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Post by mondo on Jul 11, 2023 18:37:04 GMT
Mondo 2. I disagree with you. I have taken the $-B$ into account. Vasco That's what I said. After all you got an answer $-2\pi B$. Anyway I follow your calculation and I think it is fine. I would only skip the multiplication by $-B$ and hence no need to $\cdot T$ (multiplication by a tangent vector) As for my integral - the curl has an imaginary unit in it, while the work/circulation should be all real right? So my integral won't be real. |
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Post by Admin on Jul 11, 2023 19:10:51 GMT
Mondo
The curl has no imaginary unit (see page 28 and page 483). I still disagree with you about $T$. I will amend my document to make it clear.
Vasco
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Post by Admin on Jul 11, 2023 19:30:15 GMT
Mondo
You have used $iv$ instead of $v$ when calculating the curl.
Vasco
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Post by mondo on Jul 11, 2023 20:05:47 GMT
Mondo You have used $iv$ instead of $v$ when calculating the curl. Vasco Right, thanks for spotting that. I used wolframalpha to verify my calculation Wolfram calculationHere I simulate the integral round a loop with integral over half unit circle from $[-1,1]]$. This means I should multiply the end result by $2$. But this would give a result which is exactly 2x bigger than the expected one. |
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Post by Admin on Jul 11, 2023 20:31:59 GMT
Mo;ndo
So what exactly did you integrate?
Vasco
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Post by mondo on Jul 11, 2023 20:39:52 GMT
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Post by Admin on Jul 11, 2023 20:58:34 GMT
Mondo
You have to integrate over an area not round a contour if you are using (8)on page 481
Vasco
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Post by mondo on Jul 11, 2023 21:02:29 GMT
Mondo You have to integrate over an area not round a contour if you are using (8)on page 481 Vasco But isn't that exactly what I do here? A double integral of the curl? $\int_{-1}^{1}\int_{-1}^{1}-[- \frac{(x^2-y^2)}{(x^2+y^2)^2} + \frac{(y^2-x^2)}{(x^2+y^2)^2}]dxdy$ |
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Post by mondo on Jul 12, 2023 0:41:08 GMT
Ok my integral limits are wrong - I treat it as if we integrate over a square centered at the origin. I will fix it soon..
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Post by mondo on Jul 12, 2023 7:52:58 GMT
My calculations are progressing:
$\overline{H}(z) = u + iv = \frac{-y}{x^2+y^2} + i(\frac{x}{x^2+y^2})$
According to equations on page 483 $\nabla \times \overline{H}(z) = -[\partial_x{v} + \partial_y{u}]$
According to (14) p.484 $W[\overline{H}(z),C] = \int \int_{r} [\nabla \times \overline{H}(z)]dA$
My partials are:
$\partial_x{v} = \frac{y^2 - x^2}{(x^2+y^2)^2}$
$\partial_y{u} = \frac{y^2 - x^2}{(x^2+y^2)^2}$
Exactly the same.
Now I calculate the area integral for each of partials separately, and add them together at the end, hence:
$\int \int_{R} \partial_x{v}dA = 2\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} \partial_x{v}dxdy = -\pi$
I get the exact same result for the other partial hence the total integral sums up to $-2\pi$
In the book, at the very end of page 488 author says the result is $2\pi$ - Does he already multiplied by $-B$? I don't think so.. I still must have something wrong here. Can you spot it Vasco?
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Post by Admin on Jul 12, 2023 13:12:25 GMT
Mondo
Did you notice that in (8) on page 481 that the curl is calculated using the vector field not the Polya vector field.
Whereas using my approach, integrating round the loop we use the Polya vector field.
Vasco
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Post by Admin on Jul 12, 2023 20:07:43 GMT
Mondo
I'm having second thoughts about this now. Just to let you know.
Vasco
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Post by mondo on Jul 12, 2023 20:36:04 GMT
Yes please, I am about to start my ~1h session on this.
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Post by Admin on Jul 12, 2023 20:49:15 GMT
Mondo
What I mean is that I think what I said in reply #25 may not be right.
Vasco
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Post by mondo on Jul 12, 2023 21:32:27 GMT
Mondo What I mean is that I think what I said in reply #25 may not be right. Vasco I think it actually is correct. I also spotted an error in my calculations: Should be: $\int \int_{R} -\partial_x{v}dA = 2\int_{-1}^{1} \int_{0}^{\sqrt(1-x^2)} -\partial_x{v}dydx = 2\int_{-1}^{1} -[\frac{-y}{x^2+y^2} | y=\sqrt(1-x^2)]dx = 2\int_{-1}^{1} \sqrt(1-x^2)dx = \pi$ Same for the other partial and now it summs up to $2\pi$. PS: I also plotted the vector field that I optained to make sure it looks ok: Vector fieldSorry for the confusion! Vasco, does it look right now to you? |
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