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How to derive "dark energy" density from Jorrie's Hubble radius limit
(c^4/(8 pi G))*3*(17.3e9 light years)^(-2)
Paste that in the google window, and see what you get.
When I paste that into google I get 0.5393 nanopascal, which is equivalent to
0.5393 nanojoule per cubic meter.
If the cosmological curvature constant actually arose from a real "dark energy" that would be the density of that energy. I don't know of any observational evidence that it represents any sort of real energy. Observation is entirely consistent with it simply being a very small constant vacuum curvature. But real or fictitious, that's a way to get the energy density.
A curvature is the reciprocal of an area. small curvature ↔ large area.
The most readily available handle on the cosmo curvature constant is this distance:
17.3 billion lightyears the longterm limiting value of the Hubble radius.
When you open Jorrie's calculator you immediately see that the Hubble radius is tending to a longterm limit of 17.3 billion lightyears.
The conventional Lambda in the Einstein gr equation is actually equal to 3*(17.3e9 light years)^(-2)
That is what you'd put in the Einstein equation, in some form or another, if you were using the equation in its traditional form. Aside from that factor of 3, the area is just the square of 17.3 billion lightyears.
To continue deriving the possibly fictitious "energy" density---the factor (c^4/(8 pi G)) is just a force. People familiar with "Planck units" will recognize it (up to the 8 pi factor) as the unit force belonging to that system of "natural units". If you multiply a force by a reciprocal area you get a pressure, or equivalently an energy density---IOW a nanojoules per cubic meter quantity, or nanopascals.
Here's the calculator if you want to see what I mean about limiting value of Hubble radius:
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone1A/LightCone.html
(c^4/(8 pi G))*3*(17.3e9 light years)^(-2)
Paste that in the google window, and see what you get.
When I paste that into google I get 0.5393 nanopascal, which is equivalent to
0.5393 nanojoule per cubic meter.
If the cosmological curvature constant actually arose from a real "dark energy" that would be the density of that energy. I don't know of any observational evidence that it represents any sort of real energy. Observation is entirely consistent with it simply being a very small constant vacuum curvature. But real or fictitious, that's a way to get the energy density.
A curvature is the reciprocal of an area. small curvature ↔ large area.
The most readily available handle on the cosmo curvature constant is this distance:
17.3 billion lightyears the longterm limiting value of the Hubble radius.
When you open Jorrie's calculator you immediately see that the Hubble radius is tending to a longterm limit of 17.3 billion lightyears.
The conventional Lambda in the Einstein gr equation is actually equal to 3*(17.3e9 light years)^(-2)
That is what you'd put in the Einstein equation, in some form or another, if you were using the equation in its traditional form. Aside from that factor of 3, the area is just the square of 17.3 billion lightyears.
To continue deriving the possibly fictitious "energy" density---the factor (c^4/(8 pi G)) is just a force. People familiar with "Planck units" will recognize it (up to the 8 pi factor) as the unit force belonging to that system of "natural units". If you multiply a force by a reciprocal area you get a pressure, or equivalently an energy density---IOW a nanojoules per cubic meter quantity, or nanopascals.
Here's the calculator if you want to see what I mean about limiting value of Hubble radius:
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone1A/LightCone.html
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