The solution of the Cosmological constant problem ?

[Markus Hanke]

8 hours ago, stephaneww said:

What is not correct this time or needs clarification?

There are no magnetic charges; magnetic field “lines” either extend to infinity, or form closed loops.

[stephaneww]

um, thank you, you make me doubt, maybe I didn't transcribe correctly what I was told: and with this case of electromagnetic waves, is it better? 

 

Edited January 11, 2020 by stephaneww

[Mordred]

2 hours ago, stephaneww said:

um, thank you, you make me doubt, maybe I didn't transcribe correctly what I was told: and with this case of electromagnetic waves, is it better? 

 

Far better. The Maxwell well equations for the E and B fields. Electro is former, magnetism for latter. Have the 90 degree polarization shift in the GIF on that link.  Learn the Maxwell equations first. In order to do that you will need to study the vector properties of gradient, divergences and curl.

 Those you will need to understand polarizations of the EM field.

Pay particular under math of when the dot product or cross product of two field vectors apply.

I recommend Griffiths "Introductory to Electrodynamics" if you can get a hold of it. 

Symbolism is as follows use two vectors (actual letters don't matter) pay attention to the Operator )

[math] A\cdot B [/math] now if you deal with the inner product of two vectors you return a scalar. (Yes I know I am going too far but I want to give you direction of study) you also have outer products of two vectors. 

[math] A×B [/math] for cross product. One is linear the latter is perpendicular. (Hint )

Now one thing to take the above further. The Cosmological constant has no polarizations. However one can treat gravity vs the Cosmological constant under dipole treatments without repulsive charge. If one treats the two under vector direction. So you wouldn't use Colombia force laws under that examination.

 You would apply pressure via the equations of state.

[math] w=\frac{\rho}{p}[/math] under GR pressure is vector direction in the i direction. Ie [math]T_{ijk}[/math] T for arbitrary tensor. (See Euler angles) fir subscript.

 Now to give an example of why I mentioned all the above. Lets look at a common Cosmology application. Lets describe spacetime under the Minkowskii limit. The Minkowskii tensor is an orthogonal group that follows the following vector relations between  

[math]\mu\cdot\nu=\nu\cdot\mu[/math] this relation tells us that the inner dot (linear hint) product of the vectors  [math]\mu [/math] and [math]\nu [/math] are symmetric and hermitean/orthogonal.

With the EM fields you will not find the above to be the same case. (You will be dealing with the cross product ) which is antisymmetric. 

PS you will find the above relevant to the differences between two 90 degree phase shifted fields and fields that are not phase shifted in terms of the attempts you are making in a charged application of Lambda vs gravity. Ie one situation is symmetric while the other is not.

ie

[math]E×B[/math] vs [math]\Lambda\cdot g[/math]

g for gravitational potential at an arbitrary coordinate.

 

Edited January 11, 2020 by Mordred

[stephaneww]

Thank you but we'll have to proceed slowly, it's too much to learn at once...

Let's start with this, please:

 

14 hours ago, Mordred said:

Symbolism is as follows use two vectors (actual letters don't matter) pay attention to the Operator )

A⋅B now if you deal with the inner product of two vectors you return a scalar. (Yes I know I am going too far but I want to give you direction of study) you also have outer products of two vectors. 

A×B for cross product. One is linear the latter is perpendicular. (Hint )

can I have links (wiki ideally) that detail this please (the French notations seem different from the English notations) ?

 

14 hours ago, Mordred said:

A⋅B

and 

14 hours ago, Mordred said:

A×B

 

 

edit :

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,

 

can we go faster by noticing that the electric charge and the magnetic charge are of the same sign at the summits of their respective quantum waves ?

Edited January 11, 2020 by stephaneww

[Markus Hanke]

8 hours ago, stephaneww said:

can we go faster by noticing that the electric charge and the magnetic charge are of the same sign at the summits of their respective quantum waves ?

As I’ve pointed out earlier, there is no such thing as a “magnetic charge”.

[stephaneww]

 

 

On 1/11/2020 at 7:09 AM, stephaneww said:

um, thank you, you make me doubt, maybe I didn't transcribe correctly what I was told: and with this case of electromagnetic waves, is it better? 

 

23 hours ago, Mordred said:

Far better. The Maxwell well equations for the E and B fields. Electro is former, magnetism for latter

 

1 hour ago, Markus Hanke said:

As I’ve pointed out earlier, there is no such thing as a “magnetic charge”.

ok thank you, so we must speak in terms of electric and magnetic fields (instead of charges)? 

edit : sorry , it was a stupid question. the answer is Yes, of course

Edited January 12, 2020 by stephaneww

[Mordred]

Here is a simple intro to for and cross product.

https://www.mathsisfun.com/algebra/vectors-dot-product.html

On that page is the link to cross product.

[stephaneww]

12 hours ago, Mordred said:

Here is a simple intro to for and cross product.

https://www.mathsisfun.com/algebra/vectors-dot-product.html

On that page is the link to cross product.

thank you very much

 

On 1/11/2020 at 9:53 PM, stephaneww said:

Let's start with this, please:

 

On 1/11/2020 at 7:59 AM, Mordred said:

Symbolism is as follows use two vectors (actual letters don't matter) pay attention to the Operator )

A⋅B now if you deal with the inner product of two vectors you return a scalar. (Yes I know I am going too far but I want to give you direction of study) you also have outer products of two vectors. 

A×B for cross product. One is linear the latter is perpendicular. (Hint )

can I have links (wiki ideally) that detail this please (the French notations seem different from the English notations) ?

take a look at the French notation of the vector product (=cross product):

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

(notation and understanding of the 2 notions acquired) 

what is the latex for dot please ?

links for the next step, please ?

On 1/11/2020 at 7:59 AM, Mordred said:

you will need to study the vector properties of gradient, divergences and curl.

Edited January 13, 2020 by stephaneww

[stephaneww]

I just noticed this with [math]A[/math] and [math]B[/math] from this message (mathematic solution of the cosmological constant problem) :

[math]\sqrt{A}. \sqrt{B}=C= \text{ energy density of cosmological constant}[/math]

 

[math]{\Large{\frac{\sqrt{A}}{\sqrt{B}}}}=8.73*10^{122}=\text{ exact value of vaccum catastrophe}[/math]

does anyone have any idea what it means physically ?

is it moving forward or are we going in circles ?

Edited January 13, 2020 by stephaneww

[Mordred]

cdot is the latex for the dot. I will have to take at that link later.

[stephaneww]

On 1/11/2020 at 7:59 AM, Mordred said:

With the EM fields you will not find the above to be the same case. (You will be dealing with the cross product ) which is antisymmetric. 

PS you will find the above relevant to the differences between two 90 degree phase shifted fields and fields that are not phase shifted in terms of the attempts you are making in a charged application of Lambda vs gravity. Ie one situation is symmetric while the other is not.

ie

E×B vs Λ⋅g

g for gravitational potential at an arbitrary coordinate.

..... I  think I won't have all the knowledge necessary to reach the above conclusion. 

Two questions about it : 
1. How do you go from the cross product of E through B to what appears to be a numerical value, please ?
2. what exactly does g represent and how is its value determined ?

edit : I have found for g = https://en.wikipedia.org/wiki/Gravitational_potential

 

2. is that  Λ⋅g  is simple multiplication?

Edited January 17, 2020 by stephaneww

[Mordred]

No it the inner product between Lambda and g. Your dealing with vectors. Here is cross product.

https://www.mathsisfun.com/algebra/vectors-cross-product.html

I don't know if your familiar with the right hand rule in EM theory 

https://en.m.wikipedia.org/wiki/Right-hand_rule

You can from those links how the cross product equates to the right hand rule. The dot product returns a scalar however the cross product returns a vector.

So here is a question how can one claim that the EM field and the cosmological constant are related when the latter has no cross product of two vectors and is described strictly via the inner or dot product ?

Edited January 18, 2020 by Mordred

[stephaneww]

1 hour ago, Mordred said:

No it the inner product between Lambda and g. Your dealing with vectors. Here is cross product.

I still don't understand. It's too complicated. Is there a relationship between the electromagnetic field of light in vacuum and the cosmological constant finally ?

edit :

1 hour ago, Mordred said:

You can from those links how the cross product equates to the right hand rule. The dot product returns a scalar however the cross product returns a vector.

So here is a question how can one claim that the EM field and the cosmological constant are related when the latter has no cross product of two vectors and is described strictly via the inner or dot product ?

We can't  claim that, I think.

One is a vector , the other is a scalar.

except perhaps by looking at the origin side of the cross product norm (an area) and the dimension of the cosmological constant (the inverse of an area)???? 🙄

 

that's ok for that, I already understand

1 hour ago, Mordred said:

https://www.mathsisfun.com/algebra/vectors-cross-product.html

I don't know if your familiar with the right hand rule in EM theory 

https://en.m.wikipedia.org/wiki/Right-hand_rule

Edited January 18, 2020 by stephaneww

[stephaneww]

Hello Mordred

Does my mathematical solution match the suggestion at the end of this paper's conclusion, please?

I'm not sure masses distribution corresponds to energy density.

Quote

 In accordance with Zel’dovich hypothesis [1] we suggest that physics of elementary particles is separated into low/high energy ones, the standard notion of smooth spacetime is assumed to be altered at a high energy cutoff scale Λ_∗ and a new treatment based on QFT in a fractal spacetime with negative dimension is used above that scale. This would fit in the observed value of the dark energy needed to explain the accelerated expansion of the universe if we choose highly symmetric masses distribution below that scale Λ_*

source : https://file.scirp.org/Html/5-7503727_93134.htm#ref1

edit : my question is essentially about the low/high energy cutoff as a physical interpretation of my solution. 

Edited January 23, 2020 by stephaneww

[stephaneww]

Well, I'm gonna try to make some physical sense out of this message:

Classically we say that the energy density of the quantum vacuum is:  [math] A=m_pc^2/l_p^3=\hbar.(l_p^{-2})^2.c[/math]

I, for one, found this unknown quantum energy density: [math] B=\hbar.\Lambda_{m^{-2}}^2.c[/math]

We can note that in [math] A[/math] we have [math]m_pc^2[/math] which is a mass energy while in [math] B[/math] we have [math]\Lambda[/math] which is the vacuum energy of the cosmology constant.

We have [math]\rho_ {\Lambda}.c^2=\sqrt{A.B}[/math] energy density of the cosmological constant

I think it can be said that the problem of the cosmological constant contains a formulation error:

Indeed [math] B[/math] would be more logically suited to a quantum vacuum energy density, while [math] A[/math] would be more logically suited to a mass quantum energy density in terms of definition.

Please tell me what you think of this opinion

edit:
we can also note that the values of A and B correspond to a thermal equilibrium such as [math]exp^{((lnA)/2+(lnB)/2)}=\rho_{\Lambda}.c^2[/math] by posing [math]k_B=1[/math].

but I don't know how to calculate what that's like in terms of pressure for [math]1m^3[/math].

Edited February 4, 2020 by stephaneww

[stephaneww]

2 hours ago, stephaneww said:

but I don't know how to calculate what that's like in terms of pressure for [math]1m^3[/math]

I'm an idiot sometimes: the unit J/m^3 corresponds to 1 Pa. It fits in terms of units (1kg.m^-1.s^-2 ) in S.I. but I don't know if the expected sign (the "-") is the one obtained. I don't know the sign conventions

Um, I'm not sure about my handwriting in In.
You might want to use this one for verification:

[math]\Large{exp^{\frac{\ln(A)}{2}+\frac{\ln(B)}{2}}}=\rho_{\Lambda}.c^2[/math]

[math]ln[/math] comes from statistical physics, thermodynamic and entropy
But these fields are new to me. It's more than likely poorly formulated.

Edited February 4, 2020 by stephaneww

[stephaneww]

6 hours ago, stephaneww said:

expln(A)2+ln(B)2=ρΛ.c2

ln comes from statistical physics, thermodynamic and entropy
But these fields are new to me. It's more than likely poorly formulated.

If I'm not misinterpreting: 
the universe being an isolated system in the thermodynamic sense,
this would mean that at the quantum level, for 1/2 m^3 of A + 1/2 m^3 of B, we find in a 1 m^3, by thermal equilibrium of A and B (thermodynamics?), a pressure C which is numerically equal to the energy density of the relativistic vacuum (= at the macro level).

Could this be correct please ?

Edit : we have the Relationship [math] E=k_B T[/math] where [math] E[/math]  is Energy,  [math] k_B [/math]  is Boltzmann constant and  [math] T[/math] temperature for an ideal gaz

Edited February 4, 2020 by stephaneww

[Mordred]

The formula you want is more complex as each particle species of the SM model will add their effective degrees of freedom. 

 For mixed particles of fermions and bosons you will need the Maxwell Boltzmann equations.

https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.damtp.cam.ac.uk/research/gr/members/gibbons/SPC.pdf&ved=2ahUKEwiBzsv9i7nnAhWM4J4KHZbWDlcQFjAQegQIChAB&usg=AOvVaw0REZ23epSZs3ohjVp_naU9

There is a considerable learning curve to understand statistical mechanics as applied to cosmology the above is a start

[stephaneww]

40 minutes ago, Mordred said:

There is a considerable learning curve to understand statistical mechanics as applied to cosmology the above is a start

Thank you, too complex for me again indeed. (edit: It seems to me as if I use (96) page 23, but I guess it's more complicated)

[math]m_p[/math] can't be considered the macro expression of the components you're quoting? Because that would be okay for the values and the dimension.

and for this please ? :

9 hours ago, stephaneww said:

...We can note that in A we have mpc2 which is a mass energy while in B we have Λ which is the vacuum energy of the cosmology constant. ...

I think it can be said that the problem of the cosmological constant contains a formulation error:

Indeed B would be more logically suited to a quantum vacuum energy density, while A would be more logically suited to a mass quantum energy density in terms of definition.

Please tell me what you think of this opinion

 

Edited February 5, 2020 by stephaneww

[Mordred]

Ok ask yourself this question how is vacuum defined ? (This gets back to scalar vs vector quantities)

Hint think of the equation 

[math]w=\frac{\rho}{p}[/math] for Lambda w=-1. Then consider all particles with mass also has monentum.

So I'm going to throw an equation at you

[math]-p_{\Lambda vac}=\rho_{\Lambda vac}=\frac{\Lambda}{8\pi G_n}[/math]

The proof behind the last equation uses the stress energy momentum equation of a perfect fluid (isotropic)

[math]T_{ab}=(\rho+p)U_aU_b+p g_{ab}[/math] lol I know you don't know GR but think about the last paragraph with the above. (There is a difference between mass energy density and vacuum energy density of a field.)

 

 

Edited February 5, 2020 by Mordred

[stephaneww]

Actually, I didn't understand the argument above...😓

what is [math]G_n[/math] please ? and I'm not sure I understand it any better...

[Mordred]

Newtonian Gravitational constant.

Ok mass energy density of a field is the property of a field to resist inertia change or acceleration. 

The vacuum energy density of a field is the property describing the amount of stress of a field. Ie due to quantum fluctuations

More accurately the flux of the field via the energy momentum tensor. The combination of the two will describe the action of a field via the Langrangian  (in thus case scalar field) which will correspond to the Maxwell Boltzmann equations. From this one would be able to determine the probabilistic particle number density. ( just to throw the QFT correspondence to statistical mechanics into the mix)

Edited February 5, 2020 by Mordred

[stephaneww]

1 hour ago, Mordred said:

Newtonian Gravitational constant.

ok  (you omitted a c^2: [math]\rho_{\Lambda}=\frac{c^2 \Lambda}{8 \pi G_n}[/math] )

for the "quantum energy density of the vacuum" (B), if a temperature T_B/m^3 = E_B/k_B/m^3 can be made to correspond, we have T_B/m^3 = about 10^-110 K/m^3. There would be no absolute zero due again to the fluctuations of the quantum vacuum.
(E=k_B T)

for the quantum mass density (A=10^96 kg/m^3) the mass of the current universe being 10^54 kg would be the index of a possible physical nonsense of this presentation, without it being a sufficient argument to prefer its value in K/m^3.

as you suspect, the rest of your explanation escapes me, and in particular the reject of the use of the statistical mechanics of the mix

edit : for the record, can you remind me what [math]U[/math] is, please?

Edited February 5, 2020 by stephaneww
.. and notation ok kB

[stephaneww]

I confess I've lost track a little over the time...

On 1/12/2020 at 6:40 AM, Markus Hanke said:

As I’ve pointed out earlier, there is no such thing as a “magnetic charge”.

but is it foolish to consider the vacuum of the universe as a photon bath of the CMB as a support for an electromagnetic field that would fill the vacuum of the universe ? 

 

[stephaneww]

Hmm, no reaction, it's most probably a nonsense.