电动力学 第8周作业
Chasse_neige
1. 电磁场的矢势和标势
作业：书5.1, 5.4
5.1 若把麦克斯韦方程组的所有矢量都分解为无旋的（纵场）和无散的（横场）两部分，写出 \(E\) 和 \(B\) 的这两部分在真空中所满足的方程式，并证明电场的无旋部分对应于库仑场。
分解：（\(L\) 代表纵向，\(T\) 代表横向）
\[
\begin{aligned}
\mathbf{E} &= \mathbf{E}_{T} + \mathbf{E}_{L} \\\\
\mathbf{B} &= \mathbf{B}_{T} + \mathbf{B}_{L}
\end{aligned}
\]
所以麦克斯韦方程组化为
\[
\begin{aligned}
\nabla \cdot \mathbf{E} &= \nabla \cdot (\mathbf{E}_{T} + \mathbf{E}_{L}) = \nabla \cdot \mathbf{E}_{L} = \frac{\rho}{\epsilon_{0}} \\\\
\nabla \cdot \mathbf{B} &= \nabla \cdot (\mathbf{B}_{T} + \mathbf{B}_{L}) = \nabla \cdot \mathbf{B}_{L} = 0 \\\\
\nabla \cdot \mathbf{B}_{L} &= \nabla \times \mathbf{B}_{L} = 0 \\\\
\text{可以取适当横场使得} \, \mathbf{B}_{L} &= 0 \\\\
\nabla \times \mathbf{E} &= \nabla \times (\mathbf{E}_{T} + \mathbf{E}_{L}) = \nabla \times \mathbf{E}_{T} = - \frac{\partial}{\partial t} \mathbf{B}_{T} \\\\
\nabla \times \mathbf{B} &= \nabla \times \mathbf{B}_{T} = \mu_{0} \mathbf{J} + \mu_{0} \epsilon_{0} \frac{\partial}{\partial t} \mathbf{E} = \mu_{0} (\mathbf{J}_{T} + \mathbf{J}_{L}) + \mu_{0} \epsilon_{0} \frac{\partial}{\partial t} (\mathbf{E}_{T} + \mathbf{E}_{L}) = \mu_{0} \mathbf{J}_{T} + \mu_{0} \epsilon_{0} \frac{\partial}{\partial t} \mathbf{E}_{T} \\\\
\mu_{0} \mathbf{J}_{L} + \mu_{0} \epsilon_{0} \frac{\partial}{\partial t} \mathbf{E}_{L} &= 0
\end{aligned}
\]
其中电场的无旋部分 \(\mathbf{E}_{L}\) 满足 \(\nabla \cdot \mathbf{E}_{L} = \frac{\rho}{\epsilon_{0}}\) ，即对应库仑场
5.4 设真空中矢势 \(\mathbf{A}(\mathbf{x},t)\) 可用复数傅里叶展开为
\[
\mathbf{A}(\mathbf{x},t) = \sum_k \left[ \mathbf{a}_k(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \mathbf{a}_k^*(t) e^{-i \mathbf{k} \cdot \mathbf{x}} \right]
\]
其中 \(\mathbf{a}_k^*\) 是 \(\mathbf{a}_k\) 的复共轭。
(1) 证明 \(\mathbf{a}_k\) 满足谐振子方程
\[
\frac{d^2 \mathbf{a}_k(t)}{dt^2} + k^2 c^2 \mathbf{a}_k(t) = 0
\]
证明：真空中没有自由电荷和电流，所以
\[
\nabla \times (\nabla \times \mathbf{B}) = \mu_{0} \epsilon_{0} \frac{\partial}{\partial t} (\nabla \times \mathbf{E}) = - \mu_{0} \epsilon_{0} \frac{\partial^{2}}{\partial t^{2}} \mathbf{B}
\]
\[
\nabla \times (\nabla \times (\nabla \times \mathbf{A})) = - \mu_{0} \epsilon_{0} \frac{\partial^{2}}{\partial t^{2}} \nabla \times \mathbf{A}
\]
\[
\nabla \times \left( \nabla \times (\nabla \times \mathbf{A}) + \mu_{0} \epsilon_{0} \frac{\partial^{2}}{\partial t^{2}} \mathbf{A} \right) = 0
\]
所以可以选取适当规范使得
\[
\nabla \times (\nabla \times \mathbf{A}) + \mu_{0} \epsilon_{0} \frac{\partial^{2}}{\partial t^{2}} \mathbf{A} = 0
\]
\[
\nabla \times (\nabla \times \sum_k \left[ \mathbf{a}_k(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \mathbf{a}_k^*(t) e^{-i \mathbf{k} \cdot \mathbf{x}} \right]) + \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} \sum_k \left[ \mathbf{a}_k(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \mathbf{a}_k^*(t) e^{-i \mathbf{k} \cdot \mathbf{x}} \right] = 0
\]
所以在库伦规范下
\[
\nabla^{2} \mathbf{A} = \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} \mathbf{A}
\]
所以
\[
- \mathbf{k}^{2} \mathbf{A} = \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} \mathbf{A}
\]
即对应系数
\[
\frac{d^2 \mathbf{a}_k(t)}{dt^2} + k^2 c^2 \mathbf{a}_k(t) = 0
\]
(2) 当选取规范 \(\nabla \cdot \mathbf{A} = 0, \varphi = 0\) 时，证明 \(\mathbf{k} \cdot \mathbf{a}_k = 0\)
证明：
\[
\nabla \cdot \mathbf{A} = \nabla \cdot \sum_k \left[ \mathbf{a}_k(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \mathbf{a}_k^*(t) e^{-i \mathbf{k} \cdot \mathbf{x}} \right] = 0
\]
\[
\therefore \,\, \nabla \cdot (\mathbf{a}_k(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \mathbf{a}_k^*(t) e^{-i \mathbf{k} \cdot \mathbf{x}}) = 0
\]
\[
\nabla \cdot (\mathbf{a}_k(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \mathbf{a}_k^*(t) e^{-i \mathbf{k} \cdot \mathbf{x}}) = (\nabla e^{i \mathbf{k} \cdot \mathbf{x}}) \cdot \mathbf{a}_{k} + (\nabla e^{- i \mathbf{k} \cdot \mathbf{x}}) \cdot \mathbf{a}_{k}^{*} = i e^{i \mathbf{k} \cdot \mathbf{x}} \mathbf{k} \cdot \mathbf{a}_{k} - i e^{- i \mathbf{k} \cdot \mathbf{x}} \mathbf{k} \cdot \mathbf{a}_{k}^{*} = 0
\]
对于任意 \(\mathbf{x}\) 成立，所以\(\mathbf{k} \cdot \mathbf{a}_k = 0\)
(3) 把 \(\mathbf{E}\) 和 \(\mathbf{B}\) 用 \(\mathbf{a}_k\) 和 \(\mathbf{a}_k^*\) 表示出来
\[
\mathbf{E} = - \frac{\partial}{\partial t} \mathbf{A} = - \sum_k \left[ \frac{d}{d t} \mathbf{a}_k(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \frac{d}{d t} \mathbf{a}_k^*(t) e^{-i \mathbf{k} \cdot \mathbf{x}} \right]
\]
\[
\begin{aligned}
\mathbf{B} &= \nabla \times \mathbf{A} = \sum_k \left[ (\nabla e^{i \mathbf{k} \cdot \mathbf{x}}) \times \mathbf{a}_k(t) + (\nabla e^{-i \mathbf{k} \cdot \mathbf{x}}) \times \mathbf{a}_k^*(t) \right] \\\\
&= i \sum_{k} e^{i \mathbf{k} \cdot \mathbf{x}} \mathbf{k} \times \mathbf{a}_{k} (t) - e^{-i \mathbf{k} \cdot \mathbf{x}} \mathbf{k} \times \mathbf{a}_{k}^{*} (t)
\end{aligned}
\]
2. 推迟势
作业： (a) 书5.5
5.5 设 \(\mathbf{A}\) 和 \(\phi\) 是满足洛伦兹规范的矢势和标势
(1) 引入一矢量函数 \(\mathbf{Z} (\vec{x}, t)\)（赫兹矢量），若令 \(\phi = -\nabla \cdot \mathbf{Z}\)，证明 \(\mathbf{A} = \frac{1}{c^2} \frac{\partial \mathbf{Z}}{\partial t}\)
由于 \(\nabla \cdot \mathbf{A} + \frac{1}{c^{2}} \frac{\partial}{\partial t} \phi = 0\)
\[
\nabla \cdot \mathbf{A} = \frac{1}{c^{2}} \frac{\partial}{\partial t} \nabla \cdot \mathbf{Z}
\]
所以可以选取合适的 \(\mathbf{Z}\) 使得
\[
\mathbf{A} = \frac{1}{c^2} \frac{\partial \mathbf{Z}}{\partial t}
\]
(2) 若令 \(\rho = -\nabla \cdot \mathbf{P}\) 证明 \(\mathbf{Z}\) 满足方程 \(\nabla^2 \mathbf{Z} - \frac{1}{c^2} \frac{\partial^2 \mathbf{Z}}{\partial t^2} = -c^2 \mu_0 \mathbf{P}\)，写出在真空中的推迟解
证明：
\[
\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_{0}}
\]
\[
\nabla \cdot (- \nabla \phi - \frac{\partial}{\partial t} \mathbf{A}) = \frac{\rho}{\epsilon_{0}}
\]
\[
\nabla \cdot (- \nabla (- \nabla \cdot \mathbf{Z}) - \frac{\partial}{\partial t} \frac{1}{c^2} \frac{\partial \mathbf{Z}}{\partial t}) = \frac{- \nabla \cdot \mathbf{P}}{\epsilon_{0}}
\]
\[
\nabla^{2} (\nabla \cdot \mathbf{Z}) - \frac{1}{c^{2}} \frac{\partial}{\partial t} (\nabla \cdot \mathbf{Z}) = - c^{2} \mu_{0} (\nabla \cdot \mathbf{P})
\]
所以可以选取合适的 \(\mathbf{Z}\) 以及 \(\mathbf{P}\) 使得
\[
\nabla^2 \mathbf{Z} - \frac{1}{c^2} \frac{\partial^2 \mathbf{Z}}{\partial t^2} = -c^2 \mu_0 \mathbf{P}
\]
在真空中，\(\mathbf{Z}\) 存在推迟解
\[
\mathbf{Z} (\mathbf{r} ,t) = \int \frac{c^{2} \mu_{0} \mathbf{P} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})}{4 \pi |\mathbf{r} - \mathbf{r'}|} d \tau'
\]
(3) 证明 \(\mathbf{E}\) 和 \(\mathbf{B}\) 可通过 \(\mathbf{Z}\) 用下列公式表出：
\[
\mathbf{E} = \nabla \times (\nabla \times \mathbf{Z}) - c^2 \mu_0 \mathbf{P}, \quad \mathbf{B} = \frac{1}{c^2} \frac{\partial}{\partial t} \nabla \times \mathbf{Z}
\]
证明：
\[
\mathbf{E} = - \nabla \phi - \frac{\partial}{\partial t} \mathbf{A} = \nabla (\nabla \cdot \mathbf{Z}) - \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} \mathbf{Z} = \nabla (\nabla \cdot \mathbf{Z}) - \nabla^{2} \mathbf{Z} - c^{2} \mu_{0} \mathbf{P} = \nabla \times (\nabla \times \mathbf{Z}) - c^2 \mu_0 \mathbf{P}
\]
\[
\mathbf{B} = \nabla \times \mathbf{A} = \nabla \times \frac{1}{c^2} \frac{\partial \mathbf{Z}}{\partial t} = \frac{1}{c^2} \frac{\partial}{\partial t} \nabla \times \mathbf{Z}
\]
(b) 试证，库伦规范的解满足\(\(\nabla \cdot \mathbf{A} = 0\)\)
证明：库伦规范下的解为
\[
\phi (\mathbf{r}, t) = \int \frac{\rho (\mathbf{r}, t)}{4 \pi \epsilon_{0} |\mathbf{r} - \mathbf{r'}|} d \tau'
\]
\[
\mathbf{A} (\mathbf{r}, t) = \int \frac{\mu_{0} (\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}))}{4 \pi |\mathbf{r} - \mathbf{r'}|} d \tau'
\]
所以矢势的散度为：（以下用 \(\nabla'_{r'}\) 表示仅仅对于空间部分的散度，即不考虑散度中推迟项带来的影响）
\[
\begin{aligned}
\nabla \cdot \mathbf{A} &= \nabla \cdot \int \frac{\mu_{0} (\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}))}{4 \pi |\mathbf{r} - \mathbf{r'}|} d \tau' \\\\
&= \frac{\mu_{0}}{4 \pi} \int \left( - \nabla' \frac{1}{|\mathbf{r} - \mathbf{r'}|}\right) \cdot \left(\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right) + \\\\
&\frac{1}{|\mathbf{r} - \mathbf{r'}|} \nabla \cdot \left(\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right) d \tau' \\\\
&= \frac{\mu_{0}}{4 \pi} \int - \nabla' \frac{\left(\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right)}{|\mathbf{r} - \mathbf{r'}|} + \\\\
&\frac{1}{|\mathbf{r} - \mathbf{r'}|} \nabla' \left(\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right) + \\\\
&\frac{1}{|\mathbf{r} - \mathbf{r'}|} \frac{\partial}{\partial t} \left(\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right) \cdot \nabla (t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) \, d \tau' \\\\
&= \frac{\mu_{0}}{4 \pi} \iint_{S'} - d \mathbf{S'} \cdot \frac{\left(\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right)}{|\mathbf{r} - \mathbf{r'}|} + \\\\
&\int \frac{1}{|\mathbf{r} - \mathbf{r'}|} \nabla'_{r'} \cdot \left(\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right) + \\\\
& \frac{1}{|\mathbf{r} - \mathbf{r'}|} \frac{\partial}{\partial t} \left(\mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} \nabla'_{r'} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right) \cdot (\nabla (t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) + \nabla' (t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})) \, d \tau' \\\\
&= \frac{\mu_{0}}{4 \pi} \int \frac{1}{|\mathbf{r} - \mathbf{r'}|} \left( \nabla'_{r'} \cdot \mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) - \epsilon_{0} \frac{\partial}{\partial t} {\nabla'_{r'}}^{2} \phi (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})\right) \, d \tau' \\\\
&= \frac{\mu_{0}}{4 \pi} \int \frac{1}{|\mathbf{r} - \mathbf{r'}|} (\nabla'_{r'} \cdot \mathbf{j} (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c}) + \frac{\partial}{\partial t} \rho (\mathbf{r'}, t - \frac{|\mathbf{r} - \mathbf{r'}|}{c})) \, d \tau' = 0
\end{aligned}
\]
其中最后一步是流守恒的直接结果。