Usually during the calculation of the Einstein-Hilbert action, we consider the integral of the variation of the Ricci tensor to be zero as the variation of it, at the boundary was assumed to be zero, as:
−116κπ∫d4x[δ√|g|(R+Λ)+√|g|(δgμν)Rμν+√|g|gμν(δRμν)−8πκ√|g|δgμνTμν]
κ=Gc4
Here:
gμνδRμν=Dμ[gαβΓμαβ−gαμΓβαβ]=DμδUμ
∫Md4x√|g|DμδUμ=0
Now, if we are to continue calculating this integral with the vanishing boundary, we obtain the following expression for the variation of the Ricci tensor:
gμνδRμν=gμν◻δgμν−DμDν(δgμν)
Using the above result in the action, we obtain:
Rμν−12gμνR−12gμνΛ+[gμν◻−DμDν]=8πκTμν
Now, under the vacuum conditions,
Λ=0, and
Tμν=0, we obtain a value for the Ricci scalar as:
R+δμμ◻=12δμμR+gμνDμDν
R=3◻
Now, in the case of Dust (in co-moving coordinates),
Tμν=ρg0μg0ν, we obtain the following:
R−12δμμR=8πκT+2Λ−4◻
This yields:
Tμνgμν=T=ρ=◻−2Λ8πκ
But we know that the density in the dust model has the following form:
ρ=N∑q=1γmqδ(3)[→x−→zq(S)]1√|g|
This implies:
Λ=12√|g|[◻√|det(gμν)|−8πκN∑q=1γmqδ(3)[→x−→zq(S)]]
If we replace 1/G with a scalar field ϕ, which varies from one place to another, then we obtain the following form:
Note: the D'alembertian for a scalar field is: ◻=1√|g|∂μ(√|g|∂μ), hence, κ=G/c4=1/(c4ϕ).
From Brans-Dicke theory:◻ϕ=8πT(3+2ω)
where
ω is the dimensionless Dicke coupling constant.
This yields: >
Λ≈4πTϕ(3+2ω)
Under strong coupling, i.e. ω>−1.5, Λ is positive (Λ≈2πρ3ϕ when ω=1.5). Under weak coupling, i.e. ω<−1.5, Λ is negative. Is my derivation correct? What can be the possible implications of this? I am aware that in f(R) gravity, in Einstein field equations when linearized on a de Sitter background gives rise to wave equations for the de Sitter covariant field ϕμν: (◻−2Λ)ϕ≈8πκTμν
When Λ is negative, we can observe that the equation takes the form of Klein-Gordon equations for a massive spin-zero field.