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Two vortices, of strengths $\Gamma_1$ and $\Gamma_2$, are at the points $z = z_1$ and $z = z_2$ respectively in the complex plane, where they are free to move.

So I have the complex potential

$$w(z)=\frac{-i\Gamma_1}{2\pi}\operatorname{log}(z-z_1)+\frac{-i\Gamma_2}{2\pi}\operatorname{log}(z-z_2)$$

Therefore I now consider for $z_1(t)$ and $z_2(t)$. Where here we are now letting the locations of the vortices change depending on the flow around them. To simulate this we ignore the flow caused by the vortex itself and find this differential equation.

As $w'=u-iv$ where $u$ and $v$ are the $x$ and $y$ parts of the velocity of the flow. And $\dot{z_i}=u+iv|_{z_i}$

$$\dot{z_i}=\operatorname{lim}_{z\rightarrow z_i}(w'(z)+\frac{i\Gamma_1}{2\pi(z-z_1)})$$

By computing both these and equating this leads me to conclude $\frac{\dot{z_1}}{\Gamma_2}=-\frac{\dot{z_2}}{\Gamma_1}$

Also I have shown that

$$a:=|z_1-z_2|$$ and $$Z:=\frac{\Gamma_1 z_1+\Gamma_2 z_2}{\Gamma_1+\Gamma_2}$$

Are constant. So this implies that the vortices rotate around each other. However i'm very much struggling to show that the centre is Z and how to find the angular velocity of this system? If anyone could help me i'd be very grateful.

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I wonder whether this might be better placed at physics.stackexchange.com. –  joriki Feb 20 '12 at 2:12
    
@joriki: I see where you are coming from, but this is very much a mathematical question. The physics forum seem to not like such specific maths based questions sometimes -which is a shame! –  Freeman Feb 20 '12 at 14:05
    
Then you may want to explain some of this to the mathematical laypeople :-). I studied theoretical physics, and I understand none of your "so"s and "therefore"s. Also there seem to be some typos -- presumably the lowercase and uppercase gammas are supposed to refer to the same variable? And presumably you mean $\frac{\dot{z_1}}{\Gamma_1}=\frac{\dot{z_2}}{\Gamma_2}$? –  joriki Feb 20 '12 at 14:17
    
@joriki: Whoops, sorry about the typoes.. I wrote this in notepad when the site was down. I'm sorry about the vagueness, I will attempt to correct it. –  Freeman Feb 20 '12 at 14:29
    
You've now introduced a minus sign in $\frac{\dot{z_1}}{\Gamma_2}=-\frac{\dot{z_2}}{\Gamma_2}$, but you still have $\Gamma_2$ in both denominators. Are you sure you don't want $\frac{\dot{z_1}}{\Gamma_1}=-\frac{\dot{z_2}}{\Gamma_2}$? –  joriki Feb 20 '12 at 14:31

1 Answer 1

up vote 3 down vote accepted

The ODE you got looks like $$\dot z_1=\frac{i\Gamma_2}{2\pi (z_2-z_1)},\quad \dot z_2=\frac{i\Gamma_2}{2\pi (z_1-z_2)}$$ Having observed that $|z_1-z_2|=a$ is constant and the weighted average $Z =\frac{\Gamma_1 z_1+\Gamma_2 z_2}{\Gamma_1+\Gamma_2}$ does not move either, you are almost done. What remains to observe:

  1. Since $\dot z_k$ is orthogonal to $z_k-Z$ for $k=1,2$, the distances $|z_k-Z|$ are constant in time. Thus, each $z_k$ moves in a circle around $Z$. The radius is easily found using proportions: $|z_1-Z|=\dfrac{\Gamma_2}{\Gamma_1+\Gamma_2}$ and $|z_2-Z|=\dfrac{\Gamma_1}{\Gamma_1+\Gamma_2}$.
  2. The angular velocity is $|\dot z_k|/|z_k-Z|$. On geometric grounds, it should be the same for both vortices, and the algebra confirms it: $\dfrac{\Gamma_1+\Gamma_2}{2\pi a}$.
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