Curvature Index in Flat Universe with Cosmological Constant?

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Discussion Overview

The discussion revolves around the curvature index κ in the context of a flat universe with a cosmological constant. Participants explore the implications of the Friedmann-Robertson-Walker-Lemaitre models and the relationship between density parameters and curvature.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that in a flat universe with a cosmological constant, the curvature index κ is necessarily zero, based on the condition that the density relative to critical density equals one.
  • Others propose that the curvature index κ can be expressed in terms of energy density, raising questions about the identification of the 'source' of this energy density.
  • One participant challenges the notion that the total mass/energy density of the universe can be directly linked to the curvature index, emphasizing that critical density serves as a threshold for determining the spatial curvature of the universe.
  • There is a distinction made between the definitions of total density and critical density, with a specific definition of ρκ introduced, which is noted to be a notational construct rather than a physical density.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between curvature index and energy density, indicating that multiple competing perspectives remain unresolved regarding the implications of these definitions and conditions.

Contextual Notes

The discussion includes various assumptions about density parameters and their implications for curvature, which are not fully resolved. The definitions and relationships presented may depend on specific interpretations of cosmological models.

Ranku
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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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