stephaneww

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About stephaneww

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  • Birthday 10/02/1968

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  1. 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
  2. That's his definition, but his sense physic here?
  3. 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
  4. 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]
  5. 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
  6. 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]
  7. 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
  8. 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
  9. 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
  10. 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]
  11. oops : two errors in "cosmological Force", I'll give the correction later
  12. 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
  13. hum, I forgot to say that [latex]_p[/latex] is for Planck mass
  14. oops I missed that
  15. um, nothing new in fact, it's under key : https://en.wikipedia.org/wiki/Planck_units#Base_units and it must be taken into account that: [latex]\Large{F_p=\frac{c^4}{G}}[/latex] so, we can deduce that it takes at least a relativistic context