Someone reasons: "An observer standing at the Earth's surface measures

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Relativity & Gravitation · The equivalence principle and gravity as geometry

The Geometry of Gravity, Ch The Einstein field equations and the Newtonian limit

Problem

Someone reasons: "An observer standing at the Earth's surface measures Γi00=g⊕/c2=1.09×10−16 m−1\Gamma^i{}_{00} = g_\oplus/c^2 = 1.09\times10^{-16}\ \mathrm{m^{-1}}. Define a local gravitational-field energy density by u=c48πG(Γi00)2≈5.7×1010 J m−3.u = \frac{c^4}{8\pi G}\left(\Gamma^i{}_{00}\right)^2 \approx 5.7\times10^{10}\ \mathrm{J\,m^{-3}}. This uu is built only from things measured at that one point in spacetime, so it is the local energy density of the Earth's gravitational field there." Find the error in this reasoning, using what a freely falling observer passing through the same event would compute for Γi00\Gamma^i{}_{00} and hence for uu.

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