stephaneww Posted December 8, 2018 Posted December 8, 2018 (edited) 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 Edited December 8, 2018 by stephaneww
et pet Posted December 8, 2018 Posted December 8, 2018 (edited) 16 minutes ago, stephaneww said: The thing I can't remember from school it's : what mean in physics " 1kg/1m2" in simples words 1 Kilogram Per 1 Square Meter Edited December 8, 2018 by et pet
stephaneww Posted December 8, 2018 Author Posted December 8, 2018 (edited) 2 minutes ago, et pet said: " Kilograms Per Square Meter " That's his definition, but his sense physic here? Edited December 8, 2018 by stephaneww
stephaneww Posted December 14, 2018 Author Posted December 14, 2018 (edited) On 08/12/2018 at 2:41 PM, stephaneww said: That's his definition, but his sense physic here? 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 Edited December 14, 2018 by stephaneww
stephaneww Posted December 25, 2018 Author Posted December 25, 2018 (edited) 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 Edited December 25, 2018 by stephaneww
stephaneww Posted December 26, 2018 Author Posted December 26, 2018 (edited) 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% Edited December 26, 2018 by stephaneww
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now