📂Geometry

Definition of a Straight Line (Geodesic) in Differential Geometry

Buildup^{1 2}

Let’s assume there is an object moving along a certain curve on the surface $M \subset \mathbb{R}^{3}$. Even if the line looks curved from the perspective of the entire space $\mathbb{R}^{3}$, an object moving on the surface can be thought of as moving straight ahead. Then, such a line can be defined as a straight line (geodesic) on the surface. First, let’s consider the properties of a straight curve on a plane.

1. Curvature is $0$.
2. It is the shortest distance between two points.
3. If there is any $p, q$, there exists a unique straight line connecting the two points.
4. Acceleration and tangent are parallel.

Point 4 is derived from point 1, where if $\alpha^{\prime \prime} = v^{\prime} \mathbf{T} + v^{2}\kappa \mathbf{N}$(../2070#physical-meaning4) results in $\kappa = 0$, then acceleration and tangent are parallel.

To define the concept of a straight line on a surface, let’s focus on ‘1. Curvature is 0’. Locally, the surface can be thought of as the same as the tangent plane. Therefore, from the perspective of the object on the surface, the curve does not appear to be curved if it is not curved on the tangent plane. The expression representing the curvature of a curve on the surface is as follows.

$$ \kappa \mathbf{N} = \mathbf{T}^{\prime} = \alpha^{\prime \prime} = \kappa_{n}\mathbf{n} + \kappa_{g}\mathbf{S} $$

Here, since $\mathbf{S}$ is a vector on the tangent plane, the curvature in this direction being $0$ can be said to be the same as feeling curvature $0$ on the tangent plane. Therefore, let’s define a straight line on the surface as follows.

Definition

If the geodesic curvature $\kappa_{g}$ of the unit speed curve $\gamma : I \to M$ on the surface $M$ is $0$ everywhere, $\gamma$ is called a geodesic, or straight line.

$$ \gamma \text{ is geodesic} \iff k_{g} = 0 $$

Theorem

Given the unit speed curve $\gamma$ on the surface $M$. Let’s say $\gamma$ is a unit speed curve on the surface $M$. The necessary and sufficient condition for $\gamma$ to be a geodesic is as follows.

(a)

$$ \gamma \text{ is geodesic} \iff \left[ \mathbf{n}, \mathbf{T}, \mathbf{T}^{\prime} \right] = 0 $$ If $\mathbf{x}$ is referred to as a simple surface, then $\gamma$ can be represented as $\gamma (s) = \mathbf{x} \left( \gamma^{1}(s), \gamma^{2}(s) \right)$. Then, the following holds true.

(b)

$$ \gamma \text{ is geodesic} \iff (\gamma^{k})^{\prime \prime} + \sum \limits_{i=1}^{2} \sum \limits_{j=1}^{2} {\Gamma _{ij}}^{k} (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} = 0, \quad \forall k=1,2 $$

(c)

$$ \gamma \text{ is geodesic} \iff \gamma ^{\prime \prime}\text{ is normal to } M \text{ at every point.} $$

Proof

(a)

Since $\kappa_{g} = \left[ \mathbf{n}, \mathbf{T}, \mathbf{T}^{\prime} \right]$ holds, it is established.

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(b)

First, it is established that the following expression holds.

$$ \kappa_{g}\mathbf{S} = \sum_{k=1}^{2} \left( (\gamma^{k})^{\prime \prime} + \sum\limits_{i,j}^{2} \Gamma_{ij}^{k} (\gamma^{i})^{\prime} (\gamma^{j})^{\prime} \right) \mathbf{x}_{k} $$

If $\gamma$ is a geodesic, since all the values in the brackets which are components of $\mathbf{x}_{k}$ are $0$, the above equation is a zero vector, and $\kappa_{g}$. Conversely, if $\kappa_{g}=0$, all components of the above vector are $0$, thus it holds true.

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This implies that the component of acceleration in the tangent space is $0$, meaning that $\gamma$ moves at a constant speed on the surface.

(c)

By the result in (b),

$$ \gamma ^{\prime \prime} = \kappa_{n} \mathbf{n} + \kappa_{g}\mathbf{S} = \kappa_{n}\mathbf{n} \perp M $$

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