Lagrange动力学——经典力学笔记2

Lagrange方程

Lagrange关系

对于一个包含\(n\)个质点的力学系统而言，如果我们定义\(s\)个广义坐标\(q_{\alpha},\ \alpha = 1, 2, \dots, s\)，那么有

引理（Lagrange关系）：

系统的直角坐标\(\boldsymbol{r}_{i},\ i = 1, 2, \dots, n\)和广义坐标\(q_{\alpha},\ \alpha = 1, 2, \dots, s\)之间满足如下两个关系 \[ \left\{\begin{aligned} & \dfrac{\partial \dot{\boldsymbol{r}}_{i}}{\partial \dot{q}_{\alpha}} = \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \\ & \dfrac{\partial}{\partial q_{\alpha}} \left( \dfrac{\mathrm{d} \boldsymbol{r}_{i}}{\mathrm{d} t} \right) = \dfrac{\mathrm{d} }{\mathrm{d} t} \left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \right) \\ \end{aligned}\right. \]

证明：

由于系统的直角坐标\(\boldsymbol{r}_{i},\ i = 1, 2, \dots, n\)和广义坐标\(q_{\alpha},\ \alpha = 1, 2, \dots, s\)之间满足 \[ \boldsymbol{r}_{i} = \boldsymbol{r}_{i} (q_{1}, q_{2}, \dots, q_{s};\ t),\quad i = 1, 2, \dots, n \] 因此对时间求导可得 \[ \dfrac{\mathrm{d} \boldsymbol{r}_{i}}{\mathrm{d} t} = \dfrac{\partial \boldsymbol{r}_{i}}{\partial t} + \sum\limits_{\alpha}^{s}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \dot{q}_{\alpha} \]

如果将\(\dot{\boldsymbol{r}}_{i}\)对\(\dot{q}_{\alpha}\)求偏导，那么 \[ \begin{aligned} \dfrac{\partial \dot{\boldsymbol{r}}_{i}}{\partial \dot{q}_{\alpha}} & = \dfrac{\partial}{\partial \dot{q}_{\alpha}} \left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial t} + \sum\limits_{\beta = 1}^{s}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\beta}} \dot{q}_{\beta} \right) \\ & = \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \end{aligned} \]

如果将\(\dot{\boldsymbol{r}}_{i}\)对于\(q_{\alpha}\)求偏导，那么 \[ \begin{aligned} \dfrac{\partial \dot{\boldsymbol{r}}_{i}}{\partial q_{\alpha}} & = \dfrac{\partial}{\partial q_{\alpha}} \left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial t} + \sum\limits_{\beta = 1}^{s} \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\beta}} \dot{q}_{\beta} \right) \\ & = \dfrac{\partial^{2} \boldsymbol{r}_{i}}{\partial q_{\alpha}\partial t} + \sum\limits_{\beta = 1}^{s}\dfrac{\partial^{2}\boldsymbol{r}_{i}}{\partial q_{\alpha}\partial q_{\beta}} \dot{q}_{\beta} \\ & = \dfrac{\partial}{\partial t}\left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \right) + \sum\limits_{\beta = 1}^{s} \dfrac{\partial}{\partial q_{\beta}}\left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \right) \dfrac{\partial q_{\beta}}{\partial t} \\ & = \dfrac{\mathrm{d}}{\mathrm{d} t} \left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \right) \end{aligned} \]

Lagrange方程

定理（Lagrange方程）：

对于只受完整约束，并包含\(n\)个质点的系统，如果我们用\(s\)个独立的广义坐标表示质点位置，那么有方程组 \[ \dfrac{\mathrm{d}}{\mathrm{d} t}\dfrac{\partial T}{\partial \dot{q}_{\alpha}} - \dfrac{\partial T}{\partial q_{\alpha}} = Q_{\alpha},\quad \alpha = 1, 2, \dots, s \]

证明：

d'Alembert原理的表达式为 \[ \begin{aligned} \sum\limits_{i = 1}^{n} \left( \boldsymbol{F}_{i} - m_{i}\ddot{\boldsymbol{r}}_{i} \right) \cdot \delta \boldsymbol{r}_{i} = \sum\limits_{i = 1}^{n}\left( \boldsymbol{F}_{i} \cdot\delta \boldsymbol{r}_{i} - m_{i}\ddot{\boldsymbol{r}}_{i}\right) = 0 \end{aligned} \] 我们将上式拆分为两部分计算。首先计算第一部分 \[ \begin{aligned} \sum\limits_{i = 1}^{n}\boldsymbol{F}_{i} \cdot \delta \boldsymbol{r}_{i} & = \sum\limits_{i = 1}^{n}\left( \boldsymbol{F}_{i} \cdot \sum\limits_{\alpha}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \delta q_{\alpha}\right) \\ & = \sum\limits_{\alpha = 1}^{s}\left( \sum\limits_{i = 1}^{n}\boldsymbol{F}_{i} \cdot\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}}\right) \delta q_{\alpha} \\ & = \sum\limits_{\alpha = 1}^{s}Q_{\alpha}\delta q_{\alpha} \end{aligned} \] 其中\(\displaystyle Q_{\alpha} = \sum\limits_{i = 1}^{n}\boldsymbol{F}_{i} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}}\)为广义力。第二部分为

\[ \begin{aligned} - \sum\limits_{i = 1}^{n} m_{i}\ddot{r}_{i} \cdot \delta \boldsymbol{r}_{i} & = -\sum\limits_{i = 1}^{n} \left( m_{i}\ddot{\boldsymbol{r}}_{i} \cdot \sum\limits_{\alpha}^{s}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \delta q_{\alpha} \right) \\ & = -\sum\limits_{\alpha = 1}^{s} \left( \sum\limits_{i = 1}^{n} m_{i}\ddot{\boldsymbol{r}}_{i} \cdot\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \right) \delta q_{\alpha} \end{aligned} \] 针对括号中的部分，我们有 \[ \begin{aligned} \sum\limits_{i = 1}^{n} m_{i}\ddot{\boldsymbol{r}}_{i} \cdot\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} & = \sum\limits_{i = 1}^{n}m_{i}\dfrac{\mathrm{d} \dot{\boldsymbol{r}}_{i}}{\mathrm{d} t} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \\ & = \sum\limits_{i = 1}^{n} m_{i} \dfrac{\mathrm{d}}{\mathrm{d} t} \left( \dot{\boldsymbol{r}}_{i} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \right) - \sum\limits_{i = 1}^{n}m_{i}\dot{\boldsymbol{r}_{i}} \cdot \dfrac{\mathrm{d}}{\mathrm{d} t} \left( \dfrac{\partial\boldsymbol{r}_{i}}{\partial q_{\alpha}} \right) \end{aligned} \] 将Lagrange关系带入上式可得 \[ \begin{aligned} \sum\limits_{i = 1}^{n} m_{i}\ddot{\boldsymbol{r}}_{i} \cdot\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} & = \sum\limits_{i = 1}^{n} m_{i} \dfrac{\mathrm{d}}{\mathrm{d} t} \left( \dot{\boldsymbol{r}}_{i} \cdot \dfrac{\partial \dot{\boldsymbol{r}_{i}}}{\partial \dot{q_{\alpha}}} \right) - \sum\limits_{i = 1}^{n}m_{i}\dot{\boldsymbol{r}_{i}} \cdot \dfrac{\partial}{\partial q_{\alpha}} \left( \dfrac{\mathrm{d} \boldsymbol{r}_{i}}{\mathrm{d} t} \right) \\ & = \dfrac{\mathrm{d}}{\mathrm{d} t} \dfrac{\partial}{\partial \dot{q}_{\alpha}}\left( \sum\limits_{i = 1}^{n} \dfrac{1}{2} m_{i} \dot{\boldsymbol{r}}_{i}^{2} \right) - \dfrac{\partial}{\partial q_{\alpha}} \left( \sum\limits_{i = 1}^{n}\dfrac{1}{2} m_{i} \dot{\boldsymbol{r}}_{i}^{2} \right) \end{aligned} \] 注意到动能表达式为\(\displaystyle T = \sum\limits_{i = 1}^{n}\dfrac{1}{2}m_{i}\dot{\boldsymbol{r}}_{i}^{2}\)，因此 \[ \sum\limits_{i = 1}^{n} m_{i}\ddot{\boldsymbol{r}}_{i} \cdot\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} = \dfrac{\mathrm{d}}{\mathrm{d} t} \dfrac{\partial T}{\partial \dot{q}_{\alpha}} - \dfrac{\partial T}{\partial q_{\alpha}} \] 这样第二部分最终可以表示为 \[ -\sum\limits_{i = 1}^{n}m_{i}\ddot{\boldsymbol{r}}_{i} \cdot \delta\boldsymbol{r}_{i} = - \left( \sum\limits_{\alpha = 1}^{s}\dfrac{\mathrm{d}}{\mathrm{d} t} \dfrac{\partial T}{\partial \dot{q}_{\alpha}} - \dfrac{\partial T}{\partial q_{\alpha}} \right) \delta q_{\alpha} \] 将第一部分与第二部分结合可得 \[ \sum\limits_{\alpha = 1}^{s}\left( Q_{\alpha} - \dfrac{\mathrm{d}}{\mathrm{d} t} \dfrac{\partial T}{\partial \dot{q}_{\alpha}} + \dfrac{\partial T}{\partial q_{\alpha}} \right) \delta q_{\alpha} = 0 \] 对于只受完整约束的系统，由于我们前面引入的广义坐标是独立的，因此广义虚位移\(\delta q_{\alpha}\)也应该是相互独立的，这样只有每个括号都等于0才能满足上式要求。由此我们可以得到方程组 \[ \dfrac{\mathrm{d}}{\mathrm{d} t}\dfrac{\partial T}{\partial \dot{q}_{\alpha}} - \dfrac{\partial T}{\partial q_{\alpha}} = Q_{\alpha},\quad \alpha = 1, 2, \dots, s \]

当主动力均为保守力时，存在一个势能函数\(V\left( \boldsymbol{r}_{1}, \boldsymbol{r}_{2}, \dots, \boldsymbol{r}_{n}, t \right)\)，使得作用于质点\(i\)上的主动力\(\boldsymbol{F}_{i}\)满足 \[ \boldsymbol{F}_{i} = -\nabla_{i}V = -\left(\dfrac{\partial V}{\partial x_{i}}, \dfrac{\partial V}{\partial y_{i}}, \dfrac{\partial V}{\partial z_{i}} \right) \] 这样广义力表达式为 \[ \begin{aligned} Q_{\alpha} & = \sum\limits_{i = 1}^{n}\boldsymbol{F}_{i} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \\ & = -\sum\limits_{i = 1}^{n} \left( \dfrac{\partial V}{\partial x_{i}} \dfrac{\partial x_{i}}{\partial q_{\alpha}} + \dfrac{\partial V}{\partial y_{i}}\dfrac{\partial y_{i}}{\partial q_{\alpha}} + \dfrac{\partial V}{\partial z_{i}}\dfrac{\partial z_{i}}{\partial q_{\alpha}}\right) \\ & = -\dfrac{\partial V}{\partial q_{\alpha}} \end{aligned} \] 这样原本的Lagrange方程就变为 \[ \dfrac{\mathrm{d} }{\mathrm{d} t}\dfrac{\partial T}{\partial \dot{q}_{\alpha}} - \dfrac{\partial (T - V)}{\partial q_{\alpha}} = 0,\quad \alpha = 1, 2, \dots, s \] 定义Lagrange函数 \[ L = T - V \] 这样有 \[ \dfrac{\mathrm{d} }{\mathrm{d} t}\dfrac{\partial L}{\partial \dot{q}_{\alpha}} - \dfrac{\partial L}{\partial q_{\alpha}} = 0,\quad \alpha = 1, 2, \dots, s \]

运动积分

在系统运动过程中，存在某些广义坐标\(q_{\alpha}\)和广义速度\(\dot{q}_{\alpha}\)的函数不随时间变化，我们将这些函数称为运动积分。

广义动量积分

我们先给出下面需要使用的定义

定义（广义动量）：

我们将广义坐标\(q_{\alpha}\)对应的广义动量\(p_{\alpha}\)定义为 \[ p_{\alpha} = \dfrac{\partial L}{\partial \dot{q}_{\alpha}} \]

定义（可遗坐标）：

如果对于广义坐标\(q_{\alpha}\)，Lagrange函数满足 \[ \dfrac{\partial L}{\partial q_{\alpha}} = 0 \] 那么我们将该广义坐标称为可遗坐标（或循环坐标）。

对于可遗坐标而言，由Lagrange方程显然有 \[ \dfrac{\mathrm{d} }{\mathrm{d} t} \left( \dfrac{\partial L}{\partial \dot{q}_{\alpha}} \right) = 0 \] 这表明，广义动量是不随时间变化的，或者说广义动量是守恒的。由此可得

命题：

如果广义坐标\(q_{\alpha}\)为可遗坐标，那么其对应的广义动量\(p_{\alpha}\)为守恒的，即 \[ p_{\alpha} = C \]

广义能量积分

我们先给出广义能量积分相关的结论，然后再给出推导过程

定义（广义能量）：

我们将广义能量函数定义为 \[ H = \sum\limits_{\alpha = 1}^{s} p_{\alpha}\dot{q}_{\alpha} - L \]

命题：

如果Lagrange函数\(L\)不显式含时，也就是\(L= L(q, \dot{q})\)，或者说 \[ \dfrac{\partial L}{\partial t} = 0 \] 那么广义能量为守恒的，即 \[ H = C \]

证明：

对于一般的Lagrange函数，其对于时间的全导数为 \[ \dfrac{\mathrm{d} L}{\mathrm{d} t} = \dfrac{\partial L}{\partial t} + \sum\limits_{\alpha = 1}^{s} \dfrac{\partial L}{\partial q_{\alpha}}\dot{q}_{\alpha} + \sum\limits_{\alpha = 1}^{s} \dfrac{\partial L}{\partial \dot{q}_{\alpha}} \dfrac{\mathrm{d} \dot{q}_{\alpha}}{\mathrm{d} t} \] 在主动力均为保守力的情况下，由Lagrange方程有 \[ \dfrac{\partial L}{\partial q_{\alpha}} = \dfrac{\mathrm{d}}{\mathrm{d} t}\left( \dfrac{\partial L}{\partial \dot{q}_{\alpha}} \right) \] 由此可得 \[ \begin{aligned} \dfrac{\mathrm{d} L}{\mathrm{d} t} & = \sum\limits_{\alpha = 1}^{s}\dfrac{\mathrm{d}}{\mathrm{d} t}\left( \dfrac{\partial L}{\partial \dot{q}_{\alpha}} \right) \dot{q}_{\alpha} + \sum\limits_{\alpha = 1}^{s} \dfrac{\partial L}{\partial \dot{q}_{a}} \dfrac{\mathrm{d} \dot{q}_{\alpha}}{\mathrm{d} t} \\ & = \dfrac{\partial L}{\partial t} + \dfrac{\mathrm{d} }{\mathrm{d} t}\left( \sum\limits_{\alpha = 1}^{s} \dfrac{\partial L}{\partial \dot{q}_{\alpha}} \dot{q}_{\alpha} \right) \end{aligned} \] 由于\(\displaystyle \dfrac{\partial L}{\partial \dot{q}_{\alpha}} = p_{\alpha}\)，因此可得 \[ \dfrac{\mathrm{d} }{\mathrm{d} t} \left( \sum\limits_{\alpha = 1}^{s} p_{\alpha} \dot{q}_{\alpha} - L \right) = -\dfrac{\partial L}{\partial t} \] 将广义能量的表达式代入可得 \[ \dfrac{\mathrm{d} H}{\mathrm{d} t} = -\dfrac{\partial L}{\partial t} \] 如果Lagrange函数不显式含时，那么 \[ \dfrac{\mathrm{d} H}{\mathrm{d} t} = -\dfrac{\partial L}{\partial t} = 0 \] 从而广义能量为常数 \[ H = C \]

广义能量的含义

在讨论广义能量的含义之前，我们先不加证明地给出齐次函数的Euler定理

定义（齐次函数）：

对于函数\(f(x_{1}, x_{2}, \dots, x_{n})\)而言，如果其满足 \[ f(\lambda x_{1}, \lambda x_{2}, \dots, \lambda x_{n}) = \lambda^{m}f(x_{1}, x_{2}, \dots, x_{n}) \] 其中，\(\lambda\)为任意常数，\(m\)为函数中每项的次数，那么我们将这种函数称为齐次函数。

定理（齐次函数的Euler定理）：

如果函数\(f(x_{1}, x_{2}, \dots, x_{n})\)为齐次函数，那么其满足 \[ \sum\limits_{i = 1}^{n}\dfrac{\partial f}{\partial x_{i}}x_{i} = mf \] 其中，\(m\)为函数中每项的次数。

广义能量的定义式为 \[ H = \sum\limits_{\alpha = 1}^{s} p_{\alpha}\dot{q}_{\alpha} - L \] 由于势能与广义速度无关，因此上式变为 \[ H = \sum\limits_{\alpha = 1}^{s}\dfrac{\partial T}{\partial \dot{q}_{\alpha}} \dot{q}_{\alpha} - L \] 我们分为两种情况讨论：

• 如果坐标不显式含时，即\(\boldsymbol{r}_{i} = \boldsymbol{r}_{i}(q)\)，那么速度可以表示为 \[ \dot{\boldsymbol{r}}_{i} = \sum\limits_{\alpha = 1}^{s}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \dot{q}_{\alpha} \] 从而 \[ \begin{aligned} T & = \sum\limits_{i = 1}^{n} \dfrac{1}{2} m_{i} \dot{\boldsymbol{r}}_{i} \cdot \dot{\boldsymbol{r}}_{i} \\ & = \sum\limits_{i = 1}^{n} \dfrac{1}{2}m_{i}\sum\limits_{\alpha = 1}^{s} \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}}\dot{q}_{\alpha} \cdot \sum\limits_{\beta = 1}^{s}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\beta}} \dot{q}_{\beta} \\ & = \sum\limits_{i = 1}^{n} \sum\limits_{\alpha = 1}^{s}\sum\limits_{\beta = 1}^{s} \dfrac{1}{2} m_{i}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\beta}} \dot{q}_{\alpha} \dot{q}_{\beta} \end{aligned} \] 此时不难看出，动能为广义速度的二次齐次多项式，因此由齐次函数的Euler定理可得 \[ \sum\limits_{\alpha = 1}^{s} \dfrac{\partial T}{\partial \dot{q}_{\alpha}} \dot{q}_{\alpha} = 2T \] 由此可得 \[ \begin{aligned} H & = \sum\limits_{\alpha = 1}^{s} p_{\alpha}\dot{q}_{\alpha} - L \\ & = \sum\limits_{\alpha = 1}^{s}\dfrac{\partial T}{\partial \dot{q}_{\alpha}} \dot{q}_{\alpha} - L \\ & = 2T - (T - V) \\ & = T + V \\ \end{aligned} \] 由此，在坐标不显式含时的情况下，动能\(T\)为广义速度的二次齐次多项式，而广义能量函数\(H\)即为机械能。

• 如果坐标显式含时，即\(\boldsymbol{r}_{i} = \boldsymbol{r}_{i}(q, t)\)，那么速度可以表示为 \[ \dot{\boldsymbol{r}}_{i} = \dfrac{\partial \boldsymbol{r}_{i}}{\partial t} + \sum\limits_{\alpha = 1}^{s}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \dot{q}_{\alpha} \] 因此动能表达式变为 \[ \begin{aligned} T & = \sum\limits_{i = 1}^{n} \dfrac{1}{2} m_{i} \dot{\boldsymbol{r}}_{i} \cdot \dot{\boldsymbol{r}}_{i} \\ & = \sum\limits_{i = 1}^{n} \dfrac{1}{2}m_{i} \left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial t} + \sum\limits_{\alpha = 1}^{s} \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}}\dot{q}_{\alpha} \right) \cdot \left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial t} +\sum\limits_{\beta = 1}^{s}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\beta}} \dot{q}_{\beta} \right) \\ & = \sum\limits_{i = 1}^{n} \left\{ \dfrac{1}{2}m_{i}\left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial t} \right)^{2} + \sum\limits_{\alpha = 1}^{s}m_{i}\dfrac{\partial \boldsymbol{r}_{i}}{\partial t} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}}\dot{q}_{\alpha} + \sum\limits_{\alpha = 1}^{s}\sum\limits_{\beta = 1}^{s} \dfrac{1}{2} m_{i}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\beta}} \dot{q}_{\alpha} \dot{q}_{\beta} \right\} \end{aligned} \] 上式中包含三个部分，分别为广义速度的零次、一次和二次的齐次多项式，我们分别将其记为 \[ \left\{\begin{aligned} T_{0} & = \sum\limits_{i = 1}^{n} \dfrac{1}{2}m_{i}\left( \dfrac{\partial \boldsymbol{r}_{i}}{\partial t} \right)^{2} \\ T_{1} & = \sum\limits_{i = 1}^{n} \sum\limits_{\alpha = 1}^{s}m_{i}\dfrac{\partial \boldsymbol{r}_{i}}{\partial t} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}}\dot{q}_{\alpha} \\ T_{2} & = \sum\limits_{i = 1}^{n}\sum\limits_{\alpha = 1}^{s}\sum\limits_{\beta = 1}^{s} \dfrac{1}{2} m_{i}\dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\alpha}} \cdot \dfrac{\partial \boldsymbol{r}_{i}}{\partial q_{\beta}} \dot{q}_{\alpha} \dot{q}_{\beta} \\ \end{aligned}\right. \] 这样广义动能可以表示为 \[ T = T_{0} + T_{1} + T_{2} \] 这样由齐次函数的Euler定理可得 \[ \begin{aligned} \sum\limits_{\alpha = 1}^{s} \dfrac{\partial T}{\partial \dot{q}_{\alpha}} \dot{q}_{\alpha} & = \sum\limits_{\alpha = 1}^{s} \dfrac{\partial T_{0}}{\partial \dot{q}_{\alpha}}\dot{q}_{\alpha} + \dfrac{\partial T_{1}}{\partial \dot{q}_{\alpha}}\dot{q}_{\alpha} + \dfrac{\partial T_{2}}{\partial\dot{q}_{\alpha}}\dot{q}_{\alpha} \\ & = 0T_{0} + 1T_{1} + 2T_{2} \\ \end{aligned} \] 由此可得 \[ \begin{aligned} H & = \sum\limits_{\alpha = 1}^{s} p_{\alpha}\dot{q}_{\alpha} - L \\ & = \sum\limits_{\alpha = 1}^{s}\dfrac{\partial T}{\partial \dot{q}_{\alpha}} \dot{q}_{\alpha} - L \\ & = (T_{1} + 2T_{2}) - (T_{0} + T_{1} + T_{2} - V) \\ & = T_{2} - T_{0} + V \\ \end{aligned} \] 此时广义能量函数并非机械能，但是具有机械能的量纲。

Noether定理

我们先直接给出Noether定理

定理：

对于任何一种在坐标连续变化下，系统的Hamilton量 \(\displaystyle S = \int_{t_{1}}^{t_{2}}L(q(t), \dot{q}(t), t)\mathrm{d} t\) 的不变性都有相应的运动积分。