Mcgehee transformation, conversion to polar coordinates and blowing up the singularity I am looking for any reference on the above topics as I am struggling to convert the below to polar coordinates in phase space:
The system is:
  \begin{equation*}
  x''=\frac{-\mu x}{(\mu x^2 + y^2)^{3/2}}
  \end{equation*}
  \begin{equation*}
  y''=\frac{-y}{(\mu x^2 + y^2)^{3/2}}
  \end{equation*}
With $\mu>1$ a constant adding anisotropy to the otherwise newtonian system. 
The goal is to eventually blow up the singularity at the origin. 
 A: My physics savvy friend suggested I use a lagrangian to convert this to polar coordinates, which I believe works!
Taking advantage of independence of the lagrangian expression from coordinate
system and the useful identity, derived from Newtons second law
\begin{align}
  \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}&=\frac{\partial L}{\partial r}\\
  \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}&=
  \frac{\partial L}{\partial \theta}
\end{align}
We define the lagrangian as
\begin{align*}
  L(\dot{X},X)&=K(\dot{X})-U(X)
\end{align*}
Since mass is normalized to 1, the lagrangian identites and our potential energy
expression will give us our system in new coordinates.
\begin{align*}
  x=r\cos(\theta),\;y=r\sin(\theta)
  \Rightarrow -U(r,\theta)=
  &-\frac{-1}{\sqrt{\mu r^2\cos^2(\theta)+r^2\sin^2(\theta)}}\\
  &=\frac{1}{r\sqrt{\mu\cos^2(\theta)+\sin^2(\theta)}}
\end{align*}
And we have $K=\frac{1}{2}(\dot{r}+r^2\dot(\theta)^2)$ (this is $v_r^2+v_{\theta}^2$).
Then since $K$ is constant wrt. $\dot{r}$ and
$\dot{\theta}$, we have
\begin{align*}
  \frac{\partial L}{\partial r}=&\frac{\partial U}{\partial r}=
  -r^{-2}\frac{1}{\sqrt{\mu\cos^2(\theta)+\sin^2(\theta)}}\\
  &=\frac{-1}{r^2\sqrt{\mu\cos^2(\theta)+\sin^2(\theta)}}\\
  \frac{\partial L}{\partial \theta}=&\frac{\partial U}{\partial \theta}=
  \frac{1}{2}r^{-1}\frac{(2\cos(\theta)\sin(\theta)-2\mu \cos(\theta)\sin(\theta))}{
  (\mu\cos^{2}(\theta)+\sin^{2}(\theta))^{3/2}}\\
  &=\frac{(\sin(2\theta)-\mu\sin(2\theta))}{2r
  (\mu\cos^{2}(\theta)+\sin^{2}(\theta))^{3/2}}
\end{align*}
Then by $U$ being constant wrt. $\dot{r}$ and by the lagrangian identities, we derive
our new system
\begin{align*}
  \frac{d}{dt}\frac{\partial L}{\partial r}=&\frac{d}{dt}\dot{r}=\ddot{r}\\
  &\Rightarrow \ddot{r}=\frac{-1}{r^2\sqrt{\mu\cos^2(\theta)+\sin^2(\theta)}}\\
  \frac{d}{dt}\frac{\partial L}{\partial \theta}=&\frac{d}{dt}\dot{\theta}=
  \ddot{\theta}\\
  &\Rightarrow \ddot{\theta}=\frac{(\sin(2\theta)-\mu\sin(2\theta))}{2r
  (\mu\cos^{2}(\theta)+\sin^{2}(\theta))^{3/2}}
  \\
\end{align*}
