Maxwell方程组：
$$\begin{gathered} \nabla \cdot \text{D}={\rho }_{free} \\ \nabla \times \text{H}={\text{J}}_{free}+\frac{\partial }{\partial t}\text{D} \\ \nabla \times \text{E}=-\frac{\partial }{\partial t}\text{B} \\ \nabla \cdot \text{B}=0 \end{gathered}$$

用Electric bounded charge and current简化Macroscopic方程，考虑$\text{D}$与$\text{B}$的表达式
$$\begin{gathered} \text{D}={ϵ}_0\text{E}+\text{P} \\ \text{H}=\frac{\text{B}-\text{M}}{{\mu }_0} \end{gathered}$$引入electric bounded charge density ${\rho }_{bound}^{(e)}$与electric bounded current density ${\text{J}}_{bound}^{(e)}$，定义总电荷密度与总电流密度为
$$\begin{gathered} {\rho }_{total}^{(e)}={\rho }_{free}+{\rho }_{bound}^{(e)}={\rho }_{free}-\nabla \cdot \text{P} \\ {\text{J}}_{total}^{(e)}={\text{J}}_{free}+{\text{J}}_{bound}^{(e)}={\text{J}}_{free}+\frac{\partial }{\partial t}\text{P}+\frac{1}{{\mu }_0}\nabla \times \text{M} \end{gathered}$$

由此可以得到用Electric bounded charge and current简化的Macroscopic方程：
$$\begin{gathered} \nabla \cdot \text{E}=\frac{{\rho }_{total}^{(e)}}{{ϵ}_0} \\ \nabla \times \text{B}={\mu }_0{\text{J}}_{total}^{(e)}+{\mu }_0{ϵ}_0\frac{\partial }{\partial t}\text{E} \\ \nabla \times \text{E}=-\frac{\partial }{\partial t}\text{B} \\ \nabla \cdot \text{B}=0 \end{gathered}$$

要进一步简化这个方程，可以引入标量势$\psi$与矢量势$\text{A}$，
$$\begin{gathered} \text{B}=\nabla \times \text{A} \\ \text{E}=-\nabla \psi -\frac{\partial }{\partial t}\text{A} \end{gathered}$$

下面我们分别讨论：

标量势满足的2阶PDE

考虑Maxwell方程1，$\nabla \cdot \text{E}=\frac{{\rho }_{total}^{(e)}}{{ϵ}_0}$，代入$\text{E}=-\nabla \psi -\frac{\partial }{\partial t}\text{A}$，
$$-{ϵ}_0\nabla \cdot \nabla \psi -{ϵ}_0\frac{\partial }{\partial t}\nabla \text{A}={\rho }_{total}^{(e)}$$

记$\Delta ={\nabla }^2=\nabla \cdot \nabla$，这个算子被称为Laplace算子；在Lorenz Gauge下，$\nabla \cdot \text{A}+\frac{1}{c^2}\frac{\partial }{\partial t}\psi =0$，代入上述方程，
$$\Delta \psi -\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}\psi =-\frac{1}{{ϵ}_0}{\rho }_{total}^{(e)}$$

这个方程是一个典型的波动方程，它描述的是以${\rho }_{total}^{(e)}$为源的波$\psi$的传播规律，其中$c$是波的传播速度。用$\psi$与${\rho }_{total}^{(e)}$的Fourier逆变换替换，
$$\left(\Delta -\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}\right){\mathcal{F}}^{-1}[\psi ]=-\frac{1}{{ϵ}_0}{\mathcal{F}}^{-1}[{\rho }_{total}^{(e)}]$$

假设这个方程解的形式为平面波，则
$$\begin{gathered} \left(k^2-\frac{w^2}{c^2}\right)\psi (\text{k},w)=\frac{1}{{ϵ}_0}{\rho }_{total}^{(e)}(\text{k},w) \\ \psi (\text{k},w)=\frac{{\rho }_{total}^{(e)}(\text{k},w)}{{ϵ}_0(k^2-\frac{w^2}{c^2})} \end{gathered}$$

这与我们用Fourier变换法得到的解一致；如果不要平面波解的假设，只做一般性讨论，那么
$$\left(\Delta +\frac{w^2}{c^2}\right)\psi (\text{r},w)=-\frac{1}{{ϵ}_0}{\rho }_{total}^{(e)}(\text{r},w)$$

这个方程被称为Helmholtz方程，它的解法与Poisson方程类似，都可以用Green函数法。

矢量势满足的2阶PDE

考虑Maxwell方程2，$$\begin{gathered} \nabla \times \text{B}={\mu }_0{\text{J}}_{total}^{(e)}+\frac{1}{c^2}\frac{\partial }{\partial t}\text{E} \\ \nabla \times (\nabla \times \text{A})={\mu }_0{\text{J}}_{total}^{(e)}+\frac{1}{c^2}\frac{\partial }{\partial t}\left(-\nabla \psi -\frac{\partial }{\partial t}\text{A}\right) \end{gathered}$$

其中，
$$\begin{gathered} \nabla \times (\nabla \times \text{A})=\nabla (\nabla \cdot \text{A})-\nabla \cdot \nabla \text{A} \\ \nabla \left(\nabla \cdot \text{A}+\frac{1}{c^2}\frac{\partial }{\partial t}\psi \right)-{\mu }_0{\text{J}}_{total}^{(e)}=\Delta \text{A}-\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}\text{A} \end{gathered}$$

在Lorenz Gauge下，
$$\nabla \cdot \text{A}+\frac{1}{c^2}\frac{\partial }{\partial t}\psi =0$$

于是，
$$\Delta \text{A}-\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}\text{A}=-{\mu }_0{\text{J}}_{total}^{(e)}$$

Laplace算子的含义

可以发现标量势与矢量势满足的方程是可以统一起来的，$$\underset{传播规律}{\underbrace{\left(\Delta -\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}\right)}}\underset{电磁场的势}{\underbrace{\left[\begin{array}{c}\psi \\ \text{A}\end{array}\right]}}=\underset{电磁场的源}{\underbrace{\left[\begin{array}{c}-\frac{1}{{ϵ}_0}{\rho }_{total}^{(e)}(\text{r},w)\\ -{\mu }_0{\text{J}}_{total}^{(e)}\end{array}\right]}}$$

其中描述传播规律的算子为
$$\Delta -\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}-\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}$$

在Minkowski时空$(x,y,z,ict)$中，这个算子其实就是关于四维时空坐标的Laplace算子。下面简单讨论一下$\Delta \text{A}$的计算：
$$\Delta \text{A}=\nabla (\nabla \cdot \text{A})-\nabla \times (\nabla \times \text{A})$$

考虑平面波解，用$i\text{k}$代入$\nabla$，在Fourier domain中，等式右边为
$$i\text{k}(i\text{k}\cdot \text{A})-i^2\text{k}\times (\text{k}\times \text{A})=-k^2\text{A}$$

所以用Fourier逆变换表示$\Delta \text{A}$为，
$$\Delta \text{A}=(2\pi {)}^{-4}{\int }_{-\infty }^{+\infty }-k^2\text{A}(\text{k},w)e^{i(\text{k}\cdot \text{r}-wt)}d\text{k}dw$$

假设空间坐标用笛卡尔坐标系，那么
$$\Delta \text{A}=(\Delta A_x)\hat{x}+(\Delta A_y)\hat{y}+(\Delta A_z)\hat{z}$$

但是在柱坐标与球坐标中，这个结论并不成立，我们需要用Fourier逆变换的那个一般性式子计算。

标量势与矢量势的平面波解

依然考虑平面波解，上文得到了Fourier domain中标量势的解为
$$\psi (\text{k},w)=\frac{{\rho }_{total}^{(e)}(\text{k},w)}{{ϵ}_0(k^2-\frac{w^2}{c^2})}$$

简单起见，要得到四维时空中标量势的解，可以直接计算它在Fourier domain中的解的逆变换，另外，在四维时空中，用$(\text{r},t)$表示观测位置，用$({\text{r}}^{\prime },t^{\prime })$表示电磁场的源的位置，用${\text{r}}^{\prime \prime }-\text{r}-{\text{r}}^{\prime }$作为一个辅助变量。
$$\begin{array}{rl}\psi (\text{r},t)& =(2\pi {)}^{-4}{\int }_{-\infty }^{+\infty }\psi (\text{k},w)e^{i(\text{k}\cdot \text{r}-wt)}d\text{k}dw\\ & =(2\pi {)}^{-4}{\int }_{-\infty }^{+\infty }\frac{{\rho }_{total}^{(e)}(\text{k},w)}{{ϵ}_0(k^2-\frac{w^2}{c^2})}{e}^{i(\text{k}\cdot \text{r}-wt)}d\text{k}dw\\ & =\frac{1}{16{\pi }^4{ϵ}_0}{\int }_{-\infty }^{+\infty }{\rho }_{total}^{(e)}(\text{k},w)e^{i(\text{k}\cdot \text{r}-wt)}{\int }_{-\infty }^{+\infty }\frac{e^{iw|{\text{r}}^{\prime \prime }|/c}}{4\pi |{\text{r}}^{\prime \prime }|}{e}^{-i\text{k}\cdot {\text{r}}^{\prime \prime }}d{\text{r}}^{\prime \prime }d\text{k}dw\\ & =\frac{1}{64{\pi }^5{ϵ}_0}{\int }_{-\infty }^{+\infty }\frac{e^{iw|{\text{r}}^{\prime \prime }|/c}}{|{\text{r}}^{\prime \prime }|}{e}^{-iwt}{\int }_{-\infty }^{+\infty }{\rho }_{total}^{(e)}(\text{k},w)e^{i\text{k}\cdot (\text{r}-{\text{r}}^{\prime \prime })}d\text{k}dwd{\text{r}}^{\prime \prime }\\ & =\frac{1}{8{\pi }^2{ϵ}_0}{\int }_{-\infty }^{+\infty }\frac{1}{|{\text{r}}^{\prime \prime }|}{\int }_{-\infty }^{+\infty }{\rho }_{total}^{(e)}(\text{r}-{\text{r}}^{\prime \prime },w)e^{-iw(t-|{\text{r}}^{\prime \prime }|/c)}dwd{\text{r}}^{\prime \prime }\\ & =\frac{1}{4\pi {ϵ}_0}{\int }_{-\infty }^{+\infty }\frac{{\rho }_{total}^{(e)}(\text{r}-{\text{r}}^{\prime \prime },t-|{\text{r}}^{\prime \prime }|/c)}{|{\text{r}}^{\prime \prime }|}d{\text{r}}^{\prime \prime }\end{array}$$

第三个等号用的是$\frac{1}{k^2-w^2/c^2}$的Fourier变换式，第四个等号用的是Fubini定理，第五个等号与第六个等号用的是Fourier变换的定义。矢量势的计算与之类似，最终得到的解为
$$\begin{gathered} \psi (\text{r},t)=\frac{1}{4\pi {ϵ}_0}{\int }_{-\infty }^{+\infty }\frac{{\rho }_{total}^{(e)}({\text{r}}^{\prime },t-|\text{r}-{\text{r}}^{\prime }|/c)}{|\text{r}-{\text{r}}^{\prime }|}d{\text{r}}^{\prime } \\ \text{A}(\text{r},t)=\frac{{\mu }_0}{4\pi }{\int }_{-\infty }^{+\infty }\frac{{\text{J}}_{total}^{(e)}({\text{r}}^{\prime },t-|\text{r}-{\text{r}}^{\prime }|/c)}{|\text{r}-{\text{r}}^{\prime }|}d{\text{r}}^{\prime } \end{gathered}$$