Claim: The angular velocity of any point mass of a rigid body relative to any other point mass is the same, i.e., $\vec{\omega_{i,j}} = \vec{\omega}\;\,\forall{i}\,\forall{j}$, where $\vec{\omega}$ is by definition, “the angular velocity of the rigid body”
My attempt at a proof:
It suffices to show that:
(1) $\vec{\omega_{i,j}} = \vec{\omega_{j,i}}\;\,\forall{i}\,\forall{j}\,(i \neq j)$
(2) $\vec{\omega_{j,i}} = \vec{\omega_{k,i}}\;\,\forall{i}\,\forall{j}\,\forall{k}\,(i\neq j\neq k)$
(Assume universal quantification over all unbound variables below)
Proof for (1):
$\vec{\omega_{i,j}} := \frac{1}{|\vec{r_{i,j}}|^2}\vec{r_{i,j}}\times \vec{v_{i,j}}$$\;$and $\;$$\vec{\omega_{j,i}} := \frac{1}{|\vec{r_{j,i}}|^2}\vec{r_{j,i}}\times \vec{v_{j,i}}$
where $\vec{\omega_{j,i}},\vec{r_{j,i}}$ and $\vec{v_{j,i}}$ are the orbital angular velocity, position, velocity of the $j^{th}$ point mass w.r.t the $i^{th}$ point mass of the rigid body respectively.Subtracting these two expressions and using $\vec{r_{j,i}} = -\,\vec{r_{i,j}}$$\;$,$\;$ $\vec{v_{j,i}} = -\,\vec{v_{i,j}}$$\;$,$\;$$|\vec{r_{i,j}}| = |\vec{r_{j,i}}|$ we get $\,$$\vec{\omega_{i,j}} = \vec{\omega_{j,i}}$$\,$as required.
Proof for (2):
$\vec{\omega_{k,i}} := \frac{1}{|\vec{r_{k,i}}|^2}\vec{r_{k,i}}\times \vec{v_{k,i}}$$\;$and $\;$$\vec{\omega_{j,i}} := \frac{1}{|\vec{r_{j,i}}|^2}\vec{r_{j,i}}\times \vec{v_{j,i}}$
I tried expanding the expression of$\,$$\vec{\omega_{k,i}}$ using $\vec{r_{k,i}} = \vec{r_{k,j}}+\vec{r_{j,i}}$$\,$ and similarly for velocity, $\vec{v_{k,i}} = \vec{v_{k,j}}+\vec{v_{j,i}}$$\,$ but to no avail:
$\vec{\omega_{k,i}} = \frac{1}{|\vec{r_{k,i}}|^2}(\vec{r_{k,j}}\times \vec{v_{k,j}} + \vec{r_{j,i}}\times \vec{v_{k,j}}+ \vec{r_{k,j}}\times \vec{v_{j,i}}+ \vec{r_{j,i}}\times \vec{v_{j,i}})$
Now $\vec{r_{i,j}}\perp\vec{v_{i,j}}\;\,\forall{i}\,\forall{j}(i \neq j)$ (Rigid body constraint since $|\vec{r_{i,j}}| = constant$) but that doesn’t help in simplifying these cross products either. Any guidance will be much appreciated.
EDIT(27/02/22): I have made some progress:
To help prove (2), I make the following claim:
Claim: $\vec{v_{j,i}} = \vec{\omega_{j,i}} \times \vec{r_{j,i}}$
Proof: $\vec{\omega_{j,i}} := \frac{1}{|\vec{r_{j,i}}|^2}\vec{r_{j,i}}\times \vec{v_{j,i}}$
$\implies{\vec{\omega_{j,i}} \times \vec{r_{j,i}} = \left(\frac{1}{|\vec{r_{j,i}}|^2}\vec{r_{j,i}}\times \vec{v_{j,i}}\right)\times \vec{r_{j,i}}}$
$\implies{\vec{\omega_{j,i}} \times \vec{r_{j,i}}= \vec{v_{j,i}}-(\vec{r_{j,i}}\cdot \vec{v_{j,i}}) \frac{1}{|\vec{r_{j,i}}|^2}}$
(using the vector triple product identity) Now, $\vec{r_{j,i}}\cdot \vec{v_{j,i}} = 0$ since $|\vec{r_{j,i}}| = constant$ is the rigid body constraint. Using this proves the claim.
Now, $\vec{v_{j,i}} = \vec{\omega_{j,i}} \times \vec{r_{j,i}}$ and $\vec{v_{k,i}} = \vec{\omega_{k,i}} \times \vec{r_{k,i}}$ Subtracting these two and using $\vec{v_{j,k}}=\vec{v_{j,i}}-\vec{v_{k,i}}$, $\vec{r_{k,i}}=\vec{r_{j,i}}-\vec{r_{j,k}}$ we get:
$\vec{v_{j,k}} = (\vec{\omega_{j,i}}-\vec{\omega_{k,i}})\times\vec{r_{j,i}}+\vec{\omega_{k,i}}\times\vec{r_{j,k}}$. Since $\vec{v_{j,k}} = \vec{\omega_{j,k}} \times \vec{r_{j,k}}$,
$\implies{(\vec{\omega_{j,k}}-\vec{\omega_{k,i}})\times\vec{r_{j,k}}= (\vec{\omega_{j,i}}-\vec{\omega_{k,i}})\times\vec{r_{j,i}}}$
$\implies{[(\vec{\omega_{j,i}}-\vec{\omega_{k,i}})\times\vec{r_{j,i}}]\cdot\vec{r_{j,k}}=0}$
Using the scalar triple product identity:
$\implies{(\vec{\omega_{j,i}}-\vec{\omega_{k,i}})\cdot(\vec{r_{j,i}}\times \vec{r_{j,k}})=0}$
$\implies{| \vec{\omega_{j,i}}-\vec{\omega_{k,i}}|| \vec{r_{j,i}}\times \vec{r_{j,k}}|\cos{\theta}=0}$
$\implies{ \vec{\omega_{j,i}}= \vec{\omega_{k,i}}}$ or $ \vec{r_{j,i}}\times \vec{r_{j,k}}=\vec{0}$ or $\theta = \frac{\pi}{2}$
If$\vec{r_{j,i}}\times\vec{r_{j,k}}=\vec{0}$, then $i,j,k$ must be collinear (since the angle between the operands of the cross product must be $0$ or $\pi$) $\implies{\vec{r_{j,i}} = \lambda\vec{r_{k,i}}}$ for some $\lambda$
I argue that $\lambda$ must be constant with time as $|\lambda| = \frac{| \vec{r_{j,i}}|}{| \vec{r_{k,i}}|}= constant$ and $\lambda$ cannot switch signs with time either as that would mean that the orientation of $j,k$ in space relative to $i$ changes with time which is not possible as the separation remains fixed,i.e., if $\lambda\gt 0$ then $j,k$ are on the same side of $i$, that is, either $j$ or $k$ is in the middle of the other two depending on $| \vec{r_{j,i}}|, | \vec{r_{k,i}}|$ So $\lambda$ cannot suddenly turn negative while retaining its magnitude as that would imply that $i$ is now the particle in the middle which in turn would imply that the distance between $j,k$ has changed resulting in a contradiction. Since we know now that $\lambda$ is constant with time, we have
$\vec{v_{j,i}}=\lambda\vec{v_{k,i}}$
Now,$\vec{\omega_{j,i}} := \frac{1}{|\vec{r_{j,i}}|^2}\vec{r_{j,i}}\times \vec{v_{j,i}}$
$\implies{\vec{\omega_{j,i}}=\frac{1}{\lambda^{2}|\vec{r_{k,i}}|^2}\lambda\vec{r_{k,i}}\times \lambda\vec{v_{k,i}}}= \frac{1}{|\vec{r_{k,i}}|^2}\vec{r_{k,i}}\times \vec{v_{k,i}}= \vec{\omega_{k,i}} $
Similarily, now one must show that $\theta = \frac{\pi}{2}\implies \vec{\omega_{j,i}} = \vec{\omega_{k,i}}$ or alternatively $\theta \neq \frac{\pi}{2}$ for any $i,j,k$ in the rigid body to complete the proof. But I have no idea how to do so. Any guidance will be much appreciated. Also, here is the proof for $\vec{r_{j,i}}\cdot \vec{v_{j,i}} = 0$ that I used earlier without proof:
$|\vec{r_{j,i}}|=c_{j,i}$ $\implies{\frac{d}{dt}(\vec{r_{j,i}}\cdot \vec{r_{j,i}})= \frac{d}{dt}c_{j,i}^{2}=0}$
$\implies{\vec{r_{j,i}}\cdot \vec{v_{j,i}} = 0}$
