Non inertial reference frames: Lagrangian Formulation

I have searched for the equations of motion arrived at through the Lagrangian formulation and I have seen few satisfactory derivations. There is one reference to it in John R Taylor's book on classical mechanics as an extremely hard problem(3 stars). Hence it is a good topic for our consideration.
First we begin by asking how a vector changing with time looks in an inertial reference \(\left(\frac{d\vec{r}}{dt}\right)_o\) as opposed to an non-inertial reference frame \(\left(\frac{d\vec{r}}{dt}\right)_q\) . Now consider two reference frames with basis vectors \(e_i\) and \(e^{'}_i\). So  \( \vec{r} = \sum_1^n r_i e_i \) or \(\vec{r} = \sum_1^n r^{'}_i e^{'}_i \). Assume that the former co-ordinate basis is inertial and the second is not. This means that: $$\left(\frac{d\vec{r}}{dt}\right)_o = \sum_1^n \frac{dr^{'}_i}{dt}e^{'}_i + \sum_1^n r^{'}_i \frac{de^{'}_j}{dt} = \sum_1^n \frac{dr}{dt}e_i \hspace{20mm} \text{eq.1}$$

If you look at the first term in the middle equation in eq.1 you will see a term that changes with respect to \(e^ {'}_i\).  This is the speed with repesct to the rotating frame. Consequently, eq.1 can be written as
$$\left(\frac{d\vec{r}}{dt}\right)_o = \left(\frac{d\vec{r}}{dt}\right)_q + \sum_1^n r^{'}_i \frac{de^{'}_i}{dt} \hspace{20mm} \text{eq.2}$$

But what exactly is this term \(\frac{de^{'}_i}{dt}\). We know that the rotating reference is rotating with some angular velocity \(\frac{d\Omega}{dt}\). So this term must involve angular velocity. We can expand this term in terms of its basis vectors to get that:
$$ \left(\frac{de^{'}_i}{dt}\right)_i = \left(\frac{d\Omega}{dt}\right)_{ij}e^{'}_j \hspace{20 mm} \text{eq.3} $$  Here I have merely written the ith component of the \(\frac{de^{'}_i}{dt}\) vector.  Eq.3 you will notice is actually matrix multiplication. At this juncture we could introduce levi-civita symbols and define the \( \left(\frac{d\Omega}{dt}\right)_{ij}\) to be \( \left(\frac{d\Omega}{dt}\right)_{k}\) in this manner \( \left(\frac{d\Omega}{dt}\right)_{ij}\)=\(\epsilon_{ijk}\left(\frac{d\Omega}{dt}\right)_{k}\). Hence,
$$ \left(\frac{de^{'}_i}{dt}\right)_i = \epsilon_{ijk}\left(\frac{d\Omega}{dt}\right)_{k}e^{'}_j \hspace{20mm} \text{eq.4}  $$ This is merely:
$$ \left(\frac{de^{'}_j}{dt}\right)_j =  \left(\frac{d\Omega}{dt}\right) \times e^{'}_j \hspace{20mm} \text{eq.5}$$ Let's define \(\frac{d\Omega}{dt}\) to be \(\omega\).  Consequently, eq.2 now becomes $$ \left(\frac{d\vec{r}}{dt}\right)_o = \left(\frac{d\vec{r}}{dt}\right)_q + \omega \times \vec{r} \hspace{20mm} \text{eq.6} $$

We are now finally in a position to formulate the equations of motion for non-inertial reference frames. Remember L = T - U where T is kinetic energy and U is potential energy. The Lagrangian in the non-inertial reference frame can then be written as
\begin{equation}
 L=\frac{m}{2}\dot{r}_o ^2 - U\hspace{10mm} \text{where}\hspace{10mm} \dot{r_o} = \dot{r_q} + \omega \times r
\end{equation}
\begin{eqnarray}
 \dot{r_o}\cdot \dot{r_o} =& \left(\dot{r_q} + \omega \times r\right) \cdot   \left(\dot{r_q} + \omega \times r\right) \\
  =&\dot{r_q}^2 + 2\dot{r_q}\cdot \left(\omega \times r \right) + \left(\omega \times r \right) \cdot \left(\omega \times r \right)
\end{eqnarray}
\begin{eqnarray}
\left(\omega \times r\right)_i \cdot \left(\omega \times r \right)_i =& \epsilon_{ijk} \omega_j r_k \epsilon_{ilm}\omega_l r_m\\
 =&\epsilon_{jkl}\epsilon_{ilm}  \omega_j r_k \omega_l r_m\\
 =&\left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}\right) \omega_j r_k \omega_l r_m\\
  = &\delta_{jl}\delta_{km}\omega_j r_k \omega_l r_m - \delta_{im}\delta_{kl} \omega_{i}r_k \omega_l r_m\\
=& \omega_j\omega_j r_k r_k - \omega_j r_j r_k\omega_k \hspace{20mm} \text{eq.7}\\
\end{eqnarray}
\begin{equation}
\dot{r_o}\cdot \dot{r_o} = \dot{r_q}^2 + 2\dot{r_q}\cdot \left(\omega \times r \right) + \left(\omega r\right)^2 - \left(\omega \cdot r \right)^2 \hspace{20mm} \text{eq.8}
\end{equation}

Armed with eq.8 we can now go to the Euler-Lagrange Equation and deal with the derivative.
\begin{equation}
 \frac{\partial L}{\partial \dot{r}_q} =  m \left(\dot{r}_q + \omega \times r \right)
\end{equation}

To get the Lagrangian with respect to \(r_q\) is a bit more tricky. The hardest terms to deal with are clearly ( at least for me ) the last two terms in eq.8. So instead what I do is look at eq.7 and take the derivative with respect to \(r_k\) . This gives  \(2\omega_j\left(\omega_j r_k - r_j \omega_k\right)\). If you are familiar with levi-civita symbols then you can see it is simply a component of \(-\omega \times \left(\omega \times r_q \right)\). To be more explicit:
\begin{eqnarray}
 2 \omega_j\left(\omega_j r_k - r_j \omega_k\right) =& 2\omega_j \epsilon_{ijk}\omega_j r_k \\
 =& 2 \epsilon_{kji} \omega_j \left( \omega \times r \right)_i \\
 =& -2 \omega \times \left(\omega \times r \right)
\end{eqnarray}
 We finally have that
\begin{equation}
\frac{\partial L}{\partial r_q} = m \left(-\omega \times \dot{r}  - \omega \times
\left( \omega \times r \right) \right) - \nabla U .
\end{equation}

The next post will be looking at the angular velocity matrix that we used in eq.3. It turns out that it is anti symmetric. I will try to find out what this means about angular velocity and the relationship between non-inertial and inertial reference frames.