Second order expansion of metric in free-fall

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SUMMARY

The discussion centers on the second order expansion of the metric in a freely-falling frame, specifically the expression for the interval ds². The formula presented includes terms involving the Riemann tensor and the Christoffel symbols. A key point raised is the derivation of this result, which can be approached through Taylor series expansion of the metric tensor gμν. The discussion highlights the importance of coordinate choice in simplifying the Riemann tensor to relate it to the curvature tensor Cμστν.

PREREQUISITES
  • Understanding of general relativity concepts, particularly metrics and curvature.
  • Familiarity with Taylor series expansions in the context of tensor calculus.
  • Knowledge of the Riemann tensor and its properties.
  • Proficiency in manipulating Christoffel symbols and their implications in curved spacetime.
NEXT STEPS
  • Study the derivation of the Riemann tensor from the metric tensor in detail.
  • Learn about the implications of coordinate transformations on the Christoffel symbols.
  • Explore the properties of the curvature tensor Cμστν and its relationship to the Riemann tensor.
  • Investigate the application of Taylor series in higher-order expansions of metrics in general relativity.
USEFUL FOR

The discussion is beneficial for physicists, particularly those specializing in general relativity, as well as advanced students seeking to deepen their understanding of metric expansions and curvature in curved spacetime.

InsertName
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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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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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