Feynman Path Integral Formulation

312 9 Scale Dependent Gravitational Couplings− B′ (r) 2 r 24A(r)B(r) 2 + λr2 − A′ (r)B ′ (r)r 24A(r) 2 B(r) + B′′ (r)r 22A(r)B(r) − A′ (r)r2A(r) 2 + B′ (r)r2A(r)B(r) = 8Gπr2 p(r) ,(9.31)with the ϕϕ component equal to sin 2 θ times the θθ component. Covariant energyconservation ∇ μ T μν = 0 implies[ p(r)+ρ(r)] B′ (r)2B(r) + p′ (r) =0 , (9.32)and forces a definite relationship between B(r), ρ(r) and p(r). The three field equationsand the energy conservation equation are, as usual, not independent, becauseof the Bianchi identity. It seems reasonable to attempt to solve the above equations[usually considered in the context of relativistic stellar structure (Misner, Thorneand Wheeler, 1973)] with the density ρ(r) given by the ρ m (r) of Eq. (9.20). This ofcourse raises the question of how the relativistic pressure p(r) should be chosen, anissue that the non-relativistic calculation did not have to address. One finds that covariantenergy conservation in fact completely determines the pressure in the staticcase, leading to consistent equations and solutions (note that in particular it wouldnot be consistent to take p(r)=0).Since the function B(r) drops out of the tt field equation, the latter can be integratedimmediately, givingA(r) −1 = 1 − 2MGr− λ 3 r2 − 8πGr∫ r0dxx 2 ρ(x) . (9.33)It is natural to identify c 1 = −2MG, which of course corresponds to the solutionwith a 0 = 0(p = ρ = 0). Next, the rr field equation can be solved for B(r),{∫ rB(r) =exp c 2 − dy 1 + A(y)( λ y 2 − 8πGy 2 p(y) − 1 ) }, (9.34)r 0ywith the constant c 2 again determined by the requirement that the above expressionfor B(r) reduce to the standard Schwarzschild solution for a 0 = 0(p = ρ = 0), givingc 2 = ln(1 − 2MG/r 0 − λr0 2 /3). The last task left therefore is the determination ofthe pressure p(r). One needs to solve the equation[8πGr 3 p(r)+2MG − 2p ′ 3 λr3 + 8πG ∫ ]rr 0dxx 2 ρ(x) [p(r)+ρ(r)](r)+(2r r − 2MG −3 λ r3 − 8πG ∫ ) = 0 ,r0 dxx2 ρ(x)(9.35)which is usually referred to as the equation of hydrostatic equilibrium. From nowon we will focus only the case λ = 0. The last differential equation can be solvedfor p(r),⎛⎞∫1r MGρ(z)p m (r) = √ ⎝c 3 − dz⎠ , (9.36)1 − 2MGr 0 z√1r2 − 2MGz