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Black holes decrease as they evaporate and their radius decreases as well.

So what is with a cosmological horizon?

If cosmological horizon is just a black hole centered at the opposite side of the universe, we should see the radius of the external space groving as the radius of the BH decreases.

But if the radiation decreases the horizon entropy we should see the area of the horizon decreasing.

What is the correct conclusion?

Anixx
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I think there's a misinterpretation here that's causing your confusion. There's a difference between the event horizon of a black hole and the cosmological horizon.

The event horizon of a black hole is the place inside which nothing can escape from the black hole. The cosmological horizon, on the other hand, is the place beyond which an observer cannot "retrieve information." The difference between the two is basically that the event horizon of a black hole is in the same place for observers in different locations, while the cosmological horizon differs for observers in different locations.

Also, if the cosmological horizon was the edge of a black hole, the black hole would have to be centered right where we are - which is highly unlikely!

HDE 226868
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I've read that we currently believe the expansion of the universe is decreasing over time from the current ~72km/s/Mpc toward an asymptote of ~45km/s/Mpc.

The size of the observable universe is a consequence of the expansion of space, and that due to the expansion: every observer has their own cosmological horizon, where relative to them any mass would be moving away at the speed of light, and accelerating.

Supposedly, the expansion rate is proportional to the mass within the observable universe, and as the mass density lessens the expansion rate of your universe is lessened...toward said asymptote, which is presumably when our local group has collapsed in on itself and is the last thing left in our observable universe.

All that said, if the expansion rate decreases over time, then the observable universe would be growing. I.e., at the same 13.8 billion light years; things wouldn't be doing the speed of light yet, relative to us. So the size of the observable universe would also be expanding toward an asymptote. Assuming the ~45km/s/Mpc is correct... The observable universe would be growing from a 13.8 billion light year radius to ~6666.6 Mpc = 21.7 billion light years.

I however, assume the universe is infinite, and that the microwave background is much more than microwave and made of highly redshifted light from the infinite galaxies beyond the cosmological horizon. I also assume that dark energy/matter is a dense ether of low energy/frequency photons permeating the infinite universe, and I assume these photons interact with each other and spawn mass into existence, eventually forming huge gas clouds, that become new galaxies, birthing additional mass into our observable universes, and possibly (likely in my opinion) at a rate that keeps the expansion nearly constant... Hubble's constant.

Dave Sims
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    Hi Dave, you have many correct statements, but there are however a few misconceptions here: 1) While it's true that the expansion rate decreases asymptotically to ~50 km/s/Mpc, the size given by the scale factor $a$ of the Universe goes toward $a(t) \propto e^{Ht}$, i.e. the size increases exponentially (see e.g. here). – pela Nov 27 '22 at 10:04
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  • It's also true that every observer has a "horizon" outside which stuff recedes superluminally. However, it's not a real horizon, since we easily see stuff far outside this distance (see e.g. here). 3) You right that the observable Universe approaches a maximum size in the stuff we can see. By this I mean that there is a maximum amount of stuff inside a sphere from which we will ever be able to receive a signal. Today the radius of this sphere is 63 Glyr. But because of expansion, its size will increase without bounds.
    • – pela Nov 27 '22 at 10:04

  • I don't think you're correct about DM and DE. If DM were photons, no matter how low frequency, it wouldn't be able to form the structures we see; it would be a so-called "hot dark matter", against which we have strong observational constraints. Photons do have energy and hence contribute to gravity, but for this reason it also couldn't be DE, since DE has the opposite effect than gravity.

  • There is no observational support for photons interacting and spawning huge gas clouds.

    • – pela Nov 27 '22 at 10:04
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    We don't! The observable Universe is defined as the region within which light has had the time to reach us since the Big Bang, i.e. in 13.8 Gyr. Because of expansion the radius of this region is much larger than 13.8 Glyr; today its radius is 46 Glyr. But this is not the same as the "Hubble sphere", which is the region within which matter recedes subliminally. This radius, the Hubble radius, is much smaller, namely $c/H(t)$ which today is 14.5 Glyr. – pela Nov 28 '22 at 15:05
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    In other words, any galaxy you see with a distance larger than 14.5 Glyr is receding faster than the speed of light. These galaxies have redshifts larger than ~1.5 and are easily observable. Pretty cool! – pela Nov 28 '22 at 15:06
  • I just deleted my comments to rephrase them and noticed you just replied. Sorry about that. If I'm understanding your links correctly, I had forgotten/neglected that light moves at the speed of light relative to the observer. I.e. for someone on the distant emitter that just left the Hubble sphere, the light is stuck on the Hubble sphere while they move away from it at the speed of light, but for us the light starts heading from the Hubble sphere toward us at the speed of light, and regardless of how arbitrarily far something is away, if it's emitting light, that light will eventually reach us – Dave Sims Nov 29 '22 at 04:18
  • This implies that the count of observers in the universe is equal to the number of "instances" of every emitted photon. Can you point me to empirical evidence of the speed of light being constant relative to the observer regardless of emitter velocity? Thanks again for all the info. – Dave Sims Nov 29 '22 at 04:18
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    Yes, exactly. Light moves locally at speed $c$, but if space is expanding (or contracting) between the photon and the observer (who hasn't yet observed that particular photon), the relative speed may be any value. One might think that a photon emitted by a galaxy receding at $<c$ could never reach us, but it helps to see it in the reference frame of the emitter: In this frame the photon leaves at $c$, but slowly picks up speed. – pela Nov 29 '22 at 09:59
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    If there was only matter in the Universe, there'd be no limit to how far we could see, if we just wait long enough. But due to dark energy accelerating expansion, there is a limit, called the particle horizon, which is currently around 16 Glyr away, a bit larger than the Hubble sphere. – pela Nov 29 '22 at 10:00
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    As for the evidence that $c$ is constant regardless of the emitter velocity, that's Michelson–Morley for ya :) (and many similar experiments after them) – pela Nov 29 '22 at 10:48