# Entropy of Cosmological Event Horizon

1. Aug 13, 2006

### Mike2

Has anyone got a reference to the entropy of the cosmological event horizon? Is this entropy an upper limit of the entropy inside (like a black hole) or a lower limit?

I'm entertaining the idea that a shrinking cosmological event horizon puts a shrinking upper bound on the entropy inside it. And this shrinking entropy my be causing complex structures to appear such as life. It seems curious that life appeared on earth about 4 billion years ago which is about the same time that the cosmological event horizon began to shink with an acceleration in the expansion rate of the universe.

Any help is appreciated. Thanks.

2. Aug 13, 2006

### chronon

You might be interested in Black hole versus cosmological horizon entropy by Davis, Davies and Lineweaver

For links between life and cosmological entropy, look at http://www.mso.anu.edu.au/~charley/papers/LineweaverChap_6.pdf

3. Aug 13, 2006

### Mike2

Thanks for the references.

Reading, Tamara M. Davis, P. C. W. Davies & Charles H. Lineweaver, page 3, last paragraph, they write, "We assume cosmological event horizons do have entropy proportional to their area, as Gibbons and Hawking (1977) proposed. The total entropy of a universe is then given by the entropy of the cosmological event horizon plus the entropy of the matter and radiation it encloses. In Sect. 2 and Sect. 3 we assess the loss of entropy as matter and radiation disappear over the cosmological event horizon and show that the loss of entropy is more than balanced by the increase in the horizon area."

I think there may still be some confusion here. If the cosmological event horizon grows (to increase its entropy) as matter within it crosses over the horizon, then expansion would have to slow (to increase horizon area) as the universe becomes less dense. But I thought by GR that it was the greater density that slowed expansion, not lesser density. Not only that, but if the horizon becomes more distant (to increase its entropy) as matter crosses it, then that same matter would be regained as the horizon increased so that there would never be any loss to begin with.

Last edited: Aug 13, 2006
4. Aug 16, 2006

### Mike2

On the other hand, some think that the entropy of the cosmological event horizon is an upper limit to the entropy inside. For example:

http://search.arxiv.org:8081/details.jsp?qid=1155652792672-1431196444&r=pdf/hep-th/0408170

page 16, between equations 72 and 73 reads, "Equivalent upper bound may be suggested when one uses Hawking radiation from cosmological horizon (as it was communicated to us by P.Wang)."

And also from:

http://search.arxiv.org:8081/details.jsp?qid=1155653534812-1431196444&r=pdf/astro-ph/0406099

page 3, 2nd paragraph from the top, 2nd sentence, we read, "With this assumption, the existence of a horizon constrains only the entropy inside the final Hubble volume to be less than the area of its horizon."

And with something as life itself hanging in the balance, it seems imparative to straighten out this understanding of how the entropy of the cosmological event horizon restrains the entropy inside it. Is it an upper or lower bound?

Last edited by a moderator: Apr 22, 2017
5. Aug 23, 2006

### Mike2

New paper out:

http://arxiv.org/abs/hep-th/0608120

Where the surface term of the Einstein-Hilbert action is equated to an entropy. Their claim that information about the bulk term of the E-H action can be obtained from the surface term, and visa versa, seems to suggest that the entropy of the cosmological horizon is equivalent to the entropy INSIDE it. Any comments? For if the entropy inside could be larger than that of the horizon surface, then we could not gain info on one from the other - the inside entropy could be anything, for example.

I wonder if this also means that ANY horizon has an entropy, horizon such as those due to acceleration, etc. I certainly would appreciate any comment from those who are interested in these kinds of things. Thanks.

Last edited by a moderator: Apr 22, 2017
6. Sep 6, 2006

### Mike2

I'm having trouble finding a derivation that breaks down the Einstein-Hilbert action into surface and bulk terms. Any help out there? Thanks.

7. Sep 7, 2006

### hellfire

Take a look to this paper.

8. Oct 11, 2006

### Mike2

Thank you, I read it, I have some questions...

The last paragraph of page 4 states...

"$$$A' = \,\,\int_\nu {d^4 x\,L'(\partial ^2 \phi ,\,\,\partial \phi ,\,\,\phi )\,\, = \,\,\int_\nu {d^4 x\,L(\partial \phi ,\,\,\phi )\, - \,\int_\nu {d^4 x\,\partial _a [\phi \,\frac{{\partial L}}{{\partial (\partial _a \phi )}}]\,\, \equiv \,\,A\, - \,S\,\,\,\,\,\,\,\,(3)} } }$$$

The second term S can, of course, be converted into a surface integral over the 3-dimensional boundary $$$\partial \nu$$$. If we consider a static field configuration (in some Lorentz frame) then the second term in (3) will have the integrand $$$\nabla \cdot [\phi \,(\partial L/\partial (\nabla \phi ))]$$$ which can be converted to an integral over a two dimensional surface on the boundary $$$\partial \partial \nu$$$. Taking the time integration over an interval (0,T), the second term in (3), for static field configurations, will reduce to

$$$S\,\, = \,\,\int_0^T {dt\,\int_{\partial \nu } {d^3 \nabla \, \cdot \,[\phi \,\frac{{\partial L}}{{\partial (\nabla \phi )}}]} } \,\, = \,\,T\,\int_{\partial \partial \nu } {d^2 x\,\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over n} \cdot } \,\,\phi \,\frac{{\partial L}}{{\partial (\nabla \phi )}}\,\, \equiv \,\,\int_{\partial \partial \nu } {d^2 x\,\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over n} \cdot \,P} ,\,\,\,\,\,\,\,(4)$$$

This procedure allows one to reconstruct the bulk action if the surface term in known."

My question has to do with the notation $$$\partial \partial \nu$$$ which is normally the notation for the boundary of a boundary which is identically zero always. Is this just an unfortunate use of notation? Does he not mean that the 3-dimensional surface would generally be a 2-dimensional surface that changes with time, and in the special case that the 2-dimensional surface does not change with time, then the time dependence can be pulled out separately from the 3-dimensional generalized surface to give a numeric time value multiplied by a static 2-dimensional surface? Thanks.

Also, am I correct in taking (3) above to be just the multidimensional version of the action that is the more general version of the 1-dimensional version one would get from an action produced from the lagranian in (1)?

I have more questions. But these troubles me the most for right now.

Last edited: Oct 11, 2006
9. Oct 15, 2006

### hellfire

In the static case neither $\phi$ nor $L$ depend on $t$ and thus the integral for $dt$ results in a factor $T$ that multiplies the integral over the spatial coordinates. This other integral is therefore over a 3-dimensional volume $\partial V$. Since it contains a divergence it can be written as a 2-dimensional surface integral over $\partial \partial V$.

The expression (1) is for a system with one degree of freedom $q$ and (3) is the generalization for fields in four space-time dimensions.

Last edited: Oct 15, 2006
10. Oct 16, 2006

### Chronos

I think this approach is fundamentally flawed. It is unrealistic to treat entropy as purely a surface effect. Not all entropy is concentrated on the surface of the event horizon. It may be mathematically convenient, but is not realistic.

11. Oct 16, 2006

### Mike2

If it is mathematically correct, then it cannot be fundamentally flawed. It's just different. He explains it like any other horizon which hides information from the observer. It's just like the Unruh affect where acceleration produces a horizon behind the observer and creates a temperature that the accelerating observer feels. Only in the case of universal expansion, the acceleration is away from every direction.

12. Oct 16, 2006

### Mike2

Thanks. It sounds like we are in agreement.

So let me go on to my next question. Below is equation (2) from page 4:

$$$\delta A'\,\, = \,\,\int_{P_1 }^{P_2 } {dt\left[ {\frac{{\partial L}}{{\partial q}}\,\, - \,\,\frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial \dot q}}} \right)} \right]} \,\delta q\,\, - \,\,q(\delta p)\,\mathop |\nolimits_{P_1 }^{P_2 } \,\,\,\,\,\,\,\,\,\,(2)$$$

What does it mean that $$$\delta p$$$ = 0 at the end points. I assume this means that in the case of the gravitational field there may be a nonzero gravitational field at the horizon (as preceived by the observer), but at the horizon that gravitational field cannot be changing. Is this right?

The thing that confuses me is where in time does this horizon exist? Are we talking about a snapshot of the present situation? Or is the horizon that which is preceived by the observer, but represents the situation billions of years ago? The cosmological event horizon at this very instant is not observable and has shrunk considerably from the one we observe (if it is observable at all yet). Is the entropy of the bulk equal to the entropy of the horizon that we would antisipate to exist at the moment? Or do we start with the observed cosmological horizon and the bulk consists of the accumulation of the volumes of the bulk in the distant past to the nearby volumes of bulk towards the present? If the distant past surface is effecting the nearby present bulk, then it is easy to see how the red shifting of galaxies as they near the horizon would freeze and thus there is no momentum of the gravitational field. But if it must be the antisipated present horizon that is effecting the present bulk, then how can we know that there is no change in momentum in the gravitational field there? Thanks.

13. Oct 18, 2006

### hellfire

I am afraid I cannot answer your question. Let me make a summary about what I understand and what I do not understand.

The point here is that starting from an action $A$ with a Lagrangian $L (q, \dot q)$ from which you can obtain the equations of motion holding fixed $q$ (as usual), you can compute a Lagrangian $L^{\prime} (q, \dot q, \ddot q)$ from which you can obtain the same equations of motion holding fixed the momenta $\pi = \partial L / \partial \dot q$. The action $A^{\prime}$ for this Lagrangian can be written as $A^{\prime} = A - S$, being $S$ a surface term.

In case of gravitation with $q = g_{\mu \nu}$ it seams that he can recover the Einstein-Hilbert action (that contains second order derivatives of $g$) imposing some conditions on the entropy of a Rindler horizon and on $S$. As far as I understand, this possible because it is always possible to perform a local coordinate transformation to a free falling frame where $\partial g = 0$, and this makes the bulk term $A$ vanish. However, it is unclear to me what the relation between the surface term and the horizon actually is. Moreover, I do not see how this paper relates to cosmology, basically because I do not understand the thermodynamics of cosmological horizons.

By the way, note that in page 8 he mentions that the condition of holding $\pi$ fixed allows for arbitrary scalar functions of $g_{ab}$ that can be added to $L^{\prime}$. It is argued that this is the reason for the inability of this approach to make any statement about the cosmological constant (which is actually the thing that leads to a cosmological event horizon...).

Last edited: Oct 18, 2006
14. Oct 19, 2006

### Mike2

Perhaps this is so because the value of a cosmological constant has no baring on whether a surface term can be equated to a bulk term. Whatever the cosmological constant, the above equality holds. I think that was his point.

As far as my question is concerned, I wonder if the answer depends on whether the constraint of the surface term inforces that constraint instantaneously throughout the bulk. If the information about the state of the surface term (the horizon) propagates instantaneously throughout, then we must take a time slice of the same instance throughout the universe. However, information travelling instantaneously seems to contradict relativity.

Is their any other situation that would indicate how an entropy constraint on information propagates? For example, how about the case of a black hole? If two black holes were to colide, would there be waves on the event horizon that propage slower than the speed of light? Or is there no surface tension on the surface of a black hole event horizon?

15. Nov 12, 2006

### Mike2

Heres what I read from the abstract:
Black hole versus cosmological horizon entropy
Auteur(s) / Author(s)
DAVIS Tamara M. ; DAVIES P. C. W. ; LINEWEAVER Charles H. ;

which is at:

http://cat.inist.fr/?aModele=afficheN&cpsidt=14959623

"In most cases, the loss of entropy from within the cosmological horizon is more than balanced by an increase in cosmological event horizon entropy, maintaining the validity of the generalized second law of thermodynamics."

It is my understanding that as the universe accelerates in its expansion, the distance to the cosmological event horizon shrinks, and with it the surface area of the cosmological event horizon, and with that the entropy of the cosmological event horizon. So it would seem that with accelerated expansion, the entropy of the comological event horizon should decrease, not increase. So how can they say it increases when it is shrinking? This sounds like an obvious error. What am I missing?

It is interesting to note the entropy density as the radius to the cosmological event horizon changes. It is:

$$$Entropy/Volume\,\, \propto \,\,Surface\,area/Volume\,\, = \,\,(4\pi r^2 )/(\frac{4}{3}\pi r^3 )\,\, = \,\,3/r.$$$

niavely, as r approaches zero with inflation out of control, the entropy density would increase without limit. Maybe that's why we have expanding space to begin with. But since black holes limit the entropy density, then either inflation is stopped or some other mechnism must come in to increase entropy, right?

Last edited: Nov 12, 2006
16. Nov 12, 2006

### hellfire

I have not done the calculations but in that paper (that you can find here: http://arxiv.org/abs/astro-ph/0305121) it is shown graphically that the proper distance to our cosmological event horizon increases. Moreover, it is argued that for models in which this distance decreases the area increases because of the effect of curvature. See section 2 and figure 1.

17. Nov 12, 2006

### Mike2

I agree with the second sentence. But I think that the third sentence is where they may have made a mistake. They present it as an assumption without proof. But if the total entropy of the universe is the entropy of the cosmological event horizon PLUS the entropy of the matter and radiation it encloses, then that would seem to exclude a holographic entropy bound since the entropy inside might be anything. For an entropy bound places an upper limit so that the entropy inside the horizon must be less than the entropy of the horizon. But that is not possible if the addition of the two must alway be greater than zero.

And it seems that the universe is approaching a flat curvature so that the surface area decreases with decreasing horizon distance. As I understand it, the curvature of space can make it so that the circumference is not pi times the diameter - or that the area is not 4pi*radius^2. But even in curved space the circumference (or surface area) grows with the radius. So I don't know what he's thinking. Perhaps he's thinking of the early universe where the curvature is changing rapidly with radius.

Last edited: Nov 12, 2006
18. Nov 12, 2006

### Chronos

What curved space are we talking about here? I allow that the universe may have always had a horizon, but never a radius. Pi is irrelevant.

19. Nov 13, 2006

### hellfire

The cosmological event horizon is a very strange one. The results for other horizons are based on the weak and strong energy condition (even the result of the area theorem). But the existence of a cosmological event horizon implies the violation of these energy conditions. This is the reason that it is not clear to me that things such as the holographic principle should hold for cosmological horizons also.

As I said, I have not done the calculations but I can imagine something like this:

First, in every model with a cosmological event horizon the curvature tends to be zero. This is because sooner or later they become dark-energy dominated. (Assume we do not take into consideration models that violate the dominant energy condition such as big-rip scenarios).

Take the last example of figure 1: (0.3, 1.4). In an universe like (0.3, 1.4), a sphere at a given fixed radius would increase its area because space is closed, but, however, it is less closed as time passes. The event horizon, however, decreases its radius. The contribution to the area due to the evolution to flatness is greater than the contribution due to the decrease of the radius, so that the area increases.

In the other cases of figure 1: (0.3, 0.7) and (0.3, 0.3) the problem does not show up, as the event horizon increases its radius.

It would be nice to see how this relation between area, expansion and event horizon generalizes. I will try to make some calculations when I have time.

Last edited: Nov 13, 2006
20. Nov 25, 2006

### Mike2

Well, I would think that if it holds in general, then it should hold for the whole observational universe since you can't get more general than that.

As time goes on, the dark energy density will tend towards 1 and the other densities tend to 0. What does that do to the surface area vs. radius of the observable universe?

Also, perhaps we are in a bit of "expansion confusion" as to which horizon is relevant to the loss of information having an effect. Would you care to have a discussion on another thread about "Expansion Confusion" of arXiv:astro-ph/0310808 v2. It would seem that their discussion lacks reference in just about every sentence to proper verses observed measures. Perhaps we can better the language if we discuss it.