# Perturbation to the flat space metric

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## Main Question or Discussion Point

From the geodesic equation
d2xμ/dΓ2μ00(dt/dΓ)2=0,for non-relativistic case ,where Γ is the proper time and vi<<c implying dxi/dΓ<<dt/dΓ.
Now if we assume that the metric tensor doesn't evolve with time (e,g gij≠f(t) ) then Γμ00=-1/2gμs∂g00/∂xs.
If we here assume that the metric components of the curved part is a perturbation on the flat part
Then gμϑμϑ(flat part)+hμϑ(perturbation)
After which I got stuck in calculating the inverse components of the metric tensor gϑμ which is needed in Γμ00 above.
Thank you.

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because all components of ##h## are assumed to satisfy ##|h_{\mu \nu}| <<1## and also off diagonal terms are zero. Then the inverse is just $$g^{\mu \nu} = \frac{1}{\eta_{\mu \nu} + h_{\mu \nu}} \approx \eta_{\mu \nu} - h_{\mu \nu}$$ This follows from the observation: if ##a^2 - b^2 = (a + b)(a-b) = \approx 1## then ##1 / (a+b) \approx a-b##.

because all components of ##h## are assumed to satisfy ##|h_{\mu \nu}| <<1## and also off diagonal terms are zero. Then the inverse is just $$g^{\mu \nu} = \frac{1}{\eta_{\mu \nu} + h_{\mu \nu}} \approx \eta_{\mu \nu} - h_{\mu \nu}$$ This follows from the observation: if ##a^2 - b^2 = (a + b)(a-b) = \approx 1## then ##1 / (a+b) \approx a-b##.
Yes it is but if the off diagonal terms are non-zero for the general case what will be it??

Yes it is but if the off diagonal terms are non-zero for the general case what will be it??
For the general case, consider that $$g_{\kappa \sigma}g^{\sigma \rho} = (\eta_{\kappa \sigma} + h_{\kappa \sigma})(\eta^{\sigma \rho} \pm h^{\sigma \rho}) = \delta_\kappa{}^\rho \pm \eta_{\kappa \sigma} h^{\sigma \rho} + \eta^{\sigma \rho} h_{\kappa \sigma} + \mathcal O (h^2) \approx \delta_\kappa{}^\rho \pm h_\kappa{}^\rho + h_\kappa{}^\rho$$ this will be equal to ##\delta_\kappa{}^\rho## only if we use the minus sign.

(indices are raised and lowered with ##\eta##)

From matrix formulation if matrices A,B and C are given with their inverses $$A^{-1},B{^-1 }$$and $$C^{-1}$$ and given A=B+C
If $$A^{-1}=B^{-1}+C^{-1}$$ ,then $$I=2I+BC^{-1}+CB^{-1}$$$$I_{ik}+b_{ij}c^{jk}+c_{ip}b^{pk}$$$$\delta_{i}^{k}+\eta_{ij}h^{jk}+h_{ip}\eta^{pk}$$$$\delta_{i}^{k}+2h^k_i=0$$
Similarly for $$A^{-1}=B^{-1}-C^{-1}$$ then $$CB^{-1}-BC^{-1}=I$$$$h_{ij}\eta^{jk}-\eta_{ip}h^{pk}=I_{ik}$$ $$h_i^k-h_i^k=\delta_i^k$$......(1)
Now if $$g_{ij}=\eta_{ij}+h_{ij}$$ and $$g^{ij}=\eta^{ij}-h^{ij}$$( where ##\eta^{ij}## is the inverse element of ##\eta_{ij}## and similarly for h also),to satisfy equation (1) that hij is not the inverse element of hij? Is it??

The notation ##h_{\kappa \sigma} h^{\sigma \rho}## means ##h_{\kappa \sigma} \eta^{\sigma \lambda} \eta^{\rho \mu} h_{\lambda \mu} = h_\kappa{}^\lambda h_\lambda{}^\rho##.
This means you are multiplying ##h## with itself: the ##\kappa##-th colunm of ##h## is being multiplied with the ##\rho##-th row of ##h##.

However ##\eta_{\kappa \sigma} \eta^{\sigma \rho} = \delta_\kappa{}^\rho## because ##\eta^{\sigma \rho}## is really the inverse of ##\eta_{\sigma \rho}##.

haushofer
I'm not sure what you want to calculate, but for static solutions you can always diagonalize the metric such that the line element is invariant under a time reflection. Then the inverse of the metric is easily found.