# Equivalent Characterizations of Rigid Bodies & Angular Velocity Interpretation

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 $$i, j$$?
• 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. Commented Jan 15 at 0:27
• I think the questions are sufficiently related. I almost have an answer ready. Commented Jan 15 at 0:31

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.