Answer for HW_10

1 正则变换

本题其实可以看作第二类正则变换，母函数是
$F_{2} (q_{α}, p_{α}^{'}, t) = \sum_{α} q_{α} p_{α}^{'} - f (q_{α}, t)$

采用爱因斯坦求和约定：

广义动量

\begin{matrix} (1) & p_{α}^{'} = \frac{\partial L^{'}}{\partial \dot{q_{α}}} = \frac{\partial L}{\partial \dot{q_{α}}} + \frac{\partial \dot{f}}{\partial \dot{q_{α}}} = p_{α} + \frac{\partial f}{\partial q_{α}} \end{matrix}

哈密顿量

\begin{aligned} H^{'} (p_{α}^{'}, q_{α}) & = p_{α}^{'} \dot{q_{α}} - L^{'} \\ = H + \frac{\partial f}{\partial q_{α}} \dot{q_{α}} - \frac{d f}{d t} \\ (2) & = H (p_{α}, q_{α}) - \frac{\partial f}{\partial t} \\ \overset{△}{=} H^{'} (p_{α} + \frac{\partial f}{\partial q_{α}}, q_{α}) \end{aligned}

哈密顿正则方程

证明 ${\dot{q}}_{α}^{'} = \frac{\partial H^{'}}{\partial p_{α}^{'}} ⟷ {\dot{q}}_{α} = \frac{\partial H}{\partial p_{α}}$ ：

\begin{aligned} {\dot{q}}_{α}^{'} = {\dot{q}}_{α} & = \frac{\partial H^{'}}{\partial p_{α}^{'}} |_{q} \\ = \frac{\partial H^{'}}{\partial p_{β}} \frac{\partial p_{β}}{\partial p_{α}^{'}} |_{q} = \frac{\partial H^{'}}{\partial p_{β}} δ_{α β} = \frac{\partial H^{'}}{\partial p_{α}} \\ 代 入 (2) 式 \to & = \frac{\partial H}{\partial p_{α}} - \frac{\partial}{\partial p_{α}} \frac{\partial f (q_{α}, t)}{\partial t} \\ (3) & = \frac{\partial H}{\partial p_{α}} \end{aligned}

证明 $- {\dot{p}}_{α}^{'} = \frac{\partial H^{'}}{\partial q_{α}^{'}} ⟷ - {\dot{p}}_{α} = \frac{\partial H}{\partial q_{α}}$ ：

注意到

\begin{matrix} (4) & p_{α}^{'} = p_{α} + \frac{\partial f}{\partial q_{α}} \Rightarrow \frac{\partial p_{β}^{'}}{\partial q_{α}} = \frac{\partial^{2} f}{\partial q_{α} \partial q_{β}} \end{matrix}

\begin{matrix} (5) & \frac{\partial H^{'} (p, q)}{\partial q_{α}} |_{p} = \frac{\partial H^{'} (p^{'}, q)}{\partial q_{α}} |_{p^{'}} + \frac{\partial H^{'} (p^{'}, q)}{\partial p_{β}^{'}} \frac{\partial p_{β}^{'}}{\partial q_{α}} |_{p} \end{matrix}

于是

\begin{aligned} - {\dot{p}}_{α}^{'} = - {\dot{p}}_{α} - \frac{d}{d t} \frac{\partial f}{\partial q_{α}} & = \frac{\partial H^{'}}{\partial q_{α}} |_{p^{'}} \\ 代 入 (5) 式 \to & = \frac{\partial H^{'}}{\partial q_{α}} |_{p} - \frac{\partial H^{'}}{\partial p_{β}^{'}} \frac{\partial p_{β}^{'}}{\partial q_{α}} |_{p} \\ 代 入 (2) 式 \to & = \frac{\partial H}{\partial q_{α}} - \frac{\partial}{\partial q_{α}} \frac{\partial f}{\partial t} - \frac{\partial H^{'}}{\partial p_{β}^{'}} \frac{\partial p_{β}^{'}}{\partial q_{α}} \\ 代 入 (3) (4) 式 \to & = \frac{\partial H}{\partial q_{α}} - \underset{= \frac{d}{d t} \frac{\partial f}{\partial q_{α}}}{\underset{⏟}{(\frac{\partial}{\partial q_{α}} \frac{\partial f}{\partial t} + {\dot{q}}_{β} \frac{\partial^{2} f}{\partial q_{α} \partial q_{β}})}} \end{aligned}

于是

\begin{matrix} (5) & \frac{\partial H}{\partial q_{α}} = - {\dot{p}}_{α} \end{matrix}

成立，反之同理。

2 球面摆

(1)

拉氏量

L (θ, ϕ; \dot{θ}, \dot{ϕ}) = T - V = \frac{1}{2} m l^{2} ({\dot{θ}}^{2} + \sin^{2} θ {\dot{ϕ}}^{2}) + m g l \cos θ

得到广义动量

p_{θ} = \frac{\partial L}{\partial \dot{θ}} = m l^{2} \dot{θ}, p_{ϕ} = \frac{\partial L}{\partial \dot{ϕ}} = m l^{2} \sin^{2} θ \dot{ϕ}

勒让德变换得到

\begin{matrix} (6) & H (θ, ϕ; p_{θ}, p_{ϕ}) = θ p_{θ} + ϕ p_{ϕ} - L = \frac{1}{2 m} (\frac{p_{θ}^{2}}{l^{2}} + \frac{p_{ϕ}^{2}}{l^{2} \sin^{2} θ}) - m g l \cos θ \end{matrix}

(2)

\begin{matrix} (7) & {\dot{p}}_{ϕ} = - \frac{\partial H}{\partial ϕ} = 0 \Rightarrow \sin^{2} θ \dot{ϕ} = const \end{matrix}

\begin{matrix} (8) & {\dot{p}}_{θ} = - \frac{\partial H}{\partial θ} = - \frac{p_{ϕ}^{2} \cos θ}{m l^{2} \sin^{3} θ} + m g l \sin θ \Rightarrow \ddot{θ} - {\dot{ϕ}}^{2} \cos θ \sin θ - \frac{g}{l} \sin θ = 0 \end{matrix}

(3)

自由度为 $s = 2$ ，对于可积系统，运动积分有 $s$ 个^[1]，即 $H$ ( $∵$ 不显含 $t$ ) 和 $p_{θ}$

参见：

[1] V. I. Arnold, Mathematical Methods of Classical Mechanics, chapter 10, §49 :

In order to integrate a system of $2 n$ ordinary differential equations, we must know $2 n$ first integrals. It turns out that if we are given a canonical system of differential equations, it is often sufficient to know only $n$ first integrals-each of them allows us to reduce the order of the system not just by one, but by two.

Two functions F 1 and F 2 on a symplectic manifold are in involution if their Poisson bracket is equal to zero. Liouville proved that if, in a system with n degrees of freedom (i.e., with a 2n-dimensional phase space), n independent first integrals in involution are known, then the system is integrable by quadratures.