Universe Hubble radius equal to Schwarzschild radius

[Smattering]

Hi bcrowell,

thank you so much for your answer.

When people say that our universe is approximately flat on large scales, they're referring to *spatial* flatness only. The Riemann tensor measures the curvature of spacetime, not just space. Our universe's spacetime is not even approximately flat. Spacetime curvature is how GR describes gravity. A universe with flat spacetime would be one in which there are no gravitational effects whatsoever.

Yes, I was aware that our universe can certainly not be completely flat due to the fact that we can observe lots of gravitational effects all around us. The question was only related to the curvature on very large scales.

I was also aware that the curvature described by GR is a curvature of 3+1-dimensional spacetime. This is quite obvious, because otherwise the geodesic line of an object would be independent of its relative speed, wouldn't it?

But I have to confess that until now I would have thought that a curvature of spacetime would also imply a spatial curvature in most cases. I can even believe that there might be some special cases where the spacetime has a curvature although there is no spatial curvature. But I would not have thought that the universe can be spatially flat on cosmological scales, and still the spacetime has a curvature on the same scale. But maybe I will understand this once I have read the material you suggested.

The Einstein field equations relate the curvature to the stress-energy tensor. The cosmological constant can be treated as one term in the stress-energy tensor. There are other terms as well, such as a term for dark matter and one for baryonic matter. Given the curvature of a manifold, there are theorems that in some cases uniquely determine the topology that is consistent with that curvature. This is one such case.

Yes, that makes sense. In the case of an ordinary sphere, I can easily imagine that there might be no other topology that fits to that global curvature. However, I was rather wondering how this particular curvature results in case of Einstein's static universe. Is it important that it is static, or does a non-staic cosmological model (as you mentioned it before) also have the same topology?

 

[PeterDonis]

I was aware that our universe can certainly not be completely flat due to the fact that we can observe lots of gravitational effects all around us. The question was only related to the curvature on very large scales.

You're still confusing space curvature with spacetime curvature. The gravitational effects we observe around us don't necessarily tell us that space is curved; that depends on how we split up spacetime into space and time. But the gravitational effects we observe do tell us unequivocally that spacetime is curved.

On the scale of the entire universe, as best we can tell from current observations, the universe is spatially flat, if we use the splitting of spacetime into space and time that is the "natural" one for observers who see the universe as homogeneous and isotropic. In other words, those observers would also observe the universe to be spatially flat. However, other observers in different states of motion, who would naturally split spacetime into space and time in a different way, might not observe the universe to be spatially flat. But the spacetime of the universe is curved regardless.

otherwise the geodesic line of an object would be independent of its relative speed

I'm not sure I understand what you mean by this. If you are trying to describe geodesic deviation--the fact that geodesics that are parallel at some particular point don't stay parallel--then yes, that is the definitive sign of spacetime curvature.

I have to confess that until now I would have thought that a curvature of spacetime would also imply a spatial curvature in most cases.

Whether space is curved depends on how we split up spacetime into space and time (see above). There is no absolute sense in which space is curved or flat.

 

[Smattering]

I'm not sure I understand what you mean by this. If you are trying to describe geodesic deviation--the fact that geodesics that are parallel at some particular point don't stay parallel--then yes, that is the definitive sign of spacetime curvature.

No, I was referring to something different. Let's assume I am throwing stones in a vacuum (jsut to get rid of aerodynamic effects). Let's further assume that I throw all the stone in the eaxctly same direction and angle:

If the curvature was purely spatial, then I would expect all the stones to follow the exactly same trajectory indepently of their initial velocity, wouldn't I? But in reality, the trajectories differ with the initial velocity. Thus, I conclude that the curvature cannot pe purely spatial, but it must also have some temporal component.

Whether space is curved depends on how we split up spacetime into space and time (see above). There is no absolute sense in which space is curved or flat.

O.k., this is something I was not aware of. Now I would like to have a brain with more than 3 dimensions.

I was trying to imagine a 2+1-dimensional spacetime, but still I am lacking at least one dimension to embed the curvature. ;-)

 

[martinbn]

[QUOTE="Smattering, post: 5284699, member: 576347".
I was trying to imagine a 2+1-dimensional spacetime, but still I am lacking at least one dimension to embed the curvature. ;-)[/QUOTE]

Use 2+1 flat space-time. The usual slicing in 2d planes would be an example of flat space through time. If you slice it differently say with a small bump then it will be non flat space through time. Imagine the slices are thin metal sheets. Stack them and hit them so that all get a bump but fit together (you have to be careful not to change the space-like nature of the slices).

 

[PeterDonis]

Let's assume I am throwing stones in a vacuum (jsut to get rid of aerodynamic effects). Let's further assume that I throw all the stone in the eaxctly same direction and angle:

Same direction and angle, but different velocities, all launched at the same time, correct?

If the curvature was purely spatial, then I would expect all the stones to follow the exactly same trajectory indepently of their initial velocity, wouldn't I?

No. First, we have to assume a specific splitting of spacetime into space and time, so that we can determine that the curvature is "purely spatial". Given that, stones launched at different velocities will spend different amounts of time in a region with a given spatial curvature, so the spatial curvature will bend their trajectories by different amounts.

in reality, the trajectories differ with the initial velocity.

"In reality", there is no invariant way to split up spacetime curvature into "space curvature" and "time curvature", so there's no way to tell what part of the difference is due to "space curvature" and what part is due to "time curvature". All you can say is that it's due to spacetime curvature.