wedge product

Hodge duality in exterior calculus and geometric algebra.

November 13, 2023 math and physics play , , , , , , ,

[Click here for a PDF version of this post]

This is a continuation of yesterday’s post on the relationships between the exterior derivative, and the curl operation (grad-wedge) in geometric algebra.

Hodge star vs. pseudoscalar multiplication.

We find a definition of the hodge star for basic k-forms in [2].

Definition 1.7: Hodge star.

Let \( \omega \) be a basic k-form on \(\mathbb{R}^n\). The hodge star of \( \omega \), denoted by \( {*} \omega \) is the unique \( n-k \)-form with the property
\begin{equation*}
\omega \wedge {*} \omega = dx_1 \wedge \cdots \wedge dx_n.
\end{equation*}

I find it interesting that this duality definition is completely free of any notion of metric or inner product. That isn’t the case with the hodge star definition from [3]. This is certainly an easier definition to understand.

Let’s calculate all the duals for the basic forms from \(\mathbb{R}^3\). We let \( I = dx_1 \wedge dx_2 \wedge dx_3 \), and then by inspection find all the duals satisfying
\begin{equation}\label{eqn:formAndCurl:1110}
\begin{aligned}
I &= 1 \wedge {*} 1 \\
I &= dx \wedge {*} dx \\
I &= dy \wedge {*} dy \\
I &= dz \wedge {*} dz \\
I &= (dx dy) \wedge {*} (dx dy) \\
I &= (dy dz) \wedge {*} (dy dz) \\
I &= (dz dx) \wedge {*} (dz dx) \\
I &= dx dy dz \wedge {*} (dx dy dz).
\end{aligned}
\end{equation}
Those are
\begin{equation}\label{eqn:formAndCurl:1130}
\begin{aligned}
{*} 1 &= dx dy dz \\
{*} dx &= dy dz \\
{*} dy &= dz dx \\
{*} dz &= dx dy \\
{*} (dx dy) &= dz \\
{*} (dy dz) &= dx \\
{*} (dz dx) &= dy \\
{*} (dx dy dz) &= 1.
\end{aligned}
\end{equation}

Now let’s compare this to multiplication of the \(\mathbb{R}^3\) basis vectors with the pseudoscalar \( I = \Be_1 \Be_2 \Be_3 \). We have
\begin{equation}\label{eqn:formAndCurl:1140}
\begin{aligned}
1 I &= I \\
\Be_1 I &= \Be_{1123} = \Be_{23} \\
\Be_2 I &= \Be_{2123} = \Be_{31} \\
\Be_3 I &= \Be_{3123} = \Be_{12} \\
\Be_{23} I &= \Be_{23123} = – \Be_1 \\
\Be_{31} I &= \Be_{31123} = – \Be_2 \\
\Be_{12} I &= \Be_{12123} = – \Be_3 \\
\Be_{123} I &= \Be_{123123} = -1.
\end{aligned}
\end{equation}
With differential forms, the duals of the duals of all our basic forms recovered the original, that is \( ** \omega = \omega \), but that isn’t the case if we use pseudoscalar multiplication to define duality. We see that to model the Hodge dual, we need to multiply by a grade specific pseudoscalar.

Definition 1.8: Hodge dual of an \(\mathbb{R}^3\) multivector

Let \( M \) be a \(\mathbb{R}^3\) multivector. The Hodge dual \( {*} M \) of that multivector is
\begin{equation*}
{*} M
=
\gpgrade{M}{0,1} I –
\gpgrade{M}{2,3} I.
\end{equation*}

In particular, if \( A \) is a k-blade in \(\mathbb{R}^3\), a round trip requires multiplication with different signed unit pseudoscalars.

Let’s step back and consider the \(\mathbb{R}^2\) case as well. This time we let \( i = dx_1 \wedge dx_2 \). We seek all the duals satisfying
\begin{equation}\label{eqn:formAndCurl:1180}
\begin{aligned}
i &= 1 \wedge {*} 1 \\
i &= dx \wedge {*} dx \\
i &= dy \wedge {*} dy \\
i &= (dx dy) \wedge {*} (dx dy).
\end{aligned}
\end{equation}
Those duals are
\begin{equation}\label{eqn:formAndCurl:1200}
\begin{aligned}
{*} 1 &= dx dy \\
{*} dx &= dy \\
{*} dy &= -dx \\
{*} (dx dy) &= 1 \\
\end{aligned}
\end{equation}

Now let’s compare this to multiplication of the \(\mathbb{R}^2\) basis vectors with the pseudoscalar \( i = \Be_1 \Be_2 \). We have
\begin{equation}\label{eqn:formAndCurl:1220}
\begin{aligned}
1 i &= i \\
\Be_1 i &= \Be_{112} = \Be_{2} \\
\Be_2 i &= \Be_{212} = -\Be_{1} \\
\Be_{12} i &= \Be_{1212} = -1 \\
\end{aligned}
\end{equation}

Definition 1.9: Hodge dual of \(\mathbb{R}^2\) multivector

Let \( M \) be a \(\mathbb{R}^2\) multivector. The Hodge dual \( {*} M \) of that multivector is
\begin{equation*}
{*} M
=
\gpgrade{M}{0,1} i –
\gpgrade{M}{2} i.
\end{equation*}

Neither of these grade specific duality operations are as nice as simply multiplying by a unit pseudoscalar, but if we care about correspondence with the Hodge dual (at least according to the definition in the article), then this is what we need.

Having done that, let’s now look at the Hodge dual that produces the divergence operation.

Lemma 1.13: Divergence relation to the exterior derivative.

Let \( \omega = f dx + g dy + h dz \) be a one-form in \(\mathbb{R}^3\). The exterior derivative of the Hodge dual of \( \omega \) is a divergence three-form
\begin{equation*}
d({*} \omega) = \lr{ \PD{x}{f} + \PD{y}{g} + \PD{z}{h} } dx \wedge dy \wedge dz.
\end{equation*}
The GA equivalent of this, for a vector corresponding to this one-form \( \Bf = f \Be_1 + g \Be_2 + h \Be_3 \in \mathbb{R}^3 \), is
\begin{equation*}
\spacegrad \wedge ({*} \Bf) = \lr{\spacegrad \cdot \Bf} I.
\end{equation*}

Start proof:

The dual of the one form is
\begin{equation}\label{eqn:formAndCurl:1280}
{*} \omega =
f dy \wedge dz
+ g dz \wedge dx
+ h dx \wedge dy,
\end{equation}
so the exterior derivative is
\begin{equation}\label{eqn:formAndCurl:1300}
\begin{aligned}
d({*} \omega) &=
\lr{
\PD{x}{f} dx +
\PD{y}{f} dy +
\PD{z}{f} dz
}
\wedge dy \wedge dz \\
&\quad+
\lr{
\PD{x}{g} dx +
\PD{y}{g} dy +
\PD{z}{g} dz
}
\wedge
dz \wedge dx \\
&\quad+
\lr{
\PD{x}{g} dx +
\PD{y}{g} dy +
\PD{z}{g} dz
}
\wedge
dx \wedge dy \\
&=
\lr{
\PD{x}{f} +
\PD{y}{g} +
\PD{z}{h}
}
dx \wedge dy \wedge dz.
\end{aligned}
\end{equation}
We expect that the GA equivalent of this is \( \spacegrad \wedge ({*} \Bf) = \lr{ \spacegrad \cdot \Bf} I \). Let’s check that this is the case. The dual, for a vector, is
\begin{equation}\label{eqn:formAndCurl:1320}
{*} \Bf
= \Bf I,
\end{equation}
so
\begin{equation}\label{eqn:formAndCurl:1340}
\begin{aligned}
\spacegrad \wedge ({*} \Bf)
&= \gpgradethree{ \spacegrad (\Bf I) } \\
&= \gpgradethree{ (\spacegrad \Bf) I } \\
&= \gpgradethree{ (\spacegrad \cdot \Bf + \spacegrad \wedge \Bf) I } \\
&= \lr{ \spacegrad \cdot \Bf } I.
\end{aligned}
\end{equation}

End proof.

References

[1] Vincent Bouchard. Math 215: Calculus iv: 4.4 the exterior derivative and vector calculus, 2023. URL https://sites.ualberta.ca/ vbouchar/MATH215/section_exterior_vector.html. [Online; accessed 11-November-2023].

[2] Vincent Bouchard. Math 215: Calculus iv: 4.8 hodge star, 2023. URL https://sites.ualberta.ca/ vbouchar/MATH215/section_hodge.html. [Online; accessed 13-November-2023].

[3] H. Flanders. Differential Forms With Applications to the Physical Sciences. Courier Dover Publications, 1989.

Book update. Now includes recent work on best fit solutions.

October 1, 2023 Geometric Algebra for Electrical Engineers , , , , , , , , ,

 

I’ve added a few new pages in the linear systems solution portion of my book, Geometric Algebra for Electrical Engineers.  This now includes the best fit content that was covered in my recent video and blog post on approximate solutions to linear systems.

The geometry that is associated with a Moore-Penrose or SVD-based pseudoinverse is not terribly obvious, and this result, providing the same answer, uses geometry exclusively.  I’ve included it in my book, since it’s a cool application, and not conceptually much trickier than the exact system solution.  This makes this section slightly more formal, as it now including an up front statement as a theorem — but that’s where formality ends, as I don’t formally prove the theorem.  I do, however, provide lots of examples and problems (with solutions), sufficient for the industrious to craft their own proof if desired.

The updated version of the book should be available on all amazon marketplaces within the next 3-5 days.  The free PDF version (and leanpub edition), both linked above, are already updated.

 

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

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

New video (on Google’s CensorshipTube):

Exact system.

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

Cramer’s rule.

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

Least squares solution, using geometry.

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

fig. 1. Perpendicular bivectors.

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

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

Least squares solution, using calculus.

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

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

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

New video: Elliptical motion from Newton’s law of gravitation.

September 14, 2023 math and physics play , , , , , , , , , ,

This blog post is a text version of the video below, available in a few forms:

 

We found previously that
\begin{equation}\label{eqn:solarellipse:20}
\mathbf{\hat{r}}’ = \inv{r} \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bx’ }.
\end{equation}
Somewhat remarkably, we can use this identity to demonstrate that orbits governed gravitational force are elliptical (or parabolic, or hyperbolic.) This ends up being possible because the angular momentum of the system is a conserved quantity, and this immediately introduces angular momentum into the mix in a fundamental way. In particular,
\begin{equation}\label{eqn:solarellipse:40}
\mathbf{\hat{r}}’ = \inv{m r^2} \mathbf{\hat{r}} L,
\end{equation}
where we define the angular momentum bivector as
\begin{equation}\label{eqn:solarellipse:60}
L = \Bx \wedge \Bp.
\end{equation}
Our gravitational law is
\begin{equation}\label{eqn:solarellipse:80}
m \ddt{\Bv} = – G m M \frac{\mathbf{\hat{r}}}{r^2},
\end{equation}
or
\begin{equation}\label{eqn:solarellipse:100}
-\inv{G M} \ddt{\Bv} = \frac{\mathbf{\hat{r}}}{r^2}.
\end{equation}
Combining the gravitational law with our \( \mathbf{\hat{r}} \) derivative identity, we have
\begin{equation}\label{eqn:solarellipse:120}
\begin{aligned}
\ddt{ \mathbf{\hat{r}} }
&= \inv{m} \frac{\mathbf{\hat{r}}}{r^2} L \\
&= -\inv{G m M} \ddt{\Bv} L \\
&= -\inv{G m M} \lr{ \ddt{(\Bv L)} – \ddt{L} }.
\end{aligned}
\end{equation}
Since angular momentum is a constant of motion of the system, means that
\begin{equation}\label{eqn:solarellipse:140}
\ddt{L} = 0,
\end{equation}
our equation of motion is integratable
\begin{equation}\label{eqn:solarellipse:160}
\ddt{ \mathbf{\hat{r}} } = -\inv{G m M} \ddt{(\Bv L)}.
\end{equation}
Introducing a vector valued integration constant \( -\Be \), we have
\begin{equation}\label{eqn:solarellipse:180}
\mathbf{\hat{r}} = -\inv{G m M} \Bv L – \Be.
\end{equation}
We’ve transformed our second order differential equation to a first order equation, one that does not look easy to integrate one more time. Luckily, we do not have to integrate, and can partially solve this algebraically, enough to describe the orbit in a compact fashion.

Before trying that, it’s worth quickly demonstrating that this equation is not a multivector equation, but a vector equation, since the multivector \( \Bv L \) is, in fact, vector valued.
\begin{equation}\label{eqn:solarellipse:200}
\begin{aligned}
\Bv L
&= \Bv \lr{ \Bx \wedge (m \Bv) } \\
&\propto \mathbf{\hat{v}} \lr{ \mathbf{\hat{r}} \wedge \mathbf{\hat{v}} } \\
&= \mathbf{\hat{v}} \cdot \lr{ \mathbf{\hat{r}} \wedge \mathbf{\hat{v}} } + \mathbf{\hat{v}} \wedge \lr{ \mathbf{\hat{r}} \wedge \mathbf{\hat{v}} } \\
&= \mathbf{\hat{v}} \cdot \lr{ \mathbf{\hat{r}} \wedge \mathbf{\hat{v}} } \\
&= \lr{ \mathbf{\hat{v}} \cdot \mathbf{\hat{r}} } \mathbf{\hat{v}} – \mathbf{\hat{r}},
\end{aligned}
\end{equation}
which is a vector (i.e.: a vector that is directed along the portion of \( \Bx \) that is perpendicular to \( \Bv \).)

We can reduce \ref{eqn:solarellipse:180} to a scalar equation by dotting with \( \Bx = r \mathbf{\hat{r}} \), leaving
\begin{equation}\label{eqn:solarellipse:220}
\begin{aligned}
r
&= -\inv{G m M} \gpgradezero{ \Bx \Bv L } – \Bx \cdot \Be \\
&= -\inv{G m^2 M} \gpgradezero{ \Bx \Bp L } – \Bx \cdot \Be \\
&= -\inv{G m^2 M} \gpgradezero{ \lr{ \Bx \cdot \Bp + L } L } – \Bx \cdot \Be \\
&= -\inv{G m^2 M} L^2 – \Bx \cdot \Be,
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:solarellipse:240}
r = -\frac{L^2}{G M m^2} – r e \cos\theta,
\end{equation}
or
\begin{equation}\label{eqn:solarellipse:260}
r \lr{ 1 + e \cos\theta } = -\frac{L^2}{G M m^2}.
\end{equation}
Observe that the RHS constant is a positive constant, since \( L^2 \le 0 \). This has the structure of a conic section, if we write
\begin{equation}\label{eqn:solarellipse:280}
-\frac{L^2}{G M m^2} = e d.
\end{equation}
This is an ellipse, for \( e \in [0,1) \), a parabola for \( e = 1 \), and hyperbola for \( e > 1 \) ([1] theorem 10.3.1).

fig. 1. Ellipse with e = 0.75

In fig. 1 is a plot with \( e = 0.75 \) (changing \( d \) doesn’t change the shape of the figure, just the size.)

References

[1] S.L. Salas and E. Hille. Calculus: one and several variables. Wiley New York, 1990.

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.)