Hyperbolic Distance Confusion 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.? 
 A: 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) ?

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.
