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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

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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 $(x_1 - c)^2 + y_1^2$.

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    @WillieWong Oh yeah man that's right it's just $\int \frac{dz}{Im(z)} = \int \frac{1}{\sin t} dt$. Yeah that's a good way to think about it I assumed earlier that I could shift my circle around by some real number and I have the numerator dependent on $x$? man...2011-04-18