Derivation of freely falling frame in Schwarzschild spacetime Thinking about the equivalence principle, is there a nice, simple way to show that a local, freely falling frame in Schwarzschild spacetime is described by the Minkowski metric
$$ds{}^{2}=c^{2}dt^{2}-dr^{2}-r^{2}d\theta^{2}-r^{2}\sin^{2}\theta\left(d\phi\right)^{2}.$$
I thought if I could describe a test body (moving radially inwards with an acceleration due to gravity) using the Schwarzschild metric
$$ds^{2}=\left(1-\frac{2GM}{c^{2}r}\right)c^{2}dt^{2}-\frac{dr^{2}}{1-\frac{2GM}{c^{2}r}}-r^{2}d\theta^{2}-r^{2}\sin^{2}\theta d\phi^{2},$$
the Minkowski metric would somehow pop out (with $d\theta = d\phi = 0$). I substituted $ar=-GM/r$
  into the metric ($a$
  is the acceleration due to gravity) to get
$$ds^{2}=\left(1+\frac{2ar}{c^{2}}\right)c^{2}dt^{2}-\frac{dr^{2}}{1+\frac{2ar}{c^{2}}},$$
which, on Earth for example with $2ar\ll c^{2}$, is pretty close to the Minkowski metric. Is this valid/correct?
 A: Let's start with the Schwarzschild metric
$$
\text{d}s^2 = \left(1 - \frac{r_\text{s}}{r}\right)c^2\text{d}t^2-
\left(1 - \frac{r_\text{s}}{r}\right)^{-1}\text{d}r^2 - r^2\text{d}\Omega^2,
$$
with
$$
\begin{align}
r_\text{s} &= \frac{2GM}{c^2},\\
\text{d}\Omega^2 &= \text{d}\theta^2 +\sin^2\theta\,\text{d}\varphi^2.
\end{align}
$$
Now let's introduce a new radial coordinate $\bar{r}$, defined as
$$
r = \bar{r}\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^2.
$$
After some algebra, we find that
$$
\begin{align}
\text{d}r^2 &= \left(1 - \frac{r_\text{s}}{4\bar{r}}\right)^2\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^2\text{d}\bar{r}^2,\\
\left(1 - \frac{r_\text{s}}{r}\right) &= \left(1 - \frac{r_\text{s}}{4\bar{r}}\right)^2\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{-2},
\end{align}
$$
so that the Schwarzschild metric can be written in the form
$$
\text{d}s^2 = \left(1 - \frac{r_\text{s}}{4\bar{r}}\right)^{2}\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{-2}c^2\text{d}t^2 - 
\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{4}\left(\text{d}\bar{r}^2 + \bar{r}^2\text{d}\Omega^2\right).
$$
Now, in a local frame around a coordinate $(t,r,\theta,\varphi)$, we can treat $r$ in the coefficients as constant, so that
$$
\begin{align}
\left(1 - \frac{r_\text{s}}{4\bar{r}}\right)^{2}\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{-2} &\approx \alpha^2,\\
\left(1 + \frac{r_\text{s}}{4\bar{r}}\right)^{4} &\approx \beta^2,
\end{align}
$$
with $\alpha$ and $\beta$ constants, thus
$$
\text{d}s^2 \approx c^2\alpha^2 \text{d}t^2 - \beta^2\left(\text{d}\bar{r}^2 + \bar{r}^2\text{d}\Omega^2\right).
$$
Finally, with the coordinate transformation
$$
\begin{align}
\tilde{t}&=\alpha\, t,\\
\tilde{r}&=\beta\, \bar{r},
\end{align}
$$
we obtain the familiar Minkowski metric
$$
\text{d}s^2 \approx c^2\text{d}\tilde{t}{}^2 - \left(\text{d}\tilde{r}^2 + \tilde{r}^2\text{d}\Omega^2\right).
$$
This transformation can be performed at any coordinate $(t,r,\theta,\varphi)$ along the path, so the local metric is always of Minkowski form.
A: Locally, that is for a fixed point $r,\theta,\phi,t$ you could always find global coordinates, such as the metrics will be Minkowski at this point (but not at other points). At the fixed point, the frame  will be a inertial frame (but not at the other points). Of corse, you could use this flat metrics as approximate metrics in the immediate neighborhood of the fixed point.
But if you want global coordinates describing a freely falling observer, it is not possible. If it was possible, this would mean that the Schwartzschild metrics would be a flat metric, and this is obviously false. 
An interesting case appears when you are using approximate the metrics near the horizon ($r\sim 2MG$), and in a small angular region ($\theta \sim 0$)
An example  is the Rindler metrics:
$$ds^2= \rho^2~d\omega ^2 - d\rho^2 - dX^2 - dY^2$$
with : $$\omega = \frac{t}{4MG},~\rho \sim 2 \sqrt{2MG(r - 2MG)}, ~X \sim 2MG \theta \cos \phi, ~Y \sim  2MG \theta \sin \phi$$
Setting : $$T = \rho ~ sh\omega, Z = \rho ~ ch\omega$$
We get : $$ds^2=dT ^2 - dZ^2 - dX^2 - dY^2$$
This proves, that, near the horizon ($r\sim 2MG$), and for a large black hole ($\theta \sim 0$), the metrics is almost quasi flat.
