Gravity and Strings

86 A perturbative introduction to general relativity
Once this point has been clarified, we proceed to evaluate the correction to the linear
solution for the gravitational field of a point-like massive particle Eq. (3.124) and the gravitational
field of a point-like massless particle Eqs. (3.131), (3.133), and (3.134). The general
setup used to calculate corrections is the following. From the first-order Lagrangian36 L (1) = LFP + Lmatter(ϕ) + 1
2 (χ/c)hµνL (1) µν(h) + Tmatter µν(ϕ)
(3.203)
we obtain the equations of motion
Dµν(h) − (χ/c) t (0) µν(h) + Tmatter µν(ϕ) = 0,
D (0) (ϕ) + (χ/c)D (1) (ϕ, h) = 0.
To find solutions to these equations, we expand the gravitational and matter fields
(3.204)
hµν = h (0) µν + χh (1) µν + ···, ϕ = ϕ (0) + χϕ (1) + ···, (3.205)
around a solution (h (0) ,ϕ (0) ) of the equations
Dµν(h (0) ) − (χ/c)Tmatter µν(ϕ (0) ) = 0,
D (0) (ϕ (0) ) = 0.
(3.206)
On substituting the expansion into the first-order equations of motion, taking into account
that t (0) µν(h) is quadratic in h, D (0) (ϕ) is linear in ϕ, andD (1) (h,ϕ)is linear both in h and
in ϕ, and using the above zeroth-order equations, we find, to lowest order in χ,
Dµν(h (1) ) − 1
c t (0) µν(h (0) ) = 0,
D (0) (ϕ (1) ) + 1
c D(1) (h (0) ,ϕ (0) ) = 0.
(3.207)
We are interested in h (1) in d = 4and we are going to calculate it by using the Rosenfeld
energy–momentum tensor Eq. (3.190) on the linear solution t (0)
Ros µν (h(0) ) and the energy–
momentum tensor Eq. (3.200) we found by imposing δ (1)
ɛ
gauge invariance on the linear
solution t (0)
GR µν (h(0) ) .
In d = 4 the solution Eq. (3.124) for a massive particle can be written in the simple form
h (0) χ Mc 1
µν = δµνk, k =− . (3.208)
8π |x3|
On substituting this expression into the energy–momentum tensors, we find
1 (0)
t Ros µν
c (h(0) ) =−∂µk∂νk − 3
2ηµν
2 2 + 2δµν (∂k) − ηµν + δµν k∂ k,
1 (0)
t GR µν
c (h(0) ) = ∂µk∂νk − 3
2 (∂k)2 + 2k∂µ∂νk −
2
ηµν − δµν k∂ k.
36 We again restore all factors of c.
(3.209)