A perfect fluid has stress-energy tensor

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Differential geometry of curved spacetime · Differential geometry of curved spacetime

The Geometry of Gravity, Ch Matter in curved spacetime: minimal coupling and relativistic hydrodynamics

Problem

A perfect fluid has stress-energy tensor Tμν=(ρ+p)uμuν+pgμνT^{\mu\nu} = (\rho+p)u^\mu u^\nu + pg^{\mu\nu}, with uμuμ=−1u^\mu u_\mu = -1 (geometric units, c=G=1c=G=1), and obeys ∇μTμν=0\nabla_\mu T^{\mu\nu} = 0. Let Pμν=gμν+uμuνP^{\mu\nu} = g^{\mu\nu}+u^\mu u^\nu be the projector orthogonal to uμu^\mu. ((a)) Contract ∇μTμν=0\nabla_\mu T^{\mu\nu}=0 with uνu_\nu to derive the energy equation uμ∇μρ+(ρ+p)∇μuμ=0u^\mu\nabla_\mu\rho + (\rho+p)\nabla_\mu u^\mu = 0. State where uμ∇νuμ=0u_\mu\nabla_\nu u^\mu = 0 is used. ((b)) Contract ∇μTμν=0\nabla_\mu T^{\mu\nu}=0 with PανP^\alpha{}_\nu to derive the relativistic Euler equation (ρ+p)uν∇νuα=−Pαν∇νp(\rho+p)u^\nu\nabla_\nu u^\alpha = -P^{\alpha\nu}\nabla_\nu p. ((c)) Set p=0p=0 everywhere in the results of (a) and (b). Identify the two equations obtained and show that a pressureless fluid (dust) follows geodesics as a consequence of ∇μTμν=0\nabla_\mu T^{\mu\nu}=0 alone, with no separate assumption about the motion of its constituent particles. ((d)) Now specialise to a static observer in a static metric with gtt=−e2A(r)g_{tt}=-e^{2\mathcal{A}(r)}, for which uμ=(e−A,0,0,0)u^\mu = (e^{-\mathcal{A}},0,0,0) and uν∇νuα=∂αAu^\nu\nabla_\nu u^\alpha = \partial^\alpha\mathcal{A} (spatial components only). Substitute into the Euler equation of part (b) to obtain the relativistic hydrostatic equilibrium condition, and write it for the radial direction.

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