Computing arc length in Poincare disk model of hyperbolic space

[owlpride]

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:

$$ds^2 = \frac{4}{(1-r^2)^2} dx^2$$

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

$$s = \int_a^b \sqrt{ds^2}$$

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

$$s = \int_0^r \frac{2}{1-x^2} dx = 2 arctanh(r)$$

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

Can someone help?

 

[L0r3n20]

There are some mistakes in your claims.
First of all the expression for the metric cannot be "square-rooted" so easily, in fact
$$\mathrm{d}x^2 = \mathrm{d}x^i\mathrm{d}x_i = \mathrm{d}x^2+\mathrm{d}y^2$$
then the definition of arc length 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):
$$L = \int^b_a \sqrt{\frac{\mathrm{d}x^{\mu}}{\mathrm{d}t} \frac{\mathrm{d}x^{\nu}}{\mathrm{d}t} g_{\mu\nu}} \mathrm{d}t$$
with $$x^{\mu} = \left\{x,y\right\}$$ .
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
$$r^2 = x^2 + y^2$$
Let me know if you get the right result! ;)

 

[L0r3n20]

Oh, I forgot to tell you that you can get $$x^{\mu}\left(t\right)$$ solving the geodesic equation
$$\frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} t^2} + \Gamma^{\mu}_{\nu\rho}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} t}\frac{\mathrm{d} x^{\rho}}{\mathrm{d} t} = 0$$

 

[owlpride]

Thanks, but I'm not really sure what to do with the information you just gave me. What does $$g_{\mu\nu}$$ 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.

 

[L0r3n20]

$$g_{\mu\nu}$$ is the metric of your space. In fact, $$\mathrm{d} s^2 = g_{\mu\nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}$$
In your case g is a square 2x2 diagonal matrix with the function $$\frac{4}{\left(1-r^2\right)^2}$$ as each element. You can also write $$g_{\mu\nu} = \frac{4}{\left(1-r^2\right)^2} \delta_{\mu\nu}$$. 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:
$$\Gamma^{\mu}_{\nu\rho}= \frac{1}{2} g^{\mu\lambda}\left(\partial_{\nu}g_{\lambda \rho} + \partial_{\rho}g_{\nu \lambda } - \partial_{\lambda}g_{\nu \rho}\right)$$