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分析力学笔记

         

预备知识 哈密顿正则方程

   拉格朗日方程为

\begin{equation} \frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial L}{\partial \dot q_i} = \frac{\partial L}{\partial q_i} \qquad L(q,\dot q, t) = T - V \end{equation}

   正则动量(canonical momentum) $p_i = \partial L/\partial \dot q_i $, 广义力(generalized force) $ \partial L/\partial q_i $, 拉氏方程就是广义力与正则动量的牛顿第二定律. 对于任何广义坐标, 拉格朗日方程的形式不变.

   拉格朗日变换(略) 后, 得到哈密顿正则(canonical)方程为

\begin{equation} \dot q_i = \frac{\partial H}{\partial p_i} \qquad \dot p_i = - \frac{\partial H}{\partial q_i} \end{equation}
其中 $H(p,q) = T + V$.

泊松括号

   对任意物理量 $\omega (q,p)$, 都有

\begin{equation} \dot \omega = \sum_i \left[ \frac{\partial \omega}{\partial q_i} \dot q_i + \frac{\partial \omega}{\partial p_i} \dot p_i \right] = \sum_i \left[ \frac{\partial \omega}{\partial q_i} \frac{\partial H}{\partial p_i} - \frac{\partial H}{\partial q_i} \frac{\partial \omega}{\partial p_i} \right] = \left\{\omega, H\right\} \end{equation}
量子力学中的对易算符对应泊松括号. 注意该物理量不能显含时间, 即不能是 $\omega (q,p,t)$. 所以若泊松括号消失, 则该物理量守恒! 当显含时间时
\begin{equation} \dot \omega = \left\{\omega, H\right\} + \frac{\partial \omega}{\partial t} \end{equation}
对应量子力学中的算符平均值演化方程. 注意若调换泊松括号里面的物理量, 结果取相反数.

坐标变换

   若变换到另一套广义坐标 $q'(q)$

\begin{equation} \dot q'_i = \sum_j \frac{\partial q'_i}{\partial q_j} \dot q_j \qquad \dot q_i = \sum_j \frac{\partial q_i}{\partial q'_j} \dot q'_j \end{equation}
拉格朗日量是系统的状态量, 所以 $L(q',\dot q', t) = L[q(q'),\dot q(q',\dot q'), t]$, 所以
\begin{equation} p'_i = \frac{\partial L}{\partial \dot q'_i} = \sum_k \frac{\partial L}{\partial \dot q_k} \frac{\partial \dot q_k}{\partial \dot q'_i} = \sum_k \frac{\partial q_k}{\partial q'_i} p_k \end{equation}
这就从坐标变换推出了动量变换. 对于任何广义坐标以及对应的正则动量, 哈密顿方程的形式不变(因为拉格朗日方程的形式不变, 哈密顿是由拉格朗日推出来的). 但是还有其他情况也不变, 所有使正则方程成立的坐标叫做正则坐标(canonical coordinates). 下面推导判断正则坐标的一般条件.

   对于不显含时的物理量有

\begin{equation} \dot \omega = \left\{\omega, H\right\} = \sum_i \left[ \frac{\partial \omega}{\partial q_i} \frac{\partial H}{\partial p_i} - \frac{\partial H}{\partial q_i} \frac{\partial \omega}{\partial p_i} \right] \end{equation}
现在若把 $H$ 看成是 $H[q'(q,p),p'(q,p)]$,
\begin{equation} \frac{\partial H}{\partial p_i} = \sum_k \frac{\partial H}{\partial q'_k} \frac{\partial q'_k}{\partial p_i} + \frac{\partial H}{\partial p'_k} \frac{\partial p'_k}{\partial p_i} \end{equation}
\begin{equation} \frac{\partial H}{\partial q_i} = \sum_k \frac{\partial H}{\partial q'_k} \frac{\partial q'_k}{\partial q_i} + \frac{\partial H}{\partial p'_k} \frac{\partial p'_k}{\partial q_i} \end{equation}

   代入得, 并对 $H$ 的偏微分合并同类项得

\begin{equation} \dot \omega = \left\{\omega, H\right\} = \sum_k \left[ \frac{\partial H}{\partial q'_k} \left\{\omega, q'_k\right\} + \frac{\partial H}{\partial p'_k} \left\{\omega, p'_k\right\} \right] \end{equation}
注意泊松括号是对 $q,p$ 进行偏微分, 记为 $\{ {}\}_{q,p}$. 分别代入 $\omega = q'_i, p'_i$, 得到转换坐标后的哈密顿方程的一般形式. 为了保持正则方程的形式, 必须要求
\begin{equation} \left\{q'_i, q'_k\right\} _{q,p} = 0 = \left\{p'_i, p'_k\right\} _{q,p} \end{equation}
\begin{equation} \left\{q'_i, p'_k\right\} _{q,p} = \delta_{ik} \end{equation}
这就是判断正则变换的一般条件.

   可以证明, 用任何正则坐标作为泊松括号的角标, 其值都不变. 下面是证明

\begin{equation} \left\{a, b\right\} _{q,p} = \sum_i \left( \frac{\partial a}{\partial q_i} \frac{\partial b}{\partial p_i} - \frac{\partial b}{\partial q_i} \frac{\partial a}{\partial p_i} \right) \end{equation}
其中
\begin{equation} \frac{\partial a}{\partial q_i} \frac{\partial b}{\partial p_i} = \sum_j \left( \frac{\partial a}{\partial q'_j} \frac{\partial q'_j}{\partial q_i} + \frac{\partial a}{\partial p'_j} \frac{\partial p'_j}{\partial q_i} \right) \sum_k \left( \frac{\partial b}{\partial q'_k} \frac{\partial q'_k}{\partial p_i} + \frac{\partial b}{\partial p'_k} \frac{\partial p'_k}{\partial p_i} \right) \end{equation}
\begin{equation} \frac{\partial b}{\partial q_i} \frac{\partial a}{\partial p_i} = \sum_k \left( \frac{\partial b}{\partial q'_k} \frac{\partial q'_k}{\partial q_i} + \frac{\partial b}{\partial p'_k} \frac{\partial p'_k}{\partial q_i} \right) \sum_j \left( \frac{\partial a}{\partial q'_j} \frac{\partial q'_j}{\partial p_i} + \frac{\partial a}{\partial p'_j} \frac{\partial p'_j}{\partial p_i} \right) \end{equation}

   现在我们要得到 $ \left\{a, b\right\} _{q',p'} = \sum_i \left( \frac{\partial a}{\partial q'_i} \frac{\partial b}{\partial p'_i} - \frac{\partial b}{\partial q'_i} \frac{\partial a}{\partial p'_i} \right) $, 可以把上两式代入后对 $ \frac{\partial a}{\partial q'} \frac{\partial b}{\partial p'} $ 和 $ \frac{\partial b}{\partial q'_i} \frac{\partial a}{\partial p'_i} $ 合并同类项, 得

\begin{equation} \begin{aligned} \left\{a, b\right\} _{q,p} & = \sum_{jk} \frac{\partial a}{\partial q'_j} \frac{\partial b}{\partial p'_k} \sum_i \left( \frac{\partial q'_j}{\partial q_i} \frac{\partial p'_k}{\partial p_i} - \frac{\partial p'_k}{\partial q_i} \frac{\partial q'_j}{\partial p_i} \right) \\ & -\sum_{jk} \frac{\partial b}{\partial q'_k} \frac{\partial a}{\partial p'_j} \sum_i \left( \frac{\partial q'_k}{\partial q_i} \frac{\partial p'_j}{\partial p_i} - \frac{\partial p'_j}{\partial q_i} \frac{\partial q'_k}{\partial p_i} \right) \\ &= \sum_{jk} \frac{\partial a}{\partial q'_j} \frac{\partial b}{\partial p'_k} \left\{q'_j, p'_k\right\} _{q,p} - \sum_{jk} \frac{\partial b}{\partial q'_k} \frac{\partial a}{\partial p'_j} \left\{q'_k, p'_j\right\} _{q,p} \end{aligned} \end{equation}
代入正则坐标条件, 得
\begin{equation} \begin{aligned} \left\{a, b\right\} _{q,p} & = \sum_{jk} \frac{\partial a}{\partial q'_j} \frac{\partial b}{\partial p'_k} \delta_{jk} - \sum_{jk} \frac{\partial b}{\partial q'_k} \frac{\partial a}{\partial p'_j} \delta_{jk} = \sum_j \left( \frac{\partial a}{\partial q'_j} \frac{\partial b}{\partial p'_j} - \frac{\partial b}{\partial q'_j} \frac{\partial a}{\partial p'_j} \right) \\ &= \left\{a, b\right\} _{q',p'} \end{aligned} \end{equation}

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