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Post by mondo on Aug 7, 2022 20:30:44 GMT
Hi,
I have some questions to the construction steps of a harmonic dual on page 512:
1. "To make $\Psi$ unique, let us demand that it vanish at some point $a$" - how is the vanishing at some point $a$ making it unique?
2. $\Psi(p) = \int_K (\nabla \phi) \cdot N ds$ We know that $\nabla \Phi = \bar H$, so when we take a normal of it the integral will give us $\Psi$. Just as the first eq at page 497. However I wonder how can we write $N$ vector so it always gives us a normal component of $\bar H$ to the curve $K$?
3. In the equation below author gives an alternative equation for $\Psi$ in terms of complex integration. I can't see how is $\bar{\nabla \Phi} = \nabla \Phi \cdot N$? Since $\nabla \Phi = \begin{pmatrix} \delta_x \Phi\\ \delta_y \Psi \end{pmatrix}$ and $\bar{\nabla \Psi} = \begin{pmatrix} \delta_x \Psi\\ - \delta_y \Psi \end{pmatrix}$ this suggests $N = [1, -1]$ but how was it calculated?.
4. In the following example there is $N = \frac{1}{\sqrt{x^2 + y^2}}[Y,-X]$. I understand the division is to get a unit normal vector by dividing by its length but two things:
a. The normal vector will have different coordinates than $X$ and $Y$ so why do we divide it by the length of the original vector?
b. This is connected to question 3, why in the equation for N we have $[Y, -X]$, How are we sure this will give us a normal vector to $[X,Y]$
5. Finally the integral, $\nabla \Psi \cdot N = \frac{9X^2Y - 3Y^3}{\sqrt{X^2 + Y^2}}$ hence $\int_K (\frac{9X^2Y + 3Y^3}{\sqrt{X^2 + Y^2}})t^2 dt =(3X^2Y - Y^3)t^2 $ while in the book it is just $(3X^2Y - Y^3)$. Why was $t^2$ dropped?
UPDATE: Ok, for the last question I just realized $t$ is defined from $[0-1]$ and thats why it vanish.
Thank you.
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Post by Admin on Aug 8, 2022 9:12:46 GMT
Mondo
1. This is all connected with the idea explained in the subsection 2 The Stream Function on pages 494-495. See also paragraph two on page 511.
Also it is a mathematical fact that when we integrate a function we always obtain an arbitrary constant which we must somehow define if we want a unique solution.
Example: if $dy/dx=2x$ then integrating we have $y=x^2+c$ where $c$ is an arbitrary constant. We have an infinity of solutions. One way to limit this to a unique solution is to demand that $y=3$ when $x=0$ which means that c=3 and our unique solution is $y=x^2+3$
2. Obviously the vector $N$ normal to $K$ depends on $K$. In this case we choose $K$ to be a straight line between the starting point and the end point. We can choose any curve connecting the two points because the result of the integral does not depend on $K$ but just on the start point and end point. In this particular example a straight line makes it easy to calculate $N$ since it is a vector of length 1 which is always perpendicular to the line $K$ and so $N$ is a constant vector on $K$.
3. Remember (11) on page 483? Take another look. $N$ is not involved in this second approach. $\overline{\nabla\Phi}$ is integrated with respect to $z$ here.
4. As I explained in 2 above, since $K$ is a straight line then it is easy to calculate $N$ as the constant vector perpendicular to $K$.
5. OK. To be precise it only vanishes at the starting point of $K$ and it is equal to 1 at the end point.
Vasco
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Post by mondo on Aug 9, 2022 6:35:56 GMT
1. This is all connected with the idea explained in the subsection 2 The Stream Function on pages 494-495. See also paragraph two on page 511. Also it is a mathematical fact that when we integrate a function we always obtain an arbitrary constant which we must somehow define if we want a unique solution. Example: if $dy/dx=2x$ then integrating we have $y=x^2+c$ where $c$ is an arbitrary constant. We have an infinity of solutions. One way to limit this to a unique solution is to demand that $y=3$ when $x=0$ which means that c=3 and our unique solution is $y=x^2+3$ Ok, makes sense. Thank you. 2. Obviously the vector $N$ normal to $K$ depends on $K$. In this case we choose $K$ to be a straight line between the starting point and the end point. We can choose any curve connecting the two points because the result of the integral does not depend on $K$ but just on the start point and end point. In this particular example a straight line makes it easy to calculate $N$ since it is a vector of length 1 which is always perpendicular to the line $K$ and so $N$ is a constant vector on $K$. I was rather thinking how do we guarantee the perpendicular direction. But I think the example below illustrates exactly this by showing a formula for $N$. Also, you said we can choose any curve because the result of the integral depends on two points only, namely start and end of K. That's true but in the case of complicated $K$ we need to come up with probably equally complicated $N$ to make our integrand a perpendicular component, right? 3. Remember (11) on page 483? Take another look. $N$ is not involved in this second approach. $\overline{\nabla\Phi}$ is integrated with respect to $z$ here. So $\bar{\nabla \Phi}$ is to get $H$ instead of $\bar H$ right? And then we can calculate the integral, take imaginary part of it ($F$) end hence find average value of $\Psi$. 4. As I explained in 2 above, since $K$ is a straight line then it is easy to calculate $N$ as the constant vector perpendicular to $K$. Ok so we make use of i.e a dot product that tells us $X_1X_2 + Y_1Y_2 = 0$ for perpendicular vectors. Vasco |
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Post by Admin on Aug 9, 2022 10:51:07 GMT
Mondo
2. No. You seem to have misunderstood the point. Because we can choose $K$ then if we want to calculate $\Psi$ at a point then we can choose any $K$, so we choose one that makes it easy for us to do the calculation.
3. Why do you say average value of $\Psi$? No, we calculate $\Psi$.
4. If, for example, $K$ is a line from the origin to the point $X+iY$ then $N$ is the unit vector in the direction of $X+iY$ rotated through $\pi/2$ clockwise i.e. $-i(X+iY)/(\sqrt{X^2+Y^2})=(Y-iX)/(\sqrt{X^2+Y^2})$
Vasco
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Post by mondo on Aug 10, 2022 3:44:33 GMT
2. No. You seem to have misunderstood the point. Because we can choose $K$ then if we want to calculate $\Psi$ at a point then we can choose any $K$, so we choose one that makes it easy for us to do the calculation. Yes I know, maybe my answer was confusing. I think it is clear now. 3. Why do you say average value of $\Psi$? No, we calculate $\Psi$. Right, my mistake. We calculate $\Psi$ for any path from $a$ to $p$, but if I understand correctly there is a chance the same streamline may describe completely different path say $L$ from $b$ to $q$ that happen to have the same value of $F$, flux on that path. Am I right? I am not talking about path independence here as in this case it does no matter how we get from $a$ to $p$. I am rather considering the scenario where we calculate flux in a different location on our vector field and it happens to have the same $\Psi$. 4. If, for example, $K$ is a line from the origin to the point $X+iY$ then $N$ is the unit vector in the direction of $X+iY$ rotated through $\pi/2$ clockwise i.e. $-i(X+iY)/(\sqrt{X^2+Y^2})=(Y-iX)/(\sqrt{X^2+Y^2})$ Clear. This is even better visible than my dot product example from previous post. I have one more question to the comment on page 511 right below the first statement in italic font - "Alternatively, $\Psi$ is the potential function for the dual of $\nabla\Phi$". Is that really the case? $\nabla\Phi$ is just $\bar H$, its dual is $i\bar H$ but this is not even close to $\Psi$ which requires calculation of an integral. So is this sentence a "shortcut" in thinking or mistaken? |
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Post by Admin on Aug 11, 2022 8:55:06 GMT
Mondo
3. Since the streamlines are lines along which the stream function is constant then if we calculate the integral at a new point and it has the same value as previously then the new point lies on the same streamline as the original point. I would advise you to make sure you understand all this before going forward into chapter 12 otherwise you will not understand it.
4.
Please read pages 508-511 again. These pages contain the answer to your question. What is written in the book here is correct.
Have you done the exercises for chapter 11 before moving on to chapter 12? Chapter 12 is a difficult chapter and you will not understand it if you don't do the exercises for chapter 11. Don't just read my answers but do them yourself. If you have done them then publish links to your answers here so that we can give you feedback.
Vasco
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Post by mondo on Jul 18, 2023 8:01:38 GMT
For this thread I have one more question - I wonder why in the middle of page 512, in the example author uses a parametric equation for the line $z = (X + iY)t$. At first I thought this is ensure the path is a line but it does not guarantee this, so what is the point?
Also, $ds = \sqrt{X^2 + Y^2}dt$ to make sure we move over the vector length/modulus right?
Thanks.
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Post by Admin on Jul 18, 2023 15:49:22 GMT
For this thread I have one more question - I wonder why in the middle of page 512, in the example author uses a parametric equation for the line $z = (X + iY)t$. At first I thought this is ensure the path is a line but it does not guarantee this, so what is the point? Also, $ds = \sqrt{X^2 + Y^2}dt$ to make sure we move over the vector length/modulus right? Thanks. Mondo It seems to me that it does make $K$ a line segment between 0 and $X+iY$. It also follows that $ds=\sqrt{X^2+Y^2}dt$. Why do you think it doesn't? Vasco |
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Post by mondo on Jul 19, 2023 7:00:37 GMT
For this thread I have one more question - I wonder why in the middle of page 512, in the example author uses a parametric equation for the line $z = (X + iY)t$. At first I thought this is ensure the path is a line but it does not guarantee this, so what is the point? Also, $ds = \sqrt{X^2 + Y^2}dt$ to make sure we move over the vector length/modulus right? Thanks. Mondo It seems to me that it does make $K$ a line segment between 0 and $X+iY$. It also follows that $ds=\sqrt{X^2+Y^2}dt$. Why do you think it doesn't? Vasco I don't see it, how does it stop me from taking three points $A[1,2], B[2,3], C[3,1]$ which are not collinear? |
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Post by mondo on Jul 19, 2023 7:59:40 GMT
I just realized $X$ and $Y$ are end point of the path $k$ and the parameter $t$ allows us to make this single point a function in the given region.
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Post by Admin on Jul 19, 2023 18:16:28 GMT
I just realized $X$ and $Y$ are end point of the path $k$ and the parameter $t$ allows us to make this single point a function in the given region. Mondo I don't understand what you mean. Anyway the best way to think about this is as follows: The integral is along $K$ from the origin to $p=X+iY$. This is a constant as $z$ travels along $K$. We know that $z=x+iy=(X+iY)t$. So $x=Xt$ and $y=Yt$ which means that $y/x=Y/X=constant$, which is the equation of a straight line. Also $s=|z|=|x+iy|=\sqrt{X^2+Y^2}t$ which means that $ds/dt=\sqrt{X^2+Y^2}$ which is a constant. Vasco |
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Post by mondo on Jul 19, 2023 20:05:11 GMT
I just realized $X$ and $Y$ are end point of the path $k$ and the parameter $t$ allows us to make this single point a function in the given region. Mondo I don't understand what you mean. Anyway the best way to think about this is as follows: The integral is along $K$ from the origin to $p=X+iY$. This is a constant as $z$ travels along $K$. We know that $z=x+iy=(X+iY)t$. So $x=Xt$ and $y=Yt$ which means that $y/x=Y/X=constant$, which is the equation of a straight line. Also $s=|z|=|x+iy|=\sqrt{X^2+Y^2}t$ which means that $ds/dt=\sqrt{X^2+Y^2}$ which is a constant. Vasco Yes that's what I meant - $X$ and $Y$ are constants now and we only vary $t$. Sorry I wrote that message from my phone, right before sleep and wasn't clear enough. You explained that way better. So for the first integral I got $3t^2X^2Y - t^3Y^3$ and since this is integral on a region $0 \le t \le 1$ our definite integral sums up to $3X^2Y - Y^3$ as in the book. |
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