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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,

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?


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up vote 2 down vote accepted

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 Merci de me dire de ce qui était incorrect. – user38268 Apr 17 '11 at 22:44
@David: de rien. – Willie Wong Apr 17 '11 at 22:56
@WillieWong Hi, I have done a new calculation; as the arc length is invariant under translation by a real number, I consider only points $(x_1,y_1)$ and $(x_2,y_2)$ that lie on a circle centered at the origin, so my circle is now $x^2 + y^2 = x_1^2 + y_1^2$. But still after performing the manipulations I get $(x_1^2 + y_1^2)\Big(cosh^{-1} (x_1/y_1) - cosh^{-1} (x_2/y_2)\Big)$. But the problem is the term $(x_1^2 + y_1^2)$ on the outside, while the one give in the wikipedia article has no term on the outside of $\ln(x)$ like that. – user38268 Apr 18 '11 at 3:46
Check your expression for $\rho$. In radial coordinates you should have the integral $\int \frac{1}{\sin \theta} d\theta$. I suspect you have an extra factor in the numerator. – Willie Wong Apr 18 '11 at 10:58
@WillieWong @WillieWong Hi, I have that $\rho(z) = \frac{|z|^2}{Im(z)}=\frac{x_1^2+y_1^2}{\sqrt{x_1^2+y_1^2} \sin t}$, $|f'(t)|dt=\sqrt{x_1^2 + y_1^2} dt$, and so $\frac{|z|^2}{Im(z)} dz =\frac{x_1^2+y_1^2}{\sqrt{x_1^2+y_1^2} \sin t} \sqrt{x_1^2 + y_1^2} dt$ = $(x_1^2+y_1^2) \csc(x)$, the expression in the integrand. Is this right? – user38268 Apr 18 '11 at 12:03

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