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Posted (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^3

I leave it to specialists of the forum to demonstrate equality ;)

Best regards

Stéphane
Edited by stephaneww
Posted (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 by stephaneww

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