# Trippy Geodesics

If you have ever traveled between the US and Europe, you may have wondered why your flight always seems to fly over Iceland. I mean, it seems kind of out of the way, right?

The reason why planes fly along these paths is that they are actually the shortest way to travel!

The weirdness goes back to what we talked about last time: maps cannot correctly represent the Earth. Either distances or angles or areas (or all three!) will be wrong. So, when we draw a straight line on a map, it usually isn’t straight on the sphere!

The path the plane follows is (roughly) a great circle. The easiest example of a great circle is the equator. Any other great circle is just that path rotated around. So, for instance, here’s the path your plane follows:

It turns out that the shortest path between any two points on a sphere is always along a great circle.

And that’s why you always fly over Iceland.

The name for this kind of path is a geodesic.1

There are two important interpretations for geodesics. The first is that they are the shortest path between different locations, at least locally.2 That’s why they’re useful for flying or sailing across the Earth.

A second, related interpretation is that geodesics are the “straight” lines for a curved space. When you travel along a geodesic, it feels straight to you. You keep traveling in the same direction the entire time.

In other words, geodesics are to manifolds (like the sphere), what lines are to the plane: shortest and straightest paths.

These geodesics are super important. So, let’s talk about them more generally.

First, let’s recall how to measure the length of a path. On a manifold, we always have some sort of coordinates, like $x$ and $y$ in $\mathbb{R}^2$, or latitude and longitude on a sphere. So, let’s say our path is represented as $(x(t), y(t))$. In other words, $x(t)$ gives the position in the $x$ direction at time $t$. The derivative $(x'(t), y'(t))$ gives how quickly this position is changing, and in what direction. The square root of the metric $ds^2$3 applied to this derivative then measures the speed. Finally, we can integrate the speed, $\displaystyle\int_{\textrm{start}}^{\textrm{end}} \sqrt{ds^2} dt$, in order to calculate the total length of the path.

Don’t worry too much about all that. We won’t be doing any actual calculations.

How does this apply to a sphere?

Let’s go back to the great circle route on the Mercator projection.

The sphere of the Earth, as a 2-dimensional manifold, can be visualized as this map, with some metric on it. The two coordinates we have are $\theta$, which represents the angle around the equator, i.e., longitude, and $\phi$, which represents the angle from the north pole, i.e., similar to latitude.4

The metric for the sphere with respect to these coordinates works out to be $ds^2 = r^2 d\phi^2 + r^2 \sin^2(\phi)d\theta^2$, where $r$ is the radius of the Earth. As compared to the usual, flat $\mathbb{R}^2$ metric, $ds^2 = dx^2 + dy^2$, the $\phi$ direction acts normally, but the $\theta$ direction is scaled by a $\sin^2(\phi)$ term. (The $r^2$ simply scales everything bigger or smaller, but doesn’t really change how the metric behaves.)

The important thing to note is that for $\phi \approx 0$ and $\phi \approx \pi$, i.e., near the north and south poles, $\sin^2(\phi)$ will be about zero, and so the $d\theta^2$ part of the metric will be much smaller than the similar metric on $\mathbb{R}^2$.

Since the length of a path is $\int_{\textrm{start}}^{\textrm{end}} \sqrt{ds^2} dt$, the fact that the metric is smaller near the poles means that the shortest path is encouraged to inch upwards towards the poles. In other words, looking at the Mercator projection, traveling near the poles doesn’t “cost” as much (since $\sqrt{ds^2}$ is smaller), and so the shortest paths bend towards the pole.

Of course, actually calculating what the precisely shortest path is is a bit tricky.

Fortunately, we can relegate all that to a footnote.5 In the end, the important thing to understand is that there is an equation you can solve to calculate a geodesic. And using that equation, we can show that great circles are indeed the geodesics on a sphere.

Let’s look at one more example of geodesics.

Just like the sphere is the standard example of a manifold with positive Gaussian curvature, hyperbolic space is the standard example of a manifold with negative Gaussian curvature. Remember, negative curvature means that triangles have angles of less than 180 degrees, but circles have more area and circumference than you would expect for their radius.

There are many ways to model hyperbolic space, but let’s consider the “disc model.” A disc is one mathematical word for the inside of a circle.

With this picture, hyperbolic space is just the disc with coordinates $x$ and $y$, but the metric, instead of being $ds^2 = dx^2 + dy^2$ like in flat space, is $ds^2 = \dfrac{dx^2 + dy^2}{(1-r^2)^2}$, where $r$ is the distance from the center of the disc.6

The important thing to notice is the $\dfrac{1}{(1-r^2)^2}$. If $r \approx 0$, i.e., near the center of the disc, the metric looks pretty much like the standard metric. However, if $r \approx 1$, $\dfrac{1}{(1-r^2)^2}$ is very large, and so the metric actually goes to infinity near the boundary of the disc.

In a very real sense, the boundary of the disc is infinitely far away from the center. If you calculate the length of the straight path from the center to the boundary, you essentially calculate $\displaystyle \int_{\textrm{start}}^{\textrm{end}} \sqrt{ds^2} dt = \displaystyle\int_0^1 \dfrac{1}{1-x^2} dx$, which happens to be infinity.

Now, back to geodesics. If a path in hyperbolic space is going to minimize length, it wants to get away from the boundary of the disc as quickly as possible, since the metric there is so large. If we carefully solve the geodesic equations from earlier, we find that geodesics are actually circles which are perpendicular to the boundary.

Yeah, it’s kind of weird, but “straight lines” in hyperbolic space end up looking like circles!

Of course, if you were living in hyperbolic space, you wouldn’t think of them as circles, no more than we feel like we’re traveling in a circle when we fly along a great circle route. Geodesics feel straight to someone traveling on one. It’s just our visualization that makes it look like a circle.

To finish up, you know how we drew a triangle with three right angles?

Well, now you can draw a triangle with three angles of zero degrees.7

1. The word geodesic comes from “geo,” i.e., Earth, and “desy,” i.e., measuring, since you use great circles in order to measure distances on the Earth.
2. Though you do need to be a little careful. Going the wrong way along the great circle would be the longest route.
3. For $\mathbb{R}^2$, this metric/speed is $ds^2 = dx^2 + dy^2$, which is just the Pythagorean theorem applied to the derivative.
4. Latitude is usually measured from the equator, while we’re measuring it from the north pole. It’s the same idea, though.
5.  Here’s the idea: Look at the integral for length. If we vary the path a little, this length will change. If we do this carefully, we can take the “variation” of the length, kind of like you can take the derivative of a function. If you set the variation to zero, that’s the equation that defines a geodesic. If our coordinates were called $x$ and $y$, the equation for $x$ would be $x'' = -x'^2 \Gamma_{xx}^x - 2x'y' \Gamma_{xy}^x - y'^2 \Gamma_{yy}^x$. Those $\Gamma$ are some quantities dealing with the metric and its first derivatives called the Christoffel symbols. They’re kind of annoying to calculate, but straightforward. The $x''$ is the main term. The two apostrophes mean it’s a second derivative, and so represents the acceleration. So, the geodesic equation looks somewhat similar to the $F=ma$ from physics; the Christoffel symbols kind of represent a “force” that keep the path from going straight.
6. There’s often also a factor of 4 in this metric, but that just scales the overall space. It’s not so important here.
7. Okay, technically, since the boundary of the circle is infinitely far away, the three lines never quite reach each other. On the other hand, if you have them meet just before the boundary, you can get a triangle with angles as close to zero as you want.