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Post by Admin on Jun 23, 2016 5:39:31 GMT
Vasco, I agree, but what does that imply? Can we say that we can bend the plane without stretching or compressing so long as we leave the plane unbent in one direction? By the way, did you notice my suggestion that the right hand side shape in Figure [4], p. 273 is a segment of an elliptic cylinder? Gary Gary I see the discussion in subsection 4 on page 273 to be about convincing us that it cannot be the precise shape of the surface in space that causes $E(T_1)$ and $E(T_2)$ to differ from their Euclidean value $E=0$. We can see that this is the case by showing that in general it is possible to bend the surface without changing these values. $E$ is governed by intrinsic (not extrinsic) curvature. Vasco PS I did notice your suggestion. I am still thinking about it. |
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Post by Admin on Jun 23, 2016 6:38:07 GMT
Gary In item 8 of your document you write Since $\Lambda$ is the same whatever the direction of $d\widehat{s}$ then we can choose the $d\widehat{s}$ which is parallel to $\mathbb{C}$ and this enables us to derive the result $d\widehat{s}=\frac{2}{[Nz]^2}ds$ without using (6) on page 126. From reading your other comments, I get the impression that you have misunderstood what Needham has written at the top of page 286. Vasco Vasco, I'm not sure I misunderstood it. I just didn't see the solution. I have revised. View AttachmentGary Gary I still don't understand your solution to this exercise in the text. In your version of [13b] in your document you have drawn $d\widehat{s}$ parallel to $\mathbb{C}$, but in the plane defined by $Nz0$. As I see it when $d\widehat{s}$ is parallel to $\mathbb{C}$ it must remain on the surface of the sphere $\Sigma$. It would therefore have to point along the circle through $\widehat{z}$ parallel to $\mathbb{C}$ and similarly for $ds$, which must point along the circle through $z$ in the plane of $\mathbb{C}$. All this can be seen by referring to [13a]. Also I think that we are then required to prove the relationship between $ds$ and $d\widehat{s}$ in its original form as $d\widehat{s}=\frac{2}{[Nz]^2}ds$, before using Pythagoras to substitute for $Nz$. Vasco |
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Gary
GaryVasco
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Post by Gary on Jun 23, 2016 15:40:22 GMT
I still don't understand your solution to this exercise in the text. In your version of [13b] in your document you have drawn $d\widehat{s}$ parallel to $\mathbb{C}$, but in the plane defined by $Nz0$. As I see it when $d\widehat{s}$ is parallel to $\mathbb{C}$ it must remain on the surface of the sphere $\Sigma$. It would therefore have to point along the circle through $\widehat{z}$ parallel to $\mathbb{C}$ and similarly for $ds$, which must point along the circle through $z$ in the plane of $\mathbb{C}$. All this can be seen by referring to [13a]. Also I think that we are then required to prove the relationship between $ds$ and $d\widehat{s}$ in its original form as $d\widehat{s}=\frac{2}{[Nz]^2}ds$, before using Pythagoras to substitute for $Nz$. Vasco Vasco, I was just adding a parenthetic note in the document that one had to imagine $ds$ and $d\hat{s}$ pointing out of the plane of Figure [13b] and then I saw your reply. I agree that it has to remain on or near the surface of $\Sigma$. $ds = |z|$, so $ds$ and $d\hat{s}$ would have to be tangents or secants. Secants would probably be more in the style of Newton. Once we have drawn them that way, we have triangles that share two angles on a plane that is either tangent to $\Sigma$ or cuts through it, so the triangles must be similar and that leads to the proof. I see your point about first proving the form containing $[Nz]^2$ without using (6). I'll think about that. Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 23, 2016 20:27:56 GMT
I still don't understand your solution to this exercise in the text. In your version of [13b] in your document you have drawn $d\widehat{s}$ parallel to $\mathbb{C}$, but in the plane defined by $Nz0$. As I see it when $d\widehat{s}$ is parallel to $\mathbb{C}$ it must remain on the surface of the sphere $\Sigma$. It would therefore have to point along the circle through $\widehat{z}$ parallel to $\mathbb{C}$ and similarly for $ds$, which must point along the circle through $z$ in the plane of $\mathbb{C}$. All this can be seen by referring to [13a]. Also I think that we are then required to prove the relationship between $ds$ and $d\widehat{s}$ in its original form as $d\widehat{s}=\frac{2}{[Nz]^2}ds$, before using Pythagoras to substitute for $Nz$. Vasco Vasco, I was just adding a parenthetic note in the document that one had to imagine $ds$ and $d\hat{s}$ pointing out of the plane of Figure [13b] and then I saw your reply. I agree that it has to remain on or near the surface of $\Sigma$. $ds = |z|$, so $ds$ and $d\hat{s}$ would have to be tangents or secants. Secants would probably be more in the style of Newton. Once we have drawn them that way, we have triangles that share two angles on a plane that is either tangent to $\Sigma$ or cuts through it, so the triangles must be similar and that leads to the proof. I see your point about first proving the form containing $[Nz]^2$ without using (6). I'll think about that. Gary Vasco, I have revised the document to respond to your comment on the proof without using (6). It is actually the same proof, but the order of presentation is now what Needham requested. I also added a version of [13a] with $d\hat{s}$ parallel to $\mathbb{C}$ and one additional exercise from p. 286-87. nh.ch6.notes.questions.pdf (551.2 KB) Gary |
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Post by Admin on Jun 27, 2016 16:25:49 GMT
Vasco, I agree, but what does that imply? Can we say that we can bend the plane without stretching or compressing so long as we leave the plane unbent in one direction? By the way, did you notice my suggestion that the right hand side shape in Figure [4], p. 273 is a segment of an elliptic cylinder? Gary Gary I see the discussion in subsection 4 on page 273 to be about convincing us that it cannot be the precise shape of the surface in space that causes $E(T_1)$ and $E(T_2)$ to differ from their Euclidean value $E=0$. We can see that this is the case by showing that in general it is possible to bend the surface without changing these values. $E$ is governed by intrinsic (not extrinsic) curvature. Vasco PS I did notice your suggestion. I am still thinking about it. Gary I see what you mean about the right-hand diagram in figure 4 on page 273, but do you understand what the significance of the question in footnote 3 is, because I'm struggling with it? Vasco |
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