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) * R_{1 ,}

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, 20187 yr by stephaneww
latex

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.

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, 20187 yr by stephaneww