1
$\begingroup$

I am currently reading the article Computing geodesics and minimal surfaces via graph cuts.I have difficulties with understanding the Cauchy-Crofton formula posed for 2D Riemannian space with continuously varying metric $D$:

$$\int \frac{\text{det}D}{2 (u^T_L \cdot D \cdot u_L )^{3/2}} n_c d \mathcal{L} = 2 |C|_{R}$$

where $u_L$ is the unit vector in the direction of the line L. $C_R$ is Riemannian length of contour $C$.

In particular I am interested in applying this formula for Poincare disc, with metric: $$ds^2 = l^2_{\text{AdS}} (d\rho ^2 + \text{sinh}^2 \rho d \theta^2) $$ here $l_{\text{AdS}}$ is a const that defines curvature of the space.

In order to check myself I am trying to evaluate length of a circle of radius $p_{*}$ which center coincides with the center of hyperbolic disc. (Following picture also represents my view on position and direction of unit vectors $u_{L}$):

enter image description here

These are my assumptions about this formula:

1)In my case metric D is given by metric of Poincare disc.

2) In phrase "$u_L$ is the unit vector in the direction of the line L" L is geodesic characterized by $(p; \phi); p \in (-\infty; +\infty), \phi \in (0,2\pi)$

3)By def of $u_{L}$ expression $(u^T_L \cdot D \cdot u_L) $ always equals 1. Why do we need to write it explicitly in the formula?

4) I assume that $d \mathcal{L} = dp d\phi$

5) For the circle defined above I'll have $n_c = 2, p \in (-p_{*};p_{*}), $ and $n_c = 0$ for other values of $p$ ($n_c$ is the same for all values of $\phi$)

If I plug all this information into original formula I obtain what seems to be the wrong answer.

I have also obtained length of this circle in two other ways. Both approaches give the same result which is different from aforementioned formula's result:

A) $|\gamma| = \int ds = \int \sqrt{d\rho^2 + \text{sinh}^2 \rho d\theta^2} = 2\pi \cdot \text{sinh}p_* $

B) This approach is taken from the article Integral Geometry and Holography.This is actually also Cauchy-Crofton formula, yet written in a different form: $$|\gamma| = \frac{1}{4} \int_{-p_{*}} ^{p_{*}} \int_0 ^{2\pi} 2 \text{cosh}p dp d\theta = 2\pi \cdot \text{sinh}p_* $$

Could someone give me hints about what is the correct way to understand the original formula? Or may be there are other misunderstandings which I do not realize? I would be glad to hear any feedback about the question.

$\endgroup$
1
$\begingroup$

It seems that I have found the solution to this problem.

My major misunderstanding was that I haven't realized that metric D should be metric of corresponding kinematic space. (One can read more about this concept in "Integral Geometry and Geometric Probability" by L. A. Santalo.)
For instance for Poincare disc corresponding metric of kinematic space will be that of de Sitter space.(derivation can be found here)

Suppose metric of Poincare disc is given by: $$ds^2 = d \rho^2 + \text{sinh}^2 \rho d \widetilde\theta ^2 $$ and equation of geodesic: $$\text{tanh} \rho \cdot \text{cos}(\widetilde\theta - \theta) = \text{cos} \alpha$$ enter image description here

The metric of corresponding kinematic space will be given by:

$$ds_{\text{kin}}^2 = \frac{1}{\text{sin}^2 \alpha}(-d \alpha^2 + d \theta^2)$$

Once we plug this information into original formula we will obtain:

$$\int \frac{1}{2 \text{sin}^2 \alpha} n_c d \alpha d \theta = 2 |C|_R$$

Let us define $p$ as a shortest length between center of the hyp. disc and given geodesic. Then one can use equation of geodesic in order to obtain following relation: $$\text{cosh}p dp = - \frac{d \alpha}{\text{sin}^2 \alpha}$$ Once we change variables in the integral, take care of limits of integration and remember that for a circle of fixed radius $n_c = 2, p \in(-p_*, p_*)$ and equals $0$ elsewhere we obtain just the same expression for the circumference of the circle as we had by two other approaches.

(Note)
I still don't fully understand why the original formula contains expression $(u_L^T \cdot D \cdot u_L )^{3/2}$ in it. Since the define $u_L$ as a unit vector I would expect this expression to be equal 1 all the time. Maybe writing out this part has something to do with discretization of this integral that they did later in the article. Yet I am not $100 \%$ sure about this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.