The angular velocity vector of a rigid body is defined as $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{|\vec{r}|^2}$. But I'd like to show that that's equivalent to how most people intuitively think of angular velocity.
Euler's theorem of rotations states that any rigid body motion with one point fixed is equivalent to a rotation about some axis passing through the fixed-point. So let's consider a rigid body undergoing some motion with one point fixed, and for any times $t_1$ and $t_2$ let $\vec{\theta}(t_1,t_2)$ denote the "rotation vector" of the rotation that's equivalent to the rigid body's motion between time $t_1$ and time $t_2$. For those who don't know, the rotation vector of a rotation is a vector whose magnitude is equal to the angle of the rotation and which points along the axis of the rotation; see this Wikipedia article.
Now my question is, how do we prove that the limit of $\frac{\vec{\theta}(t_1,t_2)}{t_2-t_1}$ as $t_2$ goes to $t_1$ exists, and that it's equal to the angular velocity vector?
This would all be much simpler if rotations were commutative since then the angular velocity would just equal the derivative of $\vec{\theta}(t_0,t)$ with respect to time. But since rotations are non-commutative, $\vec{\theta}(t_1,t_2)$ does not equal $\vec{\theta}(t_0,t_2)-\vec{\theta}(t_0,t_1)$ and thus the relation between angular velocity and the time derivative of $\vec{\theta}(t_0,t)$ is considerably more complicated; see this journal paper for details.
Note: This is a follow-up to my question here.
EDIT: Note that what this journal paper calls $\vec{\alpha}(t)$ would in my notation be written as $\vec{\theta}(t_0,t)$. The paper discusses the fact that the angular velocity vector $\vec{\omega}(t)$ is not equal to the time derivative of $\vec{\alpha}(t)$. This means that the limit of $\frac{\vec{\theta}(t_0,t_2)-\vec{\theta}(t_0,t_1)}{t_2-t_1}$ does not equal $\vec{\omega}(t_1)$. But my question is about proving a slightly different statement, which is that the limit of $\frac{\vec{\theta}(t_1,t_2)}{t_2-t_1}$ as $t_2$ goes to $t_1$ DOES equal $\vec{\omega}(t_1)$. Note that the expressions $\frac{\vec{\theta}(t_0,t_2)-\vec{\theta}(t_0,t_1)}{t_2-t_1}$ and $\frac{\vec{\theta}(t_1,t_2)}{t_2-t_1}$ are not equal, because $\vec{\theta}(t_0,t_1)+\vec{\theta}(t_1,t_2)$ does not equal $\vec{\theta}(t_0,t_2)$ due to the non-commutativity of rotations. So none of what I'm saying contradicts or seeks to disprove the journal paper.
Best Answer
If you take $t_0 = t_1$ in the EDIT part of your question, with $\vec{\theta}(t_0,t_1) = \vec{\theta}(t_1,t_1) = 0$, you are in the special case $\omega = \dot{\alpha}$ from Asher Peres's paper you have mentioned, which then proves your statement according to your observation from the EDIT part, because in that case, you have $$\frac{\vec{\theta}(t_0,t_2)-\vec{\theta}(t_0,t_1)}{t_2-t_1} = \frac{\vec{\theta}(t_1,t_2)}{t_2-t_1},$$ whose limit when $t_2 \to t_1$ is equal to $\dot{\alpha}(t_1) = \omega (t_1)$.
Indeed, according to Asher Peres, we have: $$ w = \dot{\alpha} + \frac{1 - \cos \alpha}{\alpha^2} (\alpha \times \dot{\alpha}) + \frac{\alpha - \sin \alpha}{\alpha^3} (\alpha \times (\alpha \times \dot{\alpha})),$$ which, for $\alpha(t) = \vec{\theta}(t_1,t)$ and $t = t_1$, and using $\alpha(t_1) = \vec{\theta}(t_1,t_1) = 0$ (see above), reduces to $$ w(t_1) = \dot{\alpha}(t_1) + 0.$$ Note that luckily $1 - \cos \alpha = \mathrm{O}(\alpha^2)$ and $\alpha - \sin \alpha = \mathrm{O}(\alpha^3)$, hence there is not problem when passing to the limit when $\alpha \to 0.$