linearized gravity

[kexue]

Why and how does from g_{ab}=\eta_{ab}+h_{ab} follow
g^{ab}=\eta^{ab}-h^{ab}?

Don't get the latex right, first equation all indices are supposed to below, second equation all are supposed to be up.

thanks

[Bob_for_short]

I think that you will arrive soon at this property.

If you work with GR, such a metric splitting does not mean introducing a flat space-time since the R is not zero whatever variable change you do.

[haushofer]

Why and how does from g_{ab}=\eta_{ab}+h_{ab} follow
g^{ab}=\eta^{ab}-h^{ab}?

Don't get the latex right, first equation all indices are supposed to below, second equation all are supposed to be up.

thanks
Because then


g^{ab}g_{bc} = \delta^a_c


up to first order. Maybe it helps to write


g_{ab} = \eta_{ab} + \epsilon h_{ab}


then we would like that g_{ab}g^{bc}=\delta_a^c . So up to first order in epsilon you can easily see that g^{bc}=\eta^{bc} - \epsilon h^{bc}
does the job for you:


g_{ab}g^{bc} = ( \eta_{ab} + \epsilon h_{ab}) (\eta^{bc} - \epsilon h^{bc})
= \delta_a^c + \epsilon(h_{ab}\eta^{bc} - h^{bc}\eta_{ab}) + O(\epsilon^2)


Now, you can raise the indices of h with eta because we are working at linear order in epsilon; every correction to it would give higher order epsilon terms. So we get


g_{ab}g^{bc} = \delta_a^c + \epsilon (h_a^c - h_a^c) + O(\epsilon^2) = \delta_a^c + O(\epsilon^2)


You could say that the minus-sign in g^{ab} is for cancelling the two factors linear in epsilon in order to get g_{ab}g^{bc}=\delta_a^c. Ofcourse, if you go up higher in order epsilon, say epsilon squared, then all the terms quadratic in epsilon have to cancel to maintain the identity g_{ab}g^{bc}=\delta_a^c.

[kexue]

thanks so much, haushofer! crystal-clear now