# Will the Big Rip tear black holes apart?

There seems to be an obvious contradiction between the predictions of the physics of black holes and the Big Rip, a predicted event about 16.7 Gyr in the future where local groups, galaxies, solar systems, planets, and even atoms will be torn apart. More loosely bound items will be torn apart before more tightly bound items, but would this include black holes, which are thought to casually prohibit escape across the event horizon?

The possibilities as I can conceive of them:

• Black hole will be evaporated close to the 16.7 Gyr mark as the expansion increases the rate of Hawking Radiation
• The cosmic even horizon will approach the black hole's event horizon, getting asymptotically closer, forever or until the black hole evaporates according to our predictions for a black hole
• The universe doesn't completely tear itself apart at all, as matter on the scale of local groups falls into itself, even though the separation between local group increases in a big rip, the local group itself becomes a black hole and experiences a local "big crunch"

That last one is a big stretch, but it was my main take-away from reading this one really weird (and now defunct) alternative physics website. I found it insightful, regardless of whether it has physical merit or not. This question has been asked in other places before, but didn't seem to have a very informed set of answers.

• the Big Rip, a predicted event A Big Rip is not predicted by current models of physics. – user4552 Sep 27 '17 at 23:44
• An answer that got deleted linked to this paper:Babichev, Dokuchaev, Eroshenko, "The Accretion of Dark Energy onto a Black Hole," arxiv.org/abs/astro-ph/0505618 – user4552 Sep 29 '17 at 18:35

The generic thing that happens is that the black hole horizons merge with the cosmological horizon.

To see why this is so, you can consider the case of deSitter Schwarzschild, described in Anon's answer:

$$dS^2 = f(r) dt^2 - {dr^2 \over f(r)} - r^2 d\Omega^2$$

$$f(r) = 1- {2a\over r} - br^2$$

This has a Nariai limit, described on the Wikipedia page, where the black hole is as large as possible, and this limit is instructive because it has the following properties:

• The black hole horizon and the cosmological horizon are symmetric.
• The space becomes entirely regular, no singularities are present
• The space can deform so that either horizon is the black hole.
• If you deform the space by adding dust between the black hole and the cosmological horizon, you link the space to the Einstein static universe with two black holes present.

The last point is the most relevant, because for two black holes, you know what happens when they meet--- they merge. For a typical observer, which hasn't fallen into a black hole, a deSitter black hole will be attracted to other deSitter black holes and merge. When they become big enough, they are a cosmological horizon, and then the falling of matter into the black hole turns into inflation of the universe into the cosmological horizon smoothly.

If you have a little black hole, and you aren't in it, unless you accelerate, this black hole will merge with the cosmological horizon in your patch. The merger process is an irreversible joining of horizons, analogous to any other such merger. This is the generic situation. To reach the Nariai limit, you need to keep two black holes on opposite sides of the Einstein static universe, and dump all the dust in the universe into them exactly symmetriclally. This is a very unstable process, a small deformation will lead the two black holes to merge into one cosmological horizon, that then eats all the dust.

• I think my brain just exploded. – Alan Rominger Aug 3 '12 at 18:33

White hole in expanding de Sitter

$$ds^2 = -dt^2 +\left(dr-\sqrt{k^2 r^2 +2M/r}\;dt\right)^2 +r^2d\Omega^2$$

Black hole in contracting de Sitter

$$ds^2 = -dt^2 +\left(dr+\sqrt{k^2 r^2 +2M/r}\;dt\right)^2 +r^2d\Omega^2$$

Schwarzschild

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

Believe it or not, they all describe the same metric space in different coordinates.

• Hey, I had just stumbled upon the Wikipedia article en.wikipedia.org/wiki/AdS_black_hole and I was going to post it here, but from what you say, it's likely equivalent to what you've posted here. The equations look likely equivalent to your last equation. This wraps up the question nicely. – Alan Rominger Apr 23 '13 at 22:07
• What does this have to do with the question? – user4552 Jan 2 '19 at 22:29

According to the Big Rip hypothesis, yes.

The hypothesis states that even spacetime would eventually be pulled apart (Dr Marcelo Disconzi interview: https://www.theguardian.com/science/2015/jul/02/not-with-a-bang-but-with-a-big-rip-how-the-world-will-end), which would destroy black holes (Vyachadav Dokuchaev of the Institute for Nuclear Research of the Russian Academy of Sciences interview: https://www.newscientist.com/article/dn7617-black-holes-cannot-escape-the-phantom-menace/).

Additionally (my own two cents), it seems Inflation has already shown spacetime acceleration is capable of pulling apart a singularity.

• even spacetime would eventually be pulled apart This is just kind of a pop-sci description. which would destroy black holes Clicking through to the article shows that they're actually describing this as very controversial. Additionally (my own two cents), it seems Inflation has already shown spacetime acceleration is capable of pulling apart a singularity. This doesn't really make much sense. Cosmological models that include inflation are no different from those that don't, in the sense that both start with a big bang singularity and both have cosmological expansion. – user4552 Jan 2 '19 at 22:33

No.

$$ds^2 = f(r) dt^2 - {dr^2\over f(r)} - r^2 d\Omega^2$$

$$f(r) = 1- {2a\over r} - b r^2$$

• Anon may be favoring your second possibility, see en.wikipedia.org/wiki/Nariai_spacetime – Mitchell Porter Aug 3 '12 at 15:54
• This is not what happens in a generic situation, the Nariai limit is special, in that the black hole is centered at all times. – Ron Maimon Aug 3 '12 at 18:20
• What is this answer? No explanations at all!? Sorry for the -1. – Cham Jan 2 '19 at 18:07