Rigid Body Motion with Fixed Point

Question

Suppose that $ T : \mathbb{R} \longmapsto \mathbb SO_{3} $ is a differentiable function. This means that $ T(t) $ is differentiable component wise. Does there exist a function $ \Omega : \mathbb{R} \longmapsto \mathbb{R}^{3} $ such that $ T'(t) \vec{x} = \Omega(t) \times T \vec{x} $ for all $ \vec{x} \in \mathbb{R} ^{3} $ and for all $ t \in \mathbb{R} $.

If so, is $ \Omega $ unique? Can $ \Omega $ be chosen to be differentiable?

Edit 1: changed $ \vec{x} $ to $ T \vec{x} $. Thanks to the example below, I think the correct formulation is what I have written above.

Edit 2: Some progress:

We know that $ T(t) T^{T}(t) = I $, and thus, $ T'(t) T(t) ^{T} + T(t) T'(t) ^{T} = 0 $. This implies that $ T'(t) T(t ) ^ { T } $ is skew-symmetric, which means that $ \det ( T'(t) T ^{T}(t) ) = 0 $. This implies that $ T'(t) $ is singular, which means that $ \ker ( T'(t) ) $ is non-trivial.

Also, we have $ T(t) \vec{x} \cdot T(t) \vec{x} = \vec{x} \cdot \vec{x} \implies 2 T (t) \vec{x} \cdot T ' (t) \vec{x} = 0 $.

Answer 1

I think that in general we don't have such a function $\Omega$. Consider for example $$ T(t) = \begin{pmatrix}\cos t & -\sin t & 0 \\ \sin t & \cos t & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$

Then $$ T'(t) = \begin{pmatrix}-\sin t & -\cos t & 0 \\ \cos t & -\sin t & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$

Now pick $$ x = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}.$$

Then we have $$ T'(t)x = \begin{pmatrix}-\sin t \\ \cos t \\ 0 \end{pmatrix}$$

and in particular $$ T'(\pi/4)x = \begin{pmatrix}-\frac1{\sqrt2} \\ \frac1{\sqrt2} \\ 0 \end{pmatrix}.$$

This is not orthogonal to $x$. But $\Omega(\pi/4) \times x$ is orthogonal to $x$ for any $\Omega(\pi/4)$. So $\Omega$ can't exist.

Answer 2

Yes, this $\Omega$ exists, is unique, and is called the Darboux vector.

Rotations form a Lie group; in particular, you can write $T(t_0+\delta t) = R(\delta_t, t_0) T(t_0)$ where $R$ is a second rotation that you can think of as a small correction to $T(t_0)$, with $\lim_{\delta t\to 0} R = I$. The precise formula for $R$ is a bit involved, due to the non-commuting nature of a rotations, but it can be shown that the infinitesimal rotation $\frac{\partial R}{\partial \delta t}\vert_{\delta t\to 0}$ is a skew-symmetric matrix (which you can write as $\Omega(t_0) \times$ for some vector $\Omega(t_0)$).

If you need formulas for computation, here is a very helpful paper I dug up when I was recently preparing for a lecture on rigid body dynamics: https://smartech.gatech.edu/bitstream/handle/1853/52144/DerivativeOfRotation.pdf?sequence=1