省流

拉格朗日方程： $\boxed{\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}-\frac{\partial L}{\partial q_j}= 0,L=T-V}$
广义动量： $\boxed{p_j=\frac{\partial L}{\partial \dot{q}_j}}$
广义动量守恒条件： $\boxed{\frac{\partial L}{\partial q_j}=0\Rightarrow p_j=Const}$

拉格朗日方程

拉格朗日关系

将真实坐标用广义坐标表示：

$\vec{r}_i = \vec{r}_i(q_1, q_2, \cdots, q_{N-k}, t)$ $\dot{\vec{r}}_i = \sum_{\alpha=1}^{N-k}\frac{\partial \vec{r}_i}{\partial q_\alpha}\dot{q}_\alpha + \frac{\partial \vec{r}_i}{\partial t}$

分别对$\dot{q}_\beta$和$q_\beta$求导：

$\begin{aligned} \frac{\partial \dot{\vec{r}}_i}{\partial \dot{q}_\beta} &= \frac{\partial}{\partial \dot{q}_\beta}\left(\sum_{\alpha=1}^{N-k}\frac{\partial \vec{r}_i}{\partial q_\alpha}\dot{q}_\alpha + \frac{\partial \vec{r}_i}{\partial t}\right)\\ &= \frac{\partial \vec{r}_i}{\partial q_\beta}\\ \frac{\partial \dot{\vec{r}}_i}{\partial q_\beta} &= \frac{\partial}{\partial q_\beta}\left(\sum_{\alpha=1}^{N-k}\frac{\partial \vec{r}_i}{\partial q_\alpha}\dot{q}_\alpha + \frac{\partial \vec{r}_i}{\partial t}\right)\\ &= \sum_{\alpha=1}^{N-k}\frac{\partial^2 \vec{r}_i}{\partial q_\beta\partial q_\alpha}\dot{q}_\alpha + \frac{\partial^2 \vec{r}_i}{\partial q_\beta\partial t}\\ &=\frac{\partial }{\partial t}\left(\frac{\partial \vec{r}_i}{\partial q_\beta}\right) + \sum_{\alpha=1}^{N-k}\dot{q}_\alpha\frac{\partial}{\partial q_\alpha}\left(\frac{\partial \vec{r}_i}{\partial q_\beta}\right)\\ &=\frac{d}{dt}\left(\frac{\partial \vec{r}_i}{\partial q_\beta}\right) \end{aligned}$

拉格朗日方程的推导

广义坐标的达朗贝尔原理：

$\sum_{j=1}^{N-k} \left(Q_j - \sum_{i=1}^n m_i\ddot{\vec{r}}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j}\right)\delta q_j = 0$

将拉格朗日关系代入，计算惯性力的部分：

$\begin{aligned} \sum_{i=1}^n m_i\ddot{\vec{r}}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j} &=\sum_{i=1}^n m_i\frac{d\dot{\vec{r}}_i}{dt}\cdot\frac{\partial \vec{r}_i}{\partial q_j}\\ &=\sum_{i=1}^n m_i\frac{d}{dt}\left(\dot{\vec{r}}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j}\right)-\sum_{i=1}^n m_i\left(\dot{\vec{r}}_i\cdot\frac{d}{dt}\frac{\partial \vec{r}_i}{\partial q_j}\right)\\ &=\sum_{i=1}^n m_i\frac{d}{dt}\left(\dot{\vec{r}}_i\cdot\frac{\partial \dot{\vec{r}}_i}{\partial \dot{q}_j}\right)-\sum_{i=1}^n m_i\left(\dot{\vec{r}}_i\cdot\frac{\partial \dot{\vec{r}}_i}{\partial q_j}\right)\\ &=\frac{d}{dt}\frac{\partial }{\partial \dot{q}_j}\left(\sum_{i=1}^n\frac12m_i |\dot{\vec{r}}_i|^2\right)-\frac{\partial }{\partial q_j}\left(\sum_{i=1}^n\frac12m_i |\dot{\vec{r}}_i|^2\right)\\ &=\frac{d}{dt}\frac{\partial T}{\partial \dot{q}_j}-\frac{\partial T}{\partial q_j}\\ \end{aligned}$

也就是说：

$Q_j=\frac{d}{dt}\frac{\partial T}{\partial \dot{q}_j}-\frac{\partial T}{\partial q_j}$

这就是欧拉-拉格朗日方程。如果广义力是保守力，则可以引入广义势能$V$，使得：

$Q_j = -\frac{\partial V}{\partial q_j}$

将其代入拉格朗日方程：

$\boxed{\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}-\frac{\partial L}{\partial q_j}= 0,L=T-V}$

从量纲上判断，$\frac{\partial L}{\partial \dot{q}_j}$是动量量纲，$\frac{\partial L}{\partial q_j}$是力量纲，这个方程的地位有点像牛顿第二定律$F=\frac{d}{dt}p$。这点在下面也会谈到。

如果你对变分原理熟悉，可以看出这其实是哈密顿作用量取极值的结果：
$\delta S = \delta \int_{t_1}^{t_2} L(q,\dot{q};t) dt = 0$
对于这样一个泛函，极值的条件为：
$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = 0$
和上面的结果是一样的。

对于自由粒子，其拉格朗日量为$L=\frac12 m v^2$，则有：
$ma = 0$
对于保守系统，其拉格朗日量为$L=T-V$，则有：
$ma= -\nabla V$

双摆的拉格朗日方程

alt

考虑双摆的广义坐标为$(\theta_1,\theta_2)$，对应的摆长为$l_1,l_2$，质量为$m_1,m_2$，真实坐标为$(x_1,y_1,x_2,y_2)$，则有：

$\begin{aligned} x_1 &= l_1\sin\theta_1\\ y_1 &= -l_1\cos\theta_1\\ x_2 &= l_1\sin\theta_1 + l_2\sin\theta_2\\ y_2 &= -l_1\cos\theta_1 - l_2\cos\theta_2 \end{aligned}$

则有：

$\begin{aligned} \dot{x}_1 &= l_1\cos\theta_1\dot{\theta}_1\\ \dot{y}_1 &= l_1\sin\theta_1\dot{\theta}_1\\ \dot{x}_2 &= l_1\cos\theta_1\dot{\theta}_1 + l_2\cos\theta_2\dot{\theta}_2\\ \dot{y}_2 &= l_1\sin\theta_1\dot{\theta}_1 + l_2\sin\theta_2\dot{\theta}_2 \end{aligned}$

动能写作：

$\begin{aligned} T &= \frac12 m_1(\dot{x}_1^2 + \dot{y}_1^2) + \frac12 m_2(\dot{x}_2^2 + \dot{y}_2^2)\\ &= \frac12 m_1\left(l_1^2\cos^2\theta_1\dot{\theta}_1^2 + l_1^2\sin^2\theta_1\dot{\theta}_1^2\right) + \frac12 m_2\left((l_1\cos\theta_1\dot{\theta}_1 + l_2\cos\theta_2\dot{\theta}_2)^2 + (l_1\sin\theta_1\dot{\theta}_1 + l_2\sin\theta_2\dot{\theta}_2)^2\right)\\ &=\frac12(m_1+m_2)l_1^2\dot{\theta}_1^2 + m_2 l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \cos(\theta_1 - \theta_2) + \frac12 m_2 l_2^2 \dot{\theta}_2^2 \end{aligned}$

势能写作：

$\begin{aligned} V &= m_1 g y_1 + m_2 g y_2\\ &= -m_1 g l_1\cos\theta_1 - m_2 g (l_1\cos\theta_1 + l_2\cos\theta_2)\\ &= -m_1 g l_1\cos\theta_1 - m_2 g l_1\cos\theta_1 - m_2 g l_2\cos\theta_2\\ &= - (m_1 + m_2) g l_1\cos\theta_1 - m_2 g l_2\cos\theta_2 \end{aligned}$

拉格朗日方程解为：

$\begin{aligned} \text{Left}&= \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}_1}\\ &= (m_1 + m_2) l_1^2 \ddot{\theta}_1 + m_2 l_1 l_2 \frac{d}{dt} \dot{\theta}_2\cos(\theta_1 - \theta_2)\\ &= (m_1 + m_2) l_1^2 \ddot{\theta}_1 + m_2 l_1 l_2 \ddot{\theta}_2\cos(\theta_1 - \theta_2)- m_2 l_1 l_2 \dot{\theta}_2\sin(\theta_1 - \theta_2)(\dot{\theta_1}-\dot{\theta_2})\\ \text{Right}&= \frac{\partial L}{\partial \theta_1}\\ &= -m_2 l_1 l_2 \dot{\theta}_1 \dot{\theta}_2 \sin(\theta_1 - \theta_2) -(m_1 + m_2) g l_1 \sin\theta_1 \\ \end{aligned}$

化简得到：

$(m_1 + m_2) l_1 \ddot{\theta}_1 + m_2 l_2 \ddot{\theta}_2\cos(\theta_1 - \theta_2)+ m_2 l_2 \dot{\theta}_2^2\sin(\theta_1 - \theta_2)+(m_1 + m_2) g \sin\theta_1=0$

对于$\theta_2$，可以得到另一个式子：

$m_2 l_2\ddot{\theta}_2 + m_2 l_1 \ddot{\theta}_1\cos(\theta_1 - \theta_2) - m_2 l_1\dot{\theta}_1^2\sin(\theta_1 - \theta_2) +m_2g\sin\theta_2 = 0$

对称性和守恒量

广义动量守恒

定义广义动量为：

$p_j = \frac{\partial L}{\partial \dot{q}_j}$

如果拉格朗日量中的势能不依赖于广义速度，则有：

$p_j = \frac{\partial T}{\partial \dot{q}_j}$

如果：

$\frac{d}{dt}\frac{\partial L}{\partial \dot{q_j}} = 0$

则可以积分出：

$p_j =\frac{\partial L}{\partial \dot{q}_j} = C_j$

其中$C_j$为常数。这个常数就是广义动量守恒量。

广义能量守恒

对拉格朗日函数求时间微分：

$\begin{aligned} \frac{dL}{dt} &= \sum_{j=1}^{N-k}\left(\frac{\partial L}{\partial q_j}\dot{q}_j + \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j\right) + \frac{\partial L}{\partial t}\\ &=\sum_{j=1}^{N-k}\left[\left(\frac{d}{dt}\frac{\partial L}{\partial \dot{q_j}}\right)\dot{q}_j + \frac{\partial L}{\partial \dot{q}_j}\frac{d}{dt}\dot{q}_j\right] + \frac{\partial L}{\partial t}\\ &=\sum_{j=1}^{N-k}\left[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\dot{q}_j\right)\right] + \frac{\partial L}{\partial t}\\ \end{aligned}$

如果拉格朗日量不显含时间，则有：

$\frac{dL}{dt} - \sum_{j=1}^{N-k}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\dot{q}_j\right)=0$

这同样是一个守恒量。定义广义能量为：

$H = \sum_{j=1}^{N-k}\frac{\partial L}{\partial \dot{q}_j}\dot{q}_j - L= \sum_{j=1}^{N-k}p_j\dot{q}_j - L$

则有：

$\frac{dL}{dt} = -\frac{dH}{dt}$

这样或许能回答“拉格朗日函数$L=T-V$的物理意义是什么”了。可以观察到代入$p=mv$后，广义能量函数就是：
$H=2T-(T-V)=T+V$

诺特定理

诺特定理指出，连续对称性和守恒定律的一一对应。

如果拉格朗日量对某个广义坐标$q_j$不显含，则对应的广义动量$p_j$守恒；
如果拉格朗日量对时间不显含，则对应的广义能量$H$守恒。

具体的数学证明过于繁琐，这里不再赘述。

例题

抛物面上的小球

抛物面$z(r)=Ar^2$上，小球的拉格朗日量在柱面坐标系下可以写为：

$\begin{aligned} T &= \frac12 m \left[\dot{r}^2 + r^2\dot{\theta}^2 + (2Ar\dot{r})^2\right] = \frac12 m \left[(1+4A^2r^2)\dot{r}^2 + r^2\dot{\theta}^2\right]\\ V &= mgz = mgAr^2 \end{aligned}$ $\boxed{L = \frac12m(1+4A^2r^2)\dot{r}^2 + \frac12mr^2\dot{\theta}^2 - mgAr^2}$

$L$不显含$\theta$，故角动量守恒：

$\frac{\partial L}{\partial \dot{\theta}} = mr^2\dot{\theta} = l = \text{const}$

利用守恒量消去$\dot{\theta}$，能量写作：

$E = \frac12m(1+4A^2r^2)\dot{r}^2 + \frac{l^2}{2mr^2} + mgAr^2$

即径向运动由有效势决定：

$\boxed{V_{eff}(r) = \frac{l^2}{2mr^2} + mgAr^2}$

圆轨道满足$V_{eff}’(r_0)=0$：

$-\frac{l^2}{mr_0^3} + 2mgAr_0 = 0\Rightarrow \dot{\theta}^2 = \frac{l^2}{m^2r_0^4} = 2gA$ $\boxed{\dot{\theta} = \sqrt{2gA},\qquad r_0 = \left(\frac{l^2}{m^2\cdot 2gA}\right)^{\frac14}}$

稳定性：圆轨道附近

$V_{eff}''(r_0) = \frac{3l^2}{mr_0^4} + 2mgA = 3(2mgA) + 2mgA = 8mgA > 0$

$V_{eff}$在$r_0$处恒为严格极小值，任意半径的圆轨道均稳定。

径向小振动：径向拉格朗日方程为

$(1+4A^2r^2)\ddot{r} + 4A^2r\dot{r}^2 - r\dot{\theta}^2 + 2gAr = 0$

令$\xi = r - r_0$并线性化，得$\ddot{\xi} + \omega_r^2\xi = 0$：

$\boxed{\omega_r^2 = \frac{8gA}{1+4A^2r_0^2}}$

频率比：$\dfrac{\omega_r}{\dot{\theta}} = \dfrac{2}{\sqrt{1+4A^2r_0^2}}$一般不是有理数，径向与角向不同频，轨道一般不闭合；仅当$4A^2r_0^2$使该比值为有理数时闭合。$r_0\to 0$时比值为$2$，回到二维谐振子的椭圆轨道。

推广：$z = Ar^n$

将抛物面推广为一般幂次曲面$z=Ar^n$。此时$\dot{z}=nAr^{n-1}\dot{r}$，故：

$\begin{aligned} v^2 &= (1+n^2A^2r^{2n-2})\dot{r}^2 + r^2\dot{\theta}^2\\ L &= \frac12m(1+n^2A^2r^{2n-2})\dot{r}^2 + \frac12mr^2\dot{\theta}^2 - mgAr^n \end{aligned}$

角动量$l=mr^2\dot{\theta}$仍守恒，有效势：

$V_{eff}(r) = \frac{l^2}{2mr^2} + mgAr^n$

圆轨道：$V_{eff}’(r_0)=0$给出

$-\frac{l^2}{mr_0^3} + mgAnr_0^{n-1} = 0\Rightarrow \frac{l^2}{m^2r_0^4} = gAnr_0^{n-2}$ $\boxed{\dot{\theta}^2 = gAn\,r_0^{\,n-2},\qquad r_0 = \left(\frac{l^2}{m^2gAn}\right)^{\frac{1}{n+2}}}$

与$n=2$不同，一般$n\neq 2$时角速度按幂次$\dfrac{n-2}{2}$随半径变化：$n>2$时外圈转得更快，$n<2$时内圈更快。给定角动量后$r_0$被唯一确定。

稳定性：

$V_{eff}''(r_0) = \frac{3l^2}{mr_0^4} + mgAn(n-1)r_0^{n-2} = mgAn(n+2)r_0^{n-2}$

当$A>0,\ n>0$时恒为正，所有$n>0$的圆轨道均稳定；$n<0$时$V_{eff}’=0$无正根，不存在圆轨道。

径向小振动：径向方程在$r_0$处线性化，得：

$\boxed{\omega_r^2 = \frac{gAn(n+2)r_0^{\,n-2}}{1+n^2A^2r_0^{\,2n-2}},\qquad \frac{\omega_r^2}{\dot{\theta}^2} = \frac{n+2}{1+n^2A^2r_0^{\,2n-2}}}$

受力视角的验算：曲面法线的水平分量与竖直分量之比等于斜率$\dfrac{N_r}{N_z}=\tan\alpha = nAr^{n-1}$。圆轨道上$N_z=mg$、$N_r=mr\dot{\theta}^2$，立即得$\dot{\theta}^2=gAnr^{n-2}$，与拉格朗日结果一致。可见约束反力随半径自适应地提供向心力分量——这正是曲面约束与有心力场的本质区别：中心吸引势$V\propto -r^{-n}$（见”有心力和散射问题”）的圆轨道稳定性要求$n<2$，而曲面约束下只需$n>0$。