# Weak gravity and the Newtonian limit

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1. Mar 31, 2015

### 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 $g^{k\mu} = \eta^{k\mu}$?

2. Mar 31, 2015

### 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.

3. Mar 31, 2015

### Staff: Mentor

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.

4. Mar 31, 2015

### unscientific

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

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?

5. Mar 31, 2015

### Staff: Mentor

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