I am trying to derive the distance between two arbitrary points in hyperbolic space; the model I'm using is the upper half plane model.
So the distance is just $\int_f \rho(z) dz$, where $\rho(z) = \frac{|z|^2}{\text{Im}(z)}$. Now I construct a circle between these two points $(x_1,y_1)$ and $(x_2, y_2)$ whose centre is on the real axis, and I arrive at the equation $$\Biggl(x – \frac{(y_1^2 – y_2^2 + x_1^2 – x_2^2)}{2(x_1 – x_2)}\Biggr)^2 + y^2 = x_1^2 + y_1^2.$$
I also know that $(x_1,y_1)$ and $(x_2,y_2)$ can be parametrised in terms of $t$, as when I do a change of variables in the integral I now have to say the integral is going from some value $t_2$ and $t_1$, where $t_k$ is the angle between the line joining this point to the center $(\frac{(y_1^2 – y_2^2 + x_1^2 – x_2^2)}{2(x_1 – x_2)},0)$ and the real axis. So after doing a lot of manipulations, you arrive at the equation : $$d\Big((x_1,y_1,),(x_2,y_2)\Big) = \ln\Bigl|\frac{y_1^2+c^2-2cx_1+x_1^2+cy_1-y_1x_1}{y_1^2+c^2+2cx_1+x_1^2+cy_1+y_1x_1} \Bigr|,$$ where $c$ is the $x$ coordinate of the center
But somehow this does not tally up with the answer given on http://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model,
I tried looking at the identity $\text{arcosh}(x)=\ln(x+\sqrt{x^2-1})$, but it does not help.
Unless I did not stuff up any of my calculations, any ideas?
Ben
Best Answer
Your formula for the circle is incorrect. You found the center correctly. But the radius, which you need to insert on the RHS, should be $(x_1 - c)^2 + y_1^2$.