# Linearized gravity

1. Nov 1, 2009

### 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

2. Nov 1, 2009

### 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.

3. Nov 1, 2009

### haushofer

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 $g_{ab}g^{bc}=\delta_a^c$.

4. Nov 1, 2009

### kexue

thanks so much, haushofer! crystal-clear now