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Computing arc length in Poincare disk model of hyperbolic space

  1. Jul 5, 2012 #1
    I am reading Thurston's book on the Geometry and Topology of 3-manifolds, and he describes the metric in the Poincare disk model of hyperbolic space as follows:

    ... the following formula for the hyperbolic metric ds^2 as a function of the Euclidean metric x^2:

    [tex]ds^2 = \frac{4}{(1-r^2)^2} dx^2[/tex]

    I don't understand how what ds^2 means or how to use this formula to compute distances and arc lengths. A naive guess is that the arc length should be given by

    [tex]s = \int_a^b \sqrt{ds^2}[/tex]

    but that doesn't seem to give me the correct answer. For example, take a point with Euclidean distance r from the origin. What is its distance in the hyperbolic metric?

    I know that the distance should be the arc length of the straight line connecting 0 to x, since the straight line through the origin is a geodesic in the hyperbolic metric. My guess would give me an arc length of

    [tex]s = \int_0^r \frac{2}{1-x^2} dx = 2 arctanh(r)[/tex]

    However, another website claims that the answer should be log(1+r)/log(1-r).

    Can someone help?
     
  2. jcsd
  3. Jul 5, 2012 #2
    There are some mistakes in your claims.
    First of all the expression for the metric cannot be "square-rooted" so easily, in fact
    [tex] \mathrm{d}x^2 = \mathrm{d}x^i\mathrm{d}x_i = \mathrm{d}x^2+\mathrm{d}y^2 [/tex]
    then the definition of arc lenght is the integral of the velocity along the curve (i.e. the trajectory), defined as (be careful there can be a sign under the square root, it depends on your convention):
    [tex] L = \int^b_a \sqrt{\frac{\mathrm{d}x^{\mu}}{\mathrm{d}t} \frac{\mathrm{d}x^{\nu}}{\mathrm{d}t} g_{\mu\nu}} \mathrm{d}t [/tex]
    with [tex] x^{\mu} = \left\{x,y\right\} [/tex] .
    So the first thing you have to do is to choose a parametrization of your coordinates and then proceed to compute the integral. Note also that
    [tex] r^2 = x^2 + y^2 [/tex]
    Let me know if you get the right result! ;)
     
  4. Jul 5, 2012 #3
    Oh, I forgot to tell you that you can get [tex] x^{\mu}\left(t\right)[/tex] solving the geodesic equation
    [tex] \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} t^2} + \Gamma^{\mu}_{\nu\rho}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} t}\frac{\mathrm{d} x^{\rho}}{\mathrm{d} t} = 0 [/tex]
     
  5. Jul 5, 2012 #4
    Thanks, but I'm not really sure what to do with the information you just gave me. What does [tex]g_{\mu\nu}[/tex] mean? Do I really need to compute the Christoffel symbols? If yes, how would I compute them from "ds^2"?

    All I care about right now is the simplest way to get from "ds^2" to arc length. If I don't absolutely need to compute all the other crap, I'd rather not.
     
    Last edited: Jul 5, 2012
  6. Jul 5, 2012 #5
    [tex] g_{\mu\nu} [/tex] is the metric of your space. In fact, [tex] \mathrm{d} s^2 = g_{\mu\nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} [/tex]
    In your case g is a square 2x2 diagonal matrix with the function [tex] \frac{4}{\left(1-r^2\right)^2} [/tex] as each element. You can also write [tex] g_{\mu\nu} = \frac{4}{\left(1-r^2\right)^2} \delta_{\mu\nu} [/tex]. This is how I'd solve the problem, I don't know if there is a faster way! =)
    The Christoffel symbols are easily computed from g using the usual form for the Levi Civita connection symbols:
    [tex] \Gamma^{\mu}_{\nu\rho}= \frac{1}{2} g^{\mu\lambda}\left(\partial_{\nu}g_{\lambda \rho} + \partial_{\rho}g_{\nu \lambda } - \partial_{\lambda}g_{\nu \rho}\right) [/tex]
     
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