During the study of the motion of a rigid body, in Arnold's book, two coordinates systems are introduced: one is fixed $k=\{O',\hat e_1',\hat e_2',\hat e_3'\}$ and one is inside the rigid body $K=\{O,\hat e_1,\hat e_2,\hat e_3\}$. We suppose that the rigid body is rotating around a line passing through $O$. We also have a transformation $B:K\to k$, which is the matrix of change of coordinates. One can prove that $B\in\operatorname{SO}(3)$.

I read the chapter concerning the rigid body but I didn't find an explicit definition of angular velocity of the rigid body. I think that the angular speed of a rigid body can be defined as the unique differentiable vector $\vec\omega(t)$ such that $\forall \vec u$ belonging to a coordinate system rotating with the rigid body, $\frac{d}{dt}\vec u=\vec \omega(t)\times \vec u$, $\forall t\in I \subseteq \mathbb R$.

Second thing I'd like to check is the following. Given a point $P$ belonging to the rigid body, variables in moving coordinate system are named $\vec Q=P-O,\vec V,\vec \Omega,\vec M$ and variables in the fixed coordinate system are named $\vec q=P-O',\vec v,\vec \omega,\vec m$ and it holds the relation $\forall \vec X\in K$, $B\vec X=\vec x\in k$.
Observing that the angular momentum of the point $P$ relative to the pole $O$ is given by $$M_O=(P-O)\times m \vec V=\vec Q\times m\cdot d\vec Q/dt=m \vec Q\times(\vec \Omega\times\vec Q)$$ the book introduces an operator $A:K\to K$ such that $A\vec\Omega=\vec M$, called inertia operator. I think this is not the standard definition of inertia operator. Infact, reading different notes, I think that one can can define the inertia operator in the following way. Given a real affine space $\mathbb A^n=(V,V(\mathbb R),-)$ the inertia operator relative to a pole $O\in V$ (for a discrete rigid body) is defined as $$A_O:V\to V\text{ such that}$$ $$A_O \vec u:=\sum_i m_i(P_i-O)\times[\vec u\times (P_i-O)].$$ So the operator defined in Arnold's textbook it's like a "punctual inertia operator" $(A_O)_i$ and instead of defining the operator on $\mathbb R^3$, it takes the restriction to the moving frame $K$. Is that right?

Thank you for your attention.


1 Answer 1


If you are referring to Arnold’s Mathematical Methods of Classical Mechanics, then you missed a whole section where he defines mathematically angular velocity (26-D in second edition).

For the inertia tensor, the equation $M=A\Omega$ can be used to prove your formula. In fact, it is the usual method of motivating you formula, how else would you derive it? Note that the ambiguity in the choice of the origin is deferred in the definition of $M$. In the case of Arnold, the solid rotates about a point by assumption. The angular momentum (and the moment of inertia tensor) is therefore defined with respect to this point of rotation.

Hope this helps.

  • $\begingroup$ Thank you for the answer. I've just checked that section. Yes the inertial operator is of course relative to a pole and in the first form is an operator of the form $A_O^{(P)}$ (I mean it's the inertial operator associated to a point $P$ of the rigid body relative to a pole $O$). $\endgroup$
    – Vajra
    Commented Dec 7, 2022 at 18:20
  • $\begingroup$ Every variable introced at first is relative to point then you can extend the operator summing over the point $P$: $A_O=\sum_{P\in \mathcal C}A_O^{(P)}$, where $\mathcal C$ is the rigid body. $\endgroup$
    – Vajra
    Commented Dec 7, 2022 at 18:22

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