Hello, I have read that, in a freely-falling frame, the metric/ interval will be of the form: ds^{2} = -c^{2}dt^{2}(1 + R_{0i0j}x^{i}x^{j}) - 2cdtdx^{i}([itex]\frac{2}{3}[/itex] R_{0jik}x^{j}x^{k}) + (dx^{i}dx^{j}(δ_{ij} - [itex]\frac{1}{3}[/itex] R_{ikjl}x^{k}x^{l}) to second order. Does anyone know where I could find a derivation of this result?
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: g_{μν} = A_{μν} + B_{μνσ}x^{σ} + C_{μσντ}x^{σ}x^{τ} + ... It's always possible to choose coordinates such that A_{μν} = η_{μν} and B_{μνσ} = 0, and in these coordinates the Christoffel symbols vanish. Then the formula for the Riemann tensor reduces to R_{μσντ} = ½(g_{μτ,σν} + g_{σν,μτ} - g_{μν,στ} - g_{στ,μν}) = C_{μστν} + C_{τνμσ} - C_{μσντ} - C_{σμτν}. If you assume C to have the same symmetry as the Riemann tensor, then this is 4C_{μστν}, showing that C_{μστν} = (1/4)R_{μσντ}