$$ \begin{gathered}\mathbf{B}=\boldsymbol{\nabla}\times\mathbf{A}\\\mathbf{E}=-\boldsymbol{\nabla}\Phi-\frac{\partial\mathbf{A}}{\partial t}\end{gathered} $$
自动满足磁场无源，电场的旋度势磁场变化率的负值。
The equations of motion for a charge $q$ in an electromagnetic field can be obtained by using the Lagrangian:
$$ L=\frac12m\mathbf{v}\cdot\mathbf{v}-q(\Phi-\mathbf{A}\cdot\mathbf{v}) $$
$$ \begin{gathered}\mathbf{A}^{\prime}=\mathbf{A}+\nabla\Lambda(\mathbf{r},t)\\\\\Phi^{\prime}=\Phi-\frac{\partial\Lambda(\mathbf{r},t)}{\partial t}\end{gathered} $$
-恰好构成时间、位矢的偏微分
$$ L^{\prime}=L+q\left[\mathbf{\dot{r}}\cdot\nabla\Lambda(\mathbf{r},t)+\frac{\partial\Lambda(\mathbf{r},t)}{\partial t}\right]=L+q\frac d{dt}\Lambda(\mathbf{r},t) $$
$$ \begin{aligned}&\boldsymbol{E'} =-\boldsymbol{\nabla}\Phi^{\prime}-\frac{\partial\boldsymbol{A}^{\prime}}{\partial t} \\&=-\boldsymbol{\nabla}\Phi-\frac{\partial\boldsymbol{A}}{\partial t} \\&=\boldsymbol{E} \\&\boldsymbol{B'} =\boldsymbol{\nabla}\times\boldsymbol{A}^{\prime} \\&=\boldsymbol{\nabla}\times\boldsymbol{A} \\&=\boldsymbol{B}\end{aligned} $$
A condition: $\nabla\cdot\varepsilon \hat{\mathrm A}=0$ → 电势$\phi=0$
矢势为无源场。
电场$\mathbf{E}=-\boldsymbol{\nabla}\Phi-\frac{\partial\mathbf{A}}{\partial t}$第一项为无旋场（纵场）对应库伦场；第二项为无源场（横场）对应感应电场。
$$ \begin{aligned}\nabla^{2}\mathbf{A}-\frac{1}{c^{2}}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}-\frac{1}{c^{2}}\frac{\partial}{\partial t}\nabla\varphi&=-\mu_{0}\boldsymbol{J} \\\nabla^{2}\varphi&=-\frac{\rho}{\varepsilon_{0}} \\(\nabla\cdot A&=0)\end{aligned} $$
标势满足的方程与静电场相同，解为库伦势，代入一式解出矢势，进而确定电磁场。
$$ \nabla\cdot A+\frac1{c^2}\frac{\partial\varphi}{\partial t}=0 $$
解决实际辐射问题。