# Schwarzchild solution with cosmological constant

1. Jun 3, 2007

### Terilien

How do we solve for it? I still don't know much about non linear equations. Unfortunately, this reduces to R=-4(cosmos constant) which is not a system thus making simplicfication difficult. I'm assuming that we can still use the previous arguments and assume that the metric coompontents Gtt and Grr are exponential functions but beyond that i'm not sure what to do...

Last edited: Jun 3, 2007
2. Jun 3, 2007

### pervect

Staff Emeritus
You might look at the Schwarzschild - de Sitter solution, for instance http://arxiv.org/abs/gr-qc/0602002. This eprint doesn't go into the "how" of the solution very much (in fact it's concerned about modelling aspects of bound systems in such a geometry). If you really want "how" and not just a solution, I'd try Wald. I believe his approach to finding the Schwarzschild solution could be easily generalized to finding the Schwarzschild solution with a cosmological constant.

3. Jun 3, 2007

### Chris Hillman

Deriving the Schwarzschild-de Sitter lambdavacuum

Yeah, it's trivial. You know that in any lambda vacuum solution, the Einstein tensor (evaluated wrt some frame field) has to have the form $k \, \operatorname{diag}(-1,1,1,1)$ where k is some undetermined constant. So start with a Schwarzschild coordinate chart (having two undetermined functions) and plug in this condition. The EFE then reduces to a simple ODE similar to the one you obtain in solving for the Schwarschild vacuum; indeed the case k=0 recovers the Schwarzschild vacuum. This determines both metric functions up to constant multiples which can be "gauged away". Now you can relate the parameter k to either the "cosmological horizon radius" or equivalently to the "cosmological constant". The resulting solution is properly called the "Schwarzschild-de Sitter lambdavacuum". Historically-minded writers sometimes call it "Kottler solution".

4. Jun 3, 2007

### Terilien

Question, will i need some knowledge of non linear equations, to generalize it?

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