📂Geometry

Definition of Normal Sections and Menelaus's Theorem

Definition¹

Let’s suppose a curve $\boldsymbol{\gamma}$ is given on a surface $M$. We denote by $\Pi$ the plane generated by the normal $\mathbf{n}(p)$ and $\boldsymbol{\gamma}^{\prime}(p) \in T_{p}M$ at $p \in M$. The normal section at $M \cap \Pi$ in the direction from $p$ to $\boldsymbol{\gamma}^{\prime}$ on $M$ is referred to as $M \cap \Pi$.

Theorem²

Let’s denote by $\boldsymbol{\gamma}(s)$ the unit-speed curve on the surface $M$, which has the normal curvature $\kappa_{n}$ at the point $p$. And let’s call $\tilde{\boldsymbol{\gamma}}$ the normal section. Then, the curvature $\tilde{\kappa}$ of the plane curve $\tilde{\boldsymbol{\gamma}}$ satisfies the following equation.

$$| \kappa_{n} |= \tilde{\kappa}$$

Description

This is known as Meusnier’s theorem. Meusnier is French, and it seems like in Papago it’s pronounced [무스니어] and [뫼니에] in Google.

The normal section is sometimes translated as a legal surface or perpendicular surface, but since it actually represents a curve on a surface, such translation is not considered proper simplification. Although the Korean Mathematical Society might refer to it as a vertical section line, simply calling it a normal section seems most appropriate.

A normal section $\tilde{\boldsymbol{\gamma}}$ appears as a spatial curve when viewed from $M$, but also as a plane curve on $\Pi$.

Proof

Auxiliary Lemma

Let’s consider $\alpha, \beta$ as a regular curve satisfying $\alpha (0) = \beta (0)$. If the velocity vectors of the two curves satisfy $\alpha^{\prime}(0) = \lambda \beta ^{\prime}(0)$ with respect to $\lambda \ne 0$, then when $t=0$, the normal curvatures $\kappa_{n}$ of the two curves are the same.

By the auxiliary lemma, the normal curvatures of the two curves $\boldsymbol{\gamma}, \tilde{\boldsymbol{\gamma}}$ are both $\kappa_{n}$. At this moment, the normal at the point $p$ of $\tilde{\boldsymbol{\gamma}}$ is $\pm \mathbf{n}$.

Moreover, according to the properties of plane curvature, the plane curvature $\tilde{k}$ of $\tilde{\boldsymbol{\gamma}}$ is as follows.

$$| \tilde{k} | = \tilde{\kappa}$$

Then, by the definitions of plane curvature and normal curvature,

$$\tilde{k} =\pm \left\langle \tilde{\boldsymbol{\gamma}}^{\prime \prime}, \mathbf{n} \right\rangle = \pm \kappa_{n}$$

Therefore,

$$\left| \kappa_{n} \right| = | \tilde{k} | = \tilde{\kappa}$$

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