关于电场的一些复变分析和拓扑思想

单位圆 $\to$ 线，$z$ 平面投影 $w$ 平面

拓扑学 (1) 平移变换 $w = z + a$

(2) 线性变换 $w = az + b$

$b$ 为平移，

$|a|$ 为平移 $e^{i\arg a}$为旋转

(3) 反演变换 $w = \frac{R^2}{z}$

(4) 分式变换 $w = \frac{az + b}{cz + d}$

单位圆 $\to$ 线

$w = i \frac{1 - z}{1 + z}$

(5) 幂数变换 $w = z^n$

$z = re^{i\theta}$, $w = \rho e^{i\varphi}$

$\rho = r^n$, $\varphi = n\theta$

(6) 对数变换 $w = e^z$

令 $z = x + iy$, $w = \rho e^{i\varphi}$

$\rho = e^x$

(7) 三角函数变换

$z = C \sin t$

$z = C \sin t = x + iy = C \sin(\xi + i\eta)$

$= C(\sin\xi \cosh\eta + i\cos\xi \cdot \sinh\eta)$

$\frac{x^2}{C^2 \cos^2\eta} + \frac{y^2}{C^2 \sin^2\eta} = 1$ （椭圆）

$\frac{x^2}{C^2 \sinh^2\xi} - \frac{y^2}{C^2 \cosh^2\xi} = 1$ （双曲线）

1. 平面静电场电势分布 $y(x,y)$, $\partial^2 y = 0$ ($\mathrm{Im} z \geq 0$) $\boxed{z(x,y) \Rightarrow w(u,v)}$

$y(x,0) =V_1 (x < 0) V_2 (x > 0)$

$r$ 由 $0$ 取到 $\infty$, $w = u + iv$, $z = x + iy$, $w = \ln z = \ln(x+iy)$

$T_1 = r e^{-i\cdot 0}$, $T_2 = r e^{i\pi}$

$w = \ln z$

$w_1 = \ln r_1 = \ln r$

$w_2 = \ln r_2 = \ln r + i\pi$

$u = \arg z$

$y(x,y) = \frac{V_1 - V_2}{\pi} \arg z = \frac{V_1 - V_2}{\pi} \arctan \frac{y}{x}$

例2. 圆柱面满足 $V = $

$V_1 (0 < \varphi < π) $

$V_2 (π < \varphi < 2π)$

$z_1 = \frac{z}{a}$

$z_2 = i \frac{1 - z_1}{1 + z_1}$

$z_3 = \ln z_2 = \ln i \frac{a - z}{a + z}$

$\varphi = V_1 + \frac{V_2 - V_1}{\pi} \cdot \mathrm{Im}\left(\ln i \frac{a - z}{a + z}\right)$, $\mathrm{Im}$ 只取主值

$z = x + iy$

$\varphi = V_1 + \frac{V_2 - V_1}{\pi} \left[\ln i(a - x - iy) - \ln(a + x + iy)\right]$

$= V_1 + \frac{V_2 - V_1}{\pi} \left(\ln[y + i(a - x)] - \ln[(a + x) + iy]\right)$

$= V_1 + \frac{V_2 - V_1}{\pi} \left[\arctan \frac{a - x}{y} - \arctan \frac{y}{x + a}\right]$

$\mathrm{Im}[\ln(x + iy)] = \mathrm{Im}[\ln z] = \mathrm{Im}[\ln|z| + i\arg z] = \mathrm{Im}[i\arg z] = \arctan \frac{y}{x}$

物理 关于电场的一些复变分析和拓扑思想