# Schwarzschild metric in terms of a falling observer's coordinates

The Schwarzschild metric tensor $$\textbf{g}(r)$$ in terms of a distant observer's coordinates $$(t,r,\Omega)$$, can be written as: $$ds^2=-\left(1-\frac{r_s}{r}\right)dt^2 + \left(1-\frac{r_s}{r}\right)^{-1}dr^2 +r^2\,d\Omega^2.$$ How would the metric tensor $$\textbf{g}(\tilde{r})$$ in terms of a inertal observer's coordinates $$(\tilde{t}, \tilde{r}, \tilde{\Omega})$$, $$x$$ units away from let's say the singularity, knowing that $$\textbf{g}(\tilde{x})$$ should reduce to the Minkowski metric since spacetime is locally flat in his coordinates? (where $$\tilde{x}$$ is how far he is in his coordinates)

I am asking this question because I'm interested in how a falling observer would see spacetime at $$r=\infty$$ away from the singularity. Someone very far from the effect of gravity will see the metric reduce to the Minkowski metric.

Habouz asked: "What's the Schwarzschild metric in terms of a falling observer's coordinates?"

The transformation rule can be found here and here:

$${\rm dt = d\tau+dr \ v} / \hat g_{\rm tt} \ , \ \ {\rm v = -c \ \sqrt{r_s/r}} \ , \ \ 1/\gamma = \rm \sqrt{1-v^2/c^2}$$

with that you transform the old coordiantes $$\rm \hat x$$ to the new ones $$\rm \bar x$$:

$$\bar g_{\mu \nu} = \sum_{\sigma, \kappa} \ \hat g_{\sigma \kappa} \ \rm \frac{\partial \hat x^{\sigma}}{\partial \bar x^{\mu}} \ \frac{\partial \hat x^{\kappa}}{\partial \bar x^{\nu}}$$

so the metric in regular Schwarzschild/Droste coordinates (with time $$\rm \hat x^0 = t$$) where the local observers are stationary with respect to the black hole:

$$\hat g_{\mu \nu} = \left( \begin{array}{cccc} \rm c^2/\gamma^2 & 0 & 0 & 0 \\ 0 & \rm -\gamma^2 & 0 & 0 \\ 0 & 0 & \rm -r^2 & 0 \\ 0 & 0 & 0 & \rm -r^2 \sin ^2 \theta \\ \end{array} \right)$$

transforms to Gullstrand/Painlevé coordinates (with time $$\rm \bar x^0 = \tau$$) where the local observers are free falling raindrops with the negative escape velocity $$\rm v$$:

$$\bar g_{\mu \nu} = \left( \begin{array}{cccc} \rm c^2/\gamma^2 & \rm v & 0 & 0 \\ \rm v & \rm -1 & 0 & 0 \\ 0 & 0 & \rm -r^2 & 0 \\ 0 & 0 & 0 & \rm -r^2 \sin ^2 \theta \\ \end{array} \right)$$

where the covariant spatial components are euclidean:

$$\bar g_{\rm i j} = \left( \begin{array}{cccc} \rm -1 & 0 & 0 \\ 0 & \rm -r^2 & 0 \\ 0 & 0 & \rm -r^2 \sin ^2 \theta \\ \end{array} \right)$$

which is just flat space in spherical $$\{ \rm r, \ \theta, \ \phi \}$$ coordinates:

Hamilton & Lisle wrote: "In the river model, space itself flows like a river through a flat background, while objects move through the river according to the rules of special relativity"

Here are two such free falling raindrops in Gullstrand/Painlevé coordinates sending signals to each other when they are half way through the black hole (at $$\rm r=r_s/2=1$$, photons in green):

If you take $${\rm dt = d} {}_T-{\rm dr \ c} / \hat g_{\rm tt}$$ or alternatively $${\rm dt = d{}_T-dr \ c \ r_s/r} / \hat g_{\rm tt}$$ instead you get the Schwarzschild metric in ingoing Eddington/Finkelstein coordinates where radially ingoing photons have a constant coordinate velocity of $$\rm dr/d{}_T=-c$$, while in Gullstrand/Painlevé coordinates the free fallers have a coordinate velocity of $$\rm dr/d\tau=v$$. In regular Schwarzschild/Droste coordinates on the other hand, at $$\rm r=r_s$$ you get $$\rm dr/dt=0$$ for all particles and photons since the bookkeeper's coordinate time $$\rm t$$ freezes at the horizon.