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Post by telemeter on Mar 10, 2020 12:48:55 GMT
At the top of p394 (of my 2012 reprint of the 1998 paperback publication), first line "It follows we could draw a picture very similar to [12]." In the second sentence is "The argument that followed from [12] therefore remains valid."
I think, these should read, "It follows we could draw a picture very similar to [11]." and "The argument that followed from [11] therefore remains valid." as the reference to [12] is out of context (and not very helpful!).
This section is evaluating the contour integral of z-bar on a loop around the origin... and the text refers to replacing 1/r with r in the argument following [12] to arrive at the result (8).
However, it is [11] that is considering a contour integral around the origin and the argument following [11] shows that with 1/z the integral is purely dependent on angle and = 2*pi*i. It is in this diagram that if you substitute A for 1/A that you are led, following the argument after [11], to conclude that the integral of z-bar will be A^2*2*pi*i or 2*i*(area enclosed) as in (8).
[12] is the general loop case showing that a contour integral of 1/z on a contour not including the origin is zero.
telemeter
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Post by Admin on Mar 10, 2020 16:07:04 GMT
At the top of p394 (of my 2012 reprint of the 1998 paperback publication), first line "It follows we could draw a picture very similar to [12]." In the second sentence is "The argument that followed from [12] therefore remains valid." I think, these should read, "It follows we could draw a picture very similar to [11]." and "The argument that followed from [11] therefore remains valid." as the reference to [12] is out of context (and not very helpful!). This section is evaluating the contour integral of z-bar on a loop around the origin... and the text refers to replacing 1/r with r in the argument following [12] to arrive at the result (8). However, it is [11] that is considering a contour integral around the origin and the argument following [11] shows that with 1/z the integral is purely dependent on angle and = 2*pi*i. It is in this diagram that if you substitute A for 1/A that you are led, following the argument after [11], to conclude that the integral of z-bar will be A^2*2*pi*i or 2*i*(area enclosed) as in (8). [12] is the general loop case showing that a contour integral of 1/z on a contour not including the origin is zero. telemeter telemeter I disagree with your analysis of the situation here. The references on page 394 to figure 12 on page 390 seem to me to make perfect sense since on page 394 we are dealing with a general loop, not a circle. Cauchy's theorem does not apply if the integrand is non-analytic and the value of the integral depends on the geometry of the loop. Vasco |
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Post by telemeter on Mar 11, 2020 10:40:03 GMT
Vasco
Perhaps, we'll have to agree to disagree. Consider this:
The very next paragraph starts "Next we ask how this formula would change if the origin were outside the loop". Therefore, the discussion (that leads to (8) before this sentence is about the situation when the origin is inside the loop (and implies the fact of it being in the loop is relevant to the argument) ...it is not about the general case or. It is merely trying to establish the result in a special case which is subsequently generalised.
Note [11] is with origin in loop, [12] is outside.
Further, one can deduce (8) from [11] by inspection (substituting A for 1/A). Going from [12] to (8) is at best not easy, firstly [12b] is no longer vertical / horizontal, secondly adding up the component from the elements between the rings in [13a] means you have to add up terms with continuously varying 'r' as r is now in both components of [12b] (r^2*dr and r*dtheta). It's unclear how to do that...and the text does not help as it is geared to the case where the dtheta term is by itself.
Then again, I could easily be missing a simple derivation from [12] and [13] which I'd be grateful to have some hints on.
telemeter
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Post by Admin on Mar 11, 2020 12:01:47 GMT
telemeter
I think we may well have to agree to disagree. I still strongly disagree with what you say. Here is why:
I think that if you read subsection 2 General Loops on pages 390-391 you will see from the last paragraph of this section:
"In [13b]...change into [13c]"
that the subsection is not just about loops which enclose the origin, but about general loops of any shape which do or do not enclose the origin, but it is still about integrating $(1/z)$. It refers to figure 13 on page 391.
The reason that Needham refers to figure 12 at the top of page 394 is because (as he points out at the bottom of page 393) he is trying to integrate $\overline{z}$ round a general loop, and it just so happens that $(1/z)$ points in the same direction as $\overline{z}$ and so we can still use figure 12 if we multiply by $r$ instead of dividing by it, to obtain $\widetilde{\Delta}$, and then use the argument that follows from [12].to obtain result (8) on page 394, which is the integral round a general loop $L$ which encloses the origin.
For the same reasons figure 12b is still vertical/horizontal for the case of $\overline{z}$.
We cannot deduce (8) from figure 11 because in the figure the contour is a circle whereas (8) is for general loops.
To be honest I think that if you still disagree with me about this then this is so fundamental to what follows in the book that we should spend some time trying to resolve it.
Vasco
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Post by telemeter on Mar 14, 2020 17:34:29 GMT
Vasco
We totally agree on all the mathematics and the general argument in the text. I am getting hung up on the details of Needham's presentation.
Admittedly, I couldn't make sense of [12] so I lunged (desperately) to I see him using [11] to just show an instance of (8) in a special case where you can immediately see that for a circle the integral becomes 2*i*area (ie 2*i*pi*A^2). I liked that because it works and it's easy.
Where I need help is to see the detailed argument from [12].
I see now you've pointed it out that the elements of [12b] are indeed vertical/ horizontal because the angle of z bar is the same as that of 1/z (can't see why I couldn't see that before)
Here's my problems...
Problem One
The horizontal/ vertical components become rdr and r^2 d(theta) respectively (because we are multiplying be r rather than dividing). As per the argument for [13a] the horizontal components will cancel. But the sum of the vertical components is no longer the sum of the signed angles, it is the sum of the r^2 d(theta)s.
If this reasoning is right, my issue becomes how to show that the sum of the vertical r^2 d(theta)s is 2*i*area enclosed. We need to be able to do this to fulfil the text statement that we can deduce this from the [12] argument. (I am happy about the "I"..it's the 2*area bit that puzzles)
I am still at a loss here, I'm afraid. I am totally happy with [15] and the argument there...I just can't see how to get there from [12].
Problem Two
The very next sentence after (8) is "Next we ask....if the origin were outside the loop".
It seems to me that if you can get to (8) from [12] as per the first para (and as you suggest) then that sentence at the start of the second para is wrong because we have just done the job (of considering a general loop away from the origin) when considering [12] (as [12]/[13] do not enclose origin). These two pieces of text are then inconsistent...and its the apparent inconsistency that trips me up.
Problem Three
The paragraph under question (bottom of 393 to 394) is clearly (by virtue of Problem Two) referring to [15a]. That is a loop around the origin. My contention is that the only loop around the origin for which you can clearly show that the integral of z bar is 2*i*area (by the methods used prior to p393) is in fact the special case of a circle. That is why reference to [11] would make perfect sense here.
If anyone can solve Problem One for me and in detail show how [12] leads to (8) for an origin enclosing loop then Problems Two and Three will vanish also. My suspicion is that no-one will be able to do this...but that's just because I cannot see it, so probably does not mean much!
I very much appreciate your help
telemeter
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Post by telemeter on Mar 16, 2020 14:25:10 GMT
I was just sitting down doing something else entirely when the answer came to me. I realised that summing the r^2 d(theta)s is exactly the point. As each r^2 d(theta) is 2*area of the area segment/triangle subtended by d(theta) (ie base rdtheta, height r)... and we already know that it must be in the i direction. Doh!
So as Vasco insisted, (8) is an obvious consequence of [12] and my Problem One is solved so Problems Two and Three dissolve.
I hereby withdraw my suggestion for an erratum as it was my own errata!
Thanks for making me persist to the answer.
telemeter.
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Post by Admin on Mar 17, 2020 7:40:16 GMT
I was just sitting down doing something else entirely when the answer came to me. I realised that summing the r^2 d(theta)s is exactly the point. As each r^2 d(theta) is 2*area of the area segment/triangle subtended by d(theta) (ie base rdtheta, height r)... and we already know that it must be in the i direction. Doh! So as Vasco insisted, (8) is an obvious consequence of [12] and my Problem One is solved so Problems Two and Three dissolve. I hereby withdraw my suggestion for an erratum as it was my own errata! Thanks for making me persist to the answer. telemeter. telemeter I know you have solved your problem yourself, but here's a link to a short document I had already written which might be of interest. Vasco |
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Post by telemeter on Mar 19, 2020 10:03:17 GMT
Thanks that addresses the directionality. My main issue was that I was not prepared to take (1) (as per your note) as an axiom but wanted to derive it from [12]..and now in retrospect it seems so clear that I can only shake my head at myself!
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