# Final temperature in a de-Sitter universe

1. Feb 12, 2004

### hellfire

Consider a de-Sitter universe with exponential expansion (cosmological constant dominated). As explained here:

http://math.ucr.edu/home/baez/end.html

such an universe does not tend to a thermal death, or to a 0° K state, but to a state with a constant temperature above 0 °K (prof. Baez estimates 10^-30 °K). After this temperature is reached, it will remain constant regardless of the susequent expansion of space. This is due to the fact that the cosmological horizon radiates in a similar way that observers in a Rindler space are immersed in an Unruh radiation.

With my very limited knowledge about QFT, I understood qualitatively the reason for the Unruh radiation (creation and annihilation operators are defined in different ways for Minkowski and Rindler spaces, which leads to different gound states of the Hamiltonian) and I found that it depends directly on the acceleration (T = a / 2 pi).

My assumption is that the same (or a similar) derivation applies for a de-Sitter universe and that the temperature does also depend on the acceleration. Correct?

But then the following question raises: if the acceleration is not constant and is growing exponentially, does this mean that the temperature in a de-Sitter universe will grow due to this effect? This seams very strange to me…

Thanks.

Last edited: Feb 12, 2004
2. Feb 12, 2004

### marcus

this is an absolutely fascinating topic, hellfire please elaborate as best as you can and fill in more detail even if tentative.

IIRC baez says "about" or "roughly" E-30 kelvin. It is only a wild guess.

Unruh radiation (arising from the sheer fact of acceleration!) does come from a Rindler horizon. The stuff left behind that can never catch up.

And that rindler horizon is like a BH event horizon. and the hawking temp is the SAME namely a/2pi.

just repeating what you already said or know.

or if somebody else can help out?

My intuition is that not. Because if one puts in a positive Lambda what one gets is a cosmological horizon that converges to a certain proper distance. in the limit AFAIK the cosm. horizon has a definite area. Everything that we every can see or will be able to see is plastered against that wall and redshifting out of existence

like the shadows on the wall of a cave

it is very much like the finite area of a BH event horizon and the area keeps in step with the reciprocal temperature

this is just quick pants-seat reaction, intuition only.
I think the temp is a limiting temp BECAUSE of the positive cosmological constant which is what gives us the "cosmological horizon"

by the way "cosmological horizon" is not the same thing as the hubble sphere
for clarification of the distinction you could look at
Lineweaver's article astro-ph/0305179
around page 5 and page 6 figure 1.

3. Feb 12, 2004

### hellfire

Hi marcus. I went through the derivation of the Unruh radiation with troubles. But there are two important points I should mention. First, I read that the Rindler space-time is a consecuence of an uniformly accelerated motion, but I am not able to find out whether an nonuniform acceleration leads to a significant different result for the radiation temperature. My impression is that it does not. Second, an obvious one: the acceleration term in T = a / 2 pi is the proper acceleration. I mention this because I am assuming that in the case of the de-Sitter space, the acceleration term which counts for the temperature is the acceleration of the scale factor... and may be I am wrong here. This is a variable value and would lead to a variable radiation temperature.

4. Feb 13, 2004

### marcus

hellfire, as I confessed in earlier post I am hoping very much
that some other people respond
currently you are the one here who knows most about this or has thought the most
it is a good question

first, what exactly does Baez mean by the cosmological horizon.
then, what is its area measured in Planck units of area
this should, I suspect, relate to Baez estimate of temperature

in terms of Planck temp the temperature should be around
$$\frac{1}{\sqrt A}$$

this is wild spec, you understand----have to start somewhere!
we have to have a more knowledgeable person to guide us

first, what exactly do they mean by the cosmol. event horizon

one often sees some figure for it like 60 billion LY
this is the current distance in the usual RW metric that
cosmologists use of the farthest object we shall ever see
no matter how long we wait

second, what is the area of the sphere with that radius

third, is that like the surface of a black hole because no info beyond that spherical bubble will ever reach us even if we wait forever for it.

fourth, if it is like the (everted, "inside out") surface of a BH then does it have a temperature like one over the sqrt of the area?

try it, nothing to lose, yr guess as gd as anyone's

5. Feb 13, 2004

### hellfire

Thanks for your answer, marcus, but now I am totally lost. You are telling me that the temperature depends on the area of the horizon, but where did you get this formula or relation from? This is new to me. Moreover, if we consider the case of the Unruh radiation: as fas as I know, the Rindler horizon has an infinite area, how does this fit here? I'm afraid this will turn out to be too complicated...

By the way, your definition of the cosmological horizon seams correct to me. In other words: space expands too fast for the light to reach any comoving point outside the cosmological horizon. Also, you are right that the cosmological horizon in a de-Sitter universe will remain constant,

$$R = c / H = \sqrt {3/ \Lambda}$$

Thus, if the formula with the dependence of the temperature with the horizon area is correct, this would give, indeed, an explanation for the constant radiation temperature in a de-Sitter space.

Nevertheless, my actual questions are: in case of a de-Sitter space, is applicable the formula T = a / 2 pi ? If yes, what exactly is 'a'? If I find some answer I'll post it.

Regards.

6. Feb 13, 2004

### marcus

Without trying to explain, first lets just see if we can reverse-engineer Baez estimate of the final temperature.
He says about E-62 on planck temp scale.

(actually he says E-30 kelvin but a kelvin is 0.7E-32 natural and its easier to calculate if G=hbar=c=k=1)

Now Lambda, by current estimate is about E-123 on planck scale

It looks like the temperature is about equal to the square root of the cosmological constant. Is this how Baez got his back of envelope estimate of the final temperature of the U.

7. Feb 13, 2004

### marcus

Let's gradually introduce a little more accuracy and narrow things down. You know what H is and the "hubble time" which is the reciprocal of H. And you can find the "critical density" from H.
And you know the dark energy density is estimated at 73 percent of critical.

So in natural units,
hubble time of 13.8 billion years = 8.06E60 planck time units.

Hubble radius c/H of 13.8 billion LY = 8.06E60 planck length units

What is the critical energy density?

(3/8pi) divided by (8.06E60)2

this works out to 1.8E-123
this is rho crit expressed in planck units of energy density
-----------------------------------

Lambda is 73 percent of that or 1.3E-123

that is the figure for the cosmological constant I always use and it seems to work well with what other people say
------------------------------------

So we start taking square root of that 1.3E-123 and we will be
getting the temperature Baez was talking about for the end of the Day.

what sort of coefficients would you like to put in
is it going to be 3 or 3pi or something else?

8. Feb 14, 2004

### marcus

Tammy Davis thesis bears on this

hellfire,
something was just posted on web that bears on this question
of the temperature (and entropy) of the cosmological horizon

http://arxiv.org/astro-ph/0402278 [Broken]

Tamara Davis thesis at U of NewSouthWales (down under)
"Fundamental Aspects of the Expansion of the Universe and Cosmic Horizons"

For example look at equation (5.5) on page 71

this is page "89" of the file because the TOC and acknowledgements and stuff are on unnumbered pages before page 1.
So you have to scroll down to where the browser says "89"
and then it is actually page 71 of the regularly numbered pages.

I know Tammy Davis work and that of her PhD thesis advisor Charles Lineweaver. He was one of those in charge of the COBE project observing the CMB and he is world class and is apt to have one or two PhD students worthy of note.

So, although its a bother scrolling thru a 130 page thesis it may
be useful to you (dont know for sure but seems to have some bearing on this temperature of horizon business)

Last edited by a moderator: May 1, 2017
9. Feb 15, 2004

### hellfire

Very nice reference marcus, thank you. Yes, it seams that there is some confusion in my original question. It is not the existence of an acceleration the fact which leads to a thermal radiation, but the existence of a cosmological (or event) horizon. My impression is therefore that the formula T = a / 2 pi is actually not applicable, but I need to go through the derivation of the Unruh radiation again and also through this paper. Regards.

10. Feb 15, 2004

### marcus

if you get some intuition about how the horizon temperature is determined then I for one would be glad to listen

(and I cannot give you any guidance about whether accleration has something to do with it or not, I was trying to reverse-engineer the temperature estimate Baez gave and relate it to Lambda somehow, without grasping the connection. maybe someone else can.

i get the feeling that more curved horizons are hotter, and dont see why)

11. Jan 28, 2008

### rashmatash

The de Sittes space has a de Sitter horizen located at distance L from any observer. The temperature is proportional to 1/(2Pi*L). Since L is constant, the temperature is constant. See for more info Birrel & Davis "quantm fields in curved spacetime" chapter 5.