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stephaneww last won the day on September 15 2019

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About stephaneww

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  1. how to explain then that rho_c * c^2 = [M L^-1 T^-2] has the dimension of a pressure? where is the error?
  2. 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.🙄
  3. 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
  4. 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+0\\ \hline 3.20e+1&8.97471e-2&7.17048e+3&9.76466e-23&6.08678e-5&1.00e+0\\ \hline 9.29e+0&5.20128e-1&1.24469e+3&2.94228e-24&2.02004e-3&1.00e+0\\ \hline 2.21e+0&2.96640e+0&2.23251e+2&9.46559e-26&6.27909e-2&1.00e+0\\ \hline 0.00e+0&1.37737e+1&6.74009e+1&8.62765e-27&6.88894e-1&1.00e+0\\ \hline -6.88e-1&3.29612e+1&5.63238e+1&6.02483e-27&9.86506e-1&1.00e+0\\ \hline -8.68e-1&4.78813e+1&5.59720e+1&5.94981e-27&9.98945e-1&1.00e+0\\ \hline -9.44e-1&6.28349e+1&5.59450e+1&5.94406e-27&9.99911e-1&1.00e+0\\ \hline -9.76e-1&7.77911e+1&5.59427e+1&5.94357e-27&9.99993e-1&1.00e+0\\ \hline -9.90e-1&9.27477e+1&5.59428e+1&5.94360e-27&9.99988e-1&1.00e+0\\ \hline \end{array}}[/tex] ... and there is this after PLANCK 2015 : https://en.wikipedia.org/wiki/Planck_(spacecraft)#2018_final_data_release
  5. 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 : [math]A=\frac{c^7}{G^2 \hbar}[/math] [math]B'=\frac{1}{(8 \pi)^2}(\frac{H^2}{c^2})^2\hbar c[/math] [math]A B'=\frac{c^8}{(8 \pi)^2 G^2}\frac{H^4}{c^4} [/math] [math]A B'=\frac{c^4 H^4 }{(8 \pi)^2 G^2} [/math] [math] 3 \sqrt{A }\sqrt{B'}=\frac{3 H^2 c^2}{8 \pi G} = \rho_c c^2 [/math] https://en.wikipedia.org/wiki/Friedmann_equations#Density_parameter (the c^2 factor for the critical density differs between the French and English versions of wikipedia, hence a small float on this point)
  6. 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.
  7. 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)
  8. 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 density of the universe for [math]H_0[/math]. We have [math] 3 \sqrt{A} \sqrt{B'}=\rho_c[/math] for a quick demonstration we'll use [math]A=\frac{c^7}{G^2 \hbar}=m_p c^2/l_p^3 \text{ } J/m^3[/math]
  9. Hi, In this thread: https://www.scienceforums.net/topic/122453-an-attempt-to-approach-a-notion-of-solubility-in-cosmology-to-explain-the-cosmological-constant/, I proposed a mathematical solution to the cosmological constant problem. However, I have not found a physical explanation. Failing that, I found a generalization of this solution to the whole universe to validate a hypothesis that had been made in this solution In this it is, it seems to me, a confirmation (and perhaps help to understand the problem of the cosmological constant?). The energy density of the quantum vacuum in Planck units is: [math] A=m_pc^2/l_p^3=\hbar(l_p^{-2})^2.c[/math] I, on the other hand, found this unknown hypothetical quantum energy density of cosmological constant : [math]B=\frac{1}{(8\pi)^2}\hbar(\Lambda_{m^{-2})^2.c}[/math] and demonstrated that the cosmological constant [math]C= \sqrt{\hbar(l_p^{-2})^2.c} \sqrt{ \frac{1}{(8\pi)^2} \hbar(\Lambda_{m^{-2 }})^2.c}=\sqrt{A} \sqrt{B}[/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 c^{-2}[/math] instead of [math]\Lambda_{m^{-2}}[/math] to get [math] B'=\frac{1}{(8\pi)^2} \hbar (H_0/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 density of the universe for [math]H_0[/math]. We have [math] 3 \sqrt{A} \sqrt{B'}=\rho_c[/math] The method of dimensional analysis for application in quantum mechanics of general relativity data operates again...
  10. Thank you. I'll look for other cases where we'd have square root energy densities
  11. I have no idea where I'm going.I don't even know if it's applicable, although this encourages me to say yes ( cosmological constant seen as a perfect gas with a negative pressure) : But it's going to be very complicated and the finalization of this project is more than uncertain. I don't know the solubility issues at all and I don't know if scalar and vector aspects are involved in this potential approach. I just want to know if at first glance it seems completely silly or if it's worth trying to find out more about it.
  12. I've got it. It's just [math]\frac{1}{t_p^2\Lambda_{s^{-2}}}[/math] or [math]\frac{1}{l_p^2\Lambda_{m^{-2}}}[/math], whatever you want. The factor is [math]8\pi[/math].
  13. done here : https://www.scienceforums.net/topic/122453-an-attempt-to-approach-a-notion-of-solubility-in-cosmology-to-explain-the-cosmological-constant/
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