Gary
GaryVasco
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Post by Gary on Sept 26, 2016 2:56:45 GMT
Vasco,
Regarding Ch 6, $\mathbb{S}$ III, subsection 3, A Conformal Map of the Pseudosphere, paragraph 2, p. 297 seems very hard to follow. I would like to suggest a paraphrase for the remainder of the paragraph beginning at sentence 4 immediately following "...separated by distance dx", but I want to make sure I am clarifying and not adding confusion. What do you think?
"$X dx$ on the pseudosphere corresponds to $dx$ on the map and the actual distance $X ds$ on the pseudosphere corresponds to apparent distance $ds$ on the map. In other words, the metric is
$\hspace{10em}d\hat{s} = X ds$."
"Since the map is conformal, an infinitesimal line segment on the pseudosphere emanating from $(x, \sigma)$ in any direction must, when mapped, be multiplied by the factor $(1/X) = e^{\sigma}$.
"See p. 284 for definitions of $d\hat{s}$ and $ds$."
Gary
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Post by Admin on Sept 28, 2016 16:20:40 GMT
Vasco, Regarding Ch 6, $\mathbb{S}$ III, subsection 3, A Conformal Map of the Pseudosphere, paragraph 2, p. 297 seems very hard to follow. I would like to suggest a paraphrase for the remainder of the paragraph beginning at sentence 4 immediately following "...separated by distance dx", but I want to make sure I am clarifying and not adding confusion. What do you think? "$X dx$ on the pseudosphere corresponds to $dx$ on the map and the actual distance $X ds$ on the pseudosphere corresponds to apparent distance $ds$ on the map. In other words, the metric is $\hspace{10em}d\hat{s} = X ds$." "Since the map is conformal, an infinitesimal line segment on the pseudosphere emanating from $(x, \sigma)$ in any direction must, when mapped, be multiplied by the factor $(1/X) = e^{\sigma}$. "See p. 284 for definitions of $d\hat{s}$ and $ds$." Gary Gary I have read the paragraph on page 297 a few times and thought about what you have suggested to paraphrase the last part of the paragraph. Firstly, why exactly do you find the original text hard to follow? Secondly, I don't think that your paraphrasing says the same thing as the original text: In my opinion it is not possible to say this, and that is why Needham writes what he writes. You could write "..the actual distance $d\widehat{s}$ corresponds to apparent distance $ds$ on the map", but this doesn't help us to arrive at the relationship between these two quantities. Needham is appealing to the ideas of the previous two chapters, in particular the "amplitwist" concept, it seems to me, in the part that begins "However, since..." Vasco. |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 28, 2016 17:37:58 GMT
Vasco, Regarding Ch 6, $\mathbb{S}$ III, subsection 3, A Conformal Map of the Pseudosphere, paragraph 2, p. 297 seems very hard to follow. I would like to suggest a paraphrase for the remainder of the paragraph beginning at sentence 4 immediately following "...separated by distance dx", but I want to make sure I am clarifying and not adding confusion. What do you think? "$X dx$ on the pseudosphere corresponds to $dx$ on the map and the actual distance $X ds$ on the pseudosphere corresponds to apparent distance $ds$ on the map. In other words, the metric is $\hspace{10em}d\hat{s} = X ds$." "Since the map is conformal, an infinitesimal line segment on the pseudosphere emanating from $(x, \sigma)$ in any direction must, when mapped, be multiplied by the factor $(1/X) = e^{\sigma}$. "See p. 284 for definitions of $d\hat{s}$ and $ds$." Gary Gary I have read the paragraph on page 297 a few times and thought about what you have suggested to paraphrase the last part of the paragraph. Firstly, why exactly do you find the original text hard to follow? Secondly, I don't think that your paraphrasing says the same thing as the original text: In my opinion it is not possible to say this, and that is why Needham writes what he writes. You could write "..the actual distance $d\widehat{s}$ corresponds to apparent distance $ds$ on the map", but this doesn't help us to arrive at the relationship between these two quantities. Needham is appealing to the ideas of the previous two chapters, in particular the "amplitwist" concept, it seems to me, in the part that begins "However, since..." Vasco. Vasco, Thank you. I have been doing some plotting, so I think I see it more clearly now. I think the reason I was having difficulty with it might have to do with line 9 of the paragraph. "an infinitesimal line-segment emanating from $(x,\sigma)$ in any direction must be multiplied by the same factor $(1/X) = e^{\sigma}$. Would it be correct to add the phrase "to obtain its value on the map" ? In other words, it was not immediately obvious to me which expressions pertained to the pseudosphere and which pertained to the map. I think I read it the same way you did, but I thought it could be more clear. I don't see why it is not possible to say "the actual distance $X ds$ on the pseudosphere corresponds to apparent distance $ds$ on the map" since the actual distance $d\hat{s} = X ds$". The terms "actual" and "apparent" were used following their usages on p. 284. Here I think they can serve as brief synonyms for "on the pseudosphere" and "on the map". When I wrote it I debated using $d\hat{s}$ versus $X ds$, but it didn't seem to matter. Perhaps your objection was to the word "corresponds", which could be read as "equals", which is obviously wrong. Maybe it was an unfortunate choice. Gary |
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Post by Admin on Sept 28, 2016 17:57:42 GMT
I can't see any problem with adding the phrase you suggest - it just clarifies.
No, it seems to me that it is something you have to prove. Just because we know that $dx$ on the map is $Xdx$ on the pseudosphere, does not mean that we can say that $d\hat{s} = X ds$ - we have to prove it. Appealing to the fact that we want a conformal(=analytic) mapping, we can say that the amplification of the mapping is $(1/X)$, the same as for $dx$. Then it follows that $d\hat{s} = X ds$ and then you can use the above phrase. It seems to me that you were using it before you had shown that it was true.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 28, 2016 18:17:11 GMT
No, it seems to me that it is something you have to prove. Just because we know that $dx$ on the map is $Xdx$ on the pseudosphere, does not mean that we can say that $d\hat{s} = X ds$ - we have to prove it. Appealing to the fact that we want a conformal(=analytic) mapping, we can say that the amplification of the mapping is $(1/X)$, the same as for $dx$. Then it follows that $d\hat{s} = X ds$ and then you can use the above phrase. It seems to me that you were using it before you had shown that it was true. Vasco, That clarifies. Thank you. It seems I have a problem with the quoting function. It often drops off part of a selection and fails to render the text properly. Gary |
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Post by Admin on Sept 28, 2016 18:29:27 GMT
No, it seems to me that it is something you have to prove. Just because we know that $dx$ on the map is $Xdx$ on the pseudosphere, does not mean that we can say that $d\hat{s} = X ds$ - we have to prove it. Appealing to the fact that we want a conformal(=analytic) mapping, we can say that the amplification of the mapping is $(1/X)$, the same as for $dx$. Then it follows that $d\hat{s} = X ds$ and then you can use the above phrase. It seems to me that you were using it before you had shown that it was true. Vasco, That clarifies. Thank you. It seems I have a problem with the quoting function. It often drops off part of a selection and fails to render the text properly. Gary Gary Not sure what happened with the quote function there. I have edited your post so that it looks OK now. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 28, 2016 21:22:58 GMT
Vasco,
The quote button malfunctioned at a time when the internet was slow, as evidenced by a sluggish cursor. That might have disrupted some formatting.
Gary
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blair
New Member
Posts: 5
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Post by blair on Apr 11, 2020 15:30:35 GMT
I realize that I am four years late for this discussion, but I also had a lot of trouble with this paragraph. It seems to me that the distance dx is defined on the map. Therefore the equivalent distance on the pseudosphere must be Xdx by the definition of the mapping. That means the two points on the pseudosphere should be (x, sigma) and (x + Xdx, sigma), as it is marked on the diagram, but unlike it says in the first sentence in the paragraph. If I am correct about that, then the rest of the paragraph then makes sense to me.
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 11, 2020 17:04:16 GMT
I realize that I am four years late for this discussion, but I also had a lot of trouble with this paragraph. It seems to me that the distance dx is defined on the map. Therefore the equivalent distance on the pseudosphere must be Xdx by the definition of the mapping. That means the two points on the pseudosphere should be (x, sigma) and (x + Xdx, sigma), as it is marked on the diagram, but unlike it says in the first sentence in the paragraph. If I am correct about that, then the rest of the paragraph then makes sense to me. blair,
I think the confusion can be at least partly cleared up by noting that $x$ is an angle (p. 296, para. 3, S1) and Xdx is a distance. I think I would have preferred to use $\theta$ in place of $x$ and write $(\theta, \sigma)$ and $(\theta + d\theta, \sigma)$ for the points on the pseudosphere.
Gary
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Post by Admin on Apr 12, 2020 6:20:44 GMT
blair
Like Gary I can see why you are confused, but I don't agree that using $\theta$ for the angle at the centre of the circle $\sigma=const.$ is a good idea because eventually the map becomes the complex plane with a point on the map being $z=x+iy=re^{i\theta}$ and so I think $\theta$ is best used in this context as it is in the rest of the book. Needham's notation using $x$ for the angle on the PS makes sense when we move to the map because $x$ on the map is just $x$ on the PS and almost all of the analysis from this point on is done on the map not the PS.
Each point on the pseudo sphere (PS) is identified by its two coordinates $(z,\sigma)$ and so it is correct in the book on the second line of the paragraph to write that the arc on the SP connects the points $(x,\sigma)$ and $(x+dx,\sigma)$. The change in the angle at the centre is $dx$ as we move from one point to the next round the circle. The distance $ds$ moved on the surface of the PS is an arc of the circle and is equal to the change in angle $(dx)$ multiplied by the radius $X$ and so is $ds=Xdx$.
On the diagram [20a] $Xdx$ represents the length of the arc of the circle. If you look carefully you will see that it is a curved line.
Vasco
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