stephaneww

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

I should have thought of that. Thank you

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

thank you Mordred

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 ?

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

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

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

Hello I have been asking myself this question since I saw the formula (II-30) on page 11 of this (unpublished) document: Indeed, [latex]d \Omega[/latex] is presented as a four-dimensional volume variation in general relativity for EINSTEIN equations in vacuum. [latex]d\Omega=\sqrt{|g|}dx^0dx^1dx^2dx^3[/latex] where [latex]x^0=ct[/latex] 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 [latex]g_0[/latex] is no longer adimensioned but takes the dimension [latex]m^{-2}[/latex], dimension of the cosmological constant. We then find a new constant [latex]X= 7.719*10^{-53}m^{-2} [/latex] that can be related to the cosmological constant [latex]\Lambda[/latex] today: volume universe [latex]V_0 = 3.48*10^{80} m^3[/latex] Hubble's time: [latex]t_0 = 4.56*10^{17} s[/latex] or [latex]H_1 = 67.66 km/s/Mpc[/latex] [latex]c* t_0 = 1.37*10^{26} m[/latex] observable universe radius : [latex]x^1=x^2=x^3=R_0 = 4.36*10^{26} m[/latex] density parameter of the cosmological constant: [latex]\Omega_{0 _ \Lambda} = 0.6889[/latex] equality (II-30) page 11 is equivalent to : [latex]1/\sqrt{g_0}=c* t_0*R_0^3/V_0=3.26*10^{25}m[/latex] hence [latex]g_0=9.386*10^{-52}m^{-2}[/latex] [latex]g_0'=g_0/8/\pi*3=1.120*10^{-52}m^{-2}[/latex] when Hubble time = [latex]t_1 = 2.91*10^{17} s[/latex] or [latex]H_1 = 70 km/s/Mpc[/latex] volume universe [latex]V_1 = 1.13*10^{80} m^3[/latex] [latex]c* t_1 = 8.25*10^{25} m[/latex] observable universe radius : [latex]x^1=x^2=x^3=R_1 = 3.00*10^{26} m[/latex] density parameter of the cosmological constant: [latex]\Omega_{1_\Lambda} = 0.2803[/latex] equality (II-30) page 11 is equivalent to : [latex]1/\sqrt{g_1}=c* t_1*R_1^3/V_1=3.08*10^{25}m[/latex] hence [latex]g_1=2.308*10^{-51}m^{-2}[/latex] [latex]g_1'=g_1/8/\pi*3=2.754*10^{-52}m^{-2}[/latex] since [latex]d \Omega[/latex] 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 [latex]X[/latex] : [latex]g_0'*\Omega_{0_\Lambda}=g_1'*\Omega_{1_\Lambda}=...=X[/latex] [latex]1.120*10^{-52}*0.6889=2.754*10^{-52}*0.2803=[/latex] [latex]X= 7.719*10^{-53}m^{-2} [/latex] A link with the cosmological constant [latex]\Lambda=1.105*10^{-52}m^{-2}[/latex] can be made as follows: [latex]\Lambda / X * 8 * \pi=[/latex] [latex]1.105*10^{-52}/7.719*10^{-53}* 8 * \pi=[/latex] [latex]36[/latex] (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.

Thank you for the changes Strange

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"

I have copy and translate this from a french forum : ________________________ The Hubble parameter defined from the scale factor of the FLRW metric is [latex]H(t) = \frac{\dot{a}(t)}{a(t)}[/latex] 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 [latex]\theta(t) = \frac{\dot{V}(t)}{V(t)}[/latex] We then simply obtain [latex]\theta(t) = 3 H(t)[/latex] Indeed, if we reason on a ball (a volume [latex]V[/latex] whose surface is a sphere), with a radius [latex]R[/latex] in co-moving coordinates, its physical volume is then [latex]V(t) = \frac{4\pi}{3} a(t)^3 R^3[/latex] and, by drifting in relation to time, [latex]\dot{V}(t) = \frac{4\pi}{3}[3 a(t)^2\dot{a}(t)] R^3[/latex] Hence the fact that [latex]\theta(t) = \frac{\dot{V}(t)}{V(t)} = 3 \frac{\dot{\dot{a}(t)}}{a(t)} = 3 H(t)[/latex] _____________________ When [latex]\Lambda[/latex] cosmological constant is expressed in [latex]s^{-2}[/latex] we have [latex]\Lambda/(3H(t)^2)=\Omega_{\Lambda(t)}[/latex], 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 [latex]H(t)[/latex] (expansion = constant Hubble effect) and that as a result, there remains [latex]3H(t)*\Omega_{\Lambda(t)}[/latex] 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. [latex]\Lambda/H(t)=3*H(t)*\Omega_{\Lambda(t)}[/latex], [latex]3*H(t)*\Omega_{\Lambda(t)}[/latex], 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

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

Thank you for everything, Mordred. If we assume, for example, that [latex]M_b * \Lambda= cst[/latex] then, when [latex]M_b[/latex] decreases [latex] \Lambda[/latex] 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?

merge then edit too late : Can we say that [latex]M_\text{b(t-1)} = M_\text{b(t)}[/latex] from close to close ?

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

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 [latex]M_b[/latex] don't change edit : of course

In abstract [latex]\Omega_b h^2=0.0224[/latex] so : [latex]\Omega_b =0.0224/h^2=0.0224/(67.4/100)^2[/latex]

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 : [latex]{\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}}[/latex] according to the abstract Planck 2018, [latex]\Omega_b = 0.0492[/latex] and I find [latex]M_b = 1.457*10^{53}kg[/latex] and [latex]\Omega_b / \Omega_m = 0.1563[/latex] for the present time. For z=0.791 I have a problem determining [latex]\Omega_b[/latex] or it's: [latex]M_b/M_{total} = 1.457*10^{53}kg/7.559*10^{53}kg=0.1927=\Omega_b[/latex] but then [latex]\Omega_b / \Omega_m = 0.2658[/latex]. why [latex]\Omega_b / \Omega_m[/latex] are différents for each time

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

oh good. it's very good link when I look with options. I use it now before the 2018 planck version +1 thank you

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

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

I also need need the ratio H/H0 to find OmegaLambda and OmegaM

It's okay with: T H0 : 14,497 T H infinite : 17.51 Omega tot : 0.9989 there's just one tiny little mistake on OmegaM