Hello

it is again a function of Hubble time t_H =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 t_H>t_Pl   (1)

and

[math] \vartheta=1 [/math] if t_H=t_Pl    (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 October 25, 20214 yr by stephaneww

ooops I forget because the question of dimensions we have:

(1) ρ_vac in m^-2 when t_H is big enough

(2) ρ_vac in kg/m³

Edited October 25, 20214 yr by stephaneww

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 October 26, 20214 yr by stephaneww

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

Edited October 26, 20214 yr by stephaneww