How to convince a student without calculus that great circles are geodesics in a sphere?

Question

how to convince or demonstrate to a high school student who does not know differential and integral calculus that the geodesics of a sphere are arcs of great circles?

Answer 1

Recall that a great circle is the intersection of the unit sphere and a plane passing through the origin. A key point is that for short arcs of great circles, the ratio of euclidean distance between endpoints to the length of the arc is close to $1$.

Let $(x,y,z)$ be the standard coordinates in $\mathbb R^3$. We work in spherical coordinates $(\phi,\theta)$ where $z=\cos\phi$. Let $c(t)$ be a path on the sphere joining north pole $A$ to south pole $B$, so that $c(0)=A$ and $c(1)=B$. Denote by $\phi(t)$ the $\phi$-coordinate of the point $c(t)$, so that $\phi(0)=0$ and $\phi(1)=\pi$.

Now suppose the path $c$ has length $|c|$ strictly less than $\pi$, and write $$|c|=\pi-\epsilon$$ where $\epsilon>0$. Let $N$ be an integer. Consider a partition $0=t_0<t_1<\cdots<t_N=1$ that breaks up $c$ into paths $c([t_i,t_{i+1}])$ of equal length $\frac{\pi-\epsilon}{N}$. Consider the corresponding sequence $\phi(t_0),\ldots,\phi(t_N)$. Choose $t_k$ which maximizes the difference $\phi(t_{k+1})-\phi(t_k)$. Then $\phi(t_{k+1}-\phi(t_k)\geq\frac{\pi}{N}$. Then we obtain $$\frac{|c([t_k,t_{k+1}])|}{\phi(t_{k+1})-\phi(t_k)}\leq\frac{\pi-\epsilon}{\pi}=1-\frac{\epsilon}{\pi}.$$

Thus the ratio is uniformly bounded way from $1$. As $N$ tends to infinity, the length of the path $c([t_k,t_{k+1}])$ tends to zero. But for such paths, the ratio of the Euclidean distance $|c(t_{k+1})-c(t_k)|$ to length of arc of great circle is bounded below by a bound that tends to $1$ by elementary plane trigonometry (requiring no calculus). Therefore for sufficiently large $N$, we will have $$\frac{c(t_{k+1})-c(t_k)}{\phi(t_{k+1})-\phi(t_k)}<1-\frac{\epsilon}{2\pi},$$ a contradiction. The contradiction shows (without calculus and using only elementary plane trigonometry) that a path joining north and south pole must have length at least $\pi$, and therefore great circles are length-minimizing.

Answer 2

Take a physical sphere such as a beach ball, and a string. Pick two points. Hold the string down with one finger at one point then stretch it to the second point. Next, holding the string tight at both points, continue to wrap the string. You find the wrapping will be a great circle. You can also try this with a rubber band.

A less "physical" approach is this. Without loss of generality after rotation, one of the two points is the North Pole. From this configuration, it is rather intuitive that the shortest path to the second point is to follow the longitudinal circle from the North Pole to the second point.

Answer 3

Answer inspired by Michał Miśkiewicz's comment on the OP.

To a high school student:

Put an ant on a basketball. Draw a tiny arrow representing the direction it should walk. The ant always just puts one foot in front of the other, trying to stay as "straight" as possible. Where will it end up?

Firstly: do you agree that it has one, and only one, path it can follow?

Now, let's draw the great circle containing this point and vector. Can the ant ever leave the great circle?

Imagine that at some time the ant leaves the circle. It has to leave to one side of the great circle or the other. However, if it leaves to one side then reflecting the path over the plane containing the great circle would give another path with identical length. So the ant would have two paths it can follow. This contradicts what we said earlier about the ant only having one path it can take. So the ant can never leave the great circle.

Answer 4

Maybe this would work? The spherical law of cosines says that, on a sphere of radius $R$, if you have a sphercial triangle with edge lengths $a$, $b$, $c$ and angles $A$, $B$, $C$, then $$\cos \tfrac{c}{R} = \cos \tfrac{a}{R} \cos \tfrac{b}{R} + \sin \tfrac{a}{R} \sin \tfrac{b}{R} \cos C. \qquad (\ast)$$ I emphasize that the definition of "spherical triangle" is "a triangle whose sides are great circles", and this relationship can be proved without knowing in advance that those great circles are geodesics; see the several proofs in the Wikipedia article.

Since $\cos C \leq 1$, we deduce that $$\cos \tfrac{c}{R} \geq \cos \tfrac{a}{R} \cos \tfrac{b}{R} - \sin \tfrac{a}{R} \sin \tfrac{b}{R} = \cos \left( \tfrac{a}{R}+\tfrac{b}{R} \right).$$ We deduce that $\tfrac{c}{R} \leq \tfrac{a}{R} + \tfrac{b}{R}$ or, in other words $c \leq a+b$.

So, if $x_1$, $x_2$ and $x_3$ are the corners of the triangle, then the distance $x_1 \to x_3$ along a great circle is $\leq$ the distance along a path $x_1 \to x_2 \to x_3$ which takes a great circle segment along each arrow.

Then, inductively, for any $x_1$, $x_2$, ..., $x_n$, the distance along a great circle $x_1 \to x_n$ is $\leq$ the distance along the path $x_1 \to x_2 \to \cdots \to x_n$ which takes a great circle segment along each arrow.

Then (wave your hands furiously) the length of a great circle path $x_1 \to x_n$ is $\leq$ the length of any path $x_1 \to x_n$.

To do better than this, you need a precise definition of arc length, which may not be somewhere you want to go.

Answer 5

Determine the shortest route from New York City USA to Rome Italy using a piece of string on a globe. Both cities are south of the 45th parallel and yet the the shortest route deviates considerably to its north. The string traces the geodesic (the great circle arc) between the two cities. It is the shortest and "straightest" line on the surface of the earth between both cities.

Similarly, in the southern hemisphere, determine the shortest route from Perth Australia to Durban South Africa. It deviates considerably to the south.

Answer 6

Since you ask specifically for a way to "convince" the student rather than to provide a proof, I would propose the following. We can assume for simplicity that the sphere has unit radius, and measure angles in degrees. Move one of the points, say $A$, to the North pole by an isometry (e.g., a suitable rotation). If there were a path from $A$ to $B$ shorter than the smaller arc of great circle connecting them, we could continue further beyond $B$ along the longitude through $A$ to $B$, to arrive to the south pole $C$ by a path of combined length less than $180^\circ$. But can one reach north pole to south pole without crossing all the latitudes? Certainly not.

Answer 7

There is a problem here with, er, circular definitions. How are you going to define a great circle? My preferred definition would be that it's a geodesic (one that's been extended as far as possible). If you want to make an argument that's at all non-vacuous, you need to give some other characterization of a great circle -- or at least a partial characterization.

One partial characterization would be that a great circle splits a sphere into two equal hemispheres. Now suppose a plane heads east from New York and flies straight and level, which is actually a physicist's definition of geodesic -- inertial motion. So now we have at least some kind of independent characterization of these two notions, and it can be non-vacuous to try to relate them.

One might expect that the plane would stay on a line of latitude. This would split the globe into two unequal parts, and would therefore not be a great circle. But then we could draw a similar path that was a mirror image of the first one, with the smaller, right-hand region being centered on some spot off the coast of Peru rather than on the south pole. Since the poles have no geometrical significance, there is no reason to prefer one of these paths to the other. Appealing to physical intuition, inertial motion should be unique. There is no reason for the plane to prefer one of these circles to the other. Therefore the line of latitude can't be a geodesic.