# Curvature of time

1. Jul 24, 2012

### JPBenowitz

So in GR spacetime is curved in the x, y, z, and t axis but suppose you are given a surface in x, y, z could you find the curvature of time by simply measuring the rate in change in curvature of the surface in the x, y, z axis?

2. Jul 24, 2012

### Ken G

No, you need the full curvature, unless you know the stress-energy distribution.

3. Jul 25, 2012

### TrickyDicky

Nope, obviously you'd be missing information. To see this more clearly consider for example a spacetime with flat 3-surface (look up FRW spatially flat universe), how would you determine the curvature of time from a flat hypersurface in that case?

4. Jul 25, 2012

### clamtrox

That's true, but how would you know what the rate of change is? You can only measure it locally, so you cannot extrapolate it through the entire universe

5. Jul 25, 2012

### TrickyDicky

No need to extrapolate anything if the curvature is assumed constant, that's not the problem, see the previous posts, this is just about confusing space and spacetime curvature.
The curvature of a submanifold doesn't determine the manifold's curvature tensor. It's the other way around.

Last edited: Jul 25, 2012
6. Jul 25, 2012

### A.T.

Even if the submanifold has only one dimension less than the manifold? Given that you have the metric of the (N-1)-submanifold, and that the dimensions must be orthogonal everywhere, can't you figure out the full metric of the N-manifold?

7. Jul 25, 2012

### TrickyDicky

Yes (always in the 4-dimensional case), this is trivial just by knowing there is three possible 3-dimensional geometries for the 4-dimensional FRW metric.
Actually what this shows is that the (n-1)-metric is an induced metric from the n manifold (up to a conformal factor) so it is the manifold what partially determines the metric of the submanifold.
To determine the metric and thus the curvature tensor for a given manifold's n dimensionality one needs the n coordinates, that is what defines the manifold one is working with.

If you add such constraint (similar to the Weyl postulate) you reduce the number of possible metrics but still don't determine it completely, as the FRW metric example shows.

Last edited: Jul 25, 2012
8. Jul 25, 2012

### Staff: Mentor

TrickyDicky is correct, and I like his example of the FRW metric since it is directly applicable to GR. Another example which is not directly applicable, but may be easier to visualize would be a flat 3D manifold. You can use a spherical coordinate system where your 2D sub-manifolds are spheres. The sub manifolds would therefore each have a uniform positive curvature despite the fact that the embedding manifold is flat.

9. Jul 25, 2012

### George Jones

Staff Emeritus
The Milne universe is another good example (similar to DaleSpam's example, and complementary to TrickyDicky's example). The spacetime of the Milne universe is (part of) flat Minkowski spacetime. The Milne universe is, however, a non-flat universe, i.e., its 3-dimensional hypersurfaces of constant time have non-zero intrinsic curvature.

10. Jul 25, 2012

### bcrowell

Staff Emeritus
You can't have curvature of time alone because time is a single dimension, and curvature doesn't exist in one-dimensional spaces. Curvature describes the failure of parallelism, and there is no notion of parallelism in one dimension.

11. Jul 25, 2012

### TrickyDicky

That is correct. But generally when people talks about curvature of time it is referring to the dependence of the spatial 3-hypersurface on the time coordinate as dictated by the scale factor. Not to a "one dimensional intrinsic curvature" that is impossible by definition.
That is certainly confusing when you approach the (spatially) flat FRW metric as a beginner if you naively think about it as "3 flat dimensions plus one curved time dimension".

12. Jul 25, 2012

### clamtrox

So if you know the curvature AND it's gradient on a 3-dimensional hypersurface, you still can't calculate the full curvature of 4-dimensional manifold? Why is that?

13. Jul 25, 2012

### bcrowell

Staff Emeritus
You're describing curvature here as if it were a vector, but actually the Riemann tensor has two indices. It doesn't have timelike and spacelike components. A vector has timelike and spacelike components. A two-index tensor has time-time, space-space, and time-space components.

It may help to note that the most familiar cosmological examples are FRW spacetimes, which are special because they have a high degree of symmetry. Because of this symmetry, there are preferred local rest frames (at rest with respect to the CMB), a preferred time coordinate (measured on a clock at rest), and a preferred way to slice up the spacetime into spacelike hypersurfaces (on surfaces of constant preferred time). It's only because of this preferred slicing that we can even unambiguously talk about spatial curvature (i.e., space-space components of the Riemann tensor). This fails in the general case. (In the example of the Milne universe, it fails because the symmetry is even higher than that of an FRW solution, so there are no preferred rest frames.)

See George's #9.

No, people who know what they're talking about don't talk about "curvature of time."

Last edited: Jul 25, 2012
14. Jul 25, 2012

### TrickyDicky

The Riemann tensor has nothing to do with any preferred frame, AFAIK.
As previously commented the general case is that of the manifold's metric inducing a metric on the submanifold, so it is always possible to talk about the spatial hypersuface curvature (or lack, of depending on the foliation choice) .

Well yes, most likely, I guess I was being over-optimistic and don't like to sound condescending.

15. Jul 26, 2012

### bcrowell

Staff Emeritus
Right.

The point about the preferred frame is that it provides a natural choice of that foliation.

BTW, my #13 should have referred to the Ricci tensor (which has 2 indices), not the Riemann tensor (which has 4). For some reason the web interface won't let me edit it.

16. Jul 26, 2012

### TrickyDicky

That's an interesting point I had never thought about. Usually the FRW preferred frame that can be obtained by the choice of a "preferred time" of the fundamental observers (those who observe no dipole of the CMB) is considered an example of spontaneous symetry breaking, and it is not AFAIK used to choose between the 3 possible geometries of the spatial hypersurfaces.
So according to you which of the 3 geometries would be "naturally chosen" based on the FRW preferred frame?
(Caveat: let's remember that frame is never considered in cosmology as a "truly" preferred frame in order not to break general covariance)
Hint: the timelike worldlines are (by convention,because the math allows them to be also contracting into the future) considered to be diverging (thus expansion) from a singularity in the past.