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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

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I think it is a very good question. +1 –  user1355 Feb 16 '11 at 15:59

4 Answers 4

up vote -1 down vote accepted

The final fate of the universe can be computed. The de Sitter spacetime cloaks every region with a cosmological horizon, which is similar in some ways to a Rindler horizon. $10^{100} years from now there will be nothing going on: Black holes will have decayed away and largely the universe is a de Sitter vacuum. There may be a smattering of neutron stars around, which as I recall these exist for a very long time. This horizon does emit quanta, though the flux of this radiation is exceedingly small. The universe will then eventually decay as the vacuum decays and the cosmological horizon retreats off “to infinity.” Nothing else much will be going on.

We consider the decay of the de Sitter vacuum by quantum means and the prospect for this as a mechanism for the production of nascent cosmologies or baby universes. The observable universe, under eternal inflation from dark energy, will asymptotically evolve to a de Sitter spacetime. This spacetime is a vacuum configuration with a cosmological constant $\Lambda$. The stationary metric for this spacetime is $$ ds^2~=~Adt^2~–~A(r)^{-1}dr^2~-~r^2d\Omega^2,~ A(r)~=~(1~-~\Lambda r^2/3) $$ A radial null geodesic with $ds^2~=~0$ and $d\Omega^2~=~0$ gives the velocity ${\dot r}~=~dr/dt$ $=~A(r)$, where this pertains to both out and in going geodesics near the cosmological horizon $r~=~\sqrt{3/\Lambda}$ as measured from $r~=~0$. The total action for the motion of a particle is $S~=~\int p_r dr$ $-~\int Hdt$. Consider the bare action of massless particles, using methods found in [1], across the horizon from $r$ to $r'$, $$ S~=~\int_r^{r'}p_rdr~=~\int_r^{r'}\int_0^{p_r}dp_rdr. $$ The radial velocity of a particle is ${\dot r}~=~dr/dt$ $=~dH/dp_r$, which enters into the action as, $$ S~=\int_r^{r'}\int_0^H{{dH'}\over{\dot r}}dr. $$ The field defines $H^\prime~=~\hbar\omega'$. The integration over frequencies is from $E$ to $E~-~\omega$, for the ADM energy. The action is properly written as $$ S~=~-\hbar\int_r^{r'}\int_E^{E-\omega}{{d\omega'}\over{\dot r}}dr, $$ where the negative sign indicates the quanta is tunneling across the horizon to escape the Hubble region with radius The radial velocity $$ {\dot r}~=~\sqrt{\Lambda/3}r $$ defines the action $$ S~=~-\hbar\int_r^{r’}\int_0^\omega{{d\omega dr}\over{\pm 1~-~\sqrt{\Lambda r^2/3} }}~=~\sqrt{3/\Lambda}tanh^{-1}(\sqrt{\Lambda/3}r) $$ The action is then the delay coordinate evaluated as $$ r^*~=~\int {{dr}\over{1~–~\Lambda r^2/3}}~=~\sqrt{3/\Lambda}tanh^{-1}(\sqrt{\lambda/3}r). $$ The domain $[0,~\sqrt{3/\Lambda})$ defines a real valued action. Since, $tanh^{-1}(x)~=~{1\over 2}ln((1~+~x)/(1~-~x))$ for $r~>~\sqrt{3/\Lambda}$ the argument of the logarithm is negative. In this case the action is $$ S~=~\sqrt{3/\Lambda}ln\Big({{\sqrt{\Lambda/3}r~+~1}\over{\sqrt{\Lambda/3}r~-~1 }}\Big)~+~i\pi\sqrt{3/\Lambda}. $$ The imaginary part represents the action for the quantum field emission as $r~\rightarrow~\infty$. The delay coordinate is defined on $[0,~\infty)$ which assures an S-matrix is defined on an unbounded causal domain, and this holds in general as well.

This action does describe the emission of photons by the cosmological horizon. A tiny production of bosons occurs which causes the horizon to slowly retreat away. Eventually the de Sitter spacetime decays away into a Minkowski spacetime as $t~\rightarrow~\infty$.

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"The final fate of the universe can be computed"---I love it! But I thought that the answer was supposed to be 42 :)+1 –  Gordon Feb 16 '11 at 18:15
    
The last to lines are not clear. There are two integral signs with limits over $\Lambda$, and the integration to obtain the last line seems wrong? –  MBN Feb 16 '11 at 18:24
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@lurscher: This answer is utter crap. He made up the physics of deSitter horizons--- this is not what happens, the "photon emission" is not emission, and it doesn't make the cosmological horizon go away. This is nonsense, -1, and please, please, unaccept it. –  Ron Maimon Sep 9 '12 at 3:04
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The action you compute is meaningless, the imaginary part you are considering is nonsense, the emissions of a deSitter horizon is given by redshifting a local Unruh temperature of the near-horizon limit as always. –  Ron Maimon Sep 9 '12 at 3:10
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This answer is problematic and nonstandard. There are parts that are correct, parts that are false and parts that are pure speculation. Its better to keep things ordered by what happens classically, semiclassically and then in specific proposals of QG. Therefore I prefer Lubos' answer –  Columbia Sep 9 '12 at 6:00

Hawking radiation from a de Sitter horizon is the cosmic microwave background radiation. Both are black body radiation. The Big Bang is a myth.

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that is an interesting viewpoint i haven't heard or read before, not sure right off if it makes sense or what direct evidence contradicts. I would appreciate if you could provide some references –  lurscher Sep 4 '12 at 16:53
    
No references, just intuition. Sorry, I don't usually operate on that level. So, I can't create a new paragraph in this format. Strange. More like a telegram. OK--consider, if de Sitter's model is correct. then there really exists a space-time horizon at a finite distance from the local observer. All inside this is called the "observable universe" At this horizon, we will see the same kinds of effect we see at the horizon of a black hole. Including Hawking radiation. This is what Bell Labs observed, not relic radiation from an imaginary, anthropocentric "big bang" Copyright Breton Carr 2012 –  breton carr Sep 9 '12 at 0:32
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Maimon: I don't think your attitude is logical. It is possible to argue against a false statement, people do it all the time, without having a full-blown argument to deflate. It's not necessary to annoying to be honest, but academia can train that talent out of the unwary. You may be the next Newton, son, but I think your real talent lies in Bunny of Love! Love and neutrinos to all. –  breton carr Sep 10 '12 at 4:30
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People sometimes forget that there are several observational support for a hot dense epoch. The point is, ok, you may assume that all CMB radiation comes from other place. But to take your idea seriously you have to be compatible with all other observations. For instance, say that you explain the black body radiation, what about the $10^{-5}$ perturbations around it with an almost Harrison-Zeldovich spectrum? And how can you explain the abundance of light elements with a radiation dominated phase? The baryonic acoustic oscillations which match the CMB perturbations? –  Sandro Vitenti Sep 21 '12 at 22:40
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To throw ideas is easy. The difficult part is to know the literature, the observations and work on the ideas to match everything already known (and more). –  Sandro Vitenti Sep 21 '12 at 22:44

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.

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but i don't see why you say the horizon is static and in equilibrium. can you elaborate? –  lurscher Sep 9 '12 at 18:20
    
@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. –  Ron Maimon Sep 9 '12 at 19:32

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.

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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 –  lurscher Feb 16 '11 at 16:16
    
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. –  Luboš Motl Feb 16 '11 at 16:49
    
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. –  Luboš Motl Feb 16 '11 at 16:51
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@lurscher: +1, and please accept this answer. The answer you accepted is completely, totally, unequivocally, wrong. –  Ron Maimon Sep 9 '12 at 3:05
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@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. –  Ron Maimon Sep 9 '12 at 3:08

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