# Proving that the relative angular velocity of any particle with respect to any other particle is the same in a rigid body

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:

$$|\vec{r_{j,i}}|=c_{j,i}$$ $$\implies{\frac{d}{dt}(\vec{r_{j,i}}\cdot \vec{r_{j,i}})= \frac{d}{dt}c_{j,i}^{2}=0}$$

$$\implies{\vec{r_{j,i}}\cdot \vec{v_{j,i}} = 0}$$