电动力学每日一题 2021/10/13 用Fourier变换法计算静止电荷产生的电场

静止点电荷
具有均匀线密度的静止电荷产生的电场
具有均匀面密度的静止电荷产生的电场

用Fourier变换法计算电场时可以在频域中解出标量势后将其变换到时域中求梯度得到电场，也可以在频域中就解出电场再将其变换到时域中；具体怎么做取决于在哪一步做Fourier逆变换更简单。

静止点电荷

假设原点处有一个带电量为qqq的静止点电荷，则空间中的电荷密度为
ρ(r,t)=qδ(r)=qδ(x)δ(y)δ(z)\rho(\textbf r,t)=q \delta(\textbf r)=q\delta(x)\delta(y)\delta(z)ρ(r,t)=qδ(r)=qδ(x)δ(y)δ(z)

用Fourier变换法计算由这个电荷密度产生的电场：

第一步，用4-D Fourier变换计算频域中的电荷密度
∫−∞+∞δ(r)e−(k⋅r)d3r=e−k⋅0=1ρ(k,w)=∫−∞+∞qδ(r)e−(k⋅r−wt)d3rdt=2πqδ(w)\int_{-\infty}^{+\infty} \delta(\textbf r)e^{-(\textbf k\cdot \textbf r)}d^3 \textbf r =e^{-\textbf k \cdot 0}=1\\ \rho(\textbf k,w)=\int_{-\infty}^{+\infty} q \delta(\textbf r)e^{-(\textbf k\cdot \textbf r - wt)}d^3\textbf r dt=2 \pi q \delta(w)∫−∞+∞δ(r)e−(k⋅r)d3r=e−k⋅0=1ρ(k,w)=∫−∞+∞qδ(r)e−(k⋅r−wt)d3rdt=2πqδ(w)

第二步，在频域中计算标量势
ψ(k,w)=ρ(k,w)ϵ0[k2−(w/c)2]=2πqδ(w)ϵ0[k2−(w/c)2]\psi(\textbf k,w)=\frac{\rho(\textbf k,w)}{\epsilon_0[k^2-(w/c)^2]}=\frac{2 \pi q \delta(w)}{\epsilon_0[k^2-(w/c)^2]}ψ(k,w)=ϵ0[k2−(w/c)2]ρ(k,w)=ϵ0[k2−(w/c)2]2πqδ(w)

第三步，用4-D Fourier逆变换计算时域中的标量势
ψ(r,t)=(2π)−4∫−∞+∞ψ(k,w)ei(k⋅r−wt)d3kdw=(2π)−4∫−∞+∞2πqδ(w)ϵ0[k2−(w/c)2]ei(k⋅r−wt)d3kdw\begin{aligned} \psi(\textbf r,t) & = (2 \pi)^{-4} \int_{-\infty}^{+\infty} \psi(\textbf k,w)e^{i(\textbf k \cdot \textbf r - wt)}d^3 \textbf k d w \\ & = (2 \pi)^{-4} \int_{-\infty}^{+\infty}\frac{2 \pi q \delta(w)}{\epsilon_0[k^2-(w/c)^2]}e^{i(\textbf k \cdot \textbf r - wt)}d^3 \textbf k d w\end{aligned}ψ(r,t)=(2π)−4∫−∞+∞ψ(k,w)ei(k⋅r−wt)d3kdw=(2π)−4∫−∞+∞ϵ0[k2−(w/c)2]2πqδ(w)ei(k⋅r−wt)d3kdw

其中
∫−∞+∞δ(w)[k2−(w/c)2]e−iwtdw=e−iwt[k2−(w/c)2]∣w=0=1k2\int_{-\infty}^{+\infty} \frac{\delta(w)}{[k^2-(w/c)^2]}e^{-iwt}dw = \left. \frac{e^{-iwt}}{[k^2-(w/c)^2]}\right|_{w=0}=\frac{1}{k^2}∫−∞+∞[k2−(w/c)2]δ(w)e−iwtdw=[k2−(w/c)2]e−iwt∣∣∣∣w=0=k21

所以
ψ(r,t)=(2π)−3(q/ϵ0)∫−∞+∞eik⋅rk2d3k\begin{aligned} \psi(\textbf r,t) & = (2 \pi)^{-3}(q/\epsilon_0) \int_{-\infty}^{+\infty} \frac{e^{i \textbf k\cdot \textbf r}}{k^2} d^3 \textbf k \end{aligned}ψ(r,t)=(2π)−3(q/ϵ0)∫−∞+∞k2eik⋅rd3k

考虑如下图所示的几何关系，则k⋅r=krcos⁡θ\textbf k \cdot \textbf r=kr \cos \thetak⋅r=krcosθ，并且
∫−∞+∞d3k=∫0π∫02π∫0+∞k2sin⁡θdkdϕdθ=2π∫0π∫0+∞k2sin⁡θdkdθ\int_{-\infty}^{+\infty} d^3 \textbf k=\int_{0}^{\pi} \int_{0}^{2\pi}\int_0^{+\infty} k^2 \sin \theta dk d \phi d \theta = 2 \pi \int_{0}^{\pi} \int_0^{+\infty} k^2 \sin \theta dk d \theta ∫−∞+∞d3k=∫0π∫02π∫0+∞k2sinθdkdϕdθ=2π∫0π∫0+∞k2sinθdkdθ

因此
ψ(r,t)=(2π)−2(q/ϵ0)∫0π∫0+∞eikrcos⁡θk2k2sin⁡θdkdθ=q4π2ϵ0∫0+∞eikrcos⁡θ−ikr∣0πdk=q4π2ϵ0∫0+∞2sin⁡(kr)krdk=q4π2ϵ0r∫−∞+∞sin⁡xxdx=q4πϵ0r\begin{aligned} \psi(\textbf r,t) &= (2 \pi)^{-2}(q/\epsilon_0) \int_{0}^{\pi} \int_0^{+\infty}\frac{e^{i kr\cos \theta}}{k^2} k^2 \sin \theta d k d \theta \\ & = \frac{q}{4 \pi^2 \epsilon_0} \int_0^{+\infty} \left. \frac{e^{ikr \cos \theta}}{-ikr} \right|_{0}^{\pi}dk \\ & = \frac{q}{4 \pi^2 \epsilon_0} \int_0^{+\infty} \frac{2\sin(kr)}{kr}dk \\ & = \frac{q}{4 \pi^2 \epsilon_0 r} \int_{-\infty}^{+\infty} \frac{\sin x}{x}dx = \frac{q}{4 \pi \epsilon_0 r} \end{aligned}ψ(r,t)=(2π)−2(q/ϵ0)∫0π∫0+∞k2eikrcosθk2sinθdkdθ=4π2ϵ0q∫0+∞−ikreikrcosθ∣∣∣∣0πdk=4π2ϵ0q∫0+∞kr2sin(kr)dk=4π2ϵ0rq∫−∞+∞xsinxdx=4πϵ0rq

第四步，计算电场
E(r,t)=−∇ψ(r,t)=q4πϵ0r^r2\textbf E(\textbf r,t)=-\nabla \psi(\textbf r,t)=\frac{q}{4 \pi \epsilon_0} \frac{\hat r}{r^2}E(r,t)=−∇ψ(r,t)=4πϵ0qr2r^

具有均匀线密度的静止电荷产生的电场

假设电荷在z-轴上均匀分布，线密度为λ0\lambda_0λ0，则空间中的电荷密度为
ρ(r,t)=λ0δ(x)δ(y)\rho(\textbf r,t)=\lambda_0 \delta(x) \delta(y)ρ(r,t)=λ0δ(x)δ(y)

用Fourier变换法计算由这个电荷密度产生的电场：

第一步，用4-D Fourier变换计算频域中的电荷密度
∫−∞+∞δ(x)δ(y)e−i(kxx+kyy+kzz)dxdydz=e−ikx0−iky02πδ(kz)=2πδ(kz)ρ(k,w)=∫−∞+∞λ0δ(x)δ(y)e−i(k⋅r−wt)d3rdt=(2π)2λ0δ(kz)δ(w)\int_{-\infty}^{+\infty} \delta(x)\delta(y)e^{-i(k_xx+k_yy+k_zz)}dxdydz=e^{-ik_x 0-ik_y0}2\pi \delta(k_z)=2 \pi \delta(k_z)\\ \rho(\textbf k,w)=\int_{-\infty}^{+\infty} \lambda_0 \delta(x)\delta(y)e^{-i(\textbf k\cdot \textbf r - wt)}d^3\textbf r dt=(2 \pi)^2 \lambda_0 \delta(k_z) \delta(w)∫−∞+∞δ(x)δ(y)e−i(kxx+kyy+kzz)dxdydz=e−ikx0−iky02πδ(kz)=2πδ(kz)ρ(k,w)=∫−∞+∞λ0δ(x)δ(y)e−i(k⋅r−wt)d3rdt=(2π)2λ0δ(kz)δ(w)

第二步，在频域中计算电场

E(k,w)=−iϵ0−1kρ+iμ0wJk2−(w/c)2=−iϵ0−1kρk2−(w/c)2=−ikρϵ0[k2−(w/c)2]\textbf E(\textbf k,w)=\frac{-i \epsilon^{-1}_0 \textbf k \rho + i \mu_0 w \textbf J}{k^2-(w/c)^2} = \frac{-i \epsilon^{-1}_0 \textbf k \rho }{k^2-(w/c)^2} =\frac{-i \textbf k \rho}{\epsilon_0[k^2-(w/c)^2]}E(k,w)=k2−(w/c)2−iϵ0−1kρ+iμ0wJ=k2−(w/c)2−iϵ0−1kρ=ϵ0[k2−(w/c)2]−ikρ

第三步，将电场变换到时域中
∬−∞+∞(kxx^+kyy^+kzz^)δ(kz)δ(w)ei(kxx+kyy+kzz)−iwt[(kx2+ky2+kz2)−(w/c)2]dzdw=(kxx^+kyy^+kzz^)ei(kxx+kyy+kzz)−iwt[(kx2+ky2+kz2)−(w/c)2]∣kz=0,w=0=(kxx^+kyy^)ei(kxx+kyy)kx2+ky2\begin{aligned} &\iint_{-\infty}^{+\infty} \frac{(k_x\hat x+k_y \hat y+k_z \hat z)\delta(k_z)\delta(w)e^{i(k_xx+k_yy+k_zz)-iwt}}{[(k_x^2+k_y^2+k_z^2)-(w/c)^2]} dzdw \\ & =\left. \frac{(k_x\hat x+k_y \hat y+k_z \hat z)e^{i(k_xx+k_yy+k_zz)-iwt}}{[(k_x^2+k_y^2+k_z^2)-(w/c)^2]} \right|_{k_z=0,w=0} \\ &= \frac{(k_x \hat x + k_y \hat y)e^{i(k_xx+k_yy)}}{k_x^2+k_y^2} \end{aligned}∬−∞+∞[(kx2+ky2+kz2)−(w/c)2](kxx^+kyy^+kzz^)δ(kz)δ(w)ei(kxx+kyy+kzz)−iwtdzdw=[(kx2+ky2+kz2)−(w/c)2](kxx^+kyy^+kzz^)ei(kxx+kyy+kzz)−iwt∣∣∣∣kz=0,w=0=kx2+ky2(kxx^+kyy^)ei(kxx+kyy)

所以
E(r,t)=(2π)−4∫−∞+∞−ikρϵ0[k2−(w/c)2]ei(k⋅r−wt)d3kdw=−i(2π)−2(λ0/ϵ0)∫−∞+∞∫−∞+∞(kxx^+kyy^)ei(kxx+kyy)kx2+ky2dkxdky\begin{aligned} \textbf E(\textbf r,t) & = (2 \pi)^{-4} \int_{-\infty}^{+\infty} \frac{-i \textbf k \rho}{\epsilon_0[k^2-(w/c)^2]}e^{i(\textbf k \cdot \textbf r-wt)}d^3 \textbf k dw \\ & = -i (2 \pi)^{-2}(\lambda_0/\epsilon_0)\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{(k_x \hat x + k_y \hat y)e^{i(k_xx+k_yy)}}{k_x^2+k_y^2} dk_xdk_y \end{aligned}E(r,t)=(2π)−4∫−∞+∞ϵ0[k2−(w/c)2]−ikρei(k⋅r−wt)d3kdw=−i(2π)−2(λ0/ϵ0)∫−∞+∞∫−∞+∞kx2+ky2(kxx^+kyy^)ei(kxx+kyy)dkxdky

考虑下面的几何关系，
k∣∣=kxx^+kyy^,r∣∣=xx^+yy^k∣∣=∣k∣∣∣,r∣∣=∣r∣∣∣,r^∣∣=r∣∣r∣∣\textbf k_{||}=k_x \hat x + k_y \hat y,\textbf r_{||}=x \hat x + y \hat y \\ k_{||} = |\textbf k_{||}|,r_{||}=|\textbf r_{||}|,\hat r_{||} = \frac{\textbf r_{||}}{r_{||}}k∣∣=kxx^+kyy^,r∣∣=xx^+yy^k∣∣=∣k∣∣∣,r∣∣=∣r∣∣∣,r^∣∣=r∣∣r∣∣

则
kx2+ky2=k∣∣2kxx^+kyy^=k∣∣=k∣∣cos⁡(ϕ)r^∣∣kxx^+kyy^=k∣∣⋅r∣∣=k∣∣r∣∣cos⁡(ϕ)∫−∞+∞∫−∞+∞dkxdky=∫02π∫0+∞k∣∣dk∣∣dϕk_x^2+k_y^2 = k_{||}^2 \\ k_x \hat x + k_y \hat y = \textbf k_{||} = k_{||} \cos(\phi) \hat r_{||}\\ k_x \hat x + k_y \hat y = \textbf k_{||} \cdot \textbf r_{||} = k_{||}r_{||}\cos(\phi) \\ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}dk_xdk_y = \int_{0}^{2\pi} \int_0^{+\infty} k_{||}d k_{||}d \phikx2+ky2=k∣∣2kxx^+kyy^=k∣∣=k∣∣cos(ϕ)r^∣∣kxx^+kyy^=k∣∣⋅r∣∣=k∣∣r∣∣cos(ϕ)∫−∞+∞∫−∞+∞dkxdky=∫02π∫0+∞k∣∣dk∣∣dϕ

所以
−i(2π)−2(λ0/ϵ0)∫−∞+∞∫−∞+∞(kxx^+kyy^)ei(kxx+kyy)kx2+ky2dkxdky=−iλ04π2ϵ0∫02π∫0+∞k∣∣cos⁡(ϕ)r^∣∣k∣∣2eik∣∣r∣∣cos⁡(ϕ)k∣∣dk∣∣dϕ=−iλ0r^∣∣4π2ϵ0∫02π∫0+∞cos⁡(ϕ)eik∣∣r∣∣cos⁡(ϕ)dk∣∣dϕ=λ0r^∣∣2πϵ0∫0+∞J1(k∣∣r∣∣)dk∣∣=λ02πϵ0r^∣∣r∣∣\begin{aligned}& -i (2 \pi)^{-2}(\lambda_0/\epsilon_0)\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\frac{(k_x \hat x + k_y \hat y)e^{i(k_xx+k_yy)}}{k_x^2+k_y^2} dk_xdk_y \\ & = -i \frac{\lambda_0}{4 \pi^2 \epsilon_0} \int_{0}^{2\pi} \int_0^{+\infty} \frac{k_{||} \cos(\phi) \hat r_{||}}{k_{||}^2}e^{ik_{||} r_{||}\cos(\phi)}k_{||}d k_{||}d \phi \\ & = -i \frac{\lambda_0 \hat r_{||}}{4 \pi^2 \epsilon_0} \int_{0}^{2\pi} \int_0^{+\infty} \cos(\phi) e^{ik_{||} r_{||}\cos(\phi)}d k_{||}d \phi \\ & =\frac{\lambda_0 \hat r_{||}}{2 \pi \epsilon_0} \int_0^{+\infty} J_1(k_{||}r_{||})d k_{||} = \frac{\lambda_0}{2 \pi \epsilon_0} \frac{\hat r_{||}}{r_{||}}\end{aligned}−i(2π)−2(λ0/ϵ0)∫−∞+∞∫−∞+∞kx2+ky2(kxx^+kyy^)ei(kxx+kyy)dkxdky=−i4π2ϵ0λ0∫02π∫0+∞k∣∣2k∣∣cos(ϕ)r^∣∣eik∣∣r∣∣cos(ϕ)k∣∣dk∣∣dϕ=−i4π2ϵ0λ0r^∣∣∫02π∫0+∞cos(ϕ)eik∣∣r∣∣cos(ϕ)dk∣∣dϕ=2πϵ0λ0r^∣∣∫0+∞J1(k∣∣r∣∣)dk∣∣=2πϵ0λ0r∣∣r^∣∣

具有均匀面密度的静止电荷产生的电场

假设电荷在xOz平面上均匀分布，面密度为σ0\sigma_0σ0，则空间中的电荷密度为
ρ(r,t)=σ0δ(y)\rho(\textbf r,t)=\sigma_0 \delta(y)ρ(r,t)=σ0δ(y)

变换到频域中，
ρ(k,w)=(2π)3σ0δ(x)δ(z)δ(w)\rho(\textbf k ,w) = (2 \pi)^3\sigma_0 \delta(x)\delta(z)\delta(w)ρ(k,w)=(2π)3σ0δ(x)δ(z)δ(w)

计算电场，
E(r,t)=(2π)−4∫−∞+∞−iϵ0−1kρ+iμ0wJk2−(w/c)2ei(k⋅r−wt)d3kdw=σ02πϵ0∫−∞+∞−ikyy^eikyyky2dky=σ02ϵ0Sign(y)y^\begin{aligned} \textbf E(\textbf r,t)& = (2 \pi)^{-4} \int_{-\infty}^{+\infty} \frac{-i \epsilon^{-1}_0 \textbf k \rho + i \mu_0 w \textbf J}{k^2-(w/c)^2}e^{i(\textbf k \cdot \textbf r-wt)}d^3 \textbf k dw \\ & = \frac{\sigma_0}{2 \pi \epsilon_0} \int_{-\infty}^{+\infty} \frac{-ik_y \hat ye^{ik_yy}}{k_y^2}dk_y = \frac{\sigma_0}{2 \epsilon_0}Sign(y) \hat y\end{aligned}E(r,t)=(2π)−4∫−∞+∞k2−(w/c)2−iϵ0−1kρ+iμ0wJei(k⋅r−wt)d3kdw=2πϵ0σ0∫−∞+∞ky2−ikyy^eikyydky=2ϵ0σ0Sign(y)y^