📂Geometry

If It's the Shortest Curve, It's a Geodesic

Theorem¹

Let’s say $\boldsymbol{\gamma}$ is a unit speed curve connecting two points $P = \boldsymbol{\gamma}(a), Q = \boldsymbol{\gamma}(b)$ on surface $M$. If $\boldsymbol{\gamma}$ is the shortest distance curve connecting $P$ and $Q$, then $\boldsymbol{\gamma}$ is a geodesic.

Explanation

The converse does not hold. In other words, a geodesic is not necessarily the shortest distance curve.

Proof

Strategy: Prove by contradiction. What needs to be shown is $\kappa_{g} = 0$, so assume $\kappa_{g} \ne 0$, and show that this leads to a contradiction.

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Assume $a \lt s_{0} \lt b$ and let $\kappa_{g}$ be the geodesic curvature of $\boldsymbol{\gamma}$. Now assume $\kappa_{g}(s_{0}) \ne 0$. Then, since $\boldsymbol{\gamma}$ is continuous, there exists $c, d$ satisfying:

• $\kappa_{g}([c,d]) \ne 0$
• $a \lt c \lt s_{0} \lt d \lt b$
• Regarding coordinate patch mapping $\mathbf{x}$, $\boldsymbol{\gamma}([c,d]) \subset \mathbf{x}(U)$

Now let’s define function $\lambda : [c,d] \to \mathbb{R}$ as follows.

$$ \lambda (c) = \lambda (d) = 0 \quad \text{and} \quad \lambda (s_{0}) \ne 0 \quad \text{and} \quad \lambda (s)\kappa_{g}(s) \ge 0 \quad \text{ for } c\le s \le d $$

Since $\mathbf{S} = \mathbf{n} \times \boldsymbol{\gamma}^{\prime}$ is in the tangent space, let’s say $\lambda (s) \mathbf{S} = \sum v^{i}(s)\mathbf{x}_{i}$ regarding some $v^{i} : [c,d] \to \mathbb{R}$.

Let’s assume it’s given as $\boldsymbol{\gamma}(s) = \mathbf{x}(\gamma^{1}(s), \gamma^{2}(s))$. Furthermore, consider the perturbation $\boldsymbol{\alpha}(s ;t)$ of $\boldsymbol{\gamma}$ for sufficiently small $t$ as below.

$$ \boldsymbol{\alpha}(s ;t) = \mathbf{x}\left( \gamma^{1}(s) + t v^{1}(s), \gamma^{2}(s) + t v^{2}(s) \right) $$

$\boldsymbol{\alpha}$ is a curve from $\boldsymbol{\gamma}(c)$ to $\boldsymbol{\gamma}(d)$, and $\boldsymbol{\alpha}(s ; 0) = \boldsymbol{\gamma}(s)$. Let the length of $\boldsymbol{\alpha}(s; t)$ be $L(t)$.

$$ L(t) = \int_{c}^{d} \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle ^{1/2} ds $$

Then $\boldsymbol{\alpha}(s, 0) = \boldsymbol{\gamma}$, and since $\boldsymbol{\gamma}$ is the shortest distance curve, $L(t)$ has its minimum value when $t = 0$. Also, since $L(0) \lt L(t^{\ast} \ne 0)$, $L^{\prime}(0) = 0$. On the other hand, if we calculate $L^{\prime}$,

$$ \begin{align*} L^{\prime}(t) =&\ \dfrac{d }{d t} \int_{c}^{d} \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle ^{1/2} ds = \int_{c}^{d} \dfrac{\partial }{\partial t}\left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle ^{1/2} ds \\[1em] =&\ \int_{c}^{d} \dfrac{1}{2} \dfrac{2\left\langle \dfrac{\partial^{2} \boldsymbol{\alpha}}{\partial t \partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle }{\left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle ^{1/2}}ds = \int_{c}^{d}\dfrac{\left\langle \dfrac{\partial^{2} \boldsymbol{\alpha}}{\partial t \partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle }{\left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle ^{1/2}}ds \end{align*} $$

At this point, when $\boldsymbol{\gamma}$ is a unit speed curve, at $t=0$,

$$ \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle = \left\langle \dfrac{\partial \boldsymbol{\gamma}}{\partial s}, \dfrac{\partial \boldsymbol{\gamma}}{\partial s} \right\rangle = 1 $$

Therefore,

$$ L^{\prime}(0) = \int_{c}^{d} \left. \left\langle \dfrac{\partial^{2} \boldsymbol{\alpha}}{\partial t \partial s}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle \right|_{t=0} ds $$

Here, since $\dfrac{d}{ds} \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial t}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle = \left\langle \dfrac{\partial ^{2} \boldsymbol{\alpha}}{\partial s \partial t}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle + \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial t}, \dfrac{\partial ^{2} \boldsymbol{\alpha}}{\partial s^{2}} \right\rangle$, substituting into the above equation,

$$ \begin{align*} L^{\prime}(0) =&\ \int_{c}^{d} \left. \dfrac{d}{ds} \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial t}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle \right|_{t=0} - \left. \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial t}, \dfrac{\partial ^{2} \boldsymbol{\alpha}}{\partial s^{2}} \right\rangle \right|_{t=0} ds \\[1em] =&\ \left[ \left. \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial t}, \dfrac{\partial \boldsymbol{\alpha}}{\partial s} \right\rangle \right|_{t=0} \right]_{c}^{d} - \int_{c}^{d}\left. \left\langle \dfrac{\partial \boldsymbol{\alpha}}{\partial t}, \dfrac{\partial ^{2} \boldsymbol{\alpha}}{\partial s^{2}} \right\rangle \right|_{t=0} ds \end{align*} $$

At this point, $\left. \dfrac{\partial \boldsymbol{\alpha}}{\partial t} \right|_{t=0} = \sum v^{i}(s) \mathbf{x}_{i} = \lambda (s) \mathbf{S}$, but since $\lambda (c) = \lambda (d)=0$, the first term is $0$. Therefore, $\left. \dfrac{\partial ^{2} \boldsymbol{\alpha}}{\partial s^{2}}\right|_{t=0} = \boldsymbol{\gamma}^{\prime \prime} = \kappa_{g}\mathbf{S} + \kappa_{n}\mathbf{n}$, and since it was $L^{\prime}(0) = 0$,

$$ \begin{align*} 0 = L^{\prime}(0) =&\ 0 - \int_{c}^{d} \left\langle \lambda (s) \mathbf{S}, \kappa_{g}(s) \mathbf{S} + \kappa_{n}(s) \mathbf{n} \right\rangle ds \\ =&\ -\int _{c}^{d} \lambda (s) \kappa_{g}(s) ds \end{align*} $$

However, we assumed $\lambda (s) \kappa_{g}(s) \gt 0$, so there is a contradiction.

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