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Post by Admin on Oct 8, 2016 13:34:13 GMT
Gary
NB When I write 'you' and 'yours' below I am just following the book.
Your speed on the pseudosphere is given as $\ln2$, and so we can write
$d\widehat{s}/dt=-\ln2$
The negative sign is because $\widehat{s}$ decreases as t increases in the case we are considering.
Since in this situation (walking down the tractrix of the pseudosphere) $d\widehat{s}=dy/y$, the above differential equation becomes
$dy/y=-\ln2dt$
Integrating this becomes
$\ln y=-t\ln2+constant$, and since $y=2$ when $t=0$ then the constant is $\ln2$ and so
$\ln(y/2)=-t\ln 2=\ln 2^{-t}$ and so
$y=2\cdot 2^{-t}$
So when $t=1$ $y=1$, when $t=2$ $y=(1/2)$, and when $t=3$ $y=(1/4)$.
Vasco
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Post by mondo on Mar 18, 2023 20:30:20 GMT
Vasco,
I have a few questions to this.
1. Figure 21 is all about a map, an euclidean space so why $d\hat{s}$ in your velocity formula and not just $ds$? Similarly in you solution you talk about walking down the tractrix.
2. In the differential equation, why dy/dy? Why this particular fraction and why $y$ which is not defined on a pseudosphere?
3. This is in regards to exercise 1. author says we will never get to point q because "to me you appear to shrink". Well appearances may be deceiving.. Disk may appear smaller and small but it remains the same size. Hence the conclusion that we never reach $q$ is not true right?
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Post by Admin on Mar 19, 2023 15:21:52 GMT
Mondo
1. I am walking on [21], but not when it is considered as a euclidean plane. So I have to calculate distances as they would be on the pseudosphere by using the metric $d\widehat{s}=ds/y$, which is (31) on page 298.
Notice that on page 299, in the last paragraph, the book explains why I am effectively walking down a tractrix.
2. The exercise asks us to calculate values on the map [21] when it is considered to be a hyperbolic plane.
3. No, I will never reach $q$ as you can see from the calculation and figure 21. If we set $y=0$ in the solution of the differential equation we can see that this can only happen as $t\rightarrow\infty$ and so we will never arrive there!
Vasco
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Post by mondo on Mar 19, 2023 23:32:34 GMT
Thank you Vasco,
1. Clear.
2. This is still confusing for me. Author never says what [21] is but to me it looks like a map of some pseudosphere right? If so your comment "[21] when it is considered to be a hyperbolic plane" does not make sense? Also, author says in the middle of page 300, "As illustrated, integration of $\frac{dy}{y}$ shows.." I wonder how was this differential equation constructed, why $\frac{dy}{y}$ and not say $ydy$?
3. Yes it makes sense form the solution of differential equation but logically it doesn't make sense. The conclusion that we never reach $q$ is based on the fact that the diameter of the disc shrinks as we move towards that point but even Author says at the beginning of p.300 "Of course you remain the same hyperbolic size as you walk, but to me you appear to shrink". So either it is an impression only or the size is actually shrinking?
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Post by Admin on Mar 21, 2023 14:52:43 GMT
Mondo Thank you Vasco, 1. Clear. 2. This is still confusing for me. Author never says what [21] is but to me it looks like a map of some pseudosphere right? If so your comment "[21] when it is considered to be a hyperbolic plane" does not make sense? Also, author says in the middle of page 300, "As illustrated, integration of $\frac{dy}{y}$ shows.." I wonder how was this differential equation constructed, why $\frac{dy}{y}$ and not say $ydy$? [21] is the Beltrami hyperbolic plane and is essentially a map of the pseudosphere. So the distances on this map can be measured as euclidean distances or as hyperbolic distances. Euclidean distances can be measured with a ruler, whereas hyperbolic distances must be measured using the metric $d\widehat{s}=ds/y$. Since $d\widehat{s}=ds/y$, if I walk down a tractrix then $ds=dy$ and so the distance walked on the pseudosphere, hyperbolic distance, is $\int d\widehat{s}=\int (dy/y)$ No, it is based on the fact that the hyperbolic distance on the map (hyperbolic plane) from any point where $y>0$ to a point where $y=0$ is infinite. This is after we have made the extensions to the hyperbolic plane as described in subsection 4 on pages 298 - 301. Vasco |
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Post by Admin on Mar 21, 2023 21:58:24 GMT
Mondo
I just edited reply #4 above to correct 2 typos in the definition of the metric. I have highlighted the corrections in red.
Vasco
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Post by mondo on Mar 22, 2023 6:51:26 GMT
Thank you Vasco, well explained! Mondo No, it is based on the fact that the hyperbolic distance on the map (hyperbolic plane) from any point where $y>0$ to a point where $y=0$ is infinite. This is after we have made the extensions to the hyperbolic plane as described in subsection 4 on pages 298 - 301. I think the proof of this fact is actually a subject of exercise #2. However it is hard to accept it since it relies on the shrinking size of the traveling object which is surreal. Yes it is prover by both exercises on page 300 but logically hard to accept. |
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Post by Admin on Mar 22, 2023 15:26:29 GMT
Mondo
It isn't hard to accept, it's extremely easy to see.
Think of yourself looking down on the earth from a large distance where you see the earth as a flat disc. If you now draw a map of the curved hemisphere as you see it, onto the flat disc of your map, the straight line distances on the map will not be the same as the distances along the curved surface of the actual earth.
This is exactly what we are doing in the case of the hyperbolic plane, but not with a sphere. In the case of the sphere of the earth, since the distances on the earth are longer than the distances on the map then a person walking down from the north pole to the south pole at a constant speed would appear to you looking at the map to be walking faster and faster as he approached the equator and then slower and slower as he approached the south pole.
This is exactly the same thing as what we have been discussing in the case of the hyperbolic plane, and it makes perfect sense because as he approaches the south pole the distances on the map are much shorter than the distances on the earth's surface.
It is completely logical.
Vasco
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Post by mondo on Mar 23, 2023 18:28:07 GMT
Yes, the sphere example makes perfect sense to me. However on the sphere we never claim infinite distances as we do on the pseudosphere.. I have a few more questions that I hope will help me understand this:
1. Why our pseudosphere rim is located at $y = 1$ instead of $y = 0$ which is more natural to position objects on?
2. On top of page 298, for the fist time author describes the vanishing process of a traveling object on the surface of pseudosphere. I see that the map on [20b] is "contructed" mainly on the relation between $x$ (an angle on the pseudosphere axis) and $x$ a plain distance on the map. Now, as the disk moves down the tractrix it rotates in regards to the axis and in the limit becomes completely flat -> a circular disk became a line segment. However, the line segment should still maintain the length of the disk diameter but on the figure [20b] we see it is shrinking and shrinking until it completely vanish. This is all based on the $\epsilon/X$ from the last paragraph of page 297. Meaning we are moving further and further away from the axis, hence $X$ is bigger and bigger and hence on the map we see that the disk in varying in size depending on its distance from the axis on pseudosphere. I get that but again the way we define the mapping is counterintuitive because in actuality the disk won't disappear but will rather become a line segment as I said earlier.
I think the map is defined in this way to have a single function (conformal) describing motions on the pseudosphere but I don't see what we gain by that other than a confusion. For example, let's say we gave somebody the map shown on page [20b] and asked a question what happens to the disk as we approach the rim of the pseudosphere? That person would like answer that the disk is shrinking at a constant rate but this is of course not what happens on the pseudosphere because there we just slide down the exact same disk, we don't change it's dimensions at all.
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Post by mondo on Mar 23, 2023 18:36:14 GMT
Just updated my response above. In general I can "adjust" my thinking to the methodology presented in the book but I barely see a point in doing it this way. Maybe I need to reread the whole chapter again..
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Post by Admin on Mar 27, 2023 16:16:05 GMT
Mondo Yes, the sphere example makes perfect sense to me. However on the sphere we never claim infinite distances as we do on the pseudosphere.. I have a few more questions that I hope will help me understand this: 1. Why our pseudosphere rim is located at $y = 1$ instead of $y = 0$ which is more natural to position objects on? 2. On top of page 298, for the fist time author describes the vanishing process of a traveling object on the surface of pseudosphere. I see that the map on [20b] is "contructed" mainly on the relation between $x$ (an angle on the pseudosphere axis) and $x$ a plain distance on the map. Now, as the disk moves down the tractrix it rotates in regards to the axis and in the limit becomes completely flat -> a circular disk became a line segment. However, the line segment should still maintain the length of the disk diameter but on the figure [20b] we see it is shrinking and shrinking until it completely vanish. This is all based on the $\epsilon/X$ from the last paragraph of page 297. Meaning we are moving further and further away from the axis, hence $X$ is bigger and bigger and hence on the map we see that the disk in varying in size depending on its distance from the axis on pseudosphere. I get that but again the way we define the mapping is counterintuitive because in actuality the disk won't disappear but will rather become a line segment as I said earlier. I think the map is defined in this way to have a single function (conformal) describing motions on the pseudosphere but I don't see what we gain by that other than a confusion. For example, let's say we gave somebody the map shown on page [20b] and asked a question what happens to the disk as we approach the rim of the pseudosphere? That person would like answer that the disk is shrinking at a constant rate but this is of course not what happens on the pseudosphere because there we just slide down the exact same disk, we don't change it's dimensions at all. 1. By choosing a different value for the constant when we solve the differential equation $\frac{dy}{d\sigma}=e^{\sigma}$ we can set the rim at any value we like. 2, I'm not sure what you are saying here. You need to make it clear whether you are referring to the pseudosphere or the map ( hyperbolic plane). The disc size remains the same on the pseudosphere but on the map it gets smaller and smaller as the disc on the pseudosphere gets nearer to the edge . Since the mapping is conformal the discs cannot map to a line segment. All infinitesimal discs are mapped to infinitesimal discs. A person outside the pseudosphere/hyperbolic plane looking at the map sees the discs as shrinking. Vasco |
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Post by Admin on Mar 27, 2023 20:16:49 GMT
Mondo
I have deleted 2 words above, They are shown in red.
Vasco
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Post by Admin on Mar 27, 2023 20:49:59 GMT
Mondo
To summarise where we are on pages 300-301, at the end of subsection 4 Beltrami's Hyperbolic Plane of section III Hyperbolic Geometry, we have found a map (figure 21) which together with the metric (31) on page 298 represents the hyperbolic plane as it is to inhabitants of the hyperbolic plane. Without the metric it is a particular map (Poincare' upper half-plane) of what the hyperbolic plane looks like to someone outside the hyperbolic plane. Other maps of the hyperbolic plane are the Poincare' disc and the Klein disc with different metrics, and which are explained on later pages of the chapter.
Vasco
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