# Calculating hyperbolic distance on the hyperboloid $x_1^2+x_2^2-x_3^2=-10000$

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?

• Is the metric given in your first formula equivalent to the Euclidean arc length? Commented Nov 10, 2018 at 0:01
• 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 $x_2 =0$ Commented Nov 10, 2018 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 $$\sinh^{-1}\frac12\approx 0.4812$$