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Posted (edited)

Hello

it is again a function of Hubble time tH =1/H as here : https://www.scienceforums.net/topic/125960-why-this-egality-is-exact-please/

you will not have the referential immediately, so don't contest with dimensional analysis please.

here his equation, you will adapt the dimensions:

[math]\rho_{\text{Vacuum at }t_H}= \frac{1}{ \vartheta} \frac{ G \text{ }m_{Pl}^2}{ (t_H\text{ }c)^2} [/math]

with :

[math] \vartheta=2 [/math] if tH>tPl   (1)

and

[math] \vartheta=1 [/math] if tH=tPl    (2)

can you please confirm that :

 

(2) give the exact numerical value ?

and that

(1) is a good approximation of the measurementsis a good approximation of mesures ?

 

thanks in advance

 

stéphane

 

 

 

 

Edited by stephaneww
Posted (edited)

ooops I forget because the question of dimensions we have:

(1) ρvac in m-2 when tH is big enough

(2) ρvac in kg/m3

Edited by stephaneww
Posted (edited)

sorry I made a mistake :

the egality (1) is in m-2

the egality (2) in Joules

precision on the question of the dimensions :

in the egality (1) it is because we have a singularity and in the egality (1), two square surfaces.

 

so with this presicions :

can you please confirm that :

the egality (2) give the exact numerical value of Planck energy?

and that

the egality (1) is a good approximation of the measurements of mesures of cosmological constant ?

 

thanks in advance

Edited by stephaneww
Posted (edited)

and I made also an error in the formula 😡:

 

there was an extra square in the denominator

note that now the equation has the dimension of an energy

img2.png

Edited by stephaneww

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