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Second order expansion of metric in free-fall

by InsertName
Tags: metric
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InsertName
#1
Dec22-12, 06:59 AM
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([itex]\frac{2}{3}[/itex] R0jikxjxk) + (dxidxjij - [itex]\frac{1}{3}[/itex] Rikjlxkxl)

to second order.

Does anyone know where I could find a derivation of this result?
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Bill_K
#2
Dec22-12, 02:40 PM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
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 = (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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