# stephaneww

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• Birthday 10/02/1968

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1. ## Is the dark energy already present as vacuum energy by nature in the equations of general relativity ?

Hello I have been asking myself this question since I saw the formula (II-30) on page 11 of this (unpublished) document: Indeed, $d \Omega$ is presented as a four-dimensional volume variation in general relativity for EINSTEIN equations in vacuum. $d\Omega=\sqrt{|g|}dx^0dx^1dx^2dx^3$ where $x^0=ct$ Assuming that this four-dimensional volume is composed of the three spatial dimensions and a temporal dimension, the following calculations can be attempted from the 2018 Planck mission results data. By making this assumption $g_0$ is no longer adimensioned but takes the dimension $m^{-2}$, dimension of the cosmological constant. We then find a new constant $X= 7.719*10^{-53}m^{-2}$ that can be related to the cosmological constant $\Lambda$ today: volume universe $V_0 = 3.48*10^{80} m^3$ Hubble's time: $t_0 = 4.56*10^{17} s$ or $H_1 = 67.66 km/s/Mpc$ $c* t_0 = 1.37*10^{26} m$ observable universe radius : $x^1=x^2=x^3=R_0 = 4.36*10^{26} m$ density parameter of the cosmological constant: $\Omega_{0 _ \Lambda} = 0.6889$ equality (II-30) page 11 is equivalent to : $1/\sqrt{g_0}=c* t_0*R_0^3/V_0=3.26*10^{25}m$ hence $g_0=9.386*10^{-52}m^{-2}$ $g_0'=g_0/8/\pi*3=1.120*10^{-52}m^{-2}$ when Hubble time = $t_1 = 2.91*10^{17} s$ or $H_1 = 70 km/s/Mpc$ volume universe $V_1 = 1.13*10^{80} m^3$ $c* t_1 = 8.25*10^{25} m$ observable universe radius : $x^1=x^2=x^3=R_1 = 3.00*10^{26} m$ density parameter of the cosmological constant: $\Omega_{1_\Lambda} = 0.2803$ equality (II-30) page 11 is equivalent to : $1/\sqrt{g_1}=c* t_1*R_1^3/V_1=3.08*10^{25}m$ hence $g_1=2.308*10^{-51}m^{-2}$ $g_1'=g_1/8/\pi*3=2.754*10^{-52}m^{-2}$ since $d \Omega$ is a variation in volume since the origin of the universe, it seems coherent to take into account the density parameter of the cosmological constant related to this volume in a spreadsheet program we have the following perfect equality and a new constant $X$ : $g_0'*\Omega_{0_\Lambda}=g_1'*\Omega_{1_\Lambda}=...=X$ $1.120*10^{-52}*0.6889=2.754*10^{-52}*0.2803=$ $X= 7.719*10^{-53}m^{-2}$ A link with the cosmological constant $\Lambda=1.105*10^{-52}m^{-2}$ can be made as follows: $\Lambda / X * 8 * \pi=$ $1.105*10^{-52}/7.719*10^{-53}* 8 * \pi=$ $36$ (exact dimensionless value calculated using a spreadsheet) If the author's reasoning is correct and the assumptions and calculations made here are correct, it could mean that the black energy is, by nature, the underlying vacuum energy in general relativity. Thank you for your attention and any comments you may have.
2. ## Could the acceleration of expansion have a simple geometric reason ?

Thank you for the changes Strange
3. ## Could the acceleration of expansion have a simple geometric reason ?

sorry, please read : in title :"... acceleration of the expansion", "comobil" instead ",commotional", "The vaccum created by the expansion in the universe would be of constant density energy and the new space created by the expansion would be quantum vaccum." instead: "The vaccum created by the expansion in the universe would be of constant density energy and the new space created by the expansion would be part of the quantum vaccum." and "...would be the automatic (geometric effect) expression on the cosmological constant, i.e. the acceleration of the expansion of the universe, due to the effect of the Hubble constant on a sphere region of space"
4. ## Could the acceleration of expansion have a simple geometric reason ?

I have copy and translate this from a french forum : ________________________ The Hubble parameter defined from the scale factor of the FLRW metric is $H(t) = \frac{\dot{a}(t)}{a(t)}$ a parameter representing the expansion rate of the "physical" volume of a compact spatial domain whose coordinates of the points of the surface are constant in co-moving coordinates (i.e., whose material content is always the same) can be defined by $\theta(t) = \frac{\dot{V}(t)}{V(t)}$ We then simply obtain $\theta(t) = 3 H(t)$ Indeed, if we reason on a ball (a volume $V$ whose surface is a sphere), with a radius $R$ in co-moving coordinates, its physical volume is then $V(t) = \frac{4\pi}{3} a(t)^3 R^3$ and, by drifting in relation to time, $\dot{V}(t) = \frac{4\pi}{3}[3 a(t)^2\dot{a}(t)] R^3$ Hence the fact that $\theta(t) = \frac{\dot{V}(t)}{V(t)} = 3 \frac{\dot{\dot{a}(t)}}{a(t)} = 3 H(t)$ _____________________ When $\Lambda$ cosmological constant is expressed in $s^{-2}$ we have $\Lambda/(3H(t)^2)=\Omega_{\Lambda(t)}$, which corresponds, if I am not mistaken, to Friedman's equation with only the cosmological constant as its component, i.e. for a spherical universe expanding under the effect of the cosmological constant and Hubble constant. It seems interesting to me to see that we can simplify by $H(t)$ (expansion = constant Hubble effect) and that as a result, there remains $3H(t)*\Omega_{\Lambda(t)}$ as a factor that could be interpreted as the acceleration of expansion (but I am far from being sure that this is a valid interpretation) As a result, matter (ordinary and dark) is diluted under this double effect and would further increase the share of dark energy in the total energy of the universe in the LambdaCDM model. The vaccum created by the expansion in the universe would be of constant density energy and the new space created by the expansion would be part of the quantum vaccum. $\Lambda/H(t)=3*H(t)*\Omega_{\Lambda(t)}$, $3*H(t)*\Omega_{\Lambda(t)}$, would be the automatic (geometric effect) expression on the cosmological constant, i.e.the acceleration of the expansion, due to the effect of the Hubble constant on a sphere region of space Thanks in advance for yours comments
5. ## Dark Energy increasing with Time:

something new on the subject : Best-Yet Measurements Deepen Cosmological Crisis https://www.scientificamerican.com/article/best-yet-measurements-deepen-cosmological-crisis/?amp https://arxiv.org/abs/1903.07603
6. ## Need help with density parameters of the LambdaCDM model

Thank you for everything, Mordred. If we assume, for example, that $M_b * \Lambda= cst$ then, when $M_b$ decreases $\Lambda$ increases inversely proportional. Could this be a part of the explanation for the fact that it seems that the acceleration of expansion is also accelerated?
7. ## Need help with density parameters of the LambdaCDM model

merge then edit too late : Can we say that $M_\text{b(t-1)} = M_\text{b(t)}$ from close to close ?
8. ## Need help with density parameters of the LambdaCDM model

\textif I understand correctly (because I didn't understand everything) for z=0.791, doing $\Omega_b=M_b/M_{total}$ is not a good approximation ? well, I redid the table with the same values but with 100 steps The difference $\Omega_b / \Omega_m$ for z=0 et z=0.124 becomes minor : 0.05% Can we say that $M_b_\text{(t-1)}/ M_b_text{(t)}$ from close to close ?
9. ## Need help with density parameters of the LambdaCDM model

for z=0.791, I used Ned Wright's calculator to determine radius of observable universe, then I find the volume, then I find total mass with the density of your table. I assume that $M_b$ don't change edit : of course
10. ## Need help with density parameters of the LambdaCDM model

In abstract $\Omega_b h^2=0.0224$ so : $\Omega_b =0.0224/h^2=0.0224/(67.4/100)^2$
11. ## Need help with density parameters of the LambdaCDM model

with : http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7-2017-1/LightCone_Ho7.html (base Planck 2015) and : H : 67.4 Omega L : 6851 Step : 20 I have : ${\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline z&Sc.fctr (a)&S&T (Gy)&D_{hor}(Gly)&H(t)&rho, kg/m^3&Temp(K)&OmegaM&OmegaL&OmegaR \\ \hline 1090.000&0.000917&1091.000&0.000370&0.0567&1.5678e+6&5.1243e-18&2.9784e+3&7.5549e-1&1.2661e-9&2.4451e-1\\ \hline 608.078&0.001642&609.078&0.000972&0.1009&6.1767e+5&7.9532e-19&1.6628e+3&8.4697e-1&8.1576e-9&1.5303e-1\\ \hline 339.033&0.002941&340.033&0.002479&0.1787&2.4879e+5&1.2903e-19&9.2829e+2&9.0838e-1&5.0282e-8&9.1628e-2\\ \hline 188.832&0.005268&189.832&0.006188&0.3153&1.0166e+5&2.1542e-20&5.1824e+2&9.4669e-1&3.0117e-7&5.3311e-2\\ \hline 104.978&0.009436&105.978&0.015217&0.5530&4.1901e+4&3.6600e-21&2.8932e+2&9.6952e-1&1.7726e-6&3.0480e-2\\ \hline 58.165&0.016902&59.165&0.037051&0.9621&1.7360e+4&6.2827e-22&1.6152e+2&9.8275e-1&1.0327e-5&1.7248e-2\\ \hline 32.030&0.030275&33.030&0.089654&1.6549&7.2141e+3&1.0849e-22&9.0173e+1&9.9024e-1&5.9802e-5&9.7027e-3\\ \hline 17.440&0.054230&18.440&0.216104&2.7998&3.0032e+3&1.8801e-23&5.0341e+1&9.9422e-1&3.4507e-4&5.4386e-3\\ \hline 9.295&0.097139&10.295&0.519591&4.6201&1.2522e+3&3.2689e-24&2.8104e+1&9.9498e-1&1.9847e-3&3.0385e-3\\ \hline 4.747&0.173998&5.747&1.245974&7.3303&5.2445e+2&5.7338e-25&1.5690e+1&9.8701e-1&1.1315e-2&1.6828e-3\\ \hline 2.209&0.311671&3.209&2.963870&10.8922&2.2448e+2&1.0505e-25&8.7592e+0&9.3735e-1&6.1759e-2&8.9217e-4\\ \hline 0.791&0.558275&1.791&6.796199&14.4931&1.0647e+2&2.3631e-26&4.8901e+0&7.2506e-1&2.7455e-1&3.8527e-4\\ \hline 0.000&1.000000&1.000&13.795986&16.6666&6.7400e+1&9.4701e-27&2.7300e+0&3.1480e-1&6.8509e-1&9.3386e-5\\ \hline -0.442&1.791233&0.558&23.084440&17.3355&5.7975e+1&7.0067e-27&1.5241e+0&7.4032e-2&9.2595e-1&1.2261e-5\\ \hline -0.662&2.961475&0.338&31.723059&17.4497&5.6279e+1&6.6028e-27&9.2184e-1&1.7384e-2&9.8260e-1&1.7413e-6\\ \hline -0.796&4.896255&0.204&40.497135&17.4966&5.5897e+1&6.5133e-27&5.5757e-1&3.8994e-3&9.9610e-1&2.3625e-7\\ \hline -0.876&8.095059&0.124&49.302428&17.5231&5.5812e+1&6.4936e-27&3.3724e-1&8.6547e-4&9.9913e-1&3.1715e-8\\ \hline -0.925&13.383695&0.075&58.114641&17.5290&5.5793e+1&6.4893e-27&2.0398e-1&1.9163e-4&9.9979e-1&4.2475e-9\\ \hline -0.955&22.127483&0.045&66.928214&17.5304&5.5789e+1&6.4882e-27&1.2338e-1&4.2410e-5&9.9995e-1&5.6857e-10\\ \hline -0.973&36.583733&0.027&75.742301&17.5306&5.5788e+1&6.4880e-27&7.4623e-2&9.3846e-6&9.9998e-1&7.6097e-11\\ \hline -0.983&60.484488&0.017&84.556288&17.5308&5.5788e+1&6.4879e-27&4.5136e-2&2.0766e-6&1.0000e+0&1.0185e-11\\ \hline -0.990&100.000000&0.010&93.370467&17.5307&5.5788e+1&6.4880e-27&2.7300e-2&4.5950e-7&9.9999e-1&1.3631e-12\\ \hline \end{array}}$ according to the abstract Planck 2018, $\Omega_b = 0.0492$ and I find $M_b = 1.457*10^{53}kg$ and $\Omega_b / \Omega_m = 0.1563$ for the present time. For z=0.791 I have a problem determining $\Omega_b$ or it's: $M_b/M_{total} = 1.457*10^{53}kg/7.559*10^{53}kg=0.1927=\Omega_b$ but then $\Omega_b / \Omega_m = 0.2658$. why $\Omega_b / \Omega_m$ are différents for each time
12. ## Need help with density parameters of the LambdaCDM model

Hello Mordred. I finally understood the use of the calculator (Planck 2015) and the equality of H with density parameters. But I have not been able to find satisfactory approximations to apply them to the data in the abstract Planck 2018. I have to wait for the update to go further. I will move on to the FLRW and GR equations in the meantime. PS: I found the button output in Latex and would know how to transform Tex into Latex edit : I have find a good approximation but I have a big problem
13. ## Need help with density parameters of the LambdaCDM model

oh good. it's very good link when I look with options. I use it now before the 2018 planck version +1 thank you
14. ## Need help with density parameters of the LambdaCDM model

For the moment, I'm troubled with the equations of state Thanks for these encouragements... but I think I'm doing a little better in cosmology than in FRW: _) If you have an answer for its current value (planck 2018) I take
15. ## Need help with density parameters of the LambdaCDM model

oh thank you. edit ok forgot this question : $\Omega_m$ is for baryonic matter or all the matter (dark+ordinary)? I have a new one : where can I find $\Omega_r$ ?