# B Gaussian Curvature and Riemmanian Geometry

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1. Mar 6, 2016

### ProfuselyQuarky

Please bear with me because I'm only in Pre-calculus and am taking basic high school physics. This is completely outside of my realm but curiosity has taken the better of me.

I just learned last week about the difference between Euclidean Geometry and Riemmanian Geometry (from another thread here, not school). I read more about the latter geometry and about how it's all curvy, which lead me to Gaussian curvatures. Apparently there are negative, 0, and positive curvatures depending on the surface (like the inside and outside of a torus), right? I think I understand the idea of these curvatures conceptually, but I'm a bit confused when it starts to go into the math-based explanations.

The formal definition that I've seen says that Gaussian curvature is the product of two principle curvatures at a given point: K = k1k2. How can a curve be determined with a single point? Is this even so, or am I desperately missing something? By the way, if you think my knowledge about this is way too limited to understand, just say so and I'll settle :)

Last edited by a moderator: Mar 6, 2016
2. Mar 6, 2016

### phyzguy

You are on a two-dimensional surface, so if you are given a point and a direction there is only one curve that passes through that point in that direction. Think of the surface of the Earth. If you are standing at a point and pick a direction, that uniquely determines your path, at least for a short distance. Each curve has a curvature, and the principal curvatures are the maximum and minimum values of that curvature. I think where you are having trouble is that after you have started off in a given direction, you can change direction as you move away from the starting point. But to determine the curvature you only need to consider very small (infinitesimal) displacements from the starting point. So don't think of the long wandering path you could take as you travel along the surface, just think of the first small step you would take as you leave the starting point. Does this help?

3. Mar 6, 2016

### Staff: Mentor

More precisely, there is only one geodesic curve that passes through that point in that direction. (For the OP, a "geodesic" is the equivalent in Riemannian geometry of a "straight line" in Euclidean geometry. On the surface of a 2-sphere, for example, a geodesic is a great circle.)

This usage of the word "curvature" may be confusing. A geodesic curve has zero path curvature, considered purely as a curve within the surface. For example, a great circle on a 2-sphere has zero path curvature, considered purely as a curve within the 2-sphere. We think of a great circle on the Earth as "curved" because we view it as a curve in the 3-dimensional Euclidean space in which the Earth is embedded; that is the way you are using "curvature" in the quote above. But the Riemannian geometry viewpoint, strictly speaking, is only supposed to look at intrinsic properties of the surface, without making any assumptions about how, or even if, the surface is embedded in some higher-dimensional space.

At least, that is how it is supposed to work; but many sources, like the Wikipedia article you link to, make use of embeddings when they really shouldn't. The reason why they really shouldn't is that, once you extend these concepts to general relativity, where we are working with 4-dimensional spacetime, we have to work purely with the intrinsic properties of the manifold, because they're all we have. Even if, as some physicists speculate, our 4-dimensional spacetime is embedded in some higher-dimensional space, we have no way of detecting that experimentally; the only observables we have to work with are observables that are purely within 4-dimensional spacetime. So the intrinsic viewpoint is the only option.

phyzguy's suggestion is a good one, but let me also give an alternative that does not make use of the embedding of the surface in a higher-dimensional space, in accordance with my comments above. Suppose I am at a point on the Earth's surface, and I draw a "circle" around that point with a small radius. By "circle" here I mean that I extend geodesics from my chosen point in all directions, mark points along each geodesic that are the same small distance from my chosen point, and draw a curve that connects all those points. Then I measure the circumference of this "circle" (the length of the curve connecting all the points) and divide it by the radius of the "circle" (the small distance that I marked out along each geodesic). Then I divide the circumference by the radius. What answer will I get?

If the surface I am on is flat, I will get exactly $2 \pi$ for the circumference divided by the radius. But if the surface I am on is curved, I will not; I will get a different answer. On a positively curved surface, like a 2-sphere, I will get an answer that is smaller than $2 \pi$; the circumference is shorter than it would be on a flat surface for the same radius. On a negatively curved surface, like a saddle, I will get an answer that is larger than $2 \pi$; the circumference is longer than it would be on a flat surface for the same radius.

Furthermore, the difference might not be the same in all directions--in other words, if I consider only the set of geodesics from my chosen point that cover a small angle $d \theta$, and the length $dL$ of the curve connecting the endpoints a small radius $r$ away only along those geodesics, and I compare $dL$ with $r d\theta$, which is the "expected" value for a flat surface, the difference between the two might be different in different directions. (On a perfect 2-sphere, it won't be, but on the actual Earth, it will be by a small amount since the actual Earth is not a perfect sphere.) The directions in which the difference is a maximum and a minimum will be the two principal directions that phyzguy talked about. And the principal curvatures will be related to the maximum and minimum values of the difference between $dL$ and $r d\theta$ (unfortunately I don't have the exact formula handy).

4. Mar 6, 2016

### ProfuselyQuarky

Ah, yes. This does help. I never thought that only very minuscule displacements would be necessary. A "long wandering path" is exactly what I was thinking of
That's what a geodesic is? I didn't realize that the definition was so simple. So the reason why a geodesic on the surface of a sphere is a great circle is because the two ends of the "line" on the circle will eventually meet each other?
Based on this, a "circle" drawn on the inside of a torus would have a circumference >2π and a "circle" drawn on the outside of the a torus would have a circumference <2π, right?

5. Mar 6, 2016

### Staff: Mentor

The great circle is a geodesic because it's the shortest distance between two points on the surface of the sphere. (There are some technicalities here but I don't think we need to go into them for this discussion.) The fact that this geodesic is a closed curve rather than an open, infinite one has to do with the global topology of the sphere, not its local geometry. The definition of a geodesic can be made purely locally; at any given point, the geodesics are the curves that minimize the distance to nearby points in all directions.

Yes, if by $2 \pi$ you mean $2 \pi$ times the radius.

6. Mar 6, 2016

### ProfuselyQuarky

Hee! Thanks so much for your thorough explanations! I've got it all now :D

7. Mar 6, 2016

### Staff: Mentor

You're welcome!

8. Mar 6, 2016

### Orodruin

Staff Emeritus
... assuming Riemannian geometry. It is no longer true in a pseudo-Riemannian geometry or on a manifold with an arbitrary affine connection which is not the Levi-Civita connection.