# The solution of the Cosmological constant problem ?

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o I may have found a very nice numerical and literary conclusion.

The simplest part first:

There is a time in the evolution of the universe when the density parameter of the cosmological constant would be greater than 1 while keeping the model $\Lambda CDM$

The end of the current universe would occur when the density parameter of the cosmological constant is equal to 9*10^24 and it has an age of about $3.04*10^{24} \text{ years}$.

The most likely scenario is that it will return to its initial state after a scenario similar to quantum cosmic inflation (to be defined because I have no expertise in this field...) before starting a new Big Bang again.

The numerical application now, a little difficult to follow I must admit:

I will use this 2018 data from the Planck collaboration
- $H_0=67.66 km/Mpc=1.874*10^{-11}s^{-1}$
- $\Omega_\Lambda=0.6889$
that give:
- $\Lambda =1.106 m^{-2}$
- $\rho_{\Lambda,0}=5.324*10^{-10}J/m^3$

So we will have:
- $D=6.118*10^{-113} J/m^3$

that gives us:

- $H_\text{p,minimum}= ( \frac {D.(8* \pi)^2} { \hbar.c^{3}})^{0.5*0.5}=6,073^{-31}s^{-1}$

- $\rho_\text{c,at H(p,minimum)}=\frac{3c^2 H_\text{p,minimum}^2}{8 \pi G}=5.985*10^{-35}J/m^3$-

-  $\Omega_\Lambda \text{at H(p,minimum)}=\frac {\rho_{\Lambda,0}} {\rho_\text{c,at H(p,minimum)}}= \frac{5.324*10^{-10}}{5.985*10^{-35}}=8,981*10^{24}$

This value is incomprehensible for the moment, it should be at most equal to 1.

the Ned Wright calculator gives an observable radius of the universe
$R_{max}=604568.8 Mpc=1.866*10^{28}m$
and
$V_{max}=2.719*10^{85}m^3$

we have for these limit values:
Total energy $E_{tot, end}$ of the current universe at the end of its life (there is only the dark energy left that bathes the entire volume of the universe):

- $E_{tot, end}=V_{max}.\rho_\text{c,at H(p,minimum)}.\Omega_\Lambda \text{at H(p,minimum)}=$
- $E_{tot, end}=1.612*10^{51}.\Omega_\Lambda \text{at H(p,minimum)}=1.448*10^{76}J$

So at the end of the universe's life there remains a huge reservoir of ready energy (1,448*10^{76} Joules) to be used for a new Bing Bang.

- $V_{max}. \rho_\text{c,at H(p,minimum)}.c.\pi=1.364*10^{85} J.m/s$
and
- $V_{max}/1.364*10^{85}= 2.719*10^{85} / 1.364*10^{85}=\frac{1}{1.994} .s.J^{-1}m^{-1}$

In other words, at the end of its life cycle, the universe would be in a quantum state of phase transition comparable to cosmic inflation, which would bring it back to a new era of Planck.

If this scenario is validated, you will certainly find a simpler drafting and probably complementary to what is developed in this post

"Nothing is lost, nothing is created, everything is transformed", Lavoisier, reformulation of a sentence by the Greek philosopher Anaxagoras

"Let there be light...."

Edited by stephaneww

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You would need a mechanism of collapse to get a new BB. The energy density may only be equivalent to zero point energy with total energy being huge but there would be no reason for a new  BB. You would be as close to the lowest energy density state as possible.

The state above would be as close to zero K as possible under zero point energy. You also wouldn't have a false vacuum state as per the inflationary models (false vacuum by Allen Guth was the first inflationary model)

Here is a related question..

Under thermodynamics of an ideal gas. Why does a gas uncontained expand ?

Now treat your universe as an ideal gas. Our universe follows the ideal gas laws. (The fluid equations of the FLRW metric employs those laws)

Edited by Mordred

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On 9/4/2019 at 7:04 AM, Mordred said:

Sounds good for a quick and dirty rough order age of the universe that assumes constant expansion. There is a quick calculation, to account for evolution of matter radiation and Lambda takes a considerable more work..

With the Hubble value given by Planck 2015 this will give an age of 14.4 Gyrs. However when you employ the full equation you can narrow that down. For starters lets just do the rough order calculation (initial estimate.

$V=frac{d}{t}[/math t=d/t H0=v/d v=dHO combine those two and substituting for velocity t=[math]\frac{d}{v}$= $\frac{d}{h*H_0}$=$\frac{1}{H_0}$

now as you know Hubble parameter isn't constant which is one reason for the discreptancy however this is a quick and dirty estimate. Practice this with the conversions first before we get into the more complex equations for the evolution of the density parameters.

a true critical flat universe would be

t=$\frac{2}{3}\frac{1}{H_0}$ were not quite a true critical flat and we have a cosmological term to deal with but practice those first.

You could use $3.16*10^7 s/years$   its not quite accurate as your not accounting for leap years lol.

just a side note no method I show you will get precisely the Planck value for age as Planck does a best fit over several different datasets including type 1A supernova measurements.

I'm not sure I understand.
Can you put down the numerical values, please?

On 9/4/2019 at 7:04 AM, Mordred said:

You could use $3.16*10^7 s/years$ its not quite accurate as your not accounting for leap years lol.

You are correct. This gives a universe age maximum of about 5.23*10^22 years

17 hours ago, Mordred said:

You would need a mechanism of collapse to get a new BB.

I have no idea at the moment.

17 hours ago, Mordred said:

energy density may only be equivalent to zero point Energy

... and what is its value to check that I'm not going to make a mistake?

Edited by stephaneww
corrections latex in quote

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2 hours ago, stephaneww said:
19 hours ago, Mordred said:

You would need a mechanism of collapse to get a new BB.

I have no idea at the moment.

Maybe because the time is sinusoidal

this hypothesis is based on the geometric representation of the geometric mean of Wikipedia.  (time go from C' to A ?)

as a reminder, it is this average that is used to solve the problem of the cosmological constant

but I don't know how to develop this hypothesis more precisely at the moment.

perhaps you can help me ?

Edited by stephaneww

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1 hour ago, stephaneww said:

but I don't know how to develop this hypothesis more precisely at the moment.

Normally the energy density of the cosmological constant should be the point M

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13 hours ago, stephaneww said:

I'm not sure I understand.
Can you put down the numerical values, please?

I will use 75 km/s/Mpc just for the variation.

$t_0=\frac{1}{H_0}$ =$\frac{1}{75 km/sec/Mpc}$

1 pc=$3.09*10^16$ metres

1Mpc=$10^6 pc=3.09*10^22 metres=3.09*10^19 km$

now substitute

$\frac{1}{H_0}=\frac{1}{75\frac{\frac{km}{sec}}{Mpc}}$

$\frac{1}{H_0}=\frac{1}{75\frac{\frac{km}{sec}}{3.09*10^19 km}}$=$\frac{1}{2.43*10^{-18}\frac{1}{sec}}=4.12*10^{17} sec$

convert that to years and you get 1.3*10^10 years but this is an empty universe. You should be able to run the other calc now for todays Planch 2015 or 2018 result and apply the 2/3 ratio.

(of course that won't quite get you the exact answer until we take into how matter, radiation and lambda evolve

9 hours ago, stephaneww said:

Normally the energy density of the cosmological constant should be the point M

in a homogenous and isotropic (uniform distribution the energy density should be the same everywhere) however when you state that under QM your already referring to the mean average. It is no different

the formula for zero point energy is

$e-\frac{1}{2}\hbar w$

one common used formula to describe this in terms of a cosmological constant was

$\Lambda=\frac{E^4_{planck}}{\hbar^3 c^4}$

Edited by Mordred

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14 hours ago, stephaneww said:
14 hours ago, stephaneww said:
16 hours ago, stephaneww said:
On 9/5/2019 at 1:27 AM, Mordred said:

You would need a mechanism of collapse to get a new BB.

I have no idea at the moment.

Maybe because the time is sinusoidal

this hypothesis is based on the geometric representation of the geometric mean of Wikipedia.  (time go from C' to A ?)

as a reminder, it is this average that is used to solve the problem of the cosmological constant

but I don't know how to develop this hypothesis more precisely at the moment.

perhaps you can help me ?

13 hours ago, stephaneww said:

Normally the energy density of the cosmological constant should be the point M

I think it is better to say that time describes a circle of center M  and that the 3D space is euclian

"Something like a Yin and yang…"

What do you think of this approach, please?

6 hours ago, Mordred said:

in a homogenous and isotropic (uniform distribution the energy density should be the same everywhere) however when you state that under QM your already referring to the mean average. It is no different

I'm not sure I understand. Can you develop it, please? (my translator stumbles upon "mean average" )

Edited by stephaneww

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Think of it as the average value.

I really don't know how you plan on applying time as a sinusoidal.

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47 minutes ago, Mordred said:

I really don't know how you plan on applying time as a sinusoidal.

I changed my mind after that: I apply time like a circle. so 3d space would be euclian, but not the space time

see :

19 hours ago, stephaneww said:

I think it is better to say that time describes a circle of center M  and that the 3D space is euclian

"Something like a Yin and yang…"

What do you think of this approach, please?

47 minutes ago, Mordred said:

Think of it as the average value.

So this is the point M on wikipedia, correct ?

Do I still have to recover the zero-point Energy ?

Edited by stephaneww

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Well the geometric mean is a type of mean average. However the geometric mean only applies to the Set of real numbers.

Zero point energy involves the set of imaginary and real numbers so wouldn't be applicable in this case. Zero point energy is the main basis behind the cosmological constant problem in terms of the quantum catastrophe. (120 orders of magnitude too high)

I have no idea how you would apply time in a circular fashion either it wouldn't match any of the symmetries of how time is treated as a vector. (Arrow of time) as one example . You would need to have non linear relations.

Which I don't see how you would apply that.

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The symmetric in this representation would be in the negative part of the Y axis????
So imaginary numbers but I don't understand the need for them

For the zero-point energy I think I may have a solution that would fit with this geometric representation.
I just need to recover it and put it in latex format.

Edited by stephaneww

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Think of a sine wave you have the positive number real part  then you also have the imaginary negative number portion.

Ie on an x, y graph your sine wave has a range of -y to +y for the amplitudes of the sine wave along the x axis. So too does the Zero point energy eigenstates. The creation and annihilation operators employ this in discrete units of quanta. It is fundamental to the ladder operations of the Schrodinger equations.

You have both positive and negative  frequency portions  the negative frequency portion is the imaginary number set while the positive frequency parts are the set of Real numbers.

Edited by Mordred

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Okay, I think I understand.

Preliminary question:

Do we also have zero point energy $\epsilon=\frac{1}{2} \frac{h}{t_p}=9.78*10^8J$ (with p for "Planck")?

Edited by stephaneww

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Yes you can use that but it's handier to use the reduced Planck value

Planck energy is

$E_p=\frac{\hbar}{t_p}$ if you use the normalized formula with the mass energy equivalence in normalized units e=mc^2 simplifies to e=m. Then you can employ the following formula for the reduced planck energy.

$E=\sqrt{\frac{\hbar c^5}{4\pi G}}=7.8*10^8 J$

See

Edited by Mordred

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Okay, but I prefer zero point energy with $\epsilon=\frac{1}{2}\frac{h}{t_p}=9.78*10^8J$ (with p for "Planck")?

Here's why:

By reducing it to an energy density we have:
$\epsilon_p / l_p^3=1/2 \frac{\hbar}{t_p}/l_p^3 = 2.316*10^{133} J/m^3$

and the energy density of the quantum vacuum
$A=m_p c^2/l_p^3=4.63* 10^{133}=$

$2 \epsilon_p / l_p^3$

In other words, with the circle we have the occupation of the part of the -y axis that completes the +y part. But you're right, I have to take into account the time arrow. i.e. go back to a sinusoidal

Moreover, with the geometric mean we had:
$C=\sqrt{AB}$
or
$C^2=AB$
or
$1/B=AC^{-2}$

Edited by stephaneww

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Fair enough your still within the correct orders of magnitude for each calculation.

A little side note. If you recall the LCDM model starts at $10^{-43} s$

That value is identical to Planck time. This isn't a coincidence the cutoff point for the temperature, time and volume are Planck units. Ie Planck length (for corresponding volume ) Planck temperature and Planck time. Before that time is the singularity condition.

Edited by Mordred

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Thanks for the note.

I just wanted to try to get back to planck time. I'm still going to try

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Start with the FLRW metric. You now have the start point for Planck time. With that you should be able to get your minimal scale factor a. (Lol within a rough approximation after all how many Planck lengths are in an Mpc hehe.) Even worse how many Planck time units in a Gyr ?

$d{s^2}=-{c^2}d{t^2}+a({t^2})[d{r^2}+{S,k}{(r)^2}d\Omega^2]$

Edited by Mordred

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outch, I hope to do it more simply with Planck's force.

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8 hours ago, stephaneww said:

outch, I hope to do it more simply with Planck's force.

um, it seems to be a dead end....

Edited by stephaneww

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hey hey hey, no, it's very simple with Planck's force and we find exactly Planck's time at the origin of the cycles.

I need to rest. The demonstration is coming soon.

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Well, we have:

On 4/27/2019 at 7:03 PM, stephaneww said:

Fp : Planck force

1 the energy density by volume of the quantum vacuum as A=Fp/lp^2 = 4.633*10^113 Joules/m^3 (formula derived from dimensional analysis in Planck units)

2 the energy density by volume of the vacuum of the cosmological constant as B= Fp* Lambda /8/ pi = 5.354*10^-10 Joules/m^3

the ratio between the two being the number in factor of 10^122 undimensioned

the value of the adimensionless factor, X, is more precisely : A/B=X=8.654*10122, X is called " The cosmological constant problem ".

and :

21 hours ago, stephaneww said:

Moreover, with the geometric mean we had

$C=\sqrt{AB}$

or

$C^2=AB$
or
$1/B=AC^{-2}$

important notes : B page 1, is different of the formula of $B=\hbar \Lambda^2 c /(8 \pi)^2$, page 2

B page 1  $=C$ (The energy density of the cosmological constant) page 2

so

$1/B=$

$\frac{(8*\pi)^2}{\hbar \Lambda^2 c}=\frac{F_p}{l_p^2}\frac{(8\pi)^2}{F_p^2 \Lambda^2}$

so

$\frac{1}{\hbar c}=\frac{1}{l_p^2}\frac{1}{F_p}$

so

$\frac{F_p}{\hbar c}=\frac{1}{l_p^2}$

so

$l_p^2=\frac{\hbar c}{ F_p}$ with $F_p=\frac{c^4}{G}$ cf wikipedia

so

$l_p^2=\frac{\hbar G}{ c^3}$ with $t_p^2=\frac{l_p^2}{ c^2}$ cf french wikipedia

so

$\frac{l_p^2}{ c^2}=\frac{\hbar G}{ c^5}$

so

$t_p=\sqrt{\frac{\hbar G}{ c^5} }$ i.e  the expected formula for the beginning of a new cycle

Edited by stephaneww

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now that the model is complete, I allow anyone interested to write this thread according to the scientific rules for a free consultation publication on arXiv. indeed it is not in my competence....

Edited by stephaneww

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On 9/4/2019 at 10:40 AM, stephaneww said:

- Vmax.ρc,at H(p,minimum).c.π=1.36410^85J.m/s
and
- $V_{max}/1.364∗10^{85}=2.719∗10^{85}/1.364∗10^{85}=1/1.994.s.J^{−1}m^{−1}$

Um, I believe there are errors on the physical dimensions.

I'll let you fix it until I have a moment to time it.

Edited by stephaneww

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On 9/9/2019 at 6:11 AM, stephaneww said:

- $V_{max}.ρ_c,at H(p,minimum).c.π=1.364∗10^{85}J.m/s$

-  $V_{max}/1.364∗10^{85}=2.719∗10^{85}/1.364∗10^{85}=1/1.994.sJ^{-1}m^{-2}$

$V_{max}/1.364∗10^{85}=2.719∗10^{85}/1.364∗10^{85}=1/1.994.J^{-1}m^2.s$

are the good units if I don't make a new mistake.

but apart from the numerical value = 1/2, I don't know how to interpret it

the other values and units are fair.

Edited by stephaneww

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