I've been experimenting with the Beltrami-Klein model of hyperbolic geometry. According to Wikipedia, in this model the geodesics are given by Euclidean line segments, and I want to prove this. Let $x^{\alpha}$ denote a geodesic with parameter $t$, then the geodesic is given by parallel transporting the tangent along itself, i.e.;

$$\frac{d^2x^{\alpha}}{dt^2}+\Gamma^{\alpha}_{\ \beta \gamma}\frac{dx^{\beta}}{dt}\frac{dx^{\gamma}}{dt}=0$$

Where $\Gamma^{\alpha}_{\ \beta \gamma}$ are the Christoffel symbols. For this model we have that: \begin{align*} &\Gamma^{1}_{\ 11}=\frac{2x}{1-x^2-y^2}\\ &\Gamma^{1}_{\ 12}=\frac{y}{1-x^2-y^2}\\ &\Gamma^{2}_{\ 12}=\frac{x}{1-x^2-y^2}\\ &\Gamma^{2}_{\ 22}=\frac{2y}{1-x^2-y^2}\\ \end{align*}

With $\Gamma^{\alpha}_{\ \beta \gamma}$ symmetric on the lower indices and all other symbols zero. This yields 2 second order ODES for the components of the geodesic:



I have no idea how to solve these equations and any help yould be much appreciated. I tried a simple case however: Consider the geodesic joining $(x,y)=(0,0)$ to $(x,y)=(\frac{1}{2},0)$. If this is a Euclidean line as stated on wikipedia, it should be given by $x=t$, $y=0$, for $0\leq t\leq \frac{1}{2}$. Plugging this into the system of ODES yields:



So clearly this is not a solution! I thought that if Euclidean line segments were geodesics then they would satisfy the geodesic equation.

I would really appreciate any help in solving the general ODES for the geodesic joining two points, as well as in understanding why the Euclidean line given above doesn't appear to be a geodesic.

  • $\begingroup$ You should not expect that the parametrization is given by $(t, 0)$. In could be $(\phi(t), 0)$ for some strictly increasing function $\phi$ and this still parametrizes straight lines. $\endgroup$
    – user99914
    Apr 14, 2018 at 3:27

1 Answer 1


I'll write $dx/dt=x'$ and so on for brevity.

You're close, in that otherwise your equations appear to be otherwise correct. Suppose that $x' \neq 0$ and $y' \neq 0$ in an interval (else rotate the coordinate axes so we don't have this problem). Then dividing by $x'$ in the first and $y'$ in the second, the equations become $$ \frac{x''}{x'} = - \frac{2(xx'+yy')}{1-x^2-y^2} \\ \frac{y''}{y'} = -\frac{2(xx'+yy')}{1-x^2-y^2}, $$ so we have $ y''/y'-x''/x' = 0 $. Integrating once, $$ \log{\lvert y'/x' \rvert} = \text{const.}, $$ or $ y'/x' = A $, say. But $$ \frac{y'}{x'} = \frac{dy}{dt} \frac{dt}{dx} = \frac{dy}{dx}, $$ which is the gradient of the line. So the line has constant gradient, so it's straight.

  • 1
    $\begingroup$ Thank you! I understand how you've solved the ODEs and appreciate the help a lot. You were right about the typo (Its now fixed in the original post). One thing im still fuzzy on though, is why doesn't my straight line substitution work, even after fixing the typo? $\endgroup$
    – J.doe
    Apr 15, 2018 at 0:44
  • $\begingroup$ If $y=0$, you have $x''/x'=-2xx'/(1-x^2)$. Integrating gives $\log x'=-\log k(1-x^2)$, so $x=\tanh(at+b)$ is the solution. A geodesic is not just a curve: it comes with a specific parametrisation. E.g., the geodesic equations in flat space are $x''=y''=0$. These only have solutions of the form $x=at+x_0,y=bt+y_0$: if instead we parametrise a straight line as $x=a(t+t^3)+x_0,y=b(t+t^3)+y_0$, this parametrisation doesn't satisfy the geodesic equations, though it describes the same line. $\endgroup$
    – Chappers
    Apr 15, 2018 at 2:20
  • $\begingroup$ The geodesic equations in fact have $g_{ij} \frac{dx^i}{dt} \frac{dx^j}{dt} $ as a first integral, so this must be constant along a solution. $\endgroup$
    – Chappers
    Apr 15, 2018 at 2:20
  • $\begingroup$ Thanks, I didn't realise geodesics came with a set parametrisation. That completely answers my question. $\endgroup$
    – J.doe
    Apr 16, 2018 at 6:59

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