Chapter 18: Of Cats and Their Most Wonderful Righting Reflex

Our second Lie Theoretic Systems Theory example is the theoretical physics of how a cat flips over to land on its feet when it falls, even though there are no torques on it and angular momentum must be conserved.

That a falling cat can right itself as it falls so that it always seems to land on its feet is something we humans have known about one of our most beloved companion and fellow animals since antiquity. This is called the cat’s righting reflex. What is more curious is that since Newton, and especially since Euler, physicists and mathematicians have had the mathematical wherewithal to study this seemingly paradoxical happenning precisely and yet it took a long time for this to be done. The motion at first seems paradoxical, for Euler’s second law shows that the total angular momentum of a system sundered from the rest of the World must be constant. The full version of Euler’s second law is of course that the nett torque $\vec{T}$ imparted on a system equals that system’s time rate of change $\d_\tau\vec{L}$ of angular momentum $\vec{L}$: $\vec{T} =\d_\tau\,\vec{L}$.

So how does a cat “spin” without torque from the outside World? James Clerk Maxwell, in particular, got himself a somewhat notorious fame at Trinity College, Cambridge for throwing cats out of windows to study this phenomenon. Maxwell would easily have had the mathematical wherewithal to understand my description below, and I find it unbelievable that he did not notice that the problem was not trivial, so why did he not study this problem seriously from a theoretical standpoint? Either he was busier with other things, or maybe, as I often do, he looked at this particular physical system and decided that it was too complex to yield fruitful theoretical results and instead that experimental investigation was likely to be the most successful. Indeed, I shall study highly “abstracted” cats below.

So it is indeed curious that the first really formal scientific study was done by physiologists more interested in the working of a cat’s vestibular system (balance sense begotten by the inner ear) than the physics issue of angular momentum, beginning in the late 1800s. However, the first such physiologist Étienne-Jules Marey (1830-1904) (famous for the development of motion photography for the study of high speed movements) states very clearly[Marey, 1894] that “First of all, the inspection of these figures [photos of falling cats] rules out the notion that the animal imparts a rotational motion on itself by thrusting against the hands of the experimenter. [This conclusion follows] because the first frames of the two series [of photos of a falling cat] show that in the first instants of it its fall, the cat as yet has no tendency to turn from one side nor the other. Its rotation only begins with the twisting of its waist.” (my translation – apologies to any French speakers amongst you – go and look at the original online!) so he clearly understood that cat’s righting motion was torque free. He was talking about a common misconception of time that the “nonconservation of angular momentum” was owing to the cat’s thrusting against whatever it fell off or from. The first studies from an angular momentum standpoint seem to have been [Rademaker & Ter Braak, 1936] and [Kane & Scher, 1969], the former still dealing more with the cat’s vestibular system. The work written up in [Kane & Scher, 1969] was part of the great body of research done in the late 1950s and early 1960s to understand how the human body would deal with the conditions it met in outer space travel. Thomas Kane in particular, experimentally showed that humans could, with training, make the righting reflex motions and flip over exactly like a cat (see this popular account in the Atlantic Magazine[Madrigal, 2011]). Étienne-Jules Marey had noted in 1894 that he had also found that dogs and chickens can make the same righting reflex motions in the right conditions, although I am skeptical of the latter, in the sense that the surrounding air can impart significant torques on chickens when the latter beat their wings, so the flipover witnessed here is unlikely to be torque free.

The resolution to the “paradox” is found in understanding that the cat is a deformable, not rigid, body. The cat can be described by a manifold $\mathcal{S}$ called the “space of shapes” and it can deform its shape to move smoothly between points of the manifold. These shapes are described in a co-ordinate frame that is fixed with respect to axes mounted on the tumbling (i.e. falling and rotating) cat. In the language of fiber bundles, the space of shapes is the base space $\mathcal{S}$, the fiber $\mathfrak{F}$ is the group $SO(3)$ (or $SO(2)\cong U(1)$) of the cat’s orientations in space. The fiber bundle description may at first glance seem to some a mathematical overkill, but once you get the hang of it it is extremely natural: there is a great deal of redundancy in the way we specify a cat’s (or other body’s) complete configuration. Talk of a body’s orientation shift is really only defineable at a particular given point in the space of shapes: an orientation shift is a homogeneous Euclidean isometry, so one can’t really compare a doubled up cat’s orientation (or position, in the more general problem) to that on a cat sitting on his or her haunches, for example. The points on the space of shapes are then equivalence classes of cat (or other body) configurations equivalent by a proper (nonreflective) isometry. From the cat’s standpoint, we could imagine the cat in deep space – together with a lifetime’s supply of food and water and exciting cat simulated reality including kind cat friends with endless hunts so that we could be assured that no feeling creatures were harmed in the course of this thought experiment. Then the cat’s orientation really would be more like a gauge field: a smooth symmetry linking configurations that, from the cat’s standpoint, are all completely physically equivalent.

The Space of Shapes and Cat Configuration Fiber Bundle

The cat’s complete configuration space is a fiber bundle $\mathcal{B}$ whose points comprise a complete specification of every point in the cat’s body. We then quotient out $\mathcal{B}/\mathfrak{F}$ those configurations which are equivalent modulo a rigid Euclidean isometry (or whatever maximal Lie subgroup $\mathfrak{F}$ of the Euclidean group links rigidly-related configurations) to smoothly project the bundle points onto the space $\mathcal{S}=\mathcal{B}/\mathfrak{F}$ of shapes. The points of $\mathcal{S}$ are a cat in a “standard” configuration. Naturally, we can shift our definition of a “standard” configuration by imparting a rigid transformation $\gamma(s)\in \mathfrak{F}$ to each and every point $s\in\mathcal{S}$ to get a new base space $\mathcal{S}^\prime=\{\gamma{s}\,s|\;s\in\mathcal{S}\}$ and our description must be physically equivalent. So there is a great deal of redundancy in our description.

Let us encode this redundancy in our description. In its raw form, a point in the bundle is some subset $S(s,\,x) \in \mathcal{B};\;S(s,\,x)\subseteq R^N$ of Euclidean space, so it is literally a set comprising up to $\beth_1=2^\aleph_0$ position vectors in $\R^3$. We project this point into an equivalence class of isometrically-equivalent sets by naming a “standard” shape $S_0(s) \in \mathcal{S}$ and an orientation $\gamma(y)\in \mathfrak{F}\subseteq SO(3)$. Here the list $s \in R^M$ of parameters locally parameterises the standard shape, so that the space of shapes can be thought of as an $M$-dimensional manifold and the up to three parameters in $x$ are local co-ordinates for the orientation $\gamma(x)\in \mathfrak{F}\subseteq SO(3)$. Thus the full shape is given by the set of position vectors:

$$\label{FullShape}S(s,\,x) = \gamma(x)\,S_0(s)$$

If the shape shifts with “time” $\tau$ then

$$\label{FullShapeDerivative}\d_\tau\,S = \gamma\,(\gamma^{-1}\,\d_\tau\,\gamma\, S_0(s(\tau)) + \d_\tau\,S_0(s(t(\tau)))\stackrel{def}{=} \gamma\,(H(\tau)\, S_0(s(\tau)) + \d_\tau\,S_0(s(t(\tau)))$$

where in this equation $\gamma$ and other transformations act pointwise on each of the position vectors in $S\in\mathcal{B}$. As we already well know, $\gamma^{-1}\,\d_\tau\,\gamma = H(\tau)\in\mathfrak{f}\subseteq \mathfrak{so}(3)$ lies in the Lie algebra $\mathfrak{f}$ of the fiber $SO(3)$ of the bundle and encodes an “infinitessimal” orientation shift. In particle and other physics, $H$ is called a gauge field or gauge potential. If we now to change our definition of standard shapes so that $S_0(s)\mapsto \tilde{S}_0(s) =\zeta(s) \,S_0(s)$ then our physical description must be unchanged, so that $S(s(\tau),\,x(\tau)) = \gamma(\tau)\,\zeta(s(\tau))^{-1}\,\tilde{S}_0(s(\tau))$. Our new orientation transformation is $\tilde{\gamma}= \gamma(\tau)\,\zeta(s(\tau))^{-1}$ On differentiating this equation with respect to $\tau$ and equating with $\eqref{FullShapeDerivative}$ we find:

$$\label{FullShapeDerivative_2}\d_\tau\,S = \tilde{\gamma}\,\left(\left(\zeta(\tau)\,H(\tau)\,\zeta(\tau)^{-1}+\zeta(\tau)\,\d_\tau\,\zeta^{-1}\right)\,\tilde{S}_0(s(\tau)) + \d_\tau\,S_0(s(\tau))\right)$$

As we know, $\zeta(\tau)\,\d_\tau\,\zeta^{-1}\in\mathfrak{f}$, so that the transformed “gauge potential” is:

$$\label{FullShapeDerivative_3}\tilde{H}(\tau)\stackrel{def}{=} \zeta(\tau)\,H(\tau)\,\zeta(\tau)^{-1}+\zeta(\tau)\,\d_\tau\,\zeta^{-1}\in\mathfrak{f}\subseteq\mathfrak{so}(3)$$

and $\eqref{FullShapeDerivative}$ holds when we put in $\tilde{S}_0$ and the transformed Lie algebra member $\tilde{H}$ defined by $\eqref{FullShapeDerivative_3}$.

This above should be well wonted to physicists who have explored gauge transformations in electromagnestism or Yang-Mills-kind particle theories. A clear primer for this concept is [Shapere & Wilczek, 1989], if you skim over the fluid dynamical parts. That’s not to say that the latter are trivial. This is where the physics enters our description. In our falling cat problem, as the cat shifts its shape, i.e. follows a piecewise $C^1$ path through $\mathcal{S}$, the conservation of angular momentum means that there must in general be a corresponding change in the cat’s orientation. Our bundle’s topology is defined by the parallel transport notion or connexion one gets by computing the shift in cat orientation that arises, owing to the conservation of angular momentum, from the following of a piecewise $C^1$ path through the space of shapes. The structure group is some connected Lie subgroup of $SO(3)$ acting on the fiber $SO(3)$ itself.

In this way [Montgomery, 1993] takes th eapproach of [Shapere & Wilczek, 1989] and reworks it to lay down a comprehensive gauge theoretic framework for thinking about this kind of problem. His cat motions are modelled on [Kane & Scher, 1969], abstracted and generalised into a standpoint that understands the cat’s orientation in space as a Yang-Mills gauge field on the cat’s configuration space. [Stewart, 2009] gives an amusing popular account of the phenomenon as a light hearted break to the reading of the Montgomery paper.

Introductory Physics and Co-ordinate Definitions

Any – rigid or not – body comprising a mass density distribution $\rho(\vec{r},\,\tau)$, i.e. a local mass density $\rho(\vec{r},\,\tau)$ at the point with position vector $\vec{r}$ at the time $\tau$ (measured relative to an inertial space) has a total angular momentum:

$$\label{AngularMomentum}\vec{L} = \int_\mathcal{\R^3}\,\rho(\vec{r},\,\tau)\,\vec{r}\times\d_\tau\vec{r}\,\d \R^3$$

and this angular momentum evolves with time following Euler’s Second Law of motion:

$$\label{EulersSecondLaw}\vec{T}=\d_\tau\vec{L}$$

where $\vec{T}$ is the torque exerted on the system by its surrounding environment. In the torque free case, $\d_\tau\vec{L}=\vec{0}$ and $\vec{L}$ is a constant of the motion, by Euler’s second law. Alternatively, conservation can be thought of in terms of Noether’s Theorem: that if a system’s Lagrangian is invariant with respect to co-ordinate transformations belonging to a Lie group $\G$, then there is one conserved quantity for each basis member of $\g=\operatorname{Lie}(\G)$. From this viewpoint, energy conservation is to be understood as the conserved quantity arising from the World’s Lagrangian’s being invariant with respect to translations in time (Nature doesn’t care about where we put our time origin $\tau=0$ and physics is independent of this, at least over timescales that are small in comparison with the Universe’s age). The three components of linear momentum are conserved because our description of the World must be invariant with respect to any shift in the origin of our co-ordinate system, and the three components of angular momentum are conserved because Nature doesn’t care whether we measure things relative to one co-ordinate frame, or a rigidly rotated version thereof: Her physics will be unchanged.

Let’s begin with an extremely simple example – maybe the simplest – that resolves the “paradox” and shows that deformable bodies can indeed re-orient themselves in space whilst conserving angular momentum. In all of the following, we shall ignore the translational kinematics and dynamics. This is easy to justify: Newton’s laws of motion in these cases simply say that the centre of mass of a system is either still (if the cat or astronaut is in outer space) or has the motion of a point particle in a uniform gravitational field if we think about the falling cat. So, we can treat this computation as being decoupled from the orientation and shape shifting calculations: after we work out the evolution of these latter, we then work out the implied position of the system’s centre of mass as a function of time. Then we can simply conclude that the system’s translation up to that time must be such that, when added to the centre of mass position vector as computed in a frame attached to the cat, the total translation is either nought or accelerates uniformly, as appropriate.

Figure 18.1: Simple Robot “Cat” Freefalling in Space

Above I have drawn a simplified “robot cat” freefalling (or “floating”) in space as in Figure 18.1 with no external torques. This system’s angular momentum must therefore be conserved. I am a very aural person, so this is good enough for me so long as I can imagine it mewing!

Now our “cat” comprises two cylindrical sections: the “forecat” ($F$), “hinder-cat” ($F$) and two legs ($L$) which can be drawn in so that they are flush with the hinder-cat’s surface. With the legs drawn in, the forecat on one hand and hinder-cat + legs assembly on the other have the same mass moment of inertia about the axis of the body. So now, here is how the cat rotates:

1. Deploy legs symmetrically, i.e. spread them out as shown in the drawing. Now the hinder cat + legs has a bigger mass moment of inertia than the fore cat. Note that, if the legs are diametrically opposite and identical and are opened out symmetrically, the cat undergoes no motion;
2. With an internal motor, the fore cat and hinder cat exert equal and opposite torques on one another to accelerate, then stop. Owing to the differences between the moments of inertia, the forecat undergoes a bigger angular displacement $\phi_F$ than the hinder cat $\phi_H$;
3. Pull the legs back in so that they are flush with the hinder-cat. Again this begets no motion if done symmetrically;
4. Use the internal motor again with an acceleration / deceleration sequence to bring the forecat and hinder cat back to their beginning alignment (i.e. with the line along the cylinders aligned). Now the two halves have the same mass moment of inertia, so when the cat is aligned again, the rotation angles are equal and opposite.

After this procedure, the cat’s shape is the same as it was at the beginning. Since the rotation angles are different in step 2, but the same in step 4, our robot cat’s angular orientation has shifted. Indeed, if $\phi_F=\phi_H = 0$ at the beginning of step 1 above, we can readily show that at the end of the above procedure that:

$$\label{RobotCat1Cycle}\phi_F=\phi_H = \left(\frac{1}{2}-\frac{I_F}{2\,I_H}\right) \Delta$$

where $\Delta$ is the angular displacement of the fore-cat in step 2. $I_F$ is the mass moment of inertia of the fore cat about the system’s long axis of symmetry and this is equal to that of the hinder cat in steps 1, 3 and 4. The hinder cat’s mass moment of inertia about this axis at step 2 is $I_H$ and $\eqref{RobotCat1Cycle}$ shows that the bigger the difference between $I_F$ and $I_H$, the bigger the angular displacement of the system’s orientation over the cycle. Also, by controlling the length of step 2 and thus of step 4, the size of the overall orientation shift can be precisely controlled. There is then a smooth cyclic evolution of shape that can shift the orientation through any angle about the system’s long axis of symmetry. One could imagine a real cat twisting its forward and hinder halves and deploying its legs to get a “natural” version of the above.

Let us now look at the cat’s configuration space. It is $\mathbb{T}^2\times\R$, i.e. the product of two circles (one for each angle $\phi_F$ and $\phi_H$ and a third parameter representing the hinder cat’s moment of inertia. This assumes of course that the hinder cat’s moment of inertia can take on any real value abd become infinitely positive or infinitely negative. A more realistic model is that the hinder cat’s moment of inertia is of the form $I_H = I_0 + I_d \sin(\theta)$ where $I_0$ is the moment of inertia of the fore cat (and hinder cat when its legs are flush with its body) so that the moment of inertia is parameterised by the angle $\theta$ and is kept within hard limits. In this case, our configuration space is the 3-torus $\mathbb{T}^3$.

Suppose the cat moves his or her hinder legs as in step 1 or step 3 above. Then a tangent vector to the path through configuration space is proportional to $(0,\,0,\,1)$ and thus the differential of this automorphism of $\mathcal{M}$ is the vector field:

$$\label{RobotCat1VectorField_R}\mathscr{R} = \partial_\theta$$

If the cat undergoes an internal twist as in step 2 or step 4 above, the tangent vector to the path through configuration space is proportional to $(I_0 + I_d\,\sin\theta,\,I_0,\,0)$, so the differential of this automorphism of $\mathcal{M}$ is the vector field:

$$\label{RobotCat1VectorField_W0}\mathscr{W}_0 =\left(1 + \kappa\,\sin\theta\right)\, \partial_{\phi_F} – \partial_{\phi_H}$$

where $\kappa = I_d / I_0$ and so we get the commutators:

$$\label{RobotCat1VectorField_XY}\begin{array}{c}\mathscr{X} =\left[\mathscr{R},\,\mathscr{W}_0\right] = I_d\,\cos\theta\, \partial_{\phi_F}; \quad \mathscr{Y} = \left[\mathscr{R},\,\mathscr{X}\right] = -\kappa\,\sin\theta\, \partial_{\phi_F};\quad -\mathscr{X} = \left[\mathscr{R},\,\mathscr{Y}\right]; \\\quad \left[\mathscr{W}_0,\,\mathscr{X}\right] = \left[\mathscr{W}_0,\,\mathscr{Y}\right] = \left[\mathscr{X},\,\mathscr{Y}\right]=0\end{array}$$

and so we have a closed Lie algebra, which, by dint of both $\mathscr{R}$’s and $\mathscr{W}_0$’s exponentiating to automorphisms of $\mathcal{M}$, must define a four dimensional Lie group of automorphisms of $\mathcal{M}$ by Theorem 11.14 and Theorem 13.4. This algebra is “neater” if we redefine $\mathscr{W} = \mathscr{W}_0 + \mathscr{Y} = \partial_{\phi_F} – \partial_{\phi_H}$; we then have:

$$\label{RobotCat1LieAlgebra}\begin{array}{c}\left[\mathscr{X},\,\mathscr{Y}\right] = 0;\quad \left[\mathscr{R},\,\mathscr{X}\right] = \mathscr{Y};\quad \left[\mathscr{R},\,\mathscr{Y}\right] = -\mathscr{X}\\\left[\mathscr{W},\,\mathscr{X}\right]=\left[\mathscr{W},\,\mathscr{R}\right]=\left[\mathscr{W},\,\mathscr{R}\right]=0\end{array}$$

which is the direct sum $\mathfrak{e}(2) \oplus \R$ of the Lie algebra $\mathfrak{e}(2)$ of the Euclidean group $E(2)$ which we met in $\eqref{CarConfigurationTangent_12}$ with the one dimensional real vector space $\R$. The exponentiation of the vector field $\mathscr{W}$ is simply a transformation of the form $T_{phi_0}:\mathcal{M}\to\mathcal{M};\,T_{\phi_0} (\phi_F,\,\phi_H,\,\theta)=(\phi_F+\phi_0,\,\phi_H-\phi_0,\,\theta);\,\phi_0\in\R$, thus the “mean” $(\phi_F+\phi_H)/2$ of the two angles stays constant. We can use a transformation of the form $\exp(\tau\,(x\,\mathscr{X}+y\,\mathscr{Y})$ to transform a cat configuration of the form $(\phi_F\,\,\phi_H,\,\theta)$ to one of the form $(\phi_F^\prime\,\,\phi_H,\,\theta)$, i.e. we can choose the fore-cat’s angle to be any value we like to achieve any mean angle $\phi_F^{\prime\prime}=(\phi_F^\prime+\phi_H)/2$, and then we can use the transformation $\exp((\phi_H-\phi_F^\prime)\mathscr{W})$ to make the two angles equal to $\phi_F^{\prime\prime}$.

A slight variation to this theme is to take the configuration space to be $\mathbb{T}^2\times\R$ so that we have the two angles $\phi_F,\,\phi_H$ and the ratio of the moments of inertia is then $e^\kappa$ where $\kappa\in\R$. This keeps the moments of inertia (realistically) positive, but it also means there is no upper limit on the hinder cat’s moment of inertial. Thus this configuration space is noncompact and $\kappa=0$ corresponds to equal moments of inertia. We then have the vector fields and Lie algebra:

$$\label{RobotCat1NoncompactLieAlgebra}\begin{array}{c}\mathscr{R} = \partial_\kappa; \quad \mathscr{X} = e^\kappa\,\partial_{\phi_F} – e^{-\kappa}\,\partial_{\phi_H};\quad \mathscr{Y} =e^\kappa\,\partial_{\phi_F} + e^{-\kappa}\,\partial_{\phi_H}\\ \left[\mathscr{R},\,\mathscr{X}\right]=\mathscr{Y};\quad \left[\mathscr{R},\,\mathscr{Y}\right]=\mathscr{X};\quad\left[\mathscr{X},\,\mathscr{Y}\right]=0\end{array}$$

Take heed how in $\eqref{RobotCat1LieAlgebra}$ there is a Lie subalgebra $\mathfrak{e}(2)$ comprising $\mathscr{R},\,\mathscr{X},\,\mathscr{Y}$ and in this algebra’s adjoint representation, the matrix of $\exp(\ad(\mathscr{R})$ is the two-dimensional rotation matrix $\exp(\phi\,\ad(\mathscr{R})) = \left(\begin{array}{cc}\cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{array}\right)$ whereas in $\eqref{RobotCat1NoncompactLieAlgebra}$ it is the “hyperbolic” rotation or “boost” $\exp(\eta\,\ad(\mathscr{R})) = \left(\begin{array}{cc}\cosh\eta&-\sinh\eta\\-\sinh\eta&\cosh\eta\end{array}\right)$, reflecting the compact or noncompactness of the configuration space in each case.

The Angular Momentum Conservation Connexion in the Space of Shapes

We now characterise the parallel transport notion that a body’s torque-free motion in space defines on the space of shapes. That is, motion in the space of shapes (shape shifting) begets a corresponding shift in the orientation, such that the body’s total angular momentum is unchanging. There is an angular momentum functional $\mathcal{L}: T(\mathcal{B}) \to \R^3$ mapping the bundle’s tangent space $T(\mathcal{B})$ (the space of “velocities”) to a total vector angular momentum in $\R^3$. Motion on the bundle at any point $S$ can only in the directions in that point’s tangent space $T_S(\mathcal{B})$ that correspond to $\d\,\mathcal{L}=\Or$ – the directions that conserve angular momentum. We split $\d\,\mathcal{L}$ into a part owing to shape shifting and a part owing to orientation shifting: these must sum to nought and thus define the change in orientation accompanying a given shape shift.

Given our Lie theoretical knowledge, this calculation will be easiest for us if we do not develop the full, conventional rotational mechanics machinery for this discussion but instead use quaternion notation; thus we represent our rotations in $SU(2)$. In this notation, the vector cross product is easy to represent as the Lie bracket of quaternions; things are much more awkward in standard vector notation. We wish to calculate first the total angular momentum of a single point mass $\d m$ belonging to an oriented shape $S\in\mathcal{B}$ with instantaneous position vector $R$. The Cartesian co-ordinate space then becomes the Lie algebra $\mathfrak{su}(2)$ of traceless skew-Hermitian matrices so that the point $X=(x,\,y,\,z)$ is represented by the traceless skew-Hermitian:

$$\label{QuaternionPoint}X =x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z = \left(\begin{array}{cc}i z&i\,x-y\\i\,x+y&-i z\end{array}\right)$$

where $\hat{s}_j$ are the $\mathfrak{su}(2)$-basis in Error: Reference not found. Quaternions act on this Lie algebra through the adjoint representation, so that the image of the point $X=(x,\,y,\,z)=x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z$ under the rotation represented by the quaternion $\gamma$ is computed by the so-called spinor map $X\mapsto\gamma\,X\,\gamma^{-1}=\gamma\,X\,\gamma^\dagger$. So now this point undergoes motion: its value evolves with time according to

$$\label{EvolvingQuaternionPoint}X_0=x(0)\,\hat{s}_x+y(0)\,\hat{s}_y+z(0)\,\hat{s}_z\mapsto Y(\tau)= \gamma(\tau)\,X(\tau)\,\gamma(\tau)^{-1}$$

where $\gamma(0)=\id$ and $\gamma(\tau)\in SU(2)$ defines the orientation shift with time and $X(\tau)$ defines the point’s shape shifting; that is, motion wholly owing to the evolution of the standard shape $S_0\in\mathcal{S}$ with time. Naturally, $X(0)=Y(0) = X_0$. The point mass’s instantaneous velocity is then:

$$\label{EvolvingQuaternionPointVelocity}V = \d_\tau\,\gamma(\tau)\,X(\tau)\,\gamma(\tau)^{-1}= \gamma\left([H(\tau),\,X] + \d_\tau\,X \right)\,\gamma^\dagger$$

where $H(\tau) = \gamma^{-1}\,\dot{\gamma}\in\mathfrak{su}(2)$. The point mass’s angular momentum about the origin is $L=\d m\,[Y,\,V] =\d m\,[\gamma\,X\,\gamma^\dagger,\,V]$ (where here the standard vector cross product is replaced by the Lie bracket in the quaternion notation):

$$\label{EvolvingQuaternionPointAngularMomentum}\d L = \d m\, \gamma\left([X,\,[H(\tau),\,X]] + [X,\,\d_\tau\,X] \right)\,\gamma^\dagger$$

Now we simply sum up the total angular momentum over the whole standard shape $S_0$ and equate it to the constant nought to write our torque-free motion condition $L=\Or$; we can equally well write this condition as $\gamma^{-1}\,L\,\gamma=\Or$ since the instantaneous $\gamma(\tau)\in SU(2)$ is the same over the whole standard shape $S_0$:

$$\label{TorqueFreeCondition}\int_{X\in S_0} \rho(X)\,\left([X,\,[H(\tau),\,X]] + [X,\,\d_\tau\,X] \right)\,\d\,S_0 = \Or$$

as is$H(\tau)\in\mathfrak{su}(2)$. Here $\rho(X)$ is the mass density distribution of the standard shape. Recall that $X$ is the position vector of points in the standard shape, not the absolute position, so the co-ordinate system is attached to the orientation “frame”. $\eqref{TorqueFreeCondition}$ is what we solve to find the time derivative $H(\tau)$ of the orientation shift that must accompany the shape shift described by the shape time dervatives $\d_\tau\,X;\;X\in S_0$. $\eqref{TorqueFreeCondition}$ may look more wonted if we switch back to a vector equation in the adjoint representation of $\mathfrak{su}(2)$, when the Lie brackets become cross products. The orientation rate of change $H\in\mathfrak{su}(2)$ becomes the angular velocity vector $\boldsymbol{\omega}$. The double bracket $[X,\,[H,\,X]$ thus becomes $-\boldsymbol{x}\times (\boldsymbol{x}\times \boldsymbol{\omega}) = (\boldsymbol{x}\otimes \boldsymbol{x} – |\boldsymbol{x}|^2\,\id)\boldsymbol{\omega}$ and $[X\,\d_\tau\,X]$ becomes $\boldsymbol{x}\times \d_\tau \boldsymbol{x}$. Thus we get:

$$\label{TorqueFreeCondition_2}(\mathbf{I}_0 – \mathbf{I})\,\boldsymbol{\omega} = \int_{\boldsymbol{x}\in S_0} \rho(\boldsymbol{x})\, \boldsymbol{x}\times \d_\tau \boldsymbol{x}\,\d\,S_0$$

where:

$$\label{MomentsOfInertia}\mathbf{I} = \int_{\boldsymbol{x}\in S_0} \rho(\boldsymbol{x}) \, \boldsymbol{x}\otimes \boldsymbol{x}\,\d\,S_0\;\quad \mathbf{I}_0 = \int_{\boldsymbol{x}\in S_0} \rho(\boldsymbol{x}) \,|\boldsymbol{x}|^2\,\d\,S_0\,\id = \mathrm{tr}(\mathbf{I})\,\id$$

are the inertia tensor and scalar moment of inertia and for the standard shape.

The Kane Scher Two Segment Cat

Now we think of a slightly more realistic cat, of the kind [Kane & Scher, 1969] worked with, shown in Figure 18.2. In anatomical words, the $x,\,z$ plane ($y=0$) is the cat’s sagital plane; the $y,\,z$ plane $x=0$ is its coronal plane and the $x,\,y$ “midriff” plane $z=0$ is called the transverse plane. Here the cat comprises two rigid forward and hinder sections as in the model of Figure 18.1 but now they do not rotate about the same, shared longitudinal axis. Instead, the cat’s sinews stretch and its backbone bends so that they both sections “open apart” by each rotating in opposite senses some angle $\alpha$ about an axis in the midriff plane that is tangent to the edge of cat’s body; Figure 18.2 shows the shape or configuration when this rotation axis is parallel to the $y$-axis.

Figure 18.2: Definitions for the Kane-Scher-Montgomery Space of Shapes

However, in general the rotation axis $R$ is any midriff plane axis tangent to the edge of the cat’s body, as in Figure 18.3. The cat’s motion is roughly such that the angle $2\,\alpha$ is constant, but the tangent point $P$ in Figure 18.3 rotates at a roughly uniform angular speed $\omega$ in the midriff plane about the co-ordinate system origin $O$.

Figure 18.3: Definitions for the Kane-Scher-Montgomery Space of Shapes (continued)

Look carefully at Figure 18.4 for a simplified definition of the motion; there you can see that the motion of each half is that which you would see if you stood a barrel on the ground before you and painted a circle to mark where the edge of its bottom touches the ground. Then you would tilt the barrel away from you through an angle $\alpha$ and roll its bottom edge so that it rolls without slipping along the painted circle (i.e. along the circumference of the untilted barrel’s bottom edge), all the while keeping the angle between the barrel’s bottom surface and the ground is constant $\alpha$. Let us hereafter call this the “Hula-Hoop Move”, for obvious reasons.

Each point on the barrel’s rim follows the teardrop shapen space curve shown in orange in Figure 18.4. The point on the barrel’s edge that is at $(1,\,0,\,0)$ at time $\tau=0$ (i.e. in contact with the ground at the point $(1,0,0)$ at $\tau=0$) follows the interesting teardrop shapen path shown in Figure 18.5.

Figure 18.5: Path of a Point on the Bottom Rim of the Top “Barrel” in Figure 18.4

The parametric equation of this motion is:

$$\label{TearDropEdgeMotion}\begin{array}{lcl}x(\tau) &=& 2\, \sin^2\left(\frac{\omega\,\tau}{2}\right) + \cos(\omega\,\tau) – 4\,\sin^2\left(\frac{\alpha}{2}\right)\,\sin^2\left(\frac{\omega\,\tau}{2}\right)\,\cos^2(\omega\,\tau) + 2\, \sin^2\left(\frac{\alpha}{2}\right)\,\cos(\omega\,\tau)\,\sin^2(\omega\,\tau)\\y(\tau)&=& 4\,\sin^2\left(\frac{\omega\,\tau}{2}\right)\,\sin(\omega\,\tau)\\z(\tau)&=& -2\,\sin\alpha\,\sin^2\left(\frac{\alpha}{2}\right)\end{array}$$

In a more catlike version of Figure 18.4, the motion looks like that shown in Figure 18.6.

Of course, the real cat does not open up as in Figure 18.6, but the cat’s litheness is legendary and its highly flexible backbone and strong, lithe, elastic muscle and sinew tissue joining the two halves of the “barrel sections” allow combinations of something like the motion of Figure 18.6, i.e. where $\phi$ alone varies in Figure 18.3 and where the cat bends its backbone so that $\alpha$ alone varies in Figure 18.3. The siting of the two modelling cylinder sections as a real cat undertakes its righting reflex motions is shown in Figure 18.7, taken from [Kane & Scher, 1969] and in Figure 18.8, taken from [Stewart, 2011].

Figure 18.7: Real Images of the Cat Righting Reflex taken from [Kane & Scher, 1969]

Figure 18.8: A Truly Beautiful Sequence Taken from [Stewart 2011]

So we see that the cat’s shape is specified relative to the sagital-coronal-transverse $xyz$ Cartesian frame of Figure 18.2 and Figure 18.3; the cat’s shape is specified by the two angles $(\alpha,\, \phi)$.

How A Cat Rotates Whilst Conserving Angular Momentum

Now it should be fairly clear how to go forward: the space of shapes is parameterized by the two angles $(\alpha,\,\phi)$ and so we work out our torque free transport conditions $\eqref{TorqueFreeCondition}$ and $\eqref{TorqueFreeCondition_2}$ as a condition on the abstracted space of shapes. There are several ways one might reasonably think of the latter. One could think of $\R\times\mathbb{T}/\sim$ where $\sim$ is the equivalence relationship that $(\alpha,\,\phi) \sim (\alpha^\prime,\,\phi^\prime)$ iff $\alpha=\alpha^\prime=0$ or $\alpha=\alpha^\prime\neq0$ and $\phi=\phi^\prime$. That is, all points of the form $(0,\phi)$ are identified to be the same point. This space of shapes is a “pinched” cylinder – I like to think of a juggler’s diabolo. For a real cat, it could be construed as $\mathcal{I}\times\mathbb{T}/\sim$, where $\mathcal{I}$ is some closed, symmetric interval $\left[-\alpha_0,\,\alpha_0\right]$ where $\alpha_0$ models the limit of the cat’s litheness and ability to bend. Or, one could take the lowest angle to be $\alpha=0$, which represents the cat’s most “closed up” state, and the configuration space is a cone. Or we might think of an “aetherial” cat which can pass through itself so we have a pinched torus $\mathbb{T}\times\mathbb{T}/\sim$ where now $(\alpha,\,\phi) \sim (\alpha^\prime,\,\phi^\prime)$ iff $\alpha=\alpha^\prime\in\{k\,\pi|\;\,k\in\mathbb{Z}\}$ or $\alpha\neq0$, $\alpha-\alpha^\prime\in\{k\,\pi|\;k\in\mathbb{Z}\}$ and $\phi=\phi^\prime$. Or we might combine the ideas of “aetherial” and hard limit, so that $0\leq\alpha\leq\pi$ so that now $(\alpha,\,\phi) \sim (\alpha^\prime,\,\phi^\prime)$ iff $\alpha=\alpha^\prime\in\{0,\,\pi\}$ or $\alpha=\alpha^\prime\neq0$ and $\phi=\phi^\prime$. Now the space of shapes is the topological suspension of a circle, i.e. it is a sphere. We can choose any of the possibilities and the discussion following is the same; let’s for simplicity assume that our abstracted space of shapes is the 2-sphere $\mathbb{S}^2$. Hereafter I shall call the Cartesian frame of Figure \ref{KaneCatShapeCoordinatesFigure_1} and Figure \ref{KaneCatShapeCoordinatesFigure_2} the “cat frame”. We define the cat’s orientation as the orientation of the the “cat frame” and this is what is defined by the rotation $\gamma$ in $\eqref{FullShape}$. We may also want to think about a bigger space of shapes for a cat that can deform its halves so that the moments of inertia change. This is the generalisation of the symmetrically-deployable legs of Figure \ref{RobotCat1Figure}. This could be a simple real parameter as it was with our simple robot cat. We assume that the shift in mass is symmetrical so that it begets no motion, or at least no rotation of the cat frame. In the following we will assume that this symmetrical mass shift can be used to change the cat from a symmetrical one to an asymmetric one. So our total space of shapes is the $\mathbb{S}^2\times\mathbb{T}$; we assume that the real parameter parameterising the shifting of mass (i.e. the generalised “legs” corresponding to Figure 18.1) is a compactified one (living a circle). So our full general cat shape is $(\alpha,\,\phi,\,\theta)$, where $\theta$ parameterises the mass shift.

Now our cat’s shape shifting comprises separate, rigid rotations of two halves. Therefore a point on such a half body in the cat frame has co-ordinates $X = \zeta_f^\dagger\,X_f\,\zeta_f$ where $X_f$ are the quaternion co-ordinates in a frame that is fixed with respect to the cat’s forward half and $\zeta_f$ is the quaternion representing the rotation the forward half relative to the cat frame. So, when we differentiate $\eqref{EvolvingQuaternionPoint}$ to find $\eqref{EvolvingQuaternionPointVelocity}$, use the cross product to find the point’s angular momentum then sum up over the whole half cat we find that the forward half cat’s contribution to $\eqref{TorqueFreeCondition}$ is

$$\label{ForwardHalfContribution}\int_{X\in S_{0,f}}\rho(X_f)\,\gamma\,\zeta_f^\dagger\,\left([X_f,\,[\zeta_f\,H(\tau)\,\zeta_f^\dagger,\,X_f]]+[X_f,\,[\zeta\,H_f(\tau)\,\zeta_f^\dagger,\,X_f]]\right)\,\zeta_f\,\gamma^\dagger\,\d\,V_f$$

where $H_f = \zeta\,\d_\tau\,\zeta^\dagger\in\mathfrak{su}(2)$ represents the angular velocity of the cat’s forward half $S_{0,f}$ relative to the cat frame and, as before, $H = \gamma^{-1}\,\d_\tau\,\gamma\in\mathfrak{su}(2)$ represents the cat frame’s absolute (relative to an inertial co-ordinate frame) angular velocity. The second term in $\eqref{ForwardHalfContribution}$ is the $[X\,\d_\tau\,X]$ term in $\eqref{TorqueFreeCondition}$. Thus, adding the analogous contribution from the cat’s hinder half, we get $\eqref{TorqueFreeCondition}$ expressed in terms of rotations of body parts relative to the cat frame:

$$\label{CatTorqueFreeCondition}\begin{array}{l}\displaystyle{\zeta_f^\dagger\,\int_{X\in S_{0,f}}\rho(X_f)\,\left([X_f,\,[\zeta_f\,H(\tau)\,\zeta_f^\dagger,\,X_f]]+[X_f,\,[\zeta\,H_f(\tau)\,\zeta_f^\dagger,\,X_f]]\right)\,\d\,V \,\zeta_f+}\\\quad\displaystyle{\zeta_h^\dagger\,\int_{X\in S_{0,h}}\rho(X_h)\,\left([X_h,\,[\zeta_h\,H(\tau)\,\zeta_h^\dagger,\,X_h]]+[X_h,\,[\zeta\,H_h(\tau)\,\zeta_h^\dagger,\,X_h]]\right)\,\d\,V\,\zeta_h = \Or}\end{array}$$

where the $*_h$ terms are the analogous terms for the cat’s hinder half. We now convert this into an equation in the adjoint representation, with the Lie algebra members expressed as $3\times1$ real column vectors and the Lie brackets as cross products and then we simplify as in $\eqref{TorqueFreeCondition_2}$, so that:

$$\label{CatTorqueFreeCondition_2}\Ad(\zeta_f)^{-1}\,(\mathbf{I}_f – \mathrm{tr}(\mathbf{I}_f)\,\id)\,(\Ad(\zeta_f)\,\boldsymbol{\omega} +\boldsymbol{\omega}_f) +\Ad(\zeta_h)^{-1}\,(\mathbf{I}_h -\mathrm{tr}(\mathbf{I}_h)\,\id)\,(\Ad(\zeta_h)\,\boldsymbol{\omega} +\boldsymbol{\omega}_h)=\Or$$

The $3\times 3$ matrices $\Ad(\zeta_f)^{\pm1},\,\Ad(\zeta_h)^{\pm1}\in\mathfrak{so}(3)$ are of course the rotation matrices that describe the motions of the idealized cat’s two halves as discussed in \S\ref{TwoSegmentCat} and the $\boldsymbol{I}_*$ symbols are the scalar moment of inertia and inertia tensor matrices for the idealized cat’s forward and hinder halves. Whence, at last:

$$\label{CatTorqueFreeCondition_3}\begin{array}{lcl}\boldsymbol{\omega} &=&\left(\Ad(\zeta_f)^{-1}\,\mathbf{I}_f \Ad(\zeta_f)+\Ad(\zeta_h)^{-1}\,\mathbf{I}_h \Ad(\zeta_h)-(\mathrm{tr}(\mathbf{I}_f)+\mathrm{tr}(\mathbf{I}_h))\,\id\right)^{-1} \bullet\\&&\quad \left((\Ad(\zeta_f)^{-1}\,\mathbf{I}_f \Ad(\zeta_f) -\mathrm{tr}(\mathbf{I}_f)\,\id)\,\boldsymbol{\omega}_f+(\Ad(\zeta_h)^{-1}\,\mathbf{I}_h \Ad(\zeta_h) -\mathrm{tr}(\mathbf{I}_h)\,\id)\,\boldsymbol{\omega}_h\right)\end{array}$$

yields the angular velocity $\boldsymbol{\omega}$ of the cat frame that must accompany rotational motions of the cat’s halves at angular velocities $\boldsymbol{\omega}_f,\,\boldsymbol{\omega}_h$ relative to the cat frame so that the whole system angular momentum is conserved. $\mathbf{I}_F$ and $\mathbf{I}_h$ are the inertia tensors for the fore and hinder half cat as computed relative to the cat frame when $\alpha=\phi=0$ and their traces are the scalar moments of inertia. That is:

$$\label{InertiaMatrix}\mathbf{I}_{j,\,k} =\int_\mathcal{V}\,\rho(x,\,y,\,z)\,x_j\,x_k\,\d\, V;\;j,\,k\in\{x,\,y,\,z\}$$

where $V$ is the volume of the forward or hinder cat, as appropriate. The reason for their name “tensor” is evident in the terms in $\eqref{CatTorqueFreeCondition_3}$: with a linear co-ordinate transformation $R=(\Ad(\zeta)^{-1}$ they transform as $\mathbf{I}\mapsto R\,\mathbf{I}\,R^{-1}$; in $\eqref{CatTorqueFreeCondition_3}$ we see the transformations that take the inertia tensors from their values when $\alpha=\phi=0$ to their values in the cat frame when $\alpha$ and $\phi$ are nonzero. We define the cat frame so that the transverse (“midriff”) plane is always symmetrically placed relative to the two halves. Recall that we are free to do this as a gauge fixing condition. Then the basic defining rotation of the two halves relative to the cat frame is:

$$\label{BasicRotation}\begin{array}{lcl}\Ad(\zeta_f) &\stackrel{def}{=}& R_0(\alpha,\,\phi)= e^{\frac{\alpha}{2}\,H(\phi)}\\H(\phi) &=&-\cos\phi \,\hat{S}_y + \sin\phi\,\hat{S}_x\\\Ad(\zeta_h)&=&\Ad(\zeta_f)^{-1} = R_0(-\alpha,\,\phi)\\\end{array}$$

where $\hat{S}_x,\,\hat{S}_y$ are as defined in $\eqref{SO3Example_10}$ of \S\ref{SO3Example}. So firstly, suppose the cat undergoes a motion where $\alpha$ is constant, but that $\phi$ alone changes at an instantaneous unit rate. Then $\boldsymbol{\omega}_f$ is the quantity $R_0(\alpha,\,\phi)^T\,\partial_\phi R_0(\alpha,\,\phi)$ resolved into superposition weights for the $\hat{S}_j$ matrix basis for the Lie algebra $\mathfrak{so}(3)$ in Example \ref{SO3Example} and $\boldsymbol{\omega}_h$ is the analogous quantity calculated for $R_0(-\alpha,\,\phi)^T\,\partial_\phi R_0(-\alpha,\,\phi)$ so that:

$$\label{AngularVelocityPhiChange}\boldsymbol{\omega}_f(\alpha,\,\phi) =R_0^T\,\partial_\phi R_0=\left(\begin{array}{c}\sin\frac{\alpha}{2}\,\cos\phi\\\sin\frac{\alpha}{2}\,\sin\phi\\-2\sin\left(\frac{\alpha}{4}\right)^2\end{array}\right);\quad\boldsymbol{\omega}_h(\alpha,\,\phi)=R_0\,\partial_\phi R_0^T\,=\left(\begin{array}{c}-\sin\frac{\alpha}{2}\,\cos\phi\\-\sin\frac{\alpha}{2}\,\sin\phi\\-2\sin\left(\frac{\alpha}{4}\right)^2\end{array}\right)$$

Now suppose that the cat undergoes pure “opening” motion, i.e. $\alpha$ varies with $\phi$ constant. Then we get:

$$\label{AngularVelocityAlphaChange}\boldsymbol{\omega}_f(\alpha,\,\phi) =-\boldsymbol{\omega}_h(\alpha,\,\phi)=R_0^T\,\partial_\alpha R_0=\frac{1}{2}\left(\begin{array}{c}\sin\phi\\-\cos\phi\\0\end{array}\right)$$

and $\boldsymbol{\omega}_f+\boldsymbol{\omega}_h=\Or$. Now we assume that both the fore and hinder cat have principal axes of inertia aligned with the cat frame when $\phi=\alpha=0$, then the $\mathbf{I}$ matrices are diagonal $\mathbf{I}_f = \operatorname{diag}\left[I_{x,\,f},\,I_{y,\,f},\,I_{z,\,f}\right]$ and $\mathbf{I}_h = \operatorname{diag}\left[I_{x,\,h},\,I_{y,\,h},\,I_{z,\,h}\right]$.

For our first scenario, we assume a “symmetrical” cat, i.e. $I_{j,\,f}=I_{j,\,h}=I_j;\;j\in\{x,\,y,\,z\}$. When we substitute $\eqref{AngularVelocityPhiChange}$ into $\eqref{CatTorqueFreeCondition_3}$ we find that when $\d_\tau\phi = 1;\;\d_\tau\alpha=0$, the angular velocity $\omega_I$ of the cat frame relative to the inertial frame has only a $z$-component $\omega_z$ where:

$$\label{SymmetricCatOmega}\omega_z=-\frac{8 \sin^2\left(\frac{\alpha }{4}\right) \left(\begin{array}{l}\cos\left(\frac{\alpha }{2}\right) ((I_x-I_y) \cos( 2 \phi)+I_x+I_y-2 I_z)+\\\qquad\qquad\cos\alpha((I_x-I_y) \cos (2 \phi)+I_x+I_y-2 I_z)+2 (I_x+I_y)\end{array}\right)}{\left(\begin{array}{l}I_x \cos (\alpha +2 \phi )+(I_x-I_y) \cos (\alpha -2 \phi )+2 \cos\alpha (I_x+I_y-2 I_z)-\\\qquad\qquad2 I_x\cos(2 \phi)+6 I_x-I_y \cos (\alpha +2 \phi )+2 I_y \cos(2 \phi)+6 I_y+4 I_z\end{array}\right)}$$

Interestingly we also find that as the symmetrical cat “opens up”, i.e. $\d_\tau\phi = 0;\;\d_\tau\alpha=1$, then there is still a $z$ component to $\omega_I$:

$$\label{SymmetricCatOmega_2}\omega_z=\frac{\frac{1}{2}\,\sin (\alpha )\,\sin (2\,\phi )\,\left(I_y-I_x\right)}{I_x \,\left(\sin ^2(\alpha ) \cos ^2(\phi )-1\right)+I_y\, \left(\sin^2(\alpha ) \sin ^2(\phi )-1\right)-I_z\, \sin ^2(\alpha )}$$

\noindent Note the $\sin (2\,\phi )$ dependence in the numerator. If the cat opens about an axis at right angles to a principal axis, there is no rotation of the cat frame. There is no cat frame rotation at all when $I_y-I_x$, something that should be intuitively obvious: the cat openning begets a translation in the center of mass only. Indeed, it is a little curious that an opening motion $\d_\tau\phi = 0;\;\d_\tau\alpha=1$ should beget any rotation at any value of $\phi$. However, one can think of the situation in where the fore and hinder cat were point masses: in this case, $I_y\neq I_x$ means that the system’s center of mass does not pass through the co-ordinate origin when $\alpha=0$ and so the system’s center of mass has nonzero orbital angular momentum when $\alpha$ varies.

$\eqref{SymmetricCatOmega}$ shows that a symmetric cat can indeed roll by its righting reflex: it is sometimes said that the cat uses its tail for its righting reflex but $\eqref{SymmetricCatOmega}$ falsifies this assertion. Indeed, as stated above, Thomas Kane showed that humans could be taught the cat righting reflex and, once taught, can do it almost as well as a cat, as shown in Figure 18.9.

Figure 18.9: Thomas Kane’s research showed astronauts could be taught the cat righting reflex. Source [Madrigal, 2011]

One of my own cats has been tailless since she was run over by a car in 2004 and has no difficulty whatsoever righting herself when she falls off things, which she does often owing to her somewhat clumsy nature – typically when she falls asleep with her head drooping too far over the edge of our bed. Even though twelve years old, she manages to rouse herself and make the righting reflex within the time it takes her to fall the 40cm or so to the ground. The symmetrical cat with $I_{j,\,F}=I_{j,\,H}=I_j;\;j\in\{x,\,y,\,z\}$ is in fact the scenario that the video below is correct for.

Figure 18.10: Summary of Analysis of the Falling Symmetrical Cat

Indeed, if we put all principal moments of inertia equal, $\eqref{SymmetricCatOmega}$ reduces to $\omega_z = 1-\cos\alpha$. This is the flipover angular velocity of the cat frame when the cat makes the hula-hoop motion at one radian per second. We could deduce this formula from the simple vector diagram of Figure 18.10. If $\alpha=0.5$, $\omega_z = 0.12$, so the cat would have to do the hula-hoop about four times to flip over. At $\alpha=1$ radian, i.e. about 60 degrees, we get $\omega_z = 0.45$, so that the cat needs to make about one whole hula-hoop to flip itself over. These figures are in keeping with Figure 18.7 and Figure 18.8, where the cat does seem to bend so that the angle between the forward and hinder half of its backbone is of the order of 120 degrees.

Now we look at the case of an asymmetrical cat. For simplicity we assume $I_{x,\,f}=I_{y,\,f} = I_{t,f}$ and $I_{x,\,h}=I_{y,\,h} = I_{t,h}$ with $I_{z,\,f}=I_{a,\,f}$ and $I_{z,\,h}=I_{a,\,h}$ arbitrary. That is, we assume that the moments of inertia about the two transverse plane axes ($x$ and $y$) are equal, so that each half of the cat is defined by a “transverse” and an “axial” moment of inertia. $\eqref{CatTorqueFreeCondition_3}$ in this simplified asymmetrical case when $\phi$ varies at one radian per unit “time”, i.e. when $\boldsymbol{\omega}_f,\,\boldsymbol{\omega}_h$ are given by $\eqref{AngularVelocityPhiChange}$, is the rather messy:

$$\label{AsymmetricCatDphiOmega}\begin{array}{lcl}\boldsymbol{\omega} &=& \frac{1}{D(\alpha)}\,\left(\begin{array}{c}A_t(\alpha)\,\cos\phi \\ A_t(\alpha)\,\sin\phi\\ A_a(\alpha)\end{array}\right)\end{array}$$

where:

$$\label{AsymmetricCatDphiOmega_2}\begin{array}{lcl}D(\alpha)&=&- (I_{a,f}-I_{t,f}) (I_{a,h}-I_{t,h})\,\cos (2 \alpha )+I_{a,f} (4 I_{t,f}+I_{a,h}+3 I_{t,h})+\\&&\qquad 4 I_{t,f}^2+3 I_{t,f}I_{a,h}+9 I_{t,f} I_{t,h}+4 I_{a,h} I_{t,h}+4 I_{t,h}^2\\\\A_t(\alpha)&=&4 \sin \left(\frac{\alpha }{2}\right) \left( (I_{a,f} I_{t,h}-I_{t,f} I_{a,h})\,\cos (\alpha )+I_{a,f}I_{t,f}+I_{t,f}^2-I_{a,h} I_{t,h}-I_{t,h}^2\right)\\\\A_a(\alpha)&=&-4 \sin ^2 \times\\&&\left(- (I_{a,f} (3 I_{a,h}+I_{t,h})+I_{t,f} (I_{a,h}-5 I_{t,h}))\,\cos \left(\frac{\alpha }{2}\right)+\right.\\&&\quad2\,(I_{t,f}I_{t,h}-I_{a,f} I_{a,h})\, \cos (\alpha ) -\\&&\qquad(I_{a,f}-I_{t,f})(I_{a,h}-I_{t,h})\,\left(\cos \left(\frac{3 \alpha }{2}\right)+\cos (2 \alpha )\right) +\\&&\quad\qquad2 I_{a,f} I_{t,f}-I_{a,f} I_{a,h}+I_{a,f} I_{t,h}+2 I_{t,f}^2+I_{t,f} I_{a,h}+\\&&\qquad\qquad\left.7 I_{t,f}I_{t,h}+2 I_{a,h} I_{t,h}+2 I_{t,h}^2\right)\end{array}$$

whereas the expression when $\alpha$ varies at one radian per unit “time”, i.e. when $\boldsymbol{\omega}_f,\,\boldsymbol{\omega}_h$ are given by $\eqref{AngularVelocityAlphaChange}$ is the rather simpler:

$$\label{AsymmetricCatDalphaOmega}\boldsymbol{\omega}=\left(\begin{array}{c}\omega_t\,\sin\phi \\ -\omega_t\,\cos\phi\\ 0\end{array}\right);\quad\omega_t=\frac{I_{t\,f}+I_{a\,h}-I_{t\,h}-I_{a\,h}}{I_{t\,f}+I_{a\,h}+I_{t\,h}+I_{a\,h}}$$

and is independent of $\alpha$. So let’s think about the “small”, two dimensional space of shapes $\mathcal{S}^2$ (the 2-sphere) where the cat’s inertia matrix elements for the two halves are constant, but differ as above. Suppose the cat undergoes the following sequence of moves, beginning from the point $\alpha=\phi=0$:

1. Open out to $\alpha = \alpha_0$ with $\phi = \phi_0$;
2. Make a “hula hoop” move so that $\phi$ varies from $\phi_0$ to $\phi_0+\varphi$;
3. Now bend back to the relaxed position, so that $\alpha$ varies from $\alpha_0$ to $\alpha=0$ without varying $\phi$.

The cat has now made a complete, closed loop in the space of shapes (recall that $(\alpha,\,\phi_1)\sim (\alpha,\,\phi_2)$ if $\alpha=0$);

By the conservation of angular momentum, the cat frame rotates relative to the inertial frame as defined by $\eqref{TorqueFreeCondition}$ and $\eqref{CatTorqueFreeCondition}$. Indeed as the cat makes the first move, the rotation is $R(\phi_0,\,\omega_t\,\alpha_0)=\exp(\omega_t\,(\sin\phi_0\,\hat{S}_x -\cos\phi_0\,\hat{S}_y)\,\alpha_0)$, where $\hat{S}_x,\,\hat{S}_y$ are the $x$ and $y$-basis member for the Lie algebra $\mathfrak{so}(3)$ as given in Example \ref{SO3Example} and $\omega_t$ is as given in $\eqref{AsymmetricCatDalphaOmega}$. For the last move, the rotation is $R(\phi_0+\varphi,\,-\omega_t\,\alpha_0)=\exp(\omega_t\,(-\sin(\phi_0+\varphi)\,\hat{S}_x+\cos(\phi_0+\varphi)\,\hat{S}_y)\,\alpha_0)$. We recall that the cat frame has rotated through an angle $\omega_t\,\alpha_0$, where $\omega_t$ is as defined in $\eqref{AsymmetricCatDalphaOmega}$ in response to the cat’s openning up through angle $\alpha_0$.

We now consider the set of all such three move sequences, i.e. closed loops in the space of shapes, where $\varphi$ is any value in some nonzero length interval $(-\Delta\phi,\,+\Delta\phi)$. Each member of this set begets a rotation $\sigma(\varphi,\,\phi_0,\,\alpha_0)\in SO(3)$ of the cat frame, and as we vary $\varphi$ the set of all the $\sigma(\varphi,\,\phi_0,\,\alpha_0)$ forms a $C^\omega$ path through $SO(3)$ as $\varphi$ varies in the interval $(-\Delta\phi,\,+\Delta\phi)$. The path passes through $\id$ at $\varphi=0$ and, although it is not a straightforward path like $\sigma_0 e^{K\,\varphi}$ it is nonetheless a $C^\omega$ path as long as we keep the interval $(-\Delta\phi,\,+\Delta\phi)$ small enough. We now wish to study the Lie group of all the cat frame rotations the cat can undertake by making loops of the above kind in its space of shapes, so we characterise the tangent to $\sigma(\varphi,\,\phi_0,\,\alpha)$ by calculating its tangent to the identity. This tangent at $\varphi=0$ is $R(\phi_0,\,-\omega_t\,\alpha_0)\,T_0 \,R(\phi_0,\,\omega_t\,\alpha_0) + \partial_\varphi R(\phi_0+\varphi,\,-\omega_t\,\alpha_0) \,R(\phi_0,\,\omega_t\,\alpha_0)$ where $T_0$ is the tangent to the path through $SO(3)$ induced by the second movement above at $\varphi=0$; from $\eqref{AsymmetricCatDphiOmega}$ we see that $T_0 = (A_t(\alpha_0)\,\cos\phi_0\,\hat{S}_x+A_t(\alpha_0)\,\sin\phi_0\,\hat{S}_y+A_a(\alpha_0)\,\hat{S}_z)/D(\alpha_0)$. Therefore, the tangent to the path $\sigma(\varphi,\,\phi_0,\,\alpha_0)\in SO(3)$ at the identity (where $\varphi=0$) in the adjoint representation with $\hat{S}_x,\,\hat{S}_y,\,\hat{S}_z$ basis is:

$$\label{CatPathTangent_2}\begin{array}{lcl}\Omega(\phi_0,\,\alpha_0) &=&\Ad(R(\phi_0,\,-\omega_t\,\alpha_0))\,\left(\begin{array}{c}\frac{A_t(\alpha_0)}{D(\alpha_0)}\,\cos\phi_0-\frac{\omega_t}{2}\,\sin\phi_0\\\frac{A_t(\alpha_0)}{D(\alpha_0)}\,\sin\phi_0+\frac{\omega_t}{2}\,\cos\phi_0\\\frac{A_a(\alpha_0)}{D(\alpha_0)}\end{array}\right)\\&&\\ &=& \exp(\omega_t\,(-\sin\phi_0\,\hat{S}_x+\cos\phi_0\,\hat{S}_y)\,\alpha_0)\,\left(\begin{array}{c}\frac{A_t(\alpha_0)}{D(\alpha_0)}\,\cos\phi_0-\frac{\omega_t}{2}\,\sin\phi_0\\\frac{A_t(\alpha_0)}{D(\alpha_0)}\,\sin\phi_0+\frac{\omega_t}{2}\,\cos\phi_0\\\frac{A_a(\alpha_0)}{D(\alpha_0)}\end{array}\right)\end{array}$$

where $A_t,\,A_a,\,D$ are given by $\eqref{AsymmetricCatDphiOmega}$ and $\omega_t$ by $\eqref{AsymmetricCatDalphaOmega}$. So now we have shown that, by making closed loops of different sizes in the shape space with given set values of $\alpha_0$ and $\phi_0$ of the angles $\alpha,\,\phi$, the cat can impart any rotation on a $C^\omega$ path through $SO(3)$ with tangent at the identity given by $\eqref{CatPathTangent_2}$. Let this path be $\zeta(\varphi,\,\phi_0\,\,\alpha_0)$, it passes through the identity at $\varphi=0$ and with tangent $\Omega(\phi_0,\,\alpha_0)$ there. It is now a simple matter to show that, by choosing three different, nonzero values $\alpha_1,\,\alpha_2,\,\alpha_3$ of $\alpha_0$ (we need nonzero $\alpha_j$ because $\Omega_I(\phi,\,0)=0$), with possibly different values $\phi_1,\,\phi_2,\,\phi_3$ of $\phi_0$ the vectors $\Omega(\phi_1,\,\alpha_1),\,\Omega(\phi_2,\,\alpha_2),\,\Omega(\phi_3,\,\alpha_3)$ can always be made linearly independent. Therefore:

$$\label{CatCanononicalCoordinates}\mu(\varphi_1,\,\varphi_2,\,\varphi_3) = \zeta(\varphi_1,\,\phi_1\,\,\alpha_1)\, \zeta(\varphi_2,\,\phi_2,\,\alpha_2)\, \zeta(\varphi_3,\,\phi_3,\,\alpha_3)$$

defines canonical co-ordinates of the second kind for some nucleus $\mathcal{K}$ in $SO(3)$, i.e. there is a combination of values $\varphi_1,\,\varphi_2,\,\varphi_3$ giving any member of $SO(3)$ in some neighbourhood $\mathcal{K}$ (in the group topology) of the identity. Thus we have shown that, by making shape space loops as defined by the three step sequence above, the cat can impart any orientation it likes in $SO(3)$ as long as:

$$\label{CatCanononicalCoordinatesDeterminant}\det\left(\,\left(\Omega(\phi_1,\,\alpha_1),\,\Omega(\phi_1,\,\alpha_2),\,\Omega(\phi_3,\,\alpha_3)\right)\,\right)\neq 0$$

the determinant can be made nonzero as long as we have at least one of $I_{t,\,f}\neq I_{t,\,h}$ or $I_{a,\,f}\neq I_{a,\,h}$. For example, with $I_{t,\,f}=1$, $I_{t,\,h} = \frac{3}{2}$, $I_{a,\,f} = \frac{1}{2}$ and $I_{a,\,h}=\frac{3}{4}$ and at $\alpha_1 = \alpha_2 = \alpha_3 = 1$ we plot the three components of $\Omega(\phi_0,\,1)$ as a function of $\phi_0$ in Figure \ref{BasisVectorsFigure}. We can get a maximally linearly independent set of vectors (lowest condition number, or ratio of highest to lowest singular value) when we choose $\phi_1\approx0.526564$, $\phi_2\approx2.622$ and $\phi_3\approx4.714$ when our three Lie algebra members corresponding to the tangents $\partial_{\varphi_j}\,\mu(\varphi_1,\,\varphi_2,\,\varphi_3),\,j\in\{1,\,2,\,3\}$ in $\eqref{CatCanononicalCoordinates}$ are approximately:

$$\label{CatLieAlgebraBasis}\begin{array}{lcl}\Omega(0.526564,\,1)&\approx&-0.116\,\hat{S}_y-0.198\,\hat{S}_z\\\Omega(2.622,\,1)&\approx& 0.100\,\hat{S}_x+0.058\,\hat{S}_y-0.198\,\hat{S}_z\\\Omega(4.714,\,1)&\approx& -0.100\,\hat{S}_x+0.058\,\hat{S}_y-0.198\,\hat{S}_z\end{array}$$

Figure 18.11: Cartesian Components of $\Omega(\phi_1,\,1)$ for $I_{t,\,F}=1$, $I_{t,\,H} = /frac{3}{2}$, $I_{a,\,F} = \frac{1}{2}$ and $I_{a,\,H}=\frac{3}{4}$; x blue, y green, z orange

Now we consider our full space of shapes $\mathbb{S}^2\times\mathbb{T}$ with points $(\alpha,\,\phi,\,\theta)$ and assume that our cat has a mass shifting parameter $\theta$ can transform the cat from being a symmetrical one with $I_{j,\,F}=I_{j,\,H}=I_j;\;j\in\{x,\,y,\,z\}$ to an asymmetrical one. So we might, for example have $I_{t,H} = I_{t,F} + \Delta_t\,\sin\theta,\,I_{a,H} = I_{a,F} + \Delta_a\,\sin\theta$, so that the mass shifting parameter $\theta$ transforms the cat from a symmetrical to an asymmetrical one. We assume that the mass shift has a symmetry analogous to the deployment of the legs in Figure 18.1 so that this shift, when $\phi=0$, begets no cat frame rotation relative to an inertial frame. This ability gives the cat abilty to re-orient itself more quicky, because the cat can then make the following sequence of moves:

1. With the mass shift parameter $\theta$ set to $\pi/2$, i.e. for maximum asymmetry, open out to $\alpha = \alpha_1$ with $\phi = \phi_0$;
2. Shift mass so that $\theta=0$ and the cat becomes symmetrical;
3. Now bend back to the relaxed position, so that $\alpha$ varies from $\alpha_1$ to $\alpha=0$ without varying $\phi$. Note that at this stage, the cat frame does not rotate owing to the symmetry;
4. Shift the mass back to $\theta$ set to $\pi/2$.

so we have made a closed loop in shape space but the only transformation imparted to the cat is the rotation $\exp(\omega_t\,(-\sin\phi_0\,\hat{S}_x + \cos\phi_0\,\hat{S}_y)\,\alpha_1)$. By thinking of the whole family of such moves, we can realise any rotation on the path $\zeta_\phi(\alpha)=\exp(\omega_t\,(\sin\Delta\phi\,\hat{S}_x + \cos\Delta\phi\,\hat{S}_y)\,\alpha$ for all $\alpha\in\R$ even though $\alpha$ may be hard limited to $\alpha>0$. This is because $\zeta_{\phi+\pi}(\alpha) =\zeta_\phi(-\alpha) = \zeta_\phi(\alpha)^{-1}$. We can also realise $\zeta_{\phi}(-\alpha)$ by the above four steps but with the maximum asymmetry at step 3 and with a symmetrical cat at the beginning of step 1 / end of step 4. One can also do a hula-hoop motion with the cat in the asymmetrical configuration; the above sequence of four moves with $\phi_0 = \pi/2$ gives a pure rotation about the $x$-axis, with $\phi_0 = 0$ a pure rotation about the $y$-axis and with the hula-hoop motion with the cat in symmetrical configuration, the cat can achieve a pure rotation about the $z$-axis. So with these three separate move sequences, the cat can pitch, yaw and roll. It is easy to see how to rotate to any orientation with sequences of these basic moves.

We now consider the cat configuration space in the light of the above. We have the space of shapes $\mathcal{S} = \R\times \mathbb{T}^1/\sim$, $\mathcal{S} = \mathbb{S}^2$, $\mathcal{S} = \mathbb{S}^2\times \mathbb{T}^1$ and at each point in $\mathcal{S}$ there is the set of all possible orientations of the cat. In the light of the above, there is a closed loop from any point in $\mathcal{S}$ such that any rotation $\gamma\in SO(3)$ is imparted to the cat frame by going around that loop. In other words, the cat configuration space can be thought of as a fibre bundle with base space $\mathcal{B} = \mathcal{S}$, a fibre $\mathcal{F}=SO(3)$ and which is locally homoemorphic to $\mathcal{S}\times SO(3)$ but which is globally a nontrivial fibre (indeed there is no nontrivial fibre bundle with base space $\mathbb{S}^2$ and fibre of dimension greater than 1, by the hairy ball theorem). The structure group of the bundle is $SO(3)$.

The same remarks as the above apply for the symmetrical cat, i.e. one with $I_{j,\,F}=I_{j,\,H}=I_j;\;j\in\{x,\,y,\,z\}$ so that its fore- and hinder halves have the same inertia matrices. However, we replace $SO(3)$ by $SO(2)\cong U(1)$ in the above, as loops through the space of shapes can only beget rotation about the $z$-axis. So now our cat configuration space is the fibre bundle with base space $\mathcal{S}$ and fibre $U(1)$ i.e. locally homeomorphic to $\mathcal{S}\times U(1)$. The thoughts of this and the last paragraph (or a more sophisticated version of them) are the grounding for the gauge theory of the falling cat studied in [Montgomery, 1993].

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