the shortest path between two points and the unit sphere and the arc of the great circle

Question

Prove that the shortest path between two points on the unit sphere is an arc of a great circle connecting them

Great Circle: the equator or any circle obtained from the equator by rotating further: latitude lines are not the great circle except the equator

I need help with starting this question, because I am not quite sure how to prove this.

Answer 1

HINT: (Edited 8/27/2021) Start with two points on the equator. Every great circle (except one) meets the shorter great circle arc joining them in at most one point. Let $\Sigma$ be the set of great circles meeting it in one point. Show that for any other curve $C$ joining the points, there must be an open set containing $\Sigma$ of great circles meeting $C$ in at least two points.

Answer 2

Two approaches are outlined in the following:

The great circle is an intersection between the sphere and plane. It is a geodesic, a displaced equator with vanishing geodesic curvature. It contains the sphere center. When it does not contain the sphere center it is a small circle... having non-zero geodesic curvature $k_g.$

Many small circles go through two points A and B of polar angles $ \theta_{1,2}$ and among them the shortest distance geodesic line runs. It is characterized by the Clairaut's Law derived in text books of differential geometry with Liouville thm/ formula for geodesic curvature $ k_g=0$, briefly outlined here using standard notation of the second and first fundamental forms of surface theory:

$$k_g= \psi' + P u'+ Q v';\; P * 2HE= 2 EF_1-FE_1-EE_2;\;Q* 2HE= EG_1-FE_2; \;$$

Where subscripts $1,2 $ are for meridian and circumferential directions.

The same can be also established by arc length (here independent variable s=arc used for priming ) minimization in Variational calculus using cylindrical coordinates $ ( r, \theta, z )$.

$$ds =\sqrt{ (r d \theta)^2 + dr^2+ dz^2 }= \sqrt{ r ^2 + r{'^2}+ z{'^2}}d \theta$$

Using Euler-Lagrange theorem of two dependent variables $ r,\theta $ from $ (r,\theta,z)$

$$\sqrt{ r ^2 + r{'^2}+ z{'^2}}-r'\cdot \frac{r'}{\sqrt{ r ^2 + r{'^2}+ z{'^2}}}-z'\cdot \frac{z'}{\sqrt{ r ^2 + r{'^2}+ z{'^2}}}= const $$

$$ \frac{r^2}{\sqrt{ r ^2 + r{'^2}+ z{'^2}}}= r\cdot \sin \psi = const = r_{min}$$

where the constant $r_{min}$ is interpreted as the small circle radius which is tangential to all red great circles obtained by rotation ofall great circles with respect to the axis of symmetry $ r=0 $ shown.

Projections of great circles on $ (x/y, y/z, z/x) $ planes are all ellipses with inclination $\alpha = \sin ^{-1}\dfrac{r_{min}}{a} $

The following relations hold $ \phi= slope, \psi = $ angle the great circle makes to meridian.

$$ \cos\phi= \frac{r}{a},\;\sin \psi \cos \phi= \frac{r_{min}}{a}$$

Answer 3

Calculus of Variations

Using latitude, $\phi$ and longitude, $\theta$, we have $$ 1=\cos^2(\phi)\left(\frac{\mathrm{d}\theta}{\mathrm{d}s}\right)^2+\left(\frac{\mathrm{d}\phi}{\mathrm{d}s}\right)^2\tag1 $$ Integrating $(1)$ with respect to $s$, taking the first variation, and integrating by parts, we get $$ \scriptsize\delta s=\int\left[\left(\color{#C00}{2\sin(\phi)\cos(\phi)\,\frac{\mathrm{d}\theta}{\mathrm{d}s}\,\frac{\mathrm{d}\phi}{\mathrm{d}s}-\cos^2(\phi)\,\frac{\mathrm{d}^2\theta}{\mathrm{d}s^2}}\right)\delta\theta-\left(\color{#090}{\sin(\phi)\cos(\phi)\,\left(\frac{\mathrm{d}\theta}{\mathrm{d}s}\right)^2+\frac{\mathrm{d}^2\phi}{\mathrm{d}s^2}}\right)\delta\phi\right]\mathrm{d}s\tag2 $$ Orthogonality requires that a curve where we have $\delta s=0$ for any $\delta\theta$ and $\delta\phi$ must satisfy $$ \color{#C00}{\frac{\mathrm{d}^2\theta}{\mathrm{d}s^2}=2\tan(\phi)\,\frac{\mathrm{d}\theta}{\mathrm{d}s}\,\frac{\mathrm{d}\phi}{\mathrm{d}s}}\tag3 $$ and $$ \color{#090}{\frac{\mathrm{d}^2\phi}{\mathrm{d}s^2}=-\sin(\phi)\cos(\phi)\,\left(\frac{\mathrm{d}\theta}{\mathrm{d}s}\right)^2}\tag4 $$

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Solving the Equations $$ \begin{align} \log\left(\frac{\mathrm{d}\theta}{\mathrm{d}s}\right) &=2\log(\sec(\phi))+\log(\cos(\epsilon))\tag{5a}\\ \frac{\mathrm{d}\theta}{\mathrm{d}s} &=\cos(\epsilon)\sec^2(\phi)\tag{5b}\\ \left(\frac{\mathrm{d}\phi}{\mathrm{d}s}\right)^2 &=1-\cos^2(\epsilon)\sec^2(\phi)\tag{5c}\\ s &=\int\frac{\mathrm{d}\phi}{\sqrt{1-\cos^2(\epsilon)\sec^2(\phi)}}\tag{5d}\\ &=\int\frac{\mathrm{d}\sin(\phi)}{\sqrt{\cos^2(\phi)-\cos^2(\epsilon)}}\tag{5e}\\ &=\int\frac{\mathrm{d}\sin(\phi)}{\sqrt{\sin^2(\epsilon)-\sin^2(\phi)}}\tag{5f}\\ &=\sin^{-1}\left(\frac{\sin(\phi)}{\sin(\epsilon)}\right)+s_0\tag{5g}\\[9pt] \end{align} $$ Explanation:
$\text{(5a)}$: divide $(3)$ by $\frac{\mathrm{d}\theta}{\mathrm{d}s}$ and integrate
$\text{(5b)}$: remove the log
$\text{(5c)}$: apply $(1)$ to $\text{(5b)}$
$\text{(5d)}$: separate the differential equation and integrate
$\text{(5e)}$: multiply the integrand by $\frac{\cos(\phi)}{\cos(\phi)}$
$\text{(5f)}$: $\cos^2(x)=1-\sin^2(x)$
$\text{(5g)}$: evaluate the integral

Solving $\text{(5g)}$ for $\sin(\phi)$ gives $$ \bbox[5px,border:2px solid #C0A000]{\sin(\phi)=\sin(s-s_0)\sin(\epsilon)}\tag6 $$ Furthermore, $$ \begin{align} \theta &=\int\frac{\cos(\epsilon)\,\mathrm{d}(s-s_0)}{1-\sin^2(s-s_0)\sin^2(\epsilon)}\tag{7a}\\ &=\int\frac{\cos(\epsilon)\,\mathrm{d}\tan(s-s_0)}{\sec^2(s-s_0)-\tan^2(s-s_0)\sin^2(\epsilon)}\tag{7b}\\ &=\int\frac{\cos(\epsilon)\,\mathrm{d}\tan(s-s_0)}{1+\tan^2(s-s_0)\cos^2(\epsilon)}\tag{7c}\\[6pt] &=\tan^{-1}(\tan(s-s_0)\cos(\epsilon))+\theta_0\tag{7d} \end{align} $$ Explanation:
$\text{(7a)}$: apply $\text{(5h)}$ to $\text{(5b)}$, separate the differential equation, and integrate
$\text{(7b)}$: multiply the integrand by $\frac{\sec^2(s-s_0)}{\sec^2(s-s_0)}$
$\text{(7c)}$: $\sec^2(x)=1+\tan^2(x)$
$\text{(7d)}$: evaluate the integral

Solving $\text{(7d)}$ for $\tan(\theta-\theta_0)$ gives $$ \bbox[5px,border:2px solid #C0A000]{\tan(\theta-\theta_0)=\tan(s-s_0)\cos(\epsilon)}\tag8 $$

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Compare with a Great Circle

Given the parameterization in $\mathbb{R}^3$ of a great circle $$ (x,y,z)=(\cos(s-s_0),\sin(s-s_0)\cos(\epsilon),\sin(s-s_0)\sin(\epsilon))\tag9 $$ we get $$ \sin(\phi)=z=\sin(s-s_0)\sin(\epsilon)\tag{10} $$ and $$ \tan(\theta-\theta_0)=\frac yx=\tan(s-s_0)\cos(\epsilon)\tag{11} $$ Equation $(10)$ matches equation $(6)$ and equation $(11)$ matches equation $(8)$. Thus, the shortest path between two points on a sphere is an arc of a great circle.

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Plotting the Solved Equations

Since $\phi\in\left[-\frac\pi2,\frac\pi2\right]$, we can get $\phi$ by applying $\sin^{-1}$ to $(6)$: $$ \bbox[5px,border:2px solid #C0A000]{\phi=\sin^{-1}(\sin(\epsilon)\sin(s-s_0))}\tag{12} $$ However, to get $\theta-\theta_0$, we cannot simply apply $\tan^{-1}$ to $(8)$. Instead, we note that $$ \begin{align} \tan((\theta-\theta_0)-(s-s_0)) &=\frac{\tan(\theta-\theta_0)-\tan(s-s_0)}{1+\tan(\theta-\theta_0)\tan(s-s_0)}\tag{13a}\\ &=\frac{\tan(s-s_0)(\cos(\epsilon)-1)}{1+\cos(\epsilon)\tan^2(s-s_0)}\tag{13b} \end{align} $$ Explanation:
$\text{(13a)}$: formula for the tangent of a difference
$\text{(13b)}$: apply $(8)$

Thus, we get $$ \bbox[5px,border:2px solid #C0A000]{\theta-\theta_0=(s-s_0)-\tan^{-1}\left(\frac{(1-\cos(\epsilon))\tan(s-s_0)}{1+\cos(\epsilon)\tan^2(s-s_0)}\right)}\tag{14} $$

Great circles with $\epsilon=\left\{\color{#DF0000}{0^{\large\circ}},\color{#FF8F00}{15^{\large\circ}},\color{#EFDF00}{30^{\large\circ}},\color{#00BF00}{45^{\large\circ}},\color{#0000FF}{60^{\large\circ}},\color{#8000FF}{75^{\large\circ}},\color{#CC00FF}{90^{\large\circ}}\right\}$ plotted using $(12)$ and $(14)$.