In General Relativity there is a helpful procedure which is called "Perturbative approach". Given a chart, and forgetting the basis, the metric is $g_{uv}$ and we can express it as: $$g_{uv}=\eta_{uv}+h_{uv}$$ From a physical point of view the explanation it's very simple: $\eta_{uv}$ represents the metric in absence of gravity while $h_{uv}$ are perturbations around it. Now, one would express Christoffel symbols $\Gamma_{uv}^{\alpha}$ in terms of the perturbation $h_{uv}$ but in order to do it one has to express the inverse metric in terms of $h_{uv}$. The inverse metric is not $$g^{uv}=\eta^{uv}+h^{uv}$$ otherwise $$g_{u\alpha}g^{\alpha v}\not=\delta_{uv}$$ Now, my question is: what is the expression of $g^{uv}$ in terms of $h^{uv}$? Be careful: you can' t neglect high powers of $h_{uv}$
Inverse metric in the perturbative approach of General Relativity
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0To first order of $h$, three is no problem with the simple definition of the inverse, and in linearised relativity, that's all that matters. – 2017-01-07
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0I am not dealing with Gravitational waves, so I can't neglect high powers of $h_{uv}$. That' s why I need the entire expression of $g^{uv}$ in terms of $h^{uv}$. – 2017-01-07
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0Oh, I misread. You said perturbative, I assumed linearised. – 2017-01-07