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?


2 Answers 2


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

for 3-D rotations, it is often useful to note that $\pmb{\omega}$ is more properly represented as a skew-symmetric 2-tensor $\pmb{\Omega}$ such that $\pmb{\Omega}\cdot \pmb{u}= \pmb{\omega}\times \pmb{u}$ for any vector $\pmb{u}$. the components of $\pmb{\Omega}$ and $\pmb{\omega}$ in any (right handed) orthonormal basis are related by $\omega_i = -\tfrac{1}{2}\epsilon_{ijk}\Omega_{jk}$ and $\Omega_{ij}=-\epsilon_{ijk}\omega_k$.

If $\pmb{e}_i$ is an intertial/fixed basis and $\pmb{e}'_i$ is the rotating basis, they are related linearly by some $\pmb{e}'_i=\pmb{R}\cdot \pmb{e}_i$ and the angular velocity tensor is $\pmb{\Omega} = \dot{\pmb{R}}\cdot \pmb{R}^{-1}$. If both bases are right-handed and orthonormal then $\pmb{R}$ is a special orthogonal tensor satisfying $\pmb{R}^T\cdot \pmb{I} \cdot \pmb{R} = \pmb{I}$ where $\pmb{I}$ is the euclidean metric/inner product (if we limit consideration to orthonormal bases then $\pmb{I}$ can be seen as the identity tensor and we often simply this relation as $\pmb{R}^T=\pmb{R}^{-1}$). All of this is simply to say that, letting $(\cdot)'$ denote coordinate vectors/matrix representations in the rotating $\pmb{e}'_i$ basis, then the Lagrangian for a free particle in 3-space is given in rotating coordinates $q'$ as

$$ L = \tfrac{1}{2}m (\dot{q}' + \Omega'\cdot q') \cdot I' \cdot (\dot{q}' + \Omega'\cdot q') \,=\, \tfrac{1}{2}m (\dot{q}' + \Omega'\cdot q')^2 $$

where the last equality assumes $\pmb{e}'_i$ is orthonormal such that $I'$ is the identity matrix. For any $n\times n$ matrix $A$ and $n$-tuple/column vector $x$, we have $\tfrac{\partial}{\partial x} (A\cdot x) = A$. It follow from the above and the chain rule, along with symmetry of $\pmb{I}$ and skew-symmetry of $\pmb{\Omega}$, that

$$ \tfrac{\partial}{\partial q'} L = m I'\cdot (\dot{q}' + \Omega'\cdot q')\cdot\Omega' = mI'\cdot \Omega'^T \cdot (\dot{q}' + \Omega'\cdot q') = -mI'\cdot \Omega' \cdot (\dot{q}' + \Omega'\cdot q') $$

Using $\Omega \cdot u = \omega\times u$ this is equivalent to the expression you are seeking:

$$ \tfrac{\partial}{\partial q'} L = -mI'\cdot( \omega' \times \dot{q}' + \omega' \times (\omega'\times q')) = mI'\cdot( \dot{q}' \times \omega' - \omega' \times (\omega'\times q')) $$

where, for your purposes, you can drop $I'$ as it is just the identity matrix.


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