# Velocity of points in a rigid body

I'm trying to derive the following statement:

Let $$\mathcal{B}$$ be a rigid body. Then there is an unique vector $$\vec{\omega}$$ such that for every pair of points $$P,Q\in \mathcal{B}$$ the following equation holds: $$\vec{v}_P-\vec{v}_Q=\vec{\omega}\times (P-Q),$$ where $$\vec{v}_P,\vec{v}_Q$$ are the velocities of $$P$$ and $$Q$$.

My proof attempt

Since $$\mathcal{B}$$ is a rigid body we have that for every pair of points $$P,Q\in \mathcal{B}$$: $$\|P-Q\|^2=0$$ $$(P-Q)\cdot (P-Q)=0$$ Differentiating with respect to time: $$(P-Q)\cdot (\vec{v}_P-\vec{v}_Q)=0$$ Since $$(P-Q)$$ is orthogonal to $$(\vec{v}_P-\vec{v}_Q)$$ there is a vector $$\vec{\omega}_{PQ}$$ (in general it could depend from $$P$$ and $$Q$$) such that: $$\vec{v}_P-\vec{v}_Q=\vec{\omega}_{PQ}\times (P-Q),$$ Now, how do I prove that $$\vec{\omega}_{PQ}=\vec{\omega}_{RS}$$ for all points $$P,Q,R,S \in \mathcal{B}$$?

• Jun 16, 2022 at 21:48
• The equation defining $\vec \omega_{PQ}$ does not define it uniquely. You cannot expect to prove that $\vec\omega_{PQ} = \vec \omega_{RS}$ with only this equation. Jun 16, 2022 at 21:59
• Thank you, and what should I do? Jun 16, 2022 at 22:00