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In a few days I will be giving a talk to (smart) high-school students on a topic which includes a brief overview on the notions of curvature and of gedesic lines. As an example, I will discuss flight paths and, more in general, shortest length arcs on the 2-dimensional sphere.

Here is my question: let $p,q$ be distinct (and, for simplicity, non-antipodal) points on the sphere $S^2$. Is there an elementary proof of the fact that the unique shortest path between $p$ and $q$ is the shortest arc on the unique great circle containing $p$ and $q$?

By ''elementary'' I mean a (possibly not completely formal, but somehow convinving) argument that could be understood by a high-school student. For example, did ancient Greeks know that the geodesics of the sphere are great circles, and, if so, how did they ``prove'' this fact?

Here is the best explanation that came into my mind so far. Let $gammacolon [0,1]to S^2$ be any path, and let $gamma(t_0)=a$. The best planar approximation of $gamma$ around $a$ is given by the orthogonal projection of $gamma$ on the tangent space $T_a S^2$. This projection is a straight line (that is, a length-miminizing path on $T_a S^2$) only if $gamma$ is supported in a great circle containing $a$. This sounds quite convincing, and with a bit of differential geometry it is not difficult to turn this argument into a proof. However, I was wondering if I could do better...

I doubt if there is a way to show this without calculus, or that Greeks knew how to do it, with possible exception of Archimedes who had already figured out how to take limits (through laborious calculations).

If the high school students are taking, or have already taken, Calculus I, and know how to differentiate, then I would use the notion of acceleration.
I would start by asking: what is a straight line in the plane? One way to characterize it is as the solution to the differential equation
$$
gamma''(t)=0,
$$
where $gamma(t)=(x(t), y(t))$. In particular, they can check that $(a+bt, c+dt)$ yields a solution.
So lines are curves with no acceleration. Thus on the sphere the analogue of a line segment, or the straightest curve possible, may be regarded as a curve with no acceleration in the sphere. More precisely, we want the acceleration vector to be orthogonal to the sphere, or parallel to the position vector. So we need to solve the equation
$$gamma''(t)||gamma(t),$$
where $gamma(t)=(x(t), y(t), z(t))$,
and one quickly checks that great circles, e.g., $(cos(t),sin(t), 0)$ yield a solution.
To drive the point home further, I would draw a straight line on a piece of paper, and then roll the paper to a cylinder to show them that the acceleration vector remains orthogonal to the surface under isometric deformations. More precisely, straight lines turn into helices $(cos(t), sin(t), h,t)$, where again one can quickly check that $gamma''(t)$ is parallel to $(cos(t), sin(t),0)$ and thus is orthogonal to the cylinder.

Ancients did not know line integrals, but they understood that the length of the curve
is the limit of the lengths of inscribed broken lines. To prove the inequality for broken lines (piecewise-great-circles), it is enough to prove that a side of a small
spherical triangle is less than the sum of two other sides. This can be done using spherical trigonometry, which the ancients did know.

The spherical cosine theorem says
$$cos a=cos bcos c+sin asin bcos A,$$
where $a,b,c$ are sides and $A$ is the angle between $b$ and $c$.
Estimating this from below (cosine is decreasing) we replace $cos A$ by $-1$, and obtain
$$cos ageq cos bcos c-sin asin b=cos(a+b),$$
therefore $aleq b+c.$

Spherical cosine theorem is ancient but relatively late (after the epoch of Archimedes), but one can also give an elementary-geometric proof of the triangle inequality for small spherical triangles. All one needs is that
three angles at the vertex of a 3-faced cone (or pyramid) satisfy the triangle inequality.

I think what is needed is a heuristic that would then motivate any further derivation you care to present. For this situation, I would offer curvature. If I have two points and a circular arc going through them, the arc length is greatest when the chord is the diameter, and approaches least when the involved circle is as large as can be managed. (Use some basic trig to back this up if needed.) On the plane, a circle passing through two points can have arbitrarily large radius. On a sphere, the largest circle is a great circle or geodesic.

You can expand on this heuristic to point out various subtleties, such as definition of arc length or area of a subregion of or on a manifold.

Gerhard "Or Just Keep It Simple" Paseman, 2019.01.14.

I don't know that these are correct or not.

1. The simplest way (in practice ) that come to my mind is the following trick. Consider two points $p$ and $q$ on sphere. connect them by an straight line (a line of powerful magnet) in $Bbb R^3$ then using some iron filings you can see the trace of iron filings on sphere as the shortest arc between these two points.


2. The second way for this purpose, I think stereographic projection from north pole $N$ would be helpful. Rotate the sphere such that projections of $p$ and $q$ under stereographic projection ($sigma(p)=p'$ and $sigma(q)=q'$) together south pole $S$ lies in a line (i.e. $Sp'q'$). then the inverse image of this line segment represents the shortest arc between $p$ and $q$ (of course non-antipodality is assumed).

For a few minutes of contrast, I suggest simply drawing the geodesics on a torus of revolution, earliest I know is Bliss 1902. The facts about them can be recovered from Clairaut's relation. The nice part about this example is that you can have one with you, such as an inflated tube used with young children in a swimming pool.

do Carmo does an example, pages 262-263, a geodesic that wraps infinitely often around the neck of a hyperboloid of one sheet, never reaching it. This happens with the torus also.

Draw the straigt line $L$ between $p$ and $q$ and parametrise your path $gamma(s)$ with $sin L$ such that $s$ is the orthonormal projection on $L$ of $gamma(s)$. (we note $p$ the projection on $L$ so $p(gamma(s))=s$.
Now take your path and rotate it around $L$. This draw a cylinder $C$ around $L$ defined as $$C={xin mathbb{R}^3:|x-p(x)|=|gamma(p(x))-p(x)|}$$ Then we use that

1. On a cylinder the shorter path is to go straigh. Indeed by symetrie you don't gain anything turning left or right.


2. The cylinder drawn from the geodesic line is the smallest one, Meaning that it is inside any other cylinder that we can construct this way.


3. If a cylinder $C_1$ is convex and smaller than another cylinder $C_2$. Then the path on $C_1$ is smaller than the path on $C_2$

EDIT : second proof.
Start with $p$ and $q$ the north and south pole. You can paramerise with the height $tin [-1,1]$, and $theta(t)$ ($z=t,x=r(t)cos(theta(t)),y=r(t)sin(theta(t))$), with $t^2+r(t)^2=1$. Then by symetrie of revolution the shorter path is to go straight. And therefore $theta = c$ is constant. Then we can conclude using that a smaller path of a geodesic is a geodesic.