# Definition and Existence of Angular Velocity

$\newcommand{\R}{\mathbf R}$ Let $f:\R\to SO(3)$ be a smooth map whose derivative never vanishes. So $f(t)$ can be thought of as the snapshot at time $t$ of a rigid pivoted at the origin and rotating smoothly about the pivot.

In my high school physics books, the angular velocity at time $t$ of a rigid body pivoted at a point was defined as a vector $\vec\omega(t)\in \R^3$ such that if $\vec p(t)$ is the position vector of a particle $P$ on the rigid body at time $t$, then the velocity of $P$ at $t$ is $\vec\omega(t)\times \vec p(t)$.

Under this definition, I want to know if we can prove that the angular velocity of a smoothly rotating rigid body always exists.

If $f:\mathbf R\to SO(3)$ is a smoothly rotating rigid body with $f(0)=Id$, and $v\in \R^3$ is a point, then define $P_v(t)=f(t)v$. (Here $f(t)$ is a linear map which is begin fed $v$).

Basically $P_v(t)$ is the position of the paritcle of the rigid body at time $t$ which was at $v$ at time $0$. So the velocity of this particle at time $t$ is $\dot P_v(t)$.

Can somebody see as to why at any given time $t$ we have a vector $\omega(t)\in \R^3$ such that $\omega(t)\times P_v(t)=\dot P_v(t)$ for all $v\in \R^3$?

Further, does $\omega(t)$ vary smoothly with $t$?

Thank you.

• Sketch of proof: (i) $\dot P_v(t) = f'(t)v$. (ii) $f(t)^Tf(t)=I\implies f'(t)=-f'(t)^T$. (iii) $f'(t)=\begin{bmatrix}0&c&-b\\-c&0&a\\b&-a&0\end{bmatrix}$. (iv) $f'(t)v=\begin{bmatrix}a\\b\\c\end{bmatrix}\times v$.
– user856
Jun 16 '17 at 14:40
• @Travis Since one single $\omega$ should work for all $v$, the $\omega$ should be unique. This is because all the points lying on the one dimensional subspace spanned by $\omega$ have $0$ velocity, and no other point has $0$ velocity. Jun 16 '17 at 15:44
• @Rahul The conclusion in (ii) only holds when $f(t) = Id$. In general we have that $f(t)^{\top} f'(t)$ is skew-symmetric. Jun 16 '17 at 15:57
• @caffeinemachine Yes, you're right, of course, I didn't notice the stipulation for general $v$. Jun 16 '17 at 15:58

By Euler's rotation theorem, any value of $f$ can be represented by a single rotation about a fixed axis,

$$f=e^A,$$

where $A$ is an antisymmetric matrix. In your application, $A$ is a continuous function of time with $A(0)=0$. Although the matrix $A$ described by Euler's theorem may not be unique, it is unique in this application because $A$ is continuous and connected to the zero matrix. The position of a point on the body is

$$P_v = fv,$$

and differentiation gives

$$\frac{dP_v}{dt} = e^A\frac{dA}{dt}v.$$

To prove what you want to prove, it's sufficient to prove it at $t=0$, where this simplifies to $$\left.\frac{dP_v}{dt}\right|_0 = \left.\frac{dA}{dt}\right|_0v.$$ Let's drop the $|_0$ symbols for convenience. The matrix $dA/dt$ is antisymmetric, and its components can be identified with the components of the angular velocity vector: $$A =\begin{pmatrix} 0 & \omega_z & -\omega_y \\ -\omega_z & 0 & \omega_x \\ \omega_y & -\omega_x & 0 \end{pmatrix}.$$ This value of the vector $\omega$ was calculated in a way that didn't depend on the choice of $v$, so that completes the proof.

Stepping back and looking at the big picture here, there are really two things that make all this work. (1) In 3 dimensions, we have an isomorphism between antisymmetric matrices and vectors. This isomorphism only works in 3 dimensions. (2) Rotations are not commutative, but infinitesimal rotations are, so that there is no $\Delta \theta$ vector, but there is a $d\theta$ vector.

Further, does $\omega(t)$ vary smoothly with $t$?

You say that $f$ is a smooth map, so yes, because all we did in order to find $\omega$ was differentiate and pick components out of a matrix. Actually all you really need is that $f$ be twice differentiable. Otherwise, for example, I can simply make up an example of rotation in the x-y plane by an angle $\theta(t)$, where $\theta''(t)$ doesn't exist at a certain time.

• Thank you for the most enlightening answer. To see why there is a smooth map $A$ taking $\mathbf R$ to the space of linear isometries of $\mathbf R^3$ such that $f=e^A$, I have the following rather non-elementary argument. We equip $SO(3)$ by a bi-invariant metric (this can be done since $SO(3)$ is compact). Then we know that that the exponential map of $SO(3)$ when $SO(3)$ is thought of as a Lie group is same as the exponential map at the identity of $SO(3)$ when $SO(3)$ is thought as a Riemannian manifold. Now the Cartan-Hadamard theorem implies that $\exp:T_I(SO(3))\to SO(3)$ is a smooth... Jun 17 '17 at 10:44
• ... covering map. Thus $f:\mathbf R\to SO(3)$ admits a smooth lifting to yield the desired smooth map $A$, and we are done. If you have a more elementary argument then can you please post it? Thanks. Jun 17 '17 at 10:45
• My reasoning is inaccurate. For me to apply the Cartan-Hadamard theorem I need to show that the sectional curvature of $SO(3)$ is always non-positive. The reverse is true. All sectional curvatures of $SO(3)$ are non-negative. In fact, the universal cover of $SO(3)$ is $S^3$ and thus what I said is clearly wrong. So can you please give a justification of the existence and smoothness of $A$? Thanks. Jun 17 '17 at 15:03
• At the very least, the exponential map is a local diffeomorphism near the origin of $T_I(SO(3))$, and thus a lift $A$ is available in a neighborhood of $(-\epsilon, \epsilon)$ of $0$ in $\mathbf R$. So we have a hanging question: Does there necessarily exist a global lift of $f$, and if not then are there suitable assumptions, like regularity, on $f$ which make a global lift exist? Jun 17 '17 at 15:23
• In fact a global lift shouldn't always be possible, as is shown by the bump function version of the counterexample given in the post here: math.stackexchange.com/questions/2313726/… So the relevant question now is that whether or not a global lift of $f:\R\to SO(3)$ to $T_I(SO(3))$ exist provided we assume that the derivative of $f$ never vanishes. Jun 17 '17 at 15:26