贡献者： addis
预备知识 电磁场标势和矢势
，
广义力，
欧拉—拉格朗日方程（经典力学），哈密顿正则方程
由于（式 6 ）
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial T}{\partial \dot q_i} - \frac{\partial T}{\partial q_i} = Q_i~.
\end{equation}
若非常规势能（或者称 “
广义势”） $U(q_1, q_2...\dot q_1, \dot q_2..., t)$ 与广义力满足
\begin{equation}
Q_i = \frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial U}{\partial \dot q_i} - \frac{\partial U}{\partial q_i} ~.
\end{equation}
且定义拉格朗日量为 $L = T - U$，则代入可得拉格朗日方程
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial L}{\partial \dot q_i} = \frac{\partial L}{\partial q_i} ~.
\end{equation}
现在证明在任意电磁场中运动的点电荷的广义势为
\begin{equation}
U = q(\Phi - \boldsymbol{\mathbf{A}} \boldsymbol\cdot \boldsymbol{\mathbf{v}} )~.
\end{equation}
其中 $\Phi$ 和 $ \boldsymbol{\mathbf{A}} $ 分别为点电荷所在位置的
电磁标势和矢势，$ \boldsymbol{\mathbf{v}} $ 为点电荷的速度。
另外还可以写出哈密顿量为
\begin{equation}
H = \boldsymbol{\mathbf{p}} \dot{ \boldsymbol{\mathbf{q}} } - L = \frac{1}{2m}( \boldsymbol{\mathbf{p}} - q \boldsymbol{\mathbf{A}} )^2 + q\Phi~.
\end{equation}
1. 直角坐标中的证明
以下用直角坐标证明（令 $x_1, x_2, x_3 = x,y,z$）
\begin{equation}
F_i = Q_i = \frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial U}{\partial \dot x_i} - \frac{\partial U}{\partial x_i} ~.
\end{equation}
以 $x$ 分量为例
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial U}{\partial \dot x} = - q \frac{\mathrm{d}}{\mathrm{d}{t}} A_x = - q \left[( \boldsymbol\nabla A_x) \boldsymbol\cdot \boldsymbol{\mathbf{v}} + \frac{\partial A_x}{\partial t} \right] ~,
\end{equation}
\begin{equation}
\frac{\partial U}{\partial x} = q \left( \frac{\partial \Phi}{\partial x} - \frac{\partial \boldsymbol{\mathbf{A}} }{\partial x} \boldsymbol\cdot \boldsymbol{\mathbf{v}} \right) ~.
\end{equation}
代入
式 6 右边得
\begin{equation}
q \left[- \frac{\partial \Phi}{\partial x} - \frac{\partial A_x}{\partial t} + v_y \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) - v_z \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right) \right] ~.
\end{equation}
根据广义洛伦兹力及电磁势的定义
\begin{equation}
\boldsymbol{\mathbf{F}} = q \left( \boldsymbol{\mathbf{E}} + \boldsymbol{\mathbf{v}} \boldsymbol\times \boldsymbol{\mathbf{B}} \right) ~,
\qquad
\boldsymbol{\mathbf{E}} = - \boldsymbol\nabla \Phi - \frac{\partial \boldsymbol{\mathbf{A}} }{\partial t} ~,
\qquad
\boldsymbol{\mathbf{B}} = \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{A}} ~.
\end{equation}
\begin{equation}\begin{aligned}
F_x &= q \left(E_x + v_y B_z - v_z B_y \right) \\
&= q \left[- \frac{\partial \Phi}{\partial x} - \frac{\partial A_x}{\partial t} + v_y \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) - v_z \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right) \right] ~.
\end{aligned}\end{equation}
对比可得
式 6 在 $x$ 方向成立，$y,z$ 分量的证明类似。另外容易证明多个点电荷组成的系统的非常规势等于每个点电荷的非常规势之和。
2. 广义坐标的证明
下面来证明对于任意广义坐标 $ \left\{q_i \right\} $，式 4 都满足式 2 。首先令变换关系为
\begin{equation}
x_i = x_i(q_1, q_2, \dots, t)~.
\end{equation}
其中 $x_i$ 包括每个质点的 $x, y, z$ 坐标，则
\begin{equation}
\frac{\partial U}{\partial \dot q_i} = \sum_j \frac{\partial U}{\partial \dot x_j} \frac{\partial \dot x_j}{\partial \dot q_i} = \sum_j \frac{\partial U}{\partial \dot x_j} \frac{\partial x_j}{\partial q_i} ~,
\end{equation}
注意第二步用到了
式 5 。对时间求全导数得
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial U}{\partial \dot q_i} = \sum_j \frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial U}{\partial \dot x_j} \frac{\partial x_j}{\partial q_i} + \sum_j \frac{\partial U}{\partial \dot x_j} \frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial x_j}{\partial q_i} ~.
\end{equation}
另外有
\begin{equation}
\frac{\partial U}{\partial q_i} = \sum_j \frac{\partial U}{\partial x_j} \frac{\partial x_j}{\partial q_i} + \sum_j \frac{\partial U}{\partial \dot x_j} \frac{\partial \dot x_j}{\partial q_i} ~,
\end{equation}
易证
式 14 和
式 15 的右边第二项相等。将以上两式代入
式 2 右边得（第二步利用了
式 6 ）
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial U}{\partial \dot x_i} - \frac{\partial U}{\partial x_i} = \sum_j \frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial U}{\partial \dot x_j} \frac{\partial x_j}{\partial q_i} - \sum_j \frac{\partial U}{\partial x_j} \frac{\partial x_j}{\partial q_i} = \sum_j F_j \frac{\partial x_j}{\partial q_i} ~,
\end{equation}
而这恰好是 $Q_i$ 的定义。证毕。
3. 哈密顿量
首先求广义动量
\begin{equation}
\boldsymbol{\mathbf{p}} = \frac{\partial L}{\partial \dot{ \boldsymbol{\mathbf{q}} }} = m \boldsymbol{\mathbf{v}} + q \boldsymbol{\mathbf{A}} ~,
\end{equation}
则哈密顿量为
\begin{equation}
H = \boldsymbol{\mathbf{p}} \dot{ \boldsymbol{\mathbf{q}} } - L = \frac{1}{2m}( \boldsymbol{\mathbf{p}} - q \boldsymbol{\mathbf{A}} )^2 + q\Phi~.
\end{equation}