2
$\begingroup$

Consider the Friedmann equation:

$$H^2+\frac{k}{a(t)^2} = \frac{\Lambda}{3}+\frac{8 \pi}{3}\rho$$

and set the parameters for dust in either flat euclidean or open hyperbolic spatial slices with a negative cosmological constant: $\rho = \frac{\rho_0}{a[t]^3},\Lambda<0, k=0$ or $k=-1$. With that, the Friedman equation takes the forms:

$$H^2 = -\frac{|\Lambda|}{3}+\frac{8 \pi}{3}\frac{\rho_0}{a(t)^3}$$

$$H^2-\frac{1}{a(t)^2} = -\frac{|\Lambda|}{3}+\frac{8 \pi}{3}\frac{\rho_0}{a(t)^3}.$$

With these parameters, the scale factor $a(t)$ takes the general form of a combination of trig functions that expands from a big bang, reaches a finite maximum $a(t_{max})$, and then contracts towards a big crunch when $-|\Lambda|$ dominates over the contributions from matter and spatial curvature.When the parameters are set such that the matter density at $a(t_{max})$ is exponentially small, $\frac{\rho_0}{a(t_{max})^3} <<1$, then the FLRW metric approximates a spatial slice of pure AdS. However, this FLRW approximation to AdS is only temporary and the metric quickly contracts away from AdS, increasing in density till the big crunch. Furthermore, even when $\rho$ is set to $0$ the FLRW coordinates can only ever cover a partial patch of AdS space. So, no matter how small $\frac{\rho_0}{a(t_{max})^3}$ is when $a(t)=a(t_{max})$ , and no matter how long the approximate AdS slice lasts, the metric:

$$ ds^2 = dt_{max}^2 + a(t_{max})^2 (\frac{dr^2}{1- k r^2}+r^2 (d\theta^2+sin(\theta)^2 d\phi^2))$$

can't globally extend across the future half of AdS space. This is in contrast to $\Lambda>0$ FLRW metrics with $k=1$ whose future half can asymptote to pure dS space with globally covering coordinates.

Because these $\Lambda<0$ FLRW distributions of dust can't fill all of AdS, is it possible to embed them within a larger asymptotic AdS vacuum? The FLRW dust distribution with $k=1$ and $\Lambda >= 0$ can be cut in half and the remaining finite, contracting 3-hemisphere can be patched to an outer Schwarzschild metric. The result is the Lemaitre-Tolman-Bondi metric.

In a similar way, is it possible to patch the boundary of a $-\Lambda$ FLRW metric to an outer asymptotic AdS metric, such that the $-\Lambda$ FLRW space seen from the outside is an infinite mass, planar black hole? Or possibly even a black hole with a negatively curved event horizon?

EDIT: Below I've pasted a graphic from page 65 of Exact Space-Times in Einstein's General Relativity by Griffiths and Poldolsky. It illustrates an FLRW metric covering only a partial patch of the AdS manifold, and also gives some intuitive sense of why there aren't any global Cauchy slices. If one were to perturb the area in the FLRW grid with dust, could the outer blank areas where the FLRW grid doesn't reach be constructed as the exterior of a non-spherical black hole?

enter image description here

Also, is AdS's inability to admit global Cauchy slices is relavent to my question? Because of the lack of Cauchy slices, even if one were to perturb an entire spatial slice of pure AdS with dust that alone is not enough to determine the future of the manifold. Independent boundary conditions need to be applied.

$\endgroup$
8
  • 1
    $\begingroup$ "$\Lambda$ FLRW space seen from the outside is an infinite mass, planar black hole". There is no "outside" for FLRW metric, since FLRW metric assumes spatial homogeneity. On the other hand, the Schwarzschild metric of black hole has an center and event horizon therefore there is "outside". You are confusing the FLRW metric with the Schwarzschild metric. $\endgroup$
    – MadMax
    Commented Jan 24 at 15:50
  • $\begingroup$ @MadMax, Pure AdS space admits an FLRW metric, but, despite the FLRW metric's infinite homogeneity, it's a fact that the metric doesn't cover the entire AdS manifold. So, from the POV of the manifold, there is an outside to the AdS FLRW metric. Also, remember, null rays can reach the AdS boundary in finite proper time, and suitable boundary conditions can let them pass through. Now, a -Λ FLRW w/ dust can approach this limit of pure AdS at the moment of maximum expansion. With this in mind, is it possible to patch this near-AdS to an outer blackhole metric with a non-spherical horizon. $\endgroup$
    – Michael C.
    Commented Jan 24 at 18:56
  • $\begingroup$ Hmm, at moment of maximum expansion $H^2 = -\frac{|\Lambda|}{3}+\frac{8 \pi}{3}\frac{\rho_0}{a(t)^3} = 0$, it's sort of an empty Milne universe. And indeed this snapshot of "at moment of maximum expansion" could be translated into the static Schwarzschild metric. However there are two lingering questions: 1) it happens only at the moment of maximum expansion, does your "patching" work for the whole spacetime? 2) Since there is no center for this quasi Milne universe, where do you want to put the center of the blackhole in the universe? $\endgroup$
    – MadMax
    Commented Jan 24 at 20:14
  • $\begingroup$ @MadMax, about your Milne comment. Referring to pg 75 of "Exact Space-Times in Einstein's General Relativity" by Griffiths and Poldolsky, the Milne solution is the limit of FLRW when k=-1, rho=0, and Lambda=0. The scenario I'm referring to has Lambda<0. Geodesics in Milne diverge while they converge in AdS. $\endgroup$
    – Michael C.
    Commented Jan 24 at 20:16
  • $\begingroup$ You missed the key words "sort of" and "quasi". $\endgroup$
    – MadMax
    Commented Jan 24 at 20:19

1 Answer 1

1
$\begingroup$

I found an answer to a variation to my own question in an article from 1997 by Jose P.S. Lemos called "Gravitational collapse to toroidal, cylindrical and planar black holes", arXiv:gr-qc/9709013.

Lemos patches a dust filled, $k=0$, $-\Lambda$ FLRW metric to an exterior AdS-Vaidya BH metric with a flat horizon that can have the topology of a plane, a cylinder, or a flat-torus. Below is the BH metric re-formulated to emphasize the planar case:

$$ ds^2 = -(\alpha z^2 - \frac{2 m(v)}{\alpha z})dv^2 + 2dvdz + z^2(dx^2+dy^2)$$

$\alpha = \frac{-\Lambda}{3}, -∞<v<∞, -∞<x<∞ ,-∞<y<∞ ,0<z<∞$

Lemos then matches that AdS-Vaidya metric to a flat interior FLRW metric: $$ds^2 = -dt^2 + a(t)^2 ( d\rho^2 + \rho^2(d\theta^2 + sin(\theta)^2 d\phi^2))$$

$$a(t)= a_{0} sin(\frac{3}{2} \sqrt{\frac{-\Lambda}{3}} t)^\frac{2}{3}$$ With the coordinates ranging over the usual values for a flat FLRW metric. I've also simplified Lemos's FLRW metric, because he allowed for toroidal/cylindrical topologies in the interior, which is not my concern. Most of the paper then finds the conditions under which $m(v)$ can be set to match both metrics at the horizon. Read the article if your interested.

It's very difficult for me to visualize this infinite wall of incoherent radiation described by AdS-Vaidya metric being focused and funneled by the AdS curvature into a gravitational collapse, and for there to be a fully infinite universe behind that wall undergoing a Big Crunch.

Lemos also refers to an earlier article by Smith and Mann called "Formation of Topological Black Holes from Gravitational Collapse", arXiv:gr-qc/9703007. This article focuses on FLRW metrics patched to negatively curved event horizons, with a scenario more resembling the Oppenheimer-Snyder dust collapse than the Vaidya metric. Smith's and Mann's scenario is closer to my original question but their non-traditional conventions on the signs of $\Lambda$ and $k$ confuse me.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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