Suppose I have an Einstein manifold (a manifold with a metric that solves the Einstein field equations. We can take $\Lambda = 0$ for simplicity in this example). The worldline of a freely falling observer would be parametrized, given some coordinates $(t, x^1, x^2, x^3)$, by:
$\gamma (\lambda) = \left( \gamma^0(\lambda), \gamma^1(\lambda), \gamma^2(\lambda), \gamma^3(\lambda) \right)$,
where $\lambda$ is some parameter (not necessarily the proper time).
My question is: how do I parametrize the manifold with the physical coordinates of this observer? By physical coordinates I mean coordinates such that the time coordinate is the time measured by the observer and the observer has fixed space coordinates.
My attempt:
My idea was to reparametrize $\gamma$:
$ \tilde{\gamma} (t) = \left( t, \tilde{\gamma}^1(t), \tilde\gamma^2(t), \tilde\gamma^3(t) \right) $,
and define new temporary coordinates $\tilde{x}^{\mu}$:
$ \tilde{x}^{\mu} = (t, \tilde{x}^1, \tilde{x}^2, \tilde{x}^3) $, with $\tilde{x}^i = x^i - \tilde{\gamma}^i(t)$.
The metric in the new coordinates is $\tilde{g}_{\mu \nu}$ an the time experienced by the observer is:
$T = \int_0^{t} \sqrt{-\tilde{g}_{00}(t', \pmb{0})} \; dt' $.
I have found a $T = T(t, \tilde{x}^i)$ and I can take:
$X^i = \tilde{x}^i$. These should be the coordinates of the freely falling observer according to my reasoning.
Is this correct? Also, I expect this system of coordinates to be locally inerzial (ie, the final metric should be Minkowskian around the point $X^{\mu} = (T, 0, 0, 0)$ ). Is there a way to see this explicitly?
