• Consider two frames $S$ and $S’$ whose coordinates are related by

$$ \vec r'(t)=\vec r(t)-\vec R(t) $$

• From this we can get

$$ \vec{a}'=\vec{a}-\vec{A} $$

where $\vec{A}=\ddot{\vec{R}}$

$$ \vec F'=\vec F-m\vec A= m \vec a' $$

• From the above we get to the definition $\vec F_\text{inertial}=-m\vec A$, so an observer in a non inertial frame $S’$ experiences that force

Rotating inertial frames

• Spherical polar coordinates $(r,\theta,\phi)$

$$ \begin{aligned} x&=r\sin\theta\cos\phi \\ y&=r\sin\theta\sin\phi \\ z&=r\cos\theta \end{aligned} $$

$$ \begin{aligned} r&=\sqrt{x^2+y^2+z^2} \\ \theta&=\cos^{-1}\left(\frac{z}{r}\right) \\ \phi&=\tan^-1\left(\frac{y}{x}\right) \end{aligned} $$

https://www.math3d.org/CQHUnA6rD

• We can also write the basis vectors:

$$ \begin{aligned} \vec e_r&=\sin\theta\,[\cos(\phi)\,\vec e_x+\sin(\phi)\,\vec{e}_y]+\cos(\theta)\,\vec{e}z \\ \vec e\theta &=\cos\theta\,[\cos(\phi)\,\vec{e}_x+\sin(\phi)\,\vec{e}_y]-\sin(\theta)\,\vec{e}z \\ \vec e\phi&=-\sin(\phi)\,\vec{e}_x+\cos(\phi)\,\vec{e}_y \end{aligned} $$

$$ \begin{aligned} \vec{e}x&=\cos\phi\,[\sin(\theta)\,\vec e_r+\cos(\theta)\,\vec e\theta]-\sin(\phi)\,\vec{e}\phi \\ \vec e_y&=\sin\phi\,[\sin(\theta)\,\vec e_r+\cos(\theta)\,\vec e\theta]+\cos(\phi)\,\vec{e}_\phi \\ \vec{e}z&=\cos(\theta)\,\vec e_r-\sin(\theta)\,\vec e\theta \end{aligned} $$

• Take 2 frames, $S$ and $S’$ which are inertial and rotating frames respectively, $\vec e_i=\{\vec e_x,\vec e_y,\vec e_z\}$, $\vec e'_i=\{\vec e'_x,\vec e'_y,\vec e'_z\}$. $\theta\equiv\theta(t)$ and $\phi\equiv\phi(t)$ we can find:

$$ \begin{aligned} \dot {\vec e_r}&=\dot \theta \vec e_\theta+\dot \phi\sin(\theta)\,\vec e_\phi \\ \dot {\vec e_\theta}&=-\dot \theta \vec e_r+\dot \phi\cos(\theta)\,\vec e_\phi \\ \dot {\vec e_\phi}&=-\dot \phi\,[\sin(\theta)\, \vec e_r+\cos(\theta)\,\vec e_\theta] \end{aligned} $$

• Using the fact that $\vec\omega=\omega \vec n$ where $\hat n$ is an axis of rotation, and $\vec \omega \equiv \vec \omega ( \theta , \phi )$. We can deduce that the general angular velocity associated with a rotation by $\theta$ about $\vec e_\phi$ and $\phi$ about $\vec e_z$ is:

$$ \vec{\omega}=\dot \theta \vec e_\phi + \dot \phi \vec e_z=\dot \phi\,[\cos(\theta)\,\vec e_r-\sin(\theta)\,\vec e_\theta]+\dot \theta\, \vec e_\phi $$

which can be written as $\vec\omega=\omega_k'\vec e'_k$ where $\omega_1'=\dot \phi \cos\theta$, $\omega_2'=-\dot \phi \sin \theta$ and $\omega_3'=\dot \theta$

$$ \vec\omega\times\vec e_r= \begin{vmatrix} \vec e_r & \vec e_\theta & \vec e_\phi \\ \dot \phi \cos \theta & -\dot \phi\sin\theta & \dot \theta \\ 1 & 0 &0 \end{vmatrix}=\dot \theta\vec e_\theta+\dot\phi\sin(\theta)\,\vec e_\phi=\dot{\vec e_r} $$

• Similarly $\dot{\vec e_\theta} =\vec \omega \times \vec e_\theta$ and $\dot{\vec e_\phi} =\vec \omega \times \vec e_\phi$. These can be summarised:

$$ \dot{\vec e_i'}=\vec\omega\times\vec e_i'=\omega'_k(\vec e'_k\times\vec e'i)=\epsilon{ijk}\vec e'_j \omega'_k $$

Accelerating rotating frames

• Consider an arbitrary vector $\vec A=A_i(t)\vec e_i=A_i'(t)\vec e_i'$ written in two frames $S$ and $S'$ with basis vector $\vec e_i$ and $\vec e'_i(t)$

$$ \frac{\text{d}\vec{A}}{\text{d}t}=\dot A_i\vec e_i=\dot A'_j\vec e_j'+A'_j\dot{ \vec e_j'}=\dot{A}'_j\vec e'_j+A_j'(\vec \omega\times\vec e'_j) $$

• From this we can deduce that

$$ \dot{\vec A}|S=\dot{\vec A}|{S'}+\vec\omega\times\vec A $$