The Newtonian limit of replaces by

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Relativistic dynamics and fields in flat spacetime · Relativistic stars and compact objects

The Geometry of Gravity, Ch The stress-energy tensor and relativistic continua

Problem

The Newtonian limit of ∂μTμν=0\partial_\mu T^{\mu\nu}=0 replaces ρ+p\rho+p by ρmc2\rho_{\rm m}c^2, dropping a term of relative order p/ρp/\rho. Test that step in two stars. At the centre of the Sun the mass density is about ρm=1.5×105 kg m−3\rho_{\rm m}=1.5\times10^{5}\ \mathrm{kg\,m^{-3}} and the pressure about p=2.5×1016 Pap=2.5\times10^{16}\ \mathrm{Pa}. At the centre of a typical neutron star the mass density is about ρm=5.0×1017 kg m−3\rho_{\rm m}=5.0\times10^{17}\ \mathrm{kg\,m^{-3}} and the pressure about p=1.0×1034 Pap=1.0\times10^{34}\ \mathrm{Pa}. Take ρ≈ρmc2\rho\approx\rho_{\rm m}c^2 and c=3.00×108 m s−1c=3.00\times10^8\ \mathrm{m\,s^{-1}}. ((a)) Estimate p/ρp/\rho at the centre of the Sun. ((b)) Estimate p/ρp/\rho at the centre of the neutron star. ((c)) In which star can pp be dropped beside ρ\rho in the factor ρ+p\rho+p that multiplies uμuνu^\mu u^\nu, as the Newtonian limit does, and in which does that step fail? Give the reason in each case.

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