3

In Schwarzschild coordinates the line element of the Schwarzschild metric is given by:

$$ds^2=\Big(1-\frac{r_s}{r}\Big)\ c^2dt^2-\Big(1-\frac{r_s}{r}\Big)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta\ d\phi^2).$$

In the asymptotic limit where $r>>r_s$ the Schwartzschild metric becomes:

$$ds^2=c^2dt^2-dr^2-r^2(d\theta^2+\sin^2\theta\ d\phi^2),$$

which is the Minkowski metric of flat spacetime.

But observations show that real astronomical objects are embedded in an expanding spatially flat FRW metric given in polar co-ordinates by:

$$ds^2=c^2dt^2-a^2(t)\ dr^2-a^2(t)\ r^2(d\theta^2+\sin^2\theta\ d\phi^2).$$

Therefore maybe the Schwarzschild metric should be given by:

$$ds^2=\Big(1-\frac{r_s}{r}\Big)\ c^2dt^2-a^2(t)\Big(1-\frac{r_s}{r}\Big)^{-1}dr^2-a^2(t)\ r^2(d\theta^2+\sin^2\theta\ d\phi^2).$$

Perhaps this metric would only be useful to describe a gravitational system whose size is comparable to the Universe itself?

Qmechanic
  • 201,751
John Eastmond
  • 5,917
  • 1
    FWIW, the Schwarzschild metric is the vacuum solution for a static, spherically symmetric spacetime. Think about that. – Alfred Centauri Sep 24 '16 at 23:02
  • Related: http://physics.stackexchange.com/q/2110/2451 and links therein. – Qmechanic Sep 24 '16 at 23:16
  • 2
    Have you tried plugging that into the Einstein tensor and figuring out what would be the stress energy tensor and the equations for a(t)? – Bob Bee Sep 25 '16 at 02:28

3 Answers3

3

Matching the Schwarzschild metric onto the metric of an expanding universe is not trivial. Einstein and Straus tried it in the 1940s but their paper, as I recall, has a mistake. A solution was given in 1956 by C. Gilbert in the MNRAS:

http://adsabs.harvard.edu/full/1956MNRAS.116..678G

As I said, the solution is not trivial.

Tony Rothman
  • 31
3

real astronomical objects are embedded in an expanding spatially flat FRW metric

Not really. If you think of the cosmos as clumps of matter on top of a FRW background, you're counting the same matter twice: once in a perfectly uniform distribution and then again in its actual clumped location.

You can start with FRW if you plan to construct a so-called swiss cheese solution by completely removing spherical regions of matter and replacing them with inhomogeneous spherically symmetric geometries with the same mass (such as Schwarzschild black holes). In that case you aren't counting anything twice, you're just treating some of the matter as homogeneous and some of it as clumped.

If you want to build the whole cosmos out of clumped matter, then you don't start with FRW. You start with a Minkowski or (anti) de Sitter vacuum, and you end up with an FRW geometry when all of the matter is added. FRW is essentially a bunch of Schwarzschild patches sewn together and then smoothed to remove the local bumps.

benrg
  • 26,103
2

Actually, if you write (1-2M/ra(t)) in your coefficients for the dt and dr terms in your metric, you have the McVittie Metric, which was proposed in 1933. Although it is an exact solution to the Einstein equations, there has been a lot of confusion in the literature about what it represents. See:

https://arxiv.org/abs/1003.4777

Tony Rothman
  • 21