Earth's atmosphere near sea level is modelled as a static perfect fluid in the weak…

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Relativistic stars and compact objects · Relativistic stars and compact objects

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

Problem

Earth's atmosphere near sea level is modelled as a static perfect fluid in the weak field of the Earth, where gtt=−e2Ag_{tt}=-e^{2\mathcal{A}} with A≈Φ/c2\mathcal{A}\approx\Phi/c^2 and Φ(r)=−GM⊕/r\Phi(r)=-GM_\oplus/r (take GM⊕/c2=4.44GM_\oplus/c^2=4.44 mm and R⊕=6.37×106R_\oplus=6.37\times10^6 m). The relativistic hydrostatic equilibrium equation is dpdr=−(ρ+p)dAdr.\frac{dp}{dr}=-(\rho+p)\frac{d\mathcal{A}}{dr}. Air near sea level has mass density ρm=1.2 kg m−3\rho_{\rm m}=1.2\ \mathrm{kg\,m^{-3}} and pressure patm=1.0×105 Pap_{\rm atm}=1.0\times10^5\ \mathrm{Pa}, so its rest-frame energy density is ρ=ρmc2≈1.08×1017 J m−3\rho=\rho_{\rm m}c^2\approx1.08\times10^{17}\ \mathrm{J\,m^{-3}}. ((a)) Compute the dimensionless ratio patm/(ρmc2)p_{\rm atm}/(\rho_{\rm m}c^2) and the ratio ∣Φ∣/c2=GM⊕/(R⊕c2)|\Phi|/c^2=GM_\oplus/(R_\oplus c^2) at the Earth's surface. ((b)) Take the formal limit p/ρ→0p/\rho\to0 and Φ/c2→0\Phi/c^2\to0 in the equilibrium equation above, keeping the gradients finite, and show it reduces to the Newtonian hydrostatic equation dp/dr=−ρm dΦ/drdp/dr=-\rho_{\rm m}\,d\Phi/dr. ((c)) State which of the two ratios found in part (a) is smaller, and hence which relativistic correction, the pressure added to the inertia or A\mathcal{A} replacing Φ/c2\Phi/c^2, is the tighter approximation for Earth's atmosphere.

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