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'll try and briefly run through some points (without mathematical detail) to see if this clears up any of your questions

• You seem to have misunderstood the 'blunder' part: the 'blunder' wasn't removing the cosmological constant, but adding it to his equations (in an ad-hoc way, at the time) in the first place.

• Today, dark energy isn't 'different' from the cosmological constant - the CC is just a possible (and the simplest) way to describe dark energy. (It's also the one that works best, despite the surrounding theoretical issues.)

• Einstein removed the CC because it was no longer needed for a static universe (and the other associated problems regarding stability, which you quoted).

Just to be clear, neither a cosmological constant nor any type of dark energy is needed for an expanding universe, but is needed for *accelerated* expansion. The universe was already expanding from the big bang. A universe where the CC is zero still expands.

## Best Answer

Look at this paper:

"In an expanding universe, what doesn't expand?" by Richard Price

Short summary

All or nothing behavior: If the binding force is greater, the object does not expand significantly. If the cosmological costant is stronger, the object expands.

Hence, as the cosmological constant grows to infinity (or minus infinity, according to the usual convention), more and more strongly bound systems are ripped apart.

However, this analysis assumes the binding force (i.e. gravity or electrodynamics) decreases with distance.

The strong force, however, increases with distance.

So the cosmological constant versus the strong force is still an interesting question.