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Post by Admin on Oct 18, 2016 17:01:31 GMT
Gary
Here is my attempt at an initially very confusing exercise. I'd be interested in your opinion.
I think the 'result', as Needham calls it, which is in italics in the book on page 309, is written in a way which seems almost designed to confuse the reader. This is because the term 'h-circle' is used in the text to mean two things:
1) The hyperbolic circle in the hyperbolic plane which is defined in the last sentence on page 301, and
2) The representation of the hyperbolic circle defined in 1) above, as a Euclidean circle in the map.
Here is a copy of the 'result' as it appears in my copy of the book (paperback, year 2000 with corrections):
Every h-circle is represented in the map by a Euclidean circle, and its h-centre is the intersection of any two h-lines orthogonal to it. Algebraically, the h-circle with h-centre $a=(x+iy)$ and h-radius $\rho$ is represented by the Euclidian circle with centre $(x+iy\cosh\rho)$ and radius $y\sinh\rho$.
In the first sentence "...its h-centre..." refers to the centre of the hyperbolic circle not the centre of the Euclidean representation of it.
Think of the h-circle represented in the map by a Euclidean circle, and the two points on this circle where it is intersected by the vertical line through its h-centre $a=x+iy$. Note: this is not the centre of the Euclidean circle.
Let's call these points $q$ at the top of the circle, and $p$ at the bottom.
If $\rho$ is the h-radius then we can use formula (35) on page 302 and write
$\mathcal{H}\{q,a\}=\rho=|\ln[\text{Im}(q)/\text{Im}(a)]|$
Since $\text{Im}(q)>\text{Im}(a)$ we can write
$\ln[\text{Im}(q)/\text{Im}(a)]=\rho$ or
$\text{Im}(q)=\text{Im}(a)e^{\rho}=ye^{\rho}$
Similarly we can write
$\mathcal{H}\{p,a\}=\rho=|\ln[\text{Im}(p)/\text{Im}(a)]|$
Since $\text{Im}(p)<\text{Im}(a)$ we can write
$\ln[\text{Im}(p)/\text{Im}(a)]=-\rho$ or
$\text{Im}(p)=\text{Im}(a)e^{-\rho}=ye^{-\rho}$
It follows that the centre of the Euclidean circle is at
$x+i[\text{Im}(p)+\text{Im}(q)]/2=x+iy(e^{\rho}+e^{-\rho})/2=x+iy\cosh\rho$.
The radius of the Euclidean circle is by definition equal to
$[\text{Im}(q)-\text{Im}(p)]/2=y(e^{\rho}-e^{-\rho})/2=y\sinh\rho$.
Vasco
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Post by Admin on Oct 23, 2016 7:58:16 GMT
Gary
I now take back what I wrote in the above post. The confusion was all on my side. Here is a clean version of my solution to the suggested exercise:
First, here is a copy of the 'result' plus suggested exercise, as it appears in my copy of the book (paperback, year 2000 with corrections):
Every h-circle is represented in the map by a Euclidean circle, and its h-centre is the intersection of any two h-lines orthogonal to it. Algebraically, the h-circle with h-centre $a=(x+iy)$ and h-radius $\rho$ is represented by the Euclidian circle with centre $(x+iy\cosh\rho)$ and radius $y\sinh\rho$.
Think of the h-circle represented in the map by a Euclidean circle, and the two points on the Euclidean circle where it is intersected by the vertical line through $a=x+iy$. Note: this is not the centre of the Euclidean circle.
Let's call these points $q$ at the top of the circle, and $p$ at the bottom.
If $\rho$ is the h-radius then we can use formula (35) on page 302 and write
$\mathcal{H}\{q,a\}=\rho=|\ln[\text{Im}(q)/\text{Im}(a)]|$
Since $\text{Im}(q)>\text{Im}(a)$ we can write
$\ln[\text{Im}(q)/\text{Im}(a)]=\rho$ or
$\text{Im}(q)=\text{Im}(a)e^{\rho}=ye^{\rho}$
Similarly we can write
$\mathcal{H}\{p,a\}=\rho=|\ln[\text{Im}(p)/\text{Im}(a)]|$
Since $\text{Im}(p)<\text{Im}(a)$ we can write
$\ln[\text{Im}(p)/\text{Im}(a)]=-\rho$ or
$\text{Im}(p)=\text{Im}(a)e^{-\rho}=ye^{-\rho}$
It follows that the centre of the Euclidean circle is at
$x+i[\text{Im}(p)+\text{Im}(q)]/2=x+iy(e^{\rho}+e^{-\rho})/2=x+iy\cosh\rho$.
The radius of the Euclidean circle is by definition equal to
$[\text{Im}(q)-\text{Im}(p)]/2=y(e^{\rho}-e^{-\rho})/2=y\sinh\rho$.
Vasco
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Oct 24, 2016 14:31:11 GMT
Gary I now take back what I wrote in the above post. The confusion was all on my side. Here is a clean version of my solution to the suggested exercise: First, here is a copy of the 'result' plus suggested exercise, as it appears in my copy of the book (paperback, year 2000 with corrections): Every h-circle is represented in the map by a Euclidean circle, and its h-centre is the intersection of any two h-lines orthogonal to it. Algebraically, the h-circle with h-centre $a=(x+iy)$ and h-radius $\rho$ is represented by the Euclidian circle with centre $(x+iy\cosh\rho)$ and radius $y\sinh\rho$.Think of the h-circle represented in the map by a Euclidean circle, and the two points on the Euclidean circle where it is intersected by the vertical line through $a=x+iy$. Note: this is not the centre of the Euclidean circle. Let's call these points $q$ at the top of the circle, and $p$ at the bottom. If $\rho$ is the h-radius then we can use formula (35) on page 302 and write $\mathcal{H}\{q,a\}=\rho=|\ln[\text{Im}(q)/\text{Im}(a)]|$ Since $\text{Im}(q)>\text{Im}(a)$ we can write $\ln[\text{Im}(q)/\text{Im}(a)]=\rho$ or $\text{Im}(q)=\text{Im}(a)e^{\rho}=ye^{\rho}$ Similarly we can write $\mathcal{H}\{p,a\}=\rho=|\ln[\text{Im}(p)/\text{Im}(a)]|$ Since $\text{Im}(p)<\text{Im}(a)$ we can write $\ln[\text{Im}(p)/\text{Im}(a)]=-\rho$ or $\text{Im}(p)=\text{Im}(a)e^{-\rho}=ye^{-\rho}$ It follows that the centre of the Euclidean circle is at $x+i[\text{Im}(p)+\text{Im}(q)]/2=x+iy(e^{\rho}+e^{-\rho})/2=x+iy\cosh\rho$. The radius of the Euclidean circle is by definition equal to $[\text{Im}(q)-\text{Im}(p)]/2=y(e^{\rho}-e^{-\rho})/2=y\sinh\rho$. Vasco Vasco, That all looks good and it clears up some of my confusion over h-circles and Euclidean circles. I'm still a little puzzled by p. 309, line 5, "while in the map we see $\tilde{p}$ miraculously tracing out a Euclidean circle!". How would we know it was Euclidean, unless we had first plotted the center? Then it might seem miraculous. I am now making the following distinctions: (1) Euclidean circles, which appear only on Euclidean planes, (2) hyperbolic lines, which appear on the map as semicircles centered on the horizon or as half lines, (3) h-circles, which appear on the map looking like Euclidean circles with offset centers (e.g. [27]), and (4) horocycles, which are Euclidean circles (or lines) that miraculously turn into something called "horocycle" (in the h-plane ?on the map) at the limit (p. 309). Do you think this is correct? Gary |
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Post by Admin on Oct 24, 2016 15:21:50 GMT
Gary
I think it's because you wouldn't expect a circle in the h-plane to be a circle in the map - would you? I certainly wouldn't! But we find that the non-concentric circles in figure 27 on page 308 are concentric circles in the h-plane.
I'll get back to you about your other questions as soon as I can?
Vasco
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Post by Admin on Oct 24, 2016 16:32:45 GMT
Gary
Yes I agree with all this and I would say "horocycles in the h-plane appear as circles in the map or as lines parallel to the horizon (which can be considered as circles centred at $y=\infty$).
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 24, 2016 16:33:07 GMT
Gary
I would like to revise what I wrote above:
I am now making the following distinctions: (1) Euclidean circles, which appear only on Euclidean planes, (2) hyperbolic lines, which appear on the map as semicircles centered on the horizon or as half lines, (3) h-circles, which appear on the map looking like Euclidean circles with offset centers (e.g. [27]), and (4) horocycles, which are Euclidean circles (or lines) that miraculously turn into something called "horocycle" (in the h-plane ?on the map) at the limit (p. 309). Do you think this is correct?
I would change this to:
(1) Euclidean circles, which appear only on Euclidean planes, such as on the hyperbolic map, (2) hyperbolic lines, which appear on the map as semicircles centered on the horizon or as half lines, (3) h-circles, which appear on the map looking like Euclidean circles, but appear on the hyperbolic plane as circles with offset centers (e.g. [27]), and (4) horocycles, which are Euclidean circles (or lines) on the map that appear at the limit as "horocycles" in the h-plane (p. 309). Do you think this is correct?
Gary
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Post by Admin on Oct 24, 2016 16:40:42 GMT
Gary
No. I see [27] as circles in the map as in your original post.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 25, 2016 3:16:10 GMT
Vasco,
I have reread the above and I find that I don't know what it means. If the h-circle appears as an Euclidean circle in the map, where can one view the h-circle? Have you been able to plot these two circles using these formulae?
Gary
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Post by Admin on Oct 25, 2016 6:04:24 GMT
Vasco, I have reread the above and I find that I don't know what it means. If the h-circle appears as an Euclidean circle in the map, where can one view the h-circle? Have you been able to plot these two circles using these formulae? Gary Gary As you probably guessed, I struggled with this for a while before I was able to do the suggested exercise. I think it's impossible to do the exercise unless you understand this, which is a summary of the argument in subsection 7 up to that point. I will write a small document/commentary which I hope will explain it as I finally managed to see it, and which enabled me to do the exercise. In the meantime if you want to do some plotting: choose a value for $a=x+iy$ and keep it fixed while you then choose a series of values for $\rho$ so that you have a set of concentric circles centred at $a$. The value for $x$ is not important so you can choose $x=0$. Now calculate the series of values $c=x+iy\cosh(\rho)$ and $r=y\sinh(\rho)$ for each value of $\rho$ and plot the circles with these centres and radii. You could if you wished plot the concentric circles and match them up using colours. The concentric circles really lie on the surface of the pseudosphere, just as the two shaded circles in figure 14b of chapter 6 lie on the sphere. Vasco Here is my commentary. |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 25, 2016 14:55:46 GMT
Vasco, I have reread the above and I find that I don't know what it means. If the h-circle appears as an Euclidean circle in the map, where can one view the h-circle? Have you been able to plot these two circles using these formulae? Gary Gary As you probably guessed, I struggled with this for a while before I was able to do the suggested exercise. I think it's impossible to do the exercise unless you understand this, which is a summary of the argument in subsection 7 up to that point. I will write a small document/commentary which I hope will explain it as I finally managed to see it, and which enabled me to do the exercise. In the meantime if you want to do some plotting: choose a value for $a=x+iy$ and keep it fixed while you then choose a series of values for $\rho$ so that you have a set of concentric circles centred at $a$. The value for $x$ is not important so you can choose $x=0$. Now calculate the series of values $c=x+iy\cosh(\rho)$ and $r=y\sinh(\rho)$ for each value of $\rho$ and plot the circles with these centres and radii. You could if you wished plot the concentric circles and match them up using colours. The concentric circles really lie on the surface of the pseudosphere, just as the two shaded circles in figure 14b of chapter 6 lie on the sphere. Vasco Here is my commentary.Vasco, Thank you. I did not realize that the hyperbolic part was still imagined on a hyperbolic plane (e.g. the tractrix) apart from the map. One would like to plot the concentric circles on the tractrix, but plotting curves on the tractrix seems to be tricky. At least I can visualize the zones of construction. If I understand it correctly now, a set of concentric circles on the tractrix will, on the map, look something like [27], p. 308. From the quote on p. 308, I gather that the circles on the map in [27] are Euclidean circles that represent h-circles (which would appear on the tractrix). Their h-centre is the point of intersection of the orthogonal h-lines (on both the tractrix and the map). Their Euclidean centres are not shown. Do you agree? Gary |
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Post by Admin on Oct 25, 2016 15:05:56 GMT
Gary
I agree.
Vasco
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