What's the equation relating general relativity and Newtonian mechanics?

To put it another way, how (which constants etc.) does general relativity prove that Earth's gravity is about 9.80N/kg on it's surface?|||'Die Grundlage der allgemeinen Relativitätstheorie', (The foundation of the General theory Relativity) was published by Einstein in 1916. He provided a field equation for a gravitational field tensor 'G' in terms of the stress-energy tensor 'T'.





G = 8πT





or in the expanded form: -





.. α .. α β


∂Γ + Γ Γ = -κ(T - ½g T)


_ μν μβ να .... μν ... μν


∂x


.. α





√-g = 1








In section 21 of this famous paper, Einstein considered Newton's theory as a first approximation. To do this he considered space-time to be essentially that found at spacial infinity, which means that only tensor components μ = ν = 4 may need to be considered!





Thus he derived the Newtonian equivalent equation: -





∇²g₄₄ = κρ





Where ρ = T₄₄ or the matter in the stress-energy tensor.





He then reduced this equation to one for the equivalent Newtonian gravitational potential 'V': -





-κ.⌠ρdτ


__ | _


8π⌡r





or units of time to match Newtonian theory





V =-K.⌠ρdτ


..... __ | _


..... c².⌡r








Where K = G = 6.7 x 10¯⁸ (10^-8 Einstein's cgs units, now 10^-11 SI units), with κ = 8πK /c²





Thus, Einstein's first order approximation gives a similar result to Newton's gravitational theory.





Since F = -dV/dr, a conservative central force field: -





Newtonian g = GM/r²





or Einstein's approximation





g =K.ρdτ


..... ___


..... c²r²





and from E = Mc² = ρ, we have with the proper time dτ = 1 second





g =K.M


..... ___


..... r²








Thus, to answer your question - to a first approximation, general relativity proves that Earth's gravitational acceleration is about 9.80N/kg on it's surface!!!!|||g=GM/R^2, where G=6.67*10-11, M-Earth mass, R-radius|||Luv a duck,gee whizz.