Usually I'd just look this kind of thing up on wikipedia, but the article with this name has been deleted due to copyright infringement, so bummer. Also my other searches haven't turned up a definition anywhere, just loads of uses. I am talking about Riemannian geometry here, for those who would like some context. Actually it came up due to a GR question I was asked and got wrong, specifically I expected that the scalar curvature of Schwarzschild spacetime would be something other than zero. Confused, I sought to better understand what it really meant for the scalar curvature of a manifold to be zero, and I have since learned it means that the volume of a geodesic ball on the manifold does not deviate from the volume of such a ball in flat space of the same dimension. Unfortunately I don't know what a geodesic ball is so this hasn't helped me. I can't think of a very intuitive definition based on the name, but it seems to be very commonly used so I assume it is something pretty simple. It's not just a regular ball with a fancy name is it?
I think it's a ball with a lot of flat sides. Such as a soccer ball...or a golf ball - but not a tennis ball or a basketball.
I think this pertains to the Gauss-Bonnet theorem. The theorem relate curvature and angles. Quoting Corollary 4.40 of 'Differential Geometry' by Kuhnel : Let U be an open subset of \(R^{2}\) and \(B \subset U\) be diffeomorphic to a closed disk. Let [te]f: U \to R^{3}[/tex] be a surface element such that f is injective. The boundary is assumed to consist of finitely many segments of geodesics (forming a geodesic n-gon) with exterior angles \(\alpha_{1},\ldots,\alpha_{n}\). Then \(\int_{f(B)}K \,dA = 2\pi - \sum_{i} \alpha_{i}\). This basically tells you how the curvature of f(B) (which we can take to be part of a sphere) alters the angles of an n-gon drawn on it. In the case of n=3 you get a triangle and for a sphere K>0 and thus the sum of the exterior angles is less than 2pi and so the internal angles of a triangle add to more than 180, as the Wiki page on spherical geometry will show. Only if K=0 do you get \(2pi = \sum \alpha_{i}\). This is related to the simplest version of the statement you refer to, where the 'ball' is a circle and the geodesic ball is just a polygon.
Emanate geodesics in every direction from a point, and on every such geodesic travel r distance. The union of all these geodesic will give you the geodesic ball (The geodesic sphere would be only the endpoints of course). I am afraid the characterization of scalar curvature you have mentioned does not work for Lorentzian signatures. Of course it is fine if you consider spacelike slices of Schwarzshild spacetime, but in general you just have to stick to the definition (angle deficit in the parallel transport of a vector along small loops). I guess the confusion about Schwarzshild spacetime curvature is because in Schwarzschild spacetime the curvature tensor (which has a lot of components) does not vanish, but the Ricci curvature (whose trace is the scalar curvature) vanishes (by the Einstein equation itself).
Ahh, sweet, that sounds like what I was after. Bugger, I kind of liked thinking about it that way. I assume small means infinitesimal? Oh, of course. So all the vacuum solutions have zero scalar curvature, how interesting... seems strange. Hmm, I wonder how similar this is to what temur said. I mean your K is the gaussian curvature, not the scalar curvature, although I think in 2D and maybe 1D as well they are the pretty much the same, but not higher dimensions. I'm not sure what the relationship is for higher dims. I do have some memory of the gauss-bonnet theorem, although I never looked at it for dimensions higher than 2. I assume it is somewhat more complicated... edit: Actually is gaussian curvature even defined in more than 2 dims? I guess perhaps not, this is what Riemannian geometry is for. I wonder if there is an equivalent theorem to gauss bonnet... hmm just wiki'd it, it is a little more gnarly and seems to only work for even dimensions... Ok, maybe I won't go there just now Please Register or Log in to view the hidden image!.
Yes, infinitesimal. That characterization does not work because with Lorentzian signature the geodesic balls have infinite volume (please check this). But some other similar idea might work. Yes. Example: In 3 dimension the Ricci curvature completely determines the Riemannian curvature (they have the same number of components), so in 2+1 gravity any vacuum solution is completely flat. The only way to make 2+1 spacetime curved is to add matter, and of course even then the vacuum part will still be flat. The point is that scalar curvature is just an average curvature, and in high dimensions there is a lot of room that space can curve staying flat on average.
