电动力学每日一题 2021/10/23 载流板产生的电磁场

载流板的辐射

载流板的辐射

先验证电荷守恒：
∂ρ∂t=−∇⋅J=−∂∂zJz=0\frac{\partial \rho}{\partial t} = -\nabla \cdot \textbf J = -\frac{\partial}{\partial z}J_z = 0∂t∂ρ=−∇⋅J=−∂z∂Jz=0

这说明虽然在这块载流板上存在交流电，但总电荷量并没有变化。

在10/19的题目中我们讨论了对含cos⁡(w0t)\cos(w_0t)cos(w0t)这一项的函数直接做Fourier变换在后续计算中会出现发散的积分，无法得到有物理意义的结果。所以我们假设w0=w′+iw′′w_0=w'+iw''w0=w′+iw′′，在最后结果中取w′′→0w'' \to 0w′′→0，
J=Js0δ(y)eiw0∗t+e−w0t2z^\textbf J = J_{s0}\delta(y) \frac{e^{iw_0^*t}+e^{-w_0 t}}{2} \hat zJ=Js0δ(y)2eiw0∗t+e−w0tz^

然后计算J\textbf JJ的Fourier变换，
J(k,w)=∫−∞+∞J(r,t)e−ik⋅rdr=Js02∫−∞+∞δ(y)eiw0∗te−ik⋅rdrdt+Js02∫−∞+∞δ(y)e−w0te−ik⋅rdrdt=2π2Js0δ(kx)δ(kz)e−iw0tz^+2π2Js0δ(kx)δ(kz)eiw0∗tz^\begin{aligned} \textbf J(\textbf k,w) & = \int_{-\infty}^{+\infty} \textbf J(\textbf r,t)e^{-i\textbf k\cdot \textbf r}d \textbf r \\ & = \frac{J_{s0}}{2}\int_{-\infty}^{+\infty} \delta(y)e^{iw_0^*t}e^{-i\textbf k\cdot \textbf r}d \textbf r dt+\frac{J_{s0}}{2}\int_{-\infty}^{+\infty} \delta(y)e^{-w_0t}e^{-i\textbf k\cdot \textbf r}d \textbf r dt \\ & = 2 \pi^2 J_{s0} \delta(k_x)\delta(k_z)e^{-iw_0 t}\hat z+2 \pi^2 J_{s0} \delta(k_x)\delta(k_z) e^{iw_0^*t} \hat z\end{aligned}J(k,w)=∫−∞+∞J(r,t)e−ik⋅rdr=2Js0∫−∞+∞δ(y)eiw0∗te−ik⋅rdrdt+2Js0∫−∞+∞δ(y)e−w0te−ik⋅rdrdt=2π2Js0δ(kx)δ(kz)e−iw0tz^+2π2Js0δ(kx)δ(kz)eiw0∗tz^

下面分别计算这两项产生的矢量势，第一项产生的矢量势为
(2π)−3∫−∞+∞2π2μ0Js0δ(kx)δ(kz)e−iw0tz^k2−(w0/c)2eik⋅rdk=μ0Js0e−iw0tz^4π∫−∞+∞eikyyky2−(w0/c)2dky=μ0Js0e−iw0tz^4ππeiw0∣y∣/c−iw0/c=iZ0Js04w0e−iw0(t−∣y∣/c)z^\begin{aligned} & (2\pi)^{-3} \int_{-\infty}^{+\infty} \frac{2 \pi^2 \mu_0 J_{s0} \delta(k_x)\delta(k_z)e^{-iw_0 t}\hat z}{k^2-(w_0/c)^2}e^{i \textbf k \cdot \textbf r}d \textbf k \\ = & \frac{\mu_0 J_{s0}e^{-iw_0 t}\hat z}{4 \pi} \int_{-\infty}^{+\infty} \frac{e^{i k_yy}}{k_y^2-(w_0/c)^2}dk_y \\ = & \frac{\mu_0 J_{s0}e^{-iw_0 t}\hat z}{4 \pi} \frac{\pi e^{iw_0|y|/c}}{-iw_0/c} = \frac{iZ_0J_{s0}}{4 w_0}e^{-iw_0(t-|y|/c)} \hat z\end{aligned}==(2π)−3∫−∞+∞k2−(w0/c)22π2μ0Js0δ(kx)δ(kz)e−iw0tz^eik⋅rdk4πμ0Js0e−iw0tz^∫−∞+∞ky2−(w0/c)2eikyydky4πμ0Js0e−iw0tz^−iw0/cπeiw0∣y∣/c=4w0iZ0Js0e−iw0(t−∣y∣/c)z^

其中
c=1μ0ϵ0,Z0=μ0ϵ0c = \sqrt{\frac{1}{\mu_0 \epsilon_0}},Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}c=μ0ϵ01,Z0=ϵ0μ0

第二项产生的矢量势为，
iZ0Js04w0eiw0∗(t−∣y∣/c)z^\frac{iZ_0J_{s0}}{4 w_0}e^{iw_0^*(t-|y|/c)} \hat z4w0iZ0Js0eiw0∗(t−∣y∣/c)z^

当w′′→0w'' \to 0w′′→0时，w0=w0∗w_0=w_0^*w0=w0∗，
A(r,t)=iZ0Js0z^4w0[eiw0∗(t−∣y∣/c)+e−iw0(t−∣y∣/c)]=Z0Js02w0sin⁡(w0(t−∣y∣/c))z^\begin{aligned}\textbf A(\textbf r ,t) & = \frac{iZ_0J_{s0} \hat z}{4 w_0} [e^{iw_0^*(t-|y|/c)}+e^{-iw_0(t-|y|/c)}] \\ & = \frac{Z_0J_{s0}}{2w_0}\sin(w_0(t-|y|/c))\hat z\end{aligned}A(r,t)=4w0iZ0Js0z^[eiw0∗(t−∣y∣/c)+e−iw0(t−∣y∣/c)]=2w0Z0Js0sin(w0(t−∣y∣/c))z^

接下来要计算Electric field与Magnetic field，就懒得写了