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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?

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  • $\begingroup$ Is the metric given in your first formula equivalent to the Euclidean arc length? $\endgroup$ Commented Nov 10, 2018 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 $x_2 =0$ $\endgroup$ Commented Nov 10, 2018 at 12:12

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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 $$ \sinh^{-1}\frac12\approx 0.4812 $$

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  • $\begingroup$ Does the hyperbolic distance vary depending on metric / model we choose ... Poincare3D , Klein .. hyperboloid model etc.. ? $\endgroup$
    – Narasimham
    Commented Nov 12, 2018 at 6:56
  • $\begingroup$ They are isometric with each other. $\endgroup$ Commented Nov 12, 2018 at 7:33

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