stephaneww

Oops. I knew it was the base, but we have to divide by 8 in more. . Do

[math]c.F_p.\Lambda/8=0.50144.kg/s^3[/math]

value of posts 2 and 3 page 5 :

https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1117499

https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1117557

53 minutes ago, Mordred said:

Fair enough however Planck power has a specific value. It would be interesting to see how you match that value with the cosmological constant.

Simply by the power surface power density where the cosmological constant of dimension L^-2 acts as a division by a area.

All right.
Much simpler approach:
Planck power = c .Fp
source:

https://en.wikipedia.org/wiki/Planck_units#Derived_units

Is that okay?

… and thank you for all your counter arguments that force me to look for valid scientific arguments for my model

If you follow me back on the subject...

Planck's force (which can be highlighted in the energy density of the cosmological constant) can be considered as a tension:

https://en.wikipedia.org/wiki/Planck_force#Planck_force_as_a_tension_constant_of_the_space_time_fabric

on French wiki we have also :

Quote

La force de Planck n'est dérivée que de la constante de gravitation universelle de Newton et de la vitesse de la lumière, qui sont constantes partout dans l’espace. Elle caractérise donc une propriété de l'espace-temps⁴.

translation :

Quote

Planck's force is derived only from Newton's universal gravitational constant and the speed of light, which are constant throughout space. It therefore characterizes a property of space-time⁴.

link 4 is :

https://arxiv.org/pdf/1408.1820.pdf  

where you will find an attempt to define power associated with Planck's force.

[math]A[/math] and [math]B[/math] on page 2 (formulation of the geometric mean) are they associable with "dipole boundaries"? 

c*pi, could it be associated with "electrical current"? 

note: the formulas and values of posts 2 and 3 on page 5 are correct contrary to what I said earlier. (post 3 use [math]h[/math] instead [math]\hbar[/math]  

the whole question for me is how to interpret W/m^2 equivalent to kg/s^3. 

...for the moment I don't have an answer

In fact the 2nd and 3rd post (+ errors in the 3rd) on page 5 are most probably strictly irrelevant.....

At least I would have learned a lot of things

+1 to last post of Mordred

You didn't see a geometric mean as a solution to the problem of the cosmological constant either.

I may have to rework the presentation (one equivalent to energy density * speed of light) to make it more obvious. But, this equality of 1/1,994 W/m^2 is not a coincidence in my opinion. The equivalence is too exact.

edit :

Um, It is very easy to show that

[math]\frac{\hbar\Lambda}{8\pi t_p^2}.c = \frac{c^4\Lambda}{8\pi G}[/math]

just replace [math]t_p[/math] by  [math]\sqrt{\frac{\hbar G}{c^5}}[/math]

2 hours ago, Mordred said:

From the first link in English

As an electromagnetic wave travels through space, energy is transferred from the source to other objects (receivers). The rate of this energy transfer 

Where is the energy transfer for the cosmological constant ? 

Intuitively I would say that, over time, the transfer of energy from the cosmological constant goes into increasing the volume of the vacuum of the universe with ultimately an increase in the total energy of the vacuum.
It should be considered that the vacuum is the source and the receiver at the same time.
But I don't know if that's an acceptable answer.

I wrote this before your last post 

53 minutes ago, Mordred said:

now a consequence is that the more negative the pressure becomes the less the energy density decreases as the universe expands however energy is created as the universe expands by Lambda so its pressure is minus its energy density

Can you specify the energies involved, please, I don't understand everything

Edit, I understand 

53 minutes ago, Mordred said:

So how would you define power in this instance ?

I'll have to scratch my head for a moment.

I have a lot of notions to learn or review.

Without any guarantee of results...

… and thanks a lot for this link :

On 9/15/2019 at 9:26 AM, Mordred said:

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

https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf&ved=2ahUKEwjljPSGl9LkAhVOqp4KHbb8DtwQFjAQegQICRAB&usg=AOvVaw0xjZcR1b8OcNFbrriXGEr2

and the definition of a dot 

Um, actually, there's a problem with your whole explanation:
Indeed, you are dealing with power / volume unit [math]W/m^3[/math] ( everything you say in this frame is OK)  whereas I am talking about surface power density (power /area surface unit) [math]W/m^2[/math]
cf: 

On 9/13/2019 at 10:59 PM, stephaneww said:

surface power density : [math] kg/s^3 or  W/m^2[/math]

or in French :

https://fr.wikipedia.org/wiki/Densité_surfacique_de_puissance

On 9/15/2019 at 6:38 PM, Mordred said:

Total energy in essence is being created as the volume increases

ok but what happens to the surface power density ? 

correction: read [math]\hbar[/math]

On 9/13/2019 at 10:59 PM, stephaneww said:

[math]\frac{1}{2}   \frac{h}{t^2}\frac{Λ}{8\pi}=0.50144.kg/s^3=1/1.994.kg/s^3[/math]

 

[math]\frac{1}{2}   \frac{\hbar}{t^2}\frac{Λ}{8\pi}=0.50144.kg/s^3=1/1.994.kg/s^3[/math]

+1 for this of course 

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.

The votes are sympathetic but I prefer positive or opposition written reactions. It's clearer for me 

3 hours ago, Mordred said:

Under that condition the usage of the term Planck force could apply for each discrete quantization of spacetime. The result will be the number of Planck length units will be increasing as space expands.

In addition, it reinforces me on the validity of the formulation of [math]B[/math] in this message :

https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1115799

edit :

It is also a good basis for popularizing Unruh's paper

13 minutes ago, Mordred said:

well the result is energy isn't being transferred so power density isn't applicable. 

Agree for my understanding currently in cosmology. Finally we have succeeded!

5 minutes ago, Mordred said:

The cosmological constant doesn't follow conservation of energy at any time. Total energy in essence is being created as the volume increases. (Though don't think of energy as a substance) it's always the ability to perform work ie a property.

On this point we agree

37 minutes ago, Mordred said:

What is a unique feature of the cosmological constant when it comes to energy conservation ? Total energy isn't conserved. As the volume increases the total energy of Lambda is increasing. It isn't being transferred 

Um, not even  transferred to vacuum temperature? (CMB's) 

for the above in your message it is being studied in detail but we agree in a first approach 

13 minutes ago, Mordred said:

When your applying Planck units you are quantizing spacetime into discrete units. The Planck length never changes it is always the same value. So as the volume increases the number of Planck length units must also increase without increasing the Planck length itself. 

These terms get misleading as space is just volume but in essence space is being created everywhere that is not gravitationally bound.

I understand this 2 points very well

13 minutes ago, Mordred said:

A better descriptive is the use of geometric expansion

I don't know but I understand the principle for having represented it on a sheet of paper 

This is due to the fact that roughly speaking the quantity of material is conserved while for vacuum, it is the density of the vacuum that is conserved in an expanding space.

Arf I didn't pay attention that you edited to add this:

45 minutes ago, Mordred said:

Under that condition the usage of the term Planck force could apply for each discrete quantization of spacetime.

so the consequence escapes me. I don't understand what that implies for what I propose post 2 and 3 on page 5...

Sorry I edited this after your next message

but it comforts me on this point:

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 ".

26 minutes ago, Mordred said:

Every location of spacetime the  vector will be in every direction.  Ie space expanding. Under that condition the usage of the term Planck force could apply for each discrete quantization of spacetime.

This time, it's definitive. I have understood that you are right on this issue.  

6 hours ago, Mordred said:

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

https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf&ved=2ahUKEwjljPSGl9LkAhVOqp4KHbb8DtwQFjAQegQICRAB&usg=AOvVaw0xjZcR1b8OcNFbrriXGEr2

Thank you. I disconnect after"the energy equation becomes".
... because I don't remember what [math]\dot a[/math] 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.

Thanks in advance for your answer

To add to the current confusion, this new measure:
https://science.sciencemag.org/content/365/6458/1134

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

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. 

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 

(https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.ir.isas.ac.jp/~cpp/teaching/cosmology/documents/cosmology01-05.pdf&ved=2ahUKEwi0pdiqk9LkAhWxJjQIHRrZCW4QFjABegQIAxAB&usg=AOvVaw37DQF5eUpSBEoC_QefsvGg)

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?

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

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 [math]a=d. \Omega_\Lambda H_0^2.m/s^{-2}[/math] with [math]d[/math] =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.

All right, thank you again. 

 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 ?

1. https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1115799

2. https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1115805

3. https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1116338

4.https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1116658

(+ spéculatives posts related to 4. )

5. https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1116906

6. https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1116978

( 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.