The distance between two distinct points in the Poincare upper half plane

Question

I'm trying to derive the distance between two distinct points in hyperbolic space and I'm working on the Poincare upper half plane. So, with the parametrization

$$\sigma(t): x=r\cos(t), y=r\sin(t),\; \alpha\leq t\leq \beta$$

the distance is just

$$\mathrm{dist}((x_{1},y_{1}),(x_{2},y_{2}))=\int_{\alpha}^{\beta}\rho(\sigma(t))|\sigma'(t)|dt=\int_{\alpha}^{\beta}\frac{dt}{\sin(t)}=\ln\left|\frac{\tan\frac{\beta}{2}}{\tan\frac{\alpha}{2}}\right|.$$

We also know that from https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model that

$$\mathrm{dist}((x_{1},y_{1}),(x_{2},y_{2}))=\mathrm{arccosh}\left(1+\frac{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}{2y_{1}y_{2}}\right).$$

And we know from https://en.wikipedia.org/wiki/Poincar%C3%A9_metric that

$$\mathrm{dist}((z_{1},z_{2}))=2\mathrm{arctanh}\frac{|z_{1}-z_{2}|}{|z_{1}-\overline{z_{2}}|}=\log\frac{|z_{1}-\overline{z_{2}}|+|z_{1}-z_{2}|}{|z_{1}-\overline{z_{2}}|-|z_{1}-z_{2}|}.$$

Also from Expression of the Hyperbolic Distance in the Upper Half Plane that the last three formulation are equal. Question is I cannot find the equal connection between my calculation and the other three Can anyone help?

Elif.

Answer 1

Here is one elementary proof:

1. The Poincare upper half plane $\mathbb{R}\times \mathbb{R}_+$ has metric $$\begin{align} (ds)^2~=~&\frac{ dz d\bar{z}}{({\rm Im}z)^2}~=~\frac{(dx)^2+(dy)^2}{y^2}~=~\frac{(dr)^2+r^2(d\theta)^2}{r^2\sin^2\theta}, \cr z~=~&x+iy~=~re^{i\theta}~=~r(\cos\theta+i\sin\theta).\end{align}\tag{1}$$

2. All geodesics are semicircles with center $c$ on the $x$-axis$^1$. By horizontal translation symmetry we may for simplicity assume from now on that $c=0$. (The final formulas will be translation invariant, and hence also work for $c\neq 0$.)

3. The distance $d(z_1,z_2)$ between two points $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$ is the arc length along the corresponding geodesic. Since $c=0$, we have $r_1=r=r_2$. By scale invariance, we may for simplicity assume from now on that $r=1$. (The final formulas will be scale invariant.)

4. By $z_1\leftrightarrow z_2$ symmetry, we may for simplicity assume from now on that $0<\theta_1<\theta_2<\pi$.

5. Therefore an infinitesimal distance is given by $$ ds~\stackrel{(1)}{=}~ \frac{d\theta}{\sin\theta}~=~d\ln\tan\frac{\theta}{2},\tag{2}$$ where $$ \begin{align}\tan\frac{\theta}{2}~=~&\frac{\sin\theta}{1+\cos\theta} ~=~\frac{1-\cos\theta}{\sin\theta}\cr ~=~&\frac{y}{1+x}~=~\frac{1-x}{y}\cr ~=~&\sqrt{\frac{1-x}{1+x}}~=~\frac{|z-1|}{|z+1|} ~=~\frac{|z-z_0|}{|z_{\pi}-z|}.\end{align}\tag{3}$$

6. Then the distance is $$\begin{align} d(z_1,z_2) ~\stackrel{(2)}{=}~&\ln\frac{\tan\frac{\theta_2}{2}}{\tan\frac{\theta_1}{2}} ~\stackrel{(3)}{=}~\ln\frac{|z_2-z_0||z_{\pi}-z_1|}{|z_1-z_0||z_{\pi}-z_2|}\cr ~=~&\ln\frac{\sin\frac{\theta_2}{2}\cos\frac{\theta_1}{2}}{\sin\frac{\theta_1}{2}\cos\frac{\theta_2}{2}}~=~\ln\frac{\sin\frac{\theta_2+\theta_1}{2}+\sin\frac{\theta_2-\theta_1}{2}}{\sin\frac{\theta_2+\theta_1}{2}-\sin\frac{\theta_2-\theta_1}{2}}\cr ~\stackrel{(5)}{=}~&\ln\frac{|z_2-\bar{z}_1|+|z_2-z_1|}{|z_2-\bar{z}_1|-|z_2-z_1|}~=~2{\rm artanh}\frac{|z_2-z_1|}{|z_2-\bar{z}_1|}\cr ~=~&\ln\frac{\sin^2\frac{\theta_2}{2}\cos^2\frac{\theta_1}{2}}{\sin\frac{\theta_1}{2}\cos\frac{\theta_1}{2}\sin\frac{\theta_2}{2}\cos\frac{\theta_2}{2}} ~=~2\ln\frac{2\sin\frac{\theta_2}{2}\cos\frac{\theta_1}{2}}{\sqrt{\sin\theta_1 \sin\theta_2}}\cr ~=~&2\ln\frac{\sin\frac{\theta_2+\theta_1}{2}+\sin\frac{\theta_2-\theta_1}{2}}{\sqrt{\sin\theta_1 \sin\theta_2}}~\stackrel{(5)}{=}~2\ln\frac{|z_2-\bar{z}_1|+|z_2-z_1|}{2\sqrt{y_1 y_2}}\cr ~\stackrel{(7)}{=}~&2\ln\left(\sqrt{q^2+1}+q\right)~=~-2\ln\left(\sqrt{q^2+1}-q\right)\cr ~=~&2{\rm arsinh}q~\stackrel{(7)}{=}~2{\rm arsinh}\frac{|z_2-z_1|}{2\sqrt{y_1y_2}}\cr ~=~&2 {\rm arcosh}\sqrt{q^2+1}~\stackrel{(7)}{=}~2 {\rm arcosh} \frac{|z_2-\bar{z}_1|}{2\sqrt{y_1y_2}}.\end{align}\tag{4}$$ Here we have used that $$\left\{ \begin{array}{rcl} |z_2-z_1|&=&2\sin\frac{\theta_2-\theta_1}{2},\cr |z_2-\bar{z}_1|&=&2\sin\frac{\theta_2+\theta_1}{2}, \end{array}\right.\tag{5} $$ so that $$ \begin{align} |z_2-\bar{z}_1|^2-|z_2-z_1|^2 ~\stackrel{(5)}{=}~& 4\sin^2\frac{\theta_2+\theta_1}{2}-4\sin^2\frac{\theta_2+\theta_1}{2}\cr ~=~& 2\cos(\theta_2\!-\!\theta_1)-2\cos(\theta_2\!+\!\theta_1)\cr ~=~&4\sin\theta_1 \sin\theta_2~=~4y_1y_2, \end{align}\tag{6} $$ and hence $$\frac{|z_2-\bar{z}_1|}{2\sqrt{y_1y_2}}~\stackrel{(6)}{=}~\sqrt{1+q^2} \quad\text{where}\quad q~:=~\frac{|z_2-z_1|}{2\sqrt{y_1y_2}}.\tag{7} $$

7. From $$ \sinh\frac{d(z_1,z_2)}{2} ~\stackrel{(4)}{=}~\frac{|z_2-z_1|}{2\sqrt{y_1y_2}}~\stackrel{(7)}{=}~q,\tag{8} $$ we also get that $$\begin{align} \cosh d(z_1,z_2)~=~&1+2\sinh^2\frac{d(z_1,z_2)}{2}~\stackrel{(8)}{=}~1+2q^2\cr ~\stackrel{(7)}{=}~&1+\frac{|z_2-z_1|^2}{2y_1y_2} ~=~\frac{(x_2-x_1)^2+y_1^2+y_2^2}{2y_1y_2}\cr ~=~&-X_1\cdot X_2 ,\end{align}\tag{9}$$ where in the last equality we introduced (dimensionless) imbedding coordinates.

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$^1$ Sketched proof that all geodesics in the Poincare upper half space are semicircles with center at $y=0$:

1. The Poincare upper half space $\mathbb{R}^d\times \mathbb{R}_+$ has metric $$ (ds)^2~=~\frac{(d\vec{x})^2+(dy)^2}{y^2}.\tag{10}$$

2. The corresponding Lagrangian is $$ L~=~\frac{\dot{\vec{x}}^2+\dot{y}^2}{2y^2},\tag{11}$$ where dot denotes differentiation wrt. to the worldline (WL) parameter $t$.

3. The corresponding momenta and energy are $$ {\rm const}~=~\vec{p}~=~\frac{\partial L}{\partial \dot{\vec{x}}}~=~ \frac{\dot{\vec{x}}}{y^2},\tag{12}$$ $$p_y~=~\frac{\partial L}{\partial \dot{y}}~=~ \frac{\dot{y}}{y^2},\tag{13}$$ and $$ {\rm const}~=~E~=~\vec{p}\cdot\dot{\vec{x}}+p_y\dot{y}-L~=~L, \tag{14}$$ where we have used Noether's theorem.

4. We therefore get an ODE for $y$ $$ \pm\dot{y}~\stackrel{(11)+(12)}{=}~y\sqrt{2L-y^2p^2}, \qquad p^2~:=~\vec{p}^2,\tag{15}$$ which can be solved $$ t~\stackrel{(15)}{=}~\mp\frac{{\rm artanh}\sqrt{1-\frac{y^2p^2}{2L}}}{\sqrt{2L}}+{\rm const},\tag{16}$$ although we don't need eq. (16) in what follows.

5. We calculate $$ \frac{d\vec{x}}{dy}~=~\frac{\dot{\vec{x}}}{\dot{y}}~\stackrel{(12)+(15)}{=}~\pm\frac{\vec{p}y}{\sqrt{2L-y^2p^2}},\tag{17}$$ which can be integrated $$ \vec{x}-\vec{c} ~\stackrel{(17)}{=}~\mp\frac{\vec{p}}{p^2}\sqrt{2L-y^2p^2}. \tag{18}$$

6. Hence the geodesic lies on a $d$-dimensional sphere $$ (\vec{x}-\vec{c})^2+y^2~\stackrel{(18)}{=}~\frac{2L}{p^2},\tag{19}$$ with center $(\vec{c},0)$ and radius $\sqrt{\frac{2L}{p^2}}$.

7. By translation symmetry, we may assume that $\vec{c}=\vec{0}$. By rotation symmetry, we may assume that $\vec{p}$ is parallel to the $x^1$-axis. $\Box$