2

Physicists describe gravity waves as "a ripple in the curvature of space-time". As a mathematical concept, isn't curvature of a coordinate system only meaningful relative to a coordinate system exhibiting a different curvature?

Alternatively, what is the context for curvature of a coordinate system? I.e., "Space-time may be curved - but with respect to what?"

related question here

Community
  • 1
jbbenni
  • 163
  • 2
    How is your differential geometry? Do you understand the curvature tensor of a Riemannian metric, for example? – Lee Mosher Sep 02 '16 at 22:57
  • 1
    One can prove that the curvature tensor of a (Riemannian/pseudo-Riemannian) manifold does not depend on the local coordinate system used, by showing that its form is preserved under a change of coordinates. Hence, curvature is an intrinsic property of a manifold. Nash’s Embedding Theorem says that one can isometrically embed a Riemannian manifold into $ \mathbb{R}^{n} $ for $ n $ large enough, so one can think of the manifold as being curved inside an ambient Euclidean space. However, one does not have to assume the existence of an ambient space, which seems rather artificial. – Transcendental Sep 02 '16 at 23:27
  • 1
    To add to the above - Gauss was happy to find that the 'natural' curvature (Gaussian curvature) of a surface was intrinsic - i.e., depended only on the (induced?) metric, and not any (other?) property of the ambient space in which the surface might lie. This was his "theorema egregium". On arbitrary riemannian manifolds, "curvature" is a rule that assigns a Gaussian curvature to sub-surfaces - in particular, it is intrinsic - it is not "with respect to." For surfaces, see http://www.math.uni.lodz.pl/~maczar/ggk/do%20Carmo,%20Differential%20Geometry%20of%20Curves%20and%20Surfaces.pdf – peter a g Sep 03 '16 at 02:37
  • 1
    The theorema egregium is in section 4.3 of the above.... this presentation is chart driven, but nice from a motivational point of view, I think, although I am certainly no expert, so others might well disagree... – peter a g Sep 03 '16 at 02:45

1 Answers1

3

To summarize: The curvature tensor of a (pseudo)Riemannian manifold $(M,g)$ is independent of the choice of (local) coordinates on $(M,g)$. This also applies to the claim that the curvature is defined once two coordinate systems are given. Actually, in modern treatments one defines the curvature tensor without using any coordinates whatsoever. Thus, you should not talk about "curvature of a coordinate system" but, rather, of $(M,g)$ itself. As of gravity waves being "ripples" in something, this sounds more like a metaphor rather than a mathematical statement. My favorite treatment of Riemannian geometry is do Carmo's "Riemannian geometry". Reading the first 4 chapters of the book will help to clear many misconceptions. One thing you will realize after you are done with Chapter 4 is that a tensor $T$ on a differentiable manifold is not some expression with several upper and lower indices (which is an impression one typically gets from reading physics literature), but $T$ becomes such after you introduce local coordinates (one local coordinate system at a point suffices)! This is analogous to the fact that (in the classical mechanics) the velocity vector of a point-object exists independently of whether or not somebody introduced coordinates in the space; the velocity vector becomes an expression of the form $(v_1,v_2,v_3)$ only after coordinates are introduced.

Moishe Kohan
  • 111,854