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## Main Question or Discussion Point

I have a question about spacial curvature. Before asking the question, I will summarize my understanding of the distinction between spatial curvature and spacetime curvature. If I have this wrong, I would appreciate corrections.

On the other hand,

Here's my question:

Is the heuristic of spatial curvature REQUIRED by general relativity, or is it merely a CONVENTION which is CONSISTENT WITH GR? I understand that mathematically, the curvature coefficient "k" provides the constant of integration for the Friedmann equations. However, the fact that a constant fits mathematically does not in itself require that constant to represent a physical reality.

So, it isn't clear to me why a universe with mass-energy > [tex]\Omega[/tex] is REQUIRED by GR to have 3-spherical positive spatial curvature. Why can't the geometry be flat?

If a universe were EXACTLY at [tex]\Omega[/tex], (being therefore geometrically flat), and also happened to be

My sense is that spatial curvature has been adopted as a "convention" in GR as a way to avoid the awkward discussion about whether the universe might be flat and finite, while simultaneously expanding or contracting. The discussion becomes awkward because this scenario implies the possibility of an "edge", or some kind of outer boundary to the universe. I understand why such a model would be less elegant than the spatial curve, "closed" approach.

So my question is, does GR REQUIRE spatial curvature if mass-energy density [tex]\neq \Omega,[/tex] or is spatial curvature more accurately described as a CONVENTION of GR?

Jon

**Spacetime**curvature occurs locally in the presence of gravity; i.e. mass-energy. Spacetime curvature causes timelike geodesics to be displaced from a "straight" line, towards a source of mass-energy, but only as function of the passage of time. At any given fixed instant in time, spacetime curvature does not imply any spacial curvature. For example, the spacelike coordinate geometry can be perfectly flat, even if located very close to an intense gravitational source such as a black hole. In that scenario, the circumference of a circle remains [tex]2\pi r[/tex].On the other hand,

**spacial curvature**occurs as a homogeneous geometry across the entire universe, depending on whether the universe is below, at, or above critical density [tex]\Omega[/tex] of mass-energy. Because spatial curvature is universe-wide, it is not possible to have positive curvature in one portion of the universe, flat in another portion, and negative curvature in yet another portion. (Or is it?) Spatial curvature (other than flat) causes the spacelike circumference of a circle [tex] \neq 2\pi r[/tex]r at an instant in time.Here's my question:

Is the heuristic of spatial curvature REQUIRED by general relativity, or is it merely a CONVENTION which is CONSISTENT WITH GR? I understand that mathematically, the curvature coefficient "k" provides the constant of integration for the Friedmann equations. However, the fact that a constant fits mathematically does not in itself require that constant to represent a physical reality.

So, it isn't clear to me why a universe with mass-energy > [tex]\Omega[/tex] is REQUIRED by GR to have 3-spherical positive spatial curvature. Why can't the geometry be flat?

If a universe were EXACTLY at [tex]\Omega[/tex], (being therefore geometrically flat), and also happened to be

*infinite*, and then as a thought experiment a single extra "test" photon were "released" (inserted) into this universe, that universe would necessarily become*above*[tex]\Omega[/tex] ever so slightly, and therefore would gain positive curvature and become necessarily*finite*in extent. Such an instantaneous transformation from infinite to finite is a bit of a mental stretch.My sense is that spatial curvature has been adopted as a "convention" in GR as a way to avoid the awkward discussion about whether the universe might be flat and finite, while simultaneously expanding or contracting. The discussion becomes awkward because this scenario implies the possibility of an "edge", or some kind of outer boundary to the universe. I understand why such a model would be less elegant than the spatial curve, "closed" approach.

So my question is, does GR REQUIRE spatial curvature if mass-energy density [tex]\neq \Omega,[/tex] or is spatial curvature more accurately described as a CONVENTION of GR?

Jon

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