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I wish to use the Gauss-Bonnet theorem to calculate $\int_{M_r} K dA$ for the surface $$M_r = \{ (x,y,z)\in\mathbb{R}^3 |z = \cos\sqrt{x^2 + y^2}, x^2 + y^2 < r^2, x,y, > 0 \}.$$ It feels natural to change to cylindrical coordinates $(\rho, \theta, z)$, so this is what I do. Namely, I set $x = \rho\cos \theta$, $y = \rho\cos\theta$ and $z = z$. Then $$M_r = \{ (\rho,\theta,z)\in\mathbb{R}^3 |z = \cos\rho, \rho < r, 0 < \theta < \pi/2 \}.$$

Then the surface looks like this (and forgive my terrible drawing skills): enter image description here

The arrows indicate the direction I need to path-integrate in. The direction of the normal will be justified presently. This surface is a level set, so its normal is given by $N = \nabla(z - \cos\rho) = (\sin\rho, 0, 1)$. So the normal points as in the picture, namely 'up'.

I now wish to use $$\int_{M_r}K dA = \sum \alpha_i - (n-2)\pi - \int k_g ds,$$ where the $\alpha_i$s denote the angles between the curves $\gamma$, $n$ the number of angles, and the final integral the path-integral of the geodesic curvature along the curves $\gamma$. So, $$ \int_{M_r}K dA = \pi/2 - \int k_g ds ,$$ since all the angles are $\pi/2$.

Now, because of symmetry we have $$ \int_{\gamma_1} k_g ds = -\int_{\gamma_3} k_g ds, $$ and so we only need to parametrize $\gamma_2$. At $\gamma_2$ we have $r = \rho$ and so we may parametrize it by $$\gamma_2(\theta) = (r, \theta, \cos r),$$ which clearly is an arclength-parametrization. Indeed, we have $$ \gamma_2'(t) = (0, 1, 0), \\ \gamma_2''(t) = (0, 0, 0). $$ Calculating the geodesic curvature is therefore easy, it's just $k_g = 0$. This is where I feel like I've made some mistake. It doesn't seem reasonable that the geodesic curvature along $\gamma_2$ is identically zero. If it is, the answer is just $$\int_{M_r} = \pi/2,$$ independent on what value $r$ takes, which seems strange to me.

Peatherfed
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1 Answers1

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You have not properly parametrized $\gamma_2$. Even if you use cylindrical coordinates, this does not mean you can map everything "to $(\rho,\theta,z)$ space". From your drawing it should be clear that $\gamma_2$ is a circunference, not a straight line as your parametrization suggests!

The correct parametrization: at $\gamma_2$ you would have $\rho = r$ and thus $z=\cos r$, constant. The $x$ and $y$ coordinates on the curve can be parametrized in terms of $\theta$: $\gamma_2(\theta) = (r \cos\theta,r\sin\theta, \cos r)$. This is the curve that you need to calculate the geodesic curvature of.

topolosaurus
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  • I don't understand. $\gamma_2$ is a straight line in $(\rho, \theta, z)$. I did the same when calculating the normal, where I worked entirely in $(\rho, \theta, z)$. – Peatherfed Oct 28 '21 at 15:07
  • That is the heart of the misconception: $\gamma_2$ is a straigh line in coordinate space, the piece of plane where your variables $\rho,\theta$ take values. You can parametrize your surface by these coordinates. In 3D space this curve is far from a straigh line, in exactly the same way that $M_r$ is very different from a flat portion of the plane.

    I would suggest spending some time unraveling what coordinate charts on surfaces are, and how curves on the parameter space correspond to curves on the surface.

     – topolosaurus Oct 28 '21 at 15:15
  • The normal calculation is also wrong. If you want to use cylindrical coordinates you cannot calculate the gradient how you are doing it. Check https://en.wikipedia.org/wiki/Gradient#Cylindrical_and_spherical_coordinates – topolosaurus Oct 28 '21 at 15:17
  • I think I get it. Would you then also disagree with how I rewrote the surface $M_r$? Namely, I completely threw out $(x,y,z)$ and replaced it with $(\rho, \theta, z)$. The whole reason I switched to cylindrical coordinates, and into "$(\rho, \theta, z)$ space" is so that the part of the circle WOULD become a straight line. – Peatherfed Oct 28 '21 at 15:29
  • If I am to be completely technically correct, then the equality $M_r = { (\rho,\theta,z)\in \mathbb{R}^3 | \cdots }$ is incorrect. The points are $(x,y,z)$ in $\mathbb{R}^3$, and I understand $\rho,\theta,z$ as functions defined on $\mathbb{R}^3$. A way to fix this would be to write, $M_r = { (x,y,z)\in\mathbb{R}^3 | cos \rho(x,y,z) = z, \rho(x,y,z) < r, \theta(x,y,z) < \pi/2 }$. – topolosaurus Oct 28 '21 at 15:47
  • The whole reason behind not being able to "translate into $(\rho,\theta,z)$-space" is not because of some arbitrary notion: you are implicitly setting up a diffeomorphism that does not preserve angles, distances, etc... and thus "doing geometry in parameter space" requires understanding precisely the relationship between what you have in $\mathbb{R}^3$ and what parameters you are using to describe it. – topolosaurus Oct 28 '21 at 15:50