2
$\begingroup$

If a fluid has the complex potential $w(z)=\frac{-\Gamma i}{2 \pi}\operatorname{log}z$ Can anyone show me how to find it's radial and transverse velocity components in polar coordinates?

They are meant to be $u_r=0$ and $u_\theta=\frac{\Gamma}{2r \pi}$

  • 0
    @Yrogirg: Yes, this is the complex potential for a vortex, probably should have mentioned that.2012-04-13

1 Answers 1

1

Oh, I've forgot, what is the complex potenial. So,

$w = \varphi + i \psi$

where $\varphi$ is a potential and $\psi$ is a stream function.

Thus, $\boldsymbol v = \text{grad} \varphi \;$:

$v_r = \frac{\partial \varphi}{\partial r} = \text{Re} \left[ \frac{\partial \, w(r e^{i \varphi})}{\partial r} \right] $

$v_{\theta} = \frac{1}{r} \frac{\partial \varphi}{\partial \theta} = \text{Re} \left[ \frac{1}{r} \frac{\partial \, w(r e^{i \varphi})}{\partial \theta} \right] $

You could use stream function $\psi$ instead of potential though.

I've used FriCAS to evaluate things to the answer:

(15) -> D(real(-G*%i/(2*%pi)*log(r*exp(%i*phi))),r)     (15)  0  (16) -> D(1/r * real(-G*%i/(2*%pi)*log(r*exp(%i*phi))),phi)              G    (16)  ──────          2%pi r 
  • 0
    I was using $\nabla \phi = (\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$ But I see now.. I proved these last year.. I guess i'll just have to remember them! Thank you so much for your help!2012-04-15