Entropy of Cosmological Event Horizon

  • Thread starter Mike2
  • Start date
  • #26
1,306
0
hellfire said:
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?
 
Last edited:
  • #27
hellfire
Science Advisor
1,047
1
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. Lets 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 dissapear 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 [itex]\rightarrow[/itex] 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 [itex]\rightarrow \infty[/itex].
 
Last edited:
  • #28
1,306
0
hellfire said:
There is no horizon that makes it impossible to see t [itex]\rightarrow[/itex] 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 [itex]\rightarrow \infty[/itex].
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?
 
Last edited:
  • #29
hellfire
Science Advisor
1,047
1
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 [Broken] 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:

[tex]a(t) = e^{Ht}[/tex]

To calculate then the comoving distance [itex]\chi[/itex] and proper distance [itex]D[/itex] 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.

[tex]\chi_{ph} = c \int^t_0 \frac{dt}{a(t)}[/tex]

[tex]D_{ph} = a(t) \chi_{ph}[/tex]

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

[tex]\chi_{eh} = c \int^{\infty}_t \frac{dt}{a(t)}[/tex]

[tex]D_{eh} = a(\infty) \chi_{eh}[/tex]

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 = [itex]\infty[/itex]. 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 = [itex]\infty[/itex] 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.
 
Last edited by a moderator:

Related Threads on Entropy of Cosmological Event Horizon

Replies
19
Views
3K
  • Last Post
Replies
4
Views
6K
Replies
2
Views
3K
Replies
1
Views
4K
  • Last Post
Replies
19
Views
4K
Replies
12
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
2K
Top