I On the assumption of an infinite universe

[dx]

In the standard framework of ideas about cosmology, is it possible to have a universe that is infinite in extent?

 

[kimbyd]

In the standard framework of ideas about cosmology, is it possible to have a universe that is infinite in extent?

The standard model is infinite in extent, but it's not a good idea to take the standard model seriously far beyond our horizon.

Whether or not the universe can be infinite in extent is currently unknown.

 

[elcaro]

In the standard framework of ideas about cosmology, is it possible to have a universe that is infinite in extent?

Since the measured value of the curvature is very near a flat universe, it indeed includes the possibility our universe is infinite in extent.

 

[kimbyd]

Since the measured value of the curvature is very near a flat universe, it indeed includes the possibility our universe is infinite in extent.

The spatial curvature of the observable universe isn't necessarily related. With the exception of a large positive curvature, any curvature value permits either finite or infinite solutions.

 

[elcaro]

The spatial curvature of the observable universe isn't necessarily related. With the exception of a large positive curvature, any curvature value permits either finite or infinite solutions.

Wouldn't a flat finite universe have an edge?

 

[ergospherical]

Wouldn't a flat finite universe have an edge?

no, for example a torus

 

[PeterDonis]

With the exception of a large positive curvature, any curvature value permits either finite or infinite solutions.

How can you have a spatially infinite universe with a small positive curvature (since you said "large" positive curvature instead of just positive curvature period).

 

[kimbyd]

How can you have a spatially infinite universe with a small positive curvature (since you said "large" positive curvature instead of just positive curvature period).

If the positive curvature was a local effect only, it could still be infinite. If it was sufficiently large, it would be hard for it to be a purely local effect.

 

[elcaro]

no, for example a torus

That is not exactly flat, unless you deform it into a pancake with a hole in it...

 

[PeterDonis]

If the positive curvature was a local effect only

Ah, so you are considering models that are not homogeneous.

 

[PeterDonis]

That is not exactly flat

A 2-D torus cannot be flat, but a 3-D torus can be. The 3-D flat torus is the spatial geometry being referred to.

 

[Orodruin]

That is not exactly flat, unless you deform it into a pancake with a hole in it...

Not true. You are thinking of the geometry on the torus induced by its typical embedding in three-dimensional Euclidean space. He is not.

 

[Orodruin]

A 2-D torus cannot be flat

Yes, it can.

 

[PeterDonis]

Yes, it can.

Ah, yes, I was forgetting the "Asteroids" arcade game.

 

[Orodruin]

Ah, yes, I was forgetting the "Asteroids" arcade game.

When I was an undergrad we used to play a lot of Go during breaks. Eventually we invented the game of toroidal Go by identifying the sides. It got very confusing, but fun. Very different game when you cannot cling to the borders.

 

[elcaro]

Not true. You are thinking of the geometry on the torus induced by its typical embedding in three-dimensional Euclidean space. He is not.

I can't curve my head around that...

 

[PeroK]

I can't curve my head around that...

A simple example of the difference between intrinsic and extrinsic curvature is the cylinder. Take a flat sheet of paper and bend it into an open cylinder. The 2D differential geometry in terms of intrinsic flatness has not changed, although it is now extrinsically curved in 3D.

 

[jbriggs444]

I can't curve my head around that...

Central to the notion of a metric space is the idea of a metric. The metric is a function that tells you how far it is from one point in the space to another (along the shortest path, of course). The 3-dimensional Euclidean metric is, of course, $d=\sqrt{(\Delta x)^2+(\Delta y)^2 + (\Delta z)^2}$ for the distance between two points given with cartesian coordinates.

If we have a the two-dimensional surface of a torus embedded in this three-dimensional space, we can measure the three dimensional path length of any path that stays on the surface. This allows us to induce a metric on the two dimensional space -- the length of the shortest path that stays on the surface.

But we are not required to use this metric. We can discard the connection to three dimensional Euclidean space and use a different metric. We can subtly shift the metric so that points on the outside of the torus are "closer" to one another and so that points on the inside of the torus are "farther apart". So that it becomes like a tube made from rolled up paper (still flat) and yet the two ends of the tube still meet so that the space is closed.

Still head-curving, though.