The Friedmann equation is given here as: \(\dot{R}^2 = \frac{8 \pi GR^{3}_{0}}{3c^2R}\rho - kc^2\) The curvature part is given as \(\mathbf{k}c^2 = \frac{2\mathcal{L}}{mR^2}\) Since normally \(\mathbf{k}c^2\) is the part that deals with curvature, then this will be the part that we will deal with specifically, because we wish to talk about the contribution of curved space to zero point energy density. Expanding the part we have described as a Langrangian, we would find \(\int \frac{2\hbar c }{mR^{2}_{H}} k\ dk +.... \int \frac{\hbar c}{mR^{2}_{H}} R^n \frac{dk}{k^{n-2}}\) This would make \(\dot{R}^2 = \frac{8 \pi G R^{2}_{0}}{3c^2} \rho - \int \frac{\hbar c}{mR^{2}_{H}} R^n \frac{dk}{k^{n-2}}\) With a cosmological constant term this would be \(\dot{R}^2 = \frac{8 \pi G R^{2}_{0}}{3c^2} \rho - \int \frac{\hbar c}{mR^{2}_{H}} R^n \frac{dk}{k^{n-2}} + \Lambda c^2 R^{2}_{H}\) As the universe expands, scalar particles are also created that are on-shell. The creation of virtual particles cannot be described by the operator equivalent of the real matter part because virtual particles are off shell. In an approach I considered, when implementing an idea that particle production happened alongside expansion, the on shell particle production part was described by: \(m\phi = m \sum_k [{A_k f_k + A^{\dagger}f_k}]\) \(\phi = \sum_k [{A_k f_k + A^{\dagger}f_k}]\) And the mass, \(m\), was often written in two different formulas to show simply that a mass formula could very well act as a coefficient on the quantum operator and determining the various masses of the particle zoo. The expansion can be found quite simply from an expansion of a langrangian where \(k\) is a wave number, we have (using Sakharov) \(\mathcal{L} = \hbar c R \int k\ dk + .... \hbar c R^n \int \frac{dk}{k^{n-2}} + C\) Where \(C\) is a renormalizing constant set to zero for flat space (Sivaram).
Also this following post should have been included: Using Sakharov, with dimensions of a length^-2, the cosmological constant itself could be related as \(\mathcal{L} = \hbar c R^n k\ dk ... \hbar c R^n \Lambda + C\) You can obtain from this a density and is given as (Sivaram) \(\rho_{vac} = \frac{1}{2} \hbar c \Lambda k^2\) Where Λ = 1/length² is cosmological constant, k is wave number. Such a length taken to Planck scales will lead to: \(\Lambda = \frac{c^3}{ħG} = 1/L(p)^2\) Giving \(\rho_{vac} = \frac{1}{2} \hbar c k^2 \frac{c^3}{\hbar G} = \frac{1}{2} k^2 \frac{c^4}{G}\) Notice how naturally the upper limits of the classical forces appears from the last solution in the equation.
