Laplace-Runge-Lenz Vector

\(\vec{F}_e = \hat{r}f(r)\). Let’s compute \(\dot{\vec{p}}\times\vec{L} = \cdots = -mf(r)r^2\left(\frac{\dot{\vec{r}}}{r} - \frac{\vec{r}\dot{r}}{r^2}\right)\). \(\frac{d}{dt}(\vec{p}\times\vec{L}) = \dot{\vec{p}}\times\vec{L} + \vec{p}\times\dot{\vec{L}} = \dot{\vec{p}}\times\vec{L} = -mf(r)r^2\left(\frac{\dot{\vec{r}}}{r}-\frac{\vec{r}\dot{r}}{r^2}\right)\).

Assume \(f(r) = -\frac{k}{r^2}\) with \(k\) positive. Then, \(\frac{d}{dt}\left(\vec{p}\times\vec{L}\right) = mk\frac{d}{dt}\left(\frac{\vec{r}}{r}\right)\). So, \(\frac{d}{dt}\left(\vec{p}\times\vec{L} - mk\frac{\vec{r}}{r}\right)=0\), hence a conserved quantity.

The vector is then horizontal. Consider, \(\vec{A}\cdot\vec{r} = Ar\cos\varphi\).

Let \(\vec{A} = \vec{p}\times\vec{L} - mk\frac{\vec{r}}{r} = \vec{\mathbb{C}}\). Then, \(\vec{A}\cdot\vec{r} = \vec{r}\cdot(\vec{p}\times\vec{L}) - mk\frac{\vec{r}\cdot\vec{r}}{r} = \vec{L}\cdot(\vec{r}\times\vec{p}) - mkr = \ell^2 - mkr\). Then, \(r = \frac{\ell^2/(mk)}{1 + \frac{A}{mk}\cos\varphi}\). To determine \(A\), we compute \(\vec{A}\cdot\vec{A} = (\vec{p}\times\vec{L} - mk\vec{r}/r)^2 = p^2\ell^2 - 2mk\frac{\ell^2}{r} + m^2k^2\) So, \(\frac{A^2}{2m\ell^2} = \frac{p^2}{2m} - \frac{k}{r} + \frac{mk^2}{2\ell^2} = E + \frac{mk^2}{2\ell^2}\) Hence, \(A = \sqrt{2m\ell^2E+m^2k^2}\). So, \(r = \frac{\ell^2/(mk)}{1 + \sqrt{1 + \frac{2E\ell^2}{mk^2}}\cos\varphi}\)