Newtonian Limit of Schwarzschild metric

Question

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 duplicates: How can we recover the Newtonian gravitational potential from the metric of general relativity? , How to get space component of weak field (linearized) metric? — Qmechanic, 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. — curio, 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. — Javier, Jul 11 '20 at 22:22
@Javier This is a very good point, could you put it in an answer? — curio, 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". — Yuan Yao, Jan 14 '21 at 05:01

score 2 · Answer 1 · answered Jan 14 '21 at 20:47

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.

Qmechanic · Answer 2 · 2020-07-11T16:58:46.820

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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.

edited Jul 11 '20 at 16:58

answered Jul 11 '20 at 16:24

Qmechanic

201,751

Claudio Saspinski · Answer 3 · 2020-12-12T21:51:44.877

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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: $dx^2 + dy^2 + dz^2$

edited Dec 12 '20 at 21:51

answered Dec 12 '20 at 20:55

Claudio Saspinski

16,319

Newtonian Limit of Schwarzschild metric

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