# Dynamics of a Rigid Body

This post is about the formulation of the dynamics of a rigid body, under rotations, as dynamics on a Lie group. I'm writing this to help myself understand some more general stuff about Lie theory as well as formulation of the theory of dislocations/elasticity.

## Dynamical quantities

We will assume that we are looking at a dynamics of a rigid body.

Mathematically, imagine that our object is given by some measure $\mu$ on $\mathbb{R}^d$ representing its mass distribution (we consider the general case with $d$ spatial dimensions; notice that the case $d = 1$ is trivial), and we assume that the center of mass of $\mu$ is at the origin, that is $\int_{\mathbb{R}^d} x~ d\mu = 0$ where $x$ is the ($\mathbb{R}^d$-valued) position function.

That our body is rigid implies that its motion can be described by a rigid motion of $\mathbb{R}^d$: at any given time $t$, there is an orientation-preserving isometric mapping $\gamma(t): \mathbb{R}^d \to \mathbb{R}^d$ such that the mass distribution of our body at time $t$ is given by the pushforward measure $\gamma(t)_\star \mu$.

For simplicity let us make the further assumption that the center of mass is time-independent, and always located at the origin. This implies at any $t$, there exists a rotation matrix $A(t) \in SO(d)$ such that $\gamma(t)(x) = A(t)\cdot x$. In this way we can think of the dynamics as entirely described by a curve $A(t)$ on the Lie group $SO(d)$. To think about the dynamics, we need to think about how $A(t)$ changes.

Consider the position $A(t+h)$, this can be obtained from further rotating $A(t)$ by $B(h) = A(t + h) A(t)^{-1}$. So we can write the derivative $A'(t) = \lim_{h \to 0} \frac{1}{h} \left( B(h) - e \right) A(t) = b(t) A(t)$ here, $b(t)$ is an element of the tangent space of the identity $e$ in $SO(d)$, in other words, an element of of the Lie algebra $so(d)$.

(Actually, a better way to think about this is to think about coordinating the tangent bundle of $SO(d)$ using right invariant vector fields, which allows us then to trivialize $\mathsf{T}SO(d) \cong SO(d) \times so(d)$. This point of view is convenient because it automatically incorporate the rotational invariance of the system (which we will assume). To wit, if we take a global change of coordinate system $x \mapsto A_0 x$ with $A_0 \in SO(d)$, then the curve becomes $\tilde{A}(t) = A(t) A_0$. But the corresponding $b(t)$ is unchanged.)

In typical Newtonian fashion, then the dynamics is dictated by how $b(t) := A'(t) A(t)^{-1}$ changes in time, depending on both $b(t)$ and $A(t)$.

## Kinematic quantities

Fix a point $y_0$ in our rigid body (in its reference configuration). Its position at time $t$ is $A(t) y_0$. Its velocity at time $t$ is $A'(t) y_0$. Therefore the total kinetic energy is given by $$K = \int_{\mathbb{R}^d} |A'(t) x|^2 ~d\mu$$ Similarly, the total angular momentum is given by $$\Omega = \int_{\mathbb{R}^d} (A(t) x) \wedge (A'(t) x) ~d\mu$$ Note that $\Omega$ is a bivector in $\mathbb{R}^d$, and hence is $d(d-1)/2$ dimensional. Our dynamics takes place on the tangent bundle $\mathsf{T}SO(d)$, and so has $d(d-1)$ degrees of freedom. The presence of the conserved total angular momentum strongly indicates that we may have enough conserved quantities to make the dynamics integrable.

## Equation of motion

First let us rewrite the expression of the kinetic energy. Let $$I(t) = \int_{\mathbb{R}^d} [A(t) x]\otimes [A(t) x] ~d\mu$$ we see that $I(t)$ is a symmetric matrix. We see that we can write the kinetic energy as a quadratic form on $b$ $$K = \mathrm{tr}( b(t)\cdot I(t) \cdot b^T(t) )$$ where $\cdot$ denote matrix multiplication.

Now, as long as $\mu$ is non-trivial, we have that the quadratic form defining $K$ is positive definite. In fact, since we have a trivialization of $\mathsf{T}SO(d)$ using the right-invariant vector fields, we see that the quadratic form $K$ can be described as that which is associated to a Riemannian metric on $SO(d)$. More precisely, we can extend the definition for the matrix $I$ to become $I: SO(d) \mapsto \mathbb{R}^{d\times d}$ by $$I(A) = \int_{\mathbb{R}^d} (Ax) \otimes (Ax) ~ d\mu$$. And using the local coordinates defined by the trivialization, we have that for $v \in \mathsf{T}_A SO(d)$ we can construct the associated $b$ and define $$\label{eq:dynmetric} \langle v,v\rangle = \mathrm{tr} (b I b^T).$$

Finally, as our mechanical system has only kinetic and no potential energy, using a Lagrangian formulation we see precisely that the motion satisfied by the system is geodesic motion on $SO(d)$ under the metric \eqref{eq:dynmetric}.

As an example, suppose $\mu$ is rotationally symmetric. Then for any rotation $\gamma$, $\gamma_\star \mu = \mu$, and since the integral for $A$ can be written as $I(A) = \int x\otimes x ~d A_\star\mu$ we see that in this case the induced metric is the invariant metric on $SO(d)$, and hence geodesic motions correspond to the one-parameter group of motions generated by elements of the Lie algebra.

In general, however, the motion is more complicated.

The $I$ we have defined is precisely the moment of inertia for our body. We can also write the angular momenta in terms of $I$: $$\Omega(t) = I(t)b(t)^T - b(t) I(t).$$

Now, we see that for a fixed $t$, the mapping $b \mapsto \Omega$ given by $\Omega = I(t) b^T - b I(t)$ is a linear mapping from $so(d)$ to the space of bi-vectors in $\mathbb{R}^d$, both of these being $d(d-1)/2$ dimensional spaces.

For $t$ fixed, we can choose $O\in SO(d)$ to diagonalize $I(t)$ with $I(t) = O^T D O$. Note that since $O$ is invertible, conjugation by $O$ ($b \mapsto Ob O^T$ etc) is linear surjection of $so(d)$ (space of bi-vectors) to itself, treating the elements as matrices. Hence as long as the map $b \mapsto Db^T - bD$ is invertible, so is the map $b\mapsto \Omega$.

Since $b$ is an element of $so(d)$, its matrix representation is an antisymmetric matrix. Therefore we see that if $D = \mathrm{diag}(\lambda_1, \lambda_2, \ldots, \lambda_d)$, the element $(Db^T - bD)_{ij} = \lambda_i b_{ji} - b_{ij} \lambda_j = - b_{ij} (\lambda_i + \lambda_j).$ Since $I(t)$ is positive definite, we have therefore the mapping above is surjective, and hence invertible.

This means that in fact we can parametrize $\mathsf{T}SO(d)$ by $SO(d) \times \Lambda^2 \mathbb{R}^d$ by doing the fibre-wise transformation from $\mathsf{T}_A SO(d) \cong so(d) \cong \Lambda^2 \mathbb{R}^d$ using the above argument. The "constant $\Omega$" curves are then the lifts of trajectories $A(t)$ (thereby showing the integrability of the dynamics).

## Three dimensions

In three dimensions the situation simplifies. (Also in two dimensions, but there $SO(2)$ is one dimensional and rotations are trivial to study.) First, in three dimensions we can replace two-forms/bivectors by one-forms/vectors by taking the Hodge dual. So instead of $b$ and $\Omega$ we can consider $\beta$ and $W$. In indices we compute $\epsilon_{ijk} I^{jl} \epsilon_{klm} \beta^m \propto I_{mi} \beta^m - \beta^i \mathrm{tr} I$ and hence the conserved angular momentum $W$ can be expressed in terms of $( I - \mathrm{tr}(I) \delta) \beta = A( I(e) - \mathrm{tr}(I(e)) \delta) A^T \beta.$ Similarly, the conserved energy can be written in terms of $\beta$ $K = \beta^l \epsilon_{iml} I^{mj} \epsilon_{ijk} \beta^k \propto \mathrm{tr} I |\beta|^2 - \beta^T I \beta = \beta^T (\mathrm{tr} I \delta - I ) \beta.$ If we denote by $J$ the matrix $$J := \mathrm{tr}(I(e)) \delta - I(e)$$ observe that since $I$ is positive definite, so is $J$. Then the conserved angular momentum can be expressed in terms of $$W = - A(t) J A^T(t) \beta(t)$$ and the conserved energy $$K = \beta^T(t) A(t) J A^T(t) \beta(t)$$ and therefore we can write $$K = ( J^{-1} A^T(t) W )^T (A^T(t) W)$$

Let us think about the meaning of the vector $A^T W$: this represent the conserved angular momentum when viewed from the body's (rotating) frame of reference. Therefore we have that when $A^T W$ is stationary, this represents a body whose angular momentum and whose angular velocity are entirely aligned.

Now, conservation of angular momentum means $W$ is a constant vector. $A^T$ being orthogonal means that $A^T W$ lives on the sphere of radius $|W|$. Conservation of total energy means that for any fixed $W$, the dynamics is such that $A^T W$ has to lie on the level sets of $K$. The geometry of the level sets can be described in terms of the eigenvalues of $J$.

### Equal eigenvalues

The simplest case has $J^{-1}$ having equal eigenvalues. In this case $K$ is constant over the entire sphere, and the conservation of energy gives no additional constraint. Going back to $W$, however, we see that since $J$ is now a multiple of the identity, we must have $W$ is proportional to $\beta(t)$, and therefore we conclude that $\beta$ is constant and we are in the situation where the body is just rotating about a fixed axis.

### Two of the same

Next, consider the case where $J^{-1}$ has an eigenvalue of multiplicity 2, which we label as $\lambda_2$, and an eigenvalue of multiplicity 1, which we label as $\lambda_1$. Regardless of which is bigger, we have that the level sets of $K$ can be described as circles orthogonal to the eigenvector of $\lambda_1$, degenerating at the two poles.

In this setting, the two poles represent fixed points of the dynamics, where $A^T W$ remain constant. One can check also that along the equator the dynamics are also stationary. At every other point the dynamics travels along the circle.

### Entirely distinct eigenvalues

In the final case where $J^{-1}$ has three distinct eigenvalues $\lambda_1 < \lambda_2 < \lambda_3$ with eigenvectors $e_1, e_2, e_3$, we see that $K$ has 6 critical points. The four corresponding to $e_1$ and $e_3$ directions are global min/max of $K$ on the sphere, and therefore in their neighborhoods the level sets are circles. The two corresponding to the $e_2$ direction are saddle points. In their neighborhoods the level sets are hyperbolas (with a degenerating pair).

This tells us that the dynamics are such that

1. The three eigen directions are fixed points of the dynamics, corresponding to rotation around the three principal axes of the body.
2. The two eigen directions with minimum/maximum moment of inertia are stable; small perturbations lead to precessing rotations.
3. The eigen direction with the intermediate moment of inertia is unstable. Generic perturbations lead to periodic motion that approaches alternatively the positive and negative eigen directions. Additionally, there are heteroclinic orbits (corresponding to the degenerate hyperbolas) connecting one such eigen direction with its antipode.
##### Willie WY Wong
###### Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.