Locally remove a gravitational field

Question

Let $K$ be an inertial frame of reference on $\mathbb{R}^3$ and $g=g(t,x)$ a nonuniform and nonstatic gravitational field. How I can choose a system of reference $\bar K$ such that mechanical effects of $g$ can be neglected?

Answer 1

In the context of GR and the equivalence principle, given a Lorentzian manifold $(M,g)$, the following comments seem relevant:

1. If the (Levi-Civita) Riemann curvature tensor does not vanish in a point $p\in M$, then there does not exist a neighborhood $U \subseteq M$ of $p$ (and a coordinate system defined on $U$) such that the metric $g_{\mu\nu}$ becomes on Minkowski-form in $U$. See also my Phys.SE answer here.

2. Locally, there exist Fermi normal coordinates along a tubular neighborhood of a geodesic.

Answer 2

I suppose you are asking about locally inertial frames?

Chapter 2.4:

Postulate (2) of general relativity implies that at each point of spacetime it is possible to choose locally inertial coordinates: $\xi^m$

Say you have coordinates $x^\mu$ and you want to transform to inertial coordinates $\xi^m$ which are in locally inertial frame $ds^2=\eta_{m n}\xi^m \xi^n$

The coordinates are differentiably related:

$d\xi^m=\frac{d\xi^m}{dx^\mu}dx^\mu$

Hence, you just have to find coordinates that satisfy:

$g_{\mu \nu} =\eta_{m n} \frac{d\xi^m}{dx^\mu} \frac{d\xi^n}{dx^\nu}$

Note: last equation comes from $ds^2$=(same in every frame)