Casimir effect and cosmological constant

[stephaneww]

Hi

I think it's okay.

That's why I didn't post it in Speculations first.

Thanks for your opinions

casimir eng(1).pdf

[swansont]

Can you explain what it is you’re trying to show? Just throwing up a bunch of equations isn’t sufficient.

[stephaneww]

I identify the cosmological and quantum vacuum energy with the Casimir effect to confirm that the cosmological constant of general relativity is the cosmological vacuum energy.

 

https://fr.wikipedia.org/wiki/Effet_Casimir#Expression_de_la_force_par_unité_de_surface

The Casimir effect having been proved experimentally as effect of the vacuum energy

[swansont]

1 minute ago, stephaneww said:

I identify the cosmological and quantum vacuum energy with the Casimir effect to confirm that the cosmological constant of general relativity is the cosmological vacuum energy.

 

https://fr.wikipedia.org/wiki/Effet_Casimir#Expression_de_la_force_par_unité_de_surface

The Casimir effect having been proved experimentally as effect of the vacuum energy

The casimir effect is, more precisely, a reduction of the electromagnetic vacuum energy owing to the presence of conducting plates. 

If you are going to equate this with the cosmological constant, you need to do more than unit analysis. A force is going to have units of force, regardless of the origin of it. You can’t e.g. say an electrostatic force is gravitational, just because they have the same units.

IOW your assumption that these are connected would just lead to circular reasoning. 

[MigL]

Casimir effect is a local reduction of the vacuum energy ( relative to the global level ) .
The Cosmological Constant is a global reduction of the vacuum energy, which is exchanged for the expansion force/acceleration.
( similar to exchange of potential for kinetic in a gravity field )





 

[stephaneww]

33 minutes ago, swansont said:

The casimir effect is, more precisely, a reduction of the electromagnetic vacuum energy owing to the presence of conducting plates. 

Quote

On suppose de plus que les plaques sont des conducteurs parfaits de conductivité électrique infinie, et qu'elles ne sont pas chargées.

tradution : It is further assumed that the plates are perfect conductors of infinite electrical conductivity, and that they are not charged.

if I understand correctly the vacuum is conductive

33 minutes ago, swansont said:

If you are going to equate this with the cosmological constant, you need to do more than unit analysis. A force is going to have units of force, regardless of the origin of it. You can’t e.g. say an electrostatic force is gravitational, just because they have the same units.

 

18 minutes ago, MigL said:

Casimir effect is a local reduction of the vacuum energy ( relative to the global level ) .
The Cosmological Constant is a global reduction of the vacuum energy, which is exchanged for the expansion force/acceleration.
( similar to exchange of potential for kinetic in a gravity field )

we have to combine the cosmological constant with the QFT quantum vacuum energy (=l_Pl^-2) to obtain the Casimir effect

 

... and dF/dS is an energy, not a force

Edited December 23, 2021 by stephaneww

[swansont]

1 hour ago, stephaneww said:

if I understand correctly the vacuum is conductive

I would say your understanding is incorrect. A gap in a conductor in a vacuum will not pass current unless there is sufficient voltage for there form an arc. The impedance of the vacuum is ~377 Ohms.

If it were a conductor, having a separation distance has no meaning.

1 hour ago, stephaneww said:

we have to combine the cosmological constant with the QFT quantum vacuum energy (=l_Pl^-2) to obtain the Casimir effect

If you can make connection it has to be via some physics, not unit analysis

 

1 hour ago, stephaneww said:

... and dF/dS is an energy, not a force

I didn’t say it was a force; I used force as an example that unit analysis doesn’t get you a connection between different situations.

 

 

[stephaneww]

16 minutes ago, swansont said:

f it were a conductor, having a separation distance has no meaning.

the separation distance is not important in my identification of the equalities: 1/L^4 becomes the cosmological constant multiplied by the QFT quantum vacuum energy (lPl^2) to obtain the Casimir effect

Edited December 23, 2021 by stephaneww

[exchemist]

1 hour ago, stephaneww said:

tradution : It is further assumed that the plates are perfect conductors of infinite electrical conductivity, and that they are not charged.

if I understand correctly the vacuum is conductive

 

we have to combine the cosmological constant with the QFT quantum vacuum energy (=l_Pl^-2) to obtain the Casimir effect

 

... and dF/dS is an energy, not a force

I'm a bit lost. Surely if S is a surface area, then dF/dS is a pressure function rather than an energy, isn't it?  

Edited December 23, 2021 by exchemist

[stephaneww]

12 minutes ago, exchemist said:

I'm a bit lost. Surely if S is a surface area, then dF/dS is a pressure function rather than an energy, isn't it?  

oops yes indeed it is a pressure or J/m3 i.e. an energy density

the units of (1) and (2) of the first post indicate it and are correct they

Edited December 23, 2021 by stephaneww

[swansont]

14 minutes ago, stephaneww said:

the separation distance is not important in my identification of the equalities: 1/L^4 becomes the cosmological constant multiplied by the QFT quantum vacuum energy (lPl^2) to obtain the Casimir effect

How does that work, physics-wise? It looks like you’re just waving your hands to get the answer you want. How does it “become” the cosmological constant? 

The casimir force can be derived by applying the conductor boundary conditions to the geometry. If L doesn’t matter then you don’t get that answer.

[stephaneww]

55 minutes ago, swansont said:

How does that work, physics-wise?

it is simply the interaction of the cosmological constant at the quantum scale (lPl-2).

 

55 minutes ago, swansont said:

How does it “become” the cosmological constant? 

It does not "become" the cosmological constant. The cosmological constant is already in the equality. It is the identification of (1) to (2) that allows to say that we are in the framework of the Casimir effect

 

55 minutes ago, swansont said:

The casimir force can be derived by applying the conductor boundary conditions to the geometry. If L doesn’t matter then you don’t get that answer.

 

 

Quote

De plus, il est plus que probable que l'effet dépende aussi de la distance L entre les plaques.

https://fr.wikipedia.org/wiki/Effet_Casimir#Expression_de_la_force_par_unité_de_surface

traduction : Moreover, it is more than likely that the effect also depends on the distance L between the plates.

"Probable" does not prohibit another way of presenting L^-4

edit : moreover I would be curious to know what L⁴ represents in nature

Edited December 23, 2021 by stephaneww

[swansont]

1 hour ago, stephaneww said:

it is simply the interaction of the cosmological constant at the quantum scale (lPl-2).

I am not aware of this interaction. Can you point to any peer-reviewed literature that says that the cosmological constant interacts with anything?

 

1 hour ago, stephaneww said:

 

It does not "become" the cosmological constant.

That’s what you said.

1 hour ago, stephaneww said:

The cosmological constant is already in the equality. It is the identification of (1) to (2) that allows to say that we are in the framework of the Casimir effect

An equality that you wrote down as an assumption.

The casimir effect requires conducting plates, which force a component of the electric field to become zero, something not true in free space.

 

[stephaneww]

1 hour ago, swansont said:

I am not aware of this interaction. Can you point to any peer-reviewed literature that says that the cosmological constant interacts with anything?

here : https://royalsocietypublishing.org/doi/10.1098/rsta.2019.0229

Quote

The theory [47] of the Casimir stress inside inhomogeneous planar materials makes one more prediction that, when widely extrapolated to cosmological scales, explains why the Casimir effect might play a role in cosmology: the convergence of the renormalization relies on dispersion. Ordinary dielectric materials are dispersive in the sense that the refractive index n depends on frequency. The Casimir effect is a broadband electromagnetic phenomenon [15] depending on the entire frequency window of the material. For large frequencies, all materials become completely transparent, n → 1. Without this feature, the renormalized stress would contain a logarithmically diverging contribution [47]. On the other hand, the ‘material’ of general relativity—the geometry of space and time—acts on all frequencies equally, as a consequence of the equivalence principle [2]. Therefore, even the renormalized ε_vac would still diverge, although significantly weaker than the unrenormalized one. The wavelength range contributing to the forces of the quantum vacuum would go to the Planck length (1.2) where, presumably, the equivalence principle ceases to hold. So ε_vac would not grow with the inverse forth power of the Planck length as in equation (1.3) that produces the wrong 120 orders of magnitudes, but significantly weaker. The logarithmic divergence [47] is not sufficient though, for the following reason.

you can read the whole article, I didn't take the time to translate it al

 

1 hour ago, swansont said:

That’s what you said.

either it was badly said or it is a problem of automatic translation

1 hour ago, swansont said:

An equality that you wrote down as an assumption.

The casimir effect requires conducting plates, which force a component of the electric field to become zero, something not true in free space.

I refer you to the whole article quoted above

 

 

1 hour ago, swansont said:

I am not aware of this interaction. Can you point to any peer-reviewed literature that says that the cosmological constant interacts with anything?

here too but I haven't access : https://www.semanticscholar.org/paper/The-vacuum-energy%3A-Casimir-effect-and-the-constant-Elizalde/e0d922a743ae3c2c5ef7cb61cfb391f1fab7fcb3

Edited December 23, 2021 by stephaneww

[Genady]

On 12/23/2021 at 10:34 AM, stephaneww said:

The Casimir effect having been proved experimentally as effect of the vacuum energy

Not proved, according to this:

"In discussions of the cosmological constant, the Casimir effect is often invoked as decisive evidence that the zero-point energies of quantum fields are “real.” On the contrary, Casimir effects can be formulated and Casimir forces can be computed without reference to zero-point energies. They are relativistic, quantum forces between charges and currents.

…

I have presented an argument that the experimental confirmation of the Casimir effect does not establish the reality of zero-point fluctuations. Casimir forces can be
calculated without reference to the vacuum ... . The vacuum-to-vacuum graphs (See Fig. 1) that define the zero-point energy do not enter the calculation of the Casimir force, which instead only involves graphs with external lines. So the concept of zero-point fluctuations is a heuristic and calculational aid in the description of the Casimir effect, but not a necessity."

Casimir effect and the quantum vacuum
R. L. Jaffe
Phys. Rev. D 72, 021301(R) – Published 12 July 2005

[exchemist]

10 minutes ago, Genady said:

Not proved, according to this:

"In discussions of the cosmological constant, the Casimir effect is often invoked as decisive evidence that the zero-point energies of quantum fields are “real.” On the contrary, Casimir effects can be formulated and Casimir forces can be computed without reference to zero-point energies. They are relativistic, quantum forces between charges and currents.

…

I have presented an argument that the experimental confirmation of the Casimir effect does not establish the reality of zero-point fluctuations. Casimir forces can be
calculated without reference to the vacuum ... . The vacuum-to-vacuum graphs (See Fig. 1) that define the zero-point energy do not enter the calculation of the Casimir force, which instead only involves graphs with external lines. So the concept of zero-point fluctuations is a heuristic and calculational aid in the description of the Casimir effect, but not a necessity."

Casimir effect and the quantum vacuum
R. L. Jaffe
Phys. Rev. D 72, 021301(R) – Published 12 July 2005

Yes, it is my understanding that the Casimir effect can be accounted for in terms of London (dispersion) forces between the plates.

However I recall reading that the dispersion forces model is actually equivalent to the vacuum fluctuation model.  I do not know any QED however, so I'm in no position to understand how this equivalence may arise. The common feature would seem to be the concept of random fluctuations, in the one case arising in the vacuum and in the other arising in the charge distribution of the electrons of the plates.

If anyone here can shed more light on this I'd be interested.  

Edited December 24, 2021 by exchemist

[stephaneww]

On 12/23/2021 at 3:34 PM, stephaneww said:

The Casimir effect having been proved experimentally as effect of the vacuum energy

41 minutes ago, Genady said:

Not proved, according to this:

"In discussions of the cosmological constant, the Casimir effect is often invoked as decisive evidence that the zero-point energies of quantum fields are “real.” On the contrary, Casimir effects can be formulated and Casimir forces can be computed without reference to zero-point energies. They are relativistic, quantum forces between charges and currents....

 

it's not a problem. it is indeed an excessive shortcut on my part. i haven't found any sources

41 minutes ago, Genady said:

In discussions of the cosmological constant, the Casimir effect is often invoked as decisive evidence that the zero-point energies of quantum fields are “real.” On the contrary, Casimir effects can be formulated and Casimir forces can be computed without reference to zero-point energies. They are relativistic, quantum forces between charges and currents.

in my proposition, cosmologycal constant and zero-point energy of quantum fields are linked to obtain the Casimir effect

Edited December 24, 2021 by stephaneww

[swansont]

21 hours ago, stephaneww said:

here : https://royalsocietypublishing.org/doi/10.1098/rsta.2019.0229

AFAICT this is proposing to use the same calculational approach for the cosmological constant as for the casimir effect, in terms of dealing with infinities, i.e. the renormalization. 

 

[stephaneww]

16 minutes ago, swansont said:

AFAICT this is proposing to use the same calculational approach for the cosmological constant as for the casimir effect, in terms of dealing with infinities, i.e. the renormalization. 

 

ok

question : do they need 2 plates for this calculational approach ? I

Edited December 24, 2021 by stephaneww

[swansont]

57 minutes ago, stephaneww said:

ok

question : do they need 2 plates for this calculational approach ? I

A single plate would feel no force. Symmetry tells you this, but also there’s no exclusion of any of the QM modes.

[stephaneww]

36 minutes ago, swansont said:

A single plate would feel no force

ok

36 minutes ago, swansont said:

Symmetry tells you this, but also there’s no exclusion of any of the QM modes.

 this is beyond what I know and understand

1 hour ago, swansont said:

AFAICT this is proposing to use the same calculational approach for the cosmological constant as for the casimir effect, in terms of dealing with infinities, i.e. the renormalization. 

uh, does he use two plates or a space geometry in this approach?

[swansont]

2 hours ago, stephaneww said:

uh, does he use two plates or a space geometry in this approach?

It’s your link. If you’re offering it as support you should understand it.

Quote

this is beyond what I know and understand

And is the crux if the issue here. You are throwing around equations without understanding the physics.

[stephaneww]

1 hour ago, swansont said:

It’s your link. If you’re offering it as support you should understand it.

There was a bit of irony in my question: I don't see how one can invoke 2 plates to talk about the Casimir effect in cosmology. So it must be something else.

 

1 hour ago, swansont said:

Quote

this is beyond what I know and understand

1 hour ago, swansont said:

And is the crux if the issue here. You are throwing around equations without understanding the physics.

Λ and l_Pl^-2 are energies expressed in m^-2. You don't need to know complex QM terms, nor QM to understand this.

Edited December 24, 2021 by stephaneww

[stephaneww]

On 12/23/2021 at 3:41 PM, swansont said:

The casimir effect is, more precisely, a reduction of the electromagnetic vacuum energy owing to the presence of conducting plates. 

Hello

In fact the solution is probably very simple :

We just have to pass the surface S on the other side of the equality of casimir effect with :
1/dS = Λ
and
dF = F_Pl = c⁴/G, Planck force

We have thus by the same trick as the change of side of the equality for the cosmological constant in the EINSTIEN equation to obtain the energy density of the cosmological constant.

We don't need physical plates in the cosmos anymore

 

Edited February 10, 2022 by stephaneww

[swansont]

Once again, it sounds like you are just rearranging variables and constants without regard for any physics interactions or processes. e.g. there is no "Planck Force" as an interaction, it's just the expression of force using Planck units.