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?


1 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.

  • $\begingroup$ Would $a_i \cfrac{\partial}{\partial r_i}\epsilon_{ijk} b_jr_k$ be the right LHS of the identity in index form? $\endgroup$
    – raptakem
    Feb 19, 2018 at 13:34
  • $\begingroup$ Almost, but not quite: The dummy index in the scalar product should be differnet from the free index, e.g. $a_l \cfrac{\partial}{\partial r_l}\epsilon_{ijk} b_jr_k$. $\endgroup$
    – Toffomat
    Feb 19, 2018 at 13:40
  • $\begingroup$ I used this method and after differentiating obtained $m (\dot{\vec{r}} + \vec{\omega} \times \vec{r}) \cdot \cfrac{\partial}{\partial \vec{r}} ( \vec{\omega} \times \vec{r})$, however after using the provided identity the answer is still off by a factor of -1. Is there something else I am missing? $\endgroup$
    – raptakem
    Feb 19, 2018 at 23:58

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