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Post by mondo on May 22, 2023 7:20:14 GMT
The last equation on page 314 calculates the Area of a triangle as $A(T) = (\pi - \alpha - \beta)$ how can we express area of a triangle as the sum of its angles?
Thank you.
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Post by Admin on May 22, 2023 9:07:49 GMT
Mondo
Think, think, think! Read the chapter again carefully because you have missed something that explains this.
I'll let you work this out for yourself. It's not an error in the book.
Vasco
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Post by mondo on May 26, 2023 6:50:21 GMT
Mondo Think, think, think! Read the chapter again carefully because you have missed something that explains this. I'll let you work this out for yourself. It's not an error in the book. Vasco But how could I have missed something in the book if that result is actually derived on the beginning of the next page (p.315)? At first I thought it comes from $E(T) = kA(T)$ but since there is no $k$ in the formula it cannot be this one. So my questions: 1. What have I missed in the book that would explain $A(T) = \pi - \alpha - \beta$? 2. Figure 33b - what kind of projection or cut of 33a is it? The book says, underneath this figure "..a triangle in $T$ in the upper half-pane"? This doesn't appear to be any plane of 33a - if it is a vertical cross section of 33a then $T$ should not be outside the semicircle (a rim of pseudosphere). 3. First equation on page 315 - why does the internal integral have a low limit of $\sqrt{1 - x^2}$ instead of $\sqrt{1-cos(\beta)^2}$. And, why the function is $\frac{dy}{y^2}$ 4. Why after the calculation of integral is done we substitute $x = cos(\theta)$? Thank you. |
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Post by Admin on May 26, 2023 12:48:55 GMT
Mondo
1. What you have missed is the statement at the beginning of paragraph 2 of subsection 3 A Conformal Map of the Pseudosphere on page 296. This applies from page 296 to the end of the chapter on page 327 - 32 pages in total:
"For simplicity's sake, henceforth we shall take the radius of the pseudosphere to be $R=1$, so our map will represent pseudospherical surfaces of curvature $k=-1$."
So this means that $k=-1$ and so $E(T)=A(T)$ throughout pages 296-327.
Vasco
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Post by Admin on May 26, 2023 14:41:39 GMT
Mondo
Figure 33b is the hyperbolic plane as described in subsections 3 and 4 on pages 296-301.
Vasco
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Post by Admin on May 26, 2023 15:04:00 GMT
Mondo
3. Have you studied double integrals in mathematics?
Look at (32) on 298 to understand where the function comes from.
4. This is just a well-known technique for solving integrals called 'Substitution'. Since $-1\leq x\leq +1$ we can let $x=\cos\theta$ and then substitute into the integral to obtain a new integral in terms of $\theta$, which is easier to evaluate than the one in terms of $x$.
Vasco
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Post by mondo on May 27, 2023 19:47:01 GMT
Mondo Figure 33b is the hyperbolic plane as described in subsections 3 and 4 on pages 296-301. Vasco But if that was the case then [33b] should resemble figures [20b] or [21]. What I mean is - hyperbolic plane should differ from pseudosphere only by distances of objects on it, which is described by a metric $d\hat{s} = \frac{ds}{y}$. However figure [33b] is nothing like [33a] - what does the semicircle at the horizon represents? If that is meant to represent a pseudosphere rim then: 1. It should be just the $X$ axis 2. The area $T$ should never leave it as it does I can't just make sense/connection of these two figures. |
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Post by Admin on May 27, 2023 20:01:49 GMT
Mondo
The semicircle is a hyperbolic line, just as in the diagrams 22,23,24,25,26,27.
Vasco
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Post by mondo on May 27, 2023 20:16:31 GMT
Mondo The semicircle is a hyperbolic line, just as in the diagrams 22,23,24,25,26,27. Vasco Ok but what does it represent now? Is it to be able to show angles $\alpha$ $\beta$ more "accurately"? |
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Post by mondo on May 27, 2023 20:24:55 GMT
From the next page I can deduce my suspicion is right - this semicircle seems to represent the "finite edge of $T$"
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Post by Admin on May 27, 2023 20:30:55 GMT
Mondo
The hyperbolic line segment between the two points on the hyperbolic line in 33b corresponds to the base of the triangle in 33a.
Vasco
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Post by mondo on May 28, 2023 4:08:43 GMT
Thank you Vasco, I think this is clear now.
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