0

When Gauss-Bonnet thm is used on a constant Gauss curvature surfaces of $ (~ K=-1/a^2,K=0,K<=+1/a^2)$ along a closed radial contours, the exterior angle rotation sum can be shown to be respectively:

$$ \Sigma \psi= 2 \pi (1+\Delta r/a),~~2 \pi (1+ 0\cdot r/a),~~2 \pi (1+\Delta z/a) $$ Here $(\Delta z,\Delta r)$ represent axial or radial difference of truncated shell surface area considered for $ (K>0, K<0) $ cases (I had also shown this before on this site).

For the central Pseudosphere aka Tracticoid we have area computation

$$ dA=2 \pi r\cdot ds= 2 \pi r \cdot {dr}/ \sin \phi= 2 \pi a \cdot dr $$

$$ A=2 \pi \int_{r_1}^{r_2} dr = 2 \pi a \cdot \Delta r$$

It can be shown by integration that area of each Nappe or segment is $4 \pi a^2 $ for non central cases as well that also verifies Isometry.

Considering Monkey saddle or Pringle's chip topography, the angle sum $ >2 \pi$ can appeal to intuition.

But at the pseudo-sphere maximum radius $(r=a)$ the angle sum of two full turnings at cuspidal edge as exactly $ 4 \pi$ is still open to interpretation.. (to me now) is counter-intuitive.

Is the tally of $2 \pi$ for each of two horns / Nappes?

Thanks for help in interpreting this definite limit when encountering the edge of singularity.

Narasimham
  • 40,495
  • By "pseudo-sphere" do you mean the tractricoid? That doesn't have constant Gaussian curvature (because of the edge). Also, Hilbert's theorem may be relevant, especially with the $2\pi$ bound on area. Maybe embedding into Euclidean space is not a good way to understand hyperbolic geometry. – mr_e_man Sep 26 '19 at 00:51
  • Could you further explain $\Delta r$, or link to your previous post? If it's an area, then it should be divided by $a^2$. – mr_e_man Sep 26 '19 at 00:56
  • Edited to clarify the same, hope OK. Validity of embedding pseudosphere into 3-space was firmly established by Eugenio Beltrami. – Narasimham Sep 26 '19 at 11:02

1 Answers1

1

I think you are mis-applying the Gauss-Bonnet theorem. It says

$$\oint k_g\,ds+\iint K_G\,dS=2\pi\,\chi$$

where $S$ is surface area of a region of the surface, $K_G$ is its Gaussian curvature, $s$ is arclength along the region's boundary, $k_g$ is its geodesic curvature (positive inward, negative outward), and $\chi$ is the region's Euler characteristic. $\oint k_g\,ds$ is equivalent to your $\Sigma\psi$.

If the $x$-axis is the tractricoid's axis of symmetry (going left to right in the wiki image), and $y=r\cos\theta,\;z=r\sin\theta$, then the area of a region $0<x_2<x<x_1$ is indeed

$$S=2\pi a(r_2-r_1).$$

But this region is not a disk! It's topologically an annulus, with Euler characteristic $\chi=0$, not $1$, so the Gauss-Bonnet theorem gives

$$\Sigma\psi=-\iint(-1/a^2)\,dS+2\pi\,0$$

$$=S/a^2=2\pi(r_2-r_1)/a.$$

The right half of the tractricoid has $x_2\to0,\;x_1\to\infty$, or equivalently $r_2\to a,\;r_1\to0$, so the area is $2\pi a^2$.

$\Sigma\psi=\Sigma\psi_2+\Sigma\psi_1$ involves the two circles at $x_2$ and $x_1$; but the circle on the right has constant geodesic curvature $-1/a$ regardless of $x_1$, so $\Sigma\psi_1=\int(-1/a)\,ds=-2\pi r_1/a$, and likewise the circle on the left has $\Sigma\psi_2=2\pi r_2/a$, so in fact we've just verified the theorem in this case.

Anyway, as $r_2\to a$, we have $\Sigma\psi_2\to2\pi$. This is to be expected, because the tractricoid's tangent plane approaches being vertical, that is, parallel to the circle's extrinsic curvature vector; so the geodesic curvature is the same as the extrinsic curvature.

mr_e_man
  • 5,364
  • $\chi=0$ here not homeomorphic to torus? – Narasimham Oct 01 '19 at 06:48
  • The Euler characteristic of an orientable surface is $\chi=2-2g-b$, where $g$ is the genus, and $b$ is the number of circular boundaries, or the number of disks removed from a surface without boundary. The annulus is a sphere ($g=0$) with two disks removed ($b=2$); thus $\chi=0$. – mr_e_man Oct 01 '19 at 18:58
  • ...Or use $\chi=V-E+F$ on a polygonization of the annulus, for example as four trapezoids between a big square and a small square inside it: $V=8,E=12,F=4$, so $\chi=0$. Surfaces with the same Euler characteristic are not necessarily homeomorphic if they have boundaries (or if they differ in orientability). – mr_e_man Oct 01 '19 at 19:14