stephaneww Posted December 4, 2015 Posted December 4, 2015 (edited) 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 *OmegaLambda \ c^2 =6Gmdark \ RHubble^3I leave it to specialists of the forum to demonstrate equality Best regardsStéphane Edited December 4, 2015 by stephaneww
stephaneww Posted December 7, 2015 Author Posted December 7, 2015 (edited) oops i made a mistake Lamba0= 3 *H0^2 *OmegaLambda \ c^2 =6Gmdark \ RHubble^3Lamba0= 3 *H0^2 *OmegaLambda \ c^2 =6Gmdark \( RHubble^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, 2015 by stephaneww
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