I've seen questions and answers dealing with similar topics, but none that seem to provide what I'm looking for.
The Schwarzschild metric (and indeed any valid metric) should reduce to the Minkowski metric over a sufficiently small, linearized region.
I am trying to do this mathematically by Taylor expanding the Schwarzschild metric terms, but struggling a bit with the math, specifically, what value of $r$ I should center it at, presumably not $0$ or $\inf$, maybe $1$? And what terms to neglect.
Can someone help me with this derivation, or at least tell me if I am on the right track?
Note: this is NOT the same as the Newtonian limit, where $r$ goes to infinity, because the locally Minkowski property should hold even at very high curvatures, including inside the horizon.
Best Answer
You can always choose locally-inertial coordinates to see that any metric (Schwarzschild, or otherwise) at a given point assumes the Minkowski form, its first derivative vanishes (therefore Christoffel connection vanishes too), but the second derivative does not vanish. The fact that the second derivative cannot be made to vanish is just another way of saying that even if you go to a local frame with $g_{\mu \nu} (p) = \eta_{\mu \nu}$ and $\partial_\alpha g_{\mu \nu} (p) = 0$ at a point $p$, there is still curvature characterized by the 20 independent components of the Riemann tensor (which has second derivative of the metric).
In this specific sense, you cannot really take a limit of Schwarzschild to Minkowski, since the former has non-zero Riemann curvature. You can read about locally inertial coordinates applicable in general (and not just for specific metrics like Schwarzschild) in Carroll's book in chapter 2.