Answer for HW6

web版: url to HW6 Answer

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1 1D 电磁波解

(a) 平面波动方程:

(\frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}}) φ = (\frac{1}{c} \frac{\partial}{\partial t} + \frac{\partial}{\partial x}) (\frac{1}{c} \frac{\partial}{\partial t} - \frac{\partial}{\partial x}) φ = 0

令 v = (\frac{1}{c} \frac{\partial}{\partial t} - \frac{\partial}{\partial x}) φ \to {\begin{cases} (\frac{1}{c} \frac{\partial}{\partial t} + \frac{\partial}{\partial x}) v = 0 \\ (\frac{1}{c} \frac{\partial}{\partial t} - \frac{\partial}{\partial x}) φ = 0 \end{cases} \to 特 征 方 程 : {\begin{cases} c d t = d x \\ c d t = - d x \end{cases} \Rightarrow {\begin{cases} ξ = x - c t \\ η = x + c t \end{cases}

$\Rightarrow$ 原方程的解为:

\begin{array}{l} \frac{\partial ψ}{\partial ξ} = 0 \\ \frac{\partial φ}{\partial η} = 0 \end{array}} \Rightarrow \frac{\partial^{2} φ}{\partial ξ \partial η} = 0 \Rightarrow φ (x, t) = \underset{右行波}{\underset{⏟}{f (x - c t)}} + \underset{右行波}{\underset{⏟}{g (x + c t)}}

(b) 球面波动方程:

\frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} φ = Δ_{3} φ = (\frac{\partial^{2}}{\partial r^{2}} + \frac{2}{r} \frac{\partial}{\partial r}) φ

设 $φ = \frac{1}{r} u (r, t)$ 代入上式得:

(\frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial r^{2}}) u = 0 \to φ (r, t) = \frac{1}{r} [\underset{会聚波}{\underset{⏟}{f (t + r / c}}) + \underset{发散波}{\underset{⏟}{g (t - r / c)}}]

对于源，会聚波可略去.

2 反常色散

此时 $n_{r} - 1 = C o n s t \cdot \frac{ω_{0}^{2} - ω^{2}}{{(ω_{0}^{2} - ω^{2})}^{2} + (ω γ)^{2}} = C o n s t \cdot f (ω)$ :

\begin{aligned} f^{'} (ω) & = - 2 ω [{(ω_{0}^{2} - ω^{2})}^{2} + (ω γ)^{2}] + (ω^{2} - ω_{0}^{2}) [2 (ω_{0}^{2} - ω^{2}) (- 2 ω) + 2 γ^{2} ω] / {没 用 的 分 母}^{2} \\ = 2 ω [{(ω_{0}^{2} - ω^{2})}^{2} - ω_{0}^{2} γ^{2}] / {没 用 的 分 母}^{2} \end{aligned}

对于 $f^{'} (ω_{1}) = f^{'} (ω_{2}) = 0$ :

\begin{aligned} | ω_{1} - ω_{2} | & = {(ω_{0}^{2} + ω_{0} γ)}^{1 / 2} - {(ω_{0}^{2} - ω_{0} γ)}^{1 / 2} \\ = ω_{0} {(1 + \frac{γ}{ω_{0}})}^{1 / 2} - ω_{0} {(1 - \frac{γ}{ω_{0}})}^{1 / 2} \approx γ (当 γ ≪ ω_{0} 时) \end{aligned}

3 介质边界的电磁波

(a) 假设 xoz 为入射面. 设入射波 ${\begin{cases} \vec{B} = B_{0} {\vec{e}}_{y} + 0 \cdot {\vec{e}}_{x} \\ \vec{E} = E_{0} {\vec{e}}_{x} + 0 \cdot {\vec{e}}_{y} \end{cases}$ , 反射波 $E_{R} = (E_{R x}, E_{R y})$ , 透射波 $E_{T} = (E_{T x}, E_{T y})$ .

由 $\nabla \times E = - \frac{\partial B}{\partial t}$ 和 $\nabla \times \frac{B}{μ} = \frac{\partial ε E}{\partial t} k_{0}, \vec{B} = \frac{\vec{n}}{c} \times \vec{E} = \sqrt{μ ε} \hat{h} \times \vec{E}$ 得到如下边界条件:

\begin{aligned} E_{0} + E_{R x} & = E_{T x} \\ 0 + E_{R y} & = E_{T y} \\ \frac{0 + B_{R x}}{μ} & = \frac{B_{T x}}{μ^{'}} \\ \frac{B_{0} + B_{R y}}{μ} & = B_{T y} / μ^{'} \\ \sqrt{\frac{ε}{μ}} E_{R y} & = \sqrt{\frac{ε'}{μ^{'}}} E_{T y} \end{aligned}

由第二式和最后一式可知 $E_{R y} = 0, E_{T y} = 0$ .

(b) 此时三支波均只有y分量, 有：

\nabla \times E = - \frac{\partial B}{\partial t} < + \infty \Rightarrow E_{0} + E_{R} = E_{T}

以及 $\nabla \times \frac{B}{μ} = \frac{\partial ε E}{\partial t} < + \infty$ 和 $\vec{B} = \sqrt{μ ε} \hat{n} \times \vec{E}$ :

\Rightarrow \sqrt{\frac{ε}{μ}} E_{0} \cos θ_{I} - \sqrt{\frac{ε}{μ}} E_{R} \cos θ_{I} = \sqrt{\frac{ε^{'}}{μ^{'}}} E_{T} \cos θ_{I I}

上述两式结合 $\frac{\sin θ_{I I}}{\sin θ_{I}} = \frac{n_{I I}}{n_{I}}$ 得到:

{\begin{aligned} \frac{E_{I}}{E_{0}} & = \frac{2 n_{I} \cos θ_{I}}{n_{I} \cos θ_{I} + \frac{μ}{μ^{'}} \sqrt{n_{I I}^{2} - n_{I}^{2} \sin^{2} θ_{I}}} \\ \frac{E_{I}}{E_{0}} & = \frac{n_{I} \cos θ_{I} - \frac{μ}{μ^{'}} \sqrt{n_{I I}^{2} - n_{I}^{2} \sin^{2} θ_{I}}}{n_{I} \cos θ_{I} + \frac{μ}{μ^{'}} \sqrt{n_{I I}^{2} - n_{I}^{2} \sin^{2} θ_{I}}} \end{aligned}

4 TM 模

假设方形波导管的轴向是 z 方向, 波动方程是:

(\frac{\partial^{2}}{\partial^{2} x} + \frac{\partial^{2}}{\partial^{2} y} + (μ ϵ ω^{2} - k^{2})) E_{z} = 0

边界条件是 ${\vec{E}}^{| |} = 0$ :

{\begin{array}{r} E_{z} = E_{y} = 0 a t x = 0, a \\ E_{z} = E_{x} = 0 a t y = 0, b \end{array}

通过分离变量 $E_{z} = X (x) Y (y)$ , 得到:

E_{z} = E_{0} \sin \frac{m π x}{a} \sin \frac{n π y}{b}

给出的截止频率在 $k = 0$ 时取得, 它是:

ω_{m n} = π c \sqrt{\frac{m^{2}}{a^{2}} + \frac{n^{2}}{b^{2}}}

注意, 如果 $m = 0 o r n = 0$ 会使得 $E_{z} = 0$ , 波模不存在. 所以对应的最低截止频率是 $ω_{11}$ .

5 课本题

5.1 窄带双色波

\begin{array}{l} \vec{A} & = {\vec{A}}_{0} [e^{i (k x - ω t + d k x - d ω t)} + e^{i (k x - ω t - d k x + d ω t)}] \\ = {\vec{A}}_{0} e^{i (k x - ω t)} [e^{d k x - i d ω t} + \overset{―}{e^{d k x - i d ω t}}] \\ = 2 {\vec{A}}_{0} \cos (d k \cdot x - d ω \cdot t) e^{i (k x - ω t)} \end{array}

(2) 从相速度对应于等相位面 $e^{i (k x - ω t)} = c o n s t .$ , 以及群速度对应波包面 $\cos (d k \cdot x - d ω \cdot t) = c o n s t .$ 得到:

ν_{phase} = ω / k, ν_{group} = d ω / d k

5.2 各向异性介质

(1) 对于 $ρ_{f} = J_{f} = 0$ 的线性磁介质:

\nabla \cdot B = 0 \Rightarrow \vec{k} \cdot \vec{B} = 0

\nabla \cdot \vec{D} = ϵ \nabla \cdot \vec{E} + \vec{E} \cdot \nabla ϵ = 0 \Rightarrow k \cdot \vec{D} = 0 (\nabla ϵ \neq 0 \Rightarrow \nabla \cdot \vec{E} \neq 0 \Rightarrow \vec{k} \cdot \vec{E} \neq 0)

\nabla \times E = - \frac{\partial B}{\partial t} \Rightarrow \vec{k} \times \vec{E} = ω \vec{B} \Rightarrow \vec{E} ⊥ \vec{B} \Rightarrow \vec{B} \cdot \vec{E} = \vec{B} \cdot \vec{D} = 0

\nabla \times \frac{B}{μ} = \frac{\partial D}{\partial t} \Rightarrow \vec{k} \times \frac{\vec{B}}{μ} = - ω \vec{D} \Rightarrow \vec{B} ⊥ \vec{D} \Rightarrow B \cdot \vec{D} = 0

(2) 由(1)已知

\vec{D} = - \frac{k}{ω} \times \frac{\vec{B}}{μ} = - \frac{1}{ω^{2} μ} \vec{k} \times (\vec{k} \times \vec{E}) = \frac{1}{ω^{2} μ} [k^{2} \vec{E} - \vec{k} (\vec{k} \cdot \vec{E})]

(3) 由于 $\vec{S} = E \times H$ , 对 $\vec{k} \times \vec{E} = ω \vec{B}$ 叉乘 $\vec{E}$ 得到:

(\vec{k} \times \vec{E}) \times \vec{E} = ω \vec{B} \times \vec{E} \Rightarrow \vec{E} (\vec{k} \cdot \vec{E}) - \vec{k} E^{2} = ω \vec{B} \times \vec{E}

∵ \vec{k} \cdot \vec{E} \neq 0 ∴ 可 见 \vec{B} \times \vec{E} 一 般 不 与 \vec{k} 同 向

5.3 半无界波导管

求解 $[\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}} + \frac{ω^{2}}{c^{2}}] (\binom{E}{B}) = 0$ , 边界条件:

{\begin{cases} E_{x} = E_{y} = 0, \partial E_{z} / \partial z = 0 & at z = 0 \\ E_{y} = E_{z} = 0, \partial E_{y} / \partial x = 0 & at x = 0, a \\ E_{x} = E_{z} = 0, \partial E_{y} / \partial y = 0 & at y = 0, b \end{cases}

设 $E_{z} = X (x) Y (y) Z (z)$ ,可得

\begin{matrix} \frac{1}{X} \frac{d^{2} X}{d x^{2}} = - k_{x}^{2} \\ \frac{1}{Y} \frac{d^{2} Y}{d y^{2}} = - k_{y}^{2} \\ \frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = - k_{z}^{2} \\ k_{x}^{2} + k_{y}^{2} + k_{z}^{2} = \frac{w^{2}}{c^{2}} \end{matrix}

可以得到答案的前四式. 最后, 由 $\nabla \cdot \vec{E} = 0$ 得到:

A_{x} \frac{m π}{a} + A_{y} \frac{n π}{b} + A_{z} k_{z} = 0

5.4 波导管** 2

ν_{m n} = \frac{ω_{m n}}{2 π} = \frac{1}{2} c \sqrt{\frac{m^{2}}{a^{2}} + \frac{n^{2}}{b^{2}}} = {\begin{cases} 1.5 \times 10^{8} \cdot \sqrt{\frac{m^{2}}{49 \times 10^{- 6}} + \frac{n^{2}}{1.6 \times 10^{- 6}}} = 1.5 \times 10^{11} \sqrt{\frac{m^{2}}{49} + \frac{n^{2}}{16}} Hz \\ 1.5 \times 10^{8} \cdot \sqrt{\frac{m^{2}}{49 \times 10^{- 6}} + \frac{n^{2}}{36 \times 10^{- 6}}} = 1.5 \times 10^{11} \sqrt{\frac{m^{2}}{49} + \frac{n^{2}}{36}} Hz \end{cases}

对于 $0.7 \times 0.4 {cm}^{2} : ν_{01} = 37.5 \times 10^{9} Hz > ν_{0} = 30 \times 10^{9} Hz, ν_{10} = 21.4 \times 10^{9} Hz < ν_{0}$ .

对于 $0.7 \times 0.6 {cm}^{2} : ν_{01} = 25 \times 10^{9} Hz < ν_{0}, ν_{10} = 21.4 \times 10^{9} Hz < ν_{0}, ν_{11} = 32.9 \times 10^{9} H z > ν_{0}$ .

\Rightarrow (1) T E_{10}, (2) T E_{10} a n d T E_{01}

6 波动物理量的时间平均

A B = \frac{\hat{A} + {\hat{A}}^{*}}{2} \frac{\hat{B} + {\hat{B}}^{*}}{2}

\begin{aligned} A B = Re [\hat{A} e^{i ω t}] \cdot Re [\hat{B} e^{i ω t}] & = \frac{1}{4} (\hat{A} e^{i ω t} + {\hat{A}}^{*} e^{- i ω t}) (\hat{B} e^{i ω t} + {\hat{B}}^{*} e^{- i ω t}) \\ = \frac{1}{4} [\hat{A} \hat{B} e^{2 i ω t} + {\hat{A}}^{*} {\hat{B}}^{*} e^{- i 2 ω t} + {\hat{A}}^{*} \hat{B} + \hat{A} {\hat{B}}^{*}] \\ = \frac{1}{2} {Re [\hat{A} \hat{B} e^{i 2 ω t}] + Re [A B^{*}]} \end{aligned}

可见当 A, B 的变化时间尺度 $τ ≫ t$ 时, $⟨ A B ⟩ = \frac{1}{2} Re [A B^{*}]$ .

7 三种介质中的 1D电磁波

在介质 2 中的相位变化记为 $θ = \vec{k} \cdot Δ \vec{x} = \frac{ω}{c / n_{2}} d$ , 假设波在介质 2 中反射 n 次(一个来回算一次), 那么:

\frac{E_{o u t, n}}{E_{i n}} = T R^{n}

其中, T 是完全透射 $1 \to 2 \to 3$ 对应的衰减:

T = T_{12} e^{i θ} T_{23} = \frac{2 n_{1}}{n_{1} + n_{2}} \frac{2 n_{2}}{n_{2} + n_{3}} e^{i θ}

R 是在介质 2 中传播一个来回的衰减:

R = R_{23} e^{i θ} R_{21} e^{i θ} = \frac{n_{2} - n_{3}}{n_{2} + n_{3}} \frac{n_{2} - n_{1}}{n_{1} + n_{2}} e^{i 2 θ}

于是

\begin{aligned} \frac{E_{o u t}}{E_{i n}} & = \sum_{n = 0}^{+ \infty} \frac{E_{o u t, n}}{E_{i n}} \\ = \sum_{n = 0}^{+ \infty} T R^{n} = \frac{T}{1 - R} \\ = \frac{4 n_{1} n_{2} e^{i θ}}{(n_{2} + n_{3}) (n_{1} + n_{2}) - (n_{2} - n_{3}) (n_{2} - n_{1}) e^{i 2 θ}} \end{aligned}

这里, 实际的 E 是上述的实部.

或者, 使用边界条件求解:

$$ \begin{aligned} E_{in} T_{12} + E_{2-} R_{21} &= E_{2+} \\ E_{in} R_{12} + E_{2-} T_{21} &= E_r \\ E_{2+}e^{i\theta} T_{23} &= E_{out} \\ E_{2+} e^{i\theta} R_{23} &= E_{2-}e^{-i\theta} \end{aligned} $$ 重新记无量纲复振幅 $E_i' = \frac{E_i}{E_{in}}$, 上式给出: $$ \begin{bmatrix} -1 & \frac{{n_2-n_1}}{n_1+n_2} & 0 & 0 \\ 0 & \frac{2n_2}{n_1+n_2} & -1 & 0 \\ e^{i\theta} \frac{2n_2}{n_2+n_3} & 0 & 0 & -1 \\ e^{i\theta} \frac{{n_2-n_3}}{n_2+n_3} & -e^{-i\theta} & 0 & 0 \end{bmatrix} \begin{bmatrix} E_{2+}' \\ E_{2-}' \\ E'_r \\E'_{out} \end{bmatrix} = \begin{bmatrix} -\frac{2n_1}{n_1+n_2} \\ -\frac{{(n_1-n_2)}}{n_1+n_2} \\ 0 \\0 \end{bmatrix} $$ MMA 给出一致的结果: ![zz_figure/Pasted image 20250527113220.png](/img/user/zz_figure/Pasted%20image%2020250527113220.png)

当然, 也可以用最基本的边界条件:

\begin{matrix} E_{i n} + E_{r} = E_{2 +} + E_{2 -} \\ n_{1} (E_{i n} - E_{r}) = n_{2} (E_{2 +} - E_{2 -}) \\ E_{2 +} e^{i θ} + E_{2 -} e^{- i θ} = E_{o u t} \\ n_{2} (E_{2 +} e^{i θ} - E_{2 -} e^{- i θ}) = n_{3} E_{o u t} \end{matrix}

会得到一样的结果.

8 Drude 模型

(a)

注意到极化强度 $\vec{P} = N \vec{p} = N α \vec{E} = (ϵ_{r} - 1) ϵ_{0} \vec{E}$ , 得到:

α = \frac{(ϵ_{r} - 1) ϵ_{0}}{N} = \frac{e^{2} / m}{ω_{0}^{2} - ω^{2} - i ω γ}

(b)

当电场驱动和阻尼可以忽略时, $ω = γ = 0$ :

\frac{α}{4 π ϵ_{0}} = \frac{e^{2} ℏ^{2}}{4 π ϵ_{0} m_{e} (ℏ ω_{0})^{2}} \approx 5.93 \times 10^{- 31} m^{3}

(c)

原子极化率可以写作 $α = 4 π ϵ_{0} a_{0}^{3}$ , 于是 $\frac{α}{4 π ϵ_{0}} = a_{0}^{3} \approx 1.48 \times 10^{- 31} m^{3}$ 与上式处于同一量级.

(d)

对于实数 $ω$ :

\begin{array}{l} k_{R} = \frac{n_{R}}{c} ω = \frac{1}{c} [ω + \frac{ω_{p}^{2}}{2} ω \frac{ω_{0}^{2} - ω^{2}}{{(ω_{0}^{2} - ω^{2})}^{2} + γ^{2} ω^{2}}] \\ ∴ \frac{ν_{group}}{c} = {[d k_{R} / d ω]}^{- 1} / c = {{[ω + \frac{ω_{e}^{2}}{2} ω \frac{ω_{0}^{2} - ω^{2}}{{(ω_{0}^{2} - ω^{2})}^{2} + γ^{2} ω^{2}}]}_{ω}^{'}}^{- 1} \\ = {\frac{ω_{p}^{2}}{2} \frac{ω_{0}^{2} - 3 ω^{2}}{{(ω_{0}^{2} - ω^{2})}^{2} + γ^{2} ω^{2}} - \frac{ω_{p}^{2}}{2} ω^{2} \frac{[2 γ^{2} - 4 (ω_{0}^{2} - ω^{2})] (ω_{0}^{2} - ω^{2})}{{[{(ω_{0}^{2} - ω^{2})}^{2} + γ^{2} ω^{2}]}^{2}} + 1}^{- 1} \\ \Rightarrow ν_{g} / c = {1 + 0.003 \frac{1 - 3 {(\frac{ω}{ω_{0}})}^{2}}{{[1 - {(\frac{ω}{ω_{0}})}^{2}]}^{2} + 0.01 {(\frac{ω}{ω_{0}})}^{2}} - 0.003 {(\frac{ω}{ω_{0}})}^{2} \frac{[0.02 - 4 (1 - {(\frac{ω}{ω_{0}})}^{2})] (1 - {(\frac{ω}{ω_{0}})}^{2})}{{{[1 - {(\frac{ω}{ω_{0}})}^{2}]}^{2} + 0.01 {(\frac{ω}{ω_{0}})}^{2}}^{2}}}^{- 1} \\ = {\begin{cases} {1.00658}^{- 1} \approx 0.993 & ω = 0.5 ω_{0} \\ {0.4}^{- 1} = 2.5 & ω = ω_{0} \\ {1.00598}^{- 1} \approx 0.994 & ω = 1.5 ω_{0} \end{cases} \end{array}

在共振区域附近, 反常色散的群速度超过了光速(为什么? 可以查阅: 反常色散的群速度).

有些科普说群速度是能量传播的速度。这是错误的。要记住，群速度只是波包运动的一阶近似而已，并不能完全表述波包的演化。群速度超过光速，并不说明能量/信息超光速，而是在这种情况下，群速度已经失去了意义。

(e)

对于 $\hat{\vec{E}} = {\hat{\vec{E}}}_{0} e^{- k_{I} x} e^{i (k_{I} x - ω t)}$ :

{\begin{cases} k_{I} L ≪ 1 透明 \\ k_{I} L ≫ 1 不透明 \end{cases}

(f)

n^{2} = ε_{r} μ_{r} = {μ_{r} (1 + \frac{N e^{2}}{m ϵ_{0}} \frac{1}{ω_{0}^{2} - ω^{2} - i ω γ}) |}_{ω_{0} = 0} = μ_{r} (1 - \frac{ω_{p e}^{2}}{ω^{2} + i ω γ}) \approx 1 - \frac{ω_{p e}^{2}}{ω^{2} + i ω γ}

(g)

由 $(7) (8) \to σ E = N q_{e}^{2} \frac{E}{m} τ \Rightarrow \frac{N e^{2}}{m} = \frac{σ}{τ}$ , 代入(f) 得到:

n^{2} = 1 - \frac{σ}{τ ϵ_{0}} \frac{1}{ω (ω + i / τ)} = 1 - \frac{σ / ϵ_{0}}{ω (ω τ + i)} = 1 - \frac{σ / ϵ_{0}}{i ω (1 - i τ ω)}

(h)

由(g)可知

\begin{aligned} n & = {1 - \frac{σ / ϵ_{0}}{ω (ω τ + i)}}^{1 / 2} = {1 + \frac{i σ / ϵ_{0}}{ω (1 - i ω τ)}}^{1 / 2} \\ =_{ω ≪ σ / ε_{0}}^{ω τ ≪ 1} \sqrt{\frac{σ}{ϵ_{0} ω}} \cdot e^{(i \frac{π}{4} + k π)} (1 - i ω τ)^{- \frac{1}{2}} \\ =_{ω τ ≪ 1}^{k = 0} \sqrt{\frac{σ}{2 ϵ_{0} ω}} (1 + i) (1 + \frac{1}{2} i ω τ) \\ \Rightarrow & n_{I} (虚 部) \approx \sqrt{\frac{σ}{2 ϵ_{0} ω}} (1 + \frac{1}{2} ω τ) \approx \sqrt{\frac{σ}{2 ϵ_{0} ω}} \\ \Rightarrow & k_{I} = \frac{n_{I} ω}{c} = \sqrt{μ_{0} ϵ_{0}} \cdot \sqrt{\frac{σ ω}{2 ϵ_{0}}} = \sqrt{μ_{0} σ ω / 2} \\ \Rightarrow & δ = 1 / k_{I} = \sqrt{2 / μ_{0} σ ω} \end{aligned}

(i)

直接套公式有:

δ = {\begin{array}{r} 6.6 \times 10^{- 6} f = 10^{8} H z \\ 8.56 \times 10^{- 3} f = 60 H z \end{array}

(j)

根据(9), 高频近似 $ω τ ≫ 1$ 给出:

n^{2} = 1 - \frac{σ}{τ ϵ_{0}} \frac{1}{ω^{2}} = 1 - \frac{ω_{p e}^{2}}{ω^{2}} \to 1

$ω τ ≫ 1$ 和 $ω ≫ ω_{p e}$ 意味着, 电磁波的时间尺度 $\frac{1}{ω}$ 远小于电子平均碰撞时间 $τ$ , 阻尼效应被忽略, 这导致电磁波的能量不能像课上那样迅速耗散到粒子中(其对应于 $ω ≪ ω_{p e}$ , 与这里不同), 因此波动得以维持和传播.

该式也适用于描述冷等离子体中电磁波的色散关系.

(k)

由 $(j)$ 可知, 条件是波的频率足够高/金属足够薄.

9 介质中的 Maxwell's 方程组

(a)

由 $\nabla \cdot \vec{j} + \partial ρ / \partial t = 0$ 得到:

\nabla \cdot {\vec{j}}_{p} + \partial ρ_{p} / \partial t = 0 w h e n \nabla \cdot {\vec{j}}_{f} + \partial ρ_{f} / \partial t = 0

而极化体电荷 $ρ_{ρ} = - \nabla \cdot \vec{P}$ , 带入上式得到:

{\vec{j}}_{p} = \partial \vec{P} / \partial t

(b)

同理 $\nabla \cdot {\vec{j}}_{m} + \partial ρ_{m} / \partial t = 0$ , 磁化电流 ${\vec{j}}_{m} = \nabla \times \vec{m}$ , 得到:

- \frac{\partial ρ_{m}}{\partial t} = \nabla \cdot \nabla \times \vec{m} = 0 \Rightarrow ρ_{m} = c o n s t .

(c)

\begin{aligned} \nabla \cdot \vec{E} = \frac{ρ}{ϵ_{0}} & \Rightarrow \nabla \cdot \vec{E} = (ρ_{f} + ρ_{m} + ρ_{p}) / ϵ_{0} = (ρ_{f} - \nabla \cdot p) / ϵ_{0} \\ \Rightarrow \nabla \cdot (ϵ_{0} E + p) = ρ_{f} \\ \Rightarrow \nabla \cdot \vec{D} = ρ_{f} \\ \nabla \times \frac{\vec{B}}{μ_{0}} = \vec{J} + \frac{\partial ϵ_{0} \vec{E}}{\partial t} & \Rightarrow \nabla \times \frac{\vec{B}}{μ_{0}} = {\vec{J}}_{f} + {\vec{J}}_{m} + {\vec{J}}_{p} + \frac{\partial ϵ_{0} \vec{E}}{\partial t} = {\vec{J}}_{f} + \nabla \times \vec{M} + \frac{\partial (ϵ_{0} \vec{E} + \vec{p})}{\partial t} \\ \Rightarrow \nabla \times (\frac{\vec{B}}{μ_{0}} - \vec{M}) = {\vec{J}}_{f} + \frac{\partial \vec{D}}{\partial t} \\ \Rightarrow \nabla \times \vec{H} = {\vec{J}}_{f} + \frac{\partial \vec{D}}{\partial t} \end{aligned}

10 补充: 没有补充

本来想写点电磁场的拉氏量, 怎么按一般步骤推导得到的来着, 但是大家有期末考试, 助教也有, 所以......

结果放在这里, 拉氏量:

\begin{aligned} L & = - \sum_{i} \frac{m_{i} δ (\vec{x} - {\vec{x}}_{i}) c^{2}}{γ} - \frac{1}{4 μ_{0} c} F^{α β} F_{α β} - \frac{j^{μ} A_{μ}}{c} \end{aligned}

以及连续性方程:

\partial_{μ} T_{E . M . f i e l d}^{μ ν} = - F_{μ}^{ν} j^{μ} = - f^{μ}

课上讲的四维洛伦兹力:

f^{μ} = F_{ν}^{μ} j^{ν} = m δ (\vec{x} - {\vec{x}}_{0}) \frac{d u^{μ}}{d t} = (\frac{1}{c} \vec{j} \cdot \vec{E}, \vec{j} \times \vec{B} + ρ \vec{E})