This post is inspired by:

In the diagram a 1 meter rod, initially horizontal, rotates 90-degrees about endpoint B & then translates 1 m to the right. The two endpoints trace out paths of combined length 2+π/2 ms. Is there a way to move the rod between these two positions with less total endpt path length? pic.twitter.com/MdwVKeF3Qq

— James Tanton (@jamestanton) February 8, 2021

## The Problem

Given a rod. Consider its *rigid motions*, meaning that the rod can move by translation or rotation or a combination thereof.
We hope to characterize "minimal length paths" where we define the length of a path as the sum of the lengths of the trajectories travelled by the two end points of the rod.
In particular, given an initial and a final configuration of the rod, we want to find out whether a minimal length path connecting the
two configurations exists, and if it does, what the path is.

As a note: we will assume the rod is oriented, or marked, so the two end points are distinguished from each other. A similar analysis can be performed for indistinguishable end points.

## Configuration space

The configuration space of the rod is given by the location of its center, together with its angle of repose.
In other words, we can imagine the rod's configuration as an element of $\mathcal{C} = \mathbb{R}^2\times\mathbb{S}$.
Throughout we will parametrize $\mathcal{C}$ by $(x,y,\theta)$.
For convenience we will assume the rod has length **two**, so that the locations of the two end points, $p_A$ and $p_B$, are given by
$(x + \cos\theta, y + \sin\theta)$ and $(x-\cos\theta, y-\sin\theta)$ respectively.

For analysis of the motion, we will need to think also about the phase space, which we can represent by $T\mathcal{C}$, the tangent bundle of our configuration space. We will coordinate $T\mathcal{C}$ by $(x,y,\theta, \dot{x}, \dot{y}, \dot{\theta})$.

## Distance measure

Let $\gamma:[0,1]\to\mathcal{C}$ be a path in configuration space, we want to measure its length in terms of the distance travelled individually by the two end points. So we can write \[ \ell(\gamma) = \int_0^1 \left| \frac{d}{dt}(x + \cos\theta, y+\sin\theta)\right| + \left|\frac{d}{dt}(x-\cos\theta, y-\sin\theta)\right| ~ds \] where the absolute value sign denotes the Euclidean norm of vectors in $\mathbb{R}^2$.

For convenience we will introduce the following notations for $F_A, F_B: T\mathcal{C} \to [0,\infty)$, \begin{align} F_A &= \sqrt{ (\dot{x} - \sin(\theta)\dot{\theta})^2 + (\dot{y} + \cos(\theta) \dot{\theta})^2} \newline F_B &= \sqrt{ (\dot{x} + \sin(\theta)\dot{\theta})^2 + (\dot{y} - \cos(\theta) \dot{\theta})^2} \end{align} So if we denote by $\bar{\gamma}$ the lift of $\gamma$ as a curve in $T\mathcal{C}$, we can analogously write \begin{equation} \ell(\gamma) = \int_0^1 F_A(\bar{\gamma}) + F_B(\bar{\gamma}) ~ds. \end{equation}

We notice that this distance function is *not* given by a Riemannian metric on the configuration space: $F_A+F_B$ is not the "length"
that corresponds to a quadratic form on $T\mathcal{C}$.
Notice additionally that each of $F_A$ and $F_B$ *individually* are the lengths of a *degenerate* Riemannian metric on $T\mathcal{C}$, which we will denote by $g_A$ and $g_B$ respectively.
A direct examination of the functions $F_A$, $F_B$, shows that each of $g_A$ and $g_B$ has a one dimensional null space, and the null spaces for the two metrics are transverse to each other.
The above discussion indicates that the function $F_A + F_B$ is

- Positively 1-homogeneous
- Convex
- Positive definite

and hence our length metric $\ell$ is *almost* a Finsler metric.
It fails in that, restricted to the fiber, the function $F_A+F_B$ is not smooth: the singular set $S$ is the union of the null spaces of $g_A$ and $g_B$ respectively, and is explicitly given by
\begin{equation}
S = S_A \cup S_B, \quad S_A = \{ \dot{x} - \sin(\theta) \dot{\theta} = 0 = \dot{y} + \cos(\theta) \dot{\theta} \},
\quad S_B = \{ \dot{x} + \sin(\theta) \dot{\theta} = 0 = \dot{y} - \cos(\theta) \dot{\theta} \}.
\end{equation}

## Geodesic Equation

Where $F_A + F_B$ is differentiable, it is in fact smooth. And so away from the singular set $S$, the geodesics can be described by the Euler-Lagrange equations. Observe that \[\begin{aligned} F_A ~dF_A &= (\dot{x} - \sin(\theta) \dot{\theta}) (d\dot{x} - \sin(\theta) ~d\dot{\theta} - \cos(\theta) \dot{\theta} ~d\theta) \newline & \qquad + (\dot{y} + \cos(\theta) \dot{\theta}) (d\dot{y} + \cos(\theta)~d\dot{\theta} - \sin(\theta) \dot{\theta} ~d\theta) \newline F_B ~dF_B &= (\dot{x} + \sin(\theta) \dot{\theta}) (d\dot{x} + \sin(\theta) ~d\dot{\theta} + \cos(\theta) \dot{\theta} ~d\theta)\newline & \qquad + (\dot{y} - \cos(\theta) \dot{\theta}) (d\dot{y} - \cos(\theta)~d\dot{\theta} + \sin(\theta) \dot{\theta} ~d\theta) \end{aligned}\] So the Euler-Lagrange Equations can then be simplified to \begin{gather} \label{eq:geod:x}\frac{d}{dt} \left( \frac{\dot{x}-\sin(\theta) \dot{\theta}}{F_A} + \frac{\dot{x}+ \sin(\theta)\dot{\theta}}{F_B} \right) = 0 \newline \label{eq:geod:y}\frac{d}{dt} \left( \frac{\dot{y}+\cos(\theta) \dot{\theta}}{F_A} + \frac{\dot{y}- \cos(\theta)\dot{\theta}}{F_B} \right) = 0 \newline \label{eq:geod:th} \frac{d}{dt} \left( \frac{\dot{\theta}}{F_A} + \frac{\dot{\theta}}{F_B} \right) + \sin(\theta) \frac{d}{dt} \left(\frac{\dot{x}}{F_B} - \frac{\dot{x}}{F_A} \right) + \cos(\theta) \frac{d}{dt} \left(\frac{\dot{y}}{F_A} - \frac{\dot{y}}{F_B} \right) = 0. \end{gather}

### Interpretation

Observe that the singular sets $S_A$ and $S_B$ corresponds to the zero-level-set of $F_A$ and $F_B$ respectively, which corresponds to
the *speeds* at which the two endpints $p_A$ and $p_B$ is moving. So as long as **both end points are moving**, we can apply the
geodesic equation formulation.

In this setting, let $v_A$ and $v_B$ denote the *unit velocity vectors* of $p_A$ and $p_B$ (in other words, the directions of travel of the
two points). Then \eqref{eq:geod:x} and \eqref{eq:geod:y} can be combined to read simply
\begin{equation}\label{eq:geod:vp}
\frac{d}{dt} (v_A + v_B) = 0.
\end{equation}
Using this notation, the equation \eqref{eq:geod:th} can be further simplified to read
\begin{equation}\label{eq:geod:vm}
(\sin\theta, - \cos\theta) \cdot \frac{d}{dt}(v_A - v_B) = 0.
\end{equation}

Notice that $v_A$ and $v_B$ must satisfy a *compatibility condition*.
The data $\dot{x}, \dot{y}, \dot{\theta}$ can be reconstructed from the unit vectors $v_A, v_B$ and the speeds $F_A$, $F_B$.
The preservation of length requires that:
\begin{equation}\label{eq:comp:cond} (F_A v_A - F_B v_B) \cdot (\cos\theta, \sin\theta) = 0. \end{equation}
Notice that by reparametrization we can always assume that $F_B = 1$, so the dynamical variable is the ratio $F_A/F_B$, which can be
solved algebraically once we found $v_A$ and $v_B$, unless $v_A$ and $v_B$ are both orthogonal to $(\cos\theta, \sin\theta)$.

### Special classes of solutions

It is instructive to think about some special solutions to \eqref{eq:geod:vp} and \eqref{eq:geod:vm}.

#### "Linear translations"

Observe that if $\frac{d}{dt} v_A = \frac{d}{dt} v_B = 0$, then the motion solves both \eqref{eq:geod:vp} and \eqref{eq:geod:vm}.

In this case the two end points are each constrained to move along a fixed line in $\mathbb{R}^2$ *with a fixed direction*.

There are two sub cases:

- The lines for $p_A$ and $p_B$ are parallel: in this case the geodesic can be extended indefinitely.
- The lines for $p_A$ and $p_B$ are not parallel: in this case the geodesic path must remain in some compact region in $\mathcal{C}$.

To explain the second point a bit more: given two lines $L_A, L_B$ in $\mathbb{R}^2$. The set of points $(p_A, p_B)\in L_A\times L_B$ with $|p_A - p_B| \leq 2$ is a compact subset, and since in configuration space $(x,y) = \frac12 (p_A + p_B)$ we are forced to conclude that the path must remain bounded $\mathcal{C}$.

Geometrically we also see an interesting feature: let $o$ denote the intersection of $L_A$ and $L_B$, which we assumed to be non-parallel. Then $o, p_A, p_B$ for a triangle. The preservation of the length of $\overline{p_Ap_B}$ means a non-trivial motion cannot carry the triangle to another one that is similar to itself originally. This means that the motion must be such that the angles at $p_A$ and $p_B$ change in complementary ways.

From the law of sines, we see that allowable motions depend on the angles.

- If both angles at $p_A$ and $p_B$ are acute, then the lengths of $\overline{op_A}$ and $\overline{op_B}$ change in opposite ways.
- If one angle is acute and one is obtuse, then the two lengths change in the same way.

This shows that if we transition from one case to the other, at the instant that the angle at $p_A$ is a right angle, necessarily that $p_B$ would be stationary. And we obtain the following result.

A further consequence of the geometric analysis is that

*non-consecutive*obtuse (including right) angles. (That is to say, either both the angles at $p_B$ and $p_A'$ are $\geq \pi/2$, or both the angle at $p_A$ and $p_B'$ are $\geq \pi/2$.)

#### Turning solutions

By inspection, another class of solutions to \eqref{eq:geod:vp} and \eqref{eq:geod:vm} happens when $v_A + v_B = 0$, and $\frac{d}{dt} v_A \perp (\sin\theta, -\cos\theta)$. This second relation requires, since $v_A \in \mathbb{S}$, that $v_A = \pm (\sin\theta, -\cos\theta)$. Examining the formula for $v_A, v_B$, we find that necessarily \[ (\dot{x}, \dot{y}) = \lambda (\sin\theta, - \cos\theta) \] with $|\lambda| < |\dot{\theta}|$.

The requirement that the velocities of $p_A$ and $p_B$ are not vanishing implies that we can take $\theta$ as a parameter for such solutions. So given any $\lambda: \mathbb{R}\to (-1,1)$, the curve \begin{equation} \gamma(s) = ( \int_0^s \lambda(t) \sin(t)~dt, -\int_0^s \lambda(t) \cos(t) ~dt, t) \end{equation} is a geodesic.

Some specific examples of these types of trajectories include:

- Purely rotating solutions: $\lambda(t) = \lambda_0$. These represent spinning the rod about a point strictly in its interior.
- Cycloid solutions: consider the case $\lambda(t) = \lambda_0 \sin(t)$. We can solve and find \begin{equation} \gamma(s) = ( \frac\lambda2 t - \frac\lambda4 \sin(2t), \frac\lambda4 \cos(2t), t) \end{equation} the trace of the center of the rod follows the cycloid.

Notice that the purely rotating solutions are *periodic*, and hence are only locally length minimizing (not globally).
Similarly, for the cycloid solutions, observe that the configuration between $\gamma(0)$ and $\gamma(\pi/2)$ can be joined by a non-turning, linear translation (parallel case), which has shorter length.
Such example are not unexpected since the configuration space $\mathcal{C}$ has non-trivial topology.

Now, for the Cycloid solutions if we ensure $|\lambda_0| < 1$ the resulting curve will not touch the singular set. What happens when $|\lambda_0| = 1$? We observe that when $s = \frac12 \pi + 2n\pi$ the point $p_A$ is stationary; and when $s = \frac32 \pi + 2n\pi$ the point $p_B$ is stationary. So again we have the corresponding curves begin and end on $S_A$ and $S_B$ respectively, similar to the non-parallel case of linear translation.

### Singular solutions

Finally, we arrive at the case where one of the $p_A$ and $p_B$ is stationary. Recall in this case the geodesic equation cannot be applied, as in this case we are either in $S_A$ or $S_B$.

I claim, however, any motion that lies entirely in $S_A$ or $S_B$ is a local minimum of the length functional.

Suppose, without loss of generality, we have a curve $\gamma$ whise lift $\bar{\gamma}$ lies entirely in $S_A$, which would then require $\dot{x} = \sin(\theta)\dot{\theta}$ and $\dot{y} = - \cos(\theta)\dot{\theta}$. If this curve is non-trivial, then we would require $\dot{\theta} \neq 0$. Hence we can parametrize by the angle $\theta$.

Without loss of generality we can assume that $p_A$ sits at the origin. And hence we must have $x(t) = -\cos(t)$ and $y(t) = -\sin(t)$, and that $F_A = 0$ and $F_B = 2$ along this curve.

Consider a small perturbation, again parametrized by the angle $\theta$, in the form $(-\cos(t) + \delta_x(t), -\sin(t) + \delta_y(t))$. For this curve, we have \[ F_A = \sqrt{\dot{\delta}_x^2 + \dot{\delta}_y^2} \] and \[ F_B = \sqrt{(2 \sin(t) + \dot{\delta}_x)^2 + (-2\cos(t) + \dot{\delta}_y)^2}. \] And we see from triangle inequality that \[ 2 = \sqrt{(2 \sin(t))^2 + (-2\cos(t))^2} \leq F_A + F_B \] showing that any perturbation will necessarily result in a curve with longer length.

## Classification

If you look at the discussion above, we have classified the geodesics away from $S_A\cup S_B$ as follows:

- If $v_A, v_B$ are both not orthogonal to the rod, then $F_A/F_B$ is uniquely determined, and the corresponding geodesic segment is a linear translation.
- If $F_A v_A = F_B v_B$ (including when both orthogonal to rod), then geodesic moves by parallel linear translation.
- If $v_A + v_B = 0$, and both orthogonal to the rod, then the geodesic moves by turning motion.

The remaining case which we have not treated is

- What happens when $v_A = v_B$, both orthogonal to the rod, and $F_A \neq F_B$?

It turns out such initial data is *not compatible* with the geodesic equations! As indicated above, the geodesic equation requires in \eqref{eq:geod:vp} that $v_A + v_B$ remains constant. If $v_A = v_B$ and they are both unit vectors, this will require $v_A$ and $v_B$ both to be constant. But if $F_A \neq F_B$ this will then violate the condition that the rod has length 1!

This degeneracy essentially comes from the fact that the level surfaces of $F_A + F_B$ is not uniformly convex.
Since $F_A + F_B$ is 1-homogeneous, in the *radial* direction the Hessian (in the fibre directions) of $F_A + F_B$ vanishes.
But when $v_A = v_B$, there is an additional degeneracy: the Hessian of $F_A + F_B$ has signature $(1,0,0)$ instead of $(1,1,0)$, as would be the case if instead of $F_A + F_B$ the length function were given by $\sqrt[p]{F_A^p + F_B^p}$ for some $p \in (1,\infty)$.