# Newtonian Limit of Schwarzschild metric

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?

• Possible duplicate: How to get space component of weak field (linearized) metric? – Qmechanic Jul 11 at 14:27
• 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. – curio Jul 11 at 14:56
• 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. – Javier Jul 11 at 22:22
• @Javier This is a very good point, could you put it in an answer? – 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.
• $\uparrow$ Yes. – Qmechanic Jul 11 at 17:07