Weak Gravity & Newtonian Limit: Letting g^kmu = eta^kmu

[unscientific]

Assume we have a free-falling particle in gravity in a static metric. Its worldline is described by:

$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$

where $|h_{\mu \nu} << 1|$.

Taken from Hobson's book:

Why did they let $g^{k\mu} = \eta^{k\mu}$?

 

[JorisL]

They only keep terms up to first order in the perturbation $h_{\mu\nu}$.
However, I assume (7.6) is the metric written as a flat metric ($\eta_{\mu\nu}$) with an additional perturbation h.

Then insert the expansion into the connection and see what you get. Remember that the Minkowski metric in the usual coordinates is diagonal and constant.

 

[PeterDonis]

Why did they let $g^{k\mu} = \eta^{k\mu}$?

Because the derivative operator itself is first order, so when you contract the metric with it, the term in $h^{\kappa \mu}$ is second order and can be dropped.

 

[unscientific]

Because the derivative operator itself is first order, so when you contract the metric with it, the term in $h^{\kappa \mu}$ is second order and can be dropped.

$$\frac{1}{2}g^{k \mu} \partial_k g_{00} = \frac{1}{2}(\eta^{k \mu} + h^{k \mu}) \partial_k g_{00}$$
$$= \frac{1}{2}\eta^{k \mu}\partial_k g_{00} + \frac{1}{2}h^{k \mu}\partial_k g_{00}$$
$$= \frac{1}{2}\eta^{k \mu}\partial_k g_{00} + \frac{1}{2}h^{k \mu} \partial_k h_{00}$$

How is the second term $0$? Or are we making the assumption that we want only terms corresponding to first order, so $h \times (\partial h)$ is perturbation x perturbation which is second order?

 

[PeterDonis]

are we making the assumption that we want only terms corresponding to first order

Yes, that's what your source meant by "valid to first order".