Distance in Poincaré disc

Question

I'm trying to prove that two points $x$ and $y$ in the Poincaré disc $\mathbb D \subset \mathbb R^2$ of radius $1$ is given by $$ d_H(x,y) = \cosh^{-1}\left(1+\frac{2|x-y|^2}{(1-|x|^2)(1-|y|)^2}\right) $$ where $|x| = \sqrt{(x^1)^2 + (x^2)^2}$ is the Euclidean length of $x \in \mathbb R^2$. The Riemannian metric on $\mathbb D$ is given by $$ g = 4\frac{(dx^1)^2 + (dx^2)^2}{(1-|x|^2)^2} $$ It suffices to consider the case where $x = 0$ and $y = (a,0)$ for some $a > 0$, by frame homogeneity of $\mathbb D$. Geodesic curves through the origin are diameter line segments of the disc, so we let $\gamma(t) = (t,0)$; when parametrized with $0 \leq t \leq a$, the resulting curve is length-minimizing. The speed of this curve in the above metric is $$ \left|\dot\gamma(t)\right| = \sqrt{g_{(t,0)}\left(\frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^1}\right)} = \sqrt{\frac{4}{\left(1-t^2\right)^2}} = \frac{2}{1-t^2} $$ whence the length of the curve $\gamma$ -- that is, the distance from the origin to the point $y=(a,0)$ -- is: $$ L_g(\gamma) = d_H(x,y) = \int_0^a \frac{2}{1-t^2}\,dt = 2\tanh^{-1}(a) $$ However, I'd like to show that this is equal to $\cosh^{-1}\left(1+\frac{2a^2}{1-a^2}\right)$, which is what we're trying to prove. Is there an identity here that I'm missing, or did I make a mistake in my calculations?

Answer 1

The idea is to write $cosh$ in function of $tanh$

Let be $tanh(x)=y$.

$cosh(2x)=cosh^2(x)+sinh^2(x)=$

$=\frac{cosh^2(x)+sinh^2(x)}{cosh(x)^2-sinh(x)^2}=$

$=\frac{1+y^2}{1-y^2}$

Then in your case if you choose

$x=tanh^{-1}(a)$ you have that

$cosh(2x)=\frac{1+tanh^2(x)}{1-(tanh(x))^2}=\frac{1+a^2}{1-a^2}=1+\frac{2a^2}{1-a^2}$

So

$2x=cosh^{-1}(1+\frac{2a^2}{1-a^2})$ that means

$2tanh^{-1}(a)=cosh^{-1}(1+\frac{2a^2}{1-a^2})$

So

$L_g(\gamma)= cosh^{-1}(1+\frac{2a^2}{1-a^2})$