Ron Maimon in many posts claimed that cosmological horizon is like a big black hole.

Black holes decrease as they evaporate and their radius decreases as well.

So what is with a cosmological horizon?

If cosmological horizon is just a black hole centered at the opposite side of the universe, we should see the radius of the external space groving as the radius of the BH decreases.

But if the radiation decreases the horizon entropy we should see the area of the horizon decreasing.

What is the correct conclusion?

  • $\begingroup$ I was thinking of asking if space is distorted around the cosmological event horizon like a BH horizon. It seems like double-talk to equate the two, and then say that we can't detect spacetime curvature (even with deep space surveys). An observer on the cosmological event horizon can claim to be in flat space and is no non-privileged relative to us. Not so if you're falling into a supermassive BH. I don't see how dark-energy can turn de Sitter space into a complimentary model with Schwarzschild Geometry. $\endgroup$ – Alan Rominger Sep 18 '12 at 3:21
  • $\begingroup$ @AlanSE: Space is not distorted near a black hole horizon--- you just have a diverging redshift. It's exactly the same as a cosmological horizon with cosmological redshift. An observer on a black hole also can claim to be in flat space, the horizon is nothing special locally. $\endgroup$ – Ron Maimon Sep 18 '12 at 4:42
  • $\begingroup$ I had a very hard time figuring out the problem--- I get it now--- you are asking about a 3-sphere painted on a part with a black spot, and you are saying this is like the cosmological horizon, and you are noticing that when the spot is larger than a hemisphere, the area radius relation inverts. The simple answer is that when you reach the hemisphere, your whole space is now empty deSitter--- there is no inversion in the area radius relationship at any radius. $\endgroup$ – Ron Maimon Sep 18 '12 at 5:36

Your question is phrased in a little bit of an annoying way, because you are speaking about the other side of the cosmological horizon in a way that pretends this is an absolute notion. But nitpick aside, the area always grows when the horizon absorbs matter, and it always shrinks when it emits matter. The solution has this property, and there is no contradiction.

The reason this might be intuitively confusing is that you can't just picture a sphere with a black spot painted on a large spot, reflecting the location of the cosmological horizon. The spherical limit is Einstein static universe, the limit when the cosmological horizon surrounds you is a deSitter inflating universe, and their geometry is different.

The matter that the horizon absorbs or emits curves the remaining region in such a way that the area is consistent, whether you have net emission or net absorption. It's only inconsistent with the painted-sphere picture, because here you think when the black-region crosses the equator, the area can't grow anymore. I don't know what to say to that, because it's not the geometry of deSitter. The best analogy in painted-sphere is to say that when the black part is one hemisphere, that's it, you're done--- the black hole is as big as it can get, and you are now in empty deSitter (empty except for the thermal stuff in equilibrium with the horizon temperature).

To see precisely what happens, consider deSitter Schwarzschild, with line element of Schwarzschild form with the metric function

$$ 1 - {r_1\over r} - {r^2\over r_0^2} $$

Assume $r_1$ is much smaller then $r_0$, and you find that the outer horizon radius is, to leading order

$$r_H = r_0 - r_1 $$

The two horizons have a local temperature which is hard to define in absolute units, because there is no infinity to compare to, but it is easy to see that the inner horizon, the black hole, is always hotter than the outer horizon, so the end result is that there is heat flow from the black hole to the cosmological horizon, and if you take it into account, $r_1$ shrinks and $r_0$ grows, and their sum remains constant. The square of the radius is the area, and it gets larger.

If you add more mass, make r_1 bigger and bigger, there is a degeneration, when the two horizons collide in r. This degeneration is not singular--- the metric is fine between the two horizons, it's just that they have the same area, so that the coordinates going out in r break down.

The result is the double-black hole Nariai solution, dS2xS2. If you now sprinkle dust on this, the two horizon areas both shrink a little (think of it as an emission of dust, although this never happens physically), and now you have a sphere capped with two black holes, As you add more dust, the two black holes shrink, and you are in the deSitter limit.

| cite | improve this answer | |
  • $\begingroup$ If r0 grows, it should be farther and farther from the observer, not closer. Because the larger radius has cosmological horizon, the farther it is. $\endgroup$ – Anixx Sep 18 '12 at 10:24
  • $\begingroup$ @Anixx: That's not true of static solutions at fixed cosmological constant. $\endgroup$ – Ron Maimon Sep 18 '12 at 13:08

I think it's better to think of the cosmic horizon as a black hole turned inside out. This means it radiates inwards (which we can interpret as the cosmic background radiation). And it grows.

| cite | improve this answer | |

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.