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Let's say two objects are at about twice the Hubble Distance from each other. One of them emits light towards the other.

When light emitted from one reaches the Hubble Distance from its origin (within less than Hubble Time, but more than half of it) the other will be more than another Hubble Distance farther from the first than it was - the distance from the point the light reached will be more than at the beginning of the travel.

Is that correct or am I perpetuating another misunderstanding concerning Space Expansion?

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Your text is a bit confusing, but the answer to the question in your title is yes. There is a distance called the cosmic event horizon. If galaxies beyond this distance emit light today, then this light will never reach us. Here's how the event horizon can be calculated:

The expansion of the universe can be described by the scale factor $a(t)$, which increases from 0 at the big bang to 1 at the present day, and to infinity as the universe gets older. The various constituents of the universe (radiation, matter, and dark energy) determine the expansion rate in a specific way, which is expressed by the Hubble parameter $$ H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}, $$ with $$ \begin{gather} H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\quad \Omega_{R,0}\approx 0, \\ \Omega_{M,0} = 0.315,\quad \Omega_{\Lambda,0} = 0.685, \quad \Omega_{K,0} = 0, \end{gather} $$ according to the latest Planck results. Now, in order to calculate distances, we need to create a coordinate system. Cosmologists use so-called co-moving coordinates, which is a coordinate grid that expands along with the universe. This means that a faraway galaxy has a fixed co-moving distance $D_c$, while its proper distance increases over time as $$ D(t) = a(t)D_c. $$ At the present-day cosmic age $t_0$, we have $a(t_0)=1$, and both distances coincide. If $\text{d}\ell$ is an infinitesimal co-moving distance, then a photon will travel in a time interval $\text{d}t$ a proper distance $$ c\,\text{d}t = a(t)\text{d}\ell, $$ so that $$ \text{d}\ell = \frac{c\,\text{d}t}{a(t)} = \frac{c\,\text{d}a}{a\,\dot{a}} = \frac{c\,\text{d}a}{a^2\,H(a)} = \frac{c}{H_0}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}. $$ Integrating this gives us the co-moving distance travelled by a photon, emitted from a source at time $t_\text{em}$ and observed by us at time $t_\text{ob}$: $$ D_c = \frac{c}{H_0}\int_{a(t_\text{em})}^{a(t_\text{ob})}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}. $$ Now, the cosmic event horizon at time $t$ is defined as follows: if a source at the cosmic event horizon emits a photon at time $t$, this photon will reach us at time $t=\infty$, so that the co-moving distance of the event horizon is $$ D_{c,\text{eh}}(t) = \frac{c}{H_0}\int_{a(t)}^{\infty}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}, $$ and the corresponding proper distance is $D_{\text{eh}}(t) = a(t)D_{c,\text{eh}}(t)$. This means that all photons emitted by galaxies inside the horizon at time $t$ will eventually reach us, but photons emitted outside the horizon are never going to reach us.

These graphs should make things clearer (click 'view image' for a larger version):

enter image description here

On the horizontal axis, we have the co-moving distance of light sources, in Gigalightyears (bottom) and the corresponding Gigaparsecs (top). The vertical axis shows the age of the universe (left) and the corresponding scale factor $a$ (right). The horizontal thick black line marks the current age of the universe (13.8 billion years). Co-moving sources have a constant co-moving distance, so that their world lines are vertical lines (the black dotted lines correspond with sources at 10, 20, 30, etc Gly). Our own world line is the thick black vertical line, and we are currently situated at the intersection of the horizontal and vertical black line.

The green curves are lines of constant recession velocity: the dark green area is the Hubble sphere, with co-moving radius $$ D_{c,H}(t) = \frac{c}{a(t)\,H(a(t))}, $$ and the brighter green areas are twice, three times and four time the Hubble sphere. The yellow lines are null geodesics, i.e. the paths of photons. The scale of the time axis is such that these photon paths are straight lines at 45° angles. The blue line marks the edge of the observable universe, i.e. the region of the universe that we can see (which currently has a radius of 46.2 billion ly). And finally, the red straight line is the cosmic event horizon.

You can see that photons inside the event horizon can cross the vertical black line, i.e. they can reach us; photons outside cannot. Also note that the Hubble sphere lies inside the event horizon (and will asymptotically approach it as $t$ goes to infinity). The current Hubble distance is 14.5 Gly, while the current distance to the event horizon is 16.7 Gly. Photons emitted today by sources that are located between these two distances will still reach us at some time in the future. So the Hubble radius itself is not a horizon.

The event horizon at $t=0$ is $D_{c,\text{eh}}(0) = 62.9$ billion ly. This means that galaxies beyond this distance have never been able to send photons to us, and will never be able to. So these galaxies will forever be outside our observable universe.

If we use proper distances instead of co-moving distances, then the graph looks like this:

enter image description here

Again, the event horizon is the parabola-like red curve, galaxies move on the dotted black lines, and photons follow the yellow paths. Here, you see that a photon emitted outside the Hubble radius but inside the event horizon will first move away from us (because the recession velocity of the source is greater than $c$), but are eventually able to overcome the expansion and start moving towards us. The resulting path of these photons is a teardrop-like shape. But photons that are emitted outside the event horizon cannot overcome the expansion, and they will always move away from us.

If you want even more details, see this post.

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  • $\begingroup$ Thanks - wonderful, detailed answer. And I was right, expecting that the event horizon lies somewhere between 1 and 2 Hubble Radii - from these calculations it seems it's at 1.15 its size. I still can't quite feel why the two approach asymptotically - in particular, why the Hubble Radius is a curve - but I bet that's a subject for a different question... $\endgroup$ – SF. Oct 30 '13 at 16:55
  • $\begingroup$ @SF. They approach because the Hubble parameter will become constant as dark energy becomes more dominant, see also this answer. Also, the Hubble radius looks curved in co-moving coordinates because it equals $c/\dot{a}$, and the expansion $\dot{a}$ was first very high, then decreased to a minimum (at around 7.7 billion years, the dashed line), and then increases again as dark energy becomes more dominant. $\endgroup$ – Pulsar Oct 30 '13 at 17:26
  • $\begingroup$ The answer to the question in the title is no. Expansion, even expansion with arbitrarily large recessional velocities, doesn't imply that some light will never reach sufficiently distant objects. There's no event horizon in open models without a cosmological constant even though they have arbitrarily high recessional velocities. (See my answer.) $\endgroup$ – benrg Sep 26 at 1:25
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No, it doesn't imply that. There are cosmological models that are spatially infinite (and so have recession velocities arbitrarily higher than $c$) in which light emitted at any time from any point in the universe will eventually reach any other point in the universe.

In fact, all open cosmological models with no dark energy have that property.

For example, in a critical-density matter-dominated model, the scale factor $a$ grows like $a(t) \sim t^{2/3}$, so the speed of light in FLRW coordinates is $c/a(t) \sim t^{-2/3}$, and $\int_{t_0}^\infty t^{-2/3}\,\mathrm dt = \infty$ for any $t_0$, so given enough time, light emitted from any point in spacetime will cross the worldlines of arbitrarily distant objects moving with the Hubble flow.

(I'm pretty sure that it will eventually reach arbitrarily distant objects not moving with the Hubble flow as well, as long as they're moving at less than the speed of light and don't accelerate away forever, but I'm too lazy to try to prove it right now.)

In the ΛCDM model that seems to describe our world, there is a limit to how far light can travel, because the scale factor eventually grows exponentially, and $\int_{t_0}^\infty e^{-kt}\,\mathrm dt$ is finite. However, the boundary between what we will and will not be able to see is not the Hubble distance, as Pulsar's answer explains in detail.

In the distant future, when space is essentially de Sitter, the Hubble distance does coincide with the event horizon. But that's the only case I'm aware of in which the Hubble distance can be assigned any straightforward physical interpretation. And even that one is philosophically dubious since de Sitter space is a time-symmetric vacuum. I don't think a civilization that arose in the distant de-Sitter-like future would have any reason to conclude that the universe is expanding, much less to invent Hubble's law and associated concepts.

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