辐射

E⃗=−∇φ−∂A⃗∂t=14πε0∫[ρ(x⃗′,tr)r^r2+ρ˙(x⃗′,tr)cr−J⃗˙(x⃗′,tr)c2r]dV′\vec{E} = -\nabla \varphi - \frac{\partial \vec{A}}{\partial t} = \frac{1}{4\pi\varepsilon_0} \int\left[\frac{\rho(\vec{x}',t_r)\hat{r}}{r^2} + \frac{\dot{\rho}(\vec{x}',t_r)}{cr} - \frac{\dot{\vec{J}}(\vec{x}',t_r)}{c^2r}\right]dV'

B⃗=∇×A⃗=μ04π∫[J⃗(x⃗′,tr)r2+J⃗˙(x⃗′,tr)cr]×r^dV′\vec{B} = \nabla \times \vec{A} = \frac{\mu_0}{4\pi} \int\left[\frac{\vec{J}(\vec{x}',t_r)}{r^2} + \frac{\dot{\vec{J}}(\vec{x}',t_r)}{cr} \right] \times \hat{r} dV'

其中，rho˙=∂ρ∂t\dot{rho} = \frac{\partial \rho}{\partial t}，J⃗˙=∂J⃗∂t\dot{\vec{J}} = \frac{\partial \vec{J}}{\partial t}
考虑在导体球壳表面上求他的总辐射功率：

P=lim⁡R→∞∮RE⃗×H⃗⋅dS⃗P = \lim_{R \to \infty} \oint_{R} \vec{E} \times \vec{H} \cdot d \vec{S}

dS⃗∝1R2d\vec{S} \propto \frac{1}{R^2}，E⃗×H⃗\vec{E} \times \vec{H}包含1R2,1R3,1R4\frac{1}{R^2},\frac{1}{R^3},\frac{1}{R^4}等项
只有ρ˙,J⃗˙\dot{\rho},\dot{\vec{J}}（电荷，电流随时间的变化）才会产生电磁辐射

*时谐场的辐射

ρ(x⃗,t)=ρ(x⃗)e−iωt,J⃗(x⃗,t)=J⃗(x⃗)e−iωt\rho(\vec{x},t) = \rho(\vec{x})e^{-i\omega t},\quad \vec{J}(\vec{x},t) = \vec{J}(\vec{x})e^{-i\omega t}

φ(x⃗,t)=φ(x⃗)e−iωt,A⃗(x⃗,t)=A⃗(x⃗)e−iωt\varphi(\vec{x},t) = \varphi(\vec{x})e^{-i\omega t},\quad \vec{A}(\vec{x},t) = \vec{A}(\vec{x})e^{-i\omega t}

电荷量守恒定律：

∇⋅J⃗+∂ρ∂t=0⇒∇⋅J⃗−iωρ=0\nabla \cdot \vec{J} + \frac{\partial \rho}{\partial t} = 0 \Rightarrow \nabla \cdot \vec{J} - i\omega\rho = 0

ρ,J⃗\rho,\vec{J}是不独立的
Lorentz规范：

∇⋅A⃗+1c2∂φ∂t=0⇒∇⋅A⃗−iωc2φ=0\nabla \cdot \vec{A} + \frac{1}{c^2}\frac{\partial \varphi}{\partial t} = 0 \Rightarrow \nabla \cdot \vec{A} - \frac{i\omega}{c^2}\varphi = 0

φ,A⃗\varphi,\vec{A}不独立
只需考虑J⃗,A⃗\vec{J},\vec{A}

A⃗(x⃗,t)=μ04π∫J⃗(x⃗′,tr)rdV′=μ04π∫J⃗(x⃗′)e−iω(t−rc)rdV′=[μ04π∫J⃗(x⃗′)eiωrcrdV′]e−iωt\begin{aligned} \vec{A}(\vec{x},t) &= \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{x}',t_r)}{r}dV'\\ &=\frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{x}')e^{-i\omega (t - \frac{r}{c})}}{r}dV'\\ &=\left[ \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{x}')e^{i\omega \frac{r}{c}}}{r}dV' \right] e^{-i\omega t}\\ \end{aligned}

其中A⃗(x⃗)=μ04π∫J⃗(x⃗′)eikrrdV′\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{x}')e^{ikr}}{r}dV',eikre^{ikr}代表由源点x⃗′\vec{x}'传播到场点x⃗\vec{x}的相移

  • 近似1：场点远离J⃗\vec{J}分布区域，∣x⃗∣>>∣x⃗′∣|\vec{x}| >> |\vec{x}'|
    记∣x⃗∣=R|\vec{x}| = R，r=∣x⃗−x⃗′∣≈∣x⃗∣−x⃗′⋅∇R=R−R^⋅x⃗′r = |\vec{x} - \vec{x}'| \approx |\vec{x}| - \vec{x}' \cdot \nabla R = R - \hat{R} \cdot \vec{x}'
    1R=1∣x⃗−x⃗′∣≈1∣x⃗∣−x⃗′⋅∇1R=1R+R^⋅x⃗′R2\frac{1}{R} = \frac{1}{|\vec{x} - \vec{x}'|} \approx \frac{1}{|\vec{x}|} -\vec{x}' \cdot \nabla \frac{1}{R} = \frac{1}{R} + \frac{\hat{R} \cdot \vec{x}'}{R^2}

A⃗(x⃗)≈μ04π∫(1R+R^⋅x⃗′R2)eik(R−R^⋅x⃗′)J⃗(x⃗′)dV′\begin{aligned} \vec{A}(\vec{x}) &\approx \frac{\mu_0}{4\pi} \int \left(\frac{1}{R} + \frac{\hat{R} \cdot \vec{x}'}{R^2} \right) e^{ik(R - \hat{R} \cdot \vec{x}')}\vec{J}(\vec{x}')dV'\\ \end{aligned}

  • 近似2：J⃗\vec{J}分布区域的尺度远远小于电磁波的波长

max{∣x⃗′∣}<<λ=2πkmax\left\{|\vec{x}'|\right\} << \lambda = \frac{2\pi}{k}

∣kx⃗′∣<<1|k\vec{x}'| << 1

e−ikx⃗′⋅R^≈1−ikx⃗′⋅R^e^{-ik \vec{x}' \cdot \hat{R}} \approx 1 - ik \vec{x}' \cdot \hat{R}

A⃗(x⃗)=μ0eikR4πR∫[1+(1R−ik)x⃗′⋅R^+⋯ ]J⃗(x⃗′)dV′\vec{A}(\vec{x}) = \frac{\mu_0 e^{ikR}}{4\pi R} \int \left[1 + (\frac{1}{R} - ik)\vec{x}' \cdot \hat{R} + \cdots \right] \vec{J}(\vec{x}')dV'

第一项：

A⃗(x⃗)=μ0eikR4πR∫J⃗(x⃗′)dV′\vec{A}(\vec{x}) = \frac{\mu_0 e^{ikR}}{4\pi R} \int \vec{J}(\vec{x}')dV'

*电偶极辐射：

∫J⃗(x⃗′)dV′=dP⃗(t)dt=−iωP⃗(t)\int \vec{J}(\vec{x}')dV' = \frac{d\vec{P}(t)}{dt} = -i\omega \vec{P}(t)

J⃗(x⃗′,t)=J⃗e−iωt,P⃗(t)=P⃗0e−iωt\vec{J}(\vec{x}',t) = \vec{J}e^{-i\omega t},\vec{P}(t) = \vec{P}_0 e^{-i\omega t}

∫J⃗(x⃗′)dV′=−iωP⃗0\int \vec{J}(\vec{x}')dV' = -i\omega \vec{P}_0

A⃗(x⃗)=iωμ0eikR4πRP⃗0\vec{A}(\vec{x}) = \frac{i\omega \mu_0 e^{ikR}}{4\pi R} \vec{P}_0

电磁场：

B⃗=∇×A⃗E⃗=−1−iωμ0ε0∇×B⃗=ick∇×B⃗\begin{aligned} \vec{B} &= \nabla \times \vec{A}\\ \vec{E} &= -\frac{1}{-i \omega \mu_0 \varepsilon_0} \nabla \times \vec{B} = i\frac{c}{k} \nabla \times \vec{B} \end{aligned}

在球坐标系下：

B⃗=−P0k34πε0c[i(kR)2+1kR]sin⁡θei(kR−ωt)ϕ^E⃗=−2P0k34πε0[1(kR)3−i(kR)2]cos⁡θei(kR−ωt)R^+P0k34πε0[1(kR)3−i(kR)2−1kR]sin⁡θei(kR−ωt)θ^\begin{aligned} \vec{B} &= -\frac{P_0k^3}{4\pi \varepsilon_0 c}\left[\frac{i}{(kR)^2}+ \frac{1}{kR}\right]\sin \theta e^{i(kR - \omega t)} \hat{\phi}\\ \vec{E} &= -\frac{2P_0k^3}{4\pi \varepsilon_0}\left[\frac{1}{(kR)^3} - \frac{i}{(kR)^2}\right]\cos \theta e^{i(kR - \omega t)} \hat{R}\\ &+\frac{P_0k^3}{4\pi \varepsilon_0}\left[\frac{1}{(kR)^3} - \frac{i}{(kR)^2} - \frac{1}{kR}\right]\sin \theta e^{i(kR - \omega t)} \hat{\theta}\\ \end{aligned}

  1. 近区：R<<λ=2πkR << \lambda = \frac{2\pi}{k}，kR<<1kR << 1

B⃗≈−iP0k4πε0cR2sin⁡θeiωtϕ^E⃗≈−2P04πε0R3cos⁡θeiωtR^+P04πε0R3sin⁡θe−iωtθ^=14πε03(R⃗⋅P⃗)R⃗−R2P⃗R5\begin{aligned} \vec{B} &\approx -\frac{iP_0k}{4\pi \varepsilon_0 cR^2}\sin \theta e^{i\omega t} \hat{\phi}\\ \vec{E} &\approx -\frac{2P_0}{4\pi \varepsilon_0 R^3}\cos \theta e^{i\omega t} \hat{R} + \frac{P_0}{4\pi \varepsilon_0 R^3}\sin \theta e^{-i\omega t} \hat{\theta}\\ &= \frac{1}{4\pi \varepsilon_0} \frac{3(\vec{R}\cdot \vec{P})\vec{R} - R^2 \vec{P}}{R^5} \end{aligned}

与静场分布类似
2. 远区：R>>λ=2πkR >> \lambda = \frac{2\pi}{k}，kR>>1kR >> 1

B⃗≈−P0k34πε0cRsin⁡θei(kR−ωt)ϕ^E⃗≈−P0k24πε0Rsin⁡θei(kR−ωt)θ^\begin{aligned} \vec{B} &\approx -\frac{P_0k^3}{4\pi \varepsilon_0 cR}\sin \theta e^{i(kR - \omega t)} \hat{\phi}\\ \vec{E} &\approx -\frac{P_0k^2}{4\pi \varepsilon_0 R}\sin \theta e^{i(kR - \omega t)} \hat{\theta}\\ \end{aligned}

以辐射场为主
H⃗=R^×E⃗η\vec{H} = \hat{R} \times \frac{\vec{E}}{\eta}，其中η\eta为真空波阻抗
平均能流：

<S⃗>=Re{E⃗×H⃗∗2}=∣P0∣2ω4sin⁡2θ32π2ε0c3R2R^∝sin⁡2θ<\vec{S}> = Re \left\{\frac{\vec{E} \times \vec{H}^*}{2}\right\} = \frac{|P_0|^2\omega^4\sin^2\theta}{32\pi^2 \varepsilon_0 c^3 R^2} \hat{R} \propto \sin^2\theta

考虑总辐射功率

P=∮<S⃗>⋅dS⃗=∣P0∣2ω412πε0c3∝ω4P = \oint <\vec{S}> \cdot d\vec{S} = \frac{|P_0|^2\omega^4}{12\pi \varepsilon_0 c^3} \propto \omega^4

  1. 过渡区：R∼λR \sim \lambda
    E⃗,B⃗\vec{E},\vec{B}介于近区和远区之间

*短天线(L<<λL<<\lambda)

这里应该有一张图

I(z,t)=I0(1−2∣z∣L)e−iωt,∣z∣<L2I(z,t) = I_0(1-\frac{2|z|}{L})e^{-i\omega t},\quad |z|<\frac{L}{2}

电偶极矩振幅：

P⃗0=−1iω∫J⃗(x⃗′)dV′=iI0L2ωe^z\vec{P}_0 = -\frac{1}{i\omega} \int \vec{J}(\vec{x}')dV' = \frac{i I_0 L}{2\omega}\hat{e}_z

总辐射功率：

P=πηI0212(Lλ)2=I02Rrad2P = \frac{\pi \eta I^2_0}{12}(\frac{L}{\lambda})^2 = \frac{I_0^2R_{rad}}{2}

其中Rrad=πη6(Lλ)2R_{rad} = \frac{\pi \eta}{6}(\frac{L}{\lambda})^2为辐射阻抗
不妨设Lλ=0.01\frac{L}{\lambda} = 0.01，Rrad=0.02ΩR_{rad} = 0.02 \Omega

*半波天线(L=λ2L = \frac{\lambda}{2})

I(z,t)=I0cos⁡kze−iωtI(z,t) = I_0 \cos kz e^{-i\omega t}

其中∣z∣≤L2|z|\leq \frac{L}{2}

A⃗(x⃗)=μ04π∫J⃗(x⃗′)eikrrdV′\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi} \int \frac{\vec{J}(\vec{x}')e^{ikr}}{r}dV'

在远场区，r≈R−zcos⁡θr \approx R - z\cos\theta

A⃗(x⃗)≈μ0I0eikR2πkRcos⁡(π2cos⁡θ)sin⁡2θe^z\vec{A}(\vec{x}) \approx \frac{\mu_0I_0e^{ikR}}{2\pi k R} \frac{\cos(\frac{\pi}{2} \cos\theta)}{\sin^2\theta} \hat{e}_z

辐射能流：

<S⃗>∝cos⁡2(π2cos⁡θ)sin⁡2θ<\vec{S}> \propto \frac{\cos^2(\frac{\pi}{2} \cos\theta)}{\sin^2\theta}

垂直于天线方向辐射最大
辐射阻抗：

Rrad≈2.44μ0c4π=73.2ΩR_{rad} \approx 2.44\frac{\mu_0c}{4\pi} = 73.2\Omega