|
|
Post by mondo on Sept 21, 2023 6:36:48 GMT
On page 551 Green function is defined as $G_p(z) = - \Phi(z)$ I understand the isotherms are the real part of $\Omega$ but why the minus sign?
The $log(z)$ function is defined as $|ln(z)| + iarg(z)$ so the real part is positive.
|
|
|
|
Post by mondo on Sept 21, 2023 7:25:23 GMT
Is it because we rely on the example from previous section where we dealt with a vortex of a strength $-2\pi$ while here we need a source of strength $2\pi$?
|
|
|
|
Post by Admin on Sept 21, 2023 9:45:58 GMT
On page 551 Green function is defined as $G_p(z) = - \Phi(z)$ I understand the isotherms are the real part of $\Omega$ but why the minus sign? The $log(z)$ function is defined as $|ln(z)| + iarg(z)$ so the real part is positive. Mondo It's because it's the dual flow that we get the minus sign. See page 509. By the way, it's not correct that $\log(z)=|\ln(z)| + i\arg(z)$. See page 98. Vasco |
|
|
|
Post by mondo on Sept 21, 2023 20:23:25 GMT
You are right, thanks. So the steps on p.551, to get $G_p$ are: we take (19), dividing it by $i$ due to the fact that (19) was calculated for a vortex and we need a source, then we calculate it's dual. Here I wonder, why do we bother with a dual and not take the equipotentials of this source? This dual opens a room for a mistake - should I take $\Psi$, $\Phi$, $-\Phi$, ...? From what I see if we don't do dual and take equipotential of the source then the difference would be only the sign of $\Phi$, does it matter for isotherm? On figure [31] we see only isotherm $G$, how would $-G$ look? As far as I see there is no direction for equipotential.
|
|
|
|
Post by mondo on Sept 24, 2023 5:41:39 GMT
Vasco, what do you think about it?
|
|
|
|
Post by Admin on Sept 26, 2023 18:50:14 GMT
Vasco, what do you think about it? Mondo Neither $\Phi$ nor $\Psi$ have a direction which can be deduced from their sign. For example the arrows drawn on streamlines are not deduced from the sign of $\Psi$. Vasco |
|
