# Why does the equation for transformation of relative acceleration from one non-inertial system to another involve absolute angular velocities?

It seems that the equation for transforming between two different expressions of the relative acceleration of a vector involves absolute angular velocities, which I find surprising.

Consider two non-inertial coordinate systems $$\alpha$$ and $$\beta$$, whose absolute angular velocity vectors are $$\boldsymbol{\Omega}^\alpha$$ and $$\boldsymbol{\Omega}^\beta$$, respectively.

The absolute velocity and acceleration of $$\boldsymbol{x}$$ (i.e. the motion as seen by an observer attached to the inertial system), is $$\dot{\boldsymbol{x}}$$ and $$\ddot{\boldsymbol{x}}$$, respectively.

We denote the velocity and acceleration relative to $$\alpha$$, i.e. that seen by an observer attached to the system $$\alpha$$, as $$\dot{\boldsymbol{x}}|_\alpha$$ and $$\ddot{\boldsymbol{x}}|_\alpha$$, respectively, and similarly for $$\beta$$. Note that $$\dot{\boldsymbol{x}}|_\alpha$$ is the time derivative of $$\boldsymbol{x}$$, holding the basis vectors of $$\alpha$$ fixed.

$$\dot{\boldsymbol{x}}$$ can be expressed in terms of $$\dot{\boldsymbol{x}}|_\alpha$$ and $$\dot{\boldsymbol{x}}|_\beta$$ as follows:

$$$$\begin{split} \dot{\boldsymbol{x}} &= \dot{\boldsymbol{x}}|_\alpha + \boldsymbol{\Omega}^\alpha\times\boldsymbol{x}\\ \dot{\boldsymbol{x}} &= \dot{\boldsymbol{x}}|_\beta + \boldsymbol{\Omega}^\beta\times\boldsymbol{x}\\ \end{split}$$$$

and $$\ddot{\boldsymbol{x}}$$ can be expressed as follows: $$$$\begin{split} \ddot{\boldsymbol{x}} &= \ddot{\boldsymbol{x}}|_\alpha + 2\boldsymbol{\Omega}^\alpha\times\dot{\boldsymbol{x}}|_\alpha + \dot{\boldsymbol{\Omega}}^\alpha\times\boldsymbol{x} + \boldsymbol{\Omega}^\alpha\times(\boldsymbol{\Omega}^\alpha\times\boldsymbol{x})\\ \ddot{\boldsymbol{x}} &= \ddot{\boldsymbol{x}}|_\beta + 2\boldsymbol{\Omega}^\beta\times\dot{\boldsymbol{x}}|_\beta + \dot{\boldsymbol{\Omega}}^\beta\times\boldsymbol{x} + \boldsymbol{\Omega}^\beta\times(\boldsymbol{\Omega}^\beta\times\boldsymbol{x}) \end{split}$$$$

We can use the above equations to express $$\dot{\boldsymbol{x}}|_\alpha$$ in terms of $$\dot{\boldsymbol{x}}|_\beta$$: $$$$\dot{\boldsymbol{x}}|_\alpha = \dot{\boldsymbol{x}}|_\beta + \boldsymbol{\Omega}^{\beta/\alpha}\times\boldsymbol{x}$$$$

and $$\ddot{\boldsymbol{x}}|_\alpha$$ in terms of $$\ddot{\boldsymbol{x}}|_\beta$$: $$$$\ddot{\boldsymbol{x}}|_\alpha = \ddot{\boldsymbol{x}}|_\beta + 2\boldsymbol{\Omega}^{\beta/\alpha}\times\dot{\boldsymbol{x}}|_{\beta} +\boldsymbol{\Omega}^{\beta/\alpha}\times\boldsymbol{x} +\boldsymbol{\Omega}^{\beta/\alpha}\times(\boldsymbol{\Omega}^{\beta/\alpha}\times\boldsymbol{x}) + (\boldsymbol{\Omega}^\beta\times\boldsymbol{\Omega}^\alpha)\times\boldsymbol{x}$$$$

where $$\boldsymbol{\Omega}^{\beta/\alpha} = \boldsymbol{\Omega}^\beta - \boldsymbol{\Omega}^\alpha$$.

The last equation (whose fairly lengthy derivation I've skipped) is the one I'm referring to in the title. Say that we know the motion relative to $$\beta$$, and the relative angular acceleration between $$\alpha$$ and $$\beta$$ ($$\boldsymbol{\Omega}^{\beta/\alpha})$$. Then this equation directly gives the relative acceleration in terms of $$\alpha$$ (without the need to transform from $$\beta$$ to an inertial frame, and from there to $$\alpha$$).

Although as it turns out, the equation also requires knowledge of the absolute angular velocities of the coordinate systems $$\alpha$$ and $$\beta$$ ($$\boldsymbol{\Omega}^\alpha$$ and $$\boldsymbol{\Omega}^\beta$$), since these appear as irreducible terms in this equation. Note that this is not the case for the equation relating the relative velocities.

Am I missing something here, and does this in fact make physical sense? How can the term $$(\boldsymbol{\Omega}^\beta\times\boldsymbol{\Omega}^\alpha)\times\boldsymbol{x}$$ be understood from a physical point of view?