Hello,

I got interested in this paper: http://arxiv.org/pdf/1511.05450v1.pdf

as I was not happy to see a non-constant cosmological constant in time, I calculated the value (10 ) with the latest data planck 2015 : https://en.wikipedia.org/wiki/Planck_(spacecraft)#2015_data_release

I found that they were a mistake from a strictly factor equal to 3 of the cosmological constant, and that this factor made the 3 really constant cosmological constant in time for H0 = 67.74; 59; 62; 65; 71; 75; 80; and 85.

So we have:

Lamba₀= 3 *H₀^2 *Omega_Lambda \ c^2 =6Gm_dark \ R_Hubble^3

I leave it to specialists of the forum to demonstrate equality

Best regards

Stéphane

Edited December 4, 201510 yr by stephaneww

oops i made a mistake

 

Lamba₀= 3 *H₀^2 *Omega_Lambda \ c^2 =6Gm_dark \ R_Hubble^3Lamba₀= 3 *H₀^2 *Omega_Lambda \ c^2 =6Gm_dark \( R_Hubble^3 *c^2)

 

the demonstration is easier with this (i made it on a french forum). sorry for the persons who had try before

 

you will have to find the relation between R m and H for a perfectly balance of the universe acceleration and gravitation first

Edited December 7, 201510 yr by stephaneww