Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have been exploring differential geometry slightly... And i'm trying to grasp how to define intrinsic curvature, from a visual/geometric viewpoint...

One formulation I got was that Intrinsic curvature can be thought in terms how the $n$-volume of an $N$-sphere in your particular $n$-dimensional space (example: a the area of a circle drawn on a $2$d space) differs from the $n$-volume of a euclidean N-sphere...

But here is the big problem I found... That definition is all wonderful and whatnot when talking about the curvature of a certain "area" of a surface...

Example, If I had a big sphere in $3$ dimensional space. I can start from a point in the sphere... walk $10$ meters off of the point and draw a circle of radius $10$... then measure the area using tiles and compare its ratio to that of a euclidean circle...

But as I reduce the radius that I'm measuring... The circle I draw out will get increasingly closer to the area of euclidean circle for the same radius...

Clearly the point curvature given this definiton will be $0$... At any given point on a sphere the curvature, by this definition, is equivalent to a flat plane...

I can't seem to find a way to define intrinsic curvature now. Anyone care to help?

share|improve this question

2 Answers 2

If $A(\rho)$ is the area of the circle around the point of interest with geodesic radius $\rho,$ then $$ K = \lim_{\rho \rightarrow 0^+} \frac{12 \left( \pi \rho^2 - A(\rho) \right)}{\pi \rho^4} $$

See, for example, http://en.wikipedia.org/wiki/Spherical_cap

The nice thing is that it all comes out the same for the hyperbolic plane, where $\cos$ is replaced by $\cosh.$

This is closely related to the Gauss-Bonnet theorem, as it must be, and was first found in 1848, see BDP

share|improve this answer

One can define curvature of a surface by a girthing polygon. One needs only construct a limited number of sides for this to get useful results. Once the effective radius has been constructed, the units of length can be adjusted.

We shall call a shortchord as the chord formed on a polygon, by two adjacent edge. So if AB and BC are consecutive edges of a polygon, then AC is the shortchord.

Let $e$ be the edge, and $s$ be the shortchord. We find a versine, by $v^2 = 4 e^2 - s^2$. The curvature of space, $c$ is then $v^2/e^4 = 1/r^2$. Let's look at an example.

Suppose we have a edge of $56$ giving a shortchord of $97$. We find $v^2=3135$, and this the curvature is $1/3137$, This corresponds to $r=\sqrt{3137} = 56$ approx.

In non-euclidean geometries, one can choose to measure along zero-curvature lines (chords of spheres and along horocycles in hyperbolic geometry), or surface arcs. While arcs are the use, it is better to use the chords and choh, since the geometry here is euclidean.

A circle in any geometry is the intersection of a euclidean space, and a non-euclidean space, so its circumference is always $\pi \cdot d$, where $d$ is the chord.

If on the other hand, one is free to select units, one can use the zero-curvature curve, if one is not adverse to triangles like $1:1:3$. The ruler is more crooked than the straight line, but makes for euclidean geometry.

Consder for example, establishing the curvature of a surface, containing a tiling. The tilings {5,12} and {12,10} have a common curvature. We can see this of {p,q}, becase where the radius is set to 1, the circumference is a {q} whose edge is the shortchord of {p}.

  • So {5,12} is a dodecagon of edge $1.618033$, giving a diameter of $6.25103$
  • A {12,10} is a decagon, of edge $1.93185165$, gives also a diameter of $6.25103$.

The versine is $v^2 = 4e^2-s^2$, gives $v^2 = -2.25103$, and since $e=1$, this is also the curvature.

Since all spheres are different sizes of the same thing (including the flat case), and likewise pseudocycles are a similar scale, one can then compare curvature against the straight case, etc. That is, any polytope {5,12} is a larger copy of the plane tiling, none the less zero-curvature simply gives size relative to some unit.

If one intends to bring figures to the same sphere, the curvature can be used to adjust size, since $C \cdot e^2$ becomes a measure of a radius (eg a sextant here, since were using chords), and any length so multiplied becomes a measure of arc, on a unit-radius (pseudo)sphere.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.