Geodesics on the sphere

Question

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 $\gamma\colon [0,1]\to S^2$ 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...

Answer 1

Imagine that the sphere is the surface of the earth, and that an earthquake happens at the south pole, point $p$. A seismic wave goes out from the point $p$, so that the wave front is a circle of latitude moving at a constant speed $c$. At some point this circle/wave will meet the point $q$ (let's say that $q$ lies within the continent of Antarctica, so the seismic wave will reach it). For any path from $p$ to $q$, flying along this path at the speed $c$, starting at the moment of the earthquake, will never overtake the wave, and will keep apace of it only if it coincides with the line of longitude going through $q$.

Underlying this physical explanation is Gauss's lemma, from which a rigorous calculus argument can be extracted. But maybe this intuition is enough to convince a high schooler, or at least make them think about why it might be true?

Answer 2

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 $$\cos a=\cos b\cos c+\sin b\sin c\cos A\,$$ 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 $$\cos a\geq \cos b\cos c-\sin b\sin c=\cos(b+c),$$ therefore $a\leq b+c.$

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.

Answer 3

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.

Answer 4

I always found the following argument the most appealing intuitively.

Imagine the sphere is hollow. Obviously the shortest path from $p$ to $q$ is a straight line through the interior of the sphere. (I always imagine just ramming a knitting needle through it.)

Now of course the question was not to find the shortest path in general, but the shortest path over the surface of the sphere. It is intuitively clear that this will be the path that follows the 'true' shortest path as closely as possible. How do we find that?

Well just imagine a light source at the center at the sphere. The path on the surface closest to the straight line through the interior is then simply the shadow of that straight line.

What remains to be seen is that this is part of a great circle but that is easy: the 'shadow path' is the intersection of the sphere with the plane defined by the three points $p$, $q$ and center.

Answer 5

Even knowing differential geometry I've always found it unsatisfying that all of the simplest proofs of this fact that I know still involve a reasonable amount of calculus. Spheres and great circles feel like fairly elementary geometric objects, so surely there is an elementary geometric argument. The following argument still involves a bunch of calculus under the hood, but it feels a bit more geometric to me.

Take a smooth curve on the sphere, and choose a small segment $C$ of the curve. There is a unique plane $P$ which contains the endpoints $p$ and $q$ of the segment and the center of the sphere. Consider the curve $C'$ which is obtained by reflecting $C$ across this plane and reversing the direction of travel, so that it goes from $q$ to $p$; note that $C'$ has the same length as $C$ because reflecting the sphere across a plane through the origin preserves distances.

Now consider the concatenation of $C$ and $C'$ - this is a small loop based at $p$. Among all loops based at $p$ which pass through $q$, the shortest one is the loop which encloses the smallest area, and an area of zero is attained if $C'$ is just $C$ with reversed orientation, i.e. $C$ is fixed under reflection about the plane $P$. But the curves fixed under reflection about a plane through the origin are precisely the great circles, so we're done.

The main idea here is that geodesics are fixed point sets of isometries, if there are lots of isometries. This idea can be used to find geodesics in homogeneous spaces, for instance.

Answer 6

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.

Answer 7

It might be instructive to show that great circles are the "straight lines" on a sphere: Start from a point on the northern hemisphere and (imagine to) walk due east until you have returned to your position. Then your your right foot will have walked further than your left one. On the southern hemisphere, it's the other way round. Hence, circles of constant latitude are not straight -- except, of course, the equator.

I think this gives an intuitive geometric understanding of straight lines, and it more or less directly generalises to show that other great circles are straight, and other curves are not. Also, it shows that lines that look straight on a map might not actually be. (You could follow that with a calculus-based approach if the pupils have the required tools.)

You still need to argue that this notion of straightness coincides with "shortest path", but that seems intuitively clear to me.

Answer 8

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).

Answer 9

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.

Answer 10

Your students likely know about Pythagoras theorem. Because of the triangle inequality, we know it is the shortest distance between 2 points in a flat space. It also turn out that it is the metric of a plane surface:

$$ds^2 = dx^2 + dy^2$$

Now for a sphere you can derive the metric equation much the same way. You simply create a small triangle at the surface of the sphere. The length element $ds^2$ will be a small arc along the latitude (squared) + a small arc along the longitude (squared). Which gives you this simple equation for the metric on the surface of a sphere:

$$ds^2 = (R \cdot d\theta)^2 + (R \cdot \sin(d\theta) \cdot d\phi)^2$$

Answer 11

Take some old globes to class, and some pushpins and some string.

First step put in tall pushpins at every latitude, wind the string tight around the tops, up off the surface, so it is in 6 or 8 sections. Measure the length of sections, add. Now try to find a shorter path, again up off the surface, with other pins. Don't tell them there's no shorter path, let them figure it out, and explain why.

Now ask whatis the difference between having it up off the surface in segments, and down on the surface in a smooth curve. Will it make a difference to finding shortest path?

Some smartie pants will say you can tunnel through the earth and get the shortest path. That's an opportunity to explain what lateral thinking is and how important it is to the history of science.

Sphere is symmetric under rotation so whatever is true of the longitude great circle will be true of every other great circle.

Answer 12

If you can reduce the problem to that of finding the shortest path between two points at the same longitude, then you can proceed as follows.

Follow some path from your point of origin to your destination at uniform speed $c.$ Let $a$ be the component of your velocity in the direction in which only your longitude is changing. Then the component of your velocity in the direction in which only your latitude is changing is $\sqrt{c^2 - a^2} < c.$ You're not gaining latitude as fast as if you moved along the meridian, but your odometer is still putting distance behind you just as fast.

Answer 13

Draw the straigt line $L$ between $p$ and $q$ and parametrise your path $\gamma(s)$ with $s\in L$ 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 $$C=\{x\in \mathbb{R}^3:\|x-p(x)\|=\|\gamma(p(x))-p(x)\|\}$$ 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 $t\in [-1,1]$, 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.

Answer 14

There is an elegant way to see geodesics on any surface.

Take a piece of paper in a rectangular shape (with width short enough). The geodesic is drawn by the line in the middle of the rectangle when you can flat at most the rectangle on the surface.

It comes from the fact that by using a rectangle and flatten at most both long edges, you induce a Killing field. The codimension 1 coincides with the fact that the geodesic is of dimension 1.

Answer 15

A couple of answers already propose using the triangle inequality, but I think they both apply it using a calculus argument. There's a simple geometric argument. Let $C$ be the shortest path from $A$ to $B$. Take the point $P$ on $C$ which is furthest from the great circle $AB$. If $P$ is not on the great circle then $AP + PB > AB$, and since the length of $C$ is at least $AP + PB$ you have a contradiction.