Feynman Path Integral Formulation

322 9 Scale Dependent Gravitational Couplings
Nevertheless these arguments are sufficiently concrete to allow specific predictions,
the principal one being a relationship between Newton’s constant G and the
nature of mass distribution in the universe. Feynman in particular argues that the fact
that the total energy of the visible universe, of approximate mass M and approximate
size R, is zero,
M − 3 GM 2
≃ 0 , (9.84)
5 R
might not be a simple numerical coincidence. After all, in order to cancel completely,
the first and second term need to be fine tuned to one part in 10 80 (the
approximate numbers of protons in the visible universe). In fact a result of this type
should not be seen as entirely surprising, since one knows for example that in the
canonical formulation of gravity the Hamiltonian constraint is precisely H = 0. If
that is indeed the case, then one would expect G to be related to cosmic quantities,
G ≃ R M . (9.85)
Another separate and puzzling result, coming from cosmology, is that the observed
matter density is very close to one, in the appropriate units, Ω ≈ 1 (more properly,
it is actually a combined density of baryonic, non-baryonic and dark energy components,
Ω t = Ω λ + Ω dm + Ω b , but we will overlook such subtle distinctions here).
For non-relativistic matter one has Ω = ρ/ρ crit with ρ crit = 3H 2 /8πG, which then
gives
Ω = 8πGρ
3H0
2 , (9.86)
where H 0 is the value of the Hubble constant today. Since H 0 is expected in general
to be time-dependent, a possible scenario would be one in which for large times
R −1 = H ∞ = lim t→∞ H(t), with H∞ 2 =
3 λ , and for which the horizon radius is R ∞ =
H∞
−1
. Then for H 0 ∼ R one has Ω ∼ 1if2MGR ≃ 1, which would again relate G to
the overall size and mass of the visible universe in accordance with Eq. (9.85).
Perhaps a slightly more convincing argument can be formulated by making use
of the Lense-Thirring effect (Lense and Thirring, 1918). There one considers a test
particle at rest at the origin, and a thin spherical shell of total mass M rotating rigidly
around it, at some distance R and at angular velocity Ω. Originally the problem was
solved assuming the fields to be weak and the accelerations to be small, but a more
complete analysis (Brill and Cohen, 1966) shows that the test mass at the origin (a
Foucault pendulum, for example) will rotate due to frame dragging with an angular
velocity given by
[
ω = Ω 1 + 3 4
]
R − 2MG −1
, (9.87)
MG(1 + β)
where β is a dimensionless constant that depends on the relative contributions of
T ij and T 00 to the shell’s gravitational mass. For small MG and β one recovers
the post-Newtonian result ω/Ω = 4MG/3R. The full solution on the other hand