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stephaneww last won the day on September 15 2019

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About stephaneww

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  1. 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
  2. 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.
  3. Hello I know that its fairly accurate value is 8.73 * 10122 when we are in the order of magnitude of 10120. But I also know that there is a value in the order of magnitude of 1060 I would like to know a value with a few decimal places and the exact exponent for the order of magnitude of 1060. Thanks for your answers
  4. 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 vac
  5. Hello. The weak point of my mathematical solution is that it does not allow for physical interpretation. Below I will try to propose an interpretation of the cosmological constant problem : The cosmological constant problem as a drag coefficient. ? Having read "the particles of a perfect gas have speeds close to those of light" and knowing that the formula of the cosmological constant can be interpreted as the expression is of a perfect fluid at negative pressure during my research on search engines I found the notion of drag coefficient . Its definition and
  6. Hi, See: https://constantecosmologique.fr/ available in English What is your opinion please?
  7. how to explain then that rho_c * c^2 = [M L^-1 T^-2] has the dimension of a pressure? where is the error?
  8. Okay, if the radiation is missing from rho, it makes the cross-reference a lot harder to check.(if I understand what you are saying). For H=67.74 (input) I have rho_c=8.61916 10^-27 kg/m^3, against rho_c=8.62765 10^-27 kg/m^3 in the calculator I'm really uncomfortable with state equations.🙄
  9. I confess I don't really understand. rho depends theoretically only on H? but we don't find it for H input nor for H ouput. I kept this value
  10. Hum, I don't known if that calculator is ok for all datas : for example H input (=67.74) isn't H output, and there's a little discrepancy I don't understand about rho. [tex]{\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline z&T (Gy)&H(t)&rho, kg/m^3&OmegaL&OmegaT \\ \hline 1.09e+3&3.70793e-4&1.55837e+6&4.61214e-18&1.28867e-9&1.00e+0\\ \hline 3.39e+2&2.48188e-3&2.47284e+5&1.16132e-19&5.11790e-8&1.00e+0\\ \hline 1.05e+2&1.52326e-2&4.16481e+4&3.29421e-21&1.80424e-6&1.00e+
  11. This is the case we find the formula of the critical density (*c^2 to have J/m^3) with a factor 3, so it works with the evolution of [math]H[/math] in time The method for finding [math]B'[/math] (Hubble constant) is the same as that used to determine the hypothetical [math]B[/math] (cosmological constant) value here: https://www.scienceforums.net/topic/122453-an-attempt-to-approach-a-notion-of-solubility-in-cosmology-to-explain-the-cosmological-constant/ In short, the method seems reproducible. My question therefore concerns the validity of this hypothesis. demonstration :
  12. Okay, thanks. But that's not what I'm asking about. My question is: can this new application of dimensional analysis of my hypothetical value [math]B[/math] in QM of the cosmological constant be validly duplicated for [math]B'[/math] with the Hubble's constant ? In view of the result, I think yes. Otherwise for the actual measurement of the critical density, it seems to me that at each measurement, we are either very slightly above or very slightly below.
  13. Hello, Mordred. I believe you're asking what we covered on pages one and two of this thread, aren't you? https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/#comments Otherwise I don't know what you're talking about for the moment, can you please specify (here I parallel the evolution of the total critical density with the approach of my previous solution : square root of Planck's energy density)
  14. oops error in the first post : Read [math]H_0^2 c^{-2}[/math] instead of [math]H_0 c^{-2}[/math] Let's consider [math]H_0[/math] the Hubble parameter (or Hubble constant) in [math]s^{-1}[/math]. We want a dimension in [math]L^{-2}[/math] to replace [math]\Lambda_{m^{-2}}[/math]. So we'll write [math]H_0^2 c^{-2}[/math] instead of [math]\Lambda_{m^{-2}}[/math] to get [math] B'=\frac{1}{(8\pi)^2} \hbar (H_0^2/c^2)^2.c[/math], "an energy density of Planck's universe for [math]H_0[/math]". Let's consider [math]\rho_c=\frac{3 c^2 H_0^2}{8\pi G}[/math] the critical
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