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 $P,Q,R,S \in \mathcal{B}$?

  • $\begingroup$ related: physics.stackexchange.com/a/680029/195949 $\endgroup$ Jun 16, 2022 at 21:48
  • $\begingroup$ 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. $\endgroup$ Jun 16, 2022 at 21:59
  • $\begingroup$ Thank you, and what should I do? $\endgroup$
    – Kandinskij
    Jun 16, 2022 at 22:00


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