stephaneww

I'm stranded: how to get the values with Planck 2018 please? for : http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html edit I understood : Hubble time is different from age of unoverse

Perfect. thank you again. After reading the table of contents and the passage concerning the cosmological parameters, I did not find it. Or else I don't understand. I have a formula that will go into the speculation section with [latex]\Omega_b[/latex] and [latex]\Omega_\Lambda[/latex] that could be compatible with this: https://www.scienceforums.net/topic/117894-dark-energy-increasing-with-time/ I'll calculate to see if the numerical values confirm Thank you again

more here and here : https://arxiv.org/abs/1811.02590

why not after all : it may be interesting. Thanks a lot. can we simplify on "short" time scales by saying Mb constant ?

Hello, I would like to have values (or how to calculate the values) of [latex]\Omega_b[/latex] (ordinary matter = baryonic) and [latex]\Omega_{\Lambda}[/latex] (density parameter of the cosmological constant) in relation to, or as a function of, the Hubble constant H at different ages of the universe. To verify this relationship: [latex]M_b* \Lambda / 2 \approx 8 Kg/m^{2}[/latex] ( -0.5% compared to 8 = contribution of neutrinos ? ) [latex]M[/latex] for "mass" thank you in advance[

Mr. Heisenberg is at the vet's for his cat. The latter said to him: Sir, I have a good news and a bad news for you. I start with which? I go out

Hello, Another way to present the problem : The final formula that we want is : [latex]M_b=\frac{2c^2}{G\sqrt{\pi\Lambda}}[/latex] It's base on the classical notion of the force and the density of cosmological constant : [latex]F=G\frac{m_1*m_2}{d^2}[/latex] for the force notion [latex]\rho_{\Lambda}c^2=\frac{c^4\Lambda}{8 \pi G}=\frac{F_p\Lambda}{8\pi}[/latex] for the density of cosmological constant where the Planck force is : [latex]F_p=\frac{c^4}{G}[/latex]. if [latex]m_1=m_2=m_p[/latex] and [latex]d=l_p[/latex] where [latex]_p[/latex] is for Planck value we have exactly with the notion of classical force : [latex]F_p=G\frac{m_p*m_p}{l_p^2}=1.210295*10^{44} N[/latex] as the Planck force appears in the density of the cosmological constant I think we can assume that, in the relativity, it's assimilable to a Force of the cosmological constant witch must be egal to [latex]1.210295*10^{44}N[/latex] now, we can try to duplicate the Planck's values for the force of the cosmological constant with the notion of classical force : the more easy is to use [latex]\Lambda[/latex] instead of [latex]1/ l_p^2[/latex] (it's constant and have the same dimension) now we need a mass of matter : in the Planck force we had [latex]m_1=m_2[/latex] so, in cosmology we need two constant mass and egal (it must be mass of dark matter/2 or (dark matter + baryonic mattter)/2 or baryonic matter/2) after computational tests the only approximate correct value is [latex]M_b/2[/latex] (we need to divide by two otherwise we have 2 mass of universe. they must be egals to duplicate the formula of the Planck force) and add the factor [latex]\pi[/latex] is needed to be ok to within 0.6% with the Planck force value: with datas of abstract data Planck 2018 [latex]F_p=F_{\Lambda}=G*(M_b/2)*(M_b/2)*\Lambda*\pi=[/latex][latex]6.67408*10^{-11}*(1.459*10^{53}/2)^2*1.091*10^{-52}*\pi=1.217*10^{44}N[/latex] if what is above is correct we can write : [latex]G*(M_b/2)^2*\Lambda*\pi=\frac{c^4}{G}=F_p[/latex] [latex](M_b/2)^2=\frac{c^4}{G^2*\Lambda*\pi}[/latex] [latex]M_b/2=\frac{c^2}{G*\sqrt{\pi*\Lambda }}[/latex] and finaly : [latex]M_b=\frac{2*c^2}{G*\sqrt{\pi*\Lambda}}[/latex]

oki thanks

did you made a mistake ? 1 joule = 1 kg*m2/s-2

The numerics values that I give in the last post aren't exactly from the data Planck mission 2018. With the numerics value from data Planck mission 2018, the difference is always 0.5%

Another remarkable relationship, about baryonic matter and factor 2, with correct dimensions this time with data mission Planck 2018 (table 2, page 15, last column): 1.in ΛCDM model : radius universe observable : [latex]R=4,358*10^{26} m[/latex] Hubble constant [latex]H_0=67.4km/s/Mpc=2.184*10^{-18}s^{-1}[/latex] baryon density : [latex]\Omega_b=0.04936[/latex] ___________________________________ [latex]A=R^2*H_0^2*\Omega_b=(4,358*10^{26})^2*(2.184*10^{-18})^2*0.04936=4,4727*10^{16} m^2 s^{-2}[/latex] = 2. in quantum mechanics : [latex]B=l_p^2/t_p^2=c^2=8.988*10^{16}m^2 s^{-2}[/latex] ___________________________________ [latex]B/A*2=1.005[/latex] the difference is 0.5% There could be something important with factor 2 on the mass of baryonic matter note: I began by calculating the surface of a sphere of radius R and the surface of a Planck sphere; this shows the factor 8pi in relativity

To start let's look for the value of [latex]M_b[/latex] with wikipedia data : [latex]\Large{M_b=\frac{2*299792458^2}{6.67408*10^{-11}* \sqrt{\pi*1.1056*10^{-52}}}}=1.445*10^{53}kg \text{ versus }1.46*10^{53}kg = \text{ data wikipedia}[/latex] Then break down the formula 1 kg / m ^ 2 to find in step its physical sense : [latex] a [/latex] is an acceleration. [latex] a=(M_b/2)*G*\Lambda=(1.445*10^{53}/2)*6.67408*10^{-11}*1.1056*10^{-52}=5.3317*10^{-10}m/s^{-2}[/latex] for a unit of mass M1, the force is F=M1*a = 5.3317*10-10 N (1N= 1kg*1m/1s2) per m2 we have a pression p=F/m2=5.3317*10-10 N/m2 (1N/m2= 1kg*1m-1*1s-2=1j/m3) The density of the vacuum energy of the cosmological constant = [latex]\Large{\rho_\Lambda*c^2=\frac{c^4 \Lambda}{8*\pi G}=\frac{299792458^4*1.1056*10^{-52}}{8*\pi*6.67408*10^{-11}}}=5.3241*10^{-10}J/m^3[/latex] The difference is 0.15% With data 2015 the difference is 0.13% I don't know how to justify theses steps, but the accuracy of the numerical value is remarkable I think it validates the definition of baryonic material mass from the relativity constants. What is your opinion, please

That's his definition, but his sense physic here?

Hi, In a post in date of November 30 I find that the mass of ordinary matter [latex]M_b[/latex], could be determinate with the constants of the relativity (abstract's data planck 2018 used and data Planck 2015 also) : [latex] \Large {M_b = \frac{2c^2}{G\sqrt{\pi \Lambda} }}[/latex] values here 1.1056*10-52 m-2 and here 1.46*1053kg I know it's very surprising, we'll prefer to have the mass of dark matter in addition... You might think it's a coincidence. However we can do this other calculation [latex](M_b/2)*G*\Lambda *1 kg/1m^2 = 5.34*10^{-10} \text{ Joules /m^3}[/latex] and it's the density of the cosmological constant. The thing I can't remember from school it's : what mean in physics " [latex]1kg/1m^2[/latex]" in simples words Another thing seems to appear : it's a explaination (limited) in classical mechanics of a description of the universe Thanks in advance for your answers

Hi, a friend had help me and deduce this formula from this work : [latex]m=\frac{2c^2}{G\sqrt{\pi\Lambda}} [/latex] m = ordinary mass matter of universe you can verify with the Planck data 2018 (values used = abstract values) you'll can notice that with Ned Wright's Javascript Cosmology Calculator and Planck data 2018 : "comoving radial distance, which goes into Hubble's law, is 14124.3 Mpc" = comoving radial distance, which goes into Hubble's law I used for Planck data 2015 (14124.0 Mpc) his formula give a 0.6% marge error maximum (data 2015 or 2018) he used [latex]F_p =c^4/G [/latex]

I realize there is a problem (given the controversy on the Hubble constant ?) I'm using 2016 data and 2016 Ned Wright's Javascript Cosmology Calculator results with H0 = 67.74 km / Mpc, OmegaM = 0.3089, Z infinite, OmegaVac = 0.6911 and a flat universe I had: "comoving radial distance, which goes into Hubble's law, is 14124.0 Mpc" against 14164 Mpc today for the radius of the observable universe hence my result of 1.44 * 10 ^ 53 kg for the ordinary mass (% of ordinary matter = 4.82%) of course if we take the wikipedia values of 2018 for the calculations, we do not find the same result

well well again an error : [latex]g_\Lambda*(M_\text{Ordinary matter}/2)*\pi=\text{"cosmological force"} = 1.209 *10^{44}N[/latex] and not [latex]1.029 * 10^{44} N[/latex] for [latex]F_p=1.2103 *10^{44}N[/latex]

oops a small error, sorry : [latex]g_\Lambda=G * ( M_\text{Ordinary matter}/2 ) * \Lambda = 5.34 * 10 ^{-10} N/Kg [/latex] and not [latex]( M_\text{Ordinary matter}/2 )^2[/latex] a possible interpretation of all this could be: half of the mass of ordinary matter of the universe attracts the other half and the cosmological constant (the acceleration of expansion) keeps the universe in equiibrium thus the cosmological constant would join the initial interpretation of Einstein: a new fundamental constant of gravitation that keeps the universe in equilibrium

It seems possible to split the previous calculation, to try to "proof" that it's the correct calculation : [latex] g_\Lambda = G * (M_\text{Ordinary Matter}/2)^2 * \Lambda = 5.34 * 10^{-10} N/Kg [/latex] [latex]g_\Lambda[/latex] it's the same numeric value that the [latex] \text{volume density of the vacuum of the cosmological constant} = 5.35 * 10^{-10} Joules/m^3 [/latex] the error on the numeric value is 0.2% [latex] g_\Lambda * (M_\text{Ordinary Matter}/2) * \pi = \text{the "cosmological force" }= 1.029 * 10^{44} N, \text{ Note : 1 N = 1 Kg m }s^{-2} [/latex] for [latex] F_p =1.2103 *10^{44}N [/latex] I am very interested in your opinions

humm... if I want to have the same interpretation that in the first post (I cut in two the mass of matter of the universe to have two masses) we can do : [latex]G*(M_\text{Ordinary Matter}/2)^2* \Lambda * \pi[/latex] in the last post (October 20) in this case we have : "cosmological Force"= 1.209 * 10^44 kg m s-2 with Mission Planck data Planck force =1.210 * 10^44 k m s-2 the error is only 0.07% between the two values until we know exactly what dark matter is, is it could be an acceptable defintion ? that you in advance for yours comments

so, I propose : [latex]\text{cosmological Force }= G * (M_\text{Ordinary matter} )^2 * \Lambda / 2 * \pi [/latex] [latex]=6.67*10^{-11}*(1.44*10^{53})^2*1.11*10^{-52}/2*\pi [/latex] [latex]=2.42 * 10^{44} *kg*m*s^{-2} [/latex] [latex]\frac{\text{cosmological Force}}{\text{Planck force}}=2[/latex] but this time I haven't an interpretation about the formula of the "cosmological Force" [latex]M_\text{Ordinary matter}= 1.44 *10^{53}kg[/latex]

oops : two errors in "cosmological Force", I'll give the correction later

Hello I noticed that the Planck Force is central for the vacuum catastrophe in quantum theory [latex]\text{volume density of quantum vacuum energy} = \frac{F_p}{l_p^2}= 4.63*10^{113} \text{ Joules / }m^3 [/latex] and in cosmology : [latex]\text{volume Density of vacuum Energy} = F_p*\Lambda / 8 / \pi = 5.35*10^{-10} \text{ Joules }/m^3 [/latex] [latex]F_p \text{ : Planck force = 1.21*10^44 kg m s^-2 , } l_p \text{ : Planck length = 1.61*}[/latex]10^-35 m [latex], \Lambda \text{ : cosmological constant } = 1.11*10^{-52} m^{-2} [/latex] so I think it could be interesting to try to define a "cosmological Force" to ending the vacuum catastrophe. based on this topic, applied to the universe , I try this without any warranty… [latex] \text{cosmological Force} = G * ({M_\text{(Ordinary matter + Dark matter)} /2})^2 * \Lambda * 8 * \pi = 6.67*10^{-11} * (4.62 * 10^{53})^2 * 1.11 * 10^{-52} * 8 * \pi = 2.42 * 10^{41} * kg*m*s^{-2} [/latex] I do not know if it is eligible: for the purposes of the calculations, I cut in two the mass of matter of the universe to have two masses we have : [latex] \frac{\text{cosmologigical Force } }{\text{Planck force}}=2[/latex] Thank you in advance for your comments edit : 1 joule = 1 kg * m^2 * s^-2 1 joule / m^3 = 1 kg * m^-1 * s^-2

hum, I forgot to say that [latex]_p[/latex] is for Planck mass

oops I missed that