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The repartition of dark energy in the vacuum is homogeneous all over the universe. The diagram below represents space with a black hole in the middle. The square is divided in small unit squares. If there was no black hole and no kind of gravity, each small unit square would contain the same amount of dark energy, and the repartition of energy density of dark energy would be homogeneous.

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Question: Is this homogeneous repartition of dark energy in the unit squares still maintained if there is a black hole in the middle, or does the repartition follow the locally curved spacetime around the black hole?

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Short answer: yes. But what do the original "straight lines" mean now? they cannot be defined in a nice way in the new metric, because the natural geometric entities now are geodesics, and a quadrilateral of four right angles does not exist (apart from possible special choice of corners). You must define your volume in a correct way (see below)

Long answer: To answer your question in precise terms we should have the metric of a universe with positive cosmological constant and a black-hole, which actually isn't hard to solve!

The solution is a nonlinear amalgam of de Sitter and Schwarzschild metrics:

$ds^2 = - \left( 1 - \frac{r_s}{r} - \frac{\Lambda}{3} r^2 \right) dt^2 + \frac{dr^2}{\left(1 - \frac{r_s}{r} - \frac{\Lambda}{3} r^2 \right)} + r^2 d\Omega $

(note: if you want the size of your black hole to be considerably smaller than your universe, so you may expand the spacial part of the metric $d\,\vec{r}^2 = d\,\vec{r}^2_{schwarzschild} - \frac{\Lambda}{3}\frac{r^2 dr^2}{\left(1 - \frac{r_s}{r} \right)^2}$ )

So now in order to find the amount of dark energy in any spatial volume, you must specify the boundary of this volume element in a specific frame of reference, because spacial volume is "not a relativistic invariant". The natural frame for us is that of an observer comoving with the black hole and asymptotically away from it. For such observer you now define the boundary of a volume $\mathcal{V}$, and compute the latter as

$\mathcal{V} = \int_\mathcal{V}\frac{r^2 drd\Omega}{\sqrt{1 - \frac{r_s}{r} - \frac{\Lambda}{3} r^2 }} $

Finally the amount of dark energy in a certain volume define appropriately is $E = \Lambda\mathcal{V}$

Important note: This coordinate system is natural to define a volume specified by specific distances for example from the black hole singularity (boundary of which defined as a surface $r(\theta,\phi)$), this is why the answer is time independent. However if you want to define the volume by the position of comoving stars in de sitter, the volume will be exponentially increasing as $e^{\Lambda t}$, and you will have to use a coordinate system different from the one presented, convenient for picking time slices, and you will, as typical for de sitter space have and exponentially increasing amount of dark energy in a volume defined this way.

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  • $\begingroup$ Very nice answer, even if I am disappointed about the result. - In your de Sitter and Schwartzschild amalgam metrics my straight line squares are meaningless, wrong and replaced by geodesics. - Could you cite some literature (preferentially on the web without paywall, or a common textbook) I could consult? $\endgroup$ – Moonraker Feb 25 '15 at 17:37
  • $\begingroup$ being replaced by geodesics is true for any non flat metric. You can check out Sean Carroll's GR book, it think it's online for free. and for the de Sitter Schwarzschild metric, wikipedia has a page on that. $\endgroup$ – Ali Moh Feb 27 '15 at 15:31

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