# I Understanding the curvature parameter k

1. Oct 20, 2016

### throneoo

1.)
According to my understanding, k is a constant in the Robertson-Walker metric and the Friedmann equations. Its value (-1,0 or 1) is determined by observation (e.g. the bending of light) and doesn't change over time, unlike the curvature density parameter $\Omega_k$. Therefore a strictly flat universe will always stay flat, while closed or open universes can change their degree of curvature over time. Is my understanding correct?

2.)
If k=0, then the universe is spatially flat. However spacetime is curved by energy/mass via the stress-energy tensor in a non-empty universe. Since spacetime has to be curved somehow, does that mean time is the only component of spacetime that is curved? Otherwise, how can a massive universe stay flat without some sort of "negative energy" to cancel out the curvature?

I have very limited understanding in general relativity

2. Oct 20, 2016

### andrewkirk

The FLRW curvature parameter is a parameter of the spacetime as a whole. So it would not mean anything to say it changes over time. It would be like saying the length of a rod changed between different parts of the rod.

The spatial curvature of the constant-time hypersurfaces of an elliptic or hyperbolic spacetime changes over time in the FLRW model, as space expands. But that spatial curvature measure is a different measure from the k parameter.
Similarly with this one. Spacetime curvature is a property of the four-dimensional spacetime manifold, not a property of any component of it. We can talk about the curvature of 3D spatial slices of the manifold ('constant-time hypersurfaces'). But we cannot talk about the curvature of any one-dimensional subset, since at least two dimensions are needed for curvature to be possible.

Last edited: Oct 20, 2016
3. Oct 20, 2016

### throneoo

Okay. So k is a parameter of spacetime, but it only characterize the curvature of 3D space?
I see. Since it doesn't make sense to talk about 1D time-curvature and given that the 3D space is flat, k=0 would correspond to a Minkowski space. How is it possible if matter/energy content is non-zero? A similar question I have is, knowing that according to the Friedmann equation an empty universe with no radiation or mass has negative curvature. But isn't an empty universe supposed to be flat?

4. Oct 20, 2016

### Staff: Mentor

But it does make sense to talk about spacetime curvature of a 4-D spacetime in which 3-D spatial slices happen to be flat, which is what andrewkirk was talking about. See below.

No, it wouldn't. Minkowski spacetime is a flat 4-D manifold, not a flat 3-D manifold. In a FLRW spacetime with k = 0, individual spatial slices of constant FLRW coordinate time are flat; but the spacetime as a whole is curved. The k = 0 FLRW spacetime just happens to be curved in such a way that you can cut flat 3-D slices out of it (as long as you cut them in a particular way).

5. Oct 20, 2016

### Staff: Mentor

Negative spatial curvature. Once again, spatial curvature, the curvature of 3-D slices cut out of the 4-D spacetime, is not the same as the curvature of the 4-D spacetime.

The empty universe is flat as a 4-D manifold; it is flat Minkowski spacetime. But the FLRW coordinates, in this case, cut 3-D slices out of that flat spacetime that are curved, because of the way they are cut. Just as you can (in certain cases) cut flat 3-D slices out of a curved 4-D spacetime, you can also cut curved 3-D slices out of a flat 4-D spacetime.

6. Oct 20, 2016

### throneoo

From what I can gather, the curvature on space also depends on how the choice of coordinate system (or how they are cut).
Would it be correct to draw such the following analogy from 3D Euclidean space?
3D Euclidean space can either be described by cartesian coordinates or spherical-polar coordinates. If I fix one of the coordinates in the former I'll always get flat slices of 2D surfaces(planes), whereas if I fix the $r$ or $\theta$ coordinates in the latter I will get curved 2D surfaces like spheres and cones, but in either case the 3D space is still Euclidean, just as an empty universe is still a flat (Minkowski) 4D manifold despite having negative spatial curvature in the FLRW coordinates.

7. Oct 20, 2016

### Staff: Mentor

Yes, the choice of coordinates determines how the spatial slices are cut out of spacetime.

Yes, in general this is a good way of understanding how choosing coordinates can change the curvature of slices.

However, I do have one correction: cones are not curved surfaces. That is, they have zero intrinsic curvature, which is the kind of curvature we are talking about here. The fact that the cone looks curved when embedded in Euclidean 3-space just means it has extrinsic curvature, which is not the same thing. One way of seeing that a cone has zero intrinsic curvature is to imagine taking a triangle drawn on a flat piece of paper and then rolling the paper up into a cone; you should be able to convince yourself that you can do so without distorting the triangle in any way, i.e., without changing the lengths of any of its sides or any of its interior angles. Whereas, if you try to wrap the flat piece of paper around a sphere, you can't do it without distorting the triangle, either by changing side lengths or angles (or both), illustrating that the sphere has nonzero intrinsic curvature.

8. Oct 23, 2016

### Chalnoth

Note that the convention of k being (-1, 0, 1) is a convention: this is allowed only if we can modify the scale factor. With this convention, the scale factor becomes the radius of curvature.

Another convention that is often used sets the scale factor $a = 1$ at the current time. With this convention, k is just a real number, and is equal to the inverse of the current radius of curvature squared, and a sign set by whether the universe is open or closed (with flat still being k=0).

In a spatially-flat universe, the space-time curvature is the expansion. If the universe isn't flat, the expansion still makes up a big part of the space-time curvature, but the spatial curvature also contributes.