Rigid body mechanics in Arnold's book

Question

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 \begin{equation} B:K\rightarrow k \end{equation} 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: \begin{equation} \boldsymbol{\Omega}=B^{-1}\boldsymbol{\omega} \end{equation} (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?

Answer 1

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.

Answer 2

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.