Gary
GaryVasco
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Post by Gary on Nov 11, 2016 3:47:48 GMT
Vasco, comments, question in the document [now revised after Vasco's comments and his version] nh.ch6.p.317.picture.pdf (73.79 KB) Gary |
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Post by Admin on Nov 11, 2016 8:04:53 GMT
Gary
The simplest answer to your question is that $d\widehat{s}=ds/y$ is the metric ((31) on page 298) associated with the map of the PUHP and the maths on page 318 is all about establishing the metric for the PD which is (44) on page 318: $d\widehat{s}=\frac{2}{1-|z|^2}ds$
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 11, 2016 15:33:05 GMT
Gary The simplest answer to your question is that $d\widehat{s}=ds/y$ is the metric ((31) on page 298) associated with the map of the PUHP and the maths on page 318 is all about establishing the metric for the PD which is (44) on page 318: $d\widehat{s}=\frac{2}{1-|z|^2}ds$ Vasco Vasco, Yes, I should have looked ahead a bit. By the new equation for PD, $d\widehat{s}$ depends only on $ds$ and $|z|$, so it doesn't change with rotation. It's a curious phenomenon. Gary |
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Post by Admin on Nov 12, 2016 8:48:10 GMT
Gary The simplest answer to your question is that $d\widehat{s}=ds/y$ is the metric ((31) on page 298) associated with the map of the PUHP and the maths on page 318 is all about establishing the metric for the PD which is (44) on page 318: $d\widehat{s}=\frac{2}{1-|z|^2}ds$ Vasco Vasco, Yes, I should have looked ahead a bit. By the new equation for PD, $d\widehat{s}$ depends only on $ds$ and $|z|$, so it doesn't change with rotation. It's a curious phenomenon. Gary Gary If you think about it, it would be a curious phenomenon if the metric did depend on the angle/arg of $z$ (which I am assuming is what you mean by rotation). In fact it would be impossible, because if it were the case then distances would change if you decided to measure angles from a different starting point - a very strange world indeed! Everything must look the same from the centre of the disc in any direction to someone outside the disc and things only change along a radius (as you'd expect from the metric). To a Poincarite standing at a point on the disc the world looks the same as it does to another Poincarite 1000 Poincarite miles away! If a Poincarite walks from any point towards the horizon (the unit circle) and stops, then it appears that nothing has changed. He might even join the flat earth society! Vasco |
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Gary
GaryVasco
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Post by Gary on Nov 12, 2016 18:17:52 GMT
Vasco, Yes, I should have looked ahead a bit. By the new equation for PD, $d\widehat{s}$ depends only on $ds$ and $|z|$, so it doesn't change with rotation. It's a curious phenomenon. Gary Gary If you think about it, it would be a curious phenomenon if the metric did depend on the angle/arg of $z$ (which I am assuming is what you mean by rotation). In fact it would be impossible, because if it were the case then distances would change if you decided to measure angles from a different starting point - a very strange world indeed! Everything must look the same from the centre of the disc in any direction to someone outside the disc and things only change along a radius (as you'd expect from the metric). To a Poincarite standing at a point on the disc the world looks the same as it does to another Poincarite 1000 Poincarite miles away! If a Poincarite walks from any point towards the horizon (the unit circle) and stops, then it appears that nothing has changed. He might even join the flat earth society! Vasco Vasco, I see your point when you say you would expect from the metric that things only change along a radius. When you say "Everything must look the same", what are you including? Do you mean angles and h-distances between points that undergo the same transformation? That is, all the rectangles in the hyperbolic plane would appear to have equal size and shape? But what I thought was curious was the difference between the Poincaré UHP and the Poincaré disc. On p. 308, $\mathbb{P}$ 4, in the realm of the PUHP, the Poincarites (who inhabit the hyperbolic plane) see all the rectangles as the same size and shape. Then in the next paragraph we find that the Poincarite sees the rotation of $z$ about $a$ as an h-circle, but we see it as an Euclidean circle (with a different center in most cases). We can also look at the map (Figure [27]) and see that from our vantage point, not all the "rectangles" have the same shape and size. In the PUHP, the metric is $d\widehat{s} = ds/y$ or $H\{a,b\} = |ln(\frac{Im\tilde{a}}{Im\tilde{b}})|$, so the location of $z$ is important and rotating $z$ about any point affects $Im(z)$ and $d\widehat{s}$. For a $z$ on an h-line or h-circle with radius 1 and center a, rotating $z$ to $(z-a) = i$ maximizes y without change in $|z|$. I'm not sure how that effects angles, except that an h-rotation must preserve them. Gary |
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Post by Admin on Nov 13, 2016 8:22:07 GMT
Gary
Sorry that phrase was misleading. I was just trying to emphasise that the metric is constant on any circle centred at the centre of the PD.
When Needham says on page 308 that a Poincarite sees all the black 'rectangles' as being the same size and shape I think the word 'black' here is important because it is only the 'rectangles' within each ring which are the same size to the Poincarite. The way I understand it, is that the Poincarite sees a set of concentric circles with equispaced rays coming out from the centre like [37b] on page 319 except that it fills the hyperbolic plane.
Vasco
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Gary
GaryVasco
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Post by Gary on Nov 13, 2016 16:20:26 GMT
When Needham says on page 308 that a Poincarite sees all the black 'rectangles' as being the same size and shape I think the word 'black' here is important because it is only the 'rectangles' within each ring which are the same size to the Poincarite. The way I understand it, is that the Poincarite sees a set of concentric circles with equispaced rays coming out from the centre like [37b] on page 319 except that it fills the hyperbolic plane. Vasco, It is true that in [27] and [37b] he is looking at the rings because he is discussing h-rotation, so I agree with you, but if the rings are actually "rectangles" to the Poincarites, that implies that their opposite sides are equal on the hyperbolic metric, so the inner side must equal the outer side by means of inversion in the successive circles. Then following the arrows pointing out from the center along $L_2$ or $L_1$, the rectangles must have the same size and shape. So to the Poincarites, all the rectangles have the same size and shape. But maybe I am putting too much confidence in the quoted "rectangle". Gary |
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Post by Admin on Nov 13, 2016 16:27:32 GMT
When Needham says on page 308 that a Poincarite sees all the black 'rectangles' as being the same size and shape I think the word 'black' here is important because it is only the 'rectangles' within each ring which are the same size to the Poincarite. The way I understand it, is that the Poincarite sees a set of concentric circles with equispaced rays coming out from the centre like [37b] on page 319 except that it fills the hyperbolic plane. Vasco, It is true that in [27] and [37b] he is looking at the rings because he is discussing h-rotation, so I agree with you, but if the rings are actually "rectangles" to the Poincarites, that implies that their opposite sides are equal on the hyperbolic metric, so the inner side must equal the outer side by means of inversion in the successive circles. Then following the arrows pointing out from the center along $L_2$ or $L_1$, the rectangles must have the same size and shape. So to the Poincarites, all the rectangles have the same size and shape. But maybe I am putting too much confidence in the quoted "rectangle". Gary Gary Notice that Needham puts quotes round the word rectangles and for me this is because they are not really rectangles at all but are the shapes shown in [37]. It's for want of a better word I think that he uses the word rectangle. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 13, 2016 20:55:57 GMT
Vasco, It is true that in [27] and [37b] he is looking at the rings because he is discussing h-rotation, so I agree with you, but if the rings are actually "rectangles" to the Poincarites, that implies that their opposite sides are equal on the hyperbolic metric, so the inner side must equal the outer side by means of inversion in the successive circles. Then following the arrows pointing out from the center along $L_2$ or $L_1$, the rectangles must have the same size and shape. So to the Poincarites, all the rectangles have the same size and shape. But maybe I am putting too much confidence in the quoted "rectangle". Gary Gary Notice that Needham puts quotes round the word rectangles and for me this is because they are not really rectangles at all but are the shapes shown in [37]. It's for want of a better word I think that he uses the word rectangle. Vasco Vasco, Consideration of the metric seems to bear out your conclusion. $d\widehat{s}$ increases with increasing |z|, so the outer circles (and rectangle sides) must be longer h-segments than the inner ones. Gary |
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