Feynman Path Integral Formulation

1.6 Gravity in d Dimensions 23
∇ 2 h 00 = −16πG d − 3
d − 2 τ 00 . (1.119)
In four dimension this is equivalent to Poisson’s equation for the Newtonian theory
∇ 2 φ = 4πGρ , (1.120)
when one identifies the metric with the Newtonian field φ in the usual way via
h 00 = −2φ (an identification which follows independently of the previous arguments,
from the weak field limit of the geodesic equation). But in three dimensions
such a correspondence is prevented by the fact that, due to the result in
Eq. (1.119) for the non-relativistic Newtonian coupling appearing in Poisson’s equation,
Eq. (1.120),
G Newton = d − 3
2(d − 2) G , (1.121)
the mass density τ 00 decouples from the gravitational field h 00 . As a result, the
linearized theory in three dimensions fails to reproduce the Newtonian theory.
One can show by an explicit construction in the linearized theory that gravitational
waves are not possible in three dimensions. To this purpose consider the wave
equation in the absence of sources,
✷h μν = 0 . (1.122)
It implies the existence of plane wave solutions of the type h μν (x) =ε μν (k)e ±ik·x
with k 2 = 0 and polarization vector ε μν (k). The transverse-traceless gauge conditions
∂ λ h λ ν = 0 h λ λ = 0 h μ0 = 0 , (1.123)
then imply for the remaining wave equation
✷h ij = 0 , (1.124)
the transversality k i ε i j = 0 and trace ε i i = 0 conditions, with i, j=1,2. The only
possible solution is then the trivial one ε ij (k)=0 and therefore h ij (x)=0. One concludes
therefore that three-dimensional gravity cannot sustain gravitational waves.
The explicit derivation also makes apparent the issue that the absence of gravitational
waves and the lack of a non-trivial Newtonian limit are not connected to each
other in the three-dimensional theory.
One can count explicitly the number of physical degrees of freedom in the d-
dimensional theory. The metric g μν has 1 2d(d + 1) independent components, the
Bianchi identity and the coordinate conditions reduce this number to 1 2d(d + 1) −
d − d = 1 2d(d − 3), which gives indeed the correct number of physical degrees of
freedom (two) corresponding to a massless spin two particle in d = 4, and no physical
degrees of freedom in d = 3 (it also gives minus one degree of freedom in d = 2,
which will be discussed later). One can compare this counting of degrees of freedom
to electromagnetism, where one has d degrees of freedom for the field A μ (x)