I Clarifying the Meaning of Radius of Curvature in Cosmology

[Arman777]

$$1 - \Omega_{tot} = \Omega_κ = \frac{-κc^2}{R_0^2H_0^2} $$

For $\Omega_κ=-0.0438$ we get a some value for $R_0$. This $R_0$ is the radius of the observable universe right ?
Not the universe ?

 

[Vanadium 50]

This R0R_0 is the radius of the observable universe right ?
Not the universe ?

Neither. It is the radius of curvature.

 

[Arman777]

Neither. It is the radius of curvature.

In 3D or 2D shpere Radius of curvature isn't equal to the radius ?

 

[Vanadium 50]

Why is that even relevant?

 

[Arman777]

Why is that even relevant?

The question ?

 

[Vanadium 50]

Yes. Why is the question "In 3D or 2D shpere Radius of curvature isn't equal to the radius ?" even relevant?

You are aware that spacetime is not "curved into" anything, and certainly not a pre-existing Euclidian space, right?
You are also aware that the size of the visible universe is mostly a function of how old it is, right? (Particularly when one ignores Dark Energy)

 

[Bandersnatch]

Here's how I've always understood it:

Taking a 2D surface of a 2-sphere as an analogue of a positively-curved 3D universe, the radius of the observable universe is the radius of the circle drawn on the surface of this sphere (x). The radius of curvature of the 2D universe is the radius of the 2-sphere (R0):

In 3D hyperspherical universe, the radius of the spherical 3D volume that constitutes our observable universe is the analogue of x. The radius of curvature is the radius of the 3-sphere on which the 3D volume is drawn, analogous to R0 above.

 

[Arman777]

Taking a 2D surface of a 2-sphere as an analogue of a positively-curved 3D universe, the radius of the observable universe is the radius of the circle drawn on the surface of this sphere (x). The radius of curvature of the 2D universe is the radius of the 2-sphere (R0):

Okay I see your point.

Its like the case of the Einstein's static universe model. In that case we have $R_0 = \frac{c} and {\sqrt\Lambda}$ and that's the radius of the universe

Yes. Why is the question "In 3D or 2D shpere Radius of curvature isn't equal to the radius ?" even relevant?

Curiosity ? Its also something that we can answer ( I guess)

You are aware that spacetime is not "curved into" anything, and certainly not a pre-existing Euclidian space, right?

Yes, indeed.

You are also aware that the size of the visible universe is mostly a function of how old it is, right? (Particularly when one ignores Dark Energy)

It actually depends on the density parameters not exactly time itself. But density parameters internally depends on time so..

 

[Bandersnatch]

So ıf the universe is curved we are actually measuring the radius of the universe ?

As long as you understand that it's the radius of the curvature of the universe, i.e. the radius of the higher-dimensional hypersphere, not anything within the 3D space.

 

[Arman777]

As long as you understand that it's the radius of the curvature of the universe, i.e. the radius of the higher-dimensional hypersphere, not anything within the 3D space.

For 2D sphere as a 2d creatures radius of the universe is not a meaningful thing. Radius of curvature has some meaning ..? I guess that is your point ?

 

[Bandersnatch]

Yes.
For the 2D flatlanders living on the surface of a sphere:
- radius of the observable universe has a meaning, and it's how far on the surface of the sphere they can see;
- radius of the universe has no meaning, since it'd be the distance to the boundary of the surface, and the surface of a sphere has no boundary;
- radius of the curvature of the universe has a meaning, since it tells one how curved the geometry of the 2D space is, but it is not a distance between any two points within that space. A flatlander can't point anywhere on the 2D surface and say 'the centre of curvature is R0 light years that way'.

 

[Arman777]

Thanks

 

[PAllen]

Also, perhaps misleadingly, radius of curvature is used for hyperbolic geometry, where even with embedding it has no direct meaning. It is best thought of as just a variant of Gaussian curvature or sectional curvature, which have intrinsic definitions. The valid idea being the magnitude quasilocal deviation from Euclidean geometry is similar to the surface of a sphere with radius of radius of curvature - even if the type of deviation is opposite of spherical.

 

[Arman777]

Yes you are right. I asked the question because in some site someone claimed that $R_0$ is the radius of the universe and gives it in terms of ly and compares it with the radius of the observable universe . And that confused me.

I mean for 2D sphere case yes radius of the universe do not mean anything so the same goes for 3d sphere and he is wrong then. Mathematiclaly it can mean something but physically it means nothing. Since its not measureable.