Entropy of Cosmological Event Horizon

[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.

 

[chronon]

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.

You might be interested in http://arxiv.org/abs/astro-ph/?0305121 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

 

[Mike2]

You might be interested in http://arxiv.org/abs/astro-ph/?0305121 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

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.

 

[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.

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?

 

[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?

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.

 

[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.

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

 

[hellfire]

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

Take a look to this paper.

 

[Mike2]

Take a look to this paper.

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.

 

[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 $\partial \partial V$.

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)?

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.

 

[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.

 

[Mike2]

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.

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.

 

[Mike2]

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.

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.

 

[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...).

 

[Mike2]

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...).

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 traveling 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?

 

[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 into increase entropy, right?

 

[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.

 

[Mike2]

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.

In this paper we explore the range of validity of the GSL. 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 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.

 

[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.

 

[hellfire]

then that would seem to exclude a holographic entropy bound since the entropy inside might be anything

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.

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.

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.

 

[Mike2]

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.

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 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.

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.

 

[Mike2]

OK, so I understand that the expansion of the universe is not increasing. It is only decreasing to a constant value. And thus the proper comic event horizon is not shrinking towards zero. So I can not think that the shrinking proper cosmic event horizon is forcing the entropy in the universe to decrease. But this horizon is the present existing horizon that we can not see because it is not within our light cone. It exists on the same time slice we are living on.

But what about the horizon(s) that we observe as we look down our light cone to the past. I suspect that it is this apparent horizon which should have an effect on us since information can not travel faster than light. How has this appearent horizon evolved in relation to the appearent size of the universe?

 

[Mike2]

OK, so I understand that the expansion of the universe is not increasing. It is only decreasing to a constant value. And thus the proper comic event horizon is not shrinking towards zero. So I can not think that the shrinking proper cosmic event horizon is forcing the entropy in the universe to decrease. But this horizon is the present existing horizon that we can not see because it is not within our light cone. It exists on the same time slice we are living on.

But what about the horizon(s) that we observe as we look down our light cone to the past. I suspect that it is this apparent horizon which should have an effect on us since information can not travel faster than light. How has this appearent horizon evolved in relation to the appearent size of the universe?

If we could see even past the CMB, how far could we see until there was a horizon blocking out view? What horizon would this be, the Hubble's sphere, the cosmological event horizon, the particle horizon? Would we be able to see things receding faster than light at the time of emittion? Thanks.

 

[hellfire]

If we could see even past the CMB, how far could we see until there was a horizon blocking out view? What horizon would this be, the Hubble's sphere, the cosmological event horizon, the particle horizon?

Consider a photon or massless particle emitted at t = 0. The position of this photon today would be located at our particle horizon. The position from which this photon was emitted in past would be located in the t = 0 surface of our past light-cone.

Would we be able to see things receding faster than light at the time of emittion? Thanks.

Yes, in the $\Lambda$-CDM model, the objects located at z > 1.65 from which we receive light today had a recession speed greater than $c$ when their light was emitted. Moreover, objects located at z > 1.4 have a current recession speed greater than $c$.

 

[Mike2]

Consider a photon or massless particle emitted at t = 0. The position of this photon today would be located at our particle horizon. The position from which this photon was emitted in past would be located in the t = 0 surface of our past light-cone.

If we can see all the way back to t=0, then you're saying that there would be no horizon to obstruct our view to the very beginning (assuming unobstructed source). But I have to wonder how Inflation would figure in. Surely we should not be able to detect anything from that time (assuming an unobstructed source)

Yes, in the $\Lambda$-CDM model, the objects located at z > 1.65 from which we receive light today had a recession speed greater than $c$ when their light was emitted. Moreover, objects located at z > 1.4 have a current recession speed greater than $c$.

Is this because expansion slowed somewhat due to gravity and the horizon (which one?) increased to encompass photons emitted by sources receding faster than light when emitted? This sounds like there are objects whose red shift and recession rate preclude us from ever having been able to see them. And this seems to be in contradiction with being able to see all the way back to t=0 (assuming non-obstructed source).

 

[hellfire]

At least in principle, or from a pure theoretical point of view, I do not see any impediment to see back to t = 0. Inflation is a period of expansion as any other, the only difference is that expansion is stronger. If we would be able to see some source before or from the beginning of inflation this would be located extremely far away today, more than 45 Gly, that is the current location of the particle horizon without considering inflation.

Objects that we cannot see in any future are located beyond our event horizon. The notion of event horizon is not related to superluminal recession speeds. The event horizon appears in models with accelerated expansion, whereas superluminal expansion takes place in every model that obeys the Hubble law.

 

[Mike2]

At least in principle, or from a pure theoretical point of view, I do not see any impediment to see back to t = 0. Inflation is a period of expansion as any other, the only difference is that expansion is stronger. If we would be able to see some source before or from the beginning of inflation this would be located extremely far away today, more than 45 Gly, that is the current location of the particle horizon without considering inflation.

(In this discussion we are ignoring any thermalizing interactions that would prevent direct contact with particles traveling at the speed of light from anytime in the past, even from the singularity, t=0) We still can't see to the edge of the existing universe. We can only see a portion of it. The universe is said to be e^60 times bigger than what we can see. Are you saying that at t=0 we can see all of the universe and that much of it has since left our view? If we could see ALL of the universe at any time, then what horizon did it cross so that we can no longer see it. If we could not see all of the universe at t=0, then does the principle that only premits partial view act itself as an horizon?

Or perhaps if we could see without obstruction to t=0 to the very singularity itself, then would that constitute an horizon of zero size and zero surface area and thus zero entropy?

I still get confused when I study the paper "Expansion Confusion" found at:
http://arxiv.org/abs/astro-ph/0310808

I look at graphs such as Fig 1, page 3, and Fig 3, page 11, and I see our light cone intersecting both the Hubble sphere and the particle horizon. And it would seem that trying to look down our light cone past the Hubble sphere we would encounter light moving towards us in space moving even faster away from us so that we can not now observe it on our light cone. This would seem to be a horizon. However, this intersection with the Hubble sphere occurs at about t=4Gyr. And we know we can see past this to the surface of last scattering at t=300,000yr. What's going on?

 

[hellfire]

First, the Hubble sphere is not an horizon. To avoid further confusion I see no reason to talk about the Hubble sphere here. Moreover, as far as I know, this intersection between light cone and particle horizon is not an horizon nor anything special that has to be considered. To understand this consider a flat and static space. Let's say the universe starts at t = 0 and we are located now at t = T, always at the same spatial position. Our past light-cone is a cone with its vertex located at our position and a base with radius R = c T. This means, we are able to see right now the light from objects located at that position on the base of the past light-cone. The particle horizon is a cone with vertex located at our postion at t = 0 and base with radius R = c T at t = T. This means that those objects from which we receive light right now are currently located at R = c T from us (assuming they did not disappear in the meanwhile...). Since space is static, this is the same distance than at t = 0 (this is not the case for an expanding space) ¿Where would the particle horizon intersect the past light-cone in the diagram you mention? A t = T/2 and a distance R/2. This location has no special meaning. Our only horizon here would be at c T.

We still can't see to the edge of the existing universe. We can only see a portion of it. The universe is said to be e^60 times bigger than what we can see. Are you saying that at t=0 we can see all of the universe and that much of it has since left our view? If we could see ALL of the universe at any time, then what horizon did it cross so that we can no longer see it. If we could not see all of the universe at t=0, then does the principle that only premits partial view act itself as an horizon?

There is no horizon that makes it impossible to see t $\rightarrow$ 0. However, for all practical purposes our horizon will be after inflation. Even after inflation the redshift is enormous and before inflation sources will be redshifted $\rightarrow \infty$.

 

[Mike2]

There is no horizon that makes it impossible to see t $\rightarrow$ 0. However, for all practical purposes our horizon will be after inflation. Even after inflation the redshift is enormous and before inflation sources will be redshifted $\rightarrow \infty$.

I'm having trouble following your reasoning.

Let's consider deSitter space for a moment. I understand that our universe is asyptotically headed towards being a deSitter space. Now in a deSitter space, space is expanding at a given rate for all time. And there is an horizon in deSitter space, etc. So if we are headed towards a deSitter space which has an horizon, when will that horizon appear in our universe? And which horizon will that be?

The cosmological event horizon is the past light cone looking back from very far in the future, from infinity. But in the Figures I refer to in post post 26, that light cone intersects the particle horizon, just as much as our present light cone. So if the event horizon crossing the particle horizon is a legitemat horizon (the deSitter space horizon), why is not our present light cone intersecting the particle horizon a legitemate horizon? Is it because nothing is leaving the particle horizon, so we see nothing crossing it?

 

[hellfire]

In my opinion you should try to understand the mathematical definitions instead of merely interpreting those diagrams. There exists two horizons, the particle horizon and the event horizon. In http://arxiv.org/astro-ph/0305179 you can find the definitions, that easily follow from radial null paths in a RW-metric.

Consider a flat de-Sitter space-time. The scale factor evolves as:

$$a(t) = e^{Ht}$$

To calculate then the comoving distance $\chi$ and proper distance $D$ to the horizons you have to make use of the following formulas.

Particle horizon: The current distance of a photon sent at t = 0 from our position. Note that this definition means that the limit that we can observe reaches t = 0.

$$\chi_{ph} = c \int^t_0 \frac{dt}{a(t)}$$

$$D_{ph} = a(t) \chi_{ph}$$

Event horizon: The distance that a photon sent at t = t from our position will reach at t = $\infty$, or, equivalently, the limit from which photons that are emitted today will never reach us in future.

$$\chi_{eh} = c \int^{\infty}_t \frac{dt}{a(t)}$$

$$D_{eh} = a(\infty) \chi_{eh}$$

You can verify that for the de-Sitter model this means:

• The event horizon is located at a fixed proper distance always, i.e. there exists a limit from which photons that are emitted today will never reach us in future, or, equivalently, there exists an upper limit on the distance that a photon sent by us today will reach in t = $\infty$. This limit is at a fixed proper distance (a fixed distance on a spatial hypersurface).
• The particle horizon always increases its proper distance. However, for t = $\infty$ the particle horizon is located at a fixed comoving distance. This means that no new comoving objects will ever enter the particle horizon and the amount of objects we will be able to observe is not infinite in an infinite de-Sitter universe. This is also mentioned in the paper you had referenced.