Gary
GaryVasco
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Post by Gary on Nov 8, 2016 23:51:16 GMT
Vasco,
I hope the answer to this is not in the posting that I haven't read yet, as I haven't attempted that exercise.
On p. 317, $\mathbb{P}$ 4, it says "If we apply an h-rotation $\mathcal{R}^{\phi}_{z}$ then $L$ is carried into another h-line $L'$ and $E$ is carried into an equidistant curve $E'$ of $L'$ ...."
I have two questions about this:
(1) How would one apply the h-rotation if one were actually attempting to plot the rotation? One might expect the old ray from the horizon to z to be the center of the new h-line, but judging from [36a], it is not.
(2) Shouldn't this rotation be conformal? It is apparently not conformal because the angle between E and L on the horizon changes orientation.
Gary
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Post by Admin on Nov 10, 2016 9:44:23 GMT
Gary
The rotation is about the point $z$ (think of it as the point in the 'middle' of [27] on page 308). Since the rotated $ds$ is perpendicular to $L'$ it must point along the radius of $L'$ and so projecting it back to the horizon, it will intersect the horizon at the centre of $L'$.
This is quite straightforward, but a bit difficult to explain in words, but here goes:
Notice that the angle between $L'$ and $E'$ appears also at the other intersection point of these two circles with the horizon, not shown in [36]a, but over to the left. Strictly speaking there is an error in [34a]. Needham has shown the situation when the rotation is $>\pi$ except that in that case we should be looking at the angle between $L'$ and $E'$ at the other intersection point over to the left.
This is easy to see if we imagine [36a] changing as $ds$ moves anticlockwise about $z$. When the angle of rotation of $ds$ is less than $\pi$ the point of intersection $P$ of $L'$ and $E'$ moves further and further to the west and then eventually when the angle of rotaion is equal to $\pi$, the point $P$ shoots off to $\infty$ in the northerly direction, and $L'$ and $E'$ come down from $\infty$ to $z$ with $E'$ to the right of $L'$ and so the sense of the angle between them is preserved. As the angle of rotation goes past $\pi$, the point $P$ now moves along the horizon in an easterly direction from $\infty$, with $L'$ now inside $E'$ with the sense of the angle between them still preserved.
I would recommend drawing a sketch on paper of [36a] for various angles of rotation: $<\pi$, $=\pi$, and $>\pi$.
To sum up: I agree that the rotation should be conformal and is conformal. Needham's diagram contains an error and as a result is misleading.
I will add this to the errata if you agree.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 10, 2016 17:55:48 GMT
Gary The rotation is about the point $z$ (think of it as the point in the 'middle' of [27] on page 308). Since the rotated $ds$ is perpendicular to $L'$ it must point along the radius of $L'$ and so projecting it back to the horizon, it will intersect the horizon at the centre of $L'$. This is quite straightforward, but a bit difficult to explain in words, but here goes: Notice that the angle between $L'$ and $E'$ appears also at the other intersection point of these two circles with the horizon, not shown in [36]a, but over to the left. Strictly speaking there is an error in [34a]. Needham has shown the situation when the rotation is $>\pi$ except that in that case we should be looking at the angle between $L'$ and $E'$ at the other intersection point over to the left. This is easy to see if we imagine [36a] changing as $ds$ moves anticlockwise about $z$. When the angle of rotation of $ds$ is less than $\pi$ the point of intersection $P$ of $L'$ and $E'$ moves further and further to the west and then eventually when the angle of rotaion is equal to $\pi$, the point $P$ shoots off to $\infty$ in the northerly direction, and $L'$ and $E'$ come down from $\infty$ to $z$ with $E'$ to the right of $L'$ and so the sense of the angle between them is preserved. As the angle of rotation goes past $\pi$, the point $P$ now moves along the horizon in an easterly direction from $\infty$, with $L'$ now inside $E'$ with the sense of the angle between them still preserved. I would recommend drawing a sketch on paper of [36a] for various angles of rotation: $<\pi$, $=\pi$, and $>\pi$. To sum up: I agree that the rotation should be conformal and is conformal. Needham's diagram contains an error and as a result is misleading. I will add this to the errata if you agree. Vasco Vasco, I think this clears it up for me, but I wonder if the direction is reversed. If we take the ds that points west and rotate it anti-clockwise, won't the point P will travel eastward and go to infinity after a rotation of Pi, when ds points east? If so, then E' on the right should be drawn inside (below) L' and the angle would be in the same orientation. After a total rotation of Pi, E' must flip to remain on the same side of L' as ds. But then wouldn't this create a new h-line (circle) that doesn't intersect the horizon? Is this a multifunction with two regions of conformality? I'll have to plot some possibilities. My diagrams are not sufficient. But it does seem to me that something is wrong with [36a]. Gary |
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Post by Admin on Nov 10, 2016 18:46:01 GMT
Gary The rotation is about the point $z$ (think of it as the point in the 'middle' of [27] on page 308). Since the rotated $ds$ is perpendicular to $L'$ it must point along the radius of $L'$ and so projecting it back to the horizon, it will intersect the horizon at the centre of $L'$. This is quite straightforward, but a bit difficult to explain in words, but here goes: Notice that the angle between $L'$ and $E'$ appears also at the other intersection point of these two circles with the horizon, not shown in [36]a, but over to the left. Needham has shown the situation when the rotation is $>\pi$ except that in that case we should be looking at the angle between $L'$ and $E'$ at the other intersection point over to the left. This is easy to see if we imagine [36a] changing as $ds$ moves anticlockwise about $z$. When the angle of rotation of $ds$ is less than $\pi$ the point of intersection $P$ of $L'$ and $E'$ moves further and further to the east and then eventually when the angle of rotation is equal to $\pi$, the point $P$ shoots off to $\infty$ in the northerly direction, and $L'$ and $E'$ come down from $\infty$ to $z$ with $E'$ to the left of $L'$ and so the sense of the angle between them is preserved. As the angle of rotation goes past $\pi$, the point $P$ now moves along the horizon in a westerly direction from $\infty$, with $L'$ now inside $E'$ with the sense of the angle between them still preserved. I would recommend drawing a sketch on paper of [36a] for various angles of rotation: $<\pi$, $=\pi$, and $>\pi$. To sum up: I agree that the rotation should be conformal and is conformal. Needham's diagram contains an error and as a result is misleading. I will add this to the errata if you agree. Vasco Vasco, I think this clears it up for me, but I wonder if the direction is reversed. If we take the ds that points west and rotate it anti-clockwise, won't the point P will travel eastward and go to infinity after a rotation of Pi, when ds points east? If so, then E' on the right should be drawn inside (below) L' and the angle would be in the same orientation. After a total rotation of Pi, E' must flip to remain on the same side of L' as ds. But then wouldn't this create a new h-line (circle) that doesn't intersect the horizon? Is this a multifunction with two regions of conformality? I'll have to plot some possibilities. My diagrams are not sufficient. But it does seem to me that something is wrong with [36a]. Gary Gary Yes, I accidentally reversed east and west (not a good idea). I've corrected in red. I hope the script is correct now. The original ideas were right I just somehow managed to fill it full of typos. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 10, 2016 23:35:25 GMT
Vasco, I think this clears it up for me, but I wonder if the direction is reversed. If we take the ds that points west and rotate it anti-clockwise, won't the point P will travel eastward and go to infinity after a rotation of Pi, when ds points east? If so, then E' on the right should be drawn inside (below) L' and the angle would be in the same orientation. After a total rotation of Pi, E' must flip to remain on the same side of L' as ds. But then wouldn't this create a new h-line (circle) that doesn't intersect the horizon? Is this a multifunction with two regions of conformality? I'll have to plot some possibilities. My diagrams are not sufficient. But it does seem to me that something is wrong with [36a]. Gary Gary Yes, I accidentally reversed east and west (not a good idea). I've corrected in red. I hope the script is correct now. The original ideas were right I just somehow managed to fill it full of typos. Vasco Vasco, That's actually a relief to know I'm not thinking in reverse. I'm in no position to criticize typos arising from reflections, inversions, or rotations. Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 16, 2016 19:12:20 GMT
Vasco, In the thread above, you said I am linking a document with source code that attempts to do that. I am a little disappointed that it contains a small error, as explained in the document. nh.ch6.fig36a.p318.pdf (113.13 KB) nh.ch6.fig36a.p318.src.pdf (118.57 KB) Gary |
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Post by Admin on Nov 16, 2016 20:38:18 GMT
Gary
I will study these two documents and get back to you. On a first quick reading of the maths document I notice that you say you are surprised that Neeedham is using the metric associated with the PUHP, and that you have labelled your figure 1 as being the PD. I don't see it that way - figure 1 for me is the PUHP.
Subsection 10 on pages 315-319 is about using a Möbius transformation to map the PUHP to the PD. At the end of paragraph 4 on page 317 just under (43) Needham says that because $D$ is conformal the construction (43) for the PUHP (figure 36a) is also valid in the PD (figure 36b) which is the mapping of 36a using $D$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 16, 2016 20:45:51 GMT
Gary I will study these two documents and get back to you. On a first quick reading of the maths document I notice that you say you are surprised that Neeedham is using the metric associated with the PUHP, and that you have labelled your figure 1 as being the PD. I don't see it that way - figure 1 for me is the PUHP. Subsection 10 on pages 315-319 is about using a Möbius transformation to map the PUHP to the PD. At the end of paragraph 4 on page 317 just under (43) Needham says that because $D$ is conformal the construction (43) for the PUHP (figure 36a) is also valid in the PD (figure 36b) which is the mapping of 36a using $D$. Vasco Vasco, Yes, I realized that it's the PUHP. I think I said that in the discussion, but perhaps I need to revise the beginning. Gary |
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Post by Admin on Nov 17, 2016 16:29:13 GMT
Vasco, In the thread above, you said I am linking a document with source code that attempts to do that. I am a little disappointed that it contains a small error, as explained in the document. View AttachmentView AttachmentGary Gary I looked at your documents, but I have not yet studied the code as I find it difficult due to unfamiliarity with the language and it takes me a long time to make sense of it. I did your construction on paper with a compass and ruler etc and found that the dotted $E$ lines do indeed not pass through the ends of $ds$. Making the angle $d\widetilde{s}$ smaller and smaller so that the disc around the fixed point $z$ is smaller and smaller seems to improve the situation. How do you calculate the rotation? Are you using a certain elliptic Möbius transformation with fixed points at your chosen values for $z$ and $\bar{z}$? Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 17, 2016 21:10:59 GMT
Vasco, In the thread above, you said I am linking a document with source code that attempts to do that. I am a little disappointed that it contains a small error, as explained in the document. Gary Gary I looked at your documents, but I have not yet studied the code as I find it difficult due to unfamiliarity with the language and it takes me a long time to make sense of it. I did your construction on paper with a compass and ruler etc and found that the dotted $E$ lines do indeed not pass through the ends of $ds$. Making the angle $d\widetilde{s}$ smaller and smaller so that the disc around the fixed point $z$ is smaller and smaller seems to improve the situation. How do you calculate the rotation? Are you using a certain elliptic Möbius transformation with fixed points at your chosen values for $z$ and $\bar{z}$? Vasco Vasco, Thank you for looking at it. I didn't intend that you should have to review the code. I just posted it for anyone who might have the interest. But how interesting that the mechanical construction found the same result as the computer construction! The rotations of L are (now) done by rotating ds by Pi/5 (with $(ds e^{i Pi/5}$) and finding the center of the new h-line L' by solving for the intersection of the extension of ds (at z) and the horizon. This is done in the function hLnAtZInMap[z, ds]. The line that solves for the center is: $\hspace{5em}$Solve[{x, y} $\epsilon$ InfiniteLine[v @ {z, z+ds}] && {x, y} $\epsilon$ InfiniteLine[v @ {-1, 1}], {x, y}]; where "v @ z" is equivalent to v[z] = {Re[z], Im[z]}. In this case, for example, it is applied to the list of two complex numbers {-1, 1} converting it to two vectors {{-1, 0}, {1, 0}}. I did not use any Möbius transformation, unless you regard the rotation itself as such. I have revised the documents a little to add this explanation of the rotations and simplify the function hLnAtZInMap[], but the result is the same. Gary |
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Post by Admin on Nov 18, 2016 6:57:02 GMT
Gary
The reason I asked about the Möbius transformation is that I was thinking that if you just rotate the whole PUHP about the point $z$ then this is not an h-rotation. In your program you are not actually transforming the h-line and e-line with an h-rotation, but rotating $ds$ and then drawing the lines to fit. I think this may be the explanation.
If this is the explanation, then the discrepancies should reduce as the radius $ds$ of the infinitesimal circle at $z$ gets smaller. This also means that the angle $d\widetilde{s}$ gets smaller of course.
Moving $z$ further away from the horizon while keeping $ds$ fixed might also be interesting.
I presume it would be easy to try this with your program?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 18, 2016 16:20:12 GMT
Gary The reason I asked about the Möbius transformation is that I was thinking that if you just rotate the whole PUHP about the point $z$ then this is not an h-rotation. In your program you are not actually transforming the h-line and e-line with an h-rotation, but rotating $ds$ and then drawing the lines to fit. I think this may be the explanation. If this is the explanation, then the discrepancies should reduce as the radius $ds$ of the infinitesimal circle at $z$ gets smaller. This also means that the angle $d\widetilde{s}$ gets smaller of course. Moving $z$ further away from the horizon while keeping $ds$ fixed might also be interesting. I presume it would be easy to try this with your program? Vasco Vasco, You may be correct about finding an h-rotation rather than an Euclidean rotation. I think I should have written the original question a bit differently. I should have asked "How would one construct an h-rotation...." rather than "How would one apply an h-rotation...." When I wrote that, I had no idea about how to go about it. Now, I think I would try to find the right pair of h-lines lines for two reflections. Yet it seems that just rotating the Euclidean semi-circles about z as I have done produces the correct angles at their intersections (because the tangents at z have the correct angles). Since the rotation takes L to another h-line L', I don't see how this could be wrong. There is only one h-line that could be constructed to pass through z at that angle from L. The result of the construction is two h-lines with the correct angle at their intersection. If the rotation is a Möbius transformation, then the angle of the rotated $ds'$ with $ds$ must be the same. Then $d\widehat{s} = ds/Im(z)$ does not change in length, and the angle between $ds$ and $L'$ is shown in [32a] to remain at Pi/2. The angles between E' and L' at the horizon are all $d\widehat{s}$ at both ends of E' by construction and I think that E' must be an arc of an Euclidean circle. So I don't see the logical error, but I will think about it some more. Moving $z$ farther from the origin and the horizon appears to have no effect. There might be an effect if $z$ were moved off the y axis, but that would involve additional tinkering under the hood. nh.ch6.fig36a.p318.larger.z.pdf (73.08 KB) Gary |
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Post by Admin on Nov 18, 2016 16:33:53 GMT
Gary
Yes, the drawing of the h-line must be correct as it is perpendicular to $ds$ at $z$, but because of the finite size of $ds$ I am thinking that the E-line does not intersect $ds$ at its end point, whereas if you rotated the E-line as well as $ds$ it would be forced to go through the end point of $ds$. It would then look right even though it might be only an approximation to the true picture.
Vasco
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