📂Geometry

Gaussian Curvature with Negative Values on Rotational Surfaces

Overview¹

The document explains rotation surfaces with negative Gaussian curvature.

Description

The curvature of the rotation surface is $K = - \dfrac{r^{\prime \prime}}{r}$, therefore $r^{\prime \prime} - a^{2}r = 0$. The solution of this differential equation is as follows:

$$r(s) = c_{1} \cosh(as) + c_{2}\sinh(as)$$

This can be written for some appropriate constant $B, b, C, c \in \mathbb{R}$ as follows:

$$r(s) = c_{1} \cosh (as) + c_{2} \sinh (as) = \begin{cases} B\cosh(as + b) & \text{if } c_{1} \gt c_{2} \\ A e^{as} & \text{if } c_{1} = c_{2} = A \\ C\sinh(as + c) & \text{if } c_{1} \lt c_{2} \end{cases}$$

By choosing the starting point of the curve $\boldsymbol{\alpha}$ that forms the rotation surface so that $b=0, c=0$, we can infer that, due to $z^{\prime} = \pm\sqrt{1 - (r^{\prime})^{2}}$, the rotation surface is divided into the following three cases:

$$\begin{equation} \begin{cases} r(s) &= A e^{as} \\ z(s) &= \displaystyle \pm \int_{0}^{s} \sqrt{1 - a^{2}A^{2}e^{2at}}dt + D \end{cases} \end{equation}$$

$$\begin{equation} \begin{cases} r(s) &= B \cosh (as) \\ z(s) &= \displaystyle \pm \int_{0}^{s} \sqrt{1 - a^{2}B^{2}\sinh^{2}(at)}dt + D \end{cases} \end{equation}$$

$$\begin{equation} \begin{cases} r(s) &= C \sinh (as) \\ z(s) &= \displaystyle \pm \int_{0}^{s} \sqrt{1 - a^{2}C^{2}\cosh^{2}(at)}dt + D \end{cases} \end{equation}$$

Example

Let’s examine the case of $(1)$. For $z(s)$ to be well-defined, it must be $A a e^{as} \le 1$. Therefore,

$$s \le \dfrac{1}{a}\ln \dfrac{1}{aA}$$

Now, let’s set $aAe^{at} = \sin \phi$.

$$a^{2}Ae^{at} dt = \cos \phi d \phi \implies dt = \dfrac{\cos\phi}{a^{2}Ae^{at}}d\phi = \dfrac{\cos\phi}{a\sin\phi}d\phi$$

$z(s)$ is as follows:

$$\begin{align*} z(s) &= \pm \int _{\sin^{-1}(Aa)}^{\sin^{-1}(Aae^{as})} \sqrt{1 - \sin^{2}\phi} \dfrac{\cos \phi}{a \sin \phi} d\phi \\ &= \pm \int _{\sin^{-1}(Aa)}^{\sin^{-1}(Aae^{as})} \cos \phi \dfrac{\cos \phi}{a \sin \phi} d\phi \\ &= \pm \dfrac{1}{a}\int _{\sin^{-1}(Aa)}^{\sin^{-1}(Aae^{as})} \dfrac{1 - \sin^{2}\phi}{\sin \phi} d\phi \\ &= \pm\dfrac{1}{a} \int _{\sin^{-1}(Aa)}^{\sin^{-1}(Aae^{as})} \dfrac{1}{\sin\phi} - \sin \phi d \phi \\ &= \pm \dfrac{1}{a} \left[ \ln (\tan \frac{\phi}{2}) + \cos \phi \right]_{\sin^{-1}(Aa)}^{\sin^{-1}(Aae^{as})} \\ \end{align*}$$

Substituting,

$$z(s) = \pm \dfrac{1}{a} \left( \ln \dfrac{\tan \left( \frac{1}{2}\sin^{-1}(Aae^{as}) \right)}{\tan \left( \frac{1}{2} \sin ^{-1}(Aa) \right)} + \sqrt{1 - (Aae^{as})^{2}} - \sqrt{1 - (Aa)^{2}}\right)$$