is it possible to define a cosmological force based on Newton force definition ?.

[stephaneww]

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

 

 

Edited October 20, 2018 by stephaneww
latex and ^2 sur lp

[stephaneww]

Quote

1.  

oops :

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

Edited October 20, 2018 by stephaneww

[stephaneww]

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]

 

 

Edited October 20, 2018 by stephaneww

[stephaneww]

 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

Edited November 4, 2018 by stephaneww

[stephaneww]

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

 

Edited November 10, 2018 by stephaneww
latex

[stephaneww]

  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

Edited November 10, 2018 by stephaneww

[stephaneww]

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]

Edited November 10, 2018 by stephaneww

[stephaneww]

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, Omega_M = 0.3089, Z infinite, Omega_Vac = 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

Edited November 24, 2018 by stephaneww

[stephaneww]

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]

Edited November 30, 2018 by stephaneww

[stephaneww]

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]

 

 

Edited January 4, 2019 by stephaneww