Flows round obstacles

Mondo

But $z$ is not the real part of $\Omega$ and $1/z$ is not the imaginary part either

Vasco

Mondo

$\Psi$ is the imaginary part of $\Omega$, and $\Phi$ is the real part of $\Omega$. [22] is a plot of the level curves of $\Phi$ and $\Psi$ in the complex plane.

It is the same kind of plot as the LHS of [21] on page 501. The RHS of [21] is a plot of these level curves under the mapping $\Omega=\Phi+i\Psi$ to the $\Phi\Psi$-plane.

So as we travel along any of the level curves on the LHS of [21], then on the RHS we travel along either a vertical or horizontal line.

Vasco.

Mondo

I have edited my reply #19 again.

Vasco

Thank you Vasco, your latest posts really helped.
I calculated vector field from $\Omega$ and put it into the Equation explorer. It ain't pretty but you can get a feel of the flow and it somehow resembled figure [22]:


Here is the formula I used (1- (x^2-y^2)/(((x^2-y^2)^2)+(2xy)^2))i - ((2xy)/(((x^2-y^2)^2)+(2xy)^2))j

Then I used $\Omega$ directly to plot streamlines. $\Psi = \frac{yx^2 + y^3 -y}{x^2+y^2}$, then I opened one of desmos links I bookmarked to type it in but I see you already did it under www.desmos.com/calculator/yqquzxuk8r. I just had to enable that plot

As to my question #3 from post #14 - It is fascinating to me how well $\Psi$ models this flow around an obstacle.

Now I have these questions:
1. Author calls points $\pm 1$ a breakdown of the grid (p.528 right below the equation for $\Omega$) why? I still don't see why these points are stagnation points - the streamline continues there in a circular fashion.
2. I am confused by the fact that a streamline at $0$, lying on a real axis, at a point $-1$ becomes a circle - so now it's value is no longer $0$? How to understand it? Going from the left there was no flux until point $-1$ and all of the sudden at this point this $0$ streamline of flux becomes a circle with non $0$ value?
3. Below figure [22] on page 503 author also says that the streamlines emanating from the dipole are deformed out of perfect circularity. But then, why is not the unit circle deformed at all? The deformation is caused by the uniform flow so the unit circle, outer most streamline of the dipole, should be the first one affected by the uniform flow.
4. I also wonder how to interpret figure [22]. From page 495 we know that there is zero flux before points lying on the same streamline. So for instance there should be no flux between points $2 + 4i$ and $100+4i$ - both lie on the same horizontal line where flux is constant.
5. Continuing with the potential function interpretation, what kind of information can we get from the potential function (orange color in desmos) $\phi = -2$ - we see it's graph is no longer a straight vertical line but a curve shifting towards $-1$ and then it becomes almost horizontal, finally crossing the origin and winding back to $-1$. How to understand this behavior?


Thank you.

Mondo

1. We are plotting different functions on the complex plane.
$\Omega$ does not represent the physical motion in the plane. The motion is represented by the vector field. It is this function that has stagnation points at $z=\pm1$ not the complex potential function.

Vasco

Mondo

They don't look identical to me. The vector field is tangent to $\Psi$, otherwise they are very different. The arrows have a different meaning in each case, and look at the size and direction of the vectors along the real axis near the stagnation points. They are not identical at all.

Flux at a point? What does that mean? What do you mean by graph of $\Omega$?

Vasco