The cosmic event horizon is the comoving distance beyond which the light signal from distant galaxy can never reach us. With dark matter, the event horizon converges to a finite value so does it mean that it is constant over time? i.e. If we can observe a galaxy now, we can always observe it in future because the comoving distance is unchanged over time and will be always within the event horizon?

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    $\begingroup$ Did you mean "With dark energy, the event horizon converges to a finite value"? $\endgroup$ – John Rennie Feb 23 '15 at 16:53
  • $\begingroup$ It must be dark energy, since dark matter has nothing to do with a constant hubble parameter. $\endgroup$ – Yukterez Feb 23 '15 at 19:16

With dark energy, the event horizon converges to a finite value so does it mean that it is constant over time?

The event horizon will converge with the hubble radius in about 16 billion years:


It will approach, but never reach, a fixed value of about 18 Gigalightyears.

If we can observe a galaxy now, we can always observe it in future because the comoving distance is unchanged over time and will be always within the event horizon?

As you can see in my animation above the scale factor will expand rapidly while the event horizon stays fixed. But the particle horizon grows even faster than the scale factor. That means that galaxies we can observe today will also be visible in the distant future since they stay within the particle horizon, but as soon as they moved out of our event horizon it will no longer be possible to send an actual signal to them, or to receive a signal sent by them after they left our event horizon.

So yes, we will always see old images of this galaxies as they were before they fled our event horizon, but we will never see what happened to them afterwards. Their signal will be redshifted to an infinite duration.

The same plot in comoving distances shows that the event horizon is shrinking compared to the coordinates, while the particle horizon (and future light cone) approaches, but never reaches, a fixed comoving coordinate (see Link).

  • $\begingroup$ I like the animation, but the one in comoving coordinates would be more helpful. It's easier to see from it how comoving objects move outside the observed light cone in the future. It also shows how the comoving cosmological horizon doesn't stay constant at all. Yeah, my advice is go with the comoving plot instead of the proper plot. Much more informative for this problem. $\endgroup$ – Jim Feb 23 '15 at 19:32
  • $\begingroup$ Just click on the animation to switch to comoving view! $\endgroup$ – Yukterez Feb 23 '15 at 19:34
  • $\begingroup$ Too much work. I like it anyway so +1, but it really would make sense to put the more useful one in the post itself. $\endgroup$ – Jim Feb 23 '15 at 19:42
  • $\begingroup$ The problem is that my comoving plot is 680 px so it would be rescaled to 630 px and look distorted due interpolation effects. If my PC were not that busy right now (calculating the conformal time plot for 3 days now) I'd rescale the other one too, but I do not have enough RAM free right now. The computer almost hung up while I resaled the proper distances version ): So maybe later. But the dashed line also shows the scale factor and how it behaves relative to the event- and particle horizon, so the proper distances are not that bad. $\endgroup$ – Yukterez Feb 23 '15 at 19:45
  • $\begingroup$ That's a more than satisfactory reason. $\endgroup$ – Jim Feb 23 '15 at 19:53

In a universe dominated by dark energy, the universe would expand forever. The equation for the comoving cosmological horizon in a universe that will expand forever is given by:


Since we assume in the most accepted model that dark energy is due to a cosmological constant, the scale factor as a function of time in a universe with only dark energy would be approximately $$a(t)\propto e^{Ht}$$ where $H$ would be the Hubble parameter and constant.

Thus, we find that the cosmological at any time $t$ in a dark energy dominated universe is:

$$d(t)\propto\frac{1}{a(t)H}=\frac{1}{\dot a(t)}=\frac{e^{-Ht}}{H}$$

In an expanding universe, this result is decidedly not constant (it definitely decreases), so the comoving distance is not unchanged and comoving objects do not remain in the observable universe (remember that at $t\to\infty$, the past light cone corresponding with what we can observe then would asymptotically overlap the event horizon, so comoving coordinates crossing outside the event horizon also cross outside of the observable universe).

The proper distance to the cosmological horizon in such a universe is constant. That means one could theoretically travel to the limits of it (limits here meaning arbitrarily close to but not at it) and then travel back in finite time, but the expansion would carry non-bound, comoving objects outside this limit eventually.

  • $\begingroup$ Your post contains two mistakes: 1) the observable universe is determined by the particle horizon, not the event horizon. Comoving objects always remain within the observable universe. 2) By definition, it takes an infinite amount of time to reach our event horizon. $\endgroup$ – Pulsar Feb 23 '15 at 18:52
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    $\begingroup$ @Pulsar Re (1): The OP asked about the event horizon, so I discussed the event horizon. Comoving objects do not always remain within the observable universe. The event horizon determines the distance that light emitted now could travel in the future. We may be able to infinitely watch as objects approach the horizon and redshift out of existence, but saying that objects always remain in the observable universe is like saying objects never fall into a black hole. Yes, of course they do, we just get to watch the final moments before that happens for the rest of eternity. $\endgroup$ – Jim Feb 23 '15 at 19:16
  • $\begingroup$ Re (2): It takes infinite time to reach the event horizon, yes. But to travel to the limits of it, $Max[r<R_{EH}]$, and back is doable in theoretically finite time $\endgroup$ – Jim Feb 23 '15 at 19:18
  • $\begingroup$ I know that the OP asked about the event horizon, but in your post you bring up the observable universe, which has nothing to do with the event horizon. The observable universe is defined by the integral $\int_0^t dt/a(t)$. Also, the redshift of objects at the event horizon is NOT infinite. You are confusing these two horizons. See this post for more information: physics.stackexchange.com/a/63780/24142 $\endgroup$ – Pulsar Feb 23 '15 at 19:48
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    $\begingroup$ @Pulsar the observable universe is bounded by the event horizon. Check out the animated plot in comoving coordinates from the other answer here. It's pretty clear that the event horizon defines the limits of what is and is not in the observable universe. Anything within the horizon is observable, has been observed, or will be observable. Anything beyond the horizon is never observable. $\endgroup$ – Jim Feb 23 '15 at 20:10

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