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.
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*}