Gary
GaryVasco
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Post by Gary on Nov 19, 2015 23:40:57 GMT
Vasco,
You wrote:
>PS your green arrows are pointing towards the centre!
Only because I cheated for the purposes of illustration. Thanks for the note. It's nice to know I wasn't seeing phantoms, but I'm going to start over and do a major revision and think it all through again. I'll get back when I'm done.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 20, 2015 5:56:01 GMT
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Post by Admin on Nov 20, 2015 7:27:15 GMT
Gary
I've had a quick look and it's certainly the sort of thing I'd expect. Do you have any particular concerns about it? I'll now have a more detailed look.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 20, 2015 16:51:30 GMT
Gary I've had a quick look and it's certainly the sort of thing I'd expect. Do you have any particular concerns about it? I'll now have a more detailed look. Vasco Vasco, I have no particular concerns. I think the code now reflects the theory accurately, but there is no claim to infallibility. I think two things made this tricky: (1) distinguishing the z in M(z) from the one in M'(z); and (2) distinguishing a z being passed to a function with a variable named "z" from an origin-centred z. M(z) normally applies to a point (z+q), whether centred at the origin or not. In this case we begin with z's on origin-centred circles recentred to -d/c, so z+q = z-d/c. We can see in the decomposition from (3) p. 124, that a point under M gets recentred to the origin before amplification and twisting, so we can pass (z+q) to M in the program. M'(z) requires a point of emanation centred at the origin and applies to an infinitesimal vector by multiplication. In the program, the amplitwist[z] function has a factor of \(1/(z-q)^2\). The z that is passed to it is the point of emanation (z+q) just as with M. Then z-q in the denominator works out to be (z+(q=(-d/c)) - (-d/c)) = z (origin-centred). It could be done differently, but it just seemed more convenient at the time. If we actually take the derivative of M, the result includes a factor of \(1/(cz+d)^2 = 1/(c^2(z+d/c)^2)\). If we pass (z+q) = (z-d/c) to the amplitwist[] function, we get \(1/c^2((z-d/c)+d/c)^2) = 1/c^2z^2\) where z is origin-centred and \(|z| = \rho\). I think this is what we want. Gary
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Post by Admin on Nov 21, 2015 12:58:06 GMT
Gary I have had a more detailed look at your solution and in particular the program that produces figure 1. I have produced a .pdf document which is attached to this post, with some thoughts which I hope are helpful. I think there is a typo in your document in part (i) on the line labelled (1) and 2 lines further down in the definition of $|M'(z)|$. Since $q=-d/c$, the three instances of $z+q$ should be $z-q$. Vasco Gary4-18.pdf (24.78 KB)
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 21, 2015 21:21:51 GMT
Gary I have had a more detailed look at your solution and in particular the program that produces figure 1. I have produced a .pdf document which is attached to this post, with some thoughts which I hope are helpful. I think there is a typo in your document in part (i) on the line labelled (1) and 2 lines further down in the definition of $|M'(z)|$. Since $q=-d/c$, the three instances of $z+q$ should be $z-q$. Vasco Vasco, Thank you for the comments. I believe the program is equivalent to what you have proposed but I have revised it to make the variable names consistent with your note and to eliminate some obsolete comments, which I think were adding to the confusion created by the similarity of variable names from the problem and the names of function variables. There is added discussion of your interesting comment on the use of Figure [1], p. 124 to understand the amplitwist in this problem. I benefited by working through that. Here is how I understand it using the z and w notation. In this problem, where we wish to preserve concentricity, M is applied to z = w + q, where |w| = \(\rho\) and q = -d/c, because M(z) automatically recentres circles to the origin under those conditions, but M' is calculated on an expression in w (A/w in your document), after recentering to the origin. The result is then applied to the \(\epsilon\) vectors, which after amplitwisting, emanate from M(z), so the essence of the effect on the \(\epsilon\) vectors is M(z) + amplitwist(w) \(\epsilon\). I think this could also be written: \(M(z) + (|A|/|w|^2)e^{i (-2arg(w)+arg(A))} \epsilon\), where \(A = (ad-bc)/c^2\) Gary
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Post by Admin on Nov 21, 2015 23:21:25 GMT
Gary
I will have a look at your revised document. I notice that you have not changed the $z+q$ to $z-q$. Is that because you don't agree that it should be changed?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 22, 2015 1:18:24 GMT
Gary I will have a look at your revised document. I notice that you have not changed the $z+q$ to $z-q$. Is that because you don't agree that it should be changed? Vasco Vasco, Thank you. It was a careless oversight caused by not reading your note carefully enough. I have changed it to z-q. Gary
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Post by Admin on Nov 22, 2015 16:49:45 GMT
Gary
The latest version of your document looks fine to me.
Vasco
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