stephaneww

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About stephaneww

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  • Birthday 10/02/1968

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    France
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    cosmology

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  1. I'm not expert enough to be affirmative. I the most I can offer you, it's the same formula (in [math]s^{-2}[/math] for [math]\Lambda[/math]) used by a french CNRS researcher page 10 of this document : http://www.cnrs.fr/publications/imagesdelaphysique/couv-PDF/IdP2008/03-Bernardeau.pdf edit : [math]\Omega_\Lambda \rho_c = \rho_\Lambda[/math] the adjustment is made by the variation of [math]\Omega_\Lambda[/math], right ? it's [math]\rho_\Lambda[/math] which is constant as [math]\Lambda[/math]
  2. I understand that if we consider a universe, almost, but not flat, it modifies the critical density and so affects my result. I don't have enough knowledge to do the calculation you propose next. However, I remember reading on Wikipedia, but without being able to find out where immediately, that the universe was so close to being flat that most cosmologists considered that it was indeed flat. Is this last proposal wrong please?
  3. edit me too : Yes, I should have said "energy critical density" ([math]J/m^3)[/math] [math]\Omega_\Lambda=\frac{\rho_\Lambda}{\rho_c}[/math] with : [math]\rho_\Lambda=\frac{c^2}{8 \pi G}\Lambda[/math] [math]kg/m^3[/math] and [math]\rho_c=\frac{3H_0^2}{8 \pi G}[/math] [math]kg/m^3[/math] so [math]\Lambda_{m^{-2}}=\frac{3H_0^2}{c^2}\Omega_\Lambda m^{-2}[/math] for the value in [math]kg/m^3[/math] if I'm not mistaken, we do : [math]\frac{F_p \Lambda_{m^{-2}}}{c^2 8 \pi}[/math] is that correct with these details ?
  4. Uh, is this for the first message ? I used: [math]\rho_c=\frac{3c^2H^2}{8 \pi G}[/math] for critical density, is that correct? and [math]\Lambda_{m^{-2}}=\frac{3H_0^2}{c^2}\Omega_\Lambda[/math] is that correct? Is there anything more to add?
  5. First : Thank you for your answer +1 Indeed, you're right about the second message: For H0=70,00 [math]\frac{\hbar*f}{V_u}=8.94*10^{-133}J/m^3[/math] The energy density of the cosmological constant would be for a universe age of 12.8 billion years : [math]6.44*10^{-10}J/m^3[/math] As we are in the speculation section, I will add this: Cosmologists are debating the value of H0. This could result in a modification of the standard model. (https://www.scienceforums.net/topic/119814-h0licow-new-measurements-of-hubble-constant-highlight-problem/) Among the hypotheses there would be this one : the cosmological constant would not be constant, it would accelerate the acceleration of expansion. In this hypothesis, its energy density would decrease over time, all other things being equal (if I am not mistaken) Now that we agree, i.e. that H0 does not give a constant result for a second message, let's address the first message if you agree : This time all values are constants, the results are constant, accurate and precise. The thing I'm not sure about is the conclusion to be drawn. Of course, there is an effort to be made to understand the "big piece", but if you wish, I can detail it to make it more accessible. oops, traduction of French Wikipédia : Following the precise calibration of the distance of the Large Magellanic Cloud 15, measurements made by Adam Riess using Large Magellanic Cloud cepheids in 2019 give a Hubble constant value of 74.03±1.42 kilometres per second per megaparsec16.The difference between this measurement and the value calculated by the Planck mission is due to the parameters of the cosmological model used for the calculations in the case of the Planck mission.
  6. oops, I ate the powers of 10 ... correction in the quote and of course in the first message, you should read [latex]2.866*10^{-19}[/latex] and not [latex]2.866*10^{19}[/latex]
  7. Hello, There is another approach mixing relativity and quantum mechanics, simpler, but it is less precise (it has a margin of error of about 5%): With [latex]H_0=67.66 . Km/Mpc = 2.193*10^{-18} s^{-1}[/latex], Hubble constant. [latex]\hbar= 1.055*10^{-34} .kg.m^2.s^{-1}[/latex] reduced Planck constant. [latex] V_u= 3.478*10^{80} .m^3[/latex] volume of the observable universe. Quantum Energy [latex]E_Q[/latex] is given by : [latex]E_Q=\hbar*f[/latex] where [latex]f[/latex] is a frequency (we will use the Hubble constant) [latex]\hbar*f=1.055*10^{-34}*2.193*10^{-18}=2.312*10^{-52} .kg.m^2.s^{-2}[/latex] Now we can find the density energy mixing relativity and quantum mechanics : [latex]\frac{\hbar*f}{V_u}=\frac{2.312*10^{-52}}{3.478*10^{80}}=6.645*10^{-133}.J/m^3[/latex] The sequence is identical to the calculation "A" of the first message: [latex]6.645*10^{-133}*4.633*10^{133}=3.079*10^{-19}.(J/m^3)^2[/latex] [latex]\sqrt{3.079*10^{-19}}=5.549.(J/m^3)[/latex] to be compared with the value of the energy density of the cosmological constant in the LambaCDM model with the same data : [latex]5.32 .J/m^3[/latex] I hope, with this message, to inspire responses. Thank you in advance. note : values about LambdaCDM model, here : https://en.wikipedia.org/wiki/Cosmological_constant#Equation and here : https://en.wikipedia.org/wiki/Observable_universe
  8. thank you again Mordred. what is [latex]r[/latex] please ? on occasion I would look at what the spherical coordinates consist of the rotation is too hard for me, I forget for now
  9. oops my formula need a correction, [latex]g=\frac{H_0^2}{c^2}\frac{16\pi^2}{9}[/latex] in the part context cited of the document, and if I don't make mistake, is not a determinant or linked to the metric tensor indeed [latex]\frac{1}{\sqrt{|g|}}=\frac{c}{H_0}\frac{3}{4\pi R^3}.R.R.R [/latex] [latex]m[/latex] always if I don't make mistake... I deduce : [latex]|g|=\frac{H_0^2}{c^2}\frac{4^2\pi^2}{3^2} m^{-2}[/latex] I always don't know if it's ok... and finally I find [latex]\Lambda=3\frac{H_0^2}{c^2}\Omega_\Lambda=3g\frac{3^2}{4^2\pi^2}\Omega_\Lambda m^{-2} [/latex] (correction made with factor 3 in addition) The question is : is it ok ? question for Mordred : I haven't understand… about the theorical frame of the document for this part... is it correct please ?
  10. edit too late sorry Hello For information, there is a version 2 of the document. all this can be simplify by [latex]1.106 * 10^{-52}m^{-2}= \Lambda= g \frac{9}{16\pi^2}\Omega_\Lambda = 3\frac{H^2}{c^2} \Omega_\Lambda[/latex] (nothing new) I hope that somebody will can explain me why this is OK and what does this mean in the context of the source document note : V0=4pi/3 R0^3
  11. Hello For information, there is a version 2 of the document. all this can be simplify by [latex]1.106 * 10^{-52}m^{-2}= \Lambda= g \frac{9}{16\pi^2} = 3\frac{H^2}{c^2} \Omega_\Lambda[/latex] (nothing new) I hope that somebody will can explain me why this is OK and what does this mean in the context of the source document
  12. Hello A clarification above all. I would use here the values of the results of the Planck mission published in 2015, in particular : [latex]H_0=67.74 km/s/Mpc[/latex] Hubble constant [latex]\Omega_{\Lambda}=0.6911[/latex] density parameter of the cosmological constant of the epoch hence [latex]\Lambda_{m^{-2}}=1.111*10^{-52}m^{-2}[/latex] cosmological constant expressed in [latex] m^{-2}[/latex] that is to say [latex]\Lambda_{s^{-2}}=9.992*10^{-36}s^{-2}[/latex] when the cosmological constant is expressed in [latex]s^{-2}[/latex] and [latex]R=4.359*10^{26}m[/latex] radius of the universe observable still with the results of the 2015 period But basically what matters is that the same reasoning can be applied as in the results of the Planck mission published in 2018. as a reminder, we obtain with the values 2015 : ____________________________________________________________ Fp : Planck force 1 the energy density by volume of the quantum vacuum as A=Fp/lp^2 = 4.633*10^113 Joules/m^3 (formula derived from dimensional analysis in Planck units) 2 the energy density by volume of the vacuum of the cosmological constant as B= Fp* Lambda /8/ pi = 5.354*10^-10 Joules/m^3 the ratio between the two being the number in factor of 10^122 undimensioned the value of the adimensionless factor, X, is more precisely : A/B=X=8.654*10122, X is called " The cosmological constant problem ". _____________________________________________________________ -Now the big piece : Still doing dimensional analysis, we can calculate an energy density of the vacuum mixing the quantum approach and the relativistic approach : that is to say [latex]\hbar=1.055*^10^{-34}*kg*m^2*s^{-1}[/latex] the reduced Planck constant. we want an energy density by volume which is kg m-1 s-2 (or from Joules/m3) based on [latex]\hbar[/latex] and values in [latex]m^{-2}[/latex] and in [latex]s^{-2}[/latex] of [latex] \Lambda[/latex] we can reasonably calculate the volume energy density of the vacuum mixing the quantum and relativistic approaches as follows and add a dimensionless factor in 8*pi : 3. [latex]D=\hbar* ({\Lambda_{m^{-2}}})^{3/2}*({\Lambda_{s^{-2}}})^{1/2}/(8*\pi)^2=6.196*10^{-133}kg*m^{-1}*s^{-2}[/latex] as [latex]\Lambda_{m^{-2}}*c^2=\Lambda_{s^{-2}}[/latex] where [latex]c[/latex] is the speed of light in vacuum we have : [latex]({\Lambda{m^{-2}}})^{1/2}*c=({\Lambda_{s^{-2}}})^{1/2}[/latex] the volume density of energy of the vacuum mixing the quantum approach and the relativistic approach becomes : [latex]D=\hbar*({\Lambda_{m^{-2}}})^2*c/(8*\pi)^2=6.196*10^{-133}kg*m^{-1}*s^{-2}[/latex] unit equivalent to [latex]Joules/m^3[/latex] -Much simpler now : A. volume density of vacuum energy mixing the quantum approach and the relativistic approach * volume density of energy from quantum vacuum = [latex]6.196*10^{-133}*4.633*10^{113}=2.866*10^{19} (Joules/m^3)^2[/latex] we find to more than a dozen decimal places when we do the calculation in a spreadsheet : [latex](2.866*10^{19} (Joules/m^3)^2)^{1/2}= 5.354*10^{-10}Joules/m^3[/latex] that is the volume energy density of the vacuum of the cosmological constant for a first appearance B. volume density of vacuum energy mixing the quantum and relativistic approaches * X "cosmological constant problem" = [latex]6.196*10^{-133}*8.654*10^{122}=5.354*10^{-10}Joules/m^3[/latex] we find again and to more than a dozen decimal places when calculating in a spreadsheet, the energy density by volume of the vacuum of the cosmological constant I am not sure how to writte to my conclusion: I would say that the the energy density density of the vacuum of the cosmological constant is central around the two modes of calculations of the quantum vacuum energy density density. The X value known as "the constant cosmologic problem" is spread around this central value