# Second order expansion of metric in free-fall

by InsertName
Tags: metric
 P: 27 Hello, I have read that, in a freely-falling frame, the metric/ interval will be of the form: ds2 = -c2dt2(1 + R0i0jxixj) - 2cdtdxi($\frac{2}{3}$ R0jikxjxk) + (dxidxj(δij - $\frac{1}{3}$ Rikjlxkxl) to second order. Does anyone know where I could find a derivation of this result?
 Sci Advisor Thanks P: 3,864 InsertName, That's an interesting formula! It's derivation is almost immediate, except I have doubts about the absence of the time coordinate, and the factors of 1/3. Expand the metric in a Taylor's series: gμν = Aμν + Bμνσxσ + Cμσντxσxτ + ... It's always possible to choose coordinates such that Aμν = ημν and Bμνσ = 0, and in these coordinates the Christoffel symbols vanish. Then the formula for the Riemann tensor reduces to Rμσντ = ½(gμτ,σν + gσν,μτ - gμν,στ - gστ,μν) = Cμστν + Cτνμσ - Cμσντ - Cσμτν. If you assume C to have the same symmetry as the Riemann tensor, then this is 4Cμστν, showing that Cμστν = (1/4)Rμσντ

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