电磁场与波的部分定理证明

最近在学习电磁场与波的时候，看到书上有些公式定理没给证明就直接在用，就尝试着推导了一下

  1. 关于1R的微分\frac{1}{R}的微分

∇(1R)=−RR3\nabla(\frac{1}{R}) = -\frac{\pmb R}{R^3}

这个式子第一眼看过去好怪，一个标量求散度之后得到了一个矢量，所以想推导一下

推导过程如下所示

设位置矢量为R=r−r′,其中r=xex+yey+zez,r′=x′ex+y′ey+z′ez设位置矢量为\pmb R = \pmb r - \pmb r',其中\pmb r = xe_x + ye_y + ze_z,\pmb r' = x'e_x + y'e_y + z'e_z

先推导∇R\pmb\nabla R

∇R=∂R∂xex+∂R∂yey+∂R∂zez=x−x′(x−x′)2+(y−y′)2+(z−z′)2ex+y−y′(x−x′)2+(y−y′)2+(z−z′)2ey+z−z′(x−x′)2+(y−y′)2+(z−z′)2ez=x−x′Rex+y−y′Rey+z−z′Rez=RR\begin{aligned} \pmb\nabla R &= \frac{\partial R}{\partial x}e_x + \frac{\partial R}{\partial y}e_y + \frac{\partial R}{\partial z}e_z \\\\ &= \frac{x - x'}{\sqrt{ {(x - x')}^2 + {(y - y')}^2 + {(z - z')}^2} }e_x + \frac{y - y'}{\sqrt{ {(x - x')}^2 + {(y - y')}^2 + {(z - z')}^2} }e_y + \frac{z - z'}{\sqrt{ {(x - x')}^2 + {(y - y')}^2 + {(z - z')}^2} }e_z \\\\ &= \frac{x - x'}{R}e_x + \frac{y - y'}{R}e_y + \frac{z - z'}{R}e_z \\\\ &= \frac{\pmb R}{R} \end{aligned}

再推导∇(1R)\pmb\nabla(\frac{1}{R})

∇(1R)=∇RR2=RR3\begin{aligned} \pmb\nabla(\frac{1}{R}) &= \frac{\pmb\nabla R}{R^2} \\\\ &= \frac{\pmb R}{R^3} \end{aligned}

  1. The Laplacian of 1R\frac{1}{R}

我们需要进行分类讨论

当R≠0R \neq 0时，有

∇2(1R)=1R2∂∂R(R2∂(1r)∂R)=1R2∂∂R(−1)=0\pmb\nabla^2(\frac{1}{R}) = \frac{1}{R^2}\frac{\partial}{\partial R}\left(R^2\frac{\partial(\frac{1}{r})}{\partial R}\right) = \frac{1}{R^2}\frac{\partial}{\partial R}(-1) = 0

当R包含原点时，对∇2(1R)\pmb\nabla^2(\frac{1}{R})进行体积分

∫V∇2(1R)dV=∫Va∇2(1R)dV+∫Vb∇2(1R)dV=∫Va∇2(1R)dV\int_{V}{\nabla}^2 (\frac{1}{R})dV=\int_{V_{a} }{\nabla}^2 (\frac{1}{R})dV+\int_{V_{b} }{\nabla}^2 (\frac{1}{R})dV=\int_{V_{a} }{\nabla}^2 (\frac{1}{R})dV

其中VbV_b不包含原点，VaV_a是以原点为秋心，半径为rr的球体，利用散度定理可知

∫Va∇2(1r)dV=∫Va∇⋅∇(1r)dV=∫S∇(1r)⋅ndS\int_{V_{a} }{\nabla}^2 (\frac{1}{r})dV=\int_{V_{a} }{\nabla}\cdot{\nabla}(\frac{1}{r})dV=\int_{S}{\nabla} (\frac{1}{r})\cdot \pmb n dS

其中n\pmb n为球体表面外法线方向的单位向量，注意其与半径方向一致，则

∫S∇(1R)⋅ndS=∫02π∫0π∇(1R)⋅nR2sin⁡θ dθ dφ=∫02π∫0π−(1R2)n⋅nR2sin⁡θ dθ dφ=−∫0π2πsin⁡θ dθ =−4π\begin{aligned} \int_{S}{\nabla}(\frac{1}{R})\cdot\pmb n dS &= \int_{0}^{2\pi}\int_{0}^{\pi}{\nabla} (\frac{1}{R})\cdot \pmb n R^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi \\\\ &= \int_{0}^{2\pi}\int_{0}^{\pi}-(\frac{1}{R^{2} })\pmb n \cdot \pmb n R^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d}\varphi \\\\&=-\int_{0}^{\pi}{2\pi}\sin \theta \,\mathrm {d} \theta \, \\\\&= -4\pi \end{aligned}

因此可得

∇2(1R)=−4πδ(R) {\nabla}^2 (\frac{1}{R})=-4\pi\delta(\pmb R)

To be continued⋯\cdots