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Post by mondo on Mar 29, 2023 7:04:33 GMT
There are some fundamentals facts about non-euclidean geometries that are taken from granted in the book. I would like to discuss them in this thread:
1. Bottom of page 269 - "In spherical geometry the angle sum is greater than $\pi$" and "In hyperbolic geometry the angle sum is less than $\pi$". It is said that these two facts where "discovered" but can we somehow justify it? Just a few lines above the book says it was discovered using a pure "logic" and Axioms (2) and (3). To me from these axioms we can draw either one of these conclusions. So what "logic" actually makes sure that on hyperbolic geometry angle sum of a triangle is $< \pi$?
Thank you.
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Post by Admin on Apr 1, 2023 8:21:25 GMT
There are some fundamentals facts about non-euclidean geometries that are taken from granted in the book. I would like to discuss them in this thread: 1. Bottom of page 269 - "In spherical geometry the angle sum is greater than $\pi$" and "In hyperbolic geometry the angle sum is less than $\pi$". It is said that these two facts where "discovered" but can we somehow justify it? Just a few lines above the book says it was discovered using a pure "logic" and Axioms (2) and (3). To me from these axioms we can draw either one of these conclusions. So what "logic" actually makes sure that on hyperbolic geometry angle sum of a triangle is $< \pi$? Thank you. Mondo It seems to me that the answers to your questions are in the pages that follow page 269. First for spherical geometry, and then hyperbolic geometry. If the parallel axiom is equivalent to saying $E(T)=0$ (figure 1b on page 268), then this is Euclidean geometry. The only other possibilities are $E(T)>0$ and $E(T)<0$, which for reasons which we see on these following pages, we decide to call Spherical geometry and Hyperbolic geometry. Vasco |
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Post by mondo on Apr 10, 2023 2:02:19 GMT
Okay I continue to reared this chapter and I have some more questions:
1. Page 270 bullet point 4 - I don't get it at all, why in spherical geometry we can define the absolute unit of length to be the side of the equilateral triangle having angle sum $1.01\pi$? Why this particular value? An why it only exists in non-Euclidean geometries?
2. Page 270 bullet point 6 - "Einstein's theory explains.. the weak gravitational field surrounding the earth corresponds to microscopic values of $k$...it would have been a different story...in the vicinity of a black hole" - but it means if I take some sphere say a globus from my room and travel with it close to the black hole then, according to this point my globus will deform into something else, am I right with my conclusion here?
3. Page 274 - I try to understand the Gauss curvature. First of all I don't understand this part "The sign of $k$ depends if the centre of curvature is in the direction of $n$ or -$n$". What is the centre of curvature? Is it the white dot on all three figures [5]? How to tell in what direction it goes? Looking at figure [5] what is the difference of the middle one and right most one that the curvature changed?
Thank you.
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Post by Admin on Apr 12, 2023 18:13:25 GMT
Mondo
I am thinking about the best way to reply to your last post. I will post something as soon as I can.
Vasco
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Post by Admin on Apr 13, 2023 15:17:10 GMT
Okay I continue to reared this chapter and I have some more questions: 1. Page 270 bullet point 4 - I don't get it at all, why in spherical geometry we can define the absolute unit of length to be the side of the equilateral triangle having angle sum $1.01\pi$? Why this particular value? An why it only exists in non-Euclidean geometries? We can do this because of (4) on page 270. But as Needham says in the next bullet point it is more natural and convenient to use $k$. Try reading this: blogs.scientificamerican.com/roots-of-unity/forget-meters-and-feet/Near a black hole the gravitational field would be much stronger and the linear dimensions of a triangle would then be a more significant fraction of $R$. The centre of curvature is defined and explained in the bottom half of page 274 and continues on to page 275. The white dot is not the centre of curvature. It is the point $p$ where we are measuring $k$. The intrinsic curvature $k$ at the point $p$ is zero on the RHS of [5] just as it is on all the surfaces in [4]. The centre of curvature is the centre of the circle of curvature at $p$. Read the explanation on page 274. Vasco |
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Post by mondo on May 13, 2023 6:12:46 GMT
Okay I continue to reared this chapter and I have some more questions: 1. Page 270 bullet point 4 - I don't get it at all, why in spherical geometry we can define the absolute unit of length to be the side of the equilateral triangle having angle sum $1.01\pi$? Why this particular value? An why it only exists in non-Euclidean geometries? We can do this because of (4) on page 270. But as Needham says in the next bullet point it is more natural and convenient to use $k$. Try reading this: blogs.scientificamerican.com/roots-of-unity/forget-meters-and-feet/I read it but it doesn't answer my question. Neither does equation (4). I don't get why this "absolute unit of length" is only in non-Euclidean geometry, as well as why author defines it to be a "side of equilateral triangle having angle sum $1.01\pi$"? And similarly in hyperbolic geometry to be another side of equilateral triangle with angle sum of $0.99\pi$? 2. Page 270 bullet point 6 - "Einstein's theory explains.. the weak gravitational field surrounding the earth corresponds to microscopic values of $k$...it would have been a different story...in the vicinity of a black hole" - but it means if I take some sphere say a globus from my room and travel with it close to the black hole then, according to this point my globus will deform into something else, am I right with my conclusion here? Near a black hole the gravitational field would be much stronger and the linear dimensions of a triangle would then be a more significant fraction of $R$. Your answer is even more cryptic, what does it mean that linear dimensions of a triangle will be more significant fraction of $R$? And continuing with my original question, how a stronger gravitational field affects the dimensions of a triangle? 3. Page 274 - I try to understand the Gauss curvature. First of all I don't understand this part "The sign of $k$ depends if the centre of curvature is in the direction of $n$ or -$n$". What is the centre of curvature? Is it the white dot on all three figures [5]? How to tell in what direction it goes? Looking at figure [5] what is the difference of the middle one and right most one that the curvature changed? The centre of curvature is defined and explained in the bottom half of page 274 and continues on to page 275. The white dot is not the centre of curvature. It is the point $p$ where we are measuring $k$. The intrinsic curvature $k$ at the point $p$ is zero on the RHS of [5] just as it is on all the surfaces in [4]. The centre of curvature is the centre of the circle of curvature at $p$. Read the explanation on page 274. Actually the center of curvature is not covered on these pages and I think it is not covered in the book at all. But it is not that difficult to get it from a Wikipedia page. However what is missing is how do we define the direction of curvature - when is it a positive direction and when negative. There is also a confusion in the book, on page 274 before figure [5] - "The sign of $k(p)$ tells us qualitatively what the surface is like in the immediate neighborhood of $p$" and then below figure [5] "The sign of $k$ depends on whether the centre of curvature is in the direction $n$ or $-n$" hence what would be the sign of the middle figure [5] if I flip it 180 degree? according to the later quote it should now become a negative $k(p)$ while previous quote says it will remain the same. |
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Post by mondo on May 14, 2023 4:09:04 GMT
I also don't get why on the bottom of page 281 author says that on a sphere the total amount of two rotations $\theta$ and $\phi$ is not $\theta + \phi$ but rather $\theta + \phi - kA$. This suggest that we need to subtract the angular excess from the rotation but why? Rotation doesn't change the area of the figure, the curvature on the sphere is constant so why do we subtract the angular excess form the total rotation?
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Post by Admin on May 15, 2023 14:52:08 GMT
I also don't get why on the bottom of page 281 author says that on a sphere the total amount of two rotations $\theta$ and $\phi$ is not $\theta + \phi$ but rather $\theta + \phi - kA$. This suggest that we need to subtract the angular excess from the rotation but why? Rotation doesn't change the area of the figure, the curvature on the sphere is constant so why do we subtract the angular excess form the total rotation? Mondo What you have written above means that you have not understood the explanation on pages 280 to 281. I would suggest you read these pages and ask questions about what is written there. This should help you understand. Vasco |
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Post by Admin on May 15, 2023 15:11:02 GMT
Mondo
The curvature $\kappa$ at $p$ is defined at the bottom of page 235 and hence the radius of curvature is $1/\kappa$. The centre of curvature is the centre of the circle whose tangent coincides with the tangent to the curve at $p$
On page 274 you seem to be confusing the meanings of $k$ and $\kappa$, but they are two very different things.
Vasco
PS: I will try to answer your other questions in reply 5 as soon as I can.
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Post by mondo on May 15, 2023 17:26:00 GMT
I also don't get why on the bottom of page 281 author says that on a sphere the total amount of two rotations $\theta$ and $\phi$ is not $\theta + \phi$ but rather $\theta + \phi - kA$. This suggest that we need to subtract the angular excess from the rotation but why? Rotation doesn't change the area of the figure, the curvature on the sphere is constant so why do we subtract the angular excess form the total rotation? Mondo What you have written above means that you have not understood the explanation on pages 280 to 281. I would suggest you read these pages and ask questions about what is written there. This should help you understand. Vasco Thanks for an answer. I have reread pages 280-282 and I can't ask any question other than why $\psi = \theta + \phi - kA$ because everything else seems to be clear. Which part/sentence of these pages answer this equation? |
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Post by Admin on May 15, 2023 18:37:29 GMT
Mondo
If you understand the last result on page 281 then this proves the result because it shows that $\mathcal{R}_q^{\phi}\circ\mathcal{R}_p^{\theta}=\mathcal{R}_r^{\psi}$. Then just use figure 10a to calculate $\psi$.
Notice, by the way, that these 3 rotations all take place about 3 different axes
Vasco
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Post by mondo on May 15, 2023 21:21:57 GMT
If you understand the last result on page 281 then this proves the result because it shows that $\mathcal{R}_q^{\phi}\circ\mathcal{R}_p^{\theta}=\mathcal{R}_r^{\psi}$. Then just use figure 10a to calculate $\psi$. This is very similar to 2D case described on page 38. We basically rotate great line $N$ around another great line $L$ via angle $\psi/2$. What is $\psi$ then? Since $\psi/2 = \frac{\phi + \theta}{2}$ then $\psi = \phi + \theta$. Which is against the equation just below figure [10] on page 282. Notice, by the way, that these 3 rotations all take place about 3 different axes Do you mean that we rotate on a sphere and then naturally this involves a rotation in 3D? |
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Post by Admin on May 15, 2023 21:45:40 GMT
Mondo If you understand the last result on page 281 then this proves the result because it shows that $\mathcal{R}_q^{\phi}\circ\mathcal{R}_p^{\theta}=\mathcal{R}_r^{\psi}$. Then just use figure 10a to calculate $\psi$. This is very similar to 2D case described on page 38. We basically rotate great line $N$ around another great line $L$ via angle $\psi/2$. What is $\psi$ then? Since $\psi/2 = \frac{\phi + \theta}{2}$ then $\psi = \phi + \theta$. Which is against the equation just below figure [10] on page 282. Why shouldn't it be different?Notice, by the way, that these 3 rotations all take place about 3 different axes Do you mean that we rotate on a sphere and then naturally this involves a rotation in 3D? Vasco |
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Post by mondo on May 15, 2023 22:10:02 GMT
Mondo This is very similar to 2D case described on page 38. We basically rotate great line $N$ around another great line $L$ via angle $\psi/2$. What is $\psi$ then? Since $\psi/2 = \frac{\phi + \theta}{2}$ then $\psi = \phi + \theta$. Which is against the equation just below figure [10] on page 282. Why shouldn't it be different?Do you mean that we rotate on a sphere and then naturally this involves a rotation in 3D? Vasco I believe it should not be different because as I mentioned in earlier post we just rotate so I don't see why we should take angular excess into account. You said it can be calculated from figure [10a] and I did using the 2D analogy from page 38 but it doesn't prove the need for angular excess, unless I overlooked something. |
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Post by Admin on May 15, 2023 22:52:40 GMT
Mondo
I worked through it in detail and it all works for me. In my opinion it is all perfectly explained on pages 281-2. I don't understand why you cannot see it
Vasco
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