Schwarzschild metric, cosmological constant

[lalbatros]

From an https://www.physicsforums.com/showthread.php?t=140501", a new question comes to me.

Is there a known generalisation of the Schwarzschild geometry when the cosmological constant is positive? Are there still black-holes in this case? Are there small modifications to the Newtonian potential in this case?

Michel

 

[pervect]

Take a look at http://arxiv.org/abs/gr-qc/0602002

A FRW universe isn't "Newtonian" at infinity, so taking the Newtonian limit doesn't really make sense in a universe with a cosmological constant.

The Newtonian limit really only makes sense when the universe is Newtonian at infinity, which implies an asymptotically flat space-time.

I don't think Newtonian potential makes any sense unless the Newtonian limit makes sense. For instance, at infinity, the Newtonian potential predicts that there is no force, which means that objects should not experience any acceleration. But the whole point of a FRW space-time in general, and the cosmological constant in particular, is that the universe is expanding (and in the case of a positive cosmological constant, that the expansion is accelerating). This means that objects are accelerating away from each other. Trying to fit a Newtonian potential into this is just not going to be productive in any manner I can envision.

 

[lalbatros]

Thanks a lot pervect!

Yesterday evening, I could derive that the cosmological term simply injects an additional term -Lr²/3 besides -rg/r. However, the paper you indicate considers many interresting check-applications. These can be useful references if I had new examples to convince myself that the bold extrapolation by GR is indeed the most logical path. If you had other similar references testing alternatives theories for GR (specially without black-holes), I would be very interrested. I would like to understand how compelling GR and its consequences are.

Now, concerning the Schwarzschild geometry with a cosmological constant. Clearly on "short distance" the additional term does not play any role. But on long distances, there is again an event-horizon. How should that be understood?

Thanks again,

Michel

 

[hellfire]

Take a look at http://arxiv.org/abs/gr-qc/0602002

A FRW universe isn't "Newtonian" at infinity, so taking the Newtonian limit doesn't really make sense in a universe with a cosmological constant.

The Newtonian limit really only makes sense when the universe is Newtonian at infinity, which implies an asymptotically flat space-time.

I don't think Newtonian potential makes any sense unless the Newtonian limit makes sense. For instance, at infinity, the Newtonian potential predicts that there is no force, which means that objects should not experience any acceleration. But the whole point of a FRW space-time in general, and the cosmological constant in particular, is that the universe is expanding (and in the case of a positive cosmological constant, that the expansion is accelerating). This means that objects are accelerating away from each other. Trying to fit a Newtonian potential into this is just not going to be productive in any manner I can envision.

You do not need to have an FRW geometry in order to have a cosmological constant. The cosmological constant is a constant added to the Einstein-Hilbert action, so the question remains whether black hole solutions are possible with such a modification of the action. I have no clue, but my wild guess is that they are not possible because the energy conditions for singularity theorems are violated.

 

[lalbatros]

hellfire,

From what I just learned, on shorth 'distance' from the center, the Schwarzschild geometry is not much affected by the cosmological term since then:

gtt = 1-rg/r -Lr²/3
grr = 1/gtt​

and obviously the Lr²/3 is easily made negligible as compared to rg/r, for small r.

Michel

 

[pervect]

You do not need to have an FRW geometry in order to have a cosmological constant. The cosmological constant is a constant added to the Einstein-Hilbert action, so the question remains whether black hole solutions are possible with such a modification of the action. I have no clue, but my wild guess is that they are not possible because the energy conditions for singularity theorems are violated.

Sloppy wording on my part. I would consider the metric for a Schwarzschild-de-Sitter space-time to be a "black hole" in a universe with a cosmological constant, though.

Read the paper I mentioned, http://arxiv.org/abs/gr-qc/0602002 and see if you agree with this point.

As far as the singularity theorems go, they may not guarantee a singularity with a cosmological constant, but I don't think they forbid one.

I would be very surprised if the cosmolgoical constant made much difference to the interior geometry, actually, but I'm relying a lot on my fallible intuition here rather than formal calculations.

 

[Chris Hillman]

Schwarzschild-de Sitter lambdadust and friends

Hi again, Michel,

From an https://www.physicsforums.com/showthread.php?t=140501", a new question comes to me.

Is there a known generalisation of the Schwarzschild geometry when the cosmological constant is positive? Are there still black-holes in this case? Are there small modifications to the Newtonian potential in this case?

I tried to add a lengthy comment to this thread yesterday, but due to chronic instability at PF in the past few days, I lost my work. So I'll keep this brief.

The solution you seek is often called the Schwarzschild-de Sitter or Schwarzschild-ADS lambdavacuum (depending on the sign of \Lambda), or sometimes the "Kottler solution". It features both an event horizon (from the massive object) and a cosmological horizon (from the \Lambda), which means that it has an interesting block diagram (aka Carter-Penrose diagram or conformal diagram; see http://www.arxiv.org/abs/gr-qc/9507019) . It also has various other interesting properties. A not very difficult exercise for an advanced undergraduate student is to compute the precession of pericenters for a particle in quasi-Keplerian orbit around the massive object, and to compare the result with that for the Schwarzschild vacuum. (If you get stuck, looks like the eprint pervect linked to provides a solution.)

I would quickly add that the Schwarzschild vacuum can be generalized in many directions, such as by including a hypothetical scalar field (two popular choices are minimally coupled or conformally coupled massless scalar fields), or an "externally applied gravitational field", and so on. Similar remarks apply to the Kerr-Newman electrovacuum generalization of the Schwarzschild vacuum (to rotating objects with an EM field). All these solution belong to still larger families, such as the Weyl vacuums (all static axisymmetric vacuums) or the Ernst electrovacuums (all stationary axisymmetric electrovacuums).

The book Exact Solutions of Einstein's Field Equations by Stephani et al. is a veritable gold mine of information of this kind, although even this massive book only scratches the surface of the huge literature on exact solutions in gtr.

Chris Hillman