1
$\begingroup$

Consider a largish but finite sphere of pressure-less homogeneous dust late in the $\Lambda$CDM cosmology. The outside metric is presumably Schwarzschild-de Sitter with no matter, the interior a Friedman metric.

Were the interior an Einstein static universe $\rho = \Lambda c^2 / 4\pi G \approx 1.1788\cdot 10^{-26}$ kg/m$^3$, about the critical density (of course). But this seems odd, since at the surface of the sphere there has to be junction conditions and here dust grains only feel a hemisphere's worth of gravitational curvature - wouldn't they start moving? Or does the cosmological constant balance this exactly - but then it seems the density should be sphere size dependent since the effect becomes larger with greater size?

(In reality such a sphere would be unstable since anything that compressed it a bit would trigger a gravitational collapse. But I am trying to understand the limits of mass systems that can be "parked" into static configurations).

$\endgroup$
1
$\begingroup$

Einstein field equations are local, so conditions of staticity with spatial homogeneity provide an unambiguous relations between matter density, cosmological constant and spatial curvature radius (obtainable from Friedmann equations by setting $\dot a \equiv 0$ and $\ddot a \equiv 0$). The global structure of the solution here does not matter. So yes, the density of dust would be $\rho = \Lambda c^2 / 4\pi G $

The junction conditions at the boundary of the dust sphere are:

  1. The metric induced at the boundary from both sides is the same. This means that the radius of the boundary sphere is the same (let us denote it $R$) in both geometries and that time component of the metric ($g_{tt}$ in static coordinates) is continuous.

  2. Static observer near the boundary at the outer, SdS side, is geodesic. This means that at the boundary gravitational attraction of the dust matter is precisely compensated by repulsion due to cosmological constant, so that test mass realeased with zero velocity in static reference frame would remain static. Quantitatively this means that in static coordinates $\frac{d}{dr}g_{tt}=0$ at the boundary. Since the free parameter of the SdS solution is the mass $M$, fixing the radius of the junction $R$ would fix the outer geometry.

For the SdS solution $g_{tt} = 1 - \frac{2M} r -\frac{\Lambda r^2} 3 $ (in units with $G=c=1$), so the second junction condition means that $$ M = \frac{\Lambda R^3 } 3 = \frac {4 \pi \rho R^3 } 3 \,. $$ Note, that while for small $R$ this mass could be interpreted as a mass of the dust inside the sphere, for larger $R$ the spatial interior is noticeably curved, so that the interior volume is larger than $\frac{4 \pi R^3 } 3$ (gravitational mass defect).

More info on the junction conditions in this model could be found in vast literature discussing Einstein–Straus model which is in a sense an inversion of the solution we are discussing: cosmological solution surrounding the void around the point mass/black hole.

But I am trying to understand the limits of mass systems …

While the above equation for $M$ does not make it obvious, there is, in fact, a limit on the mass (and consequently on $R$), because the $g_{tt}$ function must be positive at $r=R$. This means that the mass must satisfy the condition $3 M < \Lambda ^{-\frac12}$. The limiting case is the same in which the Nariai solution is obtained, and so the limiting geometry would be the Nariai geometry $dS_2 \times S_2 $ cut along the timelike geodesic of $dS_2 $ factor, glued together with static lambda–dust Einstein universe ($S_3 \times \mathbb{ R}$) cut in half along the large sphere of $S_3$. So the dust matter content of such a limiting solution is precisely the half of dust matter in the static Einstein universe.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.