The Schwarzschild metric describes the gravity of a spherically symmetric mass $M$ in spherical coordinates:

$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1-\frac{2GM}{c^2r}\right)^{-1}dr^2+r^2 \,d\Omega^2 \tag{1}$$

Naively, I would expect the classical Newtonian limit to be $\frac{2GM}{c^2r}\ll1$ (Wikipedia seems to agree), which yields

$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1+\frac{2GM}{c^2r}\right)dr^2+r^2 \,d\Omega^2 \tag{2}$$

However, the correct "Newtonian limit" as can be found for example in Carroll's Lectures, eq.(6.29), is

$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1+\frac{2GM}{c^2r}\right)\left(dr^2+r^2 \,d\Omega^2\right) \tag{3}$$

Question: Why is the first procedure of obtaining the Newtonian limit from the Schwarzschild solution incorrect?

  • $\begingroup$ Possible duplicate: How to get space component of weak field (linearized) metric? $\endgroup$ – Qmechanic Jul 11 at 14:27
  • $\begingroup$ I really don't see why this is a duplicate. I want to know why we cannot obtain the newtonian limit from the schwarzschild metric. The question you linked and the answers don't even mention the schwarzschild metric. $\endgroup$ – curio Jul 11 at 14:56
  • $\begingroup$ The "correct" one is using different coordinates, called isotropic coordinates. The first one is not incorrect, just not as useful, because it doesn't translate as easily to Newtonian 3D space. $\endgroup$ – Javier Jul 11 at 22:22
  • $\begingroup$ @Javier This is a very good point, could you put it in an answer? $\endgroup$ – curio Jul 12 at 10:39

Carroll is merely matching the Schwarzschild solution to the linearized weak field solution, treated as a consistent truncated Laurent series in $c^{-1}$, cf. this Phys.SE post. The main point is that the spatial components of the metric are subleading in an $c^{-1}$ expansion and may receive non-trivial contributions in order to maintain EFE.

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  • $\begingroup$ But why is it wrong to simply Taylor expand the coefficients in the Schwarzschild solution? $\endgroup$ – curio Jul 11 at 16:58
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jul 11 at 16:59
  • $\begingroup$ Sorry but I don't understand. Are you saying that the Taylor expansion of the Schwarzschild metric would not satisfy the EFE? $\endgroup$ – curio Jul 11 at 17:05
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Jul 11 at 17:07

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