# The solution of the Cosmological constant problem ?

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You're gonna laugh. I made a mistake earlier that we both didn't see.

The good equality is :

$(H/H_0)^2=\Omega_{\Lambda,H0} / \Omega_{\Lambda,H}$.

of course I used $\Lambda_{m^-2}=3H^2\Omega_{\Lambda}/c^2$

problem for me I didn't succes to find the value of (H/H_0)^2 with all $\Omega$

I don't want to violate the rules of the speculation section by adding a spreadsheet as an attachment. For all the formulas, I can send you the complete attachment by personal message

 H0 Lambda (m^-2) 67,9 1,12009E-052 2,20048913788313E-18 Om_Lambda,H H²/H0² OmL0/OmLH col I-col J H H/Ho 0,693 column D ² 22915,263 1,31973E-009 525109278,359169 525109278,359169 0,00E+00 5,042478732E-014 15740,128 2,79716E-009 247751629,456384 247751629,456384 0,00E+00 3,463598069E-014 10859,192 5,87676E-009 117922050,892864 117922050,892864 0,00E+00 2,389553404E-014 7520,218 1,22538E-008 56553678,767524 56553678,767524 0,00E+00 1,654815802E-014 5224,758 2,53864E-008 27298096,158564 27298096,158564 0,00E+00 1,149702323E-014 3639,803 5,23091E-008 13248165,878809 13248165,878809 0,00E+00 8,009346966E-015 2541,361 1,07300E-007 6458515,732321 6458515,732321 0,00E+00 5,592237276E-015 1777,702 2,19288E-007 3160224,400804 3160224,400804 0,00E+00 3,911813941E-015 1245,393 4,46807E-007 1551003,724449 1551003,724449 0,00E+00 2,740473769E-015 873,554 9,08142E-007 763096,590916 763096,590916 0,00E+00 1,922246088E-015 613,344 1,84215E-006 376190,862336 376190,862336 0,00E+00 1,349656810E-015 430,988 3,73081E-006 185750,656144 185750,656144 0,00E+00 9,483844126E-016 303,042 7,54619E-006 91834,453764 91834,453764 0,00E+00 6,668406293E-016 213,19 1,52475E-005 45449,9761 45449,9761 0,00E+00 4,691222793E-016 150,041 3,07832E-005 22512,301681 22512,301681 0,00E+00 3,301635907E-016 105,633 6,21061E-005 11158,330689 11158,330689 0,00E+00 2,324442691E-016 74,389 1,25232E-004 5533,723321 5533,723321 0,00E+00 1,636921865E-016 52398 2,52408E-010 2745550404 2745550404 0,00E+00 1,153012298E-013 36,917 0,00050848767592 1362,864889 1362,864889 0,00E+00 8,123545750E-017 26,017 0,001023808664586 676,884289 676,884289 0,00E+00 5,725012590E-017 18,342 0,002059870207846 336,428964 336,428964 0,00E+00 4,036137177E-017 12,938 0,004139986653113 167,391844 167,391844 0,00E+00 2,846992847E-017 9,137 0,008300915344211 83,484769 83,4847690000001 0,00E+00 2,010586925E-017 6,467 0,016570190934269 41,822089 41,822089 0,00E+00 1,423056325E-017 4,596 0,032807504311844 21,123216 21,123216 0,00E+00 1,011344808E-017 3,292 0,063946029182273 10,837264 10,837264 0,00E+00 7,244010242E-018 2,395 0,120815373015285 5,736025 5,736025 0,00E+00 5,270171485E-018 1,788 0,216769514886717 3,196944 3,196944 0,00E+00 3,934474579E-018 1,392 0,357647146254459 1,937664 1,937664 0,00E+00 3,063080880E-018 1,145 0,528594039015274 1,311025 1,311025 0,00E+00 2,519560063E-018 1 0,693 1 1 0,00E+00 2,200489138E-018 0,919 0,82054463798352 0,844561 0,844561 0,00E+00 2,022249518E-018 0,879 0,896923668301319 0,772641 0,772641 0,00E+00 1,934229952E-018 0,857 0,943564495288305 0,734449 0,734449 0,00E+00 1,885819191E-018 0,845 0,97055425230209 0,714025 0,714025 0,00E+00 1,859413322E-018 0,839 0,984485474932556 0,703921 0,703921 0,00E+00 1,846210387E-018 0,836 0,991563837824226 0,698896 0,698896 0,00E+00 1,839608919E-018 0,834 0,996325241964702 0,695556 0,695556 0,00E+00 1,835207941E-018 0,833 0,998718815257195 0,693889 0,693889 0,00E+00 1,833007452E-018 0,833 0,998718815257195 0,693889 0,693889 0,00E+00 1,833007452E-018 0,833 0,998718815257195 0,693889 0,693889 0,00E+00 1,833007452E-018 0,833 0,998718815257195 0,693889 0,693889 0,00E+00 1,833007452E-018 0,832 1,00112102440828 0,692224 0,692224 0,00E+00 1,830806963E-018 0,832 1,00112102440828 0,692224 0,692224 0,00E+00 1,830806963E-018 0,832 1,00112102440828 0,692224 0,692224 0,00E+00 1,830806963E-018
Edited by stephaneww

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I'm getting closer in my spreadsheet: I have the correct formula for $(\frac{\Omega{r,0}}{a}+\Omega_{m,0})=\frac{(1-\Omega_{\Lambda,0}-(1-\Omega_0))a^3}{1+\frac{1}{3400*a}}$.
I'm just missing the numerical value of $(1-\Omega_0)$ to be sure of me by numerical check

Edited by stephaneww

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3 hours ago, stephaneww said:

I'm just missing the numerical value of (1Ω0)  to be sure of me by numerical check

$(1-\Omega_0)$ is close to $9.0267568*10^{-5}$.

This is the best estimate I could make for $H^2/H_0^2$ for the calculation with all $\Omega$.

However, it is only valid, with an accuracy better than 1%, for about ten lines of the table above and below the value $H/H_0=1[/math,] i.e. today. edit : arf, actually it doesn't change much compared to [math]1-\Omega_0=0$

Edited by stephaneww

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On 8/23/2019 at 2:18 AM, stephaneww said:

I understand that if we consider a universe, almost, but not flat, it modifies the critical density and so affects my result.

I don't have enough knowledge to do the calculation you propose next.

However, I remember reading on Wikipedia, but without being able to find out where immediately, that the universe was so close to being flat that most cosmologists considered that it was indeed flat.

Is this last proposal wrong please?

On 8/23/2019 at 3:18 AM, Mordred said:

Not as far as the universe today but I do question it's validity at other cosmological times.

Ok,  http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html the equality with all $\Omega$ allow to show it. ( and $\Omega_0=1$ is in input parameters  )

On 8/23/2019 at 3:18 AM, Mordred said:

The problem being how the critical density evolves while Lambda doesn't.

we gave both a part of the solution :

On 8/23/2019 at 6:27 AM, Mordred said:

...

and this leads to the critical density

$\rho_c=\frac{3H^2}{8 \pi G}$ if the universe is at critical density it is flat….

note that $\rho_\Lambda=\frac{\Lambda c^2}{8 \pi G}$ giving $\Omega_\Lambda=\frac{\Lambda c^2}{3H^2}$

taking the ratios of Ω gives us the energy density that dominates example radiation to matter is Ωr/Ωm=ρrρm1a=13600a if a_0=1 then there was a time the universe was radiation dominant.

for a Lambda dominant

$\dot{a}^2=\frac{\Lambda c^2}{3}a^2$

so for the evolution of the three components we get

$\frac{H^2}{H_0^2}=\frac{\Omega_{r,0}}{a^4}+\frac{\Omega_{m,0}}{a^3}+\Omega_{\Lambda,0}+\frac{1-\Omega_0}{a^2}$

this is how its done to preserve the curvature terms and account for the radiation and matter dominant eras

hence my hesitancy on your second equation

and

Quote

You're gonna laugh. I made a mistake earlier that we both didn't see.

The good equality is :

$(H/H_0)^2=\Omega_{\Lambda,H0}/\Omega_{\Lambda,H}$

of course I used Λm2=3ΩΛ/

so, for $\Lambda$, it's $\Omega_{\Lambda,H}$ in relation to $\Omega_{\Lambda,H0}$  which reflects the evolution of $H$, the latter also affecting the critical density, both by $H^2$ ...

… certainly, with the exception ( very probably) of the "young" universe.

On 8/22/2019 at 8:13 AM, Mordred said:

You need to be careful here. Consider two fundamental questions.

1*

The critical density itself changes over cosmological time however the cosmological constant stays constant...

_____________________________________________________________________________________________________________________________________________________________

2*

The critical density is calculated without lambda and without any curvature term.

Our universe is close to flat however not precisely flat...

Now apply that to your second equation that you derived in your last post.

If ρ=ρcrit Then k=0. Our curvature term k isn't precisely 0 though close.

Remember the critical density is a calculated density not the actual density.

Consider this at z=1100 roughly CMB surface of last scattering H is roughly 22000 times the H_0 value today. Yet the cosmological constant is the same in both then and now...

Did I succeed to answer the first question of last quote in this message ( I had miss something ?) ?

For 2*, in the near past period and for the whole future, this affects $H^2/H0^2$ by about 0.1%. It is reasonable to say that, in these cases, the impact is negligible, but it cannot be denied.

Edited by stephaneww

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Yes I'm satisfied you have looked closely at both 1 and 2. You now have enough of the FLRW metric formulas that you can now toy model different density components so that you can check your modelling with regards to the cosmological problem at any time scale.

Well done

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Thank you...
But you've helped me a lot.

Edited by stephaneww

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It's a pleasure to help someone model build correctly when they are willing to learn and don't automatically ignore mainstream physics so you deserve credit for that as well

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This point is very simple for me : nothing can be built by ignoring conventional physics. New bricks or models can be added but they must make it possible to find or refine the existing ones.

I will try, by curiosity, to develop my summary description without the mathematical arguments   :

On 8/23/2019 at 6:57 AM, stephaneww said:

Aparty:

It's just an idea, but...

We take as a base a space-time where 3D space and time are linked.

Time scrolls in a positive direction, it "increases". So the 3D space also increases.

By bringing space-time back to the image of the surface of a sphere, the surface of the sphere will increase with time.

Assuming that this surface is the vacuum, the quantity of vacuum will increase over time while by the law of energy conservation the quantity of matter will remain stable.

In the end, we will have an increasing vacuum energy and a constant mass energy.

In terms of energy density, that of vacuum will prevail over that of matter.

It may be a description of this type that is equated by the model ΛCDM

You opposed :

On 8/23/2019 at 7:19 AM, Mordred said:

unfortunately that doesn't work with the mass density. Nor does it work with the radiation density. As the universe expands these both decrease at the ratios I have provided above. Were not dealing with the total energy content of either they are assumed constant by the adiabatic expansion (adiabatic means no net inflow or outflow of energy) the particle density is typically set constant at roughly 10^(90) particles.

unfortunately as the universe expands Lambda increases in total energy

...

The surface of the sphere, which is actually the empty volume of the universe, increases over time. As a result, the radiation and material and radiation densites decrease over time while the density of the vacuum is constant, but total Energy of vaccum increases due to the increase of the volume of the universe resulting from H. In other words, matter does not create matter but the vacuum can only be a vacuum. . I think that's correct

Other point :

On 8/24/2019 at 4:00 PM, stephaneww said:

I'm getting closer in my spreadsheet: I have the correct formula for $(\frac{\Omega_{r,0}}{a}+ \Omega_{m,0})=\frac{(1-\Omega_{\Lambda,0}- (1-\Omega_0)a^3}{{\frac{1}{3400*a}}}$
I'm just missing the numerical value of (1Ω0)  to be sure of me by numerical check

By numeric application of the "LightCone7" table it is easily shown that $H^2/H_0^2$ calculated with all $\Omega$ is not exactly  $H^2/H_0^2$ calculated with $H/H_0$ and the gap is large in the "young" universe and very small for the H "close" of us.  Whether it is with $\Omega_0=1$ or $\Omega_0=1-\Omega_{\Lambda}-\Omega_{m}-\Omega_{r}$ (as a reminder, the difference is very small between these two calculations).

We know $\Omega_{r}$ and $\Omega_{m}$ that we know how to measure by observation.

There remains $\Omega_{\Lambda}$ that we deduce from the other measurements and from the model $\Lambda CDM$ that we don't know how to explain, and this gap also has no explanation to my knowledge. Perhaps the origin of the dark energy originates in this unexplained gap, but this remains to be demonstrated.

Edited by stephaneww

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Are you using the same density values in the cosmo calc. It is an older version so uses the 2012 Planck results. Jorrie has an undated version but I lost the link on my old laptop been waiting for his reply on the newer version.

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Yes, because I haven't $\Omega_r$ isolated.

edit: after Gy > 54.532554, we need more than 3 decimals on $H/H0$ to have a correct calculation.

Edited by stephaneww

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Ah the number of digits becomes significant that makes sense

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😅

edit: I'm not sure I understood... Which value are you talking about ?

Edited by stephaneww

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7 hours ago, stephaneww said:

edit: after Gy > 54.532554, we need more than 3 decimals on H/H0  to have a correct calculation.

Actually, I'm not sure: I didn't check. (I speak about T(Gy) when I said " we need more than 3 decimals.." )

Edited by stephaneww

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I will try to approach the conclusion of this thread with the following.

We have:

- the energy density for quantum mechanics alone with J/m^3 is :

$A = m_p c^2/l_p^3=\hbar (l_p^{-4}) c$

$=\hbar (l_p^{-2})^2 c$

- the formula of the energy density with $\hbar$ gives with  J/m^3 in cosmology by adding the factor (8pi)^2 :

$B = \frac{1}{(8 \pi)^2} .\hbar (\Lambda_{m^{-2}})^2 .c$

The energy density of the cosmological constant $C$, is the geometric average of $A$ and $B$ (A/C=C/B ) :

$C=\sqrt{A.B}$

$=\sqrt{\hbar (l_p^{-2})^2. c.\hbar (\Lambda_{m^{-2}})^2. c /(8 \pi)^2 }$

$=\sqrt{\hbar^2 (l_p^{-2})^2.c^2 (\Lambda_{m^{-2}})^2/(8 \pi)^2 }$

$= \frac{\hbar c.\Lambda_{m^{-2}}}{l_p^2 8 \pi }$

$= \frac{F_p.\Lambda_{m^{-2}}}{8 \pi }$

where $F_p= \frac{c^4}{G}$ is Planck's force

$...= \frac{c^4 .\Lambda_{m^{-2}}}{ 8 \pi G}=\rho_\Lambda c^2$

i.e. the classical formula of the energy density of the cosmological constant

Edited by stephaneww

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At a glance it appears reasonably accurate let me spend a few days on it to go mull it over. From what I can see you have the right ratios regardless  of the choice of variables which is the essential part. However I can't see any immediate problem with it at the moment.

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There is no emergency... Since this problem lasts a few more days, it won't change much.

I forgot a step to make verification easier :

$l_p=\sqrt{\frac{\hbar G }{c^3}}$

$l_p^2=\frac{\hbar G }{c^3}$

then we have :

$\frac{\hbar.c}{l_p^2}=\frac{\hbar.c.c^3}{\hbar G}= \frac{c^4}{G}=F_p$

Edited by stephaneww

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That's does help, you might consider a symbol change for convention  reasons

$F_p$ F typically represents a force which is a vector. As there is no directional component you should consider the standardized representation for a scalar quantity which in turn also provides the correct spin statistics at spin zero. So

$\phi_p$ $\phi$ is typically used to represent scalar fields.

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34 minutes ago, Mordred said:

Fp F typically represents a force which is a vector. As there is no directional component you should consider the standardized representation for a scalar quantity which in turn also provides the correct spin statistics at spin zero. So

I admit I don't understand the justification you're putting forward.

34 minutes ago, Mordred said:

ϕp ϕ is typically used to represent scalar fields.

I don't know what it is. Not sure that it is in my current mathematical competences.

edit :

Okay, I just figured it out thanks to Wikipedia.

Quote

In mathematics and physics, a scalar field associates a scalar value to every point in a space – possibly physical space.

What is this symbol please : ϕ ?

Edited by stephaneww

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Fair enough you have two main categories of a field. Scalar and a vector though there are others such as a spinor or tensor field. Now a scalar field is the one we are describing. There is no mean average force or vector.

Any scalar field is typically represented  by the phi symbol. This is set by convention so that the author doesn't have to identify every symbol used and those familiar with the conventions can easily understand it's application.

Now we are comparing two scalar fields. The first being the quantum field. Typically this field is represented by the phi symbol for the reasons above. To avoid confusion Lambda was later chosen to represent by convention to represent the cosmological constant.

Part of this has a historical nature.One of the earlier explanations for inflation used to involve the quantum field. The convention choice to use \phi arose from this. When they discovered the cosmological  constant they needed to keep this separate so they used Einstein's notation for the cosmological constant for his steady state universe.

Part of the reason they needed two symbols is the earlier  thought was the quantum field  could account for Lambda until they realized the orders of magnitude didn't match.

What I am suggesting is to maintain that convention so any reader familiar with those symbols can readily understand your model without needing further identification

1 hour ago, stephaneww said:

What is this symbol please : ϕ ?

It's used to identify a scalar field. That is a field that has no inherent direction or flow. The two fields in this model both are scalar fields. A temperature field would also count as a scalar field.

Now a scalar field is one that at each coordinate you can apply a magnitude value without needing a direction.

So at each coordinate I can assign some magnitude and completely describe that field. If the magnitude values have an average gradient (not applicable in our case) you can also employ a gradient operator. (Gravity is often modeled in this fashion) the gradient operator being

$\triangledown$

Edited by Mordred

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48 minutes ago, Mordred said:

Any scalar field is typically represented  by the phi symbol. This is set by convention so that the author doesn't have to identify every symbol used and those familiar with the conventions can easily understand it's application.

Now we are comparing two scalar fields. The first being the quantum field. Typically this field is represented by the phi symbol for the reasons above. To avoid confusion Lambda was later chosen to represent by convention to represent the cosmological constant.

Part of this has a historical nature.One of the earlier explanations for inflation used to involve the quantum field. The convention choice to use \phi arose from this. When they discovered the cosmological  constant they needed to keep this separate so they used Einstein's notation for the cosmological constant for his steady state universe.

Part of the reason they needed two symbols is the earlier  thought was the quantum field  could account for Lambda until they realized the orders of magnitude didn't match.

What I am suggesting is to maintain that convention so any reader familiar with those symbols can readily understand your model without needing further identification

Um, I think I understand.

It would seem to say that I put my finger on something ?

48 minutes ago, Mordred said:

It's used to identify a scalar field. That is a field that has no inherent direction or flow. The two fields in this model both are scalar fields. A temperature field would also count as a scalar field.

Now a scalar field is one that at each coordinate you can apply a magnitude value without needing a direction.

So at each coordinate I can assign some magnitude and completely describe that field. If the magnitude values have an average gradient (not applicable in our case) you can also employ a gradient operator. (Gravity is often modeled in this fashion) the gradient operator being

Ok thanks

edit

if I understand you would like to use ϕ   instead $\Lambda$ ?

Edited by stephaneww

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Thank you but I'm not enought good in english to unterdestand. I'll try to find the same in french.

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Thank you for the welcome whith the link

Well, I have read and understood up to and including 1.8.1, 1.9 and 1.10.1

However, I am not yet familiar, for moment, with the equations of 1.10.1

Would I need anything else for what we're talking about ?

Thanks to the online translation.

Edited by stephaneww

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Not as of yet, that paper is handy for basic scalar field modelling but I already gave the main detail we need above.

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All right. Thank you.

Two questions before I try to sleep, to make things clear in my mind :
1- Have I proposed a correct solution to the cosmological constant problem ?
2 - if so, did it allow you to open a link with cosmic inflation via the scalar fields ?

Edited by stephaneww

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