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Post by mondo on May 16, 2023 1:38:01 GMT
Interesting, I wonder where is the discrepancy. At the bottom of page 281 author says "..then the formula for excess implies that $\psi = \theta + \phi - kA$" which suggest this $- kA$ part comes right from the formula for angular excess but I don't get how is that involved in the rotation?
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Post by Admin on May 16, 2023 6:39:38 GMT
Mondo Interesting, I wonder where is the discrepancy. At the bottom of page 281 author says "..then the formula for excess implies that $\psi = \theta + \phi - kA$" which suggest this $- kA$ part comes right from the formula for angular excess but I don't get how is that involved in the rotation? The point is that the angles of a triangle add up to $\pi$ in the Euclidean plane, but they do not add up to $\pi$ on the sphere. Vasco |
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Post by mondo on May 16, 2023 7:01:43 GMT
Mondo Interesting, I wonder where is the discrepancy. At the bottom of page 281 author says "..then the formula for excess implies that $\psi = \theta + \phi - kA$" which suggest this $- kA$ part comes right from the formula for angular excess but I don't get how is that involved in the rotation? The point is that the angles of a triangle add up to $\pi$ in the Euclidean plane, but they do not add up to $\pi$ on the sphere. Vasco Yes but I assume this diverge from $\pi$ is already included in $\phi$ and $\theta$ and what we now do is we rotate so these angles should not change upon this action, right? |
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Post by Admin on May 16, 2023 7:48:04 GMT
Mondo
No not right. Why do you assume that? Those angles in figure 10 do not add up to $\pi$.
Vasco
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Post by mondo on May 16, 2023 17:13:28 GMT
I agree, that is the essence of spherical geometry. But why do we care what is the angle sum in a triangle when all we do is rotate around points $p$ and $q$? Maybe my problem is here - I don't quite get what do we really rotate. The notation $R_N \cdot R_M$ suggest me we rotate a great circle $M$ around $N$ by some angle. Few posts above you mentioned we rotate the whole sphere what suggest that? Neither equation, description nor figure [10a] suggest that.
Thank you.
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Post by Admin on May 16, 2023 18:54:23 GMT
Mondo
From what you have written above in reply #19 it is clear to me again that you do not understand anything you have read in subsection 2 on pages 279-292. You say you understand it but you very clearly do not understand it at all.
Please read it all again and study the figures 8,9 and 10 together with the text.
Here is one example that proves you do not understand these pages:
Under figure 8 the book says "The obvious example of a direct motion is a rotation of the sphere about an axis $V$ passing through the centre."
This is how I know that the sphere is being rotated about an axis through its centre. This is so clear I don't understand how you didn't see this.
Vasco
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Post by Admin on May 16, 2023 19:30:01 GMT
Mondo
You may find that it helps if you make sure you understand subsection 3 of chapter 1 on pages 37-39, since this explains the same basic ideas in the Euclidean plane, that is, the idea of using a combination of reflections in lines to create rotations in the plane.
In chapter 6 reflections in lines are used to create rotations of the sphere.
Vasco
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Post by mondo on May 17, 2023 6:58:16 GMT
OK, I think I finally got that. The crucial part was to realize on figure 10a we are rotating the whole sphere around point $r$. Previously I somehow assumed (probably due to arrows showing rotation of triangles sides) that we rotate sides $K,L,M$ of that white triangle. So now the subtraction of $-kA$ also makes sense as we need to adjust the rotation with angular excess formed by that "triangular" rotation.
Thank you Vasco.
PS: Would you be able to address now my questions from post #5? They are much more interesting I guess.
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Post by Admin on May 17, 2023 15:08:24 GMT
OK, I think I finally got that. The crucial part was to realize on figure 10a we are rotating the whole sphere around point $r$. Previously I somehow assumed (probably due to arrows showing rotation of triangles sides) that we rotate sides $K,L,M$ of that white triangle. So now the subtraction of $-kA$ also makes sense as we need to adjust the rotation with angular excess formed by that "triangular" rotation. Thank you Vasco. PS: Would you be able to address now my questions from post #5? They are much more interesting I guess. Mondo Again your comments here convince me that you still do not understand this important issue. Vasco |
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Post by Admin on May 17, 2023 15:42:11 GMT
Mondo
To answer your first question in Reply #5: if you draw a triangle on a sphere and phone someone and tell them the 3 angles of the triangle and they draw a triangle with the same angles then their triangles will be exactly the same (congruent), whereas in the Euclidean plane the two triangles would generally not be the same (congruent).
Vasco
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Post by mondo on May 17, 2023 17:27:49 GMT
OK, I think I finally got that. The crucial part was to realize on figure 10a we are rotating the whole sphere around point $r$. Previously I somehow assumed (probably due to arrows showing rotation of triangles sides) that we rotate sides $K,L,M$ of that white triangle. So now the subtraction of $-kA$ also makes sense as we need to adjust the rotation with angular excess formed by that "triangular" rotation. Thank you Vasco. PS: Would you be able to address now my questions from post #5? They are much more interesting I guess. Mondo Again your comments here convince me that you still do not understand this important issue. Vasco Can you point out which part of my response suggests you I missed the important issue? |
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Post by Admin on May 17, 2023 18:09:47 GMT
Mondo Again your comments here convince me that you still do not understand this important issue. Vasco Can you point out which part of my response suggests you I missed the important issue? Mondo We are not rotating about points and especially $r$, also your reference to adjusting the rotation with $-kA$ is not correct since this factor appears naturally from the formula for the sum of angles of a spherical triangle. Vasco |
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Post by Admin on May 17, 2023 19:04:37 GMT
Mondo
Bullet point 6 is about how the effect of a strong gravitational field, such as that near a black hole, would affect the curvature of space and make it very different from Euclidean space. Gauss's experiment near a black hole would have shown how different geometry is near a black hole where gravity is very strong.
Vasco
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Post by Admin on May 18, 2023 6:22:02 GMT
Mondo
With regard to my reply #26 I see that Needham does say things like "rotation about $p$", but it's understood that this means rotation about a line through $p$ and the centre of the sphere.
I was just trying to be precise.
Vasco
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Post by mondo on May 19, 2023 5:22:04 GMT
Can you point out which part of my response suggests you I missed the important issue? Mondo We are not rotating about points and especially $r$, also your reference to adjusting the rotation with $-kA$ is not correct since this factor appears naturally from the formula for the sum of angles of a spherical triangle. Vasco With regard to my reply #26 I see that Needham does say things like "rotation about p ", but it's understood that this means rotation about a line through p and the centre of the sphere. I was just trying to be precise. I think we indeed rotate the whole sphere about a point. There is several things that suggest that: 1. $R_{q}^{\phi}$ is a rotation of a sphere around a point $q$ right? 2. $R_{q}^{\phi} \circ R_{q}^{\theta} = R_{q}^{\psi}$ - final rotation is rotation of a sphere around a point $r$ with an angle $\psi$ 3. Figure [10a] shows half rotation of angle $\psi/2$ around $r$ So how can [10a] not be a rotation around $r$? What do you think is rotated, around what and with what angle? also your reference to adjusting the rotation with $-kA$ is not correct since this factor appears naturally Yes, this is what I meant. Now when I think about it again I have one more doubt - in the last equation on page 281 we have "$R_{N} \cdot R_{L} = R_{r}^{\psi}$" but this is only adding two angles, not three that make up angle and yet we still need to care about the area of a white triangle as author justifies in the last paragraph of that page? |
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