Curvature Index in Flat Universe with Cosmological Constant?

In summary: 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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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?
 
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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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1. What is the Curvature Index in a Flat Universe with Cosmological Constant?

The Curvature Index in a Flat Universe with Cosmological Constant is a measure of the spatial curvature of the universe. It is used to determine whether the universe is flat, open, or closed, based on the value of the cosmological constant.

2. How is the Curvature Index calculated?

The Curvature Index is calculated using the Friedmann equation, which relates the curvature of the universe to the density of matter and energy in the universe. It takes into account the effects of the cosmological constant on the overall curvature.

3. What does a Curvature Index of 0 indicate?

A Curvature Index of 0 indicates that the universe is flat, meaning that the overall curvature is equal to 0. This suggests that the universe is infinite in size and will continue to expand forever.

4. How does the Cosmological Constant affect the Curvature Index?

The Cosmological Constant, also known as dark energy, has a significant impact on the Curvature Index. It can either increase or decrease the overall curvature of the universe, depending on its value. A positive value of the cosmological constant leads to a closed universe, while a negative value leads to an open universe.

5. What is the significance of the Curvature Index in the study of the universe?

The Curvature Index is a crucial parameter in understanding the overall geometry and evolution of the universe. It helps us determine the fate of the universe and provides insights into the nature of dark energy. It also plays a significant role in cosmological models and theories, such as the Big Bang theory.

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