# de sitter cosmologic limit

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It has been said that our universe is going to eventually become a de sitter universe. Expansion will accelerate until their relative speed become higher than the speed of light.

So i want to understand what happens after this point: so from our point of view, we see a progressively shrinking event horizon (each galaxy sees one, arguably each spacetime point sees one). Now, what is the Hawking Radiation expected from this event horizon? It would seems that the event horizon area is shrinking around us, but its actually a black hole turns outwards: the black hole is actually outside the event horizon, and the visible "well-behaved" spacetime without singularities in the inside of this horizon

In any case, intuitively (i don't have any arguments to support this) i would expect that the Hawking Radiation inside the horizon would grew larger as this horizon would shrink, but i would love to hear what is actually expected to happen at this point

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user lurscher
I think it is a very good question. +1

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user user1355

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We are already living in a nearly empty de Sitter space - the cosmological constant already represents 73% of the energy density in the Universe - and the Universe won't experience any qualitative change in the future: the percentage will just approach 100%.

However, once the space may be approximated as an empty de Sitter space, all moments of time are physically equivalent. It's because de Sitter space belongs among the so-called "maximally symmetric spacetimes" - in which each point may be mapped to any other point by a symmetry transformation (isometry).

So nothing will change qualitatively: the radius of the cosmic horizon will converge towards those 100 billion light years or so and never change again; we are not far from that point.

Yes, it is true that de Sitter space is analogous to a black hole except that the interior of the black hole is analogous to the space behind the cosmic horizon - outside the visible Universe. The de Sitter space also emits its thermal radiation, analogous to the Hawking radiation. It's a radiation emitted from the cosmic horizon "inwards". Because the interior of the de Sitter patch is compact, unlike its black hole counterpart (the exterior of the black hole), the radiation is reabsorbed by the cosmic horizon after some time and the de Sitter space no longer loses energy.

While the right theoretical description of the thermal radiation in de Sitter space is a theoretician's puzzle par excellence (we are only "pretty sure" about the semiclassical limit, and don't even know whether there exists any description that is more accurate than that), it has absolutely no impact on observable physics because the typical wavelength of the de Sitter thermal radiation is comparable to the radius of the Universe. (Note that it's true for black holes, too: the wavelength of the Hawking radiation is mostly comparable to the black hole radius.)

Such low-energy quanta are obviously unobservable in practice - and in some sense, they're probably unobservable even in theory. You should imagine that there are just $O(1)$ thermal photons emitted by the cosmic horizons inside the visible cosmos whose energy is $10^{-60}$ Planck energies per photon. From an empirical viewpoint, it's ludicrous.

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user Luboš Motl
answered Feb 16, 2011 by (10,278 points)
so the event horizon will reach an stable radius? what happens with the dark energy/cosmological constant expansion acceleration? is there an inflexion point after it starts to lower the acceleration and become zero? thanks for your answer @Lubos

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user lurscher
Hi @lurscher, the energy density in de Sitter space is a universal constant - that's why it's called its (positive) cosmological constant. By Einstein's equations, a constant energy density causes the same spacetime curvature - with the same (proper) curvature radius - which is locally interpreted as the Hubble constant. So $H=\dot a/a$ is constant, too. This can be solved by $a$ growing exponentially with the proper time of the static observer. But this exponential growth is just a coordinate effect. In other coordinates, you may see that de Sitter space is maximally symmetric.

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user Luboš Motl
There is no inflection point or anything like that: the acceleration - given by the spacetime curvature - is complete and eternal (positive) constant because it is linked to the cosmological constant. I feel that I must have written the same thing many times in the main answer so I wonder why you keep on asking the same question.

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user Luboš Motl
It may be a good idea to read e.g. the Wikipedia article on de Sitter space: en.wikipedia.org/wiki/De_Sitter_space#Static_coordinates - the space may be visualized as the hyperboloid in 4+1 dimensions with a fixed Minkowskian proper length from the origin (the right hyperboloid whose signature is 3+1); or it may be written in the static coordinates that only work inside the cosmic horizon; or it may be sliced into flat slices which is relevant for cosmology, and in this picture, you see why the distances are growing exponentially. It's always the same space.

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user Luboš Motl
@Lubos, sorry if i sound repetitive, the point i wasn't clear was about the event horizon reaching a stable radius, but now Lawrence explanation it becomes clear that the event horizon is actually receding, not contracting

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user lurscher
@Lubos, Also, what i meant with an inflection point is that if the current acceleration keeps like this, eventually points we currently see will be beyond the event horizon. If the event horizon is going to recede, is only because such acceleration is going to reduce with time. If it has increased relative to the relative inmediate past, then somewhere its going to be an inflection point

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user lurscher

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user Ron Maimon
@lurscher: The horizon is NOT receding after reaching deSitter state, it is staying put. It is not contracting either. The local temperature is given by the surface gravity of the horizon, redshifted consistently to the entire spacetime, as always. This doesn't cause the horizon to grow or shrink, and this has been well established since before 1980.

This post imported from StackExchange Physics at 2016-01-10 17:44 (UTC), posted by SE-user Ron Maimon
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The Hawking radiation of a deSitter space is given, as always, from the near-horizon metric, consistently redshifted to fill all space. Normalizing the cosmological constant appropriately:

$$ds^2 = - (1- {\Lambda\over 3} r^2) dt^2 + {dr^2 \over (1 - {\Lambda\over 3} r^2)} + r^2 d\Omega^2$$

If you flip the sign of $dt$, you can identify the metric of a 4-sphere after appropriate coordinate transformations, and from this read off the periodicity of t, which goes around the sphere. But this is not necessary. The near horizon metric is Rindler (as usual for hot horizons) and writing $r=r_0 - {6\over\Lambda} u^2$ where $r_0$ is the deSitter radius, you find:

$$ds^2 = - ( {\Lambda^2 u^2\over 9}) dt^2 + du^2$$

Which gives the imaginary time period is $\Lambda u\over 3$, so that the near horizon temperature is $3\over\Lambda u$. Extending this using the redshift factor, the temperature at the center is

$$1\over {2\pi r_0}$$

Where

$$r_0 = \sqrt{3\over\Lambda}$$

Which is the usual deSitter temperature. This temperature is locally the same everwhere, because the space is isotropic. The horizon is static, and stable in equilibrium with this thermal bath, it doesn't grow and it doesn't shrink.

This post imported from StackExchange Physics at 2016-01-10 17:45 (UTC), posted by SE-user Ron Maimon
answered Sep 9, 2012 by (7,720 points)
but i don't see why you say the horizon is static and in equilibrium. can you elaborate?

This post imported from StackExchange Physics at 2016-01-10 17:45 (UTC), posted by SE-user lurscher
@lurscher: Because the metric doesn't change in t once it reaches deSitter. The horizon stays put in r, sucking things into it, and that's the exponential inflation.

This post imported from StackExchange Physics at 2016-01-10 17:45 (UTC), posted by SE-user Ron Maimon
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