Rigid body mechanics in Arnold's book

I'm having some trouble understanding what is going on in Mathematical methods of classical mechanics (Arnold, 2nd edition) in the rigid body mechanics section (chapter 6). I don't know any other book that deals with rigid body mechanics like this. After introducing two frames of reference: the "stationary" one $$k$$ and the "moving" one $$K$$, he defines an operator $$$$B:K\rightarrow k$$$$ That turns vectors from one space into vectors of the other.

Then, after defining the angular velocity $$\boldsymbol{\omega}$$ (vector in $$k$$), he defines the angular velocity of the body as: $$$$\boldsymbol{\Omega}=B^{-1}\boldsymbol{\omega}$$$$ (vector in $$K$$). This is the first time I see such distinction. Why?

Anyways, I think it's fine up to this. The problems start to rise when he deals with rigid bodies and chooses $$K$$ as the "system rotating together with the rigid body", with respect to which the body is stationary (see Inertia operator section, page 136-137).

If it is so, then how does it have an angular momentum $$\boldsymbol{M}$$ in $$K$$ (its points should have zero velocity $$\boldsymbol{V}$$ in that system, since it's rotating together with the body). What am I missing?

Angular momentum of a rigid body is a vector that is non-zero when the body is in rotation with respect to absolute space (or later, with respect to distant stars). This vector can be represented by coordinates in the frame of the absolute space (inertial frame), or by coordinates in the frame of the body.

It seems the confusion is coming from thinking that in the body $$\mathbf{V}$$ equals $$\dot{\mathbf{Q}}$$. This is however not true. The position vector $$\mathbf{Q}$$ of a point that rotates with the body is constant. (Arnol'd allows more general motions where a point $$\mathbf{Q}$$ moves in a rotating coordinate system $$\mathbf{K}\,$$ but let's stick to the case of constant $$\mathbf{Q}\,.$$)

It is true that in space $$\mathbf{v}=\dot{\mathbf{q}}=[\omega,\mathbf{q}]$$ holds, and in the body $$\mathbf{V}=[\Omega,\mathbf{Q}]\,.$$ (This is Arnol'd's notation for the cross product). But it is not true that the "velocity in the body" $$\mathbf{V}$$ equals $$\dot{\mathbf{Q}}$$ (the derivative of a constant is zero).

When $$B$$ is the time dependent rotation matrix around the $$z$$-axis,

$$\begin{eqnarray}\label{eZRot} B=\left(\begin{array}{ccc}\cos(\alpha\,t) & -\sin(\alpha\,t) & 0 \\ \sin(\alpha\,t) & \cos(\alpha\,t) & 0\\ 0 & 0 & 1 \end{array}\right)\,, \end{eqnarray}$$ and $$\mathbf{Q}$$ is, say, the vector $$\mathbf{Q}=\left(\begin{array}{c}1\\0\\0\end{array}\right)$$ then $$\mathbf{q}=\left(\begin{array}{c}\cos(\alpha\,t)\\\sin(\alpha\,t)\\0\end{array}\right)\,,~~ \mathbf{v}=\dot{\mathbf{q}}=\alpha\left(\begin{array}{c}-\sin(\alpha\,t)\\\cos(\alpha\,t)\\0\end{array}\right)\,,~~ \mathbf{V}=B^\top\mathbf{v}=\left(\begin{array}{c}0\\\alpha\\0\end{array}\right)\,,$$ and $$\omega=\Omega=\left(\begin{array}{c}0\\ 0\\ \alpha\end{array}\right)\,.$$ The equations $$\mathbf{V}=[\Omega,\mathbf{Q}]$$ and $$\mathbf{v}=[\omega,\mathbf{q}]$$ are easily verified.

Denote by $$\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z$$ the cartesian basis vectors fixed in space and by $$\mathbf{E}_x,\mathbf{E}_y,\mathbf{E}_z$$ their counterparts that are fixed in (i.e. rotating with) the body. Then clearly, $$\mathbf{q},\mathbf{v},\omega$$ is a system of orthogonal vectors that rotates in space around the axis $$\omega=\alpha\mathbf{e}_z$$, and, $$\mathbf{Q}=\mathbf{E}_x,\quad\mathbf{V}=\alpha\mathbf{E}_y,\quad\Omega=\alpha \mathbf{E}_z$$ is a system of orthogonal vectors that is fixed in the body.

• This is what I was looking for. I figured out after asking the question that $\boldsymbol{V}$ was what you explained but it somehow didn't convince me. As for the angolar momentum, it has a fixed (in this case) pole as its reference so it does not depend on the frame (the components clearly do but the vector does not). That said, that angular momentum stuff was just a fleeting doubt. Thanks for clarifying about the velocity. Commented Jun 26, 2021 at 9:01
• Angular momentum is yet another concept to think about carefully. In space Arnold denotes it by $\mathbf{m}\,.$ He shows in Ch. 2 that $\dot{\mathbf m}$ is equal to the sum of all external torques. Therefore it is conserved if there are no external forces. In contrast, angular momentum $\mathbf M$ in the body is NOT conserved in general (even when force free). It obeys the Euler equation $\dot{\mathbf M}=[\mathbf M,\Omega]$ (Chapter 6). However, when the rotation is force free and around a principal axis then $\mathbf M$ is conserved. Commented Jun 26, 2021 at 11:25