stephaneww Posted March 28, 2021 Share Posted March 28, 2021 (edited) Hello for H0=67.66 km/s/Mpc i.e. tH = 4.5606*1017s, and ΩΛ = 0.68694 we have: ρc,tH = 3 / (8 π G tH2) = 8.5988 * 10-27 kg/m3. - by dimensional analysis, looking for the value of the cosmological constant Λ of dimension m-2 and with a mixture of relativity, deterministic Planck units we have: Λ = 8π ΩΛ ρc,tH lp / mp = 1.1026 * 10-52 m-2. [math]\Lambda = 8 \pi \Omega_{\Lambda} \rho_{c,t_H} \frac{l_p}{m_p}=1.1026 *10^{-52} m^{-2}[/math] - This is exactly the value of the cosmological constant of the ΛCDM model. The formula of the density of the" quantum vacuum" at our time tH0, seems to correspond to the formula of the density of the quantum vacuum at the time tp. This comes from a toy model, whose only values we can be sure of are tp and tH thanks to the ΛCDM model. Is this an acceptable solution to the cosmological constant problem . (???) we have the quantum field theory energy with this equality : 8π/3 ρc,tp lp / mp = 3.8281 * 1069 m-2 =lp-2 [math] \frac{ 8 \pi}{3} \Omega_{\Lambda} \rho_{c,t_p} \frac{l_p}{m_p}=3 8281.*10^{69} m^{-2}[/math] note that mp / lp is the Planck linear mass. Edited March 28, 2021 by stephaneww Link to comment Share on other sites More sharing options...
stephaneww Posted April 25, 2021 Author Share Posted April 25, 2021 (edited) Hello and Hum ... it is nice to calculate: ΩΛ,tH = 3 Λ c2 tH2 with tH= Planck time : 5.391247*10^(-44) s c = 299792458 m/s and Λ = 1.102 * 10^(-52) m-2 ΩΛ,tp= 9.60 * 10-123 we make: (1 / ΩΛ,tp)* 8 π / 3 = 8.73 * 10122 to have the value of the vacuum catastrophe: (with vacuum energy value in QFT = lp^(-2) = 3.83 *10^(69) m-2 8 π * 3.83 *10^(69) / 1.105 * 10^(-52) = 8,73 * 10^(122) basically the vacuum catastrophe would be the density parameter of dark energy at Planck time in the ΛCDM model and that would be no problem. Does it make sense or not ? Edited April 26, 2021 by stephaneww Link to comment Share on other sites More sharing options...
stephaneww Posted April 27, 2021 Author Share Posted April 27, 2021 (edited) ooops big error sorry and it's why I think have no anwsers : read : ΩΛ,tH = 1/3 Λ c2 tH2 and not ΩΛ,tH = 3 Λ c2 tH2 , tH= 1/H Hubble's constant with tH= Planck time : 5,391247*10^(-44) s c = 299792458 m/s and Λ = 1,102 * 10^(-52) m-2 ΩΛ,tp= 9,60 * 10-123 we make: (1 / ΩΛ,tp)* 8 π / 3 = 8,73 * 10122 to have the value of the vacuum catastrophe: (with vacuum energy value in QFT = lp^(-2) = 3,83 *10^(69) m-2) lp =1,616255*10^(-35) m and this small one error too : 8 π * 3,83 *10^(69) / 1,102 * 10^(-52) = 8,73 * 10^(122) and not : 8 π * 3.83 *10^(69) / 1.105 * 10^(-52) = 8,73 * 10^(122) so now are you agree with : "basically the vacuum catastrophe would be the density parameter of dark energy at Planck time in the ΛCDM model and that would be no problem." Please ? EDIT : more easy : ΩΛ,tp = 1/3 Λ c2 tp2 vacuum catastrophe = Λ / lp-2 = Λ lp2 lp = c tp so vacuum catastrophe = Λ c2 tp2 vacuum catastrophe = 3 ΩΛ,tp Edited April 28, 2021 by stephaneww Link to comment Share on other sites More sharing options...
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