stephaneww Posted April 27, 2019 Share Posted April 27, 2019 (edited) 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*10^{122}, 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/m^{3}) 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 Edited April 27, 2019 by stephaneww Link to comment Share on other sites More sharing options...

stephaneww Posted August 18, 2019 Author Share Posted August 18, 2019 (edited) 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 Edited August 18, 2019 by stephaneww Link to comment Share on other sites More sharing options...

stephaneww Posted August 18, 2019 Author Share Posted August 18, 2019 (edited) Quote ... The sequence is identical to the calculation "A" of the first message: ... [latex]\sqrt{3.079*10^{-19}}=5.549*10^{-10}. 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*10^{-10}.J/m^3[/latex] ... 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] Edited August 18, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 20, 2019 Share Posted August 20, 2019 The numbers may work in this case but consider this. The Hubble constant is only constant at a point in time. In the past it is much higher. Where as the zero point energy calculations don't vary. So although you see similarities for Hubble parameter value today you may not get the same relations at an earlier time period. I would recommend you try a few different time periods and see... 1 Link to comment Share on other sites More sharing options...

stephaneww Posted August 20, 2019 Author Share Posted August 20, 2019 (edited) 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. 25 minutes ago, stephaneww said: 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) 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. Edited August 20, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 20, 2019 Share Posted August 20, 2019 Yes your showing you can calculate the critical mass/energy density. You can do the same with the critical density formula which leads to the Hubble parameter. However this formula isn't strictly the cosmological constant. Critical density has three contributors. Radiation matter and Lambda. Link to comment Share on other sites More sharing options...

stephaneww Posted August 21, 2019 Author Share Posted August 21, 2019 (edited) On 8/20/2019 at 2:35 PM, Mordred said: Yes your showing you can calculate the critical mass/energy density. Uh, is this for the first message ? On 8/20/2019 at 2:35 PM, Mordred said: You can do the same with the critical density formula which leads to the Hubble parameter. However this formula isn't strictly the cosmological constant. Critical density has three contributors. Radiation matter and Lambda. 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? Edited August 21, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 22, 2019 Share Posted August 22, 2019 (edited) First equation shouldn't have the c^2 in the numerator https://en.wikipedia.org/wiki/Friedmann_equations#Density_parameter equation two doesn't look correct either as the Lambda energy density is just a fraction of the critical density. value should be of the order [latex]7*10^{-30} g/m^3[/latex] edit:Doh wait you wanted the energy critical density which explains the C^2 as opposed to the mass critical density, you have it correct on both just forgot that conversion Edited August 22, 2019 by Mordred Link to comment Share on other sites More sharing options...

stephaneww Posted August 22, 2019 Author Share Posted August 22, 2019 (edited) 1 hour ago, Mordred said: First equation shouldn't have the c^2 in the numerator edit:Doh wait you wanted the energy critical density which explains the C^2 as opposed to the mass critical density, you have it correct on both just forgot that conversion edit me too : Yes, I should have said "energy critical density" ([math]J/m^3)[/math] 1 hour ago, Mordred said: equation two doesn't look correct either as the Lambda energy density is just a fraction of the critical density. value should be of the order 7∗10−30g/m3 [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 ? Edited August 22, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 22, 2019 Share Posted August 22, 2019 (edited) You need to be careful here. Consider two fundamental questions. The critical density itself changes over cosmological time however the cosmological constant stays constant... The critical density is calculated without lambda and without any curvature term. Our universe is close to flat however not precisely flat... Now apply that to your second equation that you derived in your last post. If [latex]\rho=\rho_{crit}[/latex] Then k=0. Our curvature term k isn't precisely 0 though close. Remember the critical density is a calculated density not the actual density. Consider this at z=1100 roughly CMB surface of last scattering H is roughly 22000 times the H_0 value today. Yet the cosmological constant is the same in both then and now... Edited August 22, 2019 by Mordred Link to comment Share on other sites More sharing options...

stephaneww Posted August 23, 2019 Author Share Posted August 23, 2019 (edited) 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? Edited August 23, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 23, 2019 Share Posted August 23, 2019 Not as far as the universe today but I do question it's validity at other cosmological times. The problem being how the critical density evolves while Lambda doesn't. Link to comment Share on other sites More sharing options...

stephaneww Posted August 23, 2019 Author Share Posted August 23, 2019 (edited) 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 : 40 minutes ago, Mordred said: The problem being how the critical density evolves while Lambda doesn't. [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] Edited August 23, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 23, 2019 Share Posted August 23, 2019 (edited) Ok lets do this I had to review a few things, particularly with the curvature constant. [math]\frac{\dot{a}}{a}=H^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{R^2}\frac{1}{a^2}+\frac{\Lambda c^2}{3}[/math] kc^2/R^2 is the curvature term in a Lambda free universe if a value of H^2 is given there is a link between the curvature term and [math]\rho[/math] and this leads to the critical density [math]\rho_c=\frac{3H^2}{8\pi G}[/math] if the universe is at critical density it is flat. [math]\Omega=\frac{\rho}{\rho_c}[/math] so we can rewrite the FL as [math]1-\Omega=\frac{-kc^2}{R^2a^2H^2}[/math] if [math]\rho=\rho_c[/math] we have flat if [math]\rho>\rho_c[/math] universe is closed if [math]\rho<\rho_c [/math] universe is open. now we have to incorporate the fluid equations for adiabatic expansion. from the first law of thermodynamics [math]DE+PdV=0[/math] [math]\dot{v}/V=3\dot{a}{a}[/math] and [math] DE/dt=\rho c^2dV/dt+c^2Vdp/dt[/math] this gives us [math]\dot{\rho}+3\frac{\dot{a}}{a}\rho+P/c^2[/math] so [math]\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}(\rho+3\frac{P}{c^2})+\frac{\Lambda C^2}{3}[/math] given [math] P=wc^2\rho[/math] combining this equation with the following [math]\dot{\rho}+3\frac{\dot{a}}{a}\rho+P/c^2[/math] gives us [math]\rho=\rho_0a^{-3(1+w)}[/math] from this we can determine the matter density [math]\rho_m=\rho_0a^{-3}[/math] and radiation [math]\rho_r=\rho_{r,0}a^{-4}[/math] so now you can compare densities to obtain [math]\Omega[/math] note that [math]\rho_\Lambda=\frac{\Lambda c^2}{8\pi G}[/math] giving [math]\Omega_\Lambda=\frac{\Lambda c^2}{3H^2}[/math] taking the ratios of [math]\Omega[/math] gives us the energy density that dominates example radiation to matter is [math]\Omega_r/\Omega_m=\rho_r\rho_m\frac{1}{a}=\frac{1}{3600 }a[/math] if a_0=1 then there was a time the universe was radiation dominant. for a Lambda dominant [math]\dot{a}^2=\frac{\Lambda c^2}{3}a^2[/math] so for the evolution of the three components we get [math]\frac{H^2}{H^2_0}=\frac{\Omega_{r,0}}{a^4}+\frac{\Omega_{m,0}}{a^3}+\Omega_{\Lambda,0}+\frac{1-\Omega_0}{a^2}[/math] this is how its done to preserve the curvature terms and account for the radiation and matter dominant eras hence my hesitancy on your second equation Edited August 23, 2019 by Mordred Link to comment Share on other sites More sharing options...

stephaneww Posted August 23, 2019 Author Share Posted August 23, 2019 (edited) 38 minutes ago, Mordred said: this is how its done to preserve the curvature terms and account for the radiation and matter dominant eras Honestly, I have no idea..... and I understand your hesitancy on my second equation _____________________________________________________ Aparty: It's just an idea, but... We take as a base a space-time where 3D space and time are linked. Time scrolls in a positive direction, it "increases". So the 3D space also increases. By bringing space-time back to the image of the surface of a sphere, the surface of the sphere will increase with time. Assuming that this surface is the vacuum, the quantity of vacuum will increase over time while by the law of energy conservation the quantity of matter will remain stable. In the end, we will have an increasing vacuum energy and a constant mass energy. In terms of energy density, that of vacuum will prevail over that of matter. It may be a description of this type that is equated by the model [math]\Lambda CDM[/math] Edited August 23, 2019 by stephaneww 1 Link to comment Share on other sites More sharing options...

Mordred Posted August 23, 2019 Share Posted August 23, 2019 (edited) unfortunately that doesn't work with the mass density. Nor does it work with the radiation density. As the universe expands these both decrease at the ratios I have provided above. Were not dealing with the total energy content of either they are assumed constant by the adiabatic expansion (adiabatic means no net inflow or outflow of energy) the particle density is typically set constant at roughly 10^(90) particles. unfortunately as the universe expands Lambda increases in total energy Here is something I want you to consider, when it comes to vacuum densities we have numerous overlapping densities. You have the zero point energy, the VeV of the Higgs field the cosmological constant etc. In a particle/antiparticle complex scalar field of creation and annihation of a complex scalar field the energy density can reach infinity. However consider this we cannot know the absolute energy of the vacuum, we can only measure its potential differences. Think of an electric wire you can only measure the voltage across some potential difference such as a resistor... this brings back to mind an article By Unruh on the cosmological problem in that he argues that the fine tuning problem can be resolved via parametric down conversion and an inhomogeneous vacuum. https://arxiv.org/pdf/1703.00543.pdf you may find this interesting. PS your one of the few posters who has a thread in Speculation forum that sticks to the proper methodology in modelling and following the Speculation forum guidelines in making testable predictions via mathematics so I gave you a +1 for that Edited August 23, 2019 by Mordred Link to comment Share on other sites More sharing options...

stephaneww Posted August 23, 2019 Author Share Posted August 23, 2019 (edited) 3 hours ago, Mordred said: unfortunately that doesn't work with the mass density. Nor does it work with the radiation density. As the universe expands these both decrease at the ratios I have provided above. Were not dealing with the total energy content of either they are assumed constant by the adiabatic expansion (adiabatic means no net inflow or outflow of energy) the particle density is typically set constant at roughly 10^(90) particles. It is certain that if my description makes sense, I must confront it with the standard cosmological model, and for the moment, it does not work. 3 hours ago, Mordred said: unfortunately as the universe expands Lambda increases in total energy However, this point is in my brief description: 3 hours ago, stephaneww said: ...By bringing space-time back to the image of the surface of a sphere, the surface of the sphere will increase with time. Assuming that this surface is the vacuum, the quantity of vacuum will increase over time while by the law of energy conservation the quantity of matter will remain stable. In the end, we will have an increasing vacuum energy and a constant mass energy. In terms of energy density, that of vacuum will prevail over that of matter. edit : no, you're right, my description is incomplete, it doesn't describe Lambda enough 3 hours ago, Mordred said: Here is something I want you to consider, when it comes to vacuum densities we have numerous overlapping densities. You have the zero point energy, the VeV of the Higgs field the cosmological constant etc. In a particle/antiparticle complex scalar field of creation and annihation of a complex scalar field the energy density can reach infinity. I confined myself to 3 descriptions of vacuum energy density, 2 known: - the quantum vacuum in Planck units (mp.c^{2}/lp^{3}) which is very high to the point that when it is converted into density its value is incredible. - the vaccum of the relativist Lambda. 1 new vacuum mixing quantum and relativity with hbar and Lambda which is incredibly low. It seemed consistent to me to not mix them with other approaches of vacuum. 3 hours ago, Mordred said: However consider this we cannot know the absolute energy of the vacuum, we can only measure its potential differences. Think of an electric wire you can only measure the voltage across some potential difference such as a resistor... I may have something to answer on this point later, but I agree for the moment. 3 hours ago, Mordred said: this brings back to mind an article By Unruh on the cosmological problem in that he argues that the fine tuning problem can be resolved via parametric down conversion and an inhomogeneous vacuum. https://arxiv.org/pdf/1703.00543.pdf you may find this interesting. PS your one of the few posters who has a thread in Speculation forum that sticks to the proper methodology in modelling and following the Speculation forum guidelines in making testable predictions via mathematics so I gave you a +1 for that Of course, this document is too complex for me I noted however two unusual square on variables where I had never seen them before: [math]\Omega^2[/math] (197) page 25 and, more important, [math]\Delta \rho^2[/math] (A9) page 30 just as I had never seen [math](J/m^3)^2[/math] in the scientific literature. and thanks for the +1, it's very sympathetic . Edited August 23, 2019 by stephaneww Link to comment Share on other sites More sharing options...

stephaneww Posted August 23, 2019 Author Share Posted August 23, 2019 (edited) 7 hours ago, stephaneww said: Of course, this document is too complex for me I noted however two unusual square on variables where I had never seen them before: Ω2 (197) page 25 and, more important, Δρ2 (A9) page 30 just as I had never seen (J/m3)2 in the scientific literature. No, actually it's [math]\Omega^2[/math] that's closer to my [math](J/m^3)^2[/math] which corresponds to a [math]\rho^2[/math] : [math]\Omega^2=(\rho/\rho_c)^2=\rho^2/\rho_c^2[/math] if this is right in the context while [math]\Delta \rho^2=(\rho_1 - \rho_2)^2[/math] Edited August 23, 2019 by stephaneww Link to comment Share on other sites More sharing options...

stephaneww Posted August 23, 2019 Author Share Posted August 23, 2019 (edited) to make sure I'm not mistaken with the reverse : is [math]H/H_0=da/dt[/math] with [math]a=1[/math] now ? (I'm not sure of myself) Edited August 23, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 23, 2019 Share Posted August 23, 2019 (edited) a would be the differential between your H/H_0 recall that a as the scale factor is just a dimensionless ratio parameter. You set scale factor at cosmological time now at 1. a would be the differential between your H/H_0 recall that a as the scale factor is just a dimensionless ratio parameter. You set scale factor at cosmological time now at 1. Here this will help [latex]{\small\begin{array}{|c|c|c|c|c|c|}\hline T_{Ho} (Gy) & T_{H\infty} (Gy) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}[/latex] [latex]{\small\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (Gy)&R (Gly)&D_{now} (Gly)&D_{then}(Gly)&D_{hor}(Gly)&V_{gen}/c&H/Ho \\ \hline 0.001&1090.000&0.000373&0.000628&45.331596&0.041589&0.056714&21.023&22915.263\\ \hline 0.001&863.334&0.000551&0.000915&45.159913&0.052309&0.071406&18.232&15740.128\\ \hline 0.001&683.804&0.000810&0.001326&44.962398&0.065753&0.089864&15.881&10859.192\\ \hline 0.002&541.606&0.001183&0.001915&44.736079&0.082599&0.113040&13.885&7520.218\\ \hline 0.002&428.979&0.001722&0.002756&44.477683&0.103683&0.142116&12.180&5224.758\\ \hline 0.003&339.773&0.002496&0.003956&44.183524&0.130038&0.178562&10.712&3639.803\\ \hline 0.004&269.117&0.003606&0.005666&43.849475&0.162938&0.224202&9.443&2541.361\\ \hline 0.005&213.154&0.005194&0.008100&43.470902&0.203941&0.281289&8.340&1777.702\\ \hline 0.006&168.829&0.007463&0.011563&43.042568&0.254948&0.352603&7.377&1245.393\\ \hline 0.007&133.721&0.010698&0.016484&42.558633&0.318265&0.441559&6.533&873.554\\ \hline 0.009&105.913&0.015309&0.023478&42.012463&0.396668&0.552333&5.791&613.344\\ \hline 0.012&83.889&0.021873&0.033412&41.396601&0.493471&0.690005&5.138&430.988\\ \hline 0.015&66.444&0.031211&0.047518&40.702622&0.612585&0.860719&4.561&303.042\\ \hline 0.019&52.627&0.044487&0.067545&39.921133&0.758568&1.071848&4.051&213.190\\ \hline 0.024&41.683&0.063355&0.095974&39.041469&0.936624&1.332155&3.600&150.041\\ \hline 0.030&33.015&0.090158&0.136321&38.051665&1.152552&1.651928&3.200&105.633\\ \hline 0.038&26.150&0.128224&0.193578&36.938267&1.412573&2.043059&2.845&74.389\\ \hline 0.048&20.712&0.182274&0.274818&35.686105&1.722983&2.519001&2.530&52.398\\ \hline 0.061&16.405&0.258995&0.390062&34.278330&2.089532&3.094542&2.250&36.917\\ \hline 0.077&12.993&0.367873&0.553490&32.695921&2.516347&3.785220&2.002&26.017\\ \hline 0.097&10.291&0.522342&0.785104&30.917756&3.004225&4.606237&1.782&18.342\\ \hline 0.123&8.151&0.741396&1.112970&28.920472&3.547949&5.570564&1.587&12.938\\ \hline 0.155&6.456&1.051751&1.575989&26.679131&4.132295&6.685941&1.415&9.137\\ \hline 0.196&5.114&1.490772&2.226851&24.167785&4.726112&7.950210&1.265&6.467\\ \hline 0.247&4.050&2.109877&3.133394&21.362526&5.274330&9.344906&1.135&4.596\\ \hline 0.312&3.208&2.977691&4.373615&18.247534&5.688090&10.827382&1.026&3.292\\ \hline 0.394&2.541&4.180384&6.013592&14.827243&5.835394&12.323993&0.942&2.395\\ \hline 0.497&2.013&5.813076&8.053192&11.147771&5.539179&13.731340&0.888&1.788\\ \hline 0.627&1.594&7.955449&10.346218&7.320583&4.592515&14.935503&0.873&1.392\\ \hline 0.792&1.263&10.632280&12.576261&3.528946&2.795101&15.853609&0.907&1.145\\ \hline 1.000&1.000&13.787206&14.399932&0.000000&0.000000&16.472274&1.000&1.000\\ \hline 1.263&0.792&17.302122&15.660852&3.144839&3.970507&16.845080&1.161&0.919\\ \hline 1.571&0.637&20.811932&16.388509&5.639833&8.860361&17.041221&1.380&0.879\\ \hline 1.955&0.512&24.444731&16.808523&7.715938&15.083827&17.146441&1.675&0.857\\ \hline 2.433&0.411&28.147021&17.039505&9.416655&22.906412&17.198866&2.056&0.845\\ \hline 3.027&0.330&31.886775&17.163312&10.797410&32.682676&17.221760&2.540&0.839\\ \hline 3.766&0.265&35.646534&17.228668&11.913050&44.870264&17.228668&3.148&0.836\\ \hline 4.687&0.213&39.416981&17.262737&12.812202&60.047793&17.262737&3.910&0.834\\ \hline 5.832&0.171&43.192748&17.280614&13.535828&78.939747&17.280614&4.860&0.833\\ \hline 7.257&0.138&46.971377&17.289940&14.117812&102.450932&17.289940&6.044&0.833\\ \hline 9.030&0.111&50.751666&17.294639&14.585725&131.708610&17.294639&7.519&0.833\\ \hline 11.236&0.089&54.532554&17.297194&14.961819&168.115662&17.297194&9.354&0.833\\ \hline 13.982&0.072&58.313845&17.298546&15.264097&213.418728&17.298546&11.639&0.832\\ \hline 17.398&0.057&62.095516&17.299100&15.507045&269.791380&17.299100&14.482&0.832\\ \hline 21.649&0.046&65.877123&17.299503&15.702284&339.937814&17.299503&18.020&0.832\\ \hline 26.939&0.037&69.658785&17.299738&15.859189&427.223615&17.299738&22.423&0.832\\ \hline 33.521&0.030&73.440477&17.299885&15.985286&535.836513&17.299885&27.902&0.832\\ \hline 41.711&0.024&77.222357&17.299813&16.086628&670.987643&17.299813&34.719&0.832\\ \hline 51.902&0.019&81.004072&17.299891&16.168066&839.160848&17.299891&43.202&0.832\\ \hline 64.584&0.015&84.785791&17.299957&16.233514&1048.424819&17.299957&53.758&0.832\\ \hline 80.364&0.012&88.567685&17.299844&16.286113&1308.819673&17.299844&66.893&0.832\\ \hline 100.000&0.010&92.349407&17.299900&16.328381&1632.838131&17.299900&83.237&0.832\\ \hline \end{array}}[/latex] Row 1.00 is today after row 1.00 is future. Previous is past. Edited August 23, 2019 by Mordred Link to comment Share on other sites More sharing options...

stephaneww Posted August 23, 2019 Author Share Posted August 23, 2019 ok thank you Link to comment Share on other sites More sharing options...

stephaneww Posted August 24, 2019 Author Share Posted August 24, 2019 (edited) Wich are the values of [math]\Omega_{r,0}[/math] and of [math]1- \Omega_{0}[/math] for this table ? and what is the number of step you used please ? that's for the number of step ok : 50 is the number of step edit 2 : what does the value Dparticle correspond to? Edited August 24, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 24, 2019 Share Posted August 24, 2019 (edited) The radiation density was incorporated into the Seq value if I recall correctly Cuthbert was the main programmer same with the [latex]1-\Omega_0 [/latex] D particle is the particle horizon. Here is the formula derivatives they used to produce the calculations. http://cosmocalc.wikidot.com/advanced-user We used the methodologies presented by Line weaver and Davies hence the usage of Stretch which is the inverse of the scale factor. So the formulas were modified accordingly. Incorporating radiation density and matter density into Seq was essential to produce the correct calculations which we compared to Planck and WMAP datasets following the graphs of Lineweaver and Davies paper. Each program adjustment we has gotten several PH Ds to help confirm via several forums. Edited August 24, 2019 by Mordred Link to comment Share on other sites More sharing options...

stephaneww Posted August 24, 2019 Author Share Posted August 24, 2019 (edited) ok thank you is there a mistake? [math]\Omega_{\Lambda ,0}=\frac{\Lambda c^2}{3H_0^2}[/math] [math]\Omega_{\Lambda ,H}=\frac{\Lambda c^2}{3H^2}[/math] [math]H/H_0=\Omega_{\Lambda ,0}/\Omega_{\Lambda ,H}[/math], right? so : On 8/23/2019 at 3:41 AM, stephaneww said: the adjustment is made by the variation of ΩΛ , right ? Edited August 24, 2019 by stephaneww Link to comment Share on other sites More sharing options...

Mordred Posted August 24, 2019 Share Posted August 24, 2019 (edited) As you have a mixed state of contributors though the principle contributor is Lambda your better off using the full formula. Formula one that you have would only apply to a Lambda only contributor. If you were to compare to say the CMB time at z=1100 you would get inaccurate answers as the CMB is part of the matter dominant Era shortly after the radiation Era. Lambda dominant Era started roughly at universe age 7 Gyr depending on dataset used Here is the H/H_0 as a function of redshift. [latex]H=H_0\sqrt{\Omega_{r,0}(1+z)^4+\Omega_{m,o}(1+z)^3+\Omega_{k,0}(1+z)^2+\Omega_{\Lambda_0}}[/latex] [latex]\Omega_{k,o}=1-\Omega_{r,o}-\Omega_{m,o}-\Omega_{\Lambda,o}[/latex] which describes the curvature term. Though for a critical dense universe simply set to curvature density to zero. Ie ignore it in the above equation. Now you can apply any dataset. (Lol you can even toy model expansion rates of different component universes such as a matter only universe ) simply set unneeded terms to zero Notice the [latex]\Omega_\Lambda[/latex] doesn't change...the total energy of Lambda however does. Edited August 24, 2019 by Mordred Link to comment Share on other sites More sharing options...

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