Chapter 8
一、无损耗介质
由亥姆霍兹方程可得
$$
\nabla^2\vec E+k_0^2\vec E=0
$$
\[
k=\omega\sqrt{\mu\epsilon}
\]
考虑\(E_x\)分量上的亥姆霍兹方程，且在垂直于\(z\)的平面上\(E_x\)是均匀的
上述方程可化为
\[
\frac{d^2E_x}{dz^2}+k_0^2E_x=0
\]
解得
\[
E_x(z)=E_0^+(z)e^{-jkz}+E_0^-(z)e^{jkz}
\]
只考虑一个方向波的传播
\[
E_x(z)=E_0^+(z)e^{-jkz}
\]
\[
E_x(z,t)=Re[E_0^+(z)e^{-jkz}e^{j\omega t}]
\]
其中\(k\)为空间频率，\(\omega\)为时间频率
可得相速度
$$
u_p=\frac \omega k=\frac 1{\sqrt{\mu\epsilon}}
$$
波长
$$
\lambda=\frac{2\pi}k
$$
再根据电场的旋度的公式可得
\[
H_y^+(z)=\frac1{\eta}E_x^+(z)
\]
阻抗
$$
\eta=\frac \mu \epsilon
$$
二、损耗介质
由亥姆霍兹方程可得
$$
\nabla^2\vec E+k_0^2\vec E=0
$$
\[
k=\omega\sqrt{\mu\epsilon_c}
\]
\[
\epsilon_c=\epsilon'+j\epsilon''=\epsilon'-j\frac \sigma \omega
\]
令\(k=\beta-j\alpha，\gamma=jk=\alpha+j\beta\)，只考虑\(z\)轴正方向传播的波，解方程即得
\[
E_x=E_0^+e^{-jkz}=E_0^+e^{-\alpha z}e^{-j\beta z}
\]
1. 低损耗介质
低损耗状态下，\(\epsilon'>>\epsilon''\)，采用二项式展开近似
\[
\gamma=j\omega\sqrt{\mu(\epsilon'+j\epsilon'')}=j\omega\sqrt{\mu\epsilon'}(1+j\frac{\epsilon''}{\epsilon''})^{\frac12}=j\omega\sqrt{\mu\epsilon'}[1-j\frac{\epsilon''}{2\epsilon'}+\frac18(\frac{\epsilon''}{\epsilon'})^2]
\]
\[
\alpha \cong \frac{\omega \epsilon^{\prime \prime}}{2} \sqrt{\frac{\mu}{\epsilon^{\prime}}}
\]
\[
\beta \cong \omega \sqrt{\mu \epsilon^{\prime}}\left[1+\frac{1}{8}\left(\frac{\epsilon^{\prime \prime}}{\epsilon^{\prime}}\right)^{2}\right]
\]
\[
\begin{aligned} \eta_{c} & =\sqrt{\frac{\mu}{\epsilon^{\prime}}}\left(1-j \frac{\epsilon^{\prime \prime}}{\epsilon^{\prime}}\right)^{-1 / 2}\cong \sqrt{\frac{\mu}{\epsilon^{\prime}}}\left(1+j \frac{\epsilon^{\prime \prime}}{2 \epsilon^{\prime}}\right)\end{aligned}
\]
2. 良导体
即\(\epsilon'<<\epsilon''\)
$$
\gamma=j\omega\sqrt{\mu\epsilon_c}=j\omega\sqrt{\mu(\epsilon'-j\frac \sigma\omega)}\approx j\omega\sqrt{-\mu j\frac \sigma\omega}=j\omega\sqrt{\mu \frac \sigma\omega}\frac{1-j}{\sqrt2}=j\sqrt{\mu\omega\sigma}\frac{1-j}{\sqrt2}
$$
又\(\omega=2\pi f\)，
$$
\gamma=\sqrt{\pi f\mu \sigma}(j+1)
$$
$$
\alpha=\beta=\sqrt{\pi f \mu \sigma}
$$
$$
\eta_{c}=\sqrt{\frac{\mu}{\epsilon_{c}}} \cong \sqrt{\frac{j \omega \mu}{\sigma}}=(1+j) \sqrt{\frac{\pi f \mu}{\sigma}}
$$
所以\(H\)比\(E\)落后\(45^\circ\)的相位
三、群速
各频率的相速不同引起的信号失真现象称为色散。损耗电介质显然是色散媒质
\[
u_g=\frac{1}{d\beta/d\omega}
\]
电离层或等离子体的等效介电常数为
\[
\epsilon_p=\epsilon_0(1-\frac{\omega_p^2}{\omega^2})=\epsilon_p=\epsilon_0(1-\frac{f_p^2}{f^2})
\]
电路媒质中的波传播
\[
\beta=\omega\sqrt{\mu\epsilon_0}\sqrt{1-\frac{f_p^2}{f^2}}=\frac \omega c\sqrt{1-\frac{\omega_p^2}{\omega^2}}
\]
当\(\omega<=\omega_p\)时，\(\beta=0\)，此时只有\(\alpha\)，在介质中波只衰减不传播
当\(\omega>\omega_p\)时，波可以传播
\[
u_p=\frac \omega \beta=\frac c{\sqrt{1-\frac{\omega_p^2}{\omega^2}}}>=c
\]
\[
u_g=\frac {d\omega} {d\beta}=c\sqrt{1-\frac{\omega_p^2}{\omega^2}}<=c
\]
所以
$$
u_pu_g=c^2
$$
四、电磁能流和坡印廷矢量
\[
\nabla \cdot (\boldsymbol{E} \times \boldsymbol{H}) = \boldsymbol{H} \cdot (\nabla \times \boldsymbol{E}) - \boldsymbol{E} \cdot (\nabla \times \boldsymbol{H})
\]
\[
\nabla \cdot (\boldsymbol{E} \times \boldsymbol{H}) = \boldsymbol{H} \cdot \left( -\frac{\partial \boldsymbol{B}}{\partial t} \right) - \boldsymbol{E} \cdot \left( \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t} \right)
\]
进一步化简：
\[
\boldsymbol{H} \cdot \left( \frac{\partial \boldsymbol{B}}{\partial t} \right) = \boldsymbol{H} \cdot \left( \frac{\partial (\mu \boldsymbol{H})}{\partial t} \right) = \frac{1}{2} \frac{\partial \left( \mu \boldsymbol{H} \cdot \boldsymbol{H} \right)}{\partial t} = \frac{\partial}{\partial t} \left( \frac{1}{2} \mu H^2 \right)
\]
\[
\boldsymbol{E} \cdot \frac{\partial \boldsymbol{D}}{\partial t} = \boldsymbol{E} \cdot \frac{\partial \left( \epsilon \boldsymbol{E} \right)}{\partial t} = \frac{1}{2} \frac{\partial \left( \epsilon \boldsymbol{E} \cdot \boldsymbol{E} \right)}{\partial t} = \frac{\partial}{\partial t} \left( \frac{1}{2} \epsilon E^2 \right)
\]
\[
\boldsymbol{E} \cdot \boldsymbol{J} = \boldsymbol{E} \cdot \left( \sigma \boldsymbol{E} \right) = \sigma E^2
\]
原式可化为：
\[
\nabla \cdot (\boldsymbol{E} \times \boldsymbol{H}) = - \frac{\partial}{\partial t} \left( \frac{1}{2} \epsilon E^2 + \frac{1}{2} \mu H^2 \right) - \sigma E^2
\]
积分形式：
\[
\oint_{S} \left( \boldsymbol{E} \times \boldsymbol{H} \right) \cdot d \boldsymbol{s} = - \frac{\partial}{\partial t} \int_{V} \left( \frac{1}{2} \epsilon E^2 + \frac{1}{2} \mu H^2 \right) dV - \int_{V} \sigma E^2 \, dV
\]
其中：
\[
w_e = \frac{1}{2} \epsilon E^2 = \frac{1}{2} \epsilon \boldsymbol{E} \cdot \boldsymbol{E}^{*} = \text{Electric energy density,}
\]
\[
w_m = \frac{1}{2} \mu H^2 = \frac{1}{2} \mu \boldsymbol{H} \cdot \boldsymbol{H}^{*} = \text{Magnetic energy density,}
\]
\[
p_{\sigma} = \sigma E^2 = \frac{\mathbf{J}^2}{\sigma} = \sigma \boldsymbol{E} \cdot \boldsymbol{E}^{*} = \frac{\mathbf{J} \cdot \mathbf{J}^{*}}{\sigma} = \text{Ohmic power density.}
\]
定义：
\[
\boldsymbol{P} = \boldsymbol{E} \times \boldsymbol{H}
\]