I Asymptotically Flat Schwarzschild Metric

[binbagsss]

This is probably a stupid question but so as $r \to \infty$ it is clear that
$-(1-GM/r)dt^2+(1-GM/r)^{-1}dr^2 \to -dt^2 +dr^2$

However how do you consider $\lim r \to \infty (r^2d\Omega^2 )$..?

Schwarschild metric: $-(1-GM/r)dt^2+(1-GM/r)^{-1}dr^2+r^2 d\Omega^2$
flat metric : $-dt^2+dr^2+r^2 d\Omega^2$

i.e without doing this limit the result is clear, but what happens to this limit?

 

[stevendaryl]

What's important is not that r \rightarrow \infty, but that \frac{r}{GM} \rightarrow \infty. That is, we're assuming that r \gg GM, while still being finite.

 

[martinbn]

The difference of the two metrics approaches zero as $r$ goes to infinity.

 

[Angelika10]

The difference of the two metrics approaches zero as $r$ goes to infinity.

But why is that assumed?

 

[Angelika10]

Could it be not true on galaxy level? It is true in the solar system and on earth, but on galaxy level we cannot fit the rotation curves with this assumption. Could it be possible that the assumption "when r goes to infinity, then spacetime becomes minkowski" is violated for galaxies?
If that would be the case, then the gravitational potential would increase. But there are examples of such behavior in nature: the potential of an atom for example (potential well)?

 

[PeterDonis]

on galaxy level we cannot fit the rotation curves with this assumption

Yes, we can. All of the proposed models for fitting galaxy rotation curves are asymptotically flat. You have already been told this in another thread that you started yourself. Please do not repeat this misunderstanding in someone else's thread.