Obtaining Equations of Motion in a Rotating Coordinate System using Lagrangian

Question

Source: Chapter 2 of Classical Dynamics by David Tong (p14)

In the text the author attempts to derive the equations of motion for the system of a free particle in a frame rotating with angular velocity $\,\boldsymbol{\omega} = (0,0,\omega)$ about the $z$ axis. In the new frame our new/primed coordinates are related to the original coordinates by the following relations:

\begin{align}x' &= x \cos(\omega t) + y \sin( \omega t) \\ y' &= y \cos(\omega t) - x \sin( \omega t) \\ z' &= z \end{align}

By manipulating the above expression, we obtain the following form for the Lagrangian of the system: $$ \mathcal{L} = \cfrac{1}{2} m (\dot{\boldsymbol{r'}} + \boldsymbol{\omega} \times \boldsymbol{r'})^2$$

To obtain the first half of the equations of motion the following is done:

$$\frac{\partial\mathcal{L}}{\partial \boldsymbol{r'}} = m(\dot{\boldsymbol{r'}} \times \boldsymbol{\omega} - \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r'})$$

Is there a way to see this without having to expand the Lagrangian out of vector form and performing the tedious calculation that way?

Answer 1

Maybe a shortcut is to realise that $$\left(\vec a \cdot \frac{\partial}{\partial \vec r}\right) \left(\vec b \times \vec r\right) = \vec b \times \vec a$$ which you can see e.g. by writing this out in index form or realising that the differential operator is a scalar that only acts on $\vec r$.

Then the result follows by first using the chain rule for the square and then applying the formula above.

EDIT: Turns out the shortcut isn't actually al that helpful because the index structure is not applicable to the problem (the scalar product is nat between vector and derivative).

A better way to see it is as follows: Set $\vec a=\dot{\vec r} + \vec\omega\times \vec r $. Then $\mathcal L=\frac{m}{2}\vec a\cdot \vec a=\frac{m}{2} a_i a_i$. You want to compute $\frac{\partial \mathcal L}{\partial \vec r}$, i.e. $$\frac{\partial \mathcal L}{\partial \vec r_j}\equiv \partial_j \mathcal L= m a_i \partial_j a_i\,.$$ Now only the second term in $a$ contributes, and it gives $\partial_j \epsilon_{ikl}\omega_k r_l=\epsilon_{jik}\omega_k$. From here you can see that $\partial_j\mathcal L=m \left(\vec a\times\vec\omega\right)_j$, as required.