I want to calculate the hyperbolic distance on a hyperbolid defined by $x_1^2+x_2^2-x_3^2=-10000$.

From wikipedia and other sources I got the formula that the distance for two points $x$ and $y$ on the hyperboloid the distance is $$d(x,y) = \operatorname{arccosh}(\;\lvert (x_1y_1+x_2y_2-x_3y_3)\rvert\;) \tag{1}$$

If I take, for example, two points on the $xz$ plane, like $$x=(0,0,100) \quad\text{and}\quad y=(50,0,111.803)$$ the distance can also be calculated by the length of the curve $f(x) = \sqrt{x^2+10000}$, which is $$\int_0^{100}{\sqrt{1+\left(\frac{df}{dx}\right)^2}} = 50.0557 \tag{2}$$

If I try to calculate the distance with the formula above I get $$d(x,y)=\operatorname{arccosh}(\;\lvert 100\cdot 111.803\;\rvert) = 10.015 \tag{3}$$

The formula above might be only valid on the unit hyperboloid. Therefore I tried to normalize the hyperboloid. In my case: $$0.0001\cdot x_1^2+x_2^2-x_3^2=-1 \tag{4}$$

But $$d(x,y)=\operatorname{arccosh}\left(\;0.0001\cdot \lvert 100\cdot 111.803 \rvert\;\right) = 0.4812 \tag{5}$$ which is even if I scale it back somehow far away from $50.0557$.

Can someone see my mistake?

  • $\begingroup$ Is the metric given in your first formula equivalent to the Euclidean arc length? $\endgroup$ – Oscar Lanzi Nov 10 '18 at 0:01
  • $\begingroup$ I am not quite sure from where the first formula comes from. But if you look at the hyperboloid in a 3D cartesian coordinate system the arc length of a line on the hyperboloid is equivalent to the eculidean arc length. But to find a parametrization of a line on the hyperboloid and solve its integral gets difficult if it's not for this easy case where $x_2 =0$ $\endgroup$ – Erdbeer0815 Nov 10 '18 at 12:12

The hyperbolic distance on the hyperboloid model $x_1^2+x_2^2-x_3^2=-1$ is obtained by the metric $$ ds^2=dx_1^2+dx_2^2\color{red}{-}dx_3^2 $$ so the hyperbolic distance between your two points is $$ \color{blue}{\frac{1}{100}}\int_0^{\color{red}{50}}\sqrt{1\color{red}{-}f'(x_1)^2}\,\mathrm{d}x_1 $$ giving the answer $$ \sinh^{-1}\frac12\approx 0.4812 $$

  • $\begingroup$ Does the hyperbolic distance vary depending on metric / model we choose ... Poincare3D , Klein .. hyperboloid model etc.. ? $\endgroup$ – Narasimham Nov 12 '18 at 6:56
  • $\begingroup$ They are isometric with each other. $\endgroup$ – user10354138 Nov 12 '18 at 7:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.