贡献者: addis; 白玫瑰
预备知识 麦克斯韦方程组(介质)
,
矢量算符运算法则,
平面波
从麦克斯韦方程组出发,我们证明电磁场传播具有波动性。
麦克斯韦方程组为:
\begin{equation}
\begin{aligned}
& \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{D}} = \rho ~, \\
& \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{B}} = 0 ~, \\
& \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{E}} = -\frac{\partial \boldsymbol{\mathbf{B}} }{\partial t} ~, \\
& \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{H}} = \boldsymbol{\mathbf{j}} + \frac{\partial \boldsymbol{\mathbf{D}} }{\partial t} ~.
\end{aligned}
\end{equation}
将 $ \boldsymbol{\mathbf{D}} $ 和 $ \boldsymbol{\mathbf{H}} $ 转化为 $ \boldsymbol{\mathbf{E}} $ 和 $ \boldsymbol{\mathbf{B}} $:
\begin{equation}
\begin{aligned}
& \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{E}} = \frac{\rho}{\epsilon} ~, \\
& \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{B}} = 0 ~, \\
& \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{E}} = -\frac{\partial \boldsymbol{\mathbf{B}} }{\partial t} ~, \\
& \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{B}} = \mu( \boldsymbol{\mathbf{j}} + \epsilon \frac{\partial \boldsymbol{\mathbf{E}} }{\partial t}) ~.
\end{aligned}
\end{equation}
讨论在无限大各向同性均匀介质中的情况,此时 $\epsilon$ 和 $\mu$ 均为常数,并且在远离辐射源的地方,不存在自由电荷和传导电流,即 $\rho = 0$ 和 $ \boldsymbol{\mathbf{j}} = 0$,此时方程组化为:
\begin{equation}
\begin{aligned}
& \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{E}} = 0 ~, \\
& \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{B}} = 0 ~, \\
& \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{E}} = -\frac{\partial \boldsymbol{\mathbf{B}} }{\partial t} ~, \\
& \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{B}} = \epsilon \mu\frac{\partial \boldsymbol{\mathbf{E}} }{\partial t} ~.
\end{aligned}
\end{equation}
取第 3 式的旋度,并将第 4 式代入:
\begin{equation}
\boldsymbol{\nabla}\boldsymbol{\times} ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{E}} ) = \boldsymbol{\nabla} ( \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{E}} ) - \boldsymbol{\nabla}^2 \boldsymbol{\mathbf{E}} = - \frac{\partial}{\partial{t}} ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{B}} ) = -\epsilon \mu \frac{\partial^{2}}{\partial{t}^{2}} \boldsymbol{\mathbf{E}} ~.
\end{equation}
整理得:
\begin{equation}
\boldsymbol{\nabla}^2 \boldsymbol{\mathbf{E}} - \epsilon \mu \frac{\partial^2 \boldsymbol{\mathbf{E}} }{\partial t^2} = 0 ~.
\end{equation}
令:
\begin{equation}
v = \frac{1}{\sqrt{\epsilon \mu}} ~.
\end{equation}
所以我们得到:
\begin{equation}
\boldsymbol{\nabla}^2 \boldsymbol{\mathbf{E}} - \frac{1}{v^2} \frac{\partial^{2}{ \boldsymbol{\mathbf{E}} }}{\partial{t}^{2}} = 0 ~.
\end{equation}
这就是电场的波动方程。同理我们可以得到磁场的波动方程。将两式列出,我们得到了波动方程:
\begin{equation}
\begin{aligned}
& \boldsymbol{\nabla}^2 \boldsymbol{\mathbf{E}} - \frac{1}{v^2} \frac{\partial^{2}{ \boldsymbol{\mathbf{E}} }}{\partial{t}^{2}} = 0 ~, \\
& \boldsymbol{\nabla}^2 \boldsymbol{\mathbf{B}} - \frac{1}{v^2} \frac{\partial^{2}{ \boldsymbol{\mathbf{E}} }}{\partial{t}^{2}} = 0 ~.
\end{aligned}
\end{equation}
电场的各个分量分别满足三维波动方程。
它的解为平面波
\begin{equation}
\boldsymbol{\mathbf{E}} ( \boldsymbol{\mathbf{r}} , t) = \boldsymbol{\mathbf{E}} _0 \cos\left( \boldsymbol{\mathbf{k}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} - \omega t\right) ~,
\end{equation}
其中 $\omega = c \left\lvert \boldsymbol{\mathbf{k}} \right\rvert = ck$。而通解是这些平面波的任意线性组合。注意如果 $ \boldsymbol{\mathbf{E}} _0$ 中存在平行于 $ \boldsymbol{\mathbf{k}} $ 的分量,那么 $ \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{E}} \ne 0$,所以二者必须垂直,即 $ \boldsymbol{\mathbf{E}} \boldsymbol\cdot \boldsymbol{\mathbf{k}} = 0$。电场的通解可表示为
\begin{equation}
\boldsymbol{\mathbf{E}} ( \boldsymbol{\mathbf{r}} , t) = \int \boldsymbol{\mathbf{E}} _0( \boldsymbol{\mathbf{k}} ) \cos\left( \boldsymbol{\mathbf{k}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} - \omega k t\right) \,\mathrm{d}^{3}{k} ~.
\end{equation}
根据 $ \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{E}} = - \partial \boldsymbol{\mathbf{B}} /\partial t $,可求出式 9 伴随的磁场为
\begin{equation}
\boldsymbol{\mathbf{B}} ( \boldsymbol{\mathbf{r}} , t) = \boldsymbol{\mathbf{B}} _0 \cos\left( \boldsymbol{\mathbf{k}} \boldsymbol\cdot \boldsymbol{\mathbf{r}} - \omega t\right) ~.
\end{equation}
其中 $ \boldsymbol{\mathbf{B}} _0$ 的模长为 $ \left\lvert \boldsymbol{\mathbf{B}} _0 \right\rvert = \left\lvert \boldsymbol{\mathbf{E}} _0 \right\rvert /c$,于 $ \boldsymbol{\mathbf{E}} _0$ 垂直,方向满足 $ \hat{\boldsymbol{\mathbf{E}}} \boldsymbol\times \hat{\boldsymbol{\mathbf{B}}} = \hat{\boldsymbol{\mathbf{k}}} $。可见
电磁波是横波。
1. 介质中
非线性光学中一般认为介质具有 $\mu = \mu_0$,且假设 $ \boldsymbol{\nabla}\boldsymbol{\cdot} \boldsymbol{\mathbf{E}} = 0$ 仍然成立
介质中没有自由电荷或自由电流。
类似真空情况的推导过程,有
\begin{equation}
\boldsymbol{\nabla}\boldsymbol{\times} ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{E}} ) = - \frac{\partial}{\partial{t}} ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{B}} ) = -\mu_0 \frac{\partial}{\partial{t}} ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{H}} )
= -\mu_0 \frac{\partial^{2}}{\partial{t}^{2}} \boldsymbol{\mathbf{D}} ~.
\end{equation}
把电位移矢量的定义 $ \boldsymbol{\mathbf{D}} = \epsilon_0 \boldsymbol{\mathbf{E}} + \boldsymbol{\mathbf{P}} $ 代入上式,化简为
\begin{equation}
\boldsymbol{\nabla}\boldsymbol{\times} ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{E}} ) = - \frac{\partial}{\partial{t}} ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{B}} ) = -\mu_0 \frac{\partial}{\partial{t}} ( \boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\mathbf{H}} )
= -\mu_0 \frac{\partial^{2}}{\partial{t}^{2}} \boldsymbol{\mathbf{D}} ~.
\end{equation}