Is the dark energy already present as vacuum energy by nature in the equations of general relativity ?

[stephaneww]

Hello

I have been asking myself this question since I saw the formula (II-30) on page 11 of this (unpublished) document:

Indeed, [latex]d \Omega[/latex] is presented as a four-dimensional volume variation in general relativity for EINSTEIN equations in vacuum.

[latex]d\Omega=\sqrt{|g|}dx^0dx^1dx^2dx^3[/latex] where [latex]x^0=ct[/latex]

Assuming that this four-dimensional volume is composed of the three spatial dimensions and a temporal dimension, the following calculations can be attempted from the 2018 Planck mission results data.

By making this assumption [latex]g_0[/latex] is no longer adimensioned but takes the dimension [latex]m^{-2}[/latex], dimension of the cosmological constant.

We then find a new constant [latex]X= 7.719*10^{-53}m^{-2} [/latex] that can be related to the cosmological constant [latex]\Lambda[/latex]

today:
volume universe
[latex]V_0 = 3.48*10^{80} m^3[/latex]

Hubble's time:
[latex]t_0 = 4.56*10^{17} s[/latex] or [latex]H_1 = 67.66 km/s/Mpc[/latex]

[latex]c* t_0 = 1.37*10^{26} m[/latex]

observable universe radius :
[latex]x^1=x^2=x^3=R_0 = 4.36*10^{26} m[/latex]

density parameter of the cosmological constant:
[latex]\Omega_{0 _ \Lambda} = 0.6889[/latex]

equality (II-30) page 11 is equivalent to :

[latex]1/\sqrt{g_0}=c* t_0*R_0^3/V_0=3.26*10^{25}m[/latex]

hence

[latex]g_0=9.386*10^{-52}m^{-2}[/latex]

[latex]g_0'=g_0/8/\pi*3=1.120*10^{-52}m^{-2}[/latex]

when 

Hubble time =
[latex]t_1 = 2.91*10^{17} s[/latex] or [latex]H_1 = 70 km/s/Mpc[/latex] 

volume universe
[latex]V_1 = 1.13*10^{80} m^3[/latex]

[latex]c* t_1 = 8.25*10^{25} m[/latex]

observable universe radius :
[latex]x^1=x^2=x^3=R_1 = 3.00*10^{26} m[/latex]

density parameter of the cosmological constant:
[latex]\Omega_{1_\Lambda} = 0.2803[/latex]

equality (II-30) page 11 is equivalent to :

[latex]1/\sqrt{g_1}=c* t_1*R_1^3/V_1=3.08*10^{25}m[/latex]

hence

[latex]g_1=2.308*10^{-51}m^{-2}[/latex]

[latex]g_1'=g_1/8/\pi*3=2.754*10^{-52}m^{-2}[/latex]

since [latex]d \Omega[/latex] is a variation in volume since the origin of the universe, it seems coherent to take into account the density parameter of the cosmological constant related to this volume

in a spreadsheet program 

we have the following perfect equality and a new constant [latex]X[/latex] :
[latex]g_0'*\Omega_{0_\Lambda}=g_1'*\Omega_{1_\Lambda}=...=X[/latex]

[latex]1.120*10^{-52}*0.6889=2.754*10^{-52}*0.2803=[/latex]

[latex]X= 7.719*10^{-53}m^{-2} [/latex]

 

A link with the cosmological constant [latex]\Lambda=1.105*10^{-52}m^{-2}[/latex] can be made as follows:

[latex]\Lambda / X * 8 * \pi=[/latex]

[latex]1.105*10^{-52}/7.719*10^{-53}* 8 * \pi=[/latex]

[latex]36[/latex] (exact dimensionless value calculated using a spreadsheet)

If the author's reasoning is correct and the assumptions and calculations made here are correct, it could mean that the black energy is, by nature, the underlying vacuum energy in general relativity.

Thank you for your attention and any comments you may have.
 

Edited April 17, 2019 by stephaneww

[stephaneww]

Hello

For information, there is a version 2 of the document.

all this can be simplify by

[latex]1.106 * 10^{-52}m^{-2}= \Lambda= g \frac{9}{16\pi^2} = 3\frac{H^2}{c^2}  \Omega_\Lambda[/latex] (nothing new)

I hope that somebody will can explain me why this is OK and what does this mean in the context of the source document

Edited July 20, 2019 by stephaneww

[stephaneww]

edit too late sorry

 

Hello

For information, there is a version 2 of the document.

all this can be simplify by

[latex]1.106 * 10^{-52}m^{-2}= \Lambda= g  \frac{9}{16\pi^2}\Omega_\Lambda = 3\frac{H^2}{c^2}  \Omega_\Lambda[/latex] (nothing new)

I hope that somebody will can explain me why this is OK and what does this mean in the context of the source document

 

note : V₀=4pi/3 R₀^3

Edited July 20, 2019 by stephaneww

[Mordred]

First off [latex] g_{\mu\nu}[/latex] is strictly the metric tensor. The determinant being |g|.

The LHS of the EFE is strictly the metric. The RHS where the stress tensor resides is where one accounts for mass/energy. Never the LHS.

Secondly that article claims a Langrene being the Lorentz transformation. The Lorentz transform returns  a constant of proportionality. Not a Langrene. So I would place zero faith in the rest of the article. Even if it were published under peer review.

[Mordred]

Sorry the above isn't accurate the [latex]\Lambda[/latex] term is originally placed as a seperate term on the LHS as it didn't affect the covariance. It's original purpose was to keep the universe static.

In modern usage it can be moved to the RHS to act as a stress tensor as

[latex]T_{\mu\nu}=-\Lambda g_{\mu\nu}[/latex]

However it is seperate from the metric tensor

Edited July 20, 2019 by Mordred

[stephaneww]

oops my formula need a correction, 

[latex]g=\frac{H_0^2}{c^2}\frac{16\pi^2}{9}[/latex] in the part context cited of the document, and if I don't make mistake, is not a determinant or linked to the metric tensor

indeed

[latex]\frac{1}{\sqrt{|g|}}=\frac{c}{H_0}\frac{3}{4\pi R^3}.R.R.R [/latex] [latex]m[/latex]

always  if I don't make mistake...

I deduce :

[latex]|g|=\frac{H_0^2}{c^2}\frac{4^2\pi^2}{3^2} m^{-2}[/latex]

I always don't know if it's ok...

and finally I find 

[latex]\Lambda=3\frac{H_0^2}{c^2}\Omega_\Lambda=3g\frac{3^2}{4^2\pi^2}\Omega_\Lambda m^{-2} [/latex] (correction made with factor 3 in addition)

The question is : is it ok ?

question for Mordred : I haven't understand…  about the theorical frame of the document for this part... is it correct  please ?

 

Edited July 21, 2019 by stephaneww

[Mordred]

The document primarily discusses  different Langrene and the action principle of different fields..

 Action is certainly applied in every field theory. It is a huge part of QFT treatments. So yes I have no problem with the article in those  regards.

Using the determinant of g above isn't particularly  useful in determining the cosmological constant. The determinant of g will vary  according to the tensor entry values 

 

Edited July 23, 2019 by Mordred

[stephaneww]

thank you Mordred

[Mordred]

Let's add some greater detail. The purpose of the metric tensor is to define a coordinate basis.

For example

Flat space [latex]\mathbb{R}^4 [/latex] with Coordinates (t,x,y,z) or alternatively (ct,x,y,z) flat space is done in Cartesian coordinates.

[latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex]

In this case [latex] g_{\mu\nu}=\eta_{\mu\nu}[/latex]

[latex]\eta_{\mu\nu}=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex]

spherical polar coordinates [latex](x^0,x^1,x^2,x^3)=(\tau,r,\theta,\phi)[/latex]

[latex] g_{\mu,\nu} =\begin{pmatrix}-1+\frac{2M}{r}& 0 & 0& 0 \\ 0 &1+\frac{2M}{r}^{-1}& 0 & 0 \\0 & 0& r^2 & 0 \\0 & 0 &0& r^2sin^2\theta\end{pmatrix}[/latex]

line element

[latex]ds^2=-(1-\frac{2M}{r}dt)^2+(1-\frac{2M}{r})^{-1}+dr^2+r^2(d \phi^2 sin^2\phi d\theta^2)[/latex]

Where as in Minkowskii space in rotating space about the z axis.

(Sagnac effect)

[latex] g_{\mu,\nu} =\begin{pmatrix}-(1-\frac{\omega^2r^2}{c^2})& 0 & \frac{\omega r^2}{c}& 0 \\ 0 &1& 0 & 0 \\\frac{\omega r^2}{c}& 0& r^2 & 0 \\0 & 0 &0&1\end{pmatrix}[/latex]

As you can see entries depend upon coordinate basis which is the purpose of the metric tensor.

The determinant will vary depending on metric changes.

 

Edited July 28, 2019 by Mordred

[stephaneww]

thank you again Mordred. 

what is [latex]r[/latex] please ?

on occasion I would look at what the spherical coordinates consist of

the rotation is too hard for me, I forget for now

Edited August 18, 2019 by stephaneww

[Mordred]

Radius 

[stephaneww]

I should have thought of that. 

Thank you