分析力学概览

说明：这篇笔记是从下往上建立分析力学的过程，而朗道力学笔记则是从上往下产生力学体系的过程。

参考：(1)[复习]分析力学：哈密顿正则方程 - 知乎 (zhihu.com)；(2)结构力学程序算法理论基础(二)————理想约束l93919861的博客-CSDN博客理想约束的定义是什么；(3)勒让德变换 Legendre Transformation - 知乎 (zhihu.com)。

分析力学概览1.自由度、广义坐标、约束2.虚功原理、广义力3.拉格朗日方程4.对于拉格朗日方程的一点说明5.最小作用量原理（哈密顿原理）6.哈密顿正则方程

1.自由度、广义坐标、约束

$n$ $3n$ 自由度 $k$ 约束 $3n-k$ 个。每一个自由度对应一个广义坐标。

$i$ $\delta q_i$ $j$ $\delta \vec{r_j}$ ；这些位移都是微元），约束反力虚功和为零，则称这套约束为理想约束。对应的表达式为：

0 = \sum_{j = 1}^{n} {\vec{R}}_{j} \cdot δ {\vec{r}}_{j}

理想约束的概念是从实际约束中抽象得来，它反映了相当广泛的一些实际约束的主要性质。⁽²⁾

理想约束可以看作是没有耗散力的约束。例如质点被限制在光滑曲面内运动，它的虚位移也只可沿曲面。它在曲面内无论虚位移朝向哪个方向，约束反力都是垂直与位移，没有虚功。轻硬杆、不可伸长的线（似非完全约束）亦同。对于用轻杆连接的两个质点，它们的约束反力总沿杆反向，而它们的沿杆速度相同，故而亦为理想约束。

2.虚功原理、广义力

$\vec{F}$ $\vec{R}$ 是约束反力）：

\forall j, {\vec{F}}_{j} + {\vec{R}}_{j} = 0

故对应任意一组虚位移：

\sum_{j} ({\vec{F}}_{j} + {\vec{R}}_{j}) \cdot δ {\vec{r}}_{j} = 0

则对于理想约束，由上节中的定义知，有：

\sum_{j} {\vec{F}}_{j} \cdot δ {\vec{r}}_{j} = 0

此即虚功原理。在广义坐标下

δ {\vec{r}}_{j} = \sum_{i = 1}^{3 n - k} \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} δ q_{i}

带入可得

\begin{aligned} 0 = & \sum_{j} {\vec{F}}_{j} \cdot δ {\vec{r}}_{j} \\ = & \sum_{j} {\vec{F}}_{j} \cdot \sum_{i = 1}^{3 n - k} \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} δ q_{i} \\ = & \sum_{i} (\sum_{j} {\vec{F}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}}) δ q_{i} \\ = & \sum_{i} Q_{i} δ q_{i} \end{aligned}

$Q_i\equiv\sum_j \vec{F}_j\frac{\partial \vec{r}_j}{\partial q_i}$ 为广义力。由于广义位置变量是完全相互独立的，而且虚功原理要求虚位移的选取具有任意性，上式导致：

\forall i, Q_{i} = 0

即：平衡条件下，各广义力须均为零。

3.拉格朗日方程

对于运动中的系统，有

\begin{matrix} \forall j, {\vec{F}}_{j} + {\vec{R}}_{j} = m_{j} {\ddot{\vec{r}}}_{j} \\ i . e . ({\vec{F}}_{j} - m_{j} {\ddot{\vec{r}}}_{j}) + {\vec{R}}_{j} = 0 \end{matrix}

用这条等式替换上节中的虚功原理

\begin{matrix} (1) & \begin{matrix} 0 = \sum_{i} (\sum_{j} ({\vec{F}}_{j} - m_{j} {\ddot{\vec{r}}}_{j}) \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}}) δ q_{i} \\ i . e . 0 = \sum_{j} ({\vec{F}}_{j} - m_{j} {\ddot{\vec{r}}}_{j}) \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} \end{matrix} \end{matrix}

对于保守系统，

\begin{aligned} {\vec{F}}_{j} & = - (\frac{\partial V}{\partial x_{j}}, \frac{\partial V}{\partial y_{j}}, \frac{\partial V}{\partial z_{j}}) \\ \Rightarrow {\vec{F}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} & = - (\frac{\partial V}{\partial x_{j}}, \frac{\partial V}{\partial y_{j}}, \frac{\partial V}{\partial z_{j}}) \cdot (\frac{\partial x_{j}}{\partial q_{i}}, \frac{\partial y_{j}}{\partial q_{i}}, \frac{\partial z_{j}}{\partial q_{i}}) \\ = - \frac{\partial V}{\partial x_{j}} \frac{\partial x_{j}}{\partial q_{i}} - \frac{\partial V}{\partial y_{j}} \frac{\partial y_{j}}{\partial q_{i}} - \frac{\partial V}{\partial z_{j}} \frac{\partial z_{j}}{\partial q_{i}} \end{aligned}

则上(1)式右端第一项

\begin{aligned} \sum_{j} {\vec{F}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} \\ = & - \sum_{j} (\frac{\partial V}{\partial x_{j}} \frac{\partial x_{j}}{\partial q_{i}} + \frac{\partial V}{\partial y_{j}} \frac{\partial y_{j}}{\partial q_{i}} + \frac{\partial V}{\partial z_{j}} \frac{\partial z_{j}}{\partial q_{i}}) \\ (2) & = & - \frac{\partial V}{\partial q_{i}} \end{aligned}

$-m_j\sum_j\ddot{\vec{r}}_j\cdot\frac{\partial\vec{r}_j}{\partial q_i}$ ，有：

\begin{aligned} {\ddot{\vec{r}}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} & = \frac{d}{d t} ({\dot{\vec{r}}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}}) - {\dot{\vec{r}}}_{j} \cdot \frac{d}{d t} \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} \\ (3) & = \frac{d}{d t} ({\dot{\vec{r}}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}}) - {\dot{\vec{r}}}_{j} \cdot \frac{\partial {\dot{\vec{r}}}_{j}}{\partial q_{i}} \end{aligned}

$\frac{\partial\vec{r}_j}{\partial q_i}=\frac{\partial\dot{\vec{r}}_j}{\partial \dot{q_i}}$ :

$\vec{r}_j$ $q_1,...,q_{3n-k}$ $t$ $\dot{\vec{r}}_j$ $\dot{q}_1,...,\dot{q}_{3n-k}$ $t$ 的函数。⁽¹⁾

$\vec{r}_j$ $\vec{r}_j$ $q_1,...,q_{3n-k}$ 的函数，有：

\begin{matrix} {\vec{r}}_{j} = {\vec{r}}_{j} (q_{i}, . . . q_{3 n - k}) \\ {\dot{\vec{r}}}_{j} = \sum_{i} \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} {\dot{q}}_{i} \end{matrix}

$\frac{\partial \vec{r}_j}{\partial q_i}$ $\dot{q}_1,...,\dot{q}_{3n-k}$ ，故有：

\frac{\partial {\dot{\vec{r}}}_{j}}{\partial \dot{q_{i}}} = \frac{\partial {\vec{r}}_{j}}{\partial q_{i}}

将其代入(3)式：

\begin{aligned} {\ddot{\vec{r}}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} & = \frac{d}{d t} ({\dot{\vec{r}}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}}) - {\dot{\vec{r}}}_{j} \cdot \frac{\partial {\dot{\vec{r}}}_{j}}{\partial q_{i}} \\ = \frac{d}{d t} ({\dot{\vec{r}}}_{j} \cdot \frac{\partial {\dot{\vec{r}}}_{j}}{\partial \dot{q_{i}}}) - {\dot{\vec{r}}}_{j} \cdot \frac{\partial {\dot{\vec{r}}}_{j}}{\partial q_{i}} \\ (4) & = \frac{1}{2} (\frac{d}{d t} \frac{\partial {\dot{\vec{r}}}_{j}^{2}}{\partial \dot{q_{i}}} - \frac{\partial {\dot{\vec{r}}}_{j}^{2}}{\partial q_{i}}) \end{aligned}

将(2)和(4)式代入(1)式：

\begin{aligned} 0 & = \sum_{j} ({\vec{F}}_{j} - m_{j} {\ddot{\vec{r}}}_{j}) \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} \\ = \sum_{j} ({\vec{F}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}} - m_{j} {\ddot{\vec{r}}}_{j} \cdot \frac{\partial {\vec{r}}_{j}}{\partial q_{i}}) \\ = - \frac{\partial V}{\partial q_{i}} - \sum_{j} \frac{m_{j}}{2} (\frac{d}{d t} \frac{\partial {\dot{\vec{r}}}_{j}^{2}}{\partial \dot{q_{i}}} - \frac{\partial {\dot{\vec{r}}}_{j}^{2}}{\partial q_{i}}) \end{aligned}

$T=\sum_j\frac{1}{2}m_j\dot{\vec{r}}_j^{~2}$ ，代入上式有：

\begin{matrix} (5) & 0 = - \frac{\partial V}{\partial q_{i}} - (\frac{d}{d t} \frac{\partial T}{\partial \dot{q_{i}}} - \frac{\partial T}{\partial q_{i}}) \end{matrix}

$L=T-V$ $V$ $\dot{q}_1,...,\dot{q}_{3n-k}$ ，可化简上式为：

0 = \frac{\partial L}{\partial q_{i}} - \frac{d}{d t} \frac{\partial L}{\partial \dot{q_{i}}}

上式即为拉格朗日方程。

4.对于拉格朗日方程的一点说明

拉格朗日方程可以看作是牛顿第二定律在广义坐标空间里的推广。由上可得： $Q_i=-\frac{\partial V}{\partial q_i}$ ，代入(5)式有：

Q_{i} = \frac{d}{d t} \frac{\partial T}{\partial \dot{q_{i}}} - \frac{\partial T}{\partial q_{i}}

在笛卡尔空间里， $Q_i\rightarrow F_i, T\rightarrow \sum_i\frac{1}{2}m_i \dot{q}_i^2$ ，很容易导出牛顿第二定律。在广义坐标空间里无法直接套用牛顿第二定律形式的原因就在于动能不能被表达为 $\sum_i\frac{1}{2}m_i \dot{q}_i^2$ 。

5.最小作用量原理（哈密顿原理）

如果我们令作用量：

S = \int_{t_{1}}^{t_{2}} L d t

变分法概览 $L(t_1),L(t_2)$ $S$ 取极值时，有：

0 = \frac{\partial L}{\partial q_{i}} - \frac{d}{d t} \frac{\partial L}{\partial {\dot{q}}_{i}}

这就是拉格朗日方程。所以说：

力学系统运动规律的最一般表述由最小作用量原理给出。（朗道力学）

6.哈密顿正则方程

哈密顿函数勒让德变化 $p_i=\frac{\partial L}{\partial \dot{q}_i}$ $\dot{q}_i$ 为独立变量。

L (q_{i}, {\dot{q}}_{i}, t) \to H (q_{i}, p_{i}, t)

由此可知（见勒让德变换概览）：

H (q_{i}, p_{i}, t) = \sum_{i} {\dot{q}}_{i} p_{i} - L (q_{i}, {\dot{q}}_{i}, t)

对上式求微分：

d H = \sum_{i} ({\dot{q}}_{i} d p_{i} + p_{i} d {\dot{q}}_{i}) - d L

又：

\begin{aligned} d L & = \sum_{i} (\frac{\partial L}{\partial q_{i}} d q_{i} + \frac{\partial L}{\partial \dot{q_{i}}} d {\dot{q}}_{i}) + \frac{\partial L}{\partial t} \\ = \sum_{i} (\frac{\partial L}{\partial q_{i}} d q_{i} + p_{i} d {\dot{q}}_{i}) + \frac{\partial L}{\partial t} \end{aligned}

代入得

\begin{aligned} d H & = \sum_{i} ({\dot{q}}_{i} d p_{i} + p_{i} d {\dot{q}}_{i}) - \sum_{i} (\frac{\partial L}{\partial q_{i}} d q_{i} + p_{i} d {\dot{q}}_{i}) - \frac{\partial L}{\partial t} \\ = \sum_{i} ({\dot{q}}_{i} d p_{i} - \frac{\partial L}{\partial q_{i}} d q_{i}) - \frac{\partial L}{\partial t} \end{aligned}

$H$ $q_i,p_i,t$ 的函数，故有（再利用拉格朗日方程）：

\begin{matrix} \frac{\partial H}{\partial q_{i}} = - \frac{\partial L}{\partial q_{i}} = - \frac{d}{d t} \frac{\partial L}{\partial \dot{q_{i}}} = - \dot{p_{i}} \\ \frac{\partial H}{\partial p_{i}} = {\dot{q}}_{i} \\ \frac{\partial H}{\partial t} = - \frac{\partial L}{\partial t} \end{matrix}

前两式被称为哈密顿正则方程。