When will the Hubble volume coincide with the volume of the observable Universe?

The Hubble volume is the volume that corresponds to objects so far from the Earth that the space between us and them is expanding faster than the speed of light. (I.e. objects outside this volume could never again be visible to us, even in principle.)

The volume of the observable Universe extends from us to the maximum distance light could have travelled since the Universe became transparent; when It was roughly 380,000 years old.

Since c1998 we have known the expansion of the Universe is accelerating, implying that the number of galaxies within the Hubble volume is decreasing. Since the Big Bang (well infinitesimally close to it at least) we know that time has been going forwards, and thus that the observable Universe is expanding.

When do these two volumes coincide with one another and what will the corresponding maximum volume of the observable Universe be at that time? The associated calculation or link to a suitable reference would also be very much appreciated.

Supplementarily/

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very interesting article - btw i wonder if that observation happens to be also a significant validation of dark energy being a universal constant? i.e: cosmological constant. At the end it says "If dark energy is not uniform as is usually assumed, then one could see this in anisotropies of the CMB in a similar way as in the void models,", which is not completely free of any ambiguity – lurscher Jul 26 '11 at 17:25
@lurscher: Whilst I don't think it answers you query there was another independent varification of DE published recent. Here is a sciam article about it and here is the paper on which it is based. – qftme Jul 26 '11 at 17:59

The question can be reposed as asking when the Hubble distance, which is not $c t_0$ but rather $c/H = c a/\dot{a}$ at $t_0$, is equal to the horizon distance $\int_0^{t_0} a^{-1} c dt$ where $a(t)$ is the scale factor normalized as $a(t_0) = 1$. For a power law $a(t) = (t/t_0)^n$ this gives $c t_0/n = c t_0 \int_0^1 x^{-n} dx = c t_0/(1-n)$. Thus $n = 1/2$ which is the case for a radiation dominated Universe.

Since the Universe is expanding, $a(t) < 1$ for $t < t_0$, so the horizon distance is always bigger than $c t_0$.

If you take last scattering for the lower limit of the integral then there will some more complicated equation. But that's a photist approach. Neutrinos and gravitational waves don't care about last scattering at 380,000 years.

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Oops, thanks for pointing out my mistake in construing the question as asking about $ct_0$ rather than $c/H$. – Ben Crowell Jul 27 '11 at 1:08
Hi @Ned Wright, thank you kindly for taking the time to answer. I would like to just clarify the conclusion that appears to be drawn from your calculation, namely - that the observable Universe is destined to increase indefinitely, albeit with increasing redshift? This appears to contradict what I read in the aforementioned book by Penrose. On a side note, I opted for the "photist approach" purely as we're a long way from building a high-resolution neutrino or gravitational wave telescope (well, perhaps just decades for the former.) – qftme Jul 28 '11 at 11:13

Never if space keeps accelerating. When the universe accelerates, the Hubble volume decreases. When the universe decelerates it will increase. The only way the whole universe's size can be measured is when it neither decelerates or accelerates, i.e. remains constant, which is essentially impossible.

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protected by Qmechanic♦May 11 '14 at 16:29

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