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I need to clarify a detail about the particles horizon (also called the causality horizon, not to be confused with the event horizon). For simplicity, consider an euclidian static universe, with a sudden beginning at $t_{\circ} = 0$ (the "Big Bang" of that model), and a brutal ending at $t = t_{\bullet}$ (the "Big Crunch" of the model). The cosmological scale factor is constant : $a(t) = 1$ for $t_{\circ} < t < t_{\bullet}$. (think of that model as a 3D "game-universe" that you launch at some time on your computer, then quit after a while. Poor virtual creatures living in the game !)

Here's a picture I made to represent that simple universe. Distance $\mathcal{D}$ is the proper distance from the stationary observer, shown as a blue vertical word-line on the picture. The red vertical line is the world-line of a stationary particle. $\mathcal{H}_{\mathcal{E}}$ is the event horizon of the observer, while the past light-cone $\mathcal{H}_{\mathcal{C}}$ is the causality (or particles) horizon at time $t_0$ (the present of the observer). $\mathcal{A}$ and $\mathcal{B}$ are two arbitrary events.

game-universe

Now, what confuses me are the (very nice) pictures shown in the answer there : Is the cosmic horizon related to the Big Bang event?. I'm showing the pictures from that answer here for reference : horizon diagram

On these pictures, the particles horizon is depicted as an inverted (future-oriented) large cone, not as an evolving smaller (past-oriented) light-cone. Why ?

If I add this inverted light-cone on my picture above, event $\mathcal{A}$ would be outside that particles horizon. Yet it is an observable event for a time $t > t_0$ (as shown with the green dashed line).

I believe that the particles horizon shown on these pictures isn't right and should be labeled as something like "particles horizon distance" instead (not "particle horizon"). Am I wrong ?

EDIT : A variation of the first picture, with a better representation of the particle horizon (? I'm not convinced yet !) : game-universe 2

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Cham
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"Event A is outside the particle horizon" means by definition that at the time $t(\mathcal{A})$, no event of its world-line for $t \leq t(\mathcal{A})$ can be seen by the stationary observer at time $t(\mathcal{A})$, which agrees with your figure ($\mathcal{A}$ is to far from the observer on the time interval $[t_o \ t(\mathcal{A})]$). Similarly, in your picture, event $\mathcal{B}$ is inside the particle horizon because at $t = t(\mathcal{B})$, the stationary observer is able to see one point of the world-line of $\mathcal{B}$ that occurred for some $t \leq t(\mathcal{B})$. Hence there is nothing contradictory. Besides, your suggestion to label the "particle horizon" as something like "particles horizon distance" makes sense, as by definition (taken from your reference question), the "particle horizon" is "the distance - at that time - to the furthest object that can be seen". And to answer your initial question, the cone in the figures is correct, the distance to the furthest object that can be seen is necessarily increasing with the time (the more you wait, the more distant are the objects you can see).

  • I agree with your interpretation, but the question is still the same and unanswered : how should we represent the whole particle horizon hypersurface in these space-time diagrams ? A cone like what I've drawn, or an inverted cone like on the figures I've shown (which actually shows the distance as a function of time) ? In the case of my own figure, there's something puzzling : the distance to the horizon at time $t_0$ appears to be 0 (wrongly), unless we look at the base of the cone (at $t_{\circ} = 0$) ! It doesn't seem to be "local" in space-time. – Cham Sep 27 '16 at 21:44
  • I would say an inverted cone like on the figures you have shown. Why would the particles horizon at time $t_0$ be $0$ in your own figure? For me it would rather be $\mathcal{D}{\mathcal{C}}(t_0)$, as at time $t_0$, the stationary observer is able to see the event $(\mathcal{D}{\mathcal{C}}(t_0), t = 0)$ belonging to the world-line $(\mathcal{D}_{\mathcal{C}}(t_0), 0 \leq t \leq t_0)$. –  Sep 27 '16 at 22:12
  • The problem with the "inverted cone" is that event $\mathcal{A}$ would stay outside of it, even if you draw a full 45 degree cone all the way up to $t_{\bullet}$, while it is a visible event in the future of the observer at time $t_0 < t_{\bullet}$. I mean that the observer will be able to see $\mathcal{A}$ at a time $t_{\mathcal{A}} > t_0$. I don't see how to properly represent the particle horizon as an hypersurface in space-time. – Cham Sep 27 '16 at 22:18
  • Maybe the question could be reformulated as "Is the particle horizon really a light-like hyper surface, i.e. a cone in space-time ?". I'm beginning to suspect that it's actually a space-like hyper surface, or even just a simple 2-sphere of radius $c t_0$ (in the truncated flat space defined above). – Cham Sep 27 '16 at 22:29
  • It seems you misinterpret the definition "At a particular time, the particle horizon is the distance - at that time - to the furthest object that can be seen". Note at that time while you mention "in the future". There is no problem with $\mathcal{A}$ being outside the cone (not staying), its status w.r.t. the cone doesn't change with time. When you say "it is a visible event in the future", in that future, at that time, you are not dealing with $\mathcal{A}$ but with its future, for example $\mathcal{B}$, which is in the particle horizon because you can see an old version of it. –  Sep 28 '16 at 06:43
  • I made another version of the first picture. Please, see the last picture in the edited part of the question. Probably this is a better representation of the particle horizon, but I'm still not convinced yet. I'm not sure I understand it clearly. Event $\mathcal{A}$ is outside, but could be seen at time $t > t_0$, while $\mathcal{B}$ is inside but cannot be seen in any way since it is outside the event horizon. What is the usefulness of that (dotted) "particle horizon" ? – Cham Sep 28 '16 at 13:44
  • I notice that on your two figures, you have mentioned at the bottom the value $\mathcal{D}{\mathcal{C}}(t_0)$, so I guess you find some usefulness in this value. Well, if you now plot the two lines with co-ordinate $(\pm \mathcal{D}{\mathcal{C}}(t),t)$, you obtain the particle horizon cone you have drawn on your second figure. I hope this will help you understand its usefulness. –  Sep 28 '16 at 14:24
  • Yes, I do understand that the "inverted" (dotted) cone is the *evolution* of $\mathcal{D}{\mathcal{C}}(t)$ on the whole history of the central observer. But I'm still puzzled by events $\mathcal{A}$ and $\mathcal{B}$. Forgetting the event horizon, $\mathcal{A}$ is outside the particle horizon while $\mathcal{B}$ is inside, and yet both of them could be seen by the observer at some time (without the Big Crunch). The dotted $\mathcal{H}{\mathcal{C}}$ doesn't seem to be usefull in this regard. – Cham Sep 28 '16 at 14:31
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    Yes, surely both of them can be seen by the observer at some time, but not any older version of $\mathcal{A}$ at $t(\mathcal{A})$, whereas an older version of $\mathcal{B}$ can be seen at $t(\mathcal{B})$. That's the difference. –  Sep 28 '16 at 15:08