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Post by mondo on Jul 9, 2023 8:49:01 GMT
At the bottom of page 488 author says that the circulation of "B" part of above equation $\frac{i}{\overline{z} - \overline{p}}$ is $2\pi$ but I can't get it.
I try to calculate it for $\frac{i}{\overline{z}}$ as it should be the same, just centered at the origin.
$\frac{i}{\overline{z}} = \frac{y}{x^2 + y^2} + \frac{ix}{x^2+y^2}$
Next, $\nabla \times \overline{H}(z) = -(\partial_x{v} + \partial_y{u}) = -((1 - i) (x^2 - y^2))/(x^2 + y^2)^2$ Nex this should be integrant of integral (8) but I can already see it won't give us $2\pi$. So where is my mistake?
Thank you.
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Post by Admin on Jul 9, 2023 13:15:40 GMT
Mondo
Use the fact that $(z-p)$ and therefore $(\overline{z}-\overline{p})$ is a radius vector of $C$ and then look at the definition of the circulation $W$ on page 474 and use the definition of the dot product. Think geometrically about what the conjugate does to a vector and also what inversion does to a vector.
Vasco
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Post by Admin on Jul 9, 2023 17:32:29 GMT
At the bottom of page 488 author says that the circulation of "B" part of above equation $\frac{i}{\overline{z} - \overline{p}}$ is $2\pi$ but I can't get it. I try to calculate it for $\frac{i}{\overline{z}}$ as it should be the same, just centered at the origin. $\frac{i}{\overline{z}} = \frac{y}{x^2 + y^2} + \frac{ix}{x^2+y^2}$ Next, $\nabla \times \overline{H}(z) = -(\partial_x{v} + \partial_y{u}) = -((1 - i) (x^2 - y^2))/(x^2 + y^2)^2$ Nex this should be integrant of integral (8) but I can already see it won't give us $2\pi$. So where is my mistake? Thank you. Mondo Shouldn't there be a minus sign on the first term on line "$\frac{i}{\overline{z}} =$" Vasco |
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Post by mondo on Jul 9, 2023 19:08:18 GMT
Mondo Use the fact that $(z-p)$ and therefore $(\overline{z}-\overline{p})$ is a radius vector of $C$ and then look at the definition of the circulation $W$ on page 474 and use the definition of the dot product. Think geometrically about what the conjugate does to a vector and also what inversion does to a vector. Vasco Vasco, that is a good suggestion and I did that exercise: $\frac{1}{\overline{z}} = \frac{1}{r}e^{i\theta}$ so this is pointing radially away from the origin. However our vector field multiplies this by $i$ which changes the picture to a vector field rotating around origin. So from this, I can agree that curl should be $2\pi r$ times something. I just can't get it algebraically. Yes, so I had at least one mistake, but still it doesn't lead to any better result. The way I calculate it $\frac{i}{x-iy} = \frac{ix - y}{x^2+y^2} = -\frac{y}{x^2+y^2} + \frac{ix}{x^2+y^2}$ where the first term is $u$ and the second is $v$ Next I calculate partial derivatives, $\partial_x{v} = -\frac{i(x^2-y^2)}{(x^2+y^2)^2}$ and $\partial_y{u} = \frac{y^2-x^2}{(x^2+y^2)^2}$. Next we have a formula $\nabla \times \overline{H}(z) = -[\partial_x{v} + \partial_y{u}]$ which here gives $-\frac{y^2-x^2}{(x^2+y^2)^2} + \frac{i(x^2-y^2)}{(x^2+y^2)^2}$. In the next step I should calculate an integral of this over $da$ but it skip this part as it doesn't seem to go into the right direction already. PS:The integral over $da$ should I treat it as an integral in regards to $x$ and then $y$? |
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Post by mondo on Jul 9, 2023 19:15:22 GMT
I have corrected a few sign typos above.
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Post by Admin on Jul 10, 2023 8:02:50 GMT
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Post by mondo on Jul 10, 2023 18:14:00 GMT
Thank you Vasco for your docuemnt. I have a problem understanding this though. In the line with $W[\overline{H}, C]$ you end this line by saying $\int_K{\overline{H} \cdot T ds} = |\overline{z} - \overline{p}|d\theta$ is that what you really mean here? Why is it this way? |
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Post by Admin on Jul 10, 2023 18:57:32 GMT
Mondo
That's not what I say. I actually say "...where $ds=|\overline{z}-\overline{p}|d\theta$"
Vasco
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Post by mondo on Jul 10, 2023 19:15:52 GMT
Huh right, I am sorry I didn't sleep much tonight..
I am more interested into this actually $-|\overline{H}| \cdot |T| \cdot |\overline{z}-\overline{p}| d\theta = -Bd\theta$ how are these two sides equal?
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Post by Admin on Jul 10, 2023 19:28:02 GMT
Mondo
It's just algebraic substitution. Substitute for $|\overline{H}|$ and $ds$ and the LHS becomes $-Bd\theta$
Vasco
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Post by Admin on Jul 11, 2023 6:28:01 GMT
Mondo
The only really effective way to understand all this is to do the exercises for chapter 11 and most importantly publish your answers.
Vasco
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Post by mondo on Jul 11, 2023 6:51:48 GMT
I more or less understand your approach in the supplement document you shared. However:
1. Do you know what is wrong in my calculations?
2. You do $\overline{H} \cdot T$ however, I think we agreed that $\overline{H}$ is already parallel to the tangent to $C$. So why do you multiply by a unit tangent vector?
3. When I think about it logically, the circulation should be $2\pi r$ (our path) times the tangent component of a vector field to this path which is just $\frac{1}{r}$ hence $W[\overline{H},C] = 2\pi r * \frac{1}{r} = 2\pi$. Which agrees with the result. The only question is why my algebraic solution from first posts doesn't.
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Post by Admin on Jul 11, 2023 11:34:44 GMT
I more or less understand your approach in the supplement document you shared. However: 1. Do you know what is wrong in my calculations? 2. You do $\overline{H} \cdot T$ however, I think we agreed that $\overline{H}$ is already parallel to the tangent to $C$. So why do you multiply by a unit tangent vector? 3. When I think about it logically, the circulation should be $2\pi r$ (our path) times the tangent component of a vector field to this path which is just $\frac{1}{r}$ hence $W[\overline{H},C] = 2\pi r * \frac{1}{r} = 2\pi$. Which agrees with the result. The only question is why my algebraic solution from first posts doesn't. Mondo 1. What makes you think your calculations are wrong? 2, You are correct but the angle between them is not zero but $\pi$ and remember that $|\boldsymbol{\overline{H} \cdot T}|=|\boldsymbol{\overline{H}}|\cdot |\boldsymbol{T}|\cdot \cos\phi$ where $\phi$ is the angle between the two vectors, which in this case is $\pi$ as they point in opposite directions and so this gives us a change of sign. Don't forget I am using the equation on page 474 and the definition of dot product at the bottom of page 27. 3. Don't you need to integrate the curl according to (8)? Vasco |
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Post by mondo on Jul 11, 2023 18:05:15 GMT
I more or less understand your approach in the supplement document you shared. However: 1. Do you know what is wrong in my calculations? 2. You do $\overline{H} \cdot T$ however, I think we agreed that $\overline{H}$ is already parallel to the tangent to $C$. So why do you multiply by a unit tangent vector? 3. When I think about it logically, the circulation should be $2\pi r$ (our path) times the tangent component of a vector field to this path which is just $\frac{1}{r}$ hence $W[\overline{H},C] = 2\pi r * \frac{1}{r} = 2\pi$. Which agrees with the result. The only question is why my algebraic solution from first posts doesn't. Mondo 1. What makes you think your calculations are wrong? 2, You are correct but the angle between them is not zero but $\pi$ and remember that $|\boldsymbol{\overline{H} \cdot T}|=|\boldsymbol{\overline{H}}|\cdot |\boldsymbol{T}|\cdot \cos\phi$ where $\phi$ is the angle between the two vectors, which in this case is $\pi$ as they point in opposite directions and so this gives us a change of sign. Don't forget I am using the equation on page 474 and the definition of dot product at the bottom of page 27. 3. Don't you need to integrate the curl according to (8)? Vasco 1. I assumed so being scared how the curl looks like. I just tried to integrate it over an arbitrary circle in $dxdy$ but what I get is $-2\pi$ for $u$ and $i2\pi$ for $v$. 2. You say they point in the opposite direction because I think you did an unnecessary step of multiplying by $-B$. If you look at figure [12] You can see that vortex is illustrated just for $\frac{1}{\overline{z}}$. Similarly in the last equation on page 488, when author speaks about circulation $2\pi$ he refers to the term in the square bracket next to $-B$. 3. As I said in point #1, I try to but failed so far. |
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Post by Admin on Jul 11, 2023 18:31:53 GMT
Mondo
2. I disagree with you. I have taken the $-B$ into account.
Vasco
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