Answer for HW8

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1. 辐射功率交叉项

分解 $\dot{\vec{β}} = {\dot{\vec{β}}}_{∥} + {\dot{\vec{β}}}_{⊥}$ , 其中 ${\dot{\vec{β}}}_{∥} | | \vec{β}$ , 根据讲义的(8.148)得到:

\begin{aligned} {(\frac{d P}{d Ω})}_{c r o s s} & \propto {{\hat{e}}_{R} \times [({\hat{e}}_{R} - \vec{β}) \times {\dot{\vec{β}}}_{∥}]} \cdot {{\hat{e}}_{R} \times [({\hat{e}}_{R} - \vec{β}) \times {\dot{\vec{β}}}_{⊥}]} \\ ↓ (\vec{A} \times \vec{B}) \cdot (\vec{A} \times \vec{C}) = A^{2} (\vec{B} \cdot \vec{C}) - (\vec{A} \cdot \vec{C}) (\vec{A} \cdot \vec{B}) \\ = [({\hat{e}}_{R} - \vec{β}) \times {\dot{\vec{β}}}_{∥}] \cdot [({\hat{e}}_{R} - \vec{β}) \times {\dot{\vec{β}}}_{⊥}] - {{\hat{e}}_{R} \cdot [({\hat{e}}_{R} - \vec{β}) \times {\dot{\vec{β}}}_{⊥}]} {{\hat{e}}_{R} \cdot [({\hat{e}}_{R} - \vec{β}) \times {\dot{\vec{β}}}_{∥}]} \\ = ({\hat{e}}_{R} \times {\dot{\vec{β}}}_{∥}) \cdot [({\hat{e}}_{R} - \vec{β}) \times {\dot{\vec{β}}}_{⊥}] \\ = {\dot{\vec{β}}}_{⊥} \cdot [({\vec{e}}_{R} \times {\dot{\vec{β}}}_{∥}) \times ({\vec{e}}_{R} - \vec{β})] \\ ↓ (\vec{A} \times \vec{B}) \times \vec{C} =^{'} 先 中 间 再 外 边^{'} \\ = {\hat{e}}_{R} \cdot \dot{{\vec{β}}_{⊥}} (\vec{β} \cdot \dot{{\vec{β}}_{∥}} - {\hat{e}}_{R} \cdot \dot{{\vec{β}}_{∥}}) \end{aligned}

取 ${\vec{e}}_{R} = (\sin θ \cos φ, \sin θ \sin φ, \cos θ)$ , $\vec{β} = β {\vec{e}}_{z}$ , 得到

{(\frac{d P}{d Ω})}_{c r o s s} \propto {\hat{β}}_{∥} (β - \cos θ) ({\hat{β}}_{⊥ x} \cos φ + {\hat{β}}_{⊥ y} \sin φ) \sin θ

于是:

\int {(\frac{d P}{d Ω})}_{c r o s s} d Ω \propto \int_{0}^{2 π} ({\dot{β}}_{⊥ x} c o s φ + {\dot{β}}_{⊥ y} s i n φ) d φ = 0

2. 电子的生命周期(讲义 p154)

加速度满足:

\frac{e^{2}}{4 π ϵ_{0} r^{2}} = m_{e} a_{⊥}

根据拉莫公式有:

P_{⊥} = \frac{μ_{0} e^{2}}{6 π c} a_{⊥}^{2} = \frac{μ_{0} e^{2}}{6 π c} \frac{e^{4}}{16 π^{2} ϵ_{0}^{2} r^{4} m_{e}^{2}}

初始的能量是:

E = - \frac{1}{2} \frac{e^{2}}{4 π ϵ_{0} r}

有两种方式给出生命周期的估算, 第一种是:

t_{1} \approx \frac{E}{P_{⊥}} \approx 10^{- 11} s

另一种是:

P_{⊥} (r) = - \frac{d E (r)}{d t} = E^{'} (r) \frac{d r}{d t} ⟶ t_{2} = \int d t = \int_{r}^{0} f (r) d r

得到:

t_{2} = r_{0}^{3} \frac{4 π^{2} ϵ_{0}^{2} m_{e}^{2} c^{3}}{e^{4}} \approx 10^{- 11} s

3. 极相对论粒子的角最大辐射

讲义(8.133)给出:

\frac{d P}{d Ω} = \frac{q^{2} μ_{0} c}{16 π^{2}} \frac{{\dot{β}}^{2} \sin^{2} θ}{(1 - β \cos θ)}

最大辐射对应的角度由上式对 $θ$ 的导数为 0 给出, 即:

0 = \frac{d}{d θ} \frac{\sin^{2} θ}{(1 - β \cos θ)^{5}} = \frac{[2 \cos θ (1 - β \cos θ) - 5 β \sin^{2} θ] \sin θ}{(1 - β \cos θ)^{5}}

\begin{aligned} \Rightarrow c o s θ & = \frac{\sqrt{1 + 15 β^{2}} - 1}{3 β} \\ = \frac{1}{3} [\frac{4 \sqrt{1 - \frac{15}{8} (1 - β) + o (1 - β)} - 1}{1 - (1 - β)}] \\ ↓ 1 - β ≪ 1 \\ = 1 - \frac{1 - β}{4} \\ = 1 - \frac{θ^{2}}{2} + o (θ^{3}) \end{aligned}

得到:

θ = \sqrt{\frac{1 - β}{2}}

4. 课本题

7.1 零辐射的角度

为避免歧义, 将题目里面的角度 $β$ 记为 $θ$ , 根据讲义(8.114):

\frac{d P}{d Ω} \propto {| {\hat{e}}_{R} \times [({\hat{e}}_{R} - \vec{β}) \times \vec{a}] |}^{2}

假设:

\vec{a} = a {\vec{e}}_{x}, \vec{β} = β (\cos α, \sin α), {\hat{e}}_{R} = (\cos θ, \sin θ)

得到:

\begin{aligned} {\hat{e}}_{R} \times [({\hat{e}}_{R} - \vec{β}) \times \vec{a}] & = ({\hat{e}}_{R} \cdot \vec{a}) ({\hat{e}}_{R} - \vec{β}) - [{\hat{e}}_{R} \cdot ({\hat{e}}_{R} - \vec{β})] \vec{a} \\ = (a (β \sin α - \sin θ) \sin θ, a \cos θ (\sin θ - β \sin α)) = 0 \\ \to β \sin α = \sin θ \end{aligned}

7.2 同步辐射

由讲义(8.155)可知:

P_{⊥} = \frac{e^{2} μ_{0}}{6 π c} \frac{γ^{2}}{m^{2}} {(\frac{d \vec{p}}{d t})}^{2}

其中:

\frac{d \vec{p}}{d t} = e \vec{v} \times \vec{B}, E = γ m_{e} c^{2} = 10^{12} e V

得到:

P = 38 e V \cdot s^{- 1}

7.6 同步辐射 * 2

(1)

根据讲义(8.164), 电子受力为辐射阻尼+洛伦兹力: +

m_{e} \vec{a} = \frac{μ_{0} e^{2}}{6 π c} \dot{\vec{a}} - e \vec{v} \times \vec{B}

即

{\dot{v}}_{x} = \frac{μ_{0} e^{2}}{6 π m_{e} c} {\ddot{v}}_{x} - \frac{e B}{m_{e}} v_{y}

{\dot{v}}_{y} = \frac{μ_{0} e}{6 π m_{e^{2}} c} {\ddot{v}}_{y} + \frac{e B}{m_{e}} v_{x}

假定 $\tilde{v} = v_{x} + i v_{y}$ , 得到

\dot{\tilde{v}} = \frac{μ_{0} e^{2}}{6 π m_{e} c} \ddot{\tilde{v}} + i \frac{e B}{m_{e}} \tilde{v}

这是一个常系数二阶微分方程, 其特征方程是:

\frac{μ_{0} e}{6 π m_{e^{2}} c} λ^{2} - λ + i \frac{e B}{m_{e}} = 0

得到:

\begin{aligned} λ & = \frac{1 \pm \sqrt{1 - i \frac{e B}{m_{e}} \frac{2 μ_{0} e^{2}}{3 π m_{e} c}}}{\frac{μ_{0} e^{2}}{3 π m_{e} c}} \\ \overset{Δ}{=} \frac{1 \pm \sqrt{1 - 4 i ω_{0} k}}{2 k}, (k = \frac{γ}{ω_{0}^{2}} ≪ 1) \\ = \frac{1 \pm 1 \mp 2 i ω_{0} k \pm 2 ω_{0}^{2} k^{2} + o (k^{2})}{2 k} \\ = i ω_{0} - γ \end{aligned}

代入 $\tilde{v} = v_{0} e^{λ t} = v_{x} + i v_{y}$ , 得到:

v_{x} = v_{0} e^{- γ t} \cos ω_{0} t, v_{y} = v_{0} e^{- γ t} \sin ω_{0} t

积分即可得到 x, y, z.

(2)

\frac{d W}{d t} = - \frac{d \frac{1}{2} m_{e} v^{2}}{d t} = - \frac{d \frac{1}{2} m_{e} v_{0}^{2} e^{- 2 γ t}}{d t}

得到

\frac{d W}{d t} = γ m_{e} v_{0}^{2} e^{- 2 γ t}

7.9 同步辐射 * 3

(1)

P = \frac{μ_{0} q^{2}}{6 π c} \frac{γ^{2}}{m^{2}} {(\frac{d \vec{p}}{d t})}^{2} = \frac{μ_{0} q^{2}}{6 π c} \frac{γ^{2}}{m^{2}} (q^{2} v^{2} B^{2}) = \frac{q^{4} B^{2}}{6 π ϵ_{0} m^{2} c} (γ^{2} - 1)

(2)

P = - \frac{d E}{d t} = - m c^{2} \frac{d γ}{d t} = \frac{q^{4} B^{2}}{6 π ϵ_{0} m^{2} c} (γ^{2} - 1)

可以解得 $γ (t)$ , $E (t) = γ (t) m c^{2}$ .

(3)

对于非相对论性粒子, 有 $T = \frac{1}{2} m v^{2}$ , 得到:

\frac{d T}{d t} = - \frac{q^{4} B^{2}}{6 π ϵ_{0} m^{2} c^{3}} v^{2} = - \frac{q^{4} B^{2}}{3 π ϵ_{0} m^{3} c^{3}} T

解得:

T = T_{0} e x p [- \frac{B^{2} q^{4}}{3 π ϵ_{0} m^{3} c^{3}} (t - t_{0})]

5. 磁偶极辐射和电四极辐射

见讲义 9.4 节.

6. 辐射阻尼

(a)

对运动方程积分:

\int_{t - ϵ}^{t + ϵ} a d t = \int_{t - ϵ}^{t + ϵ} τ \dot{a} d t + \int_{t - ϵ}^{t + ϵ} \frac{F}{m} d t

当 $ε \to 0$ , 由于 a 和 $\frac{F}{m}$ 是有限的, 所以左侧和右侧第二项=0, 于是右侧第一项=0:

0 = \int_{t - ϵ}^{t + ϵ} τ \dot{a} d t = τ l i m_{ϵ \to 0} [a (t + ϵ) - a (t - ϵ)]

得到 a 是连续的.

(b)

在 $t < 0$ 时, 积分得到:

a = τ \dot{a} \Rightarrow a = e^{t / τ} a (0)

$0 < t < T$ 时:

a = τ \dot{a} + \frac{F}{m} \Rightarrow a = e^{t / τ} [a (0) - \frac{F}{m}] + \frac{F}{m} = e^{(t - T) / τ} [a (T) - \frac{F}{m}] + \frac{F}{m}

$t > T$ 时:

a = τ \dot{a} \Rightarrow a = e^{(t - T) / τ} a (T)

(c)

如果同时让 t < 0 和 t > T 时，a = 0, 那么 $a (T) = a (0) = 0$ , $0 < t < T$ 时的通解将给出:

[0 - \frac{F}{m}] = e^{- T / τ} [0 - \frac{F}{m}]

即 $T = 0$ 这样一个平凡解.

(d)

即 $a (T) = 0$ , 得到

[a (0) - \frac{F}{m}] = e^{- T / τ} [0 - \frac{F}{m}] \to a (0) = \frac{F_{0}}{m} (1 - e^{- T / τ})

于是通解是:

a = {\begin{cases} \frac{F_{0}}{m} [e^{t / τ} - e^{(t - T) / τ}], (t < 0) \\ \frac{F_{0}}{m} [1 - e^{(t - T) / τ}], (0 < t < T) \\ 0, (t > T) \end{cases}

积分得到速度:

\begin{aligned} v (t < 0) & = 0 + \int_{- \infty}^{t} \frac{F_{0}}{m} (1 - e^{- T / τ}) e^{t / τ} d t = \frac{F_{0} τ}{m} (1 - e^{- T / τ}) e^{t / τ} \\ v (0 < t < T) & = \frac{F_{0} τ}{m} (1 - e^{- T / τ}) + \int_{0}^{t} \frac{F_{0}}{m} [1 - e^{(t - T) / τ}] d t = \frac{F_{0} τ}{m} (1 - e^{(t - T) / τ}) + \frac{F_{0} t}{m} \\ v (t > T) & = \frac{F_{0} T}{m} \end{aligned}

速度曲线示意图(所有参量均取为 1):

left=0-5; right=5;
top=3; bottom=-3;
---
y=(1-e^{-1})e^x|x<0
y=(1-e^{x-1})+x|0<x<1
y=1|x>1|

加速度曲线:

left=0-5; right=5;
top=3; bottom=-3;
---
y=e^x-e^{x-1}|x<0
y=1-e^{x-1}|0<x<1
y=0|x>1

7. 辐射电阻

电流是:

I = - 2 \frac{d q}{d t} = 2 ω q_{0} s i n (ω t)

平均功率是:

< P >= \frac{1}{T} \int_{0}^{T} I^{2} R d t = 2 ω^{2} q_{0}^{2} R

对于电偶极辐射, 讲义(9.63)给出的平均辐射功率是

< P >= \frac{μ_{0} ω^{4} {(2 q_{0} d)}^{2}}{12 π c}

令两者相等可以得到:

R = \frac{μ_{0} ω^{2} d^{2}}{6 π c} = \frac{2 π μ_{0} c}{3} {(\frac{d}{λ})}^{2} \approx 790 {(\frac{d}{λ})}^{2} Ω

对于 $f = 300 k H z$ 的电磁波, 假设 $d = 5 c m$ , 得到:

R \approx 2 \times 10^{- 6} Ω

8. 电偶极子的辐射

本题可以视为两个在 x-y 平面以角速度 $ω$ 匀速旋转的电荷的辐射:

\vec{p} = p_{0} e^{- i ω t} ({\vec{e}}_{x} + e^{- i δ} {\vec{e}}_{y}) = \frac{\sqrt{2}}{2} p_{0} e^{- i ω t} (1 - i e^{- i δ}) {\vec{e}}_{R} + \frac{\sqrt{2}}{2} p_{0} e^{- i ω t} (1 + i e^{- i δ}) {\vec{e}}_{L}

其中 ${\vec{e}}_{R} = \frac{1}{\sqrt{2}} ({\vec{e}}_{x} + i {\vec{e}}_{y})$ 和 ${\vec{e}}_{L} = \frac{1}{\sqrt{2}} ({\vec{e}}_{x} - i {\vec{e}}_{y})$ 分别是右圆极化和左圆极化矢量. 可以预见, 辐射的角分布是两个圆极化的叠加, 因此角分布与单个圆极化不同. 而对空间角进行积分得到的总辐射应当是与 $δ$ 无关的(和圆极化一致).

根据讲义(9.54)有:

\vec{B} = \frac{μ_{0}}{4 π c} \frac{\ddot{\vec{p}} \times \vec{n}}{r} e^{i k r}, \vec{E} = c \vec{B} \times \vec{n}

于是

⟨ S ⟩ = \frac{1}{μ_{0}} < \vec{E} \times \vec{B} >= \frac{μ_{0} \vec{n}}{16 π^{2} c r^{2}} < (\ddot{\vec{p}} \times \vec{n})^{2} >

上式中

⟨ (\ddot{\vec{p}} \times \vec{n})^{2} ⟩ =< | \ddot{\vec{p}} |^{2} - (\ddot{\vec{p}} \cdot \vec{n})^{2} >= ω^{4} < | \vec{p} |^{2} - (\vec{p} \cdot \vec{n})^{2} >

< | \vec{p} |^{2} >=< p_{0}^{2} \cos^{2} ω t + p_{0}^{2} \cos^{2} (ω t + δ) >= p_{0}^{2}

选取 $\vec{n} = (\sin θ \cos ϕ, \sin θ \sin ϕ, \cos θ)$ , 得到:

\vec{p} ​ \cdot \vec{n} = R e (p_{0} ​ \sin θ e^{- i ω t} [c o s ϕ + e^{- i δ} s i n ϕ]) = p_{0} \sin θ [\cos ϕ \cos ω t + \sin ϕ \cos (ω t + δ)]

(\vec{p} \cdot \vec{n})^{2} = p_{0}^{2} \sin^{2} θ {[\cos ϕ \cos ω t + \sin ϕ \cos (ω t + δ)]}^{2}

\begin{aligned} ⟨ (\vec{p} \cdot \vec{n})^{2} ⟩ & = p_{0}^{2} \sin^{2} θ ⟨ \cos^{2} ϕ \cos^{2} ω t + \sin^{2} ϕ \cos^{2} (ω t + δ) + 2 \cos ϕ \sin ϕ \cos ω t \cos (ω t + δ) ⟩ \\ = \frac{p_{0}^{2} \sin^{2} θ}{2} (1 + \sin 2 ϕ \cos δ) \end{aligned}

即:

< S >= \frac{μ_{0} \vec{n}}{16 π^{2} c r^{2}} < (\ddot{\vec{p}} \times \vec{n})^{2} >= \frac{μ_{0} \vec{n}}{32 π^{2} c r^{2}} (2 - \sin^{2} θ - \sin^{2} θ \sin 2 ϕ \cos δ)

< \frac{d P}{d Ω} >=< S > n r^{2}

< P >= \int < \frac{d P}{d Ω} > \sin θ d θ d ϕ = \frac{μ_{0} ω^{4}}{6 π c} p_{0}^{2}

与预期一致.