I Curvature Index in Flat Universe with Cosmological Constant?

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In a flat universe with a cosmological constant, the curvature index κ is necessarily zero, as indicated by the Friedmann-Robertson-Walker-Lemaitre model when the total density equals the critical density. This flatness refers specifically to spatial curvature, not spacetime curvature. The total mass/energy density of the universe is composed of matter, radiation, and vacuum energy, and its relationship to critical density determines the universe's curvature. If the total density is less than, equal to, or greater than critical density, the universe is open, flat, or closed, respectively. The discussion also clarifies that the term ρκ is a definitional construct rather than a physical density.
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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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