Bv \), where
\begin{equation}\label{eqn:adjoint:960}
B \wedge \Be_i = \gpgradethree{ I \Bv \Be_i } = I \lr{ \Bv \cdot \Be_i }.
\end{equation}
This leaves us with a transformation rule for cross products equivalent to the adjoint relation \ref{eqn:adjoint:780}
\begin{equation}\label{eqn:adjoint:980}
\lr{ \Ba \cross \Bb } \rightarrow \lr{ \Ba \cross \Bb } \cdot \Be_i \Abs{M} \Bm^i.
\end{equation}
As intuited, the determinant weighted reciprocal frame vectors for the columns of the transformation \( M \), are the components of the adjoint. That is
\begin{equation}\label{eqn:adjoint:1000}
\lr{ \textrm{adj}\, M }^\T = \Abs{M}
\begin{bmatrix}
\Bm^1 & \Bm^2 & \Bm^3
\end{bmatrix}.
\end{equation}

Simplifying the previous adjoint matrix results.

January 17, 2024 math and physics play , , , , , , , , ,

[Click here for a PDF version of this (and the previous) post]

We previously found determinant expressions for the matrix elements of the adjoint for 2D and 3D matrices \( M \). However, we can extract additional structure from each of those results.

2D case.

Given a matrix expressed in block matrix form in terms of it’s columns
\begin{equation}\label{eqn:adjoint:500}
M =
\begin{bmatrix}
\Bm_1 & \Bm_2
\end{bmatrix},
\end{equation}
we found that the adjoint \( A \) satisfying \( M A = \Abs{M} I \) had the structure
\begin{equation}\label{eqn:adjoint:520}
A =
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} \\
& \\
\begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix}
\end{bmatrix}.
\end{equation}
We initially had wedge product expressions for each of these matrix elements, and can discover our structure by putting back those wedge products. Modulo sign, each of these matrix elemens has the form
\begin{equation}\label{eqn:adjoint:540}
\begin{aligned}
\begin{vmatrix} \Be_i & \Bm_j \end{vmatrix}
&=
\lr{ \Be_i \wedge \Bm_j } i^{-1} \\
&=
\gpgradezero{
\lr{ \Be_i \wedge \Bm_j } i^{-1}
} \\
&=
\gpgradezero{
\lr{ \Be_i \Bm_j – \Be_i \cdot \Bm_j } i^{-1}
} \\
&=
\gpgradezero{
\Be_i \Bm_j i^{-1}
} \\
&=
\Be_i \cdot \lr{ \Bm_j i^{-1} },
\end{aligned}
\end{equation}
where \( i = \Be_{12} \). The adjoint matrix is
\begin{equation}\label{eqn:adjoint:560}
A =
\begin{bmatrix}
-\lr{ \Bm_2 i } \cdot \Be_1 & -\lr{ \Bm_2 i } \cdot \Be_2 \\
\lr{ \Bm_1 i } \cdot \Be_1 & \lr{ \Bm_1 i } \cdot \Be_2 \\
\end{bmatrix}.
\end{equation}
If we use a column vector representation of the vectors \( \Bm_j i^{-1} \), we can write the adjoint in a compact hybrid geometric-algebra matrix form
\begin{equation}\label{eqn:adjoint:640}
A =
\begin{bmatrix}
-\lr{ \Bm_2 i }^\T \\
\lr{ \Bm_1 i }^\T
\end{bmatrix}.
\end{equation}

Check:

Let’s see if this works, by multiplying with \( M \)
\begin{equation}\label{eqn:adjoint:580}
\begin{aligned}
A M &=
\begin{bmatrix}
-\lr{ \Bm_2 i }^\T \\
\lr{ \Bm_1 i }^\T
\end{bmatrix}
\begin{bmatrix}
\Bm_1 & \Bm_2
\end{bmatrix} \\
&=
\begin{bmatrix}
-\lr{ \Bm_2 i }^\T \Bm_1 & -\lr{ \Bm_2 i }^\T \Bm_2 \\
\lr{ \Bm_1 i }^\T \Bm_1 & \lr{ \Bm_1 i }^\T \Bm_2
\end{bmatrix}.
\end{aligned}
\end{equation}
Those dot products have the form
\begin{equation}\label{eqn:adjoint:600}
\begin{aligned}
\lr{ \Bm_j i }^\T \Bm_i
&=
\lr{ \Bm_j i } \cdot \Bm_i \\
&=
\gpgradezero{ \lr{ \Bm_j i } \Bm_i } \\
&=
\gpgradezero{ -i \Bm_j \Bm_i } \\
&=
-i \lr{ \Bm_j \wedge \Bm_i },
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:adjoint:620}
\begin{aligned}
A M &=
\begin{bmatrix}
i \lr{ \Bm_2 \wedge \Bm_1 } & 0 \\
0 & -i \lr { \Bm_1 \wedge \Bm_2 }
\end{bmatrix} \\
&=
\Abs{M} I.
\end{aligned}
\end{equation}
We find the determinant weighted identity that we expected. Our methods are a bit schizophrenic, switching fluidly between matrix and geometric algebra representations, but provided we are careful enough, this isn’t problematic.

3D case.

Now, let’s look at the 3D case, where we assume a column vector representation of the matrix of interest
\begin{equation}\label{eqn:adjoint:660}
M =
\begin{bmatrix}
\Bm_1 & \Bm_2 & \Bm_3
\end{bmatrix},
\end{equation}
and try to simplify the expression we found for the adjoint
\begin{equation}\label{eqn:adjoint:680}
A =
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_2 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_3 & \Bm_1 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_1 & \Bm_2 \end{vmatrix}
\end{bmatrix}.
\end{equation}
As with the 2D case, let’s re-express these determinants in wedge product form. We’ll write \( I = \Be_{123} \), and find
\begin{equation}\label{eqn:adjoint:700}
\begin{aligned}
\begin{vmatrix} \Be_i & \Bm_j & \Bm_k \end{vmatrix}
&=
\lr{ \Be_i \wedge \Bm_j \wedge \Bm_k } I^{-1} \\
&=
\gpgradezero{ \lr{ \Be_i \wedge \Bm_j \wedge \Bm_k } I^{-1} } \\
&=
\gpgradezero{ \lr{
\Be_i \lr{ \Bm_j \wedge \Bm_k }
\Be_i \cdot \lr{ \Bm_j \wedge \Bm_k }
} I^{-1} } \\
&=
\gpgradezero{
\Be_i \lr{ \Bm_j \wedge \Bm_k }
I^{-1} } \\
&=
\gpgradezero{
\Be_i \lr{ \Bm_j \cross \Bm_k } I
I^{-1} } \\
&=
\Be_i \cdot \lr{ \Bm_j \cross \Bm_k }.
\end{aligned}
\end{equation}
We see that we can put the adjoint in block matrix form
\begin{equation}\label{eqn:adjoint:720}
A =
\begin{bmatrix}
\lr{ \Bm_2 \cross \Bm_3 }^\T \\
\lr{ \Bm_3 \cross \Bm_1 }^\T \\
\lr{ \Bm_1 \cross \Bm_2 }^\T \\
\end{bmatrix}.
\end{equation}

Check:

\begin{equation}\label{eqn:adjoint:740}
\begin{aligned}
A M
&=
\begin{bmatrix}
\lr{ \Bm_2 \cross \Bm_3 }^\T \\
\lr{ \Bm_3 \cross \Bm_1 }^\T \\
\lr{ \Bm_1 \cross \Bm_2 }^\T \\
\end{bmatrix}
\begin{bmatrix}
\Bm_1 & \Bm_2 & \Bm_3
\end{bmatrix} \\
&=
\begin{bmatrix}
\lr{ \Bm_2 \cross \Bm_3 }^\T \Bm_1 & \lr{ \Bm_2 \cross \Bm_3 }^\T \Bm_2 & \lr{ \Bm_2 \cross \Bm_3 }^\T \Bm_3 \\
\lr{ \Bm_3 \cross \Bm_1 }^\T \Bm_1 & \lr{ \Bm_3 \cross \Bm_1 }^\T \Bm_2 & \lr{ \Bm_3 \cross \Bm_1 }^\T \Bm_3 \\
\lr{ \Bm_1 \cross \Bm_2 }^\T \Bm_1 & \lr{ \Bm_1 \cross \Bm_2 }^\T \Bm_2 & \lr{ \Bm_1 \cross \Bm_2 }^\T \Bm_3
\end{bmatrix} \\
&=
\Abs{M} I.
\end{aligned}
\end{equation}

Essentially, we found that the rows of the adjoint matrix are each parallel to the reciprocal frame vectors of the columns of \( M \). This makes sense, as the reciprocal frame encodes a generalized inverse of sorts.

Potentials for multivector Maxwell’s equation (again.)

December 8, 2023 math and physics play , , , , , , , , , , , , , , , , ,

[Click here for the PDF version of this post.]

Motivation.

This revisits my last blog post where I covered this content in a meandering fashion. This is an attempt to re-express this in a more compact form. In particular, in a form that is amenable to include in my book. When I wrote the potential section of my book, I cheated, and didn’t try to motivate the results. My cheat was figuring out the multivector potential representation starting with STA where things are simpler, and then translating it back to a multivector representation, instead of figuring out a reasonable way to motivate things from the foundation already laid.

I’d like to eventually have a less rushed treatment of potentials in my book, where the results are not pulled out of a magic hat. Here is an attempted step in that direction. I’ve opted to put some of the motivational material in problems (with solutions at the chapter end.)

Multivector potentials.

We know from conventional electromagnetism (given no fictitious magnetic sources) that we can represent the six components of the electric and magnetic fields in terms of four scalar fields
\begin{equation}\label{eqn:mvpotentials:80}
\begin{aligned}
\BE &= -\spacegrad \phi – \PD{t}{\BA} \\
\BH &= \inv{\mu} \spacegrad \cross \BA.
\end{aligned}
\end{equation}
The conventional way of constructing these potentials makes use of the identities
\begin{equation}\label{eqn:mvpotentials:60}
\begin{aligned}
\spacegrad \cdot \lr{ \spacegrad \cross \BA } &= 0 \\
\spacegrad \cross \lr{ \spacegrad \phi } &= 0,
\end{aligned}
\end{equation}
applying those to the source free Maxwell’s equations to find representations of \( \BE, \BH \) that automatically satisfy those equations. For that conventional analysis, see section 18-6 [2] (available online), or section 10.1 [3], or section 6.4 [4]. We can also find such a potential representation using geometric algebra methods that are cross product free (problem 1.)

For Maxwell’s equations with fictitious magnetic sources, it can be shown that a potential representation of the field
\begin{equation}\label{eqn:mvpotentials:100}
\begin{aligned}
\BH &= -\spacegrad \phi_m – \PD{t}{\BF} \\
\BE &= -\inv{\epsilon} \spacegrad \cross \BF.
\end{aligned}
\end{equation}
satisfies the source-free grades of Maxwell’s equation.
See [1], and [5] for such derivations. As with the conventional source potentials, we can also apply our geometric algebra toolbox to easily find these results (problem 2.)

We have a mix of time partials and curls that is reminiscent of Maxwell’s equation itself. It’s obvious to wonder whether there is a more coherent integrated form for the potential. This is in fact the case.

Lemma 1.1: Multivector potentials.

For Maxwell’s equation with electric sources, the total field \( F \) can be expressed in multivector potential form
\begin{equation}\label{eqn:mvpotentials:520}
F = \gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } \lr{ -\phi + c \BA } }{1,2}.
\end{equation}
For Maxwell’s equation with only fictitious magnetic sources, the total field \( F \) can be expressed in multivector form
\begin{equation}\label{eqn:mvpotentials:540}
F = \gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } I \eta \lr{ -\phi_m + c \BF } }{1,2}.
\end{equation}

The reader should try to verify this themselves (problem 3.)

Using superposition, we can form a multivector potential that includes all grades.

Definition 1.1: Multivector potential.

We call \( A \), a multivector with all grades, the multivector potential, defining the total field as
\begin{equation}\label{eqn:mvpotentials:600}
\begin{aligned}
F
&=
\gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } A }{1,2} \\
&=
\lr{ \spacegrad – \inv{c} \PD{t}{} } A
–
\gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } A }{0,3}.
\end{aligned}
\end{equation}
Imposition of the constraint
\begin{equation}\label{eqn:mvpotentials:680}
\gpgrade{ \lr{ \spacegrad – \inv{c} \PD{t}{} } A }{0,3} = 0,
\end{equation}
is called the Lorentz gauge condition, and allows us to express \( F \) in terms of the potential without any grade selection filters.

Lemma 1.2: Conventional multivector potential.

Let
\begin{equation}\label{eqn:mvpotentials:620}
A = -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF }.
\end{equation}
This results in the conventional potential representation of the electric and magnetic fields
\begin{equation}\label{eqn:mvpotentials:640}
\begin{aligned}
\BE &= -\spacegrad \phi – \PD{t}{\BA} – \inv{\epsilon} \spacegrad \cross \BF \\
\BH &= -\spacegrad \phi_m – \PD{t}{\BF} + \inv{\mu} \spacegrad \cross \BA.
\end{aligned}
\end{equation}
In terms of potentials, the Lorentz gauge condition \ref{eqn:mvpotentials:680} takes the form
\begin{equation}\label{eqn:mvpotentials:660}
\begin{aligned}
0 &= \inv{c} \PD{t}{\phi} + \spacegrad \cdot (c \BA) \\
0 &= \inv{c} \PD{t}{\phi_m} + \spacegrad \cdot (c \BF).
\end{aligned}
\end{equation}

Start proof:

See problem 4.

End proof.

Problems.

Problem 1: Potentials for no-fictitious sources.

Starting with Maxwell’s equation with only conventional electric sources
\begin{equation}\label{eqn:mvpotentials:120}
\lr{ \spacegrad + \inv{c}\PD{t}{} } F = \gpgrade{J}{0,1}.
\end{equation}
Show that this may be split by grade into three equations
\begin{equation}\label{eqn:mvpotentials:140}
\begin{aligned}
\gpgrade{ \lr{ \spacegrad + \inv{c}\PD{t}{} } F}{0,1} &= \gpgrade{J}{0,1} \\
\spacegrad \wedge \BE + \inv{c}\PD{t}{} \lr{ I \eta \BH } &= 0 \\
\spacegrad \wedge \lr{ I \eta \BH } &= 0.
\end{aligned}
\end{equation}
Then use the identities \( \spacegrad \wedge \spacegrad \wedge \BA = 0 \), for vector \( \BA \) and \( \spacegrad \wedge \spacegrad \phi = 0 \), for scalar \( \phi \) to find the potential representation.

Answer

Taking grade(0,1) and (2,3) selections of Maxwell’s equation, we split our equations into source dependent and source free equations
\begin{equation}\label{eqn:mvpotentials:200}
\gpgrade{ \lr{ \spacegrad + \inv{c} \PD{t}{} } F }{0,1} = \gpgrade{J}{0,1},
\end{equation}
\begin{equation}\label{eqn:mvpotentials:220}
\gpgrade{ \lr{ \spacegrad + \inv{c} \PD{t}{} } F }{2,3} = 0.
\end{equation}

In terms of \( F = \BE + I \eta \BH \), the source free equation expands to
\begin{equation}\label{eqn:mvpotentials:240}
\begin{aligned}
0
&=
\gpgrade{
\lr{ \spacegrad + \inv{c} \PD{t}{} } \lr{ \BE + I \eta \BH }
}{2,3} \\
&=
\gpgradetwo{\spacegrad \BE}
+ \gpgradethree{I \eta \spacegrad \BH} + I \eta \inv{c} \PD{t}{\BH} \\
&=
\spacegrad \wedge \BE
+ \spacegrad \wedge \lr{ I \eta \BH }
+ I \eta \inv{c} \PD{t}{\BH},
\end{aligned}
\end{equation}
which can be further split into a bivector and trivector equation
\begin{equation}\label{eqn:mvpotentials:260}
0 = \spacegrad \wedge \BE + I \eta \inv{c} \PD{t}{\BH}
\end{equation}
\begin{equation}\label{eqn:mvpotentials:280}
0 = \spacegrad \wedge \lr{ I \eta \BH }.
\end{equation}
It’s clear that we want to write the magnetic field as a (bivector) curl, so we let
\begin{equation}\label{eqn:mvpotentials:300}
I \eta \BH = I c \BB = c \spacegrad \wedge \BA,
\end{equation}
or
\begin{equation}\label{eqn:mvpotentials:301}
\BH = \inv{\mu} \spacegrad \cross \BA.
\end{equation}

\Cref{eqn:mvpotentials:260} is reduced to
\begin{equation}\label{eqn:mvpotentials:320}
\begin{aligned}
0
&= \spacegrad \wedge \BE + I \eta \inv{c} \PD{t}{\BH} \\
&= \spacegrad \wedge \BE + \inv{c} \PD{t}{} \spacegrad \wedge \lr{ c \BA } \\
&= \spacegrad \wedge \lr{ \BE + \PD{t}{\BA} }.
\end{aligned}
\end{equation}
We can now let
\begin{equation}\label{eqn:mvpotentials:340}
\BE + \PD{t}{\BA} = -\spacegrad \phi.
\end{equation}
We sneakily adjust the sign of the gradient so that the result matches the conventional representation.

Problem 2: Potentials for fictitious sources.

Starting with Maxwell’s equation with only fictitious magnetic sources
\begin{equation}\label{eqn:mvpotentials:160}
\lr{ \spacegrad + \inv{c}\PD{t}{} } F = \gpgrade{J}{2,3},
\end{equation}
show that this may be split by grade into three equations
\begin{equation}\label{eqn:mvpotentials:180}
\begin{aligned}
\gpgrade{ \lr{ \spacegrad + \inv{c}\PD{t}{} } I F}{0,1} &= I \gpgrade{J}{2,3} \\
-\eta \spacegrad \wedge \BH + \inv{c}\PD{t}{(I \BE)} &= 0 \\
\spacegrad \wedge \lr{ I \BE } &= 0.
\end{aligned}
\end{equation}
Then use the identities \( \spacegrad \wedge \spacegrad \wedge \BF = 0 \), for vector \( \BF \) and \( \spacegrad \wedge \spacegrad \phi_m = 0 \), for scalar \( \phi_m \) to find the potential representation \ref{eqn:mvpotentials:100}.

Answer

We multiply \ref{eqn:mvpotentials:160} by \( I \) to find
\begin{equation}\label{eqn:mvpotentials:360}
\lr{ \spacegrad + \inv{c}\PD{t}{} } I F = I \gpgrade{J}{2,3},
\end{equation}
which can be split into
\begin{equation}\label{eqn:mvpotentials:380}
\begin{aligned}
\gpgrade{ \lr{ \spacegrad + \inv{c}\PD{t}{} } I F }{1,2} &= I \gpgrade{J}{2,3} \\
\gpgrade{ \lr{ \spacegrad + \inv{c}\PD{t}{} } I F }{0,3} &= 0.
\end{aligned}
\end{equation}
We expand the source free equation in terms of \( I F = I \BE – \eta \BH \), to find
\begin{equation}\label{eqn:mvpotentials:400}
\begin{aligned}
0
&= \gpgrade{ \lr{ \spacegrad + \inv{c}\PD{t}{} } \lr{ I \BE – \eta \BH } }{0,3} \\
&= \spacegrad \wedge \lr{ I \BE } + \inv{c} \PD{t}{(I \BE)} – \eta \spacegrad \wedge \BH,
\end{aligned}
\end{equation}
which has the respective bivector and trivector grades
\begin{equation}\label{eqn:mvpotentials:420}
0 = \spacegrad \wedge \lr{ I \BE }
\end{equation}
\begin{equation}\label{eqn:mvpotentials:440}
0 = \inv{c} \PD{t}{(I \BE)} – \eta \spacegrad \wedge \BH.
\end{equation}
We can clearly satisfy \ref{eqn:mvpotentials:420} by setting
\begin{equation}\label{eqn:mvpotentials:460}
I \BE = -\inv{\epsilon} \spacegrad \wedge \BF,
\end{equation}
or
\begin{equation}\label{eqn:mvpotentials:461}
\BE = -\inv{\epsilon} \spacegrad \cross \BF.
\end{equation}
Here, once again, the sneaky inclusion of a constant factor \( -1/\epsilon \) is to make the result match the conventional. Inserting this value for \( I \BE \) into our bivector equation yields
\begin{equation}\label{eqn:mvpotentials:480}
\begin{aligned}
0
&= -\inv{\epsilon} \inv{c} \PD{t}{} (\spacegrad \wedge \BF) – \eta \spacegrad \wedge \BH \\
&= -\eta \spacegrad \wedge \lr{ \PD{t}{\BF} + \BH },
\end{aligned}
\end{equation}
so we set
\begin{equation}\label{eqn:mvpotentials:500}
\PD{t}{\BF} + \BH = -\spacegrad \phi_m,
\end{equation}
and have a field representation that automatically satisfies the source free equations.

Problem 3: Total field in terms of potentials.

Prove lemma 1.1, either by direct expansion, or by trying to discover the multivector form of the field by construction.

Answer

Proof by expansion is straightforward, and left to the reader. We form the respective total electromagnetic fields \( F = \BE + I \eta H \) for each case.

We find
\begin{equation}\label{eqn:mvpotentials:560}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= -\spacegrad \phi – \PD{t}{\BA} + I \frac{\eta}{\mu} \spacegrad \cross \BA \\
&= -\spacegrad \phi – \inv{c} \PD{t}{(c \BA)} + \spacegrad \wedge (c\BA) \\
&= \gpgrade{ -\spacegrad \phi – \inv{c} \PD{t}{(c \BA)} + \spacegrad \wedge (c\BA) }{1,2} \\
&= \gpgrade{ -\spacegrad \phi – \inv{c} \PD{t}{(c \BA)} + \spacegrad (c\BA) }{1,2} \\
&= \gpgrade{ \spacegrad \lr{ -\phi + c \BA } – \inv{c} \PD{t}{(c \BA)} }{1,2} \\
&= \gpgrade{ \lr{ \spacegrad -\inv{c} \PD{t}{} } \lr{ -\phi + c \BA } }{1,2}.
\end{aligned}
\end{equation}

For the field for the fictitious source case, we compute the result in the same way, inserting a no-op grade selection to allow us to simplify, finding
\begin{equation}\label{eqn:mvpotentials:580}
\begin{aligned}
F
&= \BE + I \eta \BH \\
&= -\inv{\epsilon} \spacegrad \cross \BF + I \eta \lr{ -\spacegrad \phi_m – \PD{t}{\BF} } \\
&= \inv{\epsilon c} I \lr{ \spacegrad \wedge (c \BF)} + I \eta \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } \\
&= I \eta \lr{ \spacegrad \wedge (c \BF) + \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } } \\
&= I \eta \gpgrade{ \spacegrad \wedge (c \BF) + \lr{ -\spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} } }{1,2} \\
&= I \eta \gpgrade{ \spacegrad (c \BF) – \spacegrad \phi_m – \inv{c} \PD{t}{(c \BF)} }{1,2} \\
&= I \eta \gpgrade{ \spacegrad (-\phi_m + c \BF) – \inv{c} \PD{t}{(c \BF)} }{1,2} \\
&= I \eta \gpgrade{ \lr{ \spacegrad -\inv{c} \PD{t}{} } (-\phi_m + c \BF) }{1,2}.
\end{aligned}
\end{equation}

Problem 4: Fields in terms of potentials.

Prove lemma 1.2.

Answer

Let’s expand and then group by grade
\begin{equation}\label{eqn:mvpotentials:n}
\begin{aligned}
\lr{ \spacegrad – \inv{c} \PD{t}{} } A
&=
\lr{ \spacegrad – \inv{c} \PD{t}{} } \lr{ -\phi + c \BA + I \eta \lr{ -\phi_m + c \BF }} \\
&=
-\spacegrad \phi + c \spacegrad \BA + I \eta \lr{ -\spacegrad \phi_m + c \spacegrad \BF }
-\inv{c} \PD{t}{\phi} + c \inv{c} \PD{t}{ \BA } + I \eta \lr{ -\inv{c} \PD{t}{\phi_m} + c \inv{c} \PD{t}{\BF} } \\
&=
– \spacegrad \phi
+ I \eta c \spacegrad \wedge \BF
– c \inv{c} \PD{t}{\BA}
\quad + c \spacegrad \wedge \BA
-I \eta \spacegrad \phi_m
– c I \eta \inv{c} \PD{t}{\BF} \\
&\quad + c \spacegrad \cdot \BA
+\inv{c} \PD{t}{\phi}
\quad + I \eta \lr{ c \spacegrad \cdot \BF
+ \inv{c} \PD{t}{\phi_m} } \\
&=
– \spacegrad \phi
– \inv{\epsilon} \spacegrad \cross \BF
– \PD{t}{\BA}
\quad + I \eta \lr{
\inv{\mu} \spacegrad \cross \BA
– \spacegrad \phi_m
– \PD{t}{\BF}
} \\
&\quad + c \spacegrad \cdot \BA
+\inv{c} \PD{t}{\phi}
\quad + I \eta \lr{ c \spacegrad \cdot \BF
+ \inv{c} \PD{t}{\phi_m} }.
\end{aligned}
\end{equation}
Observing that \( F = \gpgrade{ \lr{ \spacegrad -(1/c) \partial_t } A }{1,2} = \BE + I \eta \BH \), completes the problem. If the Lorentz gauge condition is assumed, the scalar and pseudoscalar components above are obliterated, leaving just
\( F = \lr{ \spacegrad -(1/c) \partial_t } A \).

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.

[5] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.