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The solution of the Cosmological constant problem ?

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outch, read

 

"and validates the measurements [math]H_0=67.4[/math] of Planck's and DES (which are exactly identical) (and not 64.4)

Edited by stephaneww

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Well, I'm going to move forward in minefields because what I'm trying to understand about electromagnetism is very recent and I'm not sure I've assimilated everything.

 

On 9/14/2019 at 1:05 AM, Mordred said:

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

 

On 9/19/2019 at 6:47 AM, Mordred said:

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 ? 

 

On 9/19/2019 at 8:57 AM, Mordred said:

The thing to keep in mind is unlike say a pressure tank, a battery or a capacitor where energy can be stored and power is transferred at a rate to perform work on some external to the storage device the universe doesn't perform work outside of itself. All the energy is contained within our universe, there is no transfer of heat, energy or mass from outside our universe from inside to outside our universe. Our universe is in essence an isolated system (speaking thermodynamically) an isolated system cannot perform work outside of itself.

 

On 9/19/2019 at 6:47 AM, Mordred said:

 [math]power=\frac{work}{time}[/math]

I hope I'm not saying too much nonsense:

with 
[math]q_1=q_2=q_p= 1.8755459*10^{-18} coulombs \text{ : Planck charge }[/math] source
and 
[math]r=l_p \text{ : Planck length}[/math] source

according to Coulomb's law (source) we have the Coulomb force:

[math]F_c=k_e\frac{q_1q2}{r^2}=F_p[/math]

where [math]F_p[/math] is the Planck force source

we have seen that the energy density by volume of the cosmological constant can be written:

[math]\epsilon_\Lambda = \frac{F_p \Lambda}{8 \pi}[/math] source

as [math]F_c=F_p[/math] it can be safely stated that the volume energy density of the cosmological constant also has an electromagnetic character.

In order to apply Coulomb's law to two "quantum particles" of vacuum, it is necessary to assume that they have the same charge to account for the observation, namely accelerated expansion,

As soon as time "elapses" the two void charges distant from [math]r=l_p[/math] at the beginning could give way to a new vaccum charge as soon as the two initial charges are far from [math]l_p *2[/math] but this remains to be demonstrated,

Moreover, the more "vacuum particles" there will be, the more the amount of energy from the vacuum should, according to this hypothesis, increase,

it seems possible by combining Coulomb's strength and the passage of time to explain the expansion of the universe, or even the acceleration of its expansion.

All opinions are welcome, although I hope Mordred's opinion is more than welcome.

Edited by stephaneww
source's links

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... and since the vacuum of the universe is not completely empty, I let you guess what happens to some methods of measuring the energy density of the vacuum if there is dust or a galaxy in the area. In short, the resulting relative permittivity for each point in the universe will give a different measure and therefore an incorrect value of the cosmological constant. The important thing is therefore to find measurement methods that are not influenced by this phenomenon of relative permittivity of the universe vacuum. 

Edited by stephaneww

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How would you apply the coulomb force to the cosmological constant when the coulomb force is specifically the force between two charged particles ? 

 I'm not sure how you can classify an electromagnetic character from the above.

Don't confuse q for the deceleration parameter with charge.

Edited by Mordred

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of course, first, thank you for your reaction :)

55 minutes ago, Mordred said:

How would you apply the coulomb force to the cosmological constant when the coulomb force is specifically the force between two charged particles ? 

 I'm not sure how you can classify an electromagnetic character from the above.

edit 2 : I need more time to try to propose a correct answer :)

edit :

55 minutes ago, Mordred said:

Don't confuse q for the deceleration parameter with charge.

I don't understand, can you please develop ?

 

 

Edited by stephaneww

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3 hours ago, Mordred said:

How would you apply the coulomb force to the cosmological constant when the coulomb force is specifically the force between two charged particles ? 

It is the Planck force, = Coulomb force in Planck units (if you want I put the demonstration) , which applies to the cosmological constant to find the energy density of the cosmological constant , The electromagnetic character is given by two charges of Planck to the square (q1=q2=qp) of the same sign that enter into the calculation of the Coulomb force.

1321572061_loiceCoulomb.png.0ff1a69cfca13c8405a26420601401e2.pngsource :

 

 

As a reminder, electromagnetism does not happen by chance in this thread: we have already seen [math]W/m^2[/math] several times. For example, the [math]m_p/t_p^3[/math] which is of the same order of magnitude as the vacuum catastrophe but dimensioned in [math]W/m^2[/math]

I hope, without being sure, that I answered your question correctly 

 

3 hours ago, Mordred said:

I'm not sure how you can classify an electromagnetic character from the above.

edit: when we talk about Planck's charge, it means "electromagnetism" somewhere, right? 

Edited by stephaneww

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21 hours ago, Mordred said:

How would you apply the coulomb force to the cosmological constant when the coulomb force is specifically the force between two charged particles ?

The energy density of the vacuum is affected by the relative permittivity of the vacuum in a given area (local inhomogeneity), which should affect the measurement of the Hubble constant and thus ultimately the determination of the cosmological constant (itself directly a function of the Hubble constant)

Did I answer better at this point ?

Edited by stephaneww

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Nope you didn't the coulomb force is the between electrically charged particles. There is a significant difference between the behavior of the cosmological constant and a charged field of particles. An EM field has directional components. While the cosmological constant has no inherent direction.

The one link you had above uses the deceleration parameter q. This is the Freidmann deceleration equation (though often it's called the acceleration equation ) it's properly called the former. Charge can also be represented by q. However they have two distinct usages.

Now the mediator boson between two EM charged particles that carry the coulomb force is the photon. That would also affect the blackbody temperature of the CMB. The cosmological constant does not contribute to the blackbody temperature that we know of. (For one we don't even know if it involves particles) if it did then it's density would decrease as the universe expands ie not be constant...

Edited by Mordred

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6 hours ago, Mordred said:

Now the mediator boson between two EM charged particles that carry the coulomb force is the photon. That would also affect the blackbody temperature of the CMB. The cosmological constant does not contribute to the blackbody temperature that we know of. (For one we don't even know if it involves particles) if it did then it's density would decrease as the universe expands ie not be constant...

I'll never be able to answer at this point. it's too complicate for me.. :(

 

6 hours ago, Mordred said:

The one link you had above uses the deceleration parameter q. This is the Freidmann deceleration equation (though often it's called the acceleration equation ) it's properly called the former. Charge can also be represented by q. However they have two distinct usages.

The same image is on wkipedia.org at the page Coulomb's law, so I don't know how to consider your answer on this point. Can you light me please ?

 

6 hours ago, Mordred said:

An EM field has directional components. While the cosmological constant has no inherent direction.

Can I say as you that the "field of Planck force" have a directional component. While… ???

The 2 forces (Planck or Coulomb in unit's Planck) are in the energy density of the cosmology constant, no ? of course the cosmological constant have no direction. I'm agree with you on that point

 

6 hours ago, Mordred said:

Nope you didn't the coulomb force is the between electrically charged particles. There is a significant difference between the behavior of the cosmological constant and a charged field of particles.

I'm not sure of the following argument :

while searching on search engines I found this document published here which I am unable to say if it is correct and/or if it is very speculative.

I only understood this beginning of the conclusion but I don't know if it's valid once again

Quote

To summarize, we have demonstrated that the uniformand isotropic evolution of a macroscopic Coulomb system of identical particles with or without an oppositely charged background obeys equations that have the form of the FLRW cosmological equations.

it would seem from this document that Coulomb's force could be applied to particles without charge and regain the FLRW logic

Edited by stephaneww

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That particular paper is speculative and relies upon a negative charged cosmological constant with which there is no current evidence for. I wouldn't place any faith in it. 

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17 minutes ago, Mordred said:

That particular paper is speculative and relies upon a negative charged cosmological constant with which there is no current evidence for. I wouldn't place any faith in it. 

Okay, thank you very much. I was right to take all these precautions about its validity then :)

 

and what about this argument please ?

6 hours ago, stephaneww said:

 

12 hours ago, Mordred said:

An EM field has directional components. While the cosmological constant has no inherent direction.

Can I say as you that the "field of Planck force" have a directional component. While… ???

The 2 forces (Planck or Coulomb in unit's Planck) are in the energy density of the cosmology constant, no ? of course the cosmological constant have no direction. I'm agree with you on that point

 

 

Edited by stephaneww

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Any field with a directional component would describe a vector field. The cosmological constant we know is a scalar field by how galaxies seperate from one another.

 If you had a directional component then galaxies would seperate with an inherent direction they don't there seperation preserves the angles between galaxies you cannot have that in a anistropic expansion. You can only have that angle preservation with an isotropic expansion.

Google balloon analogy a try placing dots on a ballloon then measure the angles both before and after you inflate the balloon.

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um ok, but how to interpret the presence of these two forces in the energy density of the cosmological constant then ? It escapes me.

 

are there only scalar ?

Edited by stephaneww

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Why would you feel they are present in the energy density of the cosmological constant ? There is zero evidence that the coulomb force is involved.

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uh because [math]c^4/G=F_p[/math] is a component of the energy density of the cosmological constant... link 1 link 2

and that [math]F_p=F_c=k_e (q_p)^2/l_p^2[/math]  with [math]F_c[/math] Coulomb force in Planck units except that we need to have a signed charge (but as qp is squared the numerical result is always Fp)

Edited by stephaneww

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Using Planck force isn't an issue it does not require a charged field. However coulomb force does which is the issue with the cosmological constant. Planck force is a derived unit that doesn't depend on a particular field.

 The Coulomb force however is specifically for the EM field.

Edited by Mordred

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2 hours ago, Mordred said:

The Coulomb force however is specifically for the EM field.

 

Um, okay. Attempt to try to save my approach :

Quote

The orthodox interpretation of quantum mechanics presupposes division of the world into two parts: the observed quantum system and the observer. The observer does not have to be a man, it can be e.g. a computer, but it has to be "classical" i.e. one has to be able to say that a certain variable has a definite value and this statement has to have a clear meaning. 

Source , around the begining of page 76

Can the Coulomb force with Planck’s units can produce the Planck force for the observer whitout he can have access to the quantum EM field ?

Edited by stephaneww

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Um, something may be relevant this time:

on French wikipedia electromagnetic page:

Quote

Concepts
L'électromagnétisme dit classique correspond à la théorie « usuelle » de l'électromagnétisme, élaborée à partir du travail de Maxwell et Faraday. Il s'agit d'une théorie classique, car elle se fonde sur des champs continus, par opposition à la théorie quantique

translation  :

Quote

Concepts
The so-called classical electromagnetism corresponds to the "usual" theory of electromagnetism, developed from the work of Maxwell and Faraday. This is a classical theory because it is based on continuous fields, as opposed to quantum theory

it seems, if I understand correctly that classical electromagnetism cannot be directly applied to the quantum scale and if this statement found on wikipedia.fr is correct 

the QED issue remains to be resolved: are there any vector fields? My research did not allow me to find/understand the answer.

 

 

edit

Quote

From a classical perspective in the history of electromagnetism, the electromagnetic field can be regarded as a smooth, continuous field, propagated in a wavelike manner; whereas from the perspective of quantum field theory, the field is seen as quantized, being composed of individual particles.

source https://en.wikipedia.org/wiki/Electromagnetic_field

does this mean that  there is no vector field in the quantum approach ?

Edited by stephaneww

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1 hour ago, stephaneww said:

it seems, if I understand correctly that classical electromagnetism cannot be directly applied to the quantum scale and if this statement found on wikipedia.fr is correct 

the QED issue remains to be resolved: are there any vector fields? My research did not allow me to find/understand the answer.

The QED is a physical theory aimed at reconciling electromagnetism with quantum mechanics using a Lagrangian relativistic formalism. According to this theory, electrical charges interact by exchanging photons.

Edited by Kartazion

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I'm not going to be stubborn, you and Mordred say that the Coulomb's force with Planck's units approach is not possible

Thanks to both of you :)

edit:  a reason for my abandonment  on this subject has disappeared, but I fear that the reason is still valid

 

Edited by stephaneww

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36 minutes ago, stephaneww said:

I'm not going to be stubborn, you and Mordred say that the Coulomb's force with Planck's units approach is not possible

Thanks to both of you :)

I don't really know. I think you should continue.

Edited by Kartazion

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2 hours ago, stephaneww said:

Um, something may be relevant this time:

on French wikipedia electromagnetic page:

translation  :

it seems, if I understand correctly that classical electromagnetism cannot be directly applied to the quantum scale and if this statement found on wikipedia.fr is correct 

the QED issue remains to be resolved: are there any vector fields? My research did not allow me to find/understand the answer.

 

 

edit

source https://en.wikipedia.org/wiki/Electromagnetic_field

does this mean that  there is no vector field in the quantum approach ?

No vector fields apply to the quantum regime as well the treatments will differ somewhat when you account for the uncertainty principle and wavefunctions used in QM however through the principle of correspondence one must be able to produce the classical results in the macro world.

 Let's assume for a minute that the cosmological constant is charged. I would like you to consider what influence that would have on other charged particles. For example the Earth has a magnetic field. Take a look at the Van Allen belt. Now granted that is a spherical point source that arises from the Earth's magnetic core however one can certainly see that it affects the interstellar medium concerning other ionized particles.

Secondly if you have an electric field you must also have a corresponding magnetic field. These two fields have different polarities to one another. 

 In essence it would be impossible to have a charged field behave in the manner that the cosmological constant does. An EM field is polarized with two distinctive polarizations.

 The cosmological constant is a non polarized field. The spin 0 statistics has zero polarizations. While the EM field with spin 1 has two polarizations. Gravity at spin 2 has four polarizations. Hence it it easy to discern a gravity wave from an EM wave. The former is quadrupolar while the latter is dipolar. 

The cosmological constant has no known polarizations.

Now onto the term Observable. Under QM and QFT I'd something is observable it must have a quanta of action. It must must have sufficient mass/energy to induce kinematic change. Individual virtual particles by themself doesn't however an ensemble of VP can.

This is part of the distinction to what is often referred to as a real vs a virtual particle. (Though under particle physics) you can also have resonant and quasi particles. Resonant particles are unstable and have an extremely short mean lifetime. However resonant particles are measurable. Quasi and virtual particles are not measurable individually.

Coulomb force is measurable so it is observable.

Edited by Mordred

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Thank you very much (+1 :) ), it's very detailed and it makes me ask myself a lot of questions....

1 hour ago, Mordred said:

Now onto the term Observable... (2.) It must must have sufficient mass/energy to induce kinematic change....

(1.) Coulomb force is measurable so it is observable.

The first one (1.) 

What is the " format " of the force measurable above with Planck's units? A vector? A vector field? A scalar? A scalar field?

The next one (2.) :

I remember reading that Planck's force was the maximum value of a force in nature according to some physicists. Is that enough for it to have an effect on the micro scale of Planck?

Your answers could stop my enthusiasm :) 

... to be continued perhaps ...

 

Edited by stephaneww

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Think of Planck force as a theoretical maximum Observable force. Coulomb force between two electrons is a vector as force itself in any form is a vector quantity. Anytime you apply a force you are describing a vector.

Edited by Mordred

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You didn't stop me in my tracks, your answer corresponds exactly to what I expected so +1 :) 

 

6 hours ago, Mordred said:

The cosmological constant has no known polarizations.

I never said otherwise and I would never say the opposite. We are agree. 

 

 

6 hours ago, Mordred said:

No vector fields apply to the quantum regime as well the treatments will differ somewhat when you account for the uncertainty principle and wavefunctions used in QM however through the principle of correspondence one must be able to produce the classical results in the macro world.

as :

4 hours ago, Mordred said:

Planck force "is" a theoretical maximum Observable force. Coulomb force between two electrons is a vector as force itself in any form is a vector quantity. Anytime you apply a force you are describing a vector.

… And that I find the same values with my assumptions, can we conclude that in macro world, they are the same thing ? I have a doubt because the directions of the vectors is unknown.
 

 

6 hours ago, Mordred said:

Let's assume for a minute that the cosmological constant is charged. I would like you to consider what influence that would have on other charged particles. For example the Earth has a magnetic field. Take a look at the Van Allen belt. Now granted that is a spherical point source that arises from the Earth's magnetic core however one can certainly see that it affects the interstellar medium concerning other ionized particles.

Secondly if you have an electric field you must also have a corresponding magnetic field. These two fields have different polarities to one another. 

 In essence it would be impossible to have a charged field behave in the manner that the cosmological constant does. An EM field is polarized with two distinctive polarizations.

 The cosmological constant is a non polarized field. The spin 0 statistics has zero polarizations. While the EM field with spin 1 has two polarizations. Gravity at spin 2 has four polarizations. Hence it it easy to discern a gravity wave from an EM wave. The former is quadrupolar while the latter is dipolar. 

 

Okay, I'm persuaded. I still perhaps have to try to formulate a confusing thing that goes through my head.

 

Edited by stephaneww

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