# The solution of the Cosmological constant problem ?

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

Is that okay?

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

Edited by stephaneww

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

Planck power is one of those units that if it has a practical use it would involve an extreme energy transfer in one Planck time. (There is a reason why that link mentions some of the Planck constants has no practical use.)

Edited by Mordred

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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.

Edited by stephaneww

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

$c.F_p.\Lambda/8=0.50144.kg/s^3$

value of posts 2 and 3 page 5 :

Edited by stephaneww

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Have you calculated the constant value for Planck power?

You might want to do that first

$P_p=\frac{c^5 }{G}$

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I just saw here that:
$F_p=\frac{c^4 }{G}$
$P_p=\frac{c^5 }{G}= 3.62831×10^{52} W$

… and I have realized the calculation with $P_p=c.F_p$

Edited by stephaneww

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So given that extremely high value would you not say it represents a maximum power. Ie a theoretical maximum much like Planck length is the theoretical minimal observable length ?

Ie

Table 3 offers a sample of derived Planck units, some of which in fact are seldom used. As with the base units, their use is mostly confined to theoretical physics because most of them are too large or too small for empirical or practical use and there are large uncertainties in their values.

I would suggest it's value falls into that quoted criteria as being too large for any practical use.

Edited by Mordred

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I don't know if it's a maximum power. I'm not a theorist.
What I know about on this subject Planck's force is this :

Quote

Since 1993, various authors (De Sabbata & Sivaram, Massa, Kostro & Lange, Gibbons, Schiller) have argued that the Planck force is the maximum force value that can be observed in nature. This limit property is valid both for gravitational force and for any other type of force.

But I don't know if it's the same thing for the power of Planck.

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Planck units are typically extremum values they represent maximum or minimum theoretical bounds. Ie highest or lowest possible limits.

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Yes, generally. Even for Planck's mass?

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No  Planck mass isn't a fundamental limit. It's one of the few that isn't.

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However, if we do this simple calculation by dimensional analysis:
$m_p/t_p^3*2 \pi =8.72*10^{122}.kg/s^{3}$ that is the value of the vacuum catastrophe at less than 1%

But I don't know what to conclude from that...

and if I do, in my spreadsheet:

$m_p/t_p^3*1.994 \pi =8.72*10^{122}.kg/s^{3}$ it is the exact numerical value of the vacuum catastrophe

Edited by stephaneww

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Do you think that if we take 2 instead of 1,994 in post 3 on page 5, we could deduce the exact value, from the Planck units, of the cosmological constant value?

I am very tempted to answer "Yes" in the case of a flat universe

Edited by stephaneww

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Yes those equations work well. The cosmological constant value can vary a bit depending on dataset. So as long as you have a good approximation your doing good.

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Thank you very much.
I'm so happy that I put +1

edit :

Indeed, if we name X, the value of 1,994  the variation of the cosmological constant from datas, give X in the 3 equations.
Second indice, 2pi is the divisor for the Planck reduced constant.

edit  2

So $\Lambda$ would be equal to:
$\Lambda=0.5/0.5\frac{t_p^2.8 \pi}{h}=1.10242*10^{-52}m^{-2}$

Edited by stephaneww

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21 hours ago, stephaneww said:

c.Fp.Λ/8=0.50144.$kg.s^{-3}$

...And with power of Planck we have :

$\Lambda=\frac{0.5*8}{c F_p}=1.102422 *10^{-52}m^{-2}$

because the unit of 0.5 is $kg.s^{-3}$

edit :

Um, I didn't notice:

Quote

The Planck force is also associated with the equivalence of gravitational potential energy and electromagnetic energy[1] and in this context it can be understood as the force that confines a self-gravitating mass to half its Schwarzschild radius:

source :

Edited by stephaneww

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I'm going back to work to fill in the blanks and also because there's a lot of nonsense…

1*

On 8/29/2019 at 7:40 AM, Mordred said:

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

Okay, if I understand correctly, we have two scalar fields' ( I have revised the notion of partial derivative and its modern notation)

What can we do with these scalar fields $\phi_p \phi$ ? I don't understand their usefulness for calculations.

On 8/29/2019 at 7:40 AM, 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

... and I don't know how to use gradients.

2*

In this post,

there's been 2 big mistakes.

-Indeed, if

$\Lambda_{s^{-2}}=\Lambda_{m^{-2}}.c^2$

then $(\Lambda_{s^{-2}})^2/c^4=(\Lambda_{m^{-2}})^2$

The last line "$D=$", of the first post, $=B$ (demonstration of geometric mean page 2) becomes :

$D=B=h.(\Lambda_{s^{-2}})^2/c^3/(8∗\pi)^2=6.11∗10−133.kg.m^{−1}.s^{−2}$

-the second big error is on the value of $H_{p,minimum}$ , we do not fond a minimum but only $\sqrt{\Lambda_{s^{-2}}}=3.152*10^{-18}.s^{-1}$

3*

consequences for this post

$R$, $V$, $\rho_c$ are no longer extreme values,

and $\Omega_{\Lambda}=\frac{1}{3}$.

One might think that the cyclical universe is collapsing, but there is still this indicator that ,perhaps, remains valid :

$\frac{V.\rho_c.\Omega_{\Lambda}.c.\pi}{V}=\frac{1}{1.994}kg/s^3 \text{ or W/}m^2$

$=\frac{c^4.\Lambda.c.\pi}{8 . \pi G}=\frac{1}{1.994}kg/s^3$

$=\frac{F_p.\Lambda.c.\pi}{8 . \pi }=\frac{F_p.\Lambda.c}{8}=\frac{1}{1.994}kg/s^3$

But even with that, it is not certain that the universe is cyclical.

4*

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

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 agree with you on that point.

5*

I still have to learn electromagnetism to try to understand this power surface density...

Edited by stephaneww

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I started to take a closer look....

Beforehand we will assume that $\frac{1}{2} W/m^2$ is the value that accurately replaces $\frac{1}{1,994} W/m^2$.

We have the energy density of the cosmological constant =$\frac{c^4 \Lambda}{8 \pi G}. J/m^3$

Quote

The intensity of electromagnetic radiation can be expressed in W/m2

Let's follow the "intensity" link. We find :

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).

By following the velocity link, we have :

Quote

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north). Velocity is a fundamental concept in kinematics, the branch of classical mechanics that describes the motion of bodies.

Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s) or as the SI base unit of (m⋅s−1). For example, "5 metres per second" is a scalar, whereas "5 metres per second east" is a vector. If there is a change in speed, direction or both, then the object has a changing velocity and is said to be undergoing an acceleration.

6 hours ago, stephaneww said:

$\frac{c^4 \Lambda}{8 \pi G} .c.\pi=\frac{1}{2} W/m^2$

so :

$\frac{c^4 \Lambda}{8 \pi G} .c.2.\pi=1. W/m^2$ i.e

Quote

energy density (energy per unit volume) at a point in space and multiplying it by the velocity

where $c$ is the speed and $2 \pi$ is the ..

Quote

..direction of motion

so whatever point in the universe is considered, and if I don't say something stupid again, each point in the universe follows the perimeter of a circle.

as

Quote

If there is a change in speed, direction or both, then the object has a changing velocity and is said to be undergoing an acceleration.

its direction is tangent to the circle so "it is undergoing an acceleration."

I don't know if this first approach is correct.

If anyone can help me, thank you in advance

edit :
Um, there may be a problem with the

Quote

Edited by stephaneww

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13 hours ago, stephaneww said:

Okay, if I understand correctly, we have two scalar fields' ( I have revised the notion of partial derivative and its modern notation)

What can we do with these scalar fields ϕpϕ ? I don't understand their usefulness for calculations.

Λs2=Λm2.c2

Λm2)2

D=B=h.(Λs2)2/c3/(8π)2=6.1110133.kg.m1.s2

How to use the scalar field equation can get tricky. It's commonly used to model inflation as well as DE. So here is an opener to better understand the equation.

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6 hours ago, Mordred said:

How to use the scalar field equation can get tricky. It's commonly used to model inflation as well as DE.

Indeed, this presentation is far too difficult for me...

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On 9/30/2019 at 7:30 PM, stephaneww said:

On 9/30/2019 at 2:06 PM, stephaneww said:

$\frac{c^4 \Lambda}{8 \pi G}.c . \pi=\frac{1}{2} W/m^2$

so :

$\frac{c^4 \Lambda}{8 \pi G}.c .2. \pi=1 W/m^2/$ i.e.

Quote

energy density (energy per unit volume) at a point in space and multiplying it by the velocity

Edited by stephaneww
latex

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after an exchange with an electromagnetism specialist, this is false: $2 \pi$ is only a scalar

On 9/30/2019 at 7:30 PM, stephaneww said:

2π is the ..

Quote

..direction of motion

so whatever point in the universe is considered, and if I don't say something stupid again, each point in the universe follows the perimeter of a circle.

as

Quote

If there is a change in speed, direction or both, then the object has a changing velocity and is said to be undergoing an acceleration.

its direction is tangent to the circle so "it is undergoing an acceleration."

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Hello.

I don't have much free time to continue studying electromagnetism right now. But I'll get back to it as soon as I can.

Just a remark on the orders of magnitude. We are often surprised by the small value of the cosmological constant. It is directly linked to Planck's power by the relationship :

$\Lambda=\frac{4}{P_p}$ (remember the relationship  of power surface density power expressed in $W/m^2$ or $kg/s^3$)

On 9/20/2019 at 1:55 PM, stephaneww said:

$P_p=\frac{c^5}{G}$     =3.62831× $10^{52} W$

Edited by stephaneww

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for the relationship to work correctly in terms of dimensions, it is also necessary to remember that we have:

$P_p \Lambda / 4 = 1. kg/s^3$
and
$P_p = 3.62831 .kg.m^2/s^3$

Edited by stephaneww

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I don't have much time available right now.

I have only noticed that the second equality of this post:

could also be written:
energy density of the cosmological constant =

$5,243 *10^{-10}J/m^3=\frac{1.W/m^{2}}{2 \pi c}$

in other words, energy density of the cosmological constant could be defined only by $c$.

which again highlights the importance of the unit W/m^2 (=kg/s^3) in cosmology and validates the measurements $H_0=64.4$ of Planck's and DES (which are exactly identical) :

Which could extinguish the "tension on the current Hubble constant" and validate the model $\Lambda CDM$ once again

Edited by stephaneww

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