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6

One common way to visualize the "intrinsic appearance" of a simply-connected universe of constant curvature $\pm 1$ is to give the angular size of an object modeled as a geodesic arc of (sufficiently small) length $\ell$ placed at distance $d$ from one's eye. (I didn't try to run the linked applets, and am not sure if either implements this strategy.) On a ...


3

You shouldn't assume that the naive parametrization $(\text{cos}(t), \text{sin}(t))$ will solve the equation. It doesn't, as it clearly doesn't satisfy even the first differential equation. However, if you assume a more general form $x(t) = \text{cos}(f(t))$, $y(t)= \text{sin}(f(t))$, which still traces out a semicircle (but with a different speed), then it ...


3

First, we should note that a very similar question has already been asked here, and several interesting answers were given. But because the question keeps coming up, I'm going to go out on a limb and suggest that there still might be room for a more complete list of reasons why hyperbolic geometry is important in its own right. It's hard to know where to ...


2

It seems a little easier to construct the equivalent $$f(z)=\frac{z+i}{i(z-i)}$$ Given point $z$ in the complex plane, draw the points $z+i$ and $z-i$ one unit above and below $z$. You could use the obvious parallelograms with $z$, $0$, $i$, and each of those two points. Then rotate the line segment from $0$ to $z-i$ by $90°$ clockwise to get point ...


2

$C$ is related to the curvature. If the model being used is the upper half plane, the curvature is $-1$ and $C=1.$ A book that gives full detail on other choices is George Martin, The Foundations of Geometry and the Non-Euclidean Plane. The quantity that Martin calls the "distance scale" $k$ is defined in terms of a relationship among horocycles... It may ...


2

First, $\mathbb{H}^n$ is a complete Riemannian manifold of constant sectional curvature $-1$. Second, for each $\kappa<1$ there exists $\delta>0$ such that every complete Riemannian manifold $M$ whose sectional curvatures are all $\le \kappa$ is $\delta$-hyperbolic; by definition means that every geodesic triangle $T \subset M$ is $\delta$ thin, ...


2

$$ {\cosh ^{-1}} \frac{(R+ 1/R)}{2} $$ $$= {\cosh ^{-1}} \frac{ e^ {\log (R)}+ e^{-\log(R)}}{2} $$ $$ = {\cosh ^{-1}} [ \cosh (\log R)] = \log R . $$ EDIT1: BTW, why do we assume unit (abs value) Gauss curvature? Should it not appear symbolically at least in a formula ? $$ \operatorname{dist} (\langle a, r \rangle, \langle a, R \rangle) = ...


1

Compute first the inverse of $$y=\operatorname {arcosh}(x)=\frac{e^x+e^{-x}}2$$ for positive $x$'s. Whit the substitution $u=e^x$ we have $2y=u+\frac1u$. Multiplying both sides by $u$ we get the following quadratic equation: $$u^2-2yu+1=0,$$ the solutions of which are $$u^+=y+\sqrt{y^2-1}\ \text{ and } \ u^-=y-\sqrt{y^2-1}.$$ We need only the ...


1

Well, to being with, assume that $r= 1$ and $R>1$. Now, $$\begin{align*} \operatorname{dist} (\langle a, 1 \rangle, \langle a, R \rangle) &= \operatorname{arcosh} \left( 1 + \frac{ {(R - 1)}^2 }{ 2R } \right)\\ &= \operatorname{arcosh} \left(\frac{1}{2} \left( R + \frac{1}{R} \right) \right)\\ &= \ln(R) \end{align*}$$ where the last line is ...


1

The Beltrami-Klein model is an accurate depiction of what it would look like in hyperbolic space. To be a bit more precise, if you live in 3-dimensional hyperbolic space $\mathbb{H}^3$, and if $P \subset \mathbb{H}^3$ is a 2-dimensional hyperbolic plane tiled in red and white triangles with angles $\pi/2,\pi/3,\pi/7$ as in the picture shown in the question, ...


1

The equidistant lines to $\ell$ are exactly the Euclidean circles passing through the ideal endpoints $\xi_- = (-r,0)$, $\xi_+ = (+r,0)$ of the line $\ell$ (intersecting those circles with the upper half-plane, of course). The line $\ell$ is, of course, the unique such circle that meets the real line at angle $\pi/2$. The other equidistant lines meet the ...


1

When $t$ crosses an odd multiple of $\pi/2$, both $\sec(t)$ and $\tan(t)$ switch between $-\infty$ and $+\infty$. In the affine plane such occurences should be regarded as discontinuities, but the plotting routine does not know that and continues drawing. Workaround: Try not to sweep across discontinuities with your parameter range, e.g. thus.


1

The correspondence between the Poincare disc model and the hyperbolic model is via the projection from the source at $(0,0,-1)$ point. Suppose that there is a light source at the point $(0,0,-1)$. Then each ray connects an unique point of the unit disc (centered at origin, in the XY plane) to a unique point in the hyperboloid. Now try to picture the ...


1

First you need the Lorentz inner product $$\underbrace{( t_0, x_0, y_0)}_{\vec v^T} \underbrace{\pmatrix{-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}}_J \underbrace{\pmatrix{t_1 \\ x_1 \\ y_1}}_{\vec w} = -t_0 t_1 + x_0 x_1 + x_2 y_2 $$ which I'll write in shorthand as $$\vec v^T \, J \, \vec w $$ The hyperboloid model $\mathbb{H}$ is ...


1

Recall that $$ 0=E_1(E_j,E_k)= \frac{1}{y^2} \{ \Gamma_{1j}^k + \Gamma_{1k}^j \} \Rightarrow \Gamma_{1j}^k =-\Gamma_{1k}^j $$ $$ -2y^{-3}= E_2(E_2,E_2)= \frac{1}{y^2} (2\Gamma_{22}^2) \Rightarrow \Gamma_{22}^2= - \frac{1}{y} $$ By definition, $$ \Gamma_{12}^1 = -\frac{1}{y}$$ If $$ v= (-\sin^2 t, \sin\ t\cos\ t) :=fE_1+ g E_2$$ then $$ v (F(t))= y ...


1

FWIW, I seem to be settling on to the Beltrami-Klein model. It has the property that a geodesic in hyperbolic space maps onto a straight line in the model, which means that calculating intersections is cheap and easy; the only part where the spatial curvature becomes important is calculating the hyperbolic distance along the line where my intersection ...


1

The mistake is after the words "leading to" where the square root in the denominator is missing. It should be $$ \frac{r^2}{1-r^2}\cdot\frac{1}{\sqrt{r^2+r'^2}}=C_1. $$ P.S. Now it is much more fun integrating this guy :-)


1

Contrary to my initial gut feeling, this sounds highly plausible. If you assume that any body which is not moving (with constant speed) along a geodesic experiences some force, then you can simply take four test bodies, forming a regular hyperbolic tetrahedron, and consider translating that through space. As you said, a hyperbolic translation only moves ...



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