03-20
点源（续）¶
标准Poisson方程
\[\nabla^2\phi(\vec{x}) + \frac{\rho(\vec{x})}{\varepsilon_0} = 0\]
-
\(\rho(\vec{x})\)：如何表示阶跃的点源？
- 用 \(\delta\) 函数！
-
一维：
\[\rho(x) = Q\delta(x)\]
- 三维：
\[\rho(\vec{x}) = \iiint Q \, \underset{\delta^{(3)}(\vec{x})}{\underbrace{\delta(x_1) \delta(x_2) \delta(x_3)}}\]
- 一般的曲线坐标系
\[\int fff d^3t \frac{Q}{fff} \delta(t_1) \delta(t_2) \delta(t_3)\]
如何表示电偶极子？
\[\phi = \frac{Q}{4\pi\varepsilon_0} \underset{\text{很重要，以生成函数出现，也称Green函数}}{\boxed{\frac{1}{|\vec{x}|}}}\]
\[\nabla^2\phi = -\frac{P_i}{4\pi\varepsilon_0} \partial_i \frac{1}{|\vec{x}|}\]
\[\nabla^2\phi - \frac{P_i \boxed{\partial_i \delta^{(3)}(\vec{x})}}{\varepsilon_0} = 0\]
\[\rho(\vec{x}) = \int \rho(\vec{x}') \delta^{(3)}(\vec{x} - \vec{x}') dV'\]
考虑点源对电势的贡献
\[\nabla^2\phi(\vec{x}) = \int \frac{\rho(\vec{x}') dV}{\varepsilon_0} \underset{\text{Green}}{\underbrace{G(\vec{x}, \vec{x}')}}\]
\[\nabla^2 G(\vec{x}, \vec{x}') + \delta^{(3)}(\vec{x} - \vec{x}') = 0\]
注意 \(\nabla^2\) 是关于 \(\vec{x}\) 的
平移不变性 \(G(\vec{x}, \vec{x}') = G(\vec{x} - \vec{x}')\)
\[\nabla^2 G(\vec{x} - \vec{x}') + \delta^{(3)}(\vec{x} - \vec{x}') = 0\]
做 Fourier 变换
\[G(\vec{x} - \vec{x}') = \int \frac{d^3\vec{p}}{(2\pi)^3} e^{i\vec{p} \cdot (\vec{x} - \vec{x}')} \tilde{G}(\vec{p})\]
\[\delta^{(3)}(\vec{x} - \vec{x}') = \int \frac{d^3\vec{p}}{(2\pi)^3} e^{i\vec{p} \cdot (\vec{x} - \vec{x}')}\]
得到
\[-|\vec{p}|^2 \tilde{G}(\vec{p}) + 1 = 0\]
即
\[\tilde{G}(\vec{p}) = \frac{1}{|\vec{p}|^2}\]
\[G(\vec{x} - \vec{x}') = \int \frac{d^3\vec{p}}{(2\pi)^3} \frac{e^{i\vec{p} \cdot (\vec{x} - \vec{x}')}}{|\vec{p}|^2}\]
可以推出 \(G(\vec{x}, \vec{x}') \propto \frac{1}{|\vec{x} - \vec{x}'|}\)
带边界区域中静电场的求解¶
\[ \begin{cases} \nabla^2\phi + \frac{\rho}{\varepsilon_0} = 0 & \text{内部} \\ ？ & \text{边界} \end{cases} \]
假如能找到一个坐标系，使得边界是坐标面，问题会变得简单
\[\phi = \phi_0 + \hat{\phi}\]
- \(\phi_0\)：通解
- \(\hat{\phi}\)：特解
Green 第一恒等式¶
\[\int_{\partial V} \vec{v} \cdot d\vec{S} = \int_V \nabla \cdot \vec{v} dV\]
其中 \(\vec{v} = \phi \nabla \psi\)，且 \(\phi, \psi\) 为标量函数。代入得
\[\int_{\partial V} \phi \left(\nabla \psi \cdot d\vec{S}\right) = \int_V \left(\nabla \phi \cdot \nabla \psi + \phi \nabla^2 \psi\right) dV\]
引入 \(\phi_1, \psi_2\)，\(\nabla \phi_1 - \nabla \phi_2 = 0\)。令 \(u = \phi_1 - \phi_2\)
\[\int_V |\nabla u|^2 dV = -\int_{\partial V} \boxed{\underset{\text{Dirichlet}}{\underbrace{(\phi_1 - \phi_2)}} \underset{\text{Neumann}}{\underbrace{\left(\vec{E}_1 \cdot d\vec{S} - \vec{E}_2 \cdot d\vec{S}\right)}}}\]
Dirichlet 条件¶
- 理想导体（\(\vec{E} = 0\)）
- 等势体 \(\phi = C\)
- 只有 \(\vec{E}_{\perp}\)，\(\vec{E}_{\parallel} = 0\)
- \(\left|\vec{E}_{\perp}\right| \cancel{\left|d\vec{S}\right|} = \frac{\cancel{\left|d\vec{S}\right|} \sigma}{\varepsilon}\)
Neumann 条件¶
- 两个电介质，界面处有自由电荷
- 电势连续
- \(\left.\nabla \phi_{\text{I}}\right|_{\vec{n}} - \left.\nabla \phi_{\text{II}}\right|_{\vec{n}}= \sigma/\varepsilon_0\)