# Rotating coordinate frame

Hello I have a question about rotating coordinate frames. Following the book of Brizard the Lagrangian is given by $$$$L(\mathbf{r}, \mathbf{\dot{r}}) = \frac{m}{2} \vert \mathbf{\dot{r}} + \boldsymbol{\omega} \times \mathbf{r} \vert^2 - U(\mathbf{r})$$$$ Would it be possible that $$L$$ is independent of $$\mathbf{\dot{r}}$$? What does this mean? In particular, how can I end up with an equation of motion, since I would not have a term $$m\mathbf{\ddot{r}}$$ in that case, so no resulting force?

• What do you mean, $L$ is independent of $\dot{r}$? Your formula includes $\dot{r}$. – jacob1729 Jun 8 at 15:08
• No, it is not possible that $L$ would be independent of $\dot{\mathbf{r}}$. What would make you think that it is possible? – Void Jun 8 at 15:08
• There may be some kind of constant of motion which allows elimination of terms you are looking for. Even if something is present in the equation explicitly, doesn't mean value depends on it. – kakaz Jun 8 at 15:25
• What make me think of its possible was that usually coordinate frames are fixed. Then the Coriolis force must vanish and I think this would be possible with a vanishing velocity. So a non-rotating system as limit of a rotating was the idea behind it – Q.stion Jun 9 at 10:10

If you set $$L=L(\mathbf{q})$$ then your equations of motion become :
$$\frac{\partial L}{\partial q_i}=\frac d {dt}\left( \frac{\partial L}{\partial \dot q_i} \right)=0$$
But for any such function this reduces to a set of equations that fix the $$q_i$$ to some constant values (the roots of those equations).