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In rotational kinematics, there seem to be two common characterizations of a rigid body:

  1. A rigid body is any collection of particles with position vectors $\textbf x_1,\textbf x_2,...$ such that the distance $|\textbf x_i-\textbf x_j|$ between any pair of particles $i,j\space $ is conserved, i.e. $\frac{d}{dt}|\textbf x_i-\textbf x_j|=0\iff(\textbf x_i-\textbf x_j)\cdot(\dot{\textbf x}_i-\dot{\textbf x}_j)=0$
  2. A rigid body is any collection of particles whose configuration space $\mathcal C$ is isomorphic to $\text{SO}(3)$ (assuming the rigid body's center of mass is fixed in some reference frame).

My questions are:

  1. Are these two characterizations truly equivalent to each other? (proof?)
  2. In light of the orthogonality relation $(\textbf x_i-\textbf x_j)\cdot(\dot{\textbf x}_i-\dot{\textbf x}_j)=0$ in Characterization #1 above, is it true that the angular velocity vector $\boldsymbol\omega$ of the rigid body obeys the relation $\dot{\textbf x}_i-\dot{\textbf x}_j=\boldsymbol{\omega}\times(\textbf x_i-\textbf x_j)$ for all pairs of particles $i, j$?
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  • $\begingroup$ Welcome to Physics! These are both interesting questions but they should probably be asked separately; the Q&A format of this forum relies on one well-posed question per post which can be given a "best answer". I would recommend that you edit your post to omit one of these questions, and post that question separately. $\endgroup$ Commented Jan 15 at 0:27
  • $\begingroup$ I think the questions are sufficiently related. I almost have an answer ready. $\endgroup$ Commented Jan 15 at 0:31

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Your second question is equivalent to your first. Any rotation in SO(3) will have a dual vector to act in the cross product, and visa versa (the dual vector of a rotation matrix is a well known result). We can thus use your second question as the target of the proof.

The reverse direction is easy. If

$$ \dot{\vec{x}}_{i}-\dot{\vec{x}}_{j}=\vec{\omega}\times\left(\vec{x}_{i}-\vec{x}_j\right) $$

Then

$$ \left( \vec{x}_{i}-\vec{x}_{j}\right)\cdot\left(\dot{\vec{x}_{i}}-\dot{\vec{x}_{j}}\right) = 0 $$

should be obvious (a cross product is perpendicular to each of its operands). Going in the forward direction, we note that for any $\vec{x}_{i}$, $\vec{x}_{j}$, we can always find $\vec{\omega}_{ij}$ so that

$$ \dot{\vec{x}}_{i}-\dot{\vec{x}}_{j}=\vec{\omega}_{ij}\times\left(\vec{x}_{i}-\vec{x}_j\right) $$

by the prior result. This is to say, since the dot product is zero, they are perpendicular and so there must be some $\omega_{ij}$ to put into the cross product. We need to show that $\vec{\omega}_{ij} = \vec{\omega}$, the same for all $\vec{x}_{i}$, $\vec{x}_{j}$. Evaluating

$$ \left( \vec{x}_{i}-\vec{x}_{k}\right)\cdot\left(\dot{\vec{x}_{i}}-\dot{\vec{x}_{k}}\right) = 0 $$

$$ \left( \vec{x}_{i}-\vec{x}_{j}+\vec{x}_{j}-\vec{x}_{k}\right)\cdot\left(\dot{\vec{x}_{i}}-\dot{\vec{x}_{j}}+\dot{\vec{x}_{j}}-\dot{\vec{x}_{k}}\right) = 0 $$

$$ \left( \vec{x}_{i}-\vec{x}_{j}\right)\cdot\left(\dot{\vec{x}_{i}}-\dot{\vec{x}_{j}}\right) + \left( \vec{x}_{i}-\vec{x}_{j}\right)\cdot\left(\dot{\vec{x}_{j}}-\dot{\vec{x}_{k}}\right) + \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\left(\dot{\vec{x}_{i}}-\dot{\vec{x}_{j}}\right) + \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\left(\dot{\vec{x}_{j}}-\dot{\vec{x}_{k}}\right) = 0 $$

$$ 0 + \left( \vec{x}_{i}-\vec{x}_{j}\right)\cdot\left(\dot{\vec{x}_{j}}-\dot{\vec{x}_{k}}\right) + \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\left(\dot{\vec{x}_{i}}-\dot{\vec{x}_{j}}\right) + 0 = 0 $$

$$ \left( \vec{x}_{i}-\vec{x}_{j}\right)\cdot\vec{\omega}_{jk}\times\left(\vec{x}_{j}-\vec{x}_{k}\right) + \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\vec{\omega}_{ij}\times\left(\vec{x}_{i}-\vec{x}_{j}\right) = 0 $$

$$ \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\left(\vec{x}_{i}-\vec{x}_{j}\right)\times\vec{\omega}_{jk} + \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\vec{\omega}_{ij}\times\left(\vec{x}_{i}-\vec{x}_{j}\right) = 0 $$

$$ -\left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\vec{\omega}_{jk}\times\left(\vec{x}_{i}-\vec{x}_{j}\right) + \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\vec{\omega}_{ij}\times\left(\vec{x}_{i}-\vec{x}_{j}\right) = 0 $$

$$ \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\vec{\omega}_{ij}\times\left(\vec{x}_{i}-\vec{x}_{j}\right) = \left( \vec{x}_{j}-\vec{x}_{k}\right)\cdot\vec{\omega}_{jk}\times\left(\vec{x}_{i}-\vec{x}_{j}\right) $$

I don't know how much rigor you are looking for, but I would say that since $\vec{x}_{i}$, $\vec{x}_{j}$, and $\vec{x}_{k}$ are completely arbitrary, we must have that

$$ \vec{\omega}_{ij} = \vec{\omega}_{jk} = \vec{\omega} $$

the same for all vectors $\vec{x}$. If the last step is insufficient for you, I can try to show it, but I think that is one of those things that is less illuminating than the space to show it warrants.

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