※ 引述《cheesesteak (Terry)》之銘言：
: 令一向量場 F=(-y/(x^2+y^2))i+(x/(x^2+y^2))j+0 k for (x,y,z)≠(0,0,0)
: 可以找到 f(x,y,z)=-arctan(x/y)+C 使得▽f=F
: 令曲線c: r(t)=(0.5cost)i+(0.5sint)j+0 k 0≦t≦2π
: 2π
: ∮ F‧dr=∫ [(-0.5sint/0.25)*(-0.5sint)+(0.5cost/0.25)(0.5cost)] dt
: c 0
: = 2π≠0
: 我的問題是：已經找到函數f使得▽f=F，為何∮ F‧dr會不等於0?
: F是不是保守力場?
Let me do it using Residue Theorem
∮_c F‧dr = ∮_c (-y/r^2, x/r^2)‧(dx, dy)
= ∮_c (-y dx +x dy)/r^2
Denote z=x+iy
r^2=zz'
y=-i(z-z')/2, x=(z+z')/2
-y dx +x dy=i(z-z')/2*(dz+dz')/2-i(z+z')/2*(dz-dz')/2
= (i/4)[(z-z')(dz+dz')-(z+z')(dz-dz')]
= (i/2)(z dz' -z' dz)
∮_c F‧dr = (i/2)∮_c (dz'/z' -dz/z)
By Residue Theorem
If c includes 0, ∮_c F‧dr = (i/2) (-2πi-2πi) = 2π
If c does NOT include 0, ∮_c F‧dr=0
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 112.104.98.58