Undergrad Density parameter and curvature index

Click For Summary
In cosmology, when the density parameter Ω equals 1, the curvature index κ is defined as 0, indicating a flat universe. If Ω exceeds 1, κ must be positive, reflecting a closed universe, while if Ω is less than 1, κ is negative, indicating an open universe. The value of κ is intrinsically linked to Ω, meaning it cannot exist independently. The curvature index can be defined using different conventions, such as the (-1,0,1) system, which simplifies understanding its relationship with Ω. Thus, κ's value is directly influenced by the density parameter Ω.
Ranku
Messages
434
Reaction score
18
For Ω=1, κ=0. Does the value of κ simply follow from the value of Ω, or can its value have an independent existence? So if Ω>1, does κ have to be 1?
 
Space news on Phys.org
Context would be useful. I don't want to guess what you mean.
 
Ranku said:
So if Ω>1, does κ have to be 1?
Yes, by definition. Since Ω=1 is the critical density - the density at which the universe is flat - any other value of Ω necessitates that the k parameter is not 0 and has the same sign as Ω-1.
 
  • Informative
  • Like
Likes Jaime Rudas and Ranku
Define k for me.
 
The definition of k seems to have more than one option. For example the (-1,0,1) option simplifies your question.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
8K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K