电磁波与光波
约 1955 个字 6 张图片 预计阅读时间 7 分钟
电磁波谱
概念
-
可见光(visible light):\(\lambda = 4000A - 7000A\)
-
红外线(infrared):\(\lambda = 0.7 \mu m - 1 mm\)
-
用于夜视镜、辐射温度计、红外灯
-
微波(microwave):\(\lambda = 1mm - 1m\)
- 无线电波(radio waves):\(\lambda > 1m\)
- 中波(MW):\(\lambda = 3km - 50m\)
- 短波(SW):\(\lambda = 50m - 10m\)
-
超短波(extra-SW):\(\lambda < 1m\)
-
紫外线(ultraviolet):\(\lambda = 1nm - 400 nm\)
-
X-rays:\(\lambda = 0.01 - 10nm\)
-
gamma rays:\(\lambda < 10 pm\)
电磁波的产生与发射
我们可以从麦克斯韦方程组推导出电磁波的性质
积分形式:
\[
\iint \vec{E} \cdot d\vec{A} = \frac{q_{0}}{\epsilon_{0}}
\]
\[
\iint \vec{B} \cdot d \vec{A} = 0
\]
\[
\oint \vec{E} \cdot d \vec{l} = - \iint \frac{\partial \vec{B}}{\partial t} \cdot d \vec{A}
\]
\[
\oint \vec{H} \cdot d \vec{l} = i_{0} + \iint \frac{\partial \vec{D}}{\partial t} \cdot d \vec{A}
\]
微分形式(自由空间:\(\rho_{0} =0,\vec{J_{0}} = 0\)):
\[
\nabla \cdot \vec{E} = 0
\]
\[
\nabla \cdot \vec{B} = - \frac{\partial \vec{B}}{\partial t} = - \kappa_{m} \mu_{0} \frac{\partial \vec{H}}{\partial t}
\]
\[
\nabla \times \vec{H} = 0
\]
\[
\nabla \times \vec{H} = \kappa_{e} \epsilon_{0} \frac{\partial \vec{E}}{\partial t}
\]
分量形式(把上式中的\(\nabla\)展开,得到如下式子):
\[
\frac{\partial E_{x}}{\partial x} + \frac{\partial E_{y}}{\partial y} + \frac{\partial E_{z}}{\partial z} =0
\]
\[
\begin{vmatrix}
\widehat{i} && \widehat{j} && \widehat{k} \\
\frac{\partial}{\partial x} && \frac{\partial}{\partial y} && \frac{\partial}{\partial z} \\
\vec{E_{x}} && \vec{E_{y}} && \vec{E_{z}}
\end{vmatrix}
= -\kappa_{m} \mu_{0}(\frac{\partial \vec{H_{x}}}{\partial t} \widehat{i} + \frac{\partial \vec{H_{y}}}{\partial t} \widehat{j} + \frac{\partial \vec{H_{z}}}{\partial t} \widehat{k})
\]
\[
\frac{\partial \vec{H_{x}}}{\partial x} + \frac{\partial \vec{H_{y}}}{\partial y} + \frac{\partial \vec{H_{z}}}{\partial z} = 0
\]
\[
\begin{vmatrix}
\widehat{i} && \widehat{j} && \widehat{k} \\
\frac{\partial}{\partial x} && \frac{\partial}{\partial y} && \frac{\partial}{\partial z} \\
\vec{H_{x}} && \vec{H_{y}} && \vec{H_{z}}
\end{vmatrix}
= -\kappa_{m} \mu_{0}(\frac{\partial \vec{E_{x}}}{\partial t} \widehat{i} + \frac{\partial \vec{E_{y}}}{\partial t} \widehat{j} + \frac{\partial \vec{E_{z}}}{\partial t} \widehat{k})\notag
\]
平面波
首先假设单一波源,在很远的自由空间中,在球面上取一弧面,可以近似为平面波;以其传播方向为\(z\)轴,电场和磁场分别为\(x\)和\(y\)轴.
然后将上边的式子展开,可以得到以下八个方程
\[
\frac{\partial E_{x}}{\partial x} + \frac{\partial E_{y}}{\partial y} + \frac{\partial E_{z}}{\partial z} = 0 \tag{1} \\
\]
\[
\frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z} = -\kappa_{m} \mu_{0} \frac{\partial H_{x}}{\partial t} \tag{2-1}
\]
\[
\frac{\partial E_{x}}{\partial z} - \frac{\partial E_{z}}{\partial x} = -\kappa_{m} \mu_{0} \frac{\partial H_{y}}{\partial t} \tag{2-2}
\]
\[
\frac{\partial E_{y}}{\partial x} - \frac{\partial E_{x}}{\partial y} = -\kappa_{m} \mu_{0} \frac{\partial H_{z}}{\partial t} \tag{2-3}
\]
\[
\frac{\partial H_{x}}{\partial x} + \frac{\partial H_{y}}{\partial y} + \frac{\partial H_{z}}{\partial z} =0 \tag{3}
\]
\[
\frac{\partial H_{z}}{\partial y} - \frac{\partial H_{y}}{\partial z} =\kappa_{e} \epsilon_{0} \frac{\partial E_{x}}{\partial t} \tag{4-1}
\]
\[
\frac{\partial H_{x}}{\partial z} - \frac{\partial H_{z}}{\partial x} =\kappa_{e} \epsilon_{0} \frac{\partial E_{y}}{\partial t} \tag{4-2}
\]
\[
\frac{\partial H_{y}}{\partial x} - \frac{\partial H_{x}}{\partial y} =\kappa_{e} \epsilon_{0} \frac{\partial E_{z}}{\partial t} \tag{4-3}
\]
横波
首先,横波在x和y方向的电场强度和磁场强度都是一样的,不会发生变化,所以
\[
\frac{\partial E_{x}}{\partial x} = \frac{\partial E_{y}}{\partial y} = \frac{\partial H_{x}}{\partial x} = \frac{\partial H_{y}}{\partial y} = 0 \notag
\]
则由\((1)\)式,我们有
\[
\frac{\partial E_{x}}{\partial x} = 0 \notag
\]
由\((2-3)\)式,有
\[
\frac{\partial H_{z}}{\partial t} = 0 \notag
\]
同样地,由\((3)\)式可得
\[
\frac{\partial H_{z}}{\partial z} = 0 \notag
\]
由\((4-3)\),有
\[
\frac{\partial E_{z}}{\partial t} = 0 \notag
\]
所以电场和磁场在\(z\)轴的分量与时间和z轴都无关,可以设为\(constant\)
\[
E \perp k ,H \perp k
\notag
\]
电场垂直磁场
运用\(E_{z} = H_{z} = 0\),我们 有
\((2-1)\)式
\[
\frac{\partial E_{y}}{\partial z} =\kappa_{m} \mu_{0} \frac{\partial H_{x}}{\partial t} \tag{2-1} \notag
\]
\((2-2')\)式
\[
\frac{\partial E_{x}}{\partial z} =-\kappa_{m} \mu_{0} \frac{\partial H_{y}}{\partial t} \tag{2-2'}
\]
\((4-1')\)式
\[
\frac{\partial H_{y}}{\partial z} =-\kappa_{e} \epsilon_{0} \frac{\partial E_{x}}{\partial t} \tag{4-1'}
\]
\((4-2)\)式
\[
\frac{\partial H_{x}}{\partial z} =\kappa_{e} \epsilon_{0} \frac{\partial E_{y}}{\partial t} \notag
\]
上边四个式子只包含\(E_{y},E_{x},H_{y},H_{X}\),说明电场,磁场只在\(x,y\)方向有分量.
由于\(x,y\)的方向是任意的,那么我们取\(x\)的方向为电场方向,就有
\[
\frac{\partial H_{x}}{\partial z} = 0 = \frac{\partial H_{x}}{\partial t}
\notag
\]
所以磁场强度方向与电场强度方向垂直,我们就证明了\(\vec{E} \perp \vec{H}\)
Note
其实就是把\(E_{y} = 0\)带入\((2-1)\)和\((4-2)\),就得到的上边的结论.
波动方程
麦克斯韦:原来光就是电磁波
对\((2-2')\)式同时对\(t\)求偏导
\[
\frac{\partial^{2} E_{x}}{\partial z^{2}} = - \kappa_{m} \mu_{0}\frac{\partial}{\partial t}\frac{\partial H_{y}}{\partial z} = \kappa_{m}\mu_{0}K_{e}\epsilon_{0} \frac{\partial^{2} E_{x}}{\partial t^{2}} \notag
\]
同理对\((4-1')\)操作,得到如下方程
\[
\frac{\partial^{2} E_{x}}{\partial z^{2}} - \kappa_{e}\mu_{0}K_{m}\epsilon_{0} \frac{\partial^{2} E_{x}}{\partial t^{2}} = 0 \notag
\]
\[
\frac{\partial^{2} H_{y}}{\partial z^{2}} - \kappa_{e}\mu_{0}K_{m}\epsilon_{0} \frac{\partial^{2} H_{y}}{\partial t^{2}} = 0 \notag
\]
猜根,有:
\[
\left\{
\begin{matrix}
E_{x} = E_{x0}e^{i(\omega t - kz)} \\
H_{y} = H_{y0}e^{i(\omega t - kz)}
\end{matrix}
\right.
\notag
\]
\(\omega = \frac{2\pi}{T}\)是角频率,\(k = \frac{2\pi}{\lambda}\)是波矢,也叫波数
带回方程,得到
\[
k^{2} = \kappa_{e}\epsilon_{0}\kappa_m\mu_0\omega^2
\Rightarrow
k = \omega\sqrt{\kappa_e\epsilon_0\kappa_m\mu_0}
\notag
\]
又因为
\[
v = \frac{\omega}{k} = \frac{1}{\sqrt{\kappa_e\epsilon_0\kappa_m\mu_0}} \notag
\]
Note
而真空中,磁导率\(\kappa_m\)和介电常数\(\kappa_e\)都为1,所以代入计算得到\(v = c = 3.0 \times 10^8 m/s\),我们就计算出了光速
我们定义\(\sqrt{\kappa_e\kappa_m} = n\),就是折射率,所以可以推导出光学中的
\[
v= \frac{c}{n} \notag
\]
电场和磁场
由\((2-2')\)式,将我们猜根得到的\(E_{x},H_{y}\)代入,得到如下式子
\[
\begin{align*}
-ikE_{x0}e^{i(\omega t - kx)} &= -\kappa_m\mu_0i\omega H_{y0}e^{i(\omega t - kx)} \\
kE_{x0} &= \kappa_m\mu_0\omega H_{y0} \\
E_{x0} &= \kappa_m\mu_0\frac{\omega}{k}H_{y0} = \kappa_m\mu_0v H_{y0} \\
&= \kappa_m\mu_0\frac{1}{\sqrt{\kappa_e\epsilon_0\kappa_m\mu_0}}H_{y0} \\
\sqrt{\kappa_e\epsilon_0}E_{x0} &= \sqrt{\kappa_m\mu_0}H_{y0} \\
\sqrt{\kappa_e\epsilon_0}E_{x0}e^{i\phi_E} &= \sqrt{\kappa_m\mu_0}H_{y0}e^{i\phi_H}
\end{align*}
\notag
\]
通过上式我们可得以下两个方程
\[
\left\{
\begin{matrix}
\sqrt{\kappa_e\epsilon_0}E_{0} = \sqrt{\kappa_m\mu_0}H_{0} (振幅相等) \\
\phi_{E} = \phi_{H} (相位相同)
\end{matrix}
\right.
\notag
\]
真空中,\(\kappa_e = \kappa_m = 1\)
所以
\[
\sqrt{\epsilon_0}E_{0} = \sqrt{\mu_0}H_0 \\
\Rightarrow E_{0} = \frac{\mu_0H_0}{\sqrt{\epsilon_0\mu_0}} = cB_{0}(c为光速) \notag
\]
电场强度与磁感应强度
我们发现\(E_{0} = cB_{0}\),电场强度和磁感应强度之间只差了一个常数
电磁波的能流密度和动量
单位体积内电磁波的能量包括电场和磁场两部分
- 电场能量:\(U_E = \frac{1}{2}\epsilon_0E^2\)
- 磁场能量:\(U_B = \frac{1}{2}\frac{B^2}{\mu_0}\)
则单位体积内电磁波的能量:
\[
U = \iiint (\frac{1}{2}\epsilon_0E^2 + \frac{1}{2}\frac{B^2}{\mu_0})dv \notag
\]
更一般的,我们知道\(\vec{D} = \kappa_e\epsilon_0\vec{E},\vec{B} = \kappa_m\mu_0\vec{H}\),那么
\[
\begin{align*}
U = U_E + U_B &= \iiint(\frac{1}{2}\vec{D}\cdot \vec{E} + \frac{1}{2}\vec{B}\cdot\vec{H})dv \\
\frac{dU}{dt} &= \frac{d}{dt}\iiint(\frac{1}{2}\vec{D}\cdot \vec{E} + \frac{1}{2}\vec{B}\cdot\vec{H})dv \\
&= \frac{1}{2}\iiint \frac{\partial}{\partial t}(\vec{D}\cdot\vec{E} + \vec{B} \cdot\vec{H})dv
\end{align*}
\]
对积分内部展开:
\[
\begin{align*}
\frac{\partial}{\partial t}(\vec{D}\cdot\vec{E} + \vec{B} \cdot\vec{H}) &= \kappa_e\epsilon_0\frac{\partial}{\partial t}(\vec{E} \cdot \vec{E}) + \kappa_m\mu_0\frac{\partial}{\partial t}(\vec{H}\cdot\vec{H}) \notag \\
&= 2\kappa_e\epsilon_0\vec{E}\cdot\frac{\partial \vec{E}}{\partial t} + 2\kappa_m\mu_0\vec{H}\cdot\frac{\partial \vec{H}}{\partial t} \\
&= 2\vec{E}\cdot \frac{\partial \vec{D}}{\partial t} + 2\vec{H} \cdot \frac{\partial \vec{B}}{\partial t}
\end{align*}
\]
在麦克斯韦方程中:
\[
\frac{\partial \vec{D}}{\partial t} = \nabla \times \vec{H} - \vec{J_{0}}
\]
\[
\frac{\partial \vec{B}}{\partial t} = -\nabla \times \vec{E} \notag
\]
代入上式可得:
\[
\begin{align*}
&= 2\vec{E} \cdot (\nabla \times \vec{H} - \vec{J_{0}}) - 2\vec{H} \cdot (\nabla \times \vec{E}) \\
&= 2[\vec{E} \cdot (\nabla \times \vec{H}) - \vec{H} \cdot (\nabla \times \vec{E}) - \vec{J_{0}} \cdot \vec{E}] \\
&= -2\nabla \cdot (\vec{E} \times \vec{H}) - 2\vec{J_{0}}\cdot \vec{E} \notag
\end{align*}
\]
最后运用高斯定理化简:
\[
\begin{align*}
\frac{dU}{dt} &= - \iiint\nabla \cdot (\vec{E} \times \vec{H})dv - \iiint (J_{0} \cdot \vec{E})dv \notag \\
&= -\iint(\vec{E} \times \vec{H})\cdot dA - \iiint (J_{0} \cdot \vec{E})dv
\end{align*}
\]
我们现在关注第二项到底是什么意思
在欧姆定律中,我们有以下公式,\(\vec{E}\)为电场,\(\vec{K}\)为非静电力
\[
\vec{J_{0}} = \sigma(\vec{E} + \vec{K}) \notag \\
\Rightarrow \vec{E} = \frac{1}{\sigma}\vec{J_{0}} - \vec{K} = \rho J_{0} - \vec{K}
\]
把这个积分放到均匀的圆筒里边,那么我们对\(v\)积分其实就是乘以\(\Delta A \cdot \Delta l\),从而进行如下变换
\[
\begin{align*}
\iiint(J_{0} \cdot \vec{E})dv &= (J_{0} \cdot \vec{E})\Delta A \cdot \Delta l \\
&= J_{0} \cdot (\rho J_{0} - \vec{K})\Delta A \cdot \Delta l \\
&= \rho J_{0}^{2}\Delta A - J_{0}\cdot \vec{K}\Delta A \cdot \Delta l \\
&= \rho \frac{\Delta l}{\Delta A}(J_{0}\Delta A)^{2} - (J_{0}\Delta A)(\vec{K}\cdot \Delta l) \\
&= Ri_{0}^{2} - I_{0}\Delta \epsilon(\Delta \epsilon是电动势) \\
&= Q - P
\end{align*}
\notag
\]
所以我们记\(\vec{S} = \vec{E} \times \vec{H}\),为\(Poynting \ Vertor(玻印廷矢量)\),那么
\[
\frac{dU}{dt} = - \iint \vec{S}\cdot d\vec{A} - Q + P \notag
\]
理解上式
\(\frac{dU}{dt}\)是单位时间内电场能量与磁场能量之和的变化;
\(- \iint \vec{S}\cdot d\vec{A}\)是通过表面向外辐射的能量;
\(Q\)是产生的热量
Poynting Vertor
定义单位时间,单位面积内的能量流动
\[
\vec{S} = \vec{E} \times \vec{H} = \frac{\vec{E} \times \vec{B}}{\mu_0} = \frac{\vec{E}^{2}}{{\mu_0}c} \notag
\]
定义\(Z_{0} = \mu_0c = 377 \Omega\)
从而\(S = \frac{E^2}{377\Omega}\)
电磁波的强度
电磁波的强度\(I\)实际上就是\(S\)的平均值
\[
I = \langle S\rangle = \frac{\langle E^2\rangle}{Z_0} = \frac{E_{max}^2}{377\Omega}\langle sin^2(kz-wt)\rangle = \frac{1}{2}\frac{E_{max}^2}{377\Omega}
\]
电场能量密度和磁场能量密度的关系
由于\(\mu_{E} = \frac{1}{2}\epsilon_{0}E^2,\mu_B = \frac{1}{2}\frac{B^2}{\mu_0}\)
而\(B = \frac{E}{c}\),那么\(\mu_B = \frac{1}{2}\frac{E^2}{C^2\mu_0} = \frac{1}{2}\epsilon_0E^2 = \mu_E\)
所以二者能量各占一半
电磁波的能量密度可以表示为:\(\mu = \mu_E + \mu_B = \epsilon_0E^2\)
电磁波的强度
\[
I = c\langle\mu\rangle = c\epsilon_0\langle E^2\rangle = c\epsilon_0\frac{E_{max}^2}{2} = \frac{1}{2}\frac{E_{max}^{2}}{\mu_0c} = \frac{E_{max}^2}{2 \cdot 377 \Omega} = \frac{E_{rms}^2}{377\Omega}
\]
电路中的能量传输
如上图所示电路,考虑与电源正极相连的导线,导线内部存在一个电场,那么由于\(\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt} = 0\),导体外部一定存在一个方向相同的电场.
再加上一个垂直导线的电场,根据\(\vec{S} = \vec{E} \times \vec{H}\),我们可以得到能量流动的方向,一方面流向电阻,另一方面被导线消耗.
与电源负极相连的导线也是类似的
电磁波的动量
假设一个有一个力 \(\Delta \vec{F}\),这个力会让电荷做功,即\(\Delta W = \Delta F \cdot \Delta l\);而这部分功就是这个物体吸收的净能量;
\[
\Delta \vec{F} \cdot c \Delta t = (\vec{S_{in}} - \vec{S_{out}}) \cdot \Delta A \Delta t \notag
\]
故
\[
\Delta \vec{F} = \frac{1}{c}(\vec{S_{in}} - \vec{S_{out}}) \cdot \Delta A \notag
\]
矢量减
光压
单位面积上的力
\[
\vec{P} = \frac{1}{c}(\vec{S_{in}} - \vec{S_{out}}) \notag
\]
动量密度
单位体积内的动量
\[
\Delta g = \frac{F \cdot \Delta t}{\Delta Ac\Delta t} = \frac{F}{c\Delta A} = \frac{1}{c^2}(\vec{S_{in}} - \vec{S_{out}}) \notag
\]
动量密度
\(g_{in} = \frac{1}{c^2}\vec{S_{in}}\)为入射光的动量密度,\(g_{out} = \frac{1}{c^2}\vec{S_{out}}\)为反射光的动量密度
光压
对于白体,\(S_{in} = S_{out}\),故\(g_{in} = g_{out}\)
\[
P = \frac{2}{c}\vec{S_{in}} \notag
\]
对于黑体,\(S_{out} = 0\),故\(g_{in} = \frac{1}{c^2}\vec{S_{in}}\)
\[
P = \frac{1}{c}\vec{S_{in}}\notag
\]