Why is the derivative of a vector in polar form the cross product? In physics we were taught that if our frame of reference is rotating with a certain angular velocity $\omega$, and we want to find the derivative of a certain vector $v$ over time (in polar form), we need to find the cross product of $v$ and $\omega$.
Could someone give me a mathematical explanation as to why that is?
Thanks in advance!
 A: Assume that we have a frame rotating around the $z$ axis. Then we can write
$$\vec{r}(t) = R(t) \vec{r}_0(t),$$
where $\vec{r}_0(t)$ are Cartesian coordinates of a particle in the rotating frame, $\vec{r}(t)$ Cartesian coordinates of the same particle in the rest frame, and $R(t)$ describes the rotation between the frames at time $t$.
Then,
$$\dot{\vec{r}}(t) = \dot{R}(t) \vec{r}_0(t) + R(t) \dot{\vec{r}}(t)$$
and
$$\ddot{\vec{r}}(t) = \ddot{R}(t) \vec{r}_0(t) + 2 \dot{R}(t) \dot{\vec{r}}_0(t) + R(t) \ddot{\vec{r}}(t).$$
Note that this is in the rest frame. To transform this vector to the rotation frame we multiply from left with $R(t)^{-1}$:
$$\ddot{\vec{r}}_0(t) = R(t)^{-1} \ddot{R}(t) \vec{r}_0(t) + 2 R(t)^{-1} \dot{R}(t) \dot{\vec{r}}_0(t) + \ddot{\vec{r}}(t).$$
Now the thing is that the matrix $R(t)^{-1} \dot{R}(t)$ works on a vector $\vec{u}$ exactly as taking the cross-product of the angular velocity vector $\vec\omega(t)$ with $\vec{u}.$ For example, for a rotation around the $z$ axis we have
$$R(t) = \begin{pmatrix}
\cos\phi(t) & -\sin\phi(t) & 0 \\
\sin\phi(t) &  \cos\phi(t) & 0 \\
0 & 0 & 1
\end{pmatrix}$$
so
$$\dot{R}(t) = \begin{pmatrix}
- \dot{\phi}(t) \sin\phi(t) & -\dot{\phi}(t) \cos\phi(t) & 0 \\
\dot{\phi}(t) \cos\phi(t) &  -\dot{\phi}(t) \sin\phi(t) & 0 \\
0 & 0 & 0
\end{pmatrix}
= \omega(t)
\begin{pmatrix}
- \sin\phi(t) & -\cos\phi(t) & 0 \\
\cos\phi(t) &  -\sin\phi(t) & 0 \\
0 & 0 & 0
\end{pmatrix}
$$
where $\omega(t) = \dot{\phi}(t).$
This gives
$$
R(t)^{-1} \dot{R}(t) = 
\begin{pmatrix}
\cos\phi(t) & \sin\phi(t) & 0 \\
-\sin\phi(t) &  \cos\phi(t) & 0 \\
0 & 0 & 1
\end{pmatrix}
\omega(t)
\begin{pmatrix}
- \sin\phi(t) & -\cos\phi(t) & 0 \\
\cos\phi(t) &  -\sin\phi(t) & 0 \\
0 & 0 & 0
\end{pmatrix}
= \omega(t) \begin{pmatrix}
0 & -1 & 0 \\
1 &  0 & 0 \\
0 &  0 & 0
\end{pmatrix}$$
Compare this with
$$
\vec{\omega}(t) \times \vec{u} = 
\omega(t) \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} \times \begin{pmatrix}u_x \\ u_y \\ u_z\end{pmatrix}
= \omega(t) \begin{pmatrix}-u_y \\ u_x \\ 0\end{pmatrix}
= \omega(t) \begin{pmatrix}
0 & -1 & 0 \\
1 &  0 & 0 \\
0 &  0 & 0
\end{pmatrix}
\begin{pmatrix}u_x \\ u_y \\ u_z\end{pmatrix}
$$
Thus, if $\omega(t) = \dot{R}(t)$ is constant so that $\ddot{R}(t) \equiv 0$ we get
$$\ddot{\vec{r}}_0(t) = 2 R(t)^{-1} \dot{R}(t) \dot{\vec{r}}_0(t) + \ddot{\vec{r}}(t) = 2 \vec{\omega}(t) \times \dot{\vec{r}}_0(t) + \ddot{\vec{r}}(t).$$
