# 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?

• Jul 11 '20 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. Jul 11 '20 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. Jul 11 '20 at 22:22
• @Javier This is a very good point, could you put it in an answer? Jul 12 '20 at 10:39
• The answer to this post (as Javier already mentioned above) can be found in Exercise 10.9.9 in one of the best introductory textbooks "A First Course in GR". Jan 14 '21 at 5:01

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.
If $$\frac{2GM}{c^2R}<<1$$ both expressions are valid as approximations.
But the second one presents the expression $$dr^2 + r^2 d\Omega^2$$ detached. And that is the square of a generic path element in spherical polar coordinates.
Being an elementary spatial path, it can be then replaced by: $$dx^2 + dy^2 + dz^2$$
Consider the change of coordinate $$r=r'\left(1+{{\cal G}m\over 2r'c^2}\right)^2={r'}^2(1+U)^2$$ where $$U={\cal G}m/2r'c^2$$. One can check that $$1-{2{\cal G}m\over rc^2}=1-{2{\cal G}m\over r'c^2(1+U)^2} ={(1-U)^2\over (1+U)^2}$$
Moreover, $${dr\over dr'}={d\over dr'}\left[r'\left(1+{{\cal G}m\over 2r'c^2} \right)^2\right]=(1-U)(1+U)$$ so that $$dr=(1-U)(1+U)dr'$$ The Schwarzschild metric becomes \eqalign{ &ds^2=\!c^2\!\left(1-{2{\cal G}m\over rc^2}\right)dt^2 -\left(1-{2{\cal G}m\over rc^2}\right)^{-1}dr^2 +r^2d\theta^2+r^2\sin^2\theta d\varphi^2 \cr &=c^2\left(1-{2{\cal G}m\over rc^2}\right)dt^2 -{(1+U)^2\over (1-U)^2}(1-U)^2(1+U)^2dr'^2 -{r'}^2(1+U)^4\left[d\theta^2\!+\!\sin^2\theta d\varphi^2\right]\cr &=c^2\left(1-{2{\cal G}m\over rc^2}\right)dt^2-(1+U)^4 \left[d{r'}^2+{r'}^2d\theta^2\!+\!{r'}^2\sin^2\theta d\varphi^2\right] \cr &=c^2\left(1-{2{\cal G}m\over {r'}c^2}+{\cal O}(U^2)\right)dt^2 -\big(1+4U+{\cal O}(U^2)\big)\left[d{r'}^2+{r'}^2d\theta^2\! +\!{r'}^2\sin^2\theta d\varphi^2\right] \cr &=c^2\left(1-{2{\cal G}m\over {r'}c^2}\right)dt^2 -\left(1+{2{\cal G}m\over {r'}c^2}\right)\left[d{r'}^2 +{r'}^2d\theta^2\!+\!{r'}^2\sin^2\theta d\varphi^2\right] \cr } as expected.