# Elementary Construction of the Angular Velocity

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Physics books  seldom contain accurate definition of the angular velocity of a rigid body. I believe that the following construction is as simple as possible and also rigorous.

Assume we have a rigid body which is moving in the space.

Theorem. In each moment there exists a unique (pseudo)vector $\boldsymbol\omega$ such that for any two points $A,B$ of the rigid body the following formula holds
$$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol \omega\times\boldsymbol {BA},\qquad (***)$$
here $\boldsymbol v_A,\boldsymbol v_B$ are the velocities of the points $A,B$ respectively.

Definition. The vector $\boldsymbol \omega$ is called the angular velocity.

Before proving this theorem we formulate a lemma. Let $\boldsymbol e_1,\boldsymbol e_2,\boldsymbol e_3$ be a right orthonormal basis frozen into the rigid body. Introduce a vector
$$\boldsymbol \omega_*=(\dot{\boldsymbol e}_2,\boldsymbol e_3)\boldsymbol e_1+(\dot{\boldsymbol e}_3,\boldsymbol e_1)\boldsymbol e_2+(\dot{\boldsymbol e}_1,\boldsymbol e_2)\boldsymbol e_3.\qquad (*)$$

Lemma. The following formulas hold $$\dot{\boldsymbol e}_k=\boldsymbol \omega_*\times {\boldsymbol e}_k,\quad k=1,2,3.\qquad (**)$$
Indeed, differentiating the equlities $({\boldsymbol e}_k,{\boldsymbol e}_j)=\delta_{kj}$ we get
$(\dot{\boldsymbol e}_k,\boldsymbol e_j)=-(\dot{\boldsymbol e}_j,\boldsymbol e_k)$ and particularly $(\dot{\boldsymbol e}_k,\boldsymbol e_k)=0$. Due to these formulas one can substitute (*) into (**) to make sure that the lemma is correct.

Let us prove the theorem.

A) Existence. Let $O$ be the origin of some fixed frame. Differentiating the identity $\boldsymbol{OA}=\boldsymbol{OB}+\boldsymbol{BA}$ we have
$$\boldsymbol v_A=\boldsymbol v_B+\frac{d}{dt}\boldsymbol{BA}.$$
On the other hand we can write $\boldsymbol{BA}=\sum_{i=1}^3 x_i\boldsymbol e_i.$ From lemma it follows that
$$\frac{d}{dt}\boldsymbol{BA}=\sum_{i=1}^3 x_i\dot{\boldsymbol e}_i=\sum_{i=1}^3 x_i\boldsymbol\omega_*\times{\boldsymbol e}_i=\boldsymbol\omega_*\times \boldsymbol{BA}.$$ Consequently there exists at least one vector that satisfies the theorem. This vector is $\boldsymbol \omega_*.$ In the sequel we drop the subscript *.

B) Uniqueness. Assume that there is another vector $\boldsymbol \omega’$ such that $$\boldsymbol v_A=\boldsymbol v_B+\boldsymbol \omega’\times\boldsymbol {BA}.$$
Subtracting this formula from the formula (***) we obtain
$(\boldsymbol \omega-\boldsymbol \omega’)\times\boldsymbol {BA}=0.$ Since $\boldsymbol {BA}$ is an arbitrary vector we have $\boldsymbol \omega=\boldsymbol \omega’$.

The theorem is proved.

Example. Let $OXYZ$ be a fixed Cartesian frame. And the rigid body rotates around the axis $OZ$ such that $\boldsymbol e_Z=\boldsymbol e_3$ and
$$\boldsymbol e_1=\cos\phi\boldsymbol e_X+\sin\phi\boldsymbol e_Y,\quad \boldsymbol e_2=\cos\phi\boldsymbol e_Y-\sin\phi\boldsymbol e_X$$ here $\phi=\phi(t)$ is the angle of rotation of the rigid body.

Then differentiate these formulas $\dot{\boldsymbol e}_1=\dot\phi\cos\phi\boldsymbol e_Y-\dot\phi\sin\phi\boldsymbol e_X=\dot\phi \boldsymbol e_2.$ In the same way we get $\dot{\boldsymbol e}_2=-\dot\phi \boldsymbol e_1.$ Substitute these equalities in (*) to get $\boldsymbol \omega=\dot\phi \boldsymbol e_Z.$

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