Answer for HW_12

1 三维各向同性谐振子：球坐标

已知基本关系：

L = \frac{m}{2} ({\dot{r}}^{2} + r^{2} {\dot{θ}}^{2} + r^{2} {\dot{φ}}^{2} \sin^{2} θ) - V (r, θ, φ), V (r) = \frac{1}{2} k r^{2}

\begin{aligned} p_{r} & = \frac{\partial L}{\partial \dot{r}} = m \dot{r} \\ p_{θ} & = \frac{\partial L}{\partial \dot{θ}} = m r^{2} \dot{θ} \\ p_{φ} & = \frac{\partial L}{\partial \dot{φ}} = m r^{2} \dot{φ} \sin^{2} θ \end{aligned}

H = \dot{r} p_{r} + \dot{θ} p_{θ} + \dot{φ} p_{φ} - L = \frac{1}{2 m} (p_{r}^{2} + \frac{p_{θ}^{2}}{r^{2}} + \frac{p_{φ}^{2}}{r^{2} \sin^{2} θ}) + V (r, θ, φ)

于是 $H - J$ 方程为

\frac{1}{2 m} {(\frac{\partial S_{0}}{\partial r})}^{2} + \frac{1}{2} k r^{2} + \frac{1}{2 m r^{2}} {(\frac{\partial S_{0}}{\partial θ})}^{2} + \frac{1}{2 m r^{2} \sin^{2} θ} {(\frac{\partial S_{0}}{\partial φ})}^{2} = E

$φ$ 是循环坐标，所以记 $\frac{\partial S_{0}}{\partial φ}$ 为一常数 $p_{φ}$ ，分离变量得到

{(\frac{\partial S_{0}}{\partial θ})}^{2} + \frac{p_{φ}^{2}}{\sin^{2} θ} = 2 m r^{2} {E - \frac{1}{2 m} {(\frac{\partial S_{0}}{\partial r})}^{2} - \frac{1}{2} k r^{2}}

可以令 $S_{0} = p_{φ} φ + S_{1} (r) + S_{2} (θ)$ ，代入上式，有

{(\frac{d S_{2}}{d θ})}^{2} + \frac{p_{φ}^{2}}{\sin^{2} θ} = 2 m r^{2} {E - \frac{1}{2 m} {(\frac{d S_{1}}{d r})}^{2} - \frac{1}{2} k r^{2}}

上式两边分别为 $θ$ 和 $r$ 的函数，于是有

{(\frac{d S_{2}}{d θ})}^{2} + \frac{p_{φ}^{2}}{\sin^{2} θ} = β

\frac{1}{2 m} {(\frac{d S_{1}}{d r})}^{2} + \frac{1}{2} k r^{2} + \frac{β}{2 m r^{2}} = E

由以上两式可得

\begin{aligned} S_{1} = \int \sqrt{2 m [E - \frac{1}{2} k r^{2} - \frac{β}{2 m r^{2}}]} d r \\ S_{2} = \int \sqrt{β - \frac{p_{φ}^{2}}{\sin^{2} θ}} d θ \end{aligned}

\Rightarrow S = - E t + p_{φ} φ + \int \sqrt{2 m [E - \frac{1}{2} k r^{2} - \frac{β}{2 m r^{2}}]} d r + \int \sqrt{β - \frac{p_{φ}^{2}}{\sin^{2} θ}} d θ

于是可以给出三个新的运动积分

ξ = - t + \frac{\partial S_{0}}{\partial E} = - t + \int \dots ⟹ r = r (t)

Q_{β} = \frac{\partial S_{0}}{\partial β} = \int \dots ⟹ f (r (t), θ) = 0 ⟹ θ (t)

Q_{φ} = \frac{\partial S_{0}}{\partial p_{φ}} = \int \dots ⟹ g (φ, θ (t)) = 0 ⟹ φ (t)

三个坐标对时间的依赖关系都在以上三个积分表达式当中。特别的，如果选取 $θ \equiv \frac{π}{2}$ ，上述结果会退化在 x-y 平面上的椭圆的极坐标形式，并且坐标原点位于椭圆正中心。

类似地，已知基本关系：

L = \frac{m}{2} ({\dot{x}}^{2} + {\dot{y}}^{2} + {\dot{z}}^{2}) - \frac{1}{2} k r^{2}

\vec{p} = \frac{\partial L}{\partial \dot{\vec{r}}} = m \dot{\vec{r}}

H = \frac{p^{2}}{2 m} + \frac{1}{2} k r^{2}

于是 $H - J$ 方程为

\frac{1}{2 m} [{(\frac{\partial S_{0}}{\partial x})}^{2} + {(\frac{\partial S_{0}}{\partial y})}^{2} + {(\frac{\partial S_{0}}{\partial z})}^{2}] + \frac{1}{2} k (x^{2} + y^{2} + z^{2}) = E

可以令 $S_{0} = S_{x} (x) + S_{y} (y) + S_{z} (z)$ ，代入上式，有

[\frac{1}{2 m} {(\frac{d S_{x}}{d x})}^{2} + \frac{1}{2} k x^{2}] + [\frac{1}{2 m} {(\frac{d S_{y}}{d y})}^{2} + \frac{1}{2} k y^{2}] + [\frac{1}{2 m} {(\frac{d S_{z}}{d z})}^{2} + \frac{1}{2} k z^{2}] = E

于是有

[\frac{1}{2 m} {(\frac{d S_{i}}{d x_{i}})}^{2} + \frac{1}{2} k x_{i}^{2}] = E_{i}, i = x, y, z

积分得到

S_{i} (x_{i}) = \int \sqrt{2 m E_{i} - m k x_{i}^{2}} d x_{i}

选取 $E_{x}, E_{y}, E_{z}$ 作为独立运动积分得到

S = - E t + S_{0} = - (E_{x} + E_{y} + E_{z}) t + \int \sqrt{2 m E_{x} - m k x^{2}} d x + \int \sqrt{2 m E_{y} - m k y^{2}} d y + \int \sqrt{2 m E_{z} - m k z^{2}} d z

新的广义坐标(假设 $ω = \sqrt{k / m}, A_{i}^{2} = \frac{2 E_{i}}{m ω^{2}}$ )：

\begin{aligned} Q_{x} = \frac{\partial S}{\partial E_{x}} & = - t + \int \frac{m}{\sqrt{2 m E_{x} - m k x^{2}}} d x \\ = - t + \int \frac{1}{\sqrt{2 E_{x} / m - ω^{2} x^{2}}} d x \\ = - t + \frac{1}{ω} \arcsin (\frac{x}{A_{x}}) \end{aligned}

得到简谐振动方程

x = A_{x} \sin (ω t + Q_{x}) = \sqrt{\frac{2 E_{x}}{m ω^{2}}} \sin (ω t + Q_{x}) = \sqrt{\frac{2 E_{x}}{k}} \sin (\sqrt{k / m} t + Q_{x})

另外两个方向同理。如果你选取 $E$ 作为其中一个独立运动积分，比如 $E_{x}, E_{y}, E$ ，将会让计算变得复杂，但是结论是一致的，这种选择会给出：

\begin{aligned} Q_{x} = \frac{\partial S_{0}}{\partial E_{x}} = \int \frac{m}{\sqrt{2 m E_{x} - m k x^{2}}} d x - \int \frac{m}{\sqrt{2 m (E - E_{x} - E_{y}) - m k z^{2}}} d z \\ = \frac{1}{ω} \arcsin (\frac{x}{A_{x}}) - \frac{1}{ω} \arcsin (\frac{z}{A_{z}}) \\ Q_{y} = \frac{\partial S_{0}}{\partial E_{y}} = \frac{1}{ω} \arcsin (\frac{y}{A_{y}}) - \frac{1}{ω} \arcsin (\frac{z}{A_{z}}) \\ Q_{E} = - t + \frac{\partial S_{0}}{\partial E} = - t + \int \frac{m}{\sqrt{2 m E_{z} - m k z^{2}}} d z \\ = - t + \frac{1}{ω} \arcsin (\frac{x}{A_{x}}) \end{aligned}