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My understanding is that the radius of a black hole grows at a ratio to mass. If this is the case, then a black holes volume grows exponentially per unit mass. That means there should be a point where its growth caused by adding one kg consumes enough space that when you consider the average density of space there would be runaway growth.

below is a formula I believe should calculate this. It takes the Schwarzschild radius formula but finds the volume delta between two spheres and multiplies it by the average density of space $Z$ to calculate the Schwarzschild radius of a black hole with that mass. if that equals the delta between the radius used in the two spheres then at the radius of $r_2$ you would start to get runaway growth. Let me know if this is correct or even possible. The number I was getting was so large that this is essentially impossible considering the mass of the universe.

$$\frac{2GZ}{c^2} \left(\frac{4}{3} \pi r_2^3 ~-~ \frac{4}{3} \pi r_1^3 \right)~~=~~r_2-r_1$$

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    $\begingroup$ Note that the rate of accretion by a black hole is bounded by the Eddington limit. $\endgroup$ – Count Iblis May 17 '18 at 12:39
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    $\begingroup$ If this is the case, then a black holes volume grows exponentially per unit mass. No. $y=3^x$ is an exponential relationship. $y=x^3$ is not. $\endgroup$ – user4552 May 17 '18 at 15:02
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Note that the volume within the Schwarzschild radius does not increase exponentially with mass, rather as the cube of the mass. You represented it that way in the equation, but it would be good to get it right in the wording too, lest you confuse someone you're speaking with later.

Essentially, what you've stumbled across is the notion that if you had a region of matter with some average density, then there exists a radius of this region above which it would automatically be a black hole. This is true. If you have a cloud of gas with uniform density, then since the mass of this cloud would increase proportional to the radius cubed, its Schwarzschild radius would increase proportional to its actual radius cubed. While the Schwarzschild radius would start off smaller than the actual radius (thus, no problem), as the size of the cloud gets arbitrarily large, the Schwarzschild radius overtakes the actual radius and the cloud would automatically represent a black hole.

This does not mean that we need to worry about runaway black hole formation. This only holds for regions with a defined center of mass. An infinite or near-infinite region of uniform density would not form a black hole. You might disagree and call up the shell theorem. However, this does not predict a black hole because without a definitive center of mass, every point is effectively the center and, thus, the Shell theorem would state that every point feels the gravitation effect from a sphere of zero radius (and that all matter outside the sphere contributes no net force). Arbitrarily choosing a center and claiming a black hole forms around it is a mistake. It will always be cancelled by the gravitational effect of the matter found on the opposite side of an observer. I admit, space would be weird, but a beam of light would not be deflected in any particular direction by gravity; it would traverse the space perfectly fine (apart from scattering off the matter).

Since any given point in space is effectively the center of the observable universe at that point, the average mass density of space must cancel out on all sides. This means the only way a black hole could reach a "runaway point" is by accumulating overdensities of matter fast enough to get above that radius first; simply having a universe big enough to place an observer at that radius is not enough. There are two important things preventing such a black hole: first is the Eddington limit, as mentioned in one of the comments, which makes it very difficult for the black hole to accrete the mass it needs quick enough. Second is that, given the average density of the observable universe (including dark matter and not just ordinary matter, because why wouldn't you?), the Schwarzschild radius of the observable universe ends up being less than twice what its actual radius is. This is great because, as discussed you can't have a black hole form spontaneously just by having the Schwarzschild radius of something like the universe be greater than the actual radius (every point in space is the center) and you'll never have a pre-existing black hole grow to be larger than about half the radius of the observable universe because it would become out of causal contact with itself (one side of the event horizon would be at the edge of the observable universe seen from the other side). Not to mention the whole dark energy thing that has a negative energy density (but let's not complicate this further).

Short story? You're not technically wrong, but there are lots of things in place preventing what you might worry about from ever becoming a problem

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Notice that for a black-hole to actually grow it will need to get that mass from somewhere. So suppose for a collapsed core of a supernova there isn’t really much mass around it left to attract to add or fall below the Schwarzschild Radius. Hence you’re prediction of the black hole going through runaway growth is incomplete as the mass needs to be supplied from somewhere.

Notice that Black holes can gain some mass like when absorbing other nearby stars and matter but it will usually not grow to gain so much mass comparable to the universe as the matter distribution is such that the black hole will eventually reach an equilibrium state suppose orbiting another bigger mass. Thus then the black hole can now evaporate by Hawking Radiation over a large time and its mass won’t grow significantly then.

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The gravitational capture cross-section of a (non-rotating) black hole is $\frac{27 \pi G^2}{c^4}M^2$, scaling as the square of the mass. So if the black hole is surrounded by randomly moving particles with density $\rho$ and velocity $v$ the inflow would be $$\dot{M}=\frac{27 \pi G^2 \rho v}{c^4}M^2=\alpha M^2.$$

The solution to this differential equation is $$M(t)=\frac{1}{(1/M_0) - \alpha t}$$ where $M_0$ is the starting mass. This equation has the mass go to infinity when $t=1/\alpha M_0$.

If we use $\rho=10^{-17}$ kg/m$^3$ for the typical interstellar medium, set $v=10^5$ m/s for the particle velocity, for a $M_0=M_\odot$ black hole $t=3.4\times 10^{26}$ years. This is beyond the timescale ($\approx 10^{20}$ years) the galaxy is expected to kinematically dissolve. However, put in a $10^6 M_\odot$ central black hole and the timescales become similar (which is not entirely a coincidence - this is the timescale for most of the mass of the galaxy to diffuse anywhere in it).

Of course, in practice the growing black hole will deplete the particles and there will be other factors (like accretion disk light pressure) complicating things.

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