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Post by Admin on Sept 15, 2023 18:45:11 GMT
Mondo
The expression for $\tilde{\Omega}(w)$ contains the velocity, but that does not mean that the complex potential has a velocity but just that it is a function of the velocity.
Just as the formula for kinetic energy, $\frac{1}{2}mv^2$, is a function of the velocity but the kinetic energy itself does not have a velocity.
Vasco
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Post by mondo on Sept 16, 2023 5:04:05 GMT
And also why at the bottom of page 542 autjor say $[\phi] = 4v$? This is a mystery to me. Mondo It's all explained on page 541 in subsection 2. Vasco I read it several times and I don't find an answer as to why The potential difference between points $\tilde{a}$ and $\tilde{b}$ is four time the speed ($4v$). Similarly, I don't understand why underneath Figure [25] p.542 author says "The geometric signature of the critical point $a$ is the angle $\pi/4$". I remember in some other post you referenced me back to figure [22] p.503, while the angle between a streamline and equipotential look like something around $45$ degrees it is not obvious. How is author so sure about it? |
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Post by mondo on Sept 16, 2023 5:28:52 GMT
Mondo The expression for $\tilde{\Omega}(w)$ contains the velocity, but that does not mean that the complex potential has a velocity but just that it is a function of the velocity. Just as the formula for kinetic energy, $\frac{1}{2}mv^2$, is a function of the velocity but the kinetic energy itself does not have a velocity. Vasco Good analogy, thanks. I think I got the idea but we have to be careful with the speed here - let's take equation (18) p.539 as an example here, the leading $1/2$ is not the speed right? $3/2$ seems to be the speed but only far from origin, clos to the origin the speed is likely closer to $1/2$ actually and circular. |
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Post by mondo on Sept 16, 2023 22:39:51 GMT
I think I may have found an answer to my question regarding to "geometric signature of a critical point is $\pi/4$" - I think it is so because at a critical points angles are doubled by a complex potential. This is explained on page 530 with the figure [15]. Hence if the angle between a streamline and an equipotential at a point $\pm 1$ is $\pi/2$ on $\Omega$ then it must be $\pi/4$ in reality with vector field.
For this page (p.530) in the first bullet point author says that the $\Psi$ and $\Phi$ is $0$ for paths emanating from $p$ - is that because they both terminate on the unit disk and this region is mapped by $\Omega$ to $0$?
However the formula $[\phi] = 4v$ is still a bit of a mystery to me. Likewise on p.545 in (ii) author says that the strength $d$ of a dipole inserted into $C$ needs to be $[\psi/4]$ in order to have equal number of k-cells in both flows. This a magic number to me, I can't relate it to anything I read before.
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Post by Admin on Sept 17, 2023 13:59:22 GMT
Mondo
I see it in a way that to me seems much simpler than that:
Since the same streamline runs along the real axis and then round the obstacle (circle), the angle at $x=1$ between the segment of the streamline on the real axis and the tangent to the streamline on the circle, is $\pi/2$, then in the limit as we get close to the point $x=1$, the equipotential gets closer and closer to the line through $x=1$ that makes an angle of $\pi/4$ with the real axis.
If for example we had an obstacle that had a surface with an angle of say $3\pi/4$ like this \__, then this would have to be a streamline, and streamlines close by would be of a similar shape. Equipotentials in the vicinity of the corner would be at right angles to these streamlines and in the limit would be at an angle of $\frac{1}{2}(3\pi/4)$.
This also makes sense because if there were no corner the angle would be $2\pi$ and the equipotential would be at $\frac{1}{2}(2\pi)=\pi/2$, or at right angles to the streamline as it should be.
Or do you think this is the same argument as yours explained in a different way?
Vasco
PS What do you mean by "... on $\Omega$..."?
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Post by Admin on Sept 17, 2023 14:52:29 GMT
Mondo
This is because if we choose to measure the circulation and flux from a point $p$ then the integrals for $\Phi(p)$ and $\Psi(p)$ are both zero at $p$: $\int_{p}^{p}ds=0$. See figures [17] and [20] in chapter 11.
Vasco
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Post by Admin on Sept 17, 2023 18:57:20 GMT
Mondo
There was a typo in my reply #34 which I have now corrected. It is now in red.
Vasco
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Post by mondo on Sept 18, 2023 1:51:52 GMT
PS What do you mean by "... on $\Omega$..."? I meant that on the graph of complex potential $\Omega$ the angle between a streamline and an equipotential is always $\pi/2$. This was the case for the mapping on a figure [21] p.501 and even more importantly, on figure [15] page 530. In fact base on the later one, I concluded that since $\Omega$ doubles angles on a critical point, and since at this point the angle between a streamline and an equipotential is $\pi/2$ then on the regular flow it must be $\pi/4$. However now I am confused again, because author starts page 530 with a comment: "The top part of is essentially a copy of figure [12]" but figure [12] is a carbon copy of figure [22] p.503, and even more importantly of our plots in desmos which we thought are plots of $\Omega$! While on p.530 author claims that $\Omega$ is shown at the bottom of figure [15] which looks nothing like what we established earlier. Does it make sense to you? I see it in a way that to me seems much simpler than that: Since the same streamline runs along the real axis and then round the obstacle (circle), the angle at $x=1$ between the segment of the streamline on the real axis and the tangent to the streamline on the circle, is $\pi/2$, then in the limit as we get close to the point $x=1$, the equipotential gets closer and closer to the line through $x=1$ that makes an angle of $\pi/4$ with the real axis. Hmmm, I don't get that - in the limit as we move the tangent to the circle closer and closer to point $\pm 1$ it must become $\pi/2$ (a vertical line). A circle with it's canter at a real axis must cut real axis at a right angle. But beside that, the angle that we discuss is between a streamline and an equipotential at that point $\pm 1$ and I can't correlate your example with it. Mondo This is because if we choose to measure the circulation and flux from a point $p$ then the integrals for $\Phi(p)$ and $\Psi(p)$ are both zero at $p$: $\int_{p}^{p}ds=0$. See figures [17] and [20] in chapter 11. Vasco Right, but this is so obvious that I don't know why author decided to make a bullet point about it. Both work and flux are a measure over a distance so at a point they have to be $0$ Thanks |
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Post by Admin on Sept 18, 2023 6:35:48 GMT
Mondo
Just to reply to your last point. It is not correct to say that at a point work and flux are zero. At other points they are generally non-zero.
When we integrate from point $a$ to point $p$ this is defined to be the value of the work or flux at the point $p$.
Vasco
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Post by Admin on Sept 18, 2023 6:50:13 GMT
Mondo
The plots in [22] on page 503 and [12] on page 528 and the top half of [15] on page 530 are not plots of $\Omega$, they are plots of the level curves of $\Phi$ and $\Psi$.
The RHS of figure 21 on page 501 and figure 15 at the bottom of page 530 are plots of $\Omega$.
Vasco
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Post by Admin on Sept 18, 2023 7:13:13 GMT
Mondo
So when you wrote "on $\Omega$" you meant on the level curves of $\Phi$ and $\Psi$.
Vasco
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Post by Admin on Sept 18, 2023 7:37:47 GMT
Mondo
Even figures like the RHS of figure 21 on page 501 are not really "plots of $\Omega$". It is not possible to plot complex functions in the same way as we do with real functions.
The best way to describe the RHS of figure 21 is as Needham does at the bottom of page 501: "the image of the mapping $\Omega$ of the special phase portrait on the LHS of figure 21"
Vasco
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Post by mondo on Sept 18, 2023 8:41:01 GMT
Mondo Just to reply to your last point. It is not correct to say that at a point work and flux are zero. At other points they are generally non-zero. When we integrate from point $a$ to point $p$ this is defined to be the value of the work or flux at the point $p$. Vasco Yes but asking a question "what is the flux/work at a point $z$" without a broader context doesn't make a sense to me. You said when we integrate from $a$ to $p$ then work/flux at $p$is the result but point $a$ may be everywhere while the value of flux/work at $p$ is constant. So this question only makes sense when we know the starting point then we can count number of streamlines or equipotentials from one point to the other and therefore find out the integral on that distance. Mondo The plots in [22] on page 503 and [12] on page 528 and the top half of [15] on page 530 are not plots of $\Omega$, they are plots of the level curves of $\Phi$ and $\Psi$.The RHS of figure 21 on page 501 and figure 15 at the bottom of page 530 are plots of $\Omega$. Vasco Aha! We plot $\Psi$ and $\Phi$ from $\Omega$ taking real and imaginary part of it and this way we obtained [22] p503 or our desmos plots. So I thought this is a plot of $\Omega$, but looks like a real $\Omega$ plot is a different, custom type of mapping where we take a streamline/equipotential and map it to a horizontal/vertical line respectively. Confusing, but it does't make sense to me now. As for my question regarding last sentence on page 542 - $[\Phi] = 4v$ I think we need to use the first equation from p.540 for $\Omega$ and calculate potential function and then calculate it's value for $\pm 1$ and then a difference between these two should give this value. |
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Post by Admin on Sept 18, 2023 11:46:44 GMT
Mondo
I agree with you about the last sentence on page 542 and it does give the answer $[\widetilde{\Phi}]=4v$. Notice that it is $\widetilde{\Phi}$ not $\Phi$.
Vasco
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Post by mondo on Sept 19, 2023 4:46:50 GMT
Thanks for confirming.
Vasco, do you know what author is talking about in the last paragraph before subsection $4$ starts on page 546? He says "there was nothing wrong with choosing to place one of the dipoles (the one outside $D$ ) at $\infty$" I don't remember this was mentioned anywhere in the preceding sections. Do you know what is he talking about?
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