stephaneww

hello I would like to know the complete formula to have a precise value of the value of the acceleration of the expansion of the universe (m/s^2) at the Hubble radius in the Lambda-CDM model please ...if possible with an arXiv source, which would be a luxury thanks in advance

to be synthetic : S = lPl2 is a surface (= plate for the Casimir effect) 1/S = 1/lPl2 in QFT or 1/S = Λ in cosmology are energies according to their side in the equality of the Casimir effect, their physical meaning is different. I do not see anything more convincing

of course there are no plates in cosmology just as Λ was initially used to make the universe static to finally account for the acceleration of the expansion of the universe, 1/S = Λ changes the physical meaning of physical plates. The fact remains that dF / dS keeps the physical meaning of an energy density (or of a pressure if you prefer)

once again : in one side S is the surface of the plates, in the other side dF/dS is an energy density in Casimir effect egality

Let's see: with : SPl = lPl2 Planck surface 1/SPl = 1/lPl2 = zero point energy of the QFT FPl/SPl = energy density of zero point energy of the QFT the physical meaning is not in doubt here so why would putting 1/S =Λ not make physical sense ?

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 = FPl = 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

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

ok this is beyond what I know and understand uh, does he use two plates or a space geometry in this approach?

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

it's not a problem. it is indeed an excessive shortcut on my part. i haven't found any sources in my proposition, cosmologycal constant and zero-point energy of quantum fields are linked to obtain the Casimir effect

here : https://royalsocietypublishing.org/doi/10.1098/rsta.2019.0229 you can read the whole article, I didn't take the time to translate it al either it was badly said or it is a problem of automatic translation I refer you to the whole article quoted above here too but I haven't access : https://www.semanticscholar.org/paper/The-vacuum-energy%3A-Casimir-effect-and-the-constant-Elizalde/e0d922a743ae3c2c5ef7cb61cfb391f1fab7fcb3

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

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

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

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 (=lPl-2) to obtain the Casimir effect ... and dF/dS is an energy, not a force