|
|
Post by telemeter on Mar 20, 2020 12:29:20 GMT
With temerity, I raise another erratum. Very confident of this one though.
(17) slowed me for a little while because it does not follow from the text, is inconsistent with the text and yet is a perfectly valid equation. That it is valid, one can check by deriving it using the process at the bottom of p486 but of course considering 'H(z) bar ' and 'H(0) bar' rather than H(z) and H(0).
Reasons that show Needham put the wrong equation in as (17):
a) The text says taking the work at the bottom of p486 and substituting (13) you get (17). You do not...you get another equation which I will set out below. In particular , note that the equation at the bottom of p486 is about H(z). (17) is about 'H(z) bar'.
b) The text immediately after (17) says "where 'C bar' =". In (17) 'C bar" does not appear, only C appears...which makes the text strange.
c) In the paragraph after (17) it says "...(17) reduces to the first two terms of Taylor's series: H(z) = H(0) + H'(z) +..." This is untrue because (17) is about 'H(z) bar' = ...
d) At the top of p488 we are asked to use (17) to get an expression for the contour integral of H(z). This is odd because (17) is not about H(z) it is about 'H(z) bar'....and so of course you cannot substitute.
e) Looking at the equation at the top of p488 (fourth line) after the alleged substitution, you see many terms that do not appear in (17)...C bar instead of C; z s and 'z bar' s interchanged.
All this can be resolved by following the instructions in the second line of p487. This easily gives the correct equation (17):-
H(z) = H(0) + Div 'H bar' * 'z bar' /2 - Curl 'H bar' * i * 'z bar' /2 + 'C bar' * z.
This equation now solves all the above issues a) to e).
This is not a simple type setting error as the printed (17) is legitimate... just not relevant to the text.
telemeter
|
|
|
|
Post by Admin on Mar 20, 2020 14:49:19 GMT
telemeter
I think you have got this completely wrong. Did you just jump into the middle of this chapter without reading the preceding pages?
I think I have a document somewhere which explains all the things you are referring to, and I will post it here when I find it.
Vasco
|
|
|
|
Post by Admin on Mar 20, 2020 15:18:25 GMT
telemeter
Just a quick comment regarding your point b):
If we know $C$ then we know $\overline{C}$, so there is nothing strange about this.
All your other points seem to stem from this problem you have with the conjugate.
Vasco
|
|
|
|
Post by telemeter on Mar 20, 2020 17:59:32 GMT
Lol! I was reading carefully, honest!
I await your note with interest!
telemeter
|
|
|
|
Post by telemeter on Mar 20, 2020 19:26:45 GMT
I'm sorry, but something compels me to share further (maths excites me!)...
a) Note the expression for C bar on p487. Then compare the left most bracket at the bottom of p 486. Notice they are the same (bar the z) . Substitute C into that equation.
b) Look at the right hand bracket in the equation at the bottom of p486. Now review (13) on p483. Notice that LHS of (13) times -i = the right hand bracket (bar the z bar /2). Substitute the rhs of (13) * -i into that equation.
c) You now have my equation above.
That is why I propose the above amendment.
d) Note the equation on p488. for the contour integral of H(z). Compare with my equation derived in c) above. They are the same (if you contour the integrate mine of course).
Done.
To maintain the text is correct you need to derive the published (17) from the equation at the bottom of p486 by substituting from (13). Then you need to show how the published (17) is used to derive the expression for the contour integral of H(z) on p488. I tried this fruitlessly for too long.
You can derive (17) using the method at the bottom of p486 but that's another thing entirely.
telemeter
|
|
|
|
Post by Admin on Mar 20, 2020 20:29:14 GMT
telemeter Sorry but something compels me to say you are wrong.  Vasco |
|
|
|
Post by Admin on Mar 21, 2020 8:32:29 GMT
telemeter Here is a link which I think explains equation (17) on page 487It also contains an explanation of the Taylor series and a short derivation of (13). You will see that (17) is derived from the equation at the bottom of page 486 by substituting from (13), and that it is then very simple to derive the expression for the contour integral of $H(z)$ on page 488. We need to remember that div and curl are both real numbers as defined on page 483 and so $\overline{[\nabla\cdot\overline{H}+i\nabla\times\overline{H}]}=[\nabla\cdot\overline{H}-i\nabla\times\overline{H}]$ Vasco PS Your equivalent of (17) in your post above is in fact the same as (17) in the book. Just take the conjugate of both sides of yours and you obtain Needham's (17). |
|
|
|
Post by telemeter on Mar 21, 2020 17:32:48 GMT
Humph! I gracefully cede the point that (17) does work ( in its roundabout fashion!). But I go for points on simplicity in that my approach is so much more direct!
The equations aren't just conjugates of each other though cos they both have H bar in the div and curl. So we must be saying that H is analytic.
Excellent discussion, thanks.
Alas I have now finished Chapter 11 ..only one more chapter to go...then back to Penrose.
telemeter
|
|
|
|
Post by Admin on Mar 21, 2020 18:02:03 GMT
telemeter
I think you will find that they are conjugates of each other because the conjugate of $\nabla\cdot\overline{H}$ is the same expression $\nabla\cdot\overline{H}$ and the same for the curl. The values of div and curl are real numbers.
Vasco
|
|
|
|
Post by telemeter on Mar 23, 2020 12:13:36 GMT
Vasco,
Why are div and curl real? I'm being a bit thick here. I suppose u and v have to be real functions, is that right?
telemeter
|
|
|
|
Post by Admin on Mar 23, 2020 14:37:44 GMT
Vasco, Why are div and curl real? I'm being a bit thick here. I suppose u and v have to be real functions, is that right? telemeter telemeter Yes $u$ and $v$ are both real functions of $x$ and $y$ and so from the definitions of div and curl on page 483 we can see that they are both real. Vasco |
|
|
|
Post by telemeter on Apr 4, 2020 16:31:30 GMT
I had cause to revisit this and uncover my underlying confusion. In case it may be helpful to others I'll explain it.
In short I was looking at curl's components as a separable operation of the grad operator on a vector and manipulating symbols while forgetting the bigger picture that curl is always real.
Thus, considering complex vectors a and b, the conjugate of (a cross b) is absolutely not the same as the cross of the conjugates (a bar cross b bar)...it remains (a cross b).
I get there in the end...slowly.
telemeter
|
|
|
|
Post by Nate on Sept 29, 2020 10:04:59 GMT
Personally I was a bit confused by this part as well. When I read "substitution" I assume it means substitute, and when it doesn't I look hard to see if I've made a mistake or the author has. I think it would be a bit more clear if the text read: "multiplying both sides of (13) by -i and substituting..."
|
|
|
|
Post by Admin on Sept 30, 2020 9:49:55 GMT
Personally I was a bit confused by this part as well. When I read "substitution" I assume it means substitute, and when it doesn't I look hard to see if I've made a mistake or the author has. I think it would be a bit more clear if the text read: "multiplying both sides of (13) by -i and substituting..." Hi Nate I agree that this is a bit confusing and it is easy to make a mistake when doing the algebra. I find the easiest way to look at it is to obtain the equivalent of (13) on page 483 for the Polya vector field $\overline{H}$ by taking the conjugate of (13) which gives $-i\partial_x\overline{H}-\partial_y\overline{H}=\boldsymbol{\nabla\times\overline{H}}-i\boldsymbol{\nabla\cdot\overline{H}}$ since $\boldsymbol{\nabla\times\overline{H}}$ and $\boldsymbol{\nabla\cdot\overline{H}}$ are real. Then multiplying this result by $i$ we have $\partial_x\overline{H}-i\partial_y\overline{H}=\boldsymbol{\nabla\cdot\overline{H}}+i\boldsymbol{\nabla\times\overline{H}}$ Then we can substitute this into our expression at the end of page 486 after it has been rewritten for $\overline{H(z)}-\overline{H(0)}$ as $\frac{1}{2}[\partial_x\overline{H}+i\partial_y\overline{H}]\overline{z}+\frac{1}{2}[\partial_x\overline{H}-i\partial_y\overline{H}]z$ and we obtain (17). Vasco |
|
|
|
Post by Admin on Sept 30, 2020 10:33:23 GMT
Hi Nate
Another perhaps more straightforward way would be to define $\overline{C}$ as Needham does near the top of page 487 and multiply (13) by $-i$ to obtain
$\partial_xH+i\partial_yH=\boldsymbol{\nabla\cdot\overline{H}}-i\boldsymbol{\nabla\times\overline{H}}$
Substituting these into the result at the bottom of page 482 we obtain
$\displaystyle H(z)-H(0)=\overline{C}z+(\boldsymbol{\nabla\cdot\overline{H}})\frac{\overline{z}}{2}-(\boldsymbol{\nabla\times\overline{H}})\frac{i~\overline{z}}{2}$
By taking the complex conjugate of both sides we obtain (17).
Remember that the div and curl are both real quantities.
Vasco
|
|
