I Perturbation to Flat Space Metric: Geodesic Equation

[Apashanka]

From the geodesic equation
d²x^μ/dΓ²+Γ^μ₀₀(dt/dΓ)²=0,for non-relativistic case ,where Γ is the proper time and v_i<<c implying dxⁱ/dΓ<<dt/dΓ.
Now if we assume that the metric tensor doesn't evolve with time (e,g g_ij≠f(t) ) then Γ^μ₀₀=-1/2g^μs∂g₀₀/∂x^s.
If we here assume that the metric components of the curved part is a perturbation on the flat part
Then g_μϑ=η_μϑ(flat part)+h_μϑ(perturbation)
After which I got stuck in calculating the inverse components of the metric tensor g^ϑμ which is needed in Γ^μ₀₀ above.
Can anyone please help me in sort put this.
Thank you.

 

[kent davidge]

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 $1 / (a+b) \approx a-b$.

 

[Apashanka]

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??

 

[kent davidge]

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 $\eta$)

 

[Apashanka]

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 h^ij is not the inverse element of h_ij? Is it??

 

[kent davidge]

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.