stephaneww Posted April 9, 2018 Share Posted April 9, 2018 (edited) Hi, I have in the idea that the decrease of the temperature of the universe and the expansion of the universe are compensated exactly in terms of energies and propose the following calculations age of the universe (recombination) 378 000 years, with temperature CMB T1 = 3000 K, redshift : z1= 1 100, radius observable universe : R1 = 3.96*10^23 m age of the universe today : 13.797.000.000 years, with temperature CMB T0 = 2.725 K, reshift z0=1, radius observable universe R0 = 4.36*10^26 m and [latex]H_0=67.74[/latex], z1=T1/T0 z1=3000K/2.725K=1100 R0 = (1+z) * R1 , V1 = 2.59 * 10^71 m^3, V0 = 3.47 * 10^80 m^3. (Vt = volume of observable universe ) [latex]H_{t1}=H_0*\sqrt{\Omega_m(1+z)^3+\Omega_{\text{rad}}(1+z)^4+\Omega_{\Lambda} }=158284\text{ km/s/Mpc}[/latex] with : [latex]\Omega_m=0.3089[/latex] [latex]\Omega_{\text{rad}}=9*10^{-5}[/latex] [latex]\Omega_{\Lambda}=0.6911[/latex] __________________________________________________________ [latex]V_t [/latex] : Volume of observable universe Boltzmann constant : [latex]k_B = 1.380 648 52 * 10^{-23}J/K[/latex] [latex]X1=k_B*T_1/V_1=1.38*10^{-23}*3000/(2.59*10^{71})=1.60*10^{-91}J/m^3[/latex] [latex]X0=k_B*T_0/V_0=1.38*10^{23}*2.725/(3.47*10^{80})=1.09*10^{-103}J/m^3[/latex] [latex](X1/X0)/(\Omega_m(1+z)^3+\Omega_{\text{rad}}(1+z)^4+\Omega_{\Lambda})=(1.47*10^{12})/(5.46*10^8)=2697[/latex] dimensionless [latex]M_t[/latex] : "total mass" of observable universe (for its energy do [latex]M_t* c^2[/latex], the next ratio is the same for mass or energy) [latex]M_1=1.22*10^{54}kg[/latex] [latex]M_0=2.99*10^{54}kg[/latex] [latex]M_0/M_1=2.45[/latex] dimensionless and we finally find : [latex]z_1=2697/2.45=1100 [/latex] exactly (the first value) _________________________________________________________ I haven't lookked for the demonstration yet, and I'm not even sure it makes sense Thank you in advance for your opinion Edited April 9, 2018 by stephaneww latex Link to comment Share on other sites More sharing options...
Mordred Posted April 10, 2018 Share Posted April 10, 2018 You will find, if you haven't already done so that the universe temperature is roughly the inverse of the scale factor. There is several methodologies one can use to calculate the temperature at a given Z. One of the easier methods is to use the inverse or Gibbs law however one can also use the Einstein and Fermi-Dirac statistics. It isn't so much a result of energies as its a direct application of the ideal gas laws of a homogeneous and isotropic fluid. Link to comment Share on other sites More sharing options...
stephaneww Posted April 13, 2018 Author Share Posted April 13, 2018 (edited) On 10/04/2018 at 3:37 PM, Mordred said: You will find, if you haven't already done so that the universe temperature is roughly the inverse of the scale factor. Thank you, it's obvious but I didn't think that before you say it .... [latex]L_(t)=\frac{a_0}{a_t}L_0[/latex] [latex]L[/latex] : length radius universe [latex]a_0 = 1[/latex] [latex]a_t[/latex] : scale factor it's correct please ? On 10/04/2018 at 3:37 PM, Mordred said: ...There is several methodologies one can use to calculate the temperature at a given Z. One of the easier methods is to use the inverse or Gibbs law however one can also use the Einstein and Fermi-Dirac statistics. It isn't so much a result of energies as its a direct application of the ideal gas laws of a homogeneous and isotropic fluid. ok so my first post can't be right Edited April 13, 2018 by stephaneww Link to comment Share on other sites More sharing options...
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