Curvature of the earth from Theorema Egregium

Question

I recently learned Gauss's Theorema Egregium for surfaces embedded in $\mathbb{R}^3$. A TA for my class suggestively said that from this, i.e. that Gaussian curvature depends only on the first fundamental form of a surface, we may calculate the curvature of the earth without leaving its surface.

I understand that with knowledge of the first fundamental form, we may calculate the curvature, but for this question I'm not sure how to proceed, since we don't have a priori the first fundamental form of the earth given. It seems to me like we need to find that.

I saw some answers using the Gauss-Bonnet theorem, but I think that deals with total curvature, and I'm speaking of Gaussian curvature (I may be wrong here, I don't really know that theorem well).

Does it have something to do with measure triangles and angles? And if so, can someone help me relate this back to Gauss's theorem and the first fundamental form?

Another point of confusion: How do I even know what a triangle is on an arbitrary surface? A triangle is made by connecting three points with the curve that attains the shortest possible distance between those points, right? So on a plain, that's the normal line segment, but what about for arbitrary surfaces?

Answer 1

Shortest paths on an arbitrary surface, called (pre-)geodesics, are difficult to describe explicitly in general. On a sphere (a good approximation to the surface of the earth as far as geodesy is concerned), however, they're arcs of great circles.

Generally, if a geodesic triangle on a surface encloses a topological disk $T$, if $\Theta$ denotes the total interior angle of $T$, and if $K$ denotes the Gaussian curvature function, then $$ \iint_{T} K\, dA = \pi + \Theta. $$ Particularly, if $K > 0$, a geodesic triangle has interior angle greater than $\pi$, and if $K < 0$, a geodesic triangle has interior angle less than $\pi$.

On a sphere of radius $R$, we have $K = 1/R^{2}$, so a geodesic triangle of area $A$ has total interior angle $\pi + A/R^{2}$. For instance, a triangle with three right angles (one-eighth of a sphere) has area $4\pi R^{2}/8$ and total interior angle $\frac{3}{2}\pi$.

Curvature can be observed in practice: Longitude lines are geodesics, while latitude lines (except the equator) are not. If a surveyor wants to lay off one-mile (near-)square plots, the plots will fit well east-to-west (because two latitude lines are separated by a constant distance), but not well north-to-south (because two longitude lines get closer the farther from the equator one travels). Consequently, at moderate latitudes, every several miles the north-south boundaries of square plots must "jog" east or west in order for the plots to remain approximately square. The photographs below (own work) show this phenomenon in the nearly-planar plains of eastern Texas, taken from a plane.

Answer 2

One way to think about geodesics on a surface is that they seem like straight lines to a 2-dimensional observer living on the surface. Thus they will see a geodesic triangle the way we see a triangle on the Euclidean space $\mathbb{R}^2$. If the Gaussian curvature of a surface is constant(like on spheres, plains or pseudospheres) the 2-dimensional observer could draw geodesic triangles of different sizes, measure the angle deficit(or surplus), plug them in the theorem(with $k_g \equiv 0$), and calculate the Gaussian curvature.

Answer 3

Does it have something to do with measure triangles and angles? And if so, can someone help me relate this back to Gauss's theorem and the first fundamental form?

Yes. The point is that the first fundamental form allow you to measure lengths and areas in a surface. Also, it allows you to define what is the angle between two tangent vectors. Theorema Egregium tells you that all this information suffices to determine the Gaussian Curvature. For example, using the following

Theorem (Bertrand-Diquet-Puiseux): let $M$ be a regular surface. If $p \in M$, $C_\epsilon$ and $D_\epsilon$ are the polar circle and polar disk in $M$ centered in $p$ with radius $\epsilon$ (that is, the images via $\exp_p$ of the corresponding circle and disk in $T_pM$), then $$K(p) = \lim_{\epsilon \to 0} \frac{3}{\pi}\frac {2\pi \epsilon - L(C_\epsilon)}{\epsilon^3} = \lim_{\epsilon \to 0} \frac{12}{\pi} \frac{\pi \epsilon^2 - A(D_\epsilon)}{\epsilon^4},$$where $L(C_\epsilon)$ and $A(D_\epsilon)$ denote the length of the polar circle and the area of the polar disk. You can check p. $413$ in Elementary Differential Geometry by O'Neill for technical details.

Another point of confusion: How do I even know what a triangle is on an arbitrary surface? A triangle is made by connecting three points with the curve that attains the shortest possible distance between those points, right? So on a plain, that's the normal line segment, but what about for arbitrary surfaces?

The analogue of lines in the plane, for arbitrary surfaces are called geodesics: curves that locally minimize arc-length. Or equivalently, curves $\alpha$ that are auto-parallel: $D\alpha'/dt = 0$, where $D/dt$ denotes covariant derivative along $\alpha$. It happens that for planes, geodesics are lines, so we're really generalizing lines. For spheres, geodesics are great circles. Geodesics are described in coordinates by a system of differential equations: $$\ddot{u}^k + \sum \Gamma_{ij}^k \dot{u}^i \dot{u}^j = 0,$$where the $\Gamma_{ij}^k$ are the so-called Christoffel Symbols of the coordinate system (and for the usual coordinate system on $\Bbb R^2$, all of them are zero, hence the geodesics are lines). You'll learn more about geodesics as the course goes on, but I guess this'll give you an idea about it. Geodesic triangles are triangles on $M$ for which the sides can by parametrized as geodesics, and this allows us to prove theorems such as Gauss-Bonnet.