电动力学每日一题 2021/10/14

(a) Define r∣∣=xx^+yy^\textbf r_{||}=x\hat x+y\hat yr∣∣=xx^+yy^, r∣∣=x2+y2r_{||}=\sqrt{x^2+y^2}r∣∣=x2+y2,
M(r,t)=M0z^[Rect(x/Lx)Rect(y/Ly)−Circ(r∣∣/R)]Rect(z/Lz)\textbf M(\textbf r,t)=M_0\hat z[Rect(x/L_x)Rect(y/L_y)-Circ(r_{||}/R)]Rect(z/L_z)M(r,t)=M0z^[Rect(x/Lx)Rect(y/Ly)−Circ(r∣∣/R)]Rect(z/Lz)

(b)
ρbound(m)(r,t)=−∇⋅M=−∂Mz∂z=M0[Rect(x/Lx)Rect(y/Ly)−Circ(r∣∣/R)][δ(z−Lz/2)−δ(z+Lz/2)]\begin{aligned}\rho_{bound}^{(m)}(\textbf r,t) & = -\nabla \cdot \textbf M = -\frac{\partial M_z}{\partial z} \\ & = M_0[Rect(x/L_x)Rect(y/L_y)-Circ(r_{||}/R)][\delta(z-L_z/2)-\delta(z+L_z/2)]\end{aligned}ρbound(m)(r,t)=−∇⋅M=−∂z∂Mz=M0[Rect(x/Lx)Rect(y/Ly)−Circ(r∣∣/R)][δ(z−Lz/2)−δ(z+Lz/2)]

Jbound(e)=μ0−1∇×M=μ0−1M0[∇×Rect(x/Lx)Rect(y/Ly)Rect(z/Lz)z^−∇×Circ(r∣∣/R)Rect(z/Lz)z^]\begin{aligned} \textbf J_{bound}^{(e)} &= \mu_0^{-1}\nabla \times \textbf M \\ & =\mu_0^{-1} M_0[\nabla \times Rect(x/L_x)Rect(y/L_y)Rect(z/L_z)\hat z-\nabla \times Circ(r_{||}/R)Rect(z/L_z)\hat z]\end{aligned}Jbound(e)=μ0−1∇×M=μ0−1M0[∇×Rect(x/Lx)Rect(y/Ly)Rect(z/Lz)z^−∇×Circ(r∣∣/R)Rect(z/Lz)z^]

在直角坐标系中计算
∇×Rect(x/Lx)Rect(y/Ly)Rect(z/Lz)z^=μ0−1M0Rect(x/Lx)[δ(y+Ly/2)−δ(y−Ly/2)]Rect(z/Lz)x^−μ0−1M0[δ(x+Lx/2)−δ(x−Lx/2)]Rect(y/Ly)Rect(z/Lz)y^\nabla \times Rect(x/L_x)Rect(y/L_y)Rect(z/L_z)\hat z \\ =\mu_0^{-1} M_0 Rect(x/L_x)[\delta(y+L_y/2)-\delta(y-L_y/2)]Rect(z/L_z)\hat x \\ -\mu_0^{-1} M_0 [\delta(x+L_x/2)-\delta(x-L_x/2)]Rect(y/L_y)Rect(z/L_z)\hat y∇×Rect(x/Lx)Rect(y/Ly)Rect(z/Lz)z^=μ0−1M0Rect(x/Lx)[δ(y+Ly/2)−δ(y−Ly/2)]Rect(z/Lz)x^−μ0−1M0[δ(x+Lx/2)−δ(x−Lx/2)]Rect(y/Ly)Rect(z/Lz)y^

在柱坐标系中计算
∇×Circ(r∣∣/R)Rect(z/Lz)z^=−μ0−1M0δ(r∣∣−R)Rect(z/Lz)ϕ^\nabla \times Circ(r_{||}/R)Rect(z/L_z)\hat z =-\mu_0^{-1}M_0\delta(r_{||}-R)Rect(z/L_z) \hat \phi∇×Circ(r∣∣/R)Rect(z/Lz)z^=−μ0−1M0δ(r∣∣−R)Rect(z/Lz)ϕ^

综上，
Jbound(e)=μ0−1M0Rect(x/Lx)[δ(y+Ly/2)−δ(y−Ly/2)]Rect(z/Lz)x^−μ0−1M0[δ(x+Lx/2)−δ(x−Lx/2)]Rect(y/Ly)Rect(z/Lz)y^−μ0−1M0δ(r∣∣−R)Rect(z/Lz)ϕ^\textbf J_{bound}^{(e)} =\mu_0^{-1} M_0 Rect(x/L_x)[\delta(y+L_y/2)-\delta(y-L_y/2)]Rect(z/L_z)\hat x \\ -\mu_0^{-1} M_0 [\delta(x+L_x/2)-\delta(x-L_x/2)]Rect(y/L_y)Rect(z/L_z)\hat y \\ -\mu_0^{-1}M_0\delta(r_{||}-R)Rect(z/L_z) \hat \phiJbound(e)=μ0−1M0Rect(x/Lx)[δ(y+Ly/2)−δ(y−Ly/2)]Rect(z/Lz)x^−μ0−1M0[δ(x+Lx/2)−δ(x−Lx/2)]Rect(y/Ly)Rect(z/Lz)y^−μ0−1M0δ(r∣∣−R)Rect(z/Lz)ϕ^

(c)
ρbounded(m)(k,w)=∫−∞+∞ρbound(m)(r,t)e−i(k⋅r−wt)drdt=2πM0δ(w)(I1I2−I3)I4\begin{aligned} \rho^{(m)}_{bounded}(\textbf k,w)& =\int_{-\infty}^{+\infty} \rho^{(m)}_{bound}(\textbf r,t)e^{-i(\textbf k\cdot \textbf r-wt)}d \textbf rdt \\ & = 2 \pi M_0\delta(w) (I_1I_2-I_3)I_4\end{aligned}ρbounded(m)(k,w)=∫−∞+∞ρbound(m)(r,t)e−i(k⋅r−wt)drdt=2πM0δ(w)(I1I2−I3)I4

其中
I1=F[Rect(x/Lx)](kx)=sin⁡(Lxkx/2)kx/2I2=F[Rect(y/Ly)](ky)=sin⁡(Lyky/2)ky/2I3=F[Circ(r∣∣/R)](k∣∣)=2πRk∣∣J1(K∣∣R)I4=F[δ(z−Lz/2)−δ(z+Lz/2)](kz)=−2isin⁡(Lzkz/2)I_1 = \mathcal{F}[Rect(x/L_x)](k_x)=\frac{\sin(L_xk_x/2)}{k_x/2} \\ I_2 = \mathcal{F}[Rect(y/L_y)](k_y)=\frac{\sin(L_yk_y/2)}{k_y/2} \\ I_3 = \mathcal{F}[Circ(r_{||}/R)](k_{||})=\frac{2 \pi R}{k_{||}}J_1(K_{||}R) \\ I_4 = \mathcal{F}[\delta(z-L_z/2)-\delta(z+L_z/2)](k_z)=-2i \sin(L_zk_z/2)I1=F[Rect(x/Lx)](kx)=kx/2sin(Lxkx/2)I2=F[Rect(y/Ly)](ky)=ky/2sin(Lyky/2)I3=F[Circ(r∣∣/R)](k∣∣)=k∣∣2πRJ1(K∣∣R)I4=F[δ(z−Lz/2)−δ(z+Lz/2)](kz)=−2isin(Lzkz/2)

综上
ρbounded(m)(k,w)=−i4πM0δ(w)[LxLysinc(Lxkx2π)sinc(Lyky2π)−2πRJ1(Rkx2+ky2)kx2+ky2]sin⁡(Lzkz/2)\begin{aligned} \rho^{(m)}_{bounded}(\textbf k,w) = -i4 \pi M_0 \delta(w) \left[ L_xL_y sinc(\frac{L_xk_x}{2 \pi})sinc(\frac{L_yk_y}{2 \pi})-2 \pi R \frac{J_1(R\sqrt{k_x^2+k_y^2})}{\sqrt{k_x^2+k_y^2}} \right]\sin(L_zk_z/2) \end{aligned}ρbounded(m)(k,w)=−i4πM0δ(w)⎣⎡LxLysinc(2πLxkx)sinc(2πLyky)−2πRkx2+ky2J1(Rkx2+ky2)⎦⎤sin(Lzkz/2)

简直了，考试应该不会考这种题吧，照着公式两个小时也算不完啊。。。

Jbound(e)(k,w)=i2πμ0−1M0δ(w)Lzsinc(Lxkx2π)[LxLysinc(Lxkx2π)sinc(Lyky2π)(kyx^−kxy^)+2πRJ1(Rkx2+ky2)ϕ^]\textbf J^{(e)}_{bound}(\textbf k,w)=i 2 \pi \mu_0^{-1}M_0\delta(w)L_z sinc(\frac{L_xk_x}{2 \pi})\left[ L_xL_y sinc(\frac{L_xk_x}{2 \pi})sinc(\frac{L_yk_y}{2 \pi})(k_y \hat x - k_x \hat y)+2 \pi R J_1(R\sqrt{k_x^2+k_y^2}) \hat \phi \right]Jbound(e)(k,w)=i2πμ0−1M0δ(w)Lzsinc(2πLxkx)[LxLysinc(2πLxkx)sinc(2πLyky)(kyx^−kxy^)+2πRJ1(Rkx2+ky2)ϕ^]

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