2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.