from the geodesic equation for non-relativistic case where $$v_i\ll c$$

$$\frac{dx^i}{dt}\ll1,{\rm for }\ c =1$$

$$\frac{dx^i}{d\tau}\ll\frac{dt}{d\tau}$$using this the geodesic equation for proper time $\tau$ becomes $$\frac{d^2x^\mu}{d\tau^2}+\tau{^\mu}_{00}(\frac{dt}{d\tau})^2=0.$$ if $g_{ij}\ne f(t)$ then $$\tau^{\mu}_{00}=-g^{\mu s}\frac{d g_{00}}{2d x^s}.$$Now if we assume that the curved part of the metric is a perturbation to the flat part

$$g_{ij}=\eta_{ij}({\rm flat})+h_{ij}({\rm perturbation})$$

Can anyone please help in how to calculate $g^{ij}$ if off diagonal terms are non-zero (for general case)?

  • $\begingroup$ What do you exactly mean by "calculating" $g^{ij}$? Are you asking how to obtain the raised-index $g^{ij}$ given you have the lowered-index $g_{ij}$? Or, are you asking how to obtain either of them in the first place? $\endgroup$
    – user87745
    Feb 23, 2019 at 18:13
  • $\begingroup$ He is askin how to obtain either of them, i think. $\endgroup$ Feb 23, 2019 at 18:32
  • $\begingroup$ See my answer here physics.stackexchange.com/a/330277/133418 $\endgroup$
    – Avantgarde
    Feb 23, 2019 at 21:38

1 Answer 1


You're given $g_{ij}=\eta_{ij}+h_{ij}$ and you're interested in obtaining $g^{ij}$. You can do this by using the fact that $g_{ij}g^{jk}=\delta^i_k$, the Kronecker Delta function:

$$ \begin{align} g_{ij}g^{jk}&=\left(\eta_{ij}+h_{ij}\right)g^{jk}\\ &=\eta_{ij}g^{jk}+h_{ij}g^{jk}\\ &=\delta^i_k \end{align} $$ In hindsight, we can guess that this is true for $g^{ik}=\eta^{ik}-h^{ik}$: $$ \begin{align} g_{ij}g^{jk}&=\left(\eta_{ij}+h_{ij}\right)\left(\eta^{ik}-h^{ik}\right)\\ &=\eta_{ij}\left(\eta^{ik}-h^{ik}\right)+h_{ij}\left(\eta^{ik}-h^{ik}\right)\\ &=\left(\delta^i_k-h_j^k\right)+\left(h^k_j-\mathcal{O}(h^2)\right)\\ &=\delta^i_k \end{align} $$ We conclude that $g^{ik}=\eta^{ik}-h^{ik}$. If we had an explicit form for $g_{ij}$ we could also express $g^{ik}$ in an explicit form. Without it, we are done here.


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