Relationship betwen ##Ω_k## and ##k##

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Discussion Overview

The discussion revolves around the relationship between the curvature parameter ##k## and the density parameter ##\Omega_k## in cosmology. Participants explore the implications of different values of ##k## on the universe's geometry, specifically addressing hyperbolic universes and the associated calculations.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests that if ##k < 0##, then ##\Omega_k## must be positive, leading to confusion about the relationship between these parameters.
  • Another participant clarifies that ##\Omega_k## has a sign opposite to that of ##k##, stating the relationship as ##\Omega_k = -k/H_0^2##, with a note on the convention used for the scale factor.
  • A later reply acknowledges a mistake, indicating a potential misunderstanding or miscommunication regarding the earlier claims.

Areas of Agreement / Disagreement

There is some agreement on the relationship between ##\Omega_k## and ##k##, but confusion remains regarding the implications of these parameters, particularly in the context of a hyperbolic universe. The discussion does not reach a consensus.

Contextual Notes

Participants express uncertainty about the implications of their statements and the assumptions underlying the relationships discussed, particularly regarding the conventions used in cosmological calculations.

Quarlep
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##H^2(t)-8ρπG/3=-k/a^2(t)## (Lets suppose a(t)=1) then
##H^2(t)-8ρπG/3=-k## In equations ##k## can be -1,0,1.
Then If ##k<0## then ##-k>0## then
##H^2(t)-8ρπG/3>0 ##→Hyperbolic Universe. This means ##Ω_k>0## I mean If ##k## negative ##Ω_k## must be positive isn't it ? I am confused here.
In cosmology calculator says ##Ω_k=1-Ω_m-Ω_Λ##

(Its look like hyperbolic universe I am trying to understand)

ThanksΩ
 
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Yes, ##\Omega_k## has a sign opposite to ##k##. Specifically, ##\Omega_k = -k/H_0^2## (note: this assumes the we're using the convention where ##a(now) = 1##. For the convention where ##k = {-1, 0, 1}##, ##\Omega_k = -k/H_0^2a^2(now)##).
 
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Thanks
 
I made a mistake soory
 

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