# 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, $$r = \frac{\ell^2/(mk)}{1 + \sqrt{1 + \frac{2E\ell^2}{mk^2}}\cos\varphi}$$

Created: 2024-05-30 Thu 21:16

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