Proving that the angular velocity of a rigid body is the same about any point

Question

Please consider the below image which is from Rana and Joag, Classical Mechanics. They build on a proof, which I reiterate below and through it they show that angular velocity of any point B in the rigid body is the same about a point $B_0$ and then state that the angular velocity of B is same about any other point too.

From the image above and considering the rigid body constraints we get,

$\pmb{{\rho_1}\times (\dot{\rho_1}\times \dot{\rho_2} )} = \pmb{(\rho_1 . \dot{\rho_2})\dot{\rho_1}} - \pmb{(\rho_1 . \dot{\rho_1})\dot{\rho_2}} = -\pmb{(\rho_2 . \dot{\rho_1})\dot{\rho_1}}$

which implies that

$\pmb{ \dot{\rho_1} = (\frac{\dot{\rho_1}\times\dot{\rho_2}}{\dot{\rho_1} . \rho_2})}\times \rho_1 $

and similarly we can get

$\pmb{ \dot{\rho_2} = (\frac{\dot{\rho_1}\times\dot{\rho_2}}{\dot{\rho_1} . \rho_2})}\times \rho_2 $

The book says " that the above equations indicate that the vector

$\pmb{ \omega = (\frac{\dot{\rho_1}\times\dot{\rho_2}}{\dot{\rho_1} . \rho_2})}$

behaves as an angular velocity vector for $\pmb{\rho_1}$ and $\pmb{\rho_2}$ about $B_0$. The book proves that the any given point in the body say B has the same angular velocity as given by the expression but doesn't prove that why any point in the body should have the same angular velocity about any other point say $B_0'$.

This is what I want to prove mathematically rigorously. To do that I begin writing

$\pmb{ \rho_1 = \rho_1' + a}$ and similarly $\pmb{ \rho_2 = \rho_2' + a}$ and I try to find their time derivative and substitute the result into the expression for $\pmb{\omega}$ that has been calculated above. Now I will get my correct answer only if

$\pmb{\dot{a}} = 0$

Now I don't think that $\pmb{\dot{a}} = 0$ should be true because the direction of a will change even if its magnitude is constant and hence I fail to prove the result.

Could anyone help me how to proceed forward using the above expression for $\pmb{\omega}$ to prove that it is indeed the same about any point in the body. Should $\pmb{\dot{a}} = 0$ be true because if it is false then the result is false itself. But then why should be $\pmb{\dot{a}} = 0$ be true.

Or is there another way out.

Answer 1

From Euler's theorem we know that a rigid motion in odd number of dimensions with one point fixed is a pure rotation about an axis (cf. Goldstein). This allows one to write,

$$\mathbf{v}(\mathbf{r}) = \mathbf{v}_A + \omega_A \times(\mathbf{r}-\mathbf{r}_A), \quad \quad (1) $$

$$\mathbf{v}(\mathbf{r}) = \mathbf{v}_B + \omega_B \times(\mathbf{r}-\mathbf{r}_B). \quad \quad (2)$$

Now the velocity fields described by equation (1) and (2) must be the same. This means, in particular, that

$$ \omega_A \times \mathbf{r} = \omega_B \times \mathbf{r}.$$

The only way this is possible is if the vectors of angular velocity are equal, i.e. $\omega_A = \omega_B$, unless the rigid body is linear, in which case there is no way to uniquely define the angular velocity since rotations along the line which contains the rigid body do not contribute to physical motion.