This question was inspired by this one

By definition we cannot see any event happening beyond the cosmological horizon. Let us assume that the expansion rate of the universe is such that the radius of the cosmological horizon is constant and centered at the observer (perhaps this assumption is inconsistent?). Now suppose I move 100 light years in one direction to meet another observer. This second observer has a different cosmological horizon, shifted in the direction the I moved. Hence I will be able to see events that I could not see before my trip, because they were outside my cosmological horizon. Then I travel back and show my friends a video of the event. How is that possible if the event was always beyond their cosmological horizon? (My trip should not be able to overcome that limitation, correct?)

  • $\begingroup$ The only way to get a horizon at a constant distance is in an exponentially expanding universe i.e. a de Sitter geometry. This isn't so esoteric as our universe will asymptotically approach the de Sitter geometry with time. Anyhow, do you specifically want to know about the de Sitter geometry, or are you asking about a geometry like our current universe in which case the horizon distance isn't constant? $\endgroup$ – John Rennie Dec 6 '15 at 11:13
  • $\begingroup$ @JohnRennie yes, I believe a de Sitter geometry would be more illustrative to figure out where my conceptual problem is (but perhaps I am wrong) $\endgroup$ – user83548 Dec 6 '15 at 14:20
  • $\begingroup$ Can you clarify what you mean by cosmological horizon? Do you mean the event horizon or the particle horizon? See this post for a definition of both. $\endgroup$ – Pulsar Dec 8 '15 at 4:18
  • $\begingroup$ it's the same question ... far mediators send old reports then the sum of the age of the information plus its travel time must be under the EH $\endgroup$ – user46925 Dec 8 '15 at 4:29

I will assume that you mean by cosmological horizon the cosmological event horizon. Now, there are two possibilities:

  1. You travel to a region that is (still) located inside the event horizon of the Earth. Then, by definition, the light that you observe at that location will also eventually reach the Earth (if it hasn't already). In fact, that light will have reached the Earth before you have travelled back, so you'll have nothing new to tell you fellow Earthlings.

  2. You travel to a region that is located outside the event horizon of the Earth. Then, again by definition, you are able to see light that will never reach the Earth. However, nothing can reach the Earth: you can no longer travel back either, nor can you send any signals to Earth. Once you are outside the Earth's event horizon, you are cut off from Earth forever.

In other words, the region that you travelled to might have its own event horizon, but that doesn't matter. What matters is whether or not you are still inside the event horizon of the Earth. If you are, then the Earth will receive the same information as you, and before you can make it back. If you aren't, then you can no longer communicate with the Earth, let alone travel back.

A final word of caution. A different region of space doesn't necessarily have a different event horizon. The FLRW metric is an idealization, and only valid on large scales. A region that is e.g. 100 light years away is gravitationally bound to us, and in the Standard $\Lambda$CDM model it will remain gravitationally bound. Since all observers in a bound structure can send signals to each other, they all share the same cosmological event horizon.

  • 1
    $\begingroup$ @BenitoCiaro #2 is definitely possible (well, insofar as intergalactic travel is possible). Over time, all regions of space that are not gravitationally bound to us will be transported beyond our event horizon. So if you have a ship that can escape our local cluster, you will eventually cross the Earth's horizon. You'll probably be long dead by then, though. $\endgroup$ – Pulsar Dec 14 '15 at 1:20

The cosmological horizon is constantly expanding with time. When you were relocating, the expansion of the cosmological horizon was faster in the direction you were traveling and slower behind you. When you got to the other observer and stopped, you both had the same cosmological horizon. That cosmological horizon was larger than your original cosmological horizon and completely encloses your original cosmological horizon because you had to travel slower than light.

When you were returning to your original location, the expansion of the cosmological horizon was faster in the return direction and slower behind you. So, once you stopped at your original location, your original cosmological horizon and your now larger current cosmological horizon have the same center. As before your new cosmological horizon completely encloses the cosmological horizon you had while you were at the other observers location.

The other observers new cosmological horizon has a different center, but he can currently see part the universe that is currently outside your cosmological horizon (in the direction away from you), and you can currently see part the universe that is currently outside his cosmological horizon (in the direction away from him).

  • $\begingroup$ Thanks for your answer, but I explicitly stated that I am asking about the case when the cosmological horizon does not expand with time (see also comments below the main question) $\endgroup$ – user83548 Dec 7 '15 at 23:11
  • $\begingroup$ I believe that your thesis is inconsistent. If the universe was expanding fast enough to maintain a constant cosmological horizon (which it can't), your movement would not change the cosmological horizon. (Like traveling near the speed of light, a slight change of your velocity would not change your perceived speed of light.) The only change would be the redshift. The cosmological horizon itself would stay the same. Every place you could travel would have the same cosmological horizon. $\endgroup$ – Allyn Shell Dec 8 '15 at 14:53
  • $\begingroup$ So when you got to the other observer's location the cosmological horizon would be the same and there would be no events to record that were not observable from your original location. The events observed from different locations would be redshifted slightly differently, but they would be the same events. $\endgroup$ – Allyn Shell Dec 8 '15 at 15:08
  • $\begingroup$ but it well known that you can have a constant horizon when the expansion is exponential. You can even have a diminishing horizon, the famous big rip. $\endgroup$ – user83548 Dec 8 '15 at 17:30
  • $\begingroup$ @BruceSmitherson, I often see this kind of statement in print, but it is simply not true. Anything that can be seen now will always be visible (although redshifted beyond recognition). The misunderstanding comes from the fact that light from that object that is emitted "now" will never reach us, but "old" light from that object will always be visible. It will simply appear to slow down in time like an object falling into a black hole. $\endgroup$ – Allyn Shell Dec 24 '15 at 22:39

Let me try and rephrase your question to make it simpler to answer. Suppose you are a comoving observer in a de Sitter universe, then you are at the centre of a spherical event horizon with some constant radius $R$:

de Sitter

You can receive a signal from the green dot because it's inside your horizon, and you can send a signal to an observer at the blue dot because it's inside your horizon. The sum of the distances $d_1 + d_2$ can have any value less than $2R$, but doesn't that mean an observer at the blue dot can receive signals from the green dot at a distance greater than $R$ i.e. from outside their horizon?

To understand what is going on you need to consider the motion of the observer at the blue dot. There are three main cases:

  1. the blue observer is a comoving observer, in which case they are moving away from you with velocity $Hd_2$, and is being accelerated away from you by the expansion.

  2. the blue observer is momentarily stationary with respect to you, but is being accelerated away by the expansion.

  3. the blue observer remains stationary relative to you using some form or rocket motor.

Case 1 is the easy case since in the time the signal takes to reach you from the green dot the blue observer will have accelerated away. If the sum of $d_1 + d_2 > R$ then by the time you receive the green signal the blue observer will be on a trajectory that outpaces the signal you send so they'll never receive it.

Case 3 is harder because the position of the horizon will be different for a non-comoving observe. The acceleration will contract the horizon distance opposite to the blue observer and extend it in your direction. This does indeed allow the blue observer to receive signals from a distance greater than $R$ in the direction of the acceleration. The flipside is that the horizon moves inwards in the opposite direction.

Case 2 I confess I'm not sure about. I would have to do some calculations to convince myself I understood what was going on, however it seems plausible that a non-zero peculiar velocity would affect the horizon position.

All this is qualitative, and to make it concrete you'd need to calculate the geodesics of the light and the geodesics of the blue observer. I'll think on this, but I suspect this is a messy calculation to do.

  • $\begingroup$ Case 3: will the explanation be the same if observers remain at the same fixed distance because they are bounded (by some long range force?)It confuses me that bound objects seem to obey different rules, as suggested in the last paragraph of Pulsar's answer $\endgroup$ – user83548 Dec 8 '15 at 17:42
  • $\begingroup$ @brucesmitherson: I'm not totally sure what you're asking, but all that matters is that an accelerometer held by the blue observer reads a non-zero value. More technically the norm of their four-acceleration is non zero. It doesn't matter what means are used to accomplish this. The metric for an accelerating observer is not the same as for a comoving observer i.e. the geometry of the universe observed by the accelerating observer is not de Sitter. $\endgroup$ – John Rennie Dec 8 '15 at 17:46

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