角动量（有时称为动量矩或旋转动量）是线性动量的旋转类比。它是一个重要的物理量，因为它是一个守恒量——一个封闭系统的总角动量保持不变。
中心势由 $U(r)=\frac{1}{2} k r^2$ 给出，它表示一个三维谐振子。该系统在球坐标系中的拉格朗日量为：
$$ L=\frac{1}{2} m\left(\dot{r}^2+r^2 \dot{\theta}^2+r^2 \sin ^2 \theta \dot{\phi}^2\right)-\frac{1}{2} k r^2 $$
由于势是中心的，角动量是守恒的。设 $\lambda=m r^2 \dot{\phi}$ 为角动量的大小。能量由下式给出：
$$ E=\frac{1}{2} m\left(\dot{r}^2+r^2 \dot{\theta}^2+r^2 \sin ^2 \theta \dot{\phi}^2\right)+\frac{1}{2} k r^2 $$
为了简单起见，我们考虑赤道平面内的运动 ($\theta=\pi / 2, \dot{\theta}=0$)。那么：
$$ E=\frac{1}{2} m\left(\dot{r}^2+r^2 \dot{\phi}^2\right)+\frac{1}{2} k r^2 $$
代入 $\dot{\phi}=\lambda /\left(m r^2\right)$，我们得到：
$$ E=\frac{1}{2} m \dot{r}^2+\frac{\lambda^2}{2 m r^2}+\frac{1}{2} k r^2 $$
求解 $\dot{r}$：
$$ \dot{r}=\sqrt{\frac{2}{m}\left(E-\frac{\lambda^2}{2 m r^2}-\frac{1}{2} k r^2\right)} $$
这与给定的方程 (1 ) 一致。
现在，让我们求解这个微分方程。
$$ \begin{gathered} \frac{d r}{d t}=\sqrt{\frac{2}{m}\left(E-\frac{\lambda^2}{2 m r^2}-\frac{1}{2} k r^2\right)} \\ \frac{d r}{\sqrt{\frac{2}{m}\left(E-\frac{\lambda^2}{2 m r^2}-\frac{1}{2} k r^2\right)}}=d t \end{gathered} $$
设 $\omega=\sqrt{k / m}$，那么 $k=m \omega^2$。
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$$ \begin{gathered} E=\frac{1}{2} m \dot{r}^2+\frac{\lambda^2}{2 m r^2}+\frac{1}{2} m \omega^2 r^2 \\ \dot{r}^2=\frac{2 E}{m}-\frac{\lambda^2}{m^2 r^2}-\omega^2 r^2 \end{gathered} $$
设 $r^2=u$。那么 $2 r \dot{r}=\dot{u}$，并且 $\dot{r}=\dot{u} /(2 \sqrt{u})$。
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