Although I don't know if he intended that paper to be the main focus of the discussion there. It seems like he was emphasizing a more basic concept (the idea of an event horizon) in that thread.

You are right it is interesting. I would welcome a discussion (of the cosmology aspects of it) here. Smoot's paper is both a core cosmology paper, dealing with basic cosmo issues like dark energy in an innovative way, and a "beyond standard" topic for discussion because it invokes the proposed entropic force. The regular posters at Cosmo forum are the ones most interested and best equipped to figure out and discuss the cosmology dimension of it, I think.

After a quick reading, I think the arguments are quite persuasive. I need to read through it more carefully though. This whole recent slew of papers using the holographic principle and entropic forces has certainly convinced me that this is a very promising area for research to go in to.

But it might be a blow against the paper that the prediction [itex]\Omega_{\Lambda}=2/3[/itex] is not within WMAP 2 sigma range. WMAP 7 has [itex]\Omega_{\Lambda}=0.734\pm 0.029[/itex]

I know have a few questions about what's done in this paper. Hopefully someone can shed some insight.

"At this horizon, there is a horizon temperature [itex]T_{\beta}[/itex] which we can estimate as
[tex] T_{\beta} = \frac{\hbar}{k}\frac{H}{2\pi}[/tex]

Really? Why proportional to H? Is it because cH gives the only natural acceleration scale, and you need an acceleration for the Unruh temperature?

Then in section 4, my first problem is that they say that surface terms can contribute to the stress tensor. Really? I've never seen that done before. but I'll take their word. Then he sets the surface term equal to [tex]\frac{a_s}{d_H}[/tex]. Again, I am assuming they make this choice because it is of the right dimension?

Then... "We would anticipate that the integral of the trace of the intrinsic curvature would be of order [tex]6(2H^2 + \dot{H})[/tex]. Huh?? How in the world do you anticipate that?

Finally... [itex]P = -\frac{2}{3}\rho_{c}[/itex]. I agree with that. But they say if you use the surface terms version you get
[tex]
P = -\frac{1}{\pi}\rho_c(1 + \frac{1}{4}\frac{\dot{H}}{H^2}).
[/tex]
Really? I get a rubbish answer when I try to use this formula. Anyone else want to give it a go?

I'm going to try to answer. I will look up "de Sitter temperature" first. I suspect you know more than I do about a good bit of what is involved here, so I will just hazard some ideas in a tentative way.

Our universe is not exactly de Sitter but it resembles de Sitter. So you could estimate its horizon temp by just using the temp of a roughly similar de Sitter universe. I think they make it plain that is what they are doing.

So let's look at why a de Sitter universe has a temperature, and what the formula for it is, and see if it doesn't come out to be what they said. I could be quite wrong but I'll give it a try now.

Let's see. One thing we know about deS universe is that H is constant.
And in the proper units H = sqrt(Lambda).

Lambda is curvature---the reciprocal of area. And the H in that deS universe is the reciprocal of time. So the units make sense and in fact it is an equation that is, I think, true in planck units.

But in our universe H is ALMOST constant. It is about 71 now and declining and approaching something like 61 asymptotically (H_{now}*sqrt(0.73) in the usual astronomer units). So why don't we use OUR H and treat it as if it were a deS universe H, and get the deS temperature, and treat that as an estimate. Because we are almost a deS and approaching one asymptotically.

So let's see how you find the deS temperature. I think it is basically just the BekenHawk temp of a black hole event horizon except you use the cosmic event horizon instead. Is this making sense?

OK this seems to be coming together. Here is a pedagogical source http://arxiv.org/abs/hep-th/0205177 . What we need is on page 3, around equation 3.6.
DeSitter space has a characteristic length L = sqrt (3/Lambda)

and a characteristic temperature T_{dS} = 1/(2 pi L)

Apparently Hawking and Gibbon derived this around 1977 and later proved that an observer in deS space would actually feel that temperature.

And we know that the constant Hubble rate in deS is H = sqrt (Lambda)

So the characteristic deS length is L = sqrt(3)/H

so the temperature T_{dS} = 1/(2 pi L) = H/(2 pi sqrt(3))

that is probably what Easson Frampton Smoot said it was except for that misbegotten factor of sqrt(3).

Ok, now this is confusing. Let's just look at the beginning of section 3. Here is my understanding of their argument.

1) Assume our universe is dS
2) Assume the horizon has a temperature related to the de Sitter temperature.
3) Equate this to an acceleration by the Unruh relationship.
4) "From this viewpoint, the dark energy is non-existent"

However, in 1) they are assuming dS, which is empty space with just cosmological constant (right?). Cosmological constant implies accelerated expansion, so 3) is no surprise, and 4) is does not seem like a logical conclusion.

Therefore for this to be logically consistent, the assumption 1) must not exist. Therefore it is my interpretation that the justifcation for 1) comes from section 4 about the boundary terms - "It is easy to show that if H^2 is highly dominant... the solution is simply de Sitter. Then 1) is now a justified assumption and the rest follows.

In other words, our universe acts like de Sitter because of boundary terms in the EFE, not because of cosmological constant.

That's how I understand it, too.
I read the paper a second time (I admit, still just skimming through it), and find it even more suspicious.
They use H to derive the acceleration. Worse, in this statement

which seems to be quite central to their reasoning, they're talking about the Hubble horizon. I get the impression that they're serious about that, as they then state their eq. 12, which seems to be the basis of their argument.

My point is: There's no such thing as a Hubble Horizon. The scale 1/H is irrelevant, except in a de Sitter universe. It has no physical meaning, and must not be used as the cause for a real physical effect.

My reasoning:
Let the mass/pressure term vanish. Then, spacetime is empty. Empty spacetime is minkowski, it has no horizons. You may, however, treat it as a FRW spacetime with arbitrary H, that's just a coordinate transformation. By virtue of eq. 12, said arbitrary H will then cause your spacetime to have an acceleration term. An arbitrary acceleration, to be sure.

That's a major assault against general covariance, without even an attempt to justify it.
Not acceptable.

Yes. It does make sense to me. With the emphasis that it acts like a de Sitter, not that obeys the de Sitter model with a positive Lambda.

If you would scatter sparsely some observers in a de Sitter, so that you could record data, the redshift-distance data would be like from our universe, because our universe acts approximately like deS. (But yet we have no positive Lambda, according to their idea: there is a different reason for it acting the way it does.)

I'm interested in how you follow the application of the Unruh relation.

Any help making this concrete would be appreciated.
They get the horizon temperature already, T = 4 x 10^{-30} kelvin.
That comes from the de Sitter approximation (acting like) plus work by Hawking and Horowitz.

Now let's look at the horizon from the standpoint of an observer at rest wrt background.
The horizon is rushing at him with speed 1.4 c! Is that right?
Because the distance from us to the horizon is 15.7 Gly, and a galaxy at that distance would be receding at rate 1.4c.
And yet we know that the horizon is not getting significantly farther away from us.
It's asymptotic proper distance is 16.4, which is almost the same as 15.7. So it is not getting significantly farther, think of it as standing still. And all these galaxies are falling headlong through it!

Now this horizon must have something like a "surface gravity" that is similar to the surface gravity of a black hole event horizon----something proportional to the temperature. By the Unruh formula? And we know the temperature T, we already estimated that. So we just apply the Unruh formula to find the acceleration of the horizon?

Is this the progression of ideas you sketched in your post, in points 2)--4)? And does it make sense?

So we have a horizon which is physically meaningful for us and which stays at the same distance from us.
We cannot see anything fall through----their redshifted bodies pile up on the boundary and gradually fade away but do not ever pass thru.
And yet for someone who is falling thru the boundary, there is a mild acceleration as he is falling thru. Acceleration on the order of a nanometer per second per second.

I would welcome your pointing out any mistakes you see.

http://arxiv.org/abs/1003.0668 Entropic dark energy and sourced Friedmann equations
Ulf H. Danielsson
7 pages
(Submitted on 2 Mar 2010)
"In this paper we show that a recent attempt to derive dark energy as an entropic force suffers from the same problems as earlier attempts motivated by holography. The remedy is again the introduction of source terms."

I checked Easson Frampton Smoot's equation (7) and their assertion that it was in agreement with observation.

Using temperature and the Unruh relation they conclude, after some cancelation, that the acceleration of the CEH is cH. This works out to be 0.69 nm per second^{2}.

Now (this may involve some circularity!) I tried to get a handle on the time derivatives of the scalefactor a(t). First I found that q = (1 + 3w)/2 = - 0.595. (I am postponing rounding off.)
Does anyone have an official estimate of q, something that has been fit to data?

What I'm betting on is that this value of deceleration FITS THE DATA on distance and redshift. Otherwise they would have noticed---if there was some glaring discrepancy.
q is something that should reflect curve-fitting at some point. (But I couldn't find a paper with an empirical value of q based on data.)

From that value of q, I calculated a"(t)/a(t) = - H^{2}q
which equals the second time derivative a"(t) because the present value of a(t) is normalized to 1.

a"(present) = - H^{2}q works out to be 0.48 nanometers per second^{2}.

So this is all backofenvelope but it comes down to 0.48 being similar to 0.69.

Smoot et al seem to be in good shape on this point: both ways to calculate the acceleration of the CEH give about the same answer, that being about half a nanometer per second^{2}.