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

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
\label{eqn:twoForceStaticsProblem:20}
\BF_s + \BF_r + m \Bg = 0.

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
\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}

Here the gravitational direction has been pointed along the x-axis.

Our equation to solve is now
\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.

This we can solve as a set of scalar equations, one for each of the $$\Be_1$$ and $$\Be_2$$ directions
\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}

Our solution is
\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}

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
\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}

If we take cross products with each of the unit vectors, we find
\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}

or
\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}

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
\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}

Our equation to solve is now
\label{eqn:twoForceStaticsProblem:60}
f_r \Be_1 e^{i\alpha} + f_s \Be_1 e^{i\beta} + m g \Be_1 = 0.

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
\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}

Writing the wedges as grade two selections, and noting that $$e^{i\theta} \Be_1 = \Be_1 e^{-i\theta }$$, we have
\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}

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
\label{eqn:twoForceStaticsProblem:120}
f_r e^{i\alpha} + f_s e^{i\beta} + m g = 0.

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
\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}

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.)

## Geometric Algebra for Electrical Engineers, version V0.1.16-19.

February 7, 2023 math and physics play

A new version, V0.1.16-19, of the book has now been posted on Amazon and all the other usual places.

This version has solutions for most of the chapter I problems, increasing the size of the book by about 13 pages.

I plan to work through the chapter II and III problems too, and add those in a later version.  If that increases the size of the book too much (i.e.: forcing a price increase), I may opt to omit the problem solutions from the print versions, leaving solutions in the free PDF version, and in the Leanpub (pay what you want) edition.

## Motivation

Maverick posed the following question on the bivector discord

“I saw your blog post on curvilinear coordinates in geometric calculus. I saw your derivation of the volume of a sphere using this technique and decided for practice by doing a surface integral to calculate the area of sphere using the quantity $\partial{\theta} \wedge \partial{\phi}~ dA$ is there a way to integrate this without simply taking the magnitude of this quantity and then integrating or are we limited to only integrating quantities that are 1 dimensional like scalars and pseudoscalars”.

My initial response was that, sure, we should be able to compute bivector and trivector valued integrals. However, in retrospect, the reality is a bit more subtle.

We aren’t limited to using the magnitudes of the differential forms, but not all multivector integral are interesting.
In the original blog post, I must have computed the area of the circle using a bivector valued area element, or the volume of a sphere using a trivector valued volume element. However, if I did the volume that way, I probably cheated and computed 8 times the value of the first octant volume (which is positive), vs. the entire integral, which is zero.  [EDIT: the original post, now linked above, has been corrected.]

Let’s compute the circular area and circumference, and the spherical volume and surface area using multivector valued integrals, and see where we end up having to resort to scalar integrals.

## Circular example.

The polar parameterization of points in circular region is
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:20}
\Bx = r \Be_1 e^{i\theta},

where $$i = \Be_1 \Be_2$$.
Our differentials are
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:40}
\begin{aligned}
d\Bx_r &= \Be_1 e^{i\theta} \,dr \\
d\Bx_\theta &= r \Be_2 e^{i\theta} \,d\theta.
\end{aligned}

Our “volume” element, is a 2D pseudoscalar
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:60}
\begin{aligned}
dA &= d\Bx_r \wedge d\Bx_\theta \\
&= r \gpgradetwo{ \Be_1 e^{i\theta} \Be_2 e^{i\theta} } \, dr d\theta \\
&= r \gpgradetwo{ \Be_1 \Be_2 e^{-i\theta} e^{i\theta} } \, dr d\theta \\
&= r i \, dr d\theta.
\end{aligned}

This, as I probably pointed out in my previous blog post, can be integrated to find the area of the circle
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:80}
\begin{aligned}
A &= \int_{r = 0}^R \int_{\theta = 0}^{2\pi} r i \, dr d\theta \\
&= \frac{R^2}{2} 2 \pi i \\
&= \pi R^2 i.
\end{aligned}

However, we got lucky, as the two-form area element was strictly positive (i.e.: the Jacobean for a polar change of coordinates is strictly positive.)

However, we can’t find the circumference of a circle my integrating $$d\Bx_\theta$$ around that circular path, because $$d\Bx_\theta$$ has an orientation, and we will get zero (given the symmetry of the problem) if we integrate all the way around
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:100}
\begin{aligned}
\int_{\theta = 0}^{2\pi} d\Bx_\theta &= \int_{\theta = 0}^{2\pi} r \Be_2 e^{i\theta} \,d\theta \\
&= r \Be_2 \evalrange{ \frac{e^{i\theta}}{i} }{0}{2\pi} \\
&= \frac{r \Be_2}{i} \times 0.
\end{aligned}

If we want the circumference of a circle, we have to sum all the contributions of $$d\Bx_\theta$$ that are colinear with $$\thetacap = \Be_2 e^{i\theta}$$
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:120}
\begin{aligned}
C &= \int_{\theta = 0}^{2\pi} \thetacap \cdot \, d\Bx_\theta \\
&= \int_{\theta = 0}^{2\pi} \thetacap \cdot \lr{ r \thetacap \, d\theta } \\
&= 2 \pi r.
\end{aligned}

This is a plain old boring scalar integral, because the vector valued path integral isn’t terribly interesting.

## Spherical example.

For a spherical parameterization, our position vector is
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:140}
\Bx = r \Be_1 e^{i \phi} \sin\theta + r \Be_3 \cos\theta,

so the differentials are
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:160}
\begin{aligned}
d\Bx_r &= \lr{ \Be_1 e^{i \phi} \sin\theta + \Be_3 \cos\theta } \,dr = \rcap \, dr \\
d\Bx_\theta &= \lr{ r \Be_1 e^{i \phi} \cos\theta – r \Be_3 \sin\theta }\, d\theta = r \thetacap \,d\theta \\
d\Bx_\phi &= r \Be_2 e^{i \phi} \sin\theta \, d\phi = r \sin\theta \phicap.
\end{aligned}

The oriented area element on the surface of the sphere is
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:180}
\begin{aligned}
dA &= d\Bx_\theta \wedge d\Bx_\phi \\
&= r^2 \gpgradetwo{ \lr{ \Be_1 e^{i \phi} \cos\theta – \Be_3 \sin\theta } \Be_2 e^{i \phi} \sin\theta } \,d\theta d\phi \\
&= r^2 \sin\theta \lr{ i \cos\theta – \Be_{32} e^{i \phi} \sin\theta } \,d\theta d\phi .
\end{aligned}

Integrating this over the surface will give us zero, with the first integrand killed by the $$\theta$$ integral, and the second by the $$\phi$$ integral. As pointed out in the original question, we must integrate the absolute value of this two-form in order to find the surface area of the sphere, just as we had to integrate the absolute value of $$d\Bx_\theta$$ for the circle to find the circumference.

Let’s perform that integration to verify that we get the expected result. We will first simplify our bivector valued oriented area element. Observe that $$dA \wedge \rcap = dA \rcap \propto I$$, so $$dA \propto \rcap I$$. We should be able to simplify our expression for $$dA$$ by factoring out an $$\rcap$$ term
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:200}
\begin{aligned}
dA &= r^2 \sin\theta \lr{ \Be_{1233} \cos\theta – \Be_{1132} e^{i \phi} \sin\theta } \,d\theta d\phi \\
&= r^2 \sin\theta I \lr{ \Be_3 \cos\theta + \Be_1 e^{i \phi} \sin\theta } \,d\theta d\phi \\
&= r^2 \sin\theta I \rcap \,d\theta d\phi.
\end{aligned}

The spherical scalar area is
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:220}
\begin{aligned}
A
&= \int_{\theta = 0}^{\pi} \int_{\phi = 0}^{2 \pi} \Abs{ r^2 \sin\theta I \rcap } \,d\theta d\phi \\
&= r^2 \int_{\theta = 0}^{\pi} \int_{\phi = 0}^{2 \pi} \Abs{ \sin\theta } \,d\theta d\phi \\
&= 2 r^2 \int_{\theta = 0}^{\pi/2} \int_{\phi = 0}^{2\pi} \sin\theta \,d\theta d\phi \\
&= 2 r^2 (2 \pi) \\
&= 4 \pi r^2.
\end{aligned}

Observe that to find the volume of the sphere, we also cannot just integrate the trivector valued volume element directly either. That oriented volume element is
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:240}
\begin{aligned}
dV
&= d\Bx_r \wedge dA \\
&= \rcap\, dr dA \\
&= r^2 \sin\theta I \,dr d\theta d\phi.
\end{aligned}

This integrand is positive above the azimuthal plane, and negative below, so will give us zero if we integrate over the entire $$\theta \in [0, \pi]$$ region. So, if we want to find the volume of a sphere, we also must use an absolute integrand.
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:260}
\begin{aligned}
V
&= \int_{r = 0}^{R} \int_{\theta = 0}^{\pi} \int_{\phi = 0}^{2 \pi} \Abs{ r^2 \sin\theta I } \,dr d\theta d\phi \\
&= 2 \int_{r = 0}^{R} \int_{\theta = 0}^{\pi/2} \int_{\phi = 0}^{2\pi} r^2 \sin\theta \,dr d\theta d\phi \\
&= 2 \frac{R^3}{3} (2 \pi) \\
&= \frac{4}{3} \pi R^3.
\end{aligned}

Had the sign of our volume element been invariant over the entire integration region, as it was for the circular area computation (but not the circular boundary computation), we could have computed this as a pseudoscalar integral. For example, if we wanted to know what the oriented volume of the first quadrant of the sphere was, we could compute that directly, as
\label{eqn:circularAndSphericalAreaVolumeAndBoundaries:280}
\begin{aligned}
V_1 &= \int_{r = 0}^{R} \int_{\theta = 0}^{\pi/2} \int_{\phi = 0}^{\pi/2} r^2 \sin\theta I \,dr d\theta d\phi \\
&= \inv{6} \pi R^3 I,
\end{aligned}

but if this volume integral is extended to the entire spherical region, the result is zero, not $$(4/3) \pi R^3 I$$.

Only when our multivector integrand doesn’t change sign over the integration region, can we directly integrate without taking absolute values.
Again, this should not be too surprising.
This is why, in conventional scalar calculus, we generally must take the absolute value of our change of variable Jacobians, when we compute area or volume computations.