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Ranku
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For Ω=1, κ=0. Does the value of κ simply follow from the value of Ω, or can its value have an independent existence? So if Ω>1, does κ have to be 1?
Yes, by definition. Since Ω=1 is the critical density - the density at which the universe is flat - any other value of Ω necessitates that the k parameter is not 0 and has the same sign as Ω-1.Ranku said:So if Ω>1, does κ have to be 1?
See the discussion in the General Metric section of https://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric, for example.mathman said:Define k for me.
The density parameter, denoted as Ω, is a measure of the overall density of the universe. It is the ratio of the actual density of the universe to the critical density, which is the density needed for the universe to be flat.
The density parameter is calculated by dividing the actual density of the universe by the critical density. The actual density can be estimated by measuring the amount of matter and energy in the universe, while the critical density can be calculated using the Hubble constant and the gravitational constant.
A density parameter of Ω=1 means that the actual density of the universe is equal to the critical density. This indicates that the universe is flat and will continue to expand forever at a decreasing rate.
The curvature index, denoted as k, is a measure of the curvature of the universe. It can have three possible values: k=0 for a flat universe, k=1 for a closed universe, and k=-1 for an open universe.
The curvature index affects the evolution of the universe by determining its ultimate fate. A flat universe with k=0 will continue to expand forever, while a closed universe with k=1 will eventually collapse in on itself. An open universe with k=-1 will also continue to expand forever, but at an increasing rate.