Physics Asked by Nicholas Lee on January 9, 2021
Observations:
I have read that for a free-falling observer within the event
horizon of a black hole that all lines of sight will end at the
singularity which is black.
I also look up and see that the sky is mostly black.
I also know that by measuring the red-shifts of the galaxies that
they are all accelelerating towards that blackness. (rather than
accelerating away from a big-bang event, which makes less intuitive
sense)
The red-shifts are greatest for those galaxies closest to the
singularity (i.e. furthest away from us).
I therefore conclude from these consistent observations that all the matter in at least the finite observable region of the universe in which we are located, is accelerating due to gravity towards the singularity of a super-massive black hole.
In the spirit of good science I will happy accept being wrong, but I need to understand why I am wrong.
Predictions from the theory:
The light from the universe outside the event horizon of the black hole should be visible as a small disc or point light source in the direction facing away from the singularity. Of course, if this happens to be inconveniently located on the far side of the milky way’s core from us, then we will be unable to observe it.
Further Speculations:
It may be that in regions of the universe far ouside the influence of this black hole (e.g. >50 billon light years away) that the sky is white hot due to Olber’s paradox.
The black hole may have been formed from a large quantity of anti-matter, which might explain the local perponderance of ordinary matter.
The microwave background radiation may be a result of Hawking radiation at the event horizon (which we are viewing fromn the inside), or, depending upon the non-euclidian geometry, we may be observing the result of matter being destroyed at the singularity.
This is a case of an unwisely chosen simile taken waaaay too far. This idea, that the entire universe could be inside the event horizon of not a supermassive, but rather a superduperultrahypermegastupendouslymassive black hole, is usually introduced in introductory classes about general relativity. The instructor in this case is trying to make clear that, contrary to a fairly popular misconception, the event horizon of a black hole is locally flat. That is, there are no CGI-fireworks, nor any kind of hard "surface", nor anything else particularly special, in the immediate vicinity of the event horizon. The only special thing that happens is a long distance effect, like noticing that every direction now points off in the distance towards the singularity.
The simile is also used to point out that, at the event horizon, even second-order, nearly local effects (that is, curvature of spacetime, or tidal effects in other words) become less pronounced the more massive the black hole is. (As an aside, this also explains why Hawking radiation is more intense for smaller black holes) So... as the simile suggests, if the black hole were massive enough, we might not even be able to detect it.
The key, though is massive enough. First of all, the whole beauty of the Einstein curvature tensor (the left side of Einstein's equation) is that it is Lorentz invariant, so it can be calculated in any reference frame, including one that is hypothetically based inside an event horizon.
The curvature can still be unambiguously calculated, so when you suggest that it may be only an optical illusion, you are also suggesting that all the scientists who do that type of large-scale curvature calculation (not me personally) are totally incompetent. Just so you know. I would suggest not mentioning that at any conferences on cosmology. One of the enduring mysteries of modern cosmology is that the large-scale curvature of the Universe seems to be open (like the 3-space-plus-one-time dimensional analog of a saddle or Pringle potato chip) and not flat (like Euclidean geometry) or closed (like a sphere). The last is what we would calculate if the visible Universe were inside a black hole.
So, for the visible Universe to be inside an event horizon, the Cosmic Acceleration we have seen thus far would have to actually just be one small, contrarian region inside an even larger event horizon of globally closed curvature. Just to make the event horizon radius 13.7 giga-lightyears (a bare minimum starting point that excludes all manner of things that make the real situation many orders of magnitude worse*), you would need over 8E52 kg of mass in the singularity. This would require over 5E79 protons, where I have heard that the entire visible Universe only has about 10^80 particles, total, and I think I heard that there are about 10^18 photons for every proton, or maybe even all other particles. Somebody can look that up if they want to, but it's definitely a big number. The upshot is that there would have to be an amount of mass, all crammed into one singularity, that would render the total mass of every single thing we can see a barely detectable rounding error. Monkeying with all those dark matter and even dark energy theories is less of a leap than that.
Your prediction doesn't actually predict anything, since you account for either its presence or its absence.
Speculation 1: Olber's Paradox is already solved for accepted theories of cosmology, so pointing out that your theory can also resolve it is nice but doesn't score any points.
Speculation 2: Are you suggesting that the singularity is where all the antimatter to match the Universe's matter went? Remember the singularity dwarfs the visible Universe. That still doesn't explain the asymmetry, it only pushes the question back by one logical step: Why did the antimatter go into the big singularity and not the matter?
Speculation 3: Hawking radiation for the big singularity's event horizon lends whole new meaning to the term negligible. See my previous aside. Also, we can't observe matter being destroyed at the singularity. That's information flowing the wrong way. Also, that negates the previous assertion that the sky is black because it's towards the singularity.
*Like cosmic expansion, just how small our contrarian region is, compared to the whole event horizon, and probably some other, subtler things.
Correct answer by Andrew on January 9, 2021
The light from the universe outside the event horizon of the black hole should be visible as a small disc or point light source in the direction facing away from the singularity.
This will happen even before you reached the event horizon. You do not need to cross the horizon to see this effect. At ergosphere distance the BH will cover half of the sky and closer to the BH the BH will cover greater part.
So one can assume that we are just outside the event horizon. And this is true: if the space is de Sitter we are just outside the cosmic event horizon, whether it is BH or not. For a distant observer we seem to be sticked to this surface and time-diliated.
As this "BH" evaporates (via de Sitter/Hawking radiation), the diameter of the cosmic horizon will become smaller and the temperature will rise.
But we will never cross the horizon: we will be ripped before. This is called "Big Rip" scenario. We will experience the same what experiences an observer trying to reach a black hole: he will finally see a great explosion when the BH radius will reach zero.
Answered by Anixx on January 9, 2021
If we only consider the observable universe without the surrounding matter then yes, we are living in a black hole (which doesn't need a singularity.
For the Schwarzschild radius we have:
$$r_s=frac{2GM}{c^2}$$
Filling in:
$$r_s=2*6,7*10^{-11}*10^{53}*9*{10}^{-16}=1,2*10^{26}(m)$$
Now for M I only counted the non-dark matter neither the dark energy, which makes M in fact much bigger. Say 20 times. This makes $r_s=2,4*10^{27}(m)$.
Now the radius of the visible universe is. $$9,6*10^{15}*4,6*10^{10}=4,25*10^{26}(m)$$
From which it follows we are (under the unreal assumption that our visible universe is all there is) living inside a black hole.
Off course all the matter around the observable Universe prevent this black hole to really exist, and light rays can travel throughout the entire Universe (except of course if they meet a real black hole).
Answered by Deschele Schilder on January 9, 2021
Not when space is infinite with a homogeneous matter distribution, since there is no center where the matter would gravitate towards.
I think this is what descheleschilder is trying to say in his answer. However, one cannot use the Schwarzschild metric according to an answer in: Supermassive black holes with the density of the Universe?
Finally, since gravity is communicated with the speed of light, the curvature within a certain radius has to be causally connected for it to add up. The expansion of the universe could prevent a cosmic black hole.
Answered by Geert VS on January 9, 2021
I have read that for a free-falling observer within the event horizon of a black hole that all lines of sight will end at the singularity which is black.
That's not true. Black hole singularities can be spacelike or timelike. Spacelike singularities are in the future, and can't be seen at all. Timelike singularities can be seen, but it's not clear what they'd look like, and there's no situation in which one would occupy your whole 360-degree field of view. It also seems probable that timelike singularities can't exist in realistic black holes.
I also look up and see that the sky is mostly black.
It's mostly glowing at around 2.7 kelvin, and that light is redshifted blackbody radiation from a plasma, not from a singularity.
In the distant future, if ΛCDM is correct, Hawking/Unruh radiation from the cosmological horizon will replace the CMBR, but that's still from an event horizon, not a singularity. (And there's no singularity behind the cosmological horizon anyway.)
I also know that by measuring the red-shifts of the galaxies that they are all accelelerating towards that blackness. (rather than accelerating away from a big-bang event, which makes less intuitive sense)
The cosmological horizon is somewhat like an inside-out black hole horizon, and it's reasonable enough to say that galaxies accelerate toward it, rather than away from each other, but that's not an alternate cosmological model, it's the same model described in different words.
The red-shifts are greatest for those galaxies closest to the singularity (i.e. furthest away from us).
There's no singularity, and none of the galaxies are any closer to anything than we are. From the perspective of an astronomer in one of those galaxies, we're redshifted and accelerating toward their cosmological horizon.
The light from the universe outside the event horizon of the black hole should be visible as a small disc or point light source in the direction facing away from the singularity. Of course, if this happens to be inconveniently located on the far side of the milky way's core from us, then we will be unable to observe it.
The problem with any model of this sort is that the CMBR is isotropic to about one part in 10,000. If the "window to the outside" was hidden by the Milky Way, the CMBR temperature would still vary with angular distance from the "window".
The black hole may have been formed from a large quantity of anti-matter, which might explain the local preponderance of ordinary matter.
Antimatter is pair-produced with ordinary matter in equal quantities and in the same location. There's no plausible mechanism that could sift the matter and antimatter into clumps separated by astronomical distances.
Answered by benrg on January 9, 2021
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