# Relation between black hole mass and radius, and our universe's [duplicate]

Is there a graph of known black holes, with their estimated mass in the X axis and their estimated radius in the Y axis? If so, where can we find it? I would like to know if a black whole with the whole estimated mass of our universe would have the estimated radius of our universe (which means our universe could be a black hole, that's why light can't escape it and it looks "finite").

## marked as duplicate by StephenG, Jan Doggen, J. Chomel, Sir Cumference, James KMar 14 '18 at 21:33

• Today I found an article proposing this hypothesis in 1972: nature.com.sci-hub.cc/nature/journal/v240/n5379/pdf/… – Rodrigo Mar 12 '17 at 1:59
• sci-hub.tw/http://www.nature.com/nature/journal/v240/n5379/pdf/… – Rodrigo Feb 18 '18 at 1:03
• As your basic idea is that the universe is inside a black hole I think that this question and answer explains why we're not in a black hole. There is also a link there to a question on Physics SE that goes into this as well. – StephenG Mar 8 '18 at 22:47
• @StephenG How can my question be a duplicate, if it was asked before? I'll look at the links you provided, thank you. – Rodrigo Mar 8 '18 at 23:45
• @StephenG I said that maybe our Universe IS a black hole, not "is inside one", I don't think it's the same thing. And the results are all in the same order of magnitude. Assuming all those numbers are at best good approaches, and that our Universe as a black hole don't need to follow the same rules black holes inside our Universe follow, and also considering that dark energy was discovered "just yesterday", I don't think this hypothesis should yet be discarded. – Rodrigo Mar 9 '18 at 0:08

The Schwarzschild radius of a black hole is probably the closest we can get to your question.

$$r_s = (2G/c^2) \cdot m \mbox{, with }\ 2G/c^2 = 2.95\ \mbox{km}/\mbox{solar mass}.$$ This means, that the Schwarzschild radius for a given mass is proportional to that mass. The radius shouldn't be taken too literal in the physical sense, because space is highly non-euclidean close to a black hole.

Present (light-travel) radius of the visible universe, as seen from the earth: $$13.81 \cdot 10^9\ \mbox{lightyears} = 13.81 \cdot 10^9 * 9.4607 * 10^{12}\ \mbox{km} = 1.3065\cdot 10^{23}\ \mbox{km}.$$ So we need $$1.3065\cdot 10^{23}\ \mbox{km} / 2.95\ \mbox{km} =4.429 \cdot 10^{22}$$ solar masses to get a black hole of the light-travel Schwarzschild radius of the visible universe, pretty close (by order of magnitude) to the number of stars estimated for the visible universe.

The Wikipedia author(s) gets a similar result: "The observable universe's mass has a Schwarzschild radius of approximately 10 billion light years".

• WOW! Thank you! That's really amazing! It looks like we live inside a black hole, perhaps... – Rodrigo Jan 16 '14 at 0:34
• I was a little surprised myself. First I thought, a black hole should be much smaller, but it isn't. But there may be solutions of the field equation of general relativity leading to a comparable "radius" besides that of a black hole. – Gerald Jan 16 '14 at 0:45
• @Gerald So the average density of a black hole with the diameter of the universe, would be about the same as the density of our universe, almost vacuum? – this Jan 16 '14 at 18:47
• Yes. If we think of the black hole being replaced by space filled with the same mass with Euclidean metric, to get a reasonable definition of density. – Gerald Jan 16 '14 at 20:46
• That's not as surprising as it looks at the first instant after thinking a bit: A snapshot of the universe would make light travel roughly a circle with the diameter of the observable universe, much the same as a black hole with Schwarzschild "diameter" 2/3 of the diameter of the circle. – Gerald Jan 16 '14 at 21:07

According to the standard ΛCDM cosmological model, the observable universe has a density of about $\rho = 2.5\!\times\!10^{-27}\;\mathrm{kg/m^3}$, with a cosmological consant of about $\Lambda = 1.3\!\times\!10^{-52}\;\mathrm{m^{-2}}$, is very close to spatially flat, and has a current proper radius of about $r = 14.3\,\mathrm{Gpc}$.

From this, we can conclude that the total mass of the observable universe is about $$M = \frac{4}{3}\pi r^3\rho \sim 9.1\!\times\!10^{53}\,\mathrm{kg}\text{.}$$ Sine the universe at large is nonrotating and uncharged, it's natural to compare this to a Schwarzschild black hole. The Schwarzschild radius of such a black hole is $$R_s = \frac{2GM}{c^2}\sim 44\,\mathrm{Gpc}.$$ Well! Larger that the observable universe.

But the Schwarzschild spacetime has zero cosmological constant, whereas ours is positive, so we should instead compare this to a Schwarzschild-de Sitter black hole. The SdS metric is related to the Schwarzchild one by $$1-\frac{R_s}{r}\quad\mapsto\quad1 - \frac{R_s}{r} - \frac{1}{3}\Lambda r^2,$$ and for our values we have $9\Lambda(GM/c^2)^2 \sim 520$. This quantity is important because the black hole event horizon and the cosmological horizon become close in $r$-coordinate when it is close to $1$, a condition that creates a maximum possible mass for an SdS black hole for a given positive cosmological constant. For our $\Lambda$, that extremal limit gives $M_\text{Nariai} \sim 4\!\times\!10^{52}\,\mathrm{kg}$, smaller than the mass of the observable universe.

In conclusion, the mass of the observable universe cannot make a black hole.

Well, we don't fully comprehend black matter, do we? And it was just "yesterday" that we discovered the "black energy", wasn't it?

If GTR with cosmological constant is right, we don't need to "fully comprehend" it to know its gravitational effect, which is what the calculation is based on. If GTR is wrong, which is of course quite possible, then we could be living in some analogue of a black hole. But then it's rather unclear what theory of gravity you wish for us to use to try to answer the question. There's no remotely competitive theory that's even approaching general acceptance.

From the perspective of our huge ignorance, I think that 14.3Gpc and 44Gpc are not even one order of magnitude apart, which I consider a good approximation.

Actually, the point of that calculation was to show that it's at least prima facie plausible. The Schwarzschild radius calculation doesn't rule out the black hole--quite the opposite. However, it's also not appropriate for reasons I explained above. The more relevant one actually does have mass more than one order of magnitude apart, and shows inconsistency. So if GTR with Λ is correct, it's unlikely because the ΛCDM error bars aren't that bad.

However, even if we still treat it as "close enough", that does not by itself imply what you want. The question of what kind of black hole all the mass of the observable universe would make, if any, is quite different from whether or not we're living in one. The black hypothetical needs to be larger still.

The biggest point of uncertainly, though, is the cosmological constant, even if GTR is otherwise correct. If we're allowed to have very different conditions outside our hypothetical black hole, then we could still have one, but then we get into very speculative physics at best, and just complete guesswork at worst.

So treat the above answer as conditional on the mainstream physics; if that's not what you want, then there can be no general answer besides "we don't know". And that's always a possibility, although not a very interesting one.

• Well, we don't fully comprehend black matter, do we? And it was just "yesterday" that we discovered the "black energy", wasn't it? From the perspective of our huge ignorance, I think that 14.3Gpc and 44Gpc are not even one order of magnitude apart, which I consider a good approximation. The same applies to the figures 9.1×10^53kg and 4×10^52kg. It's not IMPOSSIBLE that we actually live inside a black hole floating in another bigger universe... Sounds like a beautiful mythology to me. – Rodrigo Jan 16 '14 at 2:03