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On the hyperboloid model for hyperbolic space, the hyperbolic distance between the two points is NOT the Euclidean length of the geodesic connecting those two points. So my question is, if hyperbolic distance doesn't measure that, then what DOES it measure.?

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    $\begingroup$ Surely this information is included in the definition of the hyperboloid model? What definition are you using and with what exactly are you having difficulties? $\endgroup$ – Will R Aug 10 '16 at 1:59
  • $\begingroup$ I don't agree to close this question about the meaning of what we do in one of the most fascinating mathematical contexts (hyperbolic geometry). I have tried to prove in my answer that it is possible to give a "supplement of soul" there. $\endgroup$ – Jean Marie Aug 10 '16 at 4:13
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A possible answer to your question is that the quantity it measures is the length of an optical path. This optical analogy, more generally, helps to understand intuitively "why" lines (geodesics) in hyperbolic geometry are certain circles.

This analogy will be presented in the half-plane version $\mathbb{H}$ of the Poincaré hyperbolic space ($Im(z) >0$). For a description in the Poincaré disk $\mathbb{D}$ defined by ($\left|Z\right|<1$), have a look at the following recent article: (http://arxiv.org/pdf/0908.2584). A description of the correspondence between $\mathbb{H}$ and $\mathbb{D}$ can be found in an Appendix at the end of this answer.

We are going to provide a "line" in $\mathbb{H}$, i.e., a half-circle centered on the $x$ axis, a physical meaning as "natural optical path" (natural = minimal) in a medium with varying optical index (something that one can encounter in optical fibers, e.g.); the optical index we are going to take is
$$n=\dfrac{1}{y},$$

in inverse ratio with the ordinate of the current point of the trajectory.

In this medium, how can we compute the trajectory of a ray path issued from $(x_0,y_0)$ with an incidence angle $i_0$ with respect to the vertical direction (see figure) ?

enter image description here

Let us consider the medium as "stratified" into an infinite number of horizontal lines, each one separating two media with resp. optical indices $\dfrac{1}{y+dy}$ and $\dfrac{1}{y}$. Let us apply Snell's law :

$$\dfrac{1}{y+dy}\sin{(i+di)}=\dfrac{1}{y}\sin{(i)}.$$

or

$$\sin{(i+di)}=\left(1+\dfrac{dy}{y}\right)\sin{(i)}.$$

Let us make a first order approximation of the LHS:

$$\sin(i)+\cos{(i)} di=\sin{(i)}+\dfrac{dy}{y}\sin(i)$$

Thus

$$\dfrac{\cos{(i)}}{\sin{(i)}} di=\dfrac{dy}{y}.$$

which can be integrated as follows:

$$ln(|\sin{(i)}|)=ln(y)+K.$$

Setting $K=-ln(R)$, we obtain

$$\sin{(i)}=\dfrac{y}{R} \ \ \text{with initial setting} \ \ \sin{(i_0)}=\dfrac{y_0}{R}.$$

When you look at the picture, it is clear that

$$y=R \sin({i})$$

represents an arc of circle with radius $R$ centered on the $x$-axis, as desired.

Appendix: Correspondence between the two models $\mathbb{H}$ and $\mathbb{D}$

Let us take the generic letters $z$ for $\mathbb{H}$ and $Z$ for $\mathbb{D}$.

The (bijective) correspondence between these two models is done through the homographic transform:

\begin{equation} Z=\dfrac{z-i}{z+i} \end{equation} A simple geometrical explanation; the upper half plane can be considered as the set of points $z$ that from $i$ than from $-i$, a fact that can be writtenas follows :

$$\left|z-i\right|<\left|z-(-i)\right| \ \ \Leftrightarrow \ \ \left|\dfrac{z-i}{z+i}\right|<1 \ \ \Leftrightarrow \ \ \left|Z\right|<1,$$

this last property being characteristic of the unit disk.

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  • $\begingroup$ @Eric Terry Is what I wrote understandable ? $\endgroup$ – Jean Marie Aug 10 '16 at 23:08
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    $\begingroup$ Interesting. I wonder if an argument along these lines could help with the question math.stackexchange.com/questions/3106000/… Maybe you can shed some light -and in passing grab that bounty :-) $\endgroup$ – MASL Feb 18 at 2:55

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