# The solution of the Cosmological constant problem ?

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Um, again a mistake at the top of  page 4, sorry :

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

...

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

...

$D=6.118∗10^{-133}J/m^3$, -133 is the good exponant

the next value was correct :

$=6.073*10^{-31}s^{-1}$

Edited by stephaneww

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there was still one mistake left at the end of first post page 4 😓

please read : ( $\Omega_\Lambda, at H(p,minimum)$ missed )

$E_{tot,end}.c.\pi=1.446*10^{76}.c.\pi=1.364*10^{85}J.m/s$

and :

$1.364*10^{85}/V_{max}=1.3364*10^{85}/2.719*10^{85}=1/1.994 \frac{J}{m^3}m/s$, in another word a energy density that displaces.

hoping I didn't leave any more mistakes to be left behind...

Edited by stephaneww

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I wouldn't sleep so dumb tonight:
$J/m^3 . m/s$,  is also the unit of the surface power density : $kg/s^3$ or  $W/m^2$

I let you do the dimensional analysis that goes well with $1/2, \hbar,t_p, \Lambda,8 \pi$ based on the first Wikipedia link in the article about the surface power density (electromagnetic radiation)

Then follow the link Particle model and quantum theory

With the factor 1/2, you should find the value $1/1,994 kg/s^3$

We should have, with the values from the first post on page 4, and second post page 5 :

$\frac{1}{2}\frac{h}{t^2}\frac{\Lambda}{8\pi}=0.50144. kg/s^3=1/1.994.kg/s^3$

But as I have very little knowledge of quantum mechanics, and nothing in electromagnetism, I don't know how to interpret this result

A little help is welcom, thanks in advance

Edited by stephaneww

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Surface power density isn't applicable in this application. The cosmological constant is a scalar field there is no force involved. A force requires a vector field. The EM field is an example as you have two charges.

The cosmological constant doesn't have a charge nor inherent vector direction.

Edited by Mordred

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Um, which post are you talking about, please?

both ?

Do you remember how to write the energy density of the cosmological constant expressed with Planck's force. Is it impossible to deal with that?

There's no way to say that Planck's field =$\phi_p=1/2 .h$ is involved ?

Finally, there is no way to say that $\Lambda$ in $m^{-2}$ is the division by the surface ?

Quote

Intensity can be found by taking the energy density (energy per unit volume) at a point in space and multiplying it by the velocity at which the energy is moving. The resulting vector has the units of power divided by area (i.e., surface power density).

Edited by stephaneww

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Any attempt to apply force or power to the cosmological constant will be wrong. The field only has magnitude only values.

To have a force you require a magnitude and a direction. There is no direction in Lambda. Expansion applies equally in all directions ie no preferred direction (isotropic) and no preferred location (ie centre) homogeneous.

The EM field has a preferred direction anisotropic and a preferred location ie the source inhomogeneous.

In the case of gravity ie a planet the planet is the source so has a preferred direction and a preferred location.

That is not the case with Lambda nor the Planck field

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I must admit that all this is beyond my current knowledge. There remains that perfect equality that I still do not understand.

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Ok let's try a simple example. Earlier this thread I posted an article detailing scalar and vector fields. In that paper they gave an example of the atmosphere.

So start with an atmosphere that has no motion or wind. It has a temperature value everywhere.

That is a scalar field

Now add wind currents, you now have an average direction or flow of movement

A vector field. A vector field can apply force, it can produce work such as on a windmill. There is an average power in watts that can be applied.

Edited by Mordred

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Okay, let me try this  :
here .https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1117499  we have also

$E_{end}/V_{max}=\frac{1.448∗10^{76}J}{2.179*10^{85}m^3}=5.243*10^{-10}J/m^3$ (the Density Energy of cosmological constant, I just realized that.)

$5.243*10^{-10}J/m^3.c.\pi=1/1.994.J/m^{-3}.m/s$ in another word the energy Density of the cosmological constant at the singularity of the Big Bang multipiled by a velocity $(c.\pi)$ in $m/s$.

Doesn't that fit that definition ?

Quote

Intensity can be found by taking the energy density (energy per unit volume) at a point in space and multiplying it by the velocity at which the energy is moving. The resulting vector has the units of power divided by area (i.e., surface power density).

Edited by stephaneww

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No because velocity is a vector. It has a direction. I think the problem your having is how expansion works. There is no Centre of the universe nor does it expand  in any direction from the Centre.

Also the singularity isn't at any particular direction from us if we were to look into the past. Assuming we can see far enough back in time you would not be able to point at any particular direction and say the BB occurred  in that direction.

Just like the CMB it surounds us in every direction we look.

Use this as an analogy. Draw dots on the surface of a balloon. As the balloon expands  (don't think about the interior of the Balloon only the surface) you will notice the angles between the dots do not change as the surface expands.

If you have an expansion with a direction then this would not be the case. The angles would change.

Now you might think you can use recessive velocity of Hubble's law but this velocity depends on the observers location. You change that location and every value also changes including vector directions. It is not a true velocity but an apparent velocity.

Just like a persons observable universe will change depending on location. So will any directional components when measuring stellar objects in terms of its recessive velocity also change.

That would not be the case of there was a net flow. Each observer location would be able to measure the net flow.

Edited by Mordred

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3 minutes ago, Mordred said:

No because velocity is a vector. It has a direction.

if the velocity vector is the time arrow (everywhere the same in all directions) it still can't work?

18 minutes ago, Mordred said:

I think the problem your having is how expansion works. There is no Centre of the universe nor does it expand  in any direction from the Centre.

Also the singularity isn't at any particular direction from us if we were to look into the past. Assuming we can see far enough back in time you would not be able to point at any particular direction and say the BB occurred  in that direction

I knew that and I understand that, no problem, we agree.

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time arrow although can be described as a vector does not apply any force. It is simply a rate of change. It doesn't have substance of any form or energy to apply the work performed to apply a force.

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Ok, it's a dead end, so I'm starting to doubt everything...

When you have a moment, could you confirm that my posts or parts of posts listed below are, in your opinion, of a scientific nature or not, please ?

(+ spéculatives posts related to 4. )

( I have a very big doubt about 6., you didn't say anything about it… )

Now,  only my first post page 5 is ok in my opinion.

Edit : I hadn't seen your next edit:

3 hours ago, Mordred said:

Use this as an analogy. Draw dots on the surface of a balloon. As the balloon expands  (don't think about the interior of the Balloon only the surface) you will notice the angles between the dots do not change as the surface expands.

If you have an expansion with a direction then this would not be the case. The angles would change.

Now you might think you can use recessive velocity of Hubble's law but this velocity depends on the observers location. You change that location and every value also changes including vector directions. It is not a true velocity but an apparent velocity.

Just like a persons observable universe will change depending on location. So will any directional components when measuring stellar objects in terms of its recessive velocity also change.

That would not be the case of there was a net flow. Each observer location would be able to measure the net flow.

But I wonder: So It'is true also for the value of the cosmological constant ? It depend of your location also, if  I well understand ? Indeed, I thought it had the same value everywhere, no? Its value is independent of H0, so.... I have to admit, I'm missing something.

Edited by stephaneww

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Correct it's the same value everywhere. So that mean there is no potential difference from one location to another.

Now think about gravity on a planet the closer you get to the Centre of mass the greater the potential. So you have a vector quantity that applies a force.

However if mass/energy has no potential difference then there is no flow or force involved  even with gravity.

Another example a copper wire has electrons yet when there is no potential difference there is no current. In this case voltage is just another word describing potential difference.

Now you agree that Lambda is uniform in potential difference everywhere. Place a planet in that field. How can there be a force applied in a specific direction if all sides of the planet has equal density ?

If every star, galaxy and large scale structure is surrounded by Lambda that has equal density everywhere you cannot have a force in a direction. As every facing has the same density.

Edited by Mordred

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All right, thank you again.

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Um, sorry to insist once again. I have one more point to clarify.
The cosmological constant is sometimes presented as "a repulsive gravity". In other words, we have an acceleration speed of the expansion that depends on the distance in a given direction. The simplified version I was given is $a=d. \Omega_\Lambda H_0^2.m/s^{-2}$ with $d$ =distance. The greater the distance from the "centre" (equivalent to the centre of gravity in classical mechanics) the greater the acceleration speed. So, by making the analogy with this quote:

23 hours ago, Mordred said:

Now think about gravity on a planet the closer you get to the Centre of mass the greater the potential. So you have a vector quantity that applies a force.

can we say we have a vector quantity that applies a force?

But it is also possible that I didn't understand this correctly :

23 hours ago, Mordred said:

However if mass/energy has no potential difference then there is no flow or force involved  even with gravity.

Edited by stephaneww
error i have change v by a

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A more accurate expression is via the vacuum pressure term. In this gravity is treated as positive pressure while lambda is treated as negative pressure via the equations of state.

The relation is $p=-p$

Now what we term as gravity is in actuality spacetime curvature.

$w=\frac{\frac{1}{2}\dot{\phi}^2-V (\phi)}{\frac{1}{2}\dot{\phi}^2+V (\phi)}$

This correlates the field potential energy to the energy density of a scalar field. Zero spacetime curvature with the stress tensor $T_{\mu\nu}=0$ has no spacetime potential difference. So the potential energy term in the last expression is $V\phi=0$ this gives w=1 via $w=\frac{p}{\rho}$

The cosmological constant however has an equation of state w=-1. The overdose above is the velocity time derivative. If you have two overdots that is the acceleration time derivative.

This article will provide the key thermodynamic relations

Edited by Mordred

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Well, once again, your explanation is beyond my competence. I just have to admit, I said one more nonsense.
Thank you for your patience and time.

I will wait patiently for you to validate or not the series of posts linked above if you have some time to devote to it

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Actually the last post isn't nonsense you were just missing the detail of the pressure relations involved in the descriptive of the cosmological constant acting as a repulsive gravity effect.

I simply supplied the details of how that is the case

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18 minutes ago, Mordred said:

Actually the last post isn't nonsense you were just missing the detail of the pressure relations involved in the descriptive of the cosmological constant acting as a repulsive gravity effect.

That's because I don't master this important part of cosmology. The link you gave does not display the equations correctly in my browser. edit : it's ok in another browser

And you reassure me of my current knowledge of cosmology. Thank you.

But I don't have a clear enough answer in my field of understanding to this question:

2 hours ago, stephaneww said:

So, by making the analogy with this quote:

On 9/14/2019 at 7:33 AM, Mordred said:

Now think about gravity on a planet the closer you get to the Centre of mass the greater the potential. So you have a vector quantity that applies a force.

can we say we have a vector quantity that applies a force?

Edited by stephaneww

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Well as French is your native language if you google equations of state cosmology pdf. (Include the pdf) in your language you should be able to pull up some decent articles. If it includes the ideal gas laws your on the right track.

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I use an excellent translator (DeepL.com) who gives very relevant translations and allows alternatives if there is a problem of meaning. I also manage to understand thanks to my basic knowledge of English, even if my translations into English are not always perfect.

The problem lies more in my incompetence in the tensors in general relativity and the understanding of some notations in the equations. I only have a 30-year-old bachelor's degree.

Edited by stephaneww

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Well the good news is you can find classical treatments for the equations of state.

Here is one that keeps it in a Newtonian fashion

By the way I rewarded you some reputation points as your diligent in following proper physics in your modelling and constantly striving to learn as you go.

It's a rare pleasure finding a thread in speculations forum where the OP is interested in learning how to properly model as well as expanding his skill set in physics.

Your skills have improved immensely since you first joined this forum well done

Edited by Mordred

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On this forum, you are my only interlocutor so all the credit for my progress goes to you

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6 hours ago, Mordred said:

Well the good news is you can find classical treatments for the equations of state.

Thank you. I disconnect after"the energy equation becomes".
... because I don't remember what $\dot a$ means.

On the other hand, I believe I understood that I managed to convince you that we had a "repulsive force" for the cosmological constant.
Therefore, is that enough for you to go back on your counter-argument quoted below ?

On 9/14/2019 at 12:24 AM, Mordred said:

Surface power density isn't applicable in this application. The cosmological constant is a scalar field there is no force involved. A force requires a vector field. The EM field is an example as you have two charges.

The cosmological constant doesn't have a charge nor inherent vector direction.

Edited by stephaneww

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