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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 $u_\theta=\frac{\Gamma}{2r \pi}$

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Are you completely sure $u_r = 0$ ? – Yrogirg Apr 13 '12 at 16:18
@Yrogirg: Yes, this is the complex potential for a vortex, probably should have mentioned that. – Freeman Apr 13 '12 at 16:20
up vote 1 down vote accepted

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)

   (16)  ──────
         2%pi r
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btw, one could swap Re and $\partial$, that is first differentiate, then take the real part. – Yrogirg Apr 14 '12 at 4:58
Thank you, but could you explain in a little more detail how you found those expressions for $v_r,v_{\theta}$ and how they are implied from $v=\nabla \phi$ as I am unsure, thanks so much! – Freeman Apr 15 '12 at 12:26
Quick explanation --- That's just an expression for gradient in polar coordinates. Personally, I usually consult tables to look up for $\nabla$ or $\Delta$ in various coordinates. See for example… the part about cylindrical coordinates. Polar coordinates are just cylindrical without $z$. Check also… – Yrogirg Apr 15 '12 at 14:59
Actually, If you thought $\text{grad} \varphi \;\;$ to be $\left( \frac{\partial \varphi}{\partial r}, \frac{\partial \varphi}{\partial \theta} \right) \;\;\;$ then it doesn't make sence due to inconsistent units. $\frac{\partial}{\partial r} \;\;$ divides by meters, while $\frac{\partial}{\partial \theta} \;\;$ is dimensionless. – Yrogirg Apr 15 '12 at 15:13
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! – Freeman Apr 15 '12 at 15:33

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