# Locally-Flat Minkowski limit of Schwarzschild Metric

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.

• If you send $M \rightarrow 0$ which is the same as $r_s \rightarrow 0$, you will recover the Minkowski metric in spherical coordinates. Which parameter are you Taylor-expanding it in? Commented Oct 31, 2021 at 20:43
• Have you tried it for a completely generic metric? The idea is that for any given metric and any given point where the metric is nonsingular, you can always choose a coordinate system in which the linear terms are zero when expanding about that point in powers of the coordinates. That's a technical version of what we mean when we say that any metric is locally flat. Commented Oct 31, 2021 at 21:11
• Try reading section 3.4 of Harvey Reall's notes on general relativity: damtp.cam.ac.uk/user/hsr1000/part3_gr_lectures.pdf. But fundamentally, you take the metric at a given point p and do a coordinate transformation that diagonalizes it, such that you recover minkowsky. Then you can taylor expand the metric atound it using some small parameter that takes $p$ to $x= p + \varepsilon X$ Commented Oct 31, 2021 at 21:32
• @Prof.Legolasov he mentions in the question body that it is valid for any metric Commented Oct 31, 2021 at 21:33
• @FilipeMiguel ah, ok. Then the question title is very misleading. Commented Oct 31, 2021 at 21:45

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).