The solution of the Cosmological constant problem ?

[Mordred]

Does a coulomb of force sound like the Planck force ? Perhaps you should look up the values of each they differ significantly in value. Secondly if you have two electrons the coulomb force between then isn't unknown in the direction. Remember your rules for two charges of the same polarity as opposed to two opposite charges.

 The Planck force is a theoretical upper limit to the maximum Observable force. Such a force would only occur at cosmological horizons such as a black holes event horizon. 

The vector form of Coulomb's law is simply the scalar definition of the law with the direction given by the unit vector, r̂₂₁, parallel with the line from charge q₂ to charge q₁.^[26] If both charges have the same sign (like charges) then the product q₁q₂ is positive and the direction of the force on q₁ is given by r̂₂₁; the charges repel each other. If the charges have opposite signs then the product q₁q₂ is negative and the direction of the force on q₁ is given by −r̂₂₁ = r̂₁₂; the charges attract each other.

https://en.m.wikipedia.org/wiki/Coulomb's_law

 

 

Edited December 17, 2019 by Mordred

[stephaneww]

13 hours ago, Mordred said:

Does a coulomb of force sound like the Planck force ?

yes, I have again forgot the problem of the charge …  

13 hours ago, Mordred said:

Secondly if you have two electrons the coulomb force between then isn't unknown in the direction.

do you want to mean  that the direction of Planck force is unknown as second difference ?

13 hours ago, Mordred said:

Perhaps you should look up the values of each they differ significantly in value.

uh, in this particular case, their values are equal, right?

On 12/16/2019 at 8:13 AM, stephaneww said:

[math]F_p=F_c=k_e(q_p)^2/l_p^2[/math]

Edited December 17, 2019 by stephaneww

[stephaneww]

hum, if we reason in terms of pressure:

[math]P=\frac{F}{S}[/math] for the matter (we will then assume that the two charges of matter are opposite for the charge of Coulomb force, gravity opposes at the force of the pressure on a surface in a Newtonian context) 
For the pressure of the cosmological constant [math]P_\Lambda=-\frac{F}{S}[/math] the two vacuum charges are the same? (this would only be a question of a convention on the charges that could be reversed). 

Hence the often encountered presentation of a cosmological constant as an "antigravity" and the parallel between the two laws in inverse square of distance

...Coulomb force  (F=P*S) would be positive for matter because two "charges of atom of matter" would be opposite (before by convention, identical) and 

Coulomb force  woukd be negative for vaccum because two "charges of vaccum at Planck scale" would be identical (before by convention, opposed) ????

but I wonder if I'm not displacing my problem ????

https://www.toutcalculer.com/mecanique/pression-force-surface.php (the opposite of pressure comes from G in a Coulomb force context with "two opposite charges")  

and the pressure of the cosmological constant is negative in a Coulomb force context with "two identical charges". 

https://fr.wikipedia.org/wiki/Constante_cosmologique#Interprétation_physique

(sorry but  I haven't found a source in English) 

 

Edited December 17, 2019 by stephaneww

[stephaneww]

 

2 hours ago, stephaneww said:

before by convention, opposed

ie electrically neutral at macro scale with this convention

Edited December 18, 2019 by stephaneww

[stephaneww]

erased : I haven't enought think

Edited December 18, 2019 by stephaneww

[Strange]

8 hours ago, stephaneww said:

erased : I haven't enought think

Just wanted to say that "I haven't enough think" is an excellent phrase and should be used more.  

[stephaneww]

43 minutes ago, Strange said:

Just wanted to say that "I haven't enough think" is an excellent phrase and should be used more.  

lol, yes

 

edit,

 

I have finish thinking 😊 : What do you think of that? Could this be interesting for a string theory?

[math]\Lambda^2/16=G^2/c^{10}[/math] with this numerical value of  [math]\Lambda=1.10242*10^{-52}[/math] calulated here

note : we would have 10 length dimensions for  [math]c[/math] 

Edited December 18, 2019 by stephaneww

[stephaneww]

I have just forgot to add the [math]\Lambda^2[/math] appears here in [math]B[/math] to give a mathemathic solution of the problem of the cosmological constant

 

come bak to the Coumb force.. :

18 hours ago, stephaneww said:

On 12/17/2019 at 7:07 AM, Mordred said:

Secondly if you have two electrons the coulomb force between then isn't unknown in the direction.

do you want to mean  that the direction of Planck force is unknown as second difference ?

Ok , you talk about, identical charges, I believe it's not a problem

Edited December 18, 2019 by stephaneww

[stephaneww]

I may have found something to support my idea. I don't usually go on viXav, but after reading the beginning of this paper, I do not think that everything should be rejected.

The equalities (7) and (9) show that there is "equality" between [math]F_G[/math] and [math]F_C[/math], cf (10) :[math]F_G/F_C=1[/math] But this assumes, for the Coulomb force, that [math]N_1[/math] and [math]N_2[/math] are opposite charges, in other words, if I understand, it reverses this usual convention between the signs of the two charges, for the determination of an attractive or repulsive effect, (with our convention we have [math]F_G/F_C=-1[/math] )

On 12/17/2019 at 7:07 AM, Mordred said:

Quote

The vector form of Coulomb's law is simply the scalar definition of the law with the direction given by the unit vector, r̂₂₁, parallel with the line from charge q₂ to charge q₁.^[26] If both charges have the same sign (like charges) then the product q₁q₂ is positive and the direction of the force on q₁ is given by r̂₂₁; the charges repel each other. If the charges have opposite signs then the product q₁q₂ is negative and the direction of the force on q₁ is given by −r̂₂₁ = r̂₁₂; the charges attract each other.

https://en.m.wikipedia.org/wiki/Coulomb's_law

.

 

On 12/17/2019 at 7:07 AM, Mordred said:

Remember your rules for two charges of the same polarity as opposed to two opposite charges.

with the convention of the paper we would have,for F_G ,  the product N1.N2 is positive, for F_C ,  the product N1.N2 is negative and the direction of the force on N1 is given by r̂21; the charges repel each other. (opposite to the attraction F_G/F_C=-1, from (10) )

 

Edited December 19, 2019 by stephaneww

[stephaneww]

I put the equalites beween (7) and  (9) here and try to be more clear :

[math]F_G=F_C=N_1 N_2\Large{ \frac{ \hbar c}{r^2}}[/math],

[math]N_1[/math] and [math]N_2[/math] are quantifications of masses and charges

but with our conventions, as [math]N_1[/math] have the same sign than [math]N_2[/math] , [math]F_G[/math]=[math]-F_G[/math], ie [math]F_G[/math] is attractive and [math]F_C[/math] is repulsive. 

to have the equivalence between  [math]F_G[/math]  and  [math]F_G[/math] we need to inverse our charges signs convention (+ -)  in (+ +) or (- -),

so our charges of same sign will become (- +) or (+ -)

Edited December 19, 2019 by stephaneww

[stephaneww]

I have contacted the author, who agrees that the issue of the signs of the charges needs to be corrected, while saying that it is trivial.  

 I'm agree with him

Edited December 19, 2019 by stephaneww

[stephaneww]

🤬🤬🤬🤬

8 hours ago, stephaneww said:

N1 and N2 are quantifications of masses and charges

but with our conventions, as N1 have the same sign than N2 , FG =-FC , ie FG is attractive and FC is repulsive. 

to have the equivalence between  FG   and  FG  we need to inverse our charges signs convention (+ -)  in (+ +) or (- -),

so our charges of same sign will become (- +) or (+ -) 

correction in this quote… sorry
 

Quote

 

but with our conventions, as N1 have the same sign than N2 , FG =-FC , ie FG is attractive and FC is repulsive, so -FC is attractive too.

 to have  same sign for  FG  and  ...[math]F_C[/math].. we need to inverse our charges signs of our convention (+ -)  in (+ +) or (- -) for the two charges, signs of N1 and N2 are  absolute values for FG

so our charges of same sign will become (- +) or (+ -) 

 

I hope that's ok this time 😴

Edited December 19, 2019 by stephaneww

[stephaneww]

what happens to that statement then?

On 12/14/2019 at 6:34 AM, stephaneww said:

as Fc=Fp 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,

 in the electrically neutral quantum vacuum my  ++ (or (- -) ) problem, becomes (+ -) and the two charges repel each other with the paper convention mentioned above.

Let us now consider two planck particles N1 and N2 whose Planck charges [math]e_p[/math] are of opposite sign with their two masses [math]m_p[/math],
(electrically neutral on a macro scale)
we set [math]d=1m[/math] or any other value:

[math] F_C=k_e (e_p^{+}e_p^{-})/d^2=-3,16152*10^{-26} N /d^2[/math]

[math] F_G=G (m_p m_p)/d^2[/math]

[math] {F_C}= {F_G} = {k_e (e_p^{+}e_p^{-})}= {G m_p^2}[/math]

[math]-3,16152*10^{-26} N=(-1)*G  m_p^2[/math]

[math]-3,16152*10^{-26}/G  \text{ } kg=(-1)  m_p^2[/math]

the -1 indicates that one will have a repulsive force with the   new convention

so

[math] m_p= 2,17647*10^{-8} kg[/math] ie the exact value of planck's mass 

planck's mass is found from his charge using Coulomb force and gravitational force.

The Coulomb force tells us that this force is repulsive between the two charges according to what was initially proposed.

 

 

 

Edited December 19, 2019 by stephaneww

[stephaneww]

of course I have forgot something to be clear:

FG=FC=N1N2 [math]\frac{\hbar c}{r^2}[/math]

N1 and N2 are quantifications of masses and charges

but with our conventions, as N1 have the same sign than N2 , FG =-FC , ie FG is attractive and FC is repulsive, so -FC is attractive too.

_____________________________

in the new convention of the paper quoted :

 to have  same sign for  FG  and  ...FC.. we need to inverse our charges signs of our convention (+ -)  in (+ +) or (- -) for the two charges, signs of N1 and N2 are  absolute values for FG

so our charges of same sign will become (- +) or (+ -) 

if  N1 have the same sign than N2 , FG =FC , ie FG is attractive and FC is attractive too,

and -FC is repulsive for two opposite  sign charges, with FG  attractive

 

Edited December 19, 2019 by stephaneww

[Mordred]

Well quite frankly I wouldn't trust that reference or take it too far. The author is drawing conclusions based on relations between two equations to start with. That would be like stating plumbing is the same as electricity as the fundamental formulas has similarities in ratios of change.

Gravity is only attractive never repulsive, gravity is symmetric in its stress tensor while the EM stress tensor is antisymmetric.

Gravity has spin 2 while EM has spin 1. Those differences the Author ignores he doesn't look into the gauge groups of each field or look at the polarization differences. Nor does he note that you can shield an EM field but not a gravitational field 

 

Edited December 20, 2019 by Mordred

[Markus Hanke]

On 12/20/2019 at 3:03 AM, Mordred said:

Gravity has spin 2 while EM has spin 1. Those differences the Author ignores he doesn't look into the gauge groups of each field or look at the polarization differences. Nor does he note that you can shield an EM field but not a gravitational field 

Plus of course, the dynamics of EM are linear, while those of gravity are non-linear.

[stephaneww]

To be honest, I am unable to argue alone with my knowledge of the two posts above.

I note however that the author of this document (first version already seen diagonally [and approved(?)] by Mordred here) has known developments, notably eq(XI-34) page 73 in V3 which claims as me that Fc=Fg (making the same mistake on the sign of the charges as the viXra document) ie by adopting my "new personal convention" on the signs of the charges.

I am not able to understand what allows the author to arrive at eq(XI-34) page 73. Only a diagonal or in-depth examination, by one or both of you, will be able to say that the author's detailed argument is consistent in responding to your objections. 

 

Edited December 22, 2019 by stephaneww

[stephaneww]

1 hour ago, stephaneww said:

Only a diagonal or in-depth examination, by one or both of you, will be able to say that the author's detailed argument is consistent in responding to your objections. 

or not, of course...

[stephaneww]

about this:

On 12/14/2019 at 6:34 AM, stephaneww said:

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,

who can explain to me why Dirac's "sea" of negative-energy electrons is false? In other words, what are the facts that make it incorrect and cannot be applied to my quotation above?

Quote

 

Solutions of the Dirac equation contained^{[clarification needed]} negative energy quantum states. As a result, an electron could always radiate energy and fall into a negative energy state. Even worse, it could keep radiating infinite amounts of energy because there were infinitely many negative energy states available. To prevent this unphysical situation from happening, Dirac proposed that a "sea" of negative-energy electrons fills the universe, already occupying all of the lower-energy states so that, due to the Pauli exclusion principle, no other electron could fall into them. Sometimes, however, one of these negative-energy particles could be lifted out of this Dirac sea to become a positive-energy particle. But, when lifted out, it would leave behind a hole in the sea that would act exactly like a positive-energy electron with a reversed charge. These holes were interpreted as "negative-energy electrons" by Paul Dirac and by mistake he identified them with protons in his 1930 paper A Theory of Electrons and Protons^[3] However, these "negative-energy electrons" turned out to be positrons, and not protons.

This picture implied an infinite negative charge for the universe—a problem of which Dirac was aware. Dirac tried^{[clarification needed]} to argue that we would perceive this as the normal state of zero charge. Another difficulty was the difference in masses of the electron and the proton. Dirac tried^{[clarification needed]} to argue that this was due to the electromagnetic interactions with the sea, until Hermann Weyl proved that hole theory was completely symmetric between negative and positive charges. Dirac also predicted a reaction e⁻ + p⁺ → γ + γ , where an electron and a proton annihilate to give two photons. Robert Oppenheimer and Igor Tamm proved that this would cause ordinary matter to disappear too fast. A year later, in 1931, Dirac modified his theory and postulated the positron, a new particle of the same mass as the electron. The discovery of this particle the next year removed the last two objections to his theory.

Within Dirac's theory, the problem of infinite charge of the universe remains. Some bosons also have antiparticles, but since bosons do not obey the Pauli exclusion principle (only fermions do), hole theory does not work for them. A unified interpretation of antiparticles is now available in quantum field theory, which solves both these problems by describing antimatter as negative energy states of the same underlying matter field i.e. particles moving backwards in time.^[4]

 

source :  https://en.wikipedia.org/wiki/Antiparticle#Dirac_hole_theory

 

Thank you in advance

 

edit

I don't fully understand this paragraph and its implications. 

Quote

Within Dirac's theory, the problem of infinite charge of the universe remains. Some bosons also have antiparticles, but since bosons do not obey the Pauli exclusion principle (only fermions do), hole theory does not work for them. A unified interpretation of antiparticles is now available in quantum field theory, which solves both these problems by describing antimatter as negative energy states of the same underlying matter field i.e. particles moving backwards in time.^[4]

 

Edited December 29, 2019 by stephaneww

[Kartazion]

This means that the Dirac Sea has as much antielectron as there is room in the negative energy (problematic).

Since there were infinitely many negative energy states available, and thanks to the Pauli exclusion principle, we can fill the negative energy space with all possible antielectron states.

If an electron appears on the surface (positive energy) then it creates a hole in negative energy which is identified as positron.

However for the bosons I do not know how it works with the antimatter.

[Mordred]

The Dirac sea was originally developed when there were only two known particles. The electron and the proton. The negative energy states which make up the sea were filled with the positive energy states. However the theory didn't consider spin zero particles ie bosons which you can have any number of bosons of the same state in the same space. This in itself meant there is no limit of bosons that could fill a hole.

 Some bosons such as the photon are it's own antiparticle as it is charge neutral. 

 It was the problem of bosons and advent of QFT that made the Dirac sea or hole theory problematic and essentially became useless. In essence you could have an infinite number of bosons in every hole.

Edited December 30, 2019 by Mordred

[stephaneww]

Um, I'm not sure I got it all figured out.

Here's what I'm getting at:

11 minutes ago, Mordred said:

 It was the problem of bosons and advent of QFT that made the Dirac sea or hole theory problematic and essentially became useless. In essence you could have an infinite number of bosons in every hole.

Thank you both.

[Kartazion]

2 hours ago, Mordred said:

However the theory didn't consider spin zero particles ie bosons which you can have any number of bosons of the same state in the same space. 

You also meant spin 1, right? Or did you specify that the scalar boson of spin 0 is the Higgs boson?
I understand that there can be as many bosons as wanted in the same state in the same space, because it does not respect the Pauli exclusion.
 

2 hours ago, Mordred said:

It was the problem of bosons and advent of QFT that made the Dirac sea or hole theory problematic and essentially became useless.

This means that the QFT gets to explain it better and differently?

[Mordred]

Any integer spin either 1 or zero as most bosons are spin 1 with the exception of Higgs and possibly graviton spin 2 assuming gravitons exist.

[stephaneww]

Hello,

New question: can the reasoning of this post be used for charges of a plane electromagnetic wave that vibrates in phase, using the classical conventions for the plane/axis (?) of electrical charges and then for the plane/axis of magnetic charges. 
If I understood well in this case the sign of the electric charge would be identical to the sign of the magnetic charge. With these assumptions and it would seem, in my opinion, that we can apply the reasoning of the post cited above. What is not correct this time or needs clarification?

Intuitively I would add that it could have something to do with the speed of light and the power density of the relation seen here which gives the energy density of the cosmological constant 

Edited January 10, 2020 by stephaneww