Curvature Index in Flat Universe with Cosmological Constant?

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SUMMARY

The curvature index κ in a flat universe with a cosmological constant is definitively zero. In a Friedmann-Robertson-Walker-Lemaitre universe, if the density relative to critical density satisfies the equation 1 = Ω = Ω_r + Ω_m + Ω_Λ, then κ equals zero, indicating flat spatial curvature. The total mass/energy density of the universe is expressed as ρ_total = ρ_m + ρ_r + ρ_Λ, with critical density defined as ρ_crit = 3H²/(8πG). If ρ_total equals ρ_crit, the universe is flat.

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Is the curvature index κ necessarily zero in a flat universe with cosmological constant?
 
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Ranku said:
Is the curvature index κ necessarily zero in a flat universe with cosmological constant?
Yes. If a Friedmann-Robertson-Walker-Lemaitre universes has density (relative to critical density) ##1 = \Omega = \Omega_r + \Omega_r +\Omega_\Lambda##, then it is flat and ##\kappa = 0##. ("Flat" refers to spatial curvature (of 3-dimensional hypersurfaces), not to spacetime curvature.)
 
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George Jones said:
Yes. If a Friedmann-Robertson-Walker-Lemaitre universes has density (relative to critical density) ##1 = \Omega = \Omega_r + \Omega_r +\Omega_\Lambda##, then it is flat and ##\kappa = 0##. ("Flat" refers to spatial curvature (of 3-dimensional hypersurfaces), not to spacetime curvature.)
To extend the discussion, the curvature index ##\kappa## can also cast in terms of its energy density
##\rho####\kappa#### =- \frac{3k}{8πGa^2}##. Can we identify the 'source' of ##\rho####\kappa##? Is it the matter density in the universe?
 
Ranku said:
To extend the discussion, the curvature index ##\kappa## can also cast in terms of its energy density
##\rho####\kappa#### =- \frac{3k}{8πGa^2}##. Can we identify the 'source' of ##\rho####\kappa##? Is it the matter density in the universe?
No.

The total mass/energy density of the universe ##\rho_{total} = \rho_m + \rho_r + \rho_\Lambda##, where ##\rho_m## is the density of matter, ##\rho_r## is the density of radiation, ##\rho_\Lambda## is the density of the vacuum.

Critical density ##\rho_{crit}## is defined by
$$\rho_{crit} := \frac{3H^2}{8\pi G}.$$
If ##\rho_{total} < \rho_{crit}## then the universe is open with negative spatial curvature, if ##\rho_{total} = \rho_{crit}## then the universe is open and flat (zero spatial curvature), and If ##\rho_{total} > \rho_{crit}## then the universe is closed with positive spatial curvature. Note that I have not said anything about whether the universe expands forever or recollapses.

Now define ##\rho_\kappa## by ##\rho_\kappa := \rho_{crit} -\rho_{total}##. This is just a definition, not a physical density, Sean Carroll writes "don't forget this just notational sleight of hand."
 
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