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Years ago I read a book about black holes. It said it is theoretically possible to make a black hole from ordinary-density materials, like tap water. It would take a lot -- about the size of a galaxy IIRC. But it implies no singularity, so I guess you could be inside a black hole and things would seem normal. Can anyone confirm this?

EDIT: What I envisioned is water in the shape of a disk, rotating like a galaxy, and mass is added to it until it becomes a black hole. If you have water about 1 km thick, then at a radius of about 10^20 km the Swarzchild radius will be bigger. I'm wondering what happens at that point, viewed from the inside.

EDIT2: The example above would be about 100x the diameter of the Milky Way. So re-envision it as 100km thick and the same diameter as our galaxy. Secondly, the density of this "galaxy" is about 10^18 times that of a typical galaxy. So you would have to put the mass of a billion billion galaxies into one galaxy in order to witness the creation of a black hole from within.

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    $\begingroup$ It implies no singularity?? $\endgroup$
    – Qmechanic
    Nov 16 '21 at 15:34
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    $\begingroup$ I suspect that a star made of 2H to 1O large enough to collapse into a black hole would ignite oxygen fusion in its core and supernova first. It certainly would be a star, not a ball of water, the O-H bonds can't stand up to that kind of pressure. $\endgroup$
    – g s
    Nov 16 '21 at 18:20
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    $\begingroup$ Is it possible you were reading about analog models of gravity? $\endgroup$ Nov 17 '21 at 22:02
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    $\begingroup$ You may enjoy playing with vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator $\endgroup$
    – PM 2Ring
    Nov 18 '21 at 8:48
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    $\begingroup$ Best way to find out: scratch your head till you recover the title of that book and ask the author. Your memories are too flaky for a reasonable guess at what you actually read. $\endgroup$ Dec 26 '21 at 11:03
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This is indeed possible. To "create" a black hole, you either can compress some object to its Schwarzschild radius or stack up more matter to it. When it reaches a critical size, it will become a black hole.

To see why this is the case, let's look at the formulae for Schwarzschild radius

$$r_s=\frac{2GM}{c^2}\tag1$$

and the volume of a sphere

$$M=\rho\frac43\pi r^3\implies r=\sqrt[\uproot 2\scriptstyle3]{\frac{3M}{4\rho\pi}}\tag2$$

You can see that $r_s\propto M$, but $r\propto\sqrt[\scriptstyle 3] M$. So as you increase the mass of an object (by "adding stuff" to it), the Schwarzschild radius will increase faster than its actual radius. So there will be a certain radius at which the object will be smaller than its Schwarzschild radius - it will collapse into a black hole.

However, that does mean that there will be a singularity. When that radius is reached, the object won't just stay how it is. It will collapse to a singularity.


See The Black Hole Tipping Point by minutephysics on YouTube.

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    $\begingroup$ In particular, for density $\rho$ we need $r=\sqrt{\frac{3}{8\pi G\rho}}c$, or equivalently $M=\sqrt{\frac{3}{8\pi G\rho}}\frac{c^3}{2G}$. For water, you'd need about $136$ million solar masses, a lot less than a galaxy. $\endgroup$
    – J.G.
    Nov 16 '21 at 15:51
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    $\begingroup$ @safesphere Are you saying that a sufficient mass of water in a given volume won't collapse into a black hole? If you are not saying that, then I don't think you should say this answer is incorrect. It is giving order of magnitude estimates rather than a rigorous calculation, which is a completely acceptable method in physics to tackle this kind of qualitative question. $\endgroup$
    – Andrew
    Nov 18 '21 at 2:26
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    $\begingroup$ This answer is essentially a dimensional analysis estimate, which I think is pretty clear to most readers. By the standard in your comment, we should only solve Fermi problems through detailed numerical simulations. $\endgroup$
    – Andrew
    Nov 18 '21 at 13:56
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    $\begingroup$ @safesphere I'll stop responding since I don't want to be annoying. But maybe to be more productive you should post your own answer since you don't find the current ones satisfactory. Of course, the Schwarzschild solution itself is only an approximation to a realistic astrophysical situation, so any answer is going to involve some idealization and approximation. However it certainly would be interesting to see a more refined treatment of this thought experiment taking into account all of the subtleties you mentioned. $\endgroup$
    – Andrew
    Nov 18 '21 at 16:47
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    $\begingroup$ What J.G. said. The Schwarzschild radius of that 135.7 million solar mass BH is only 2.68 AU. $\endgroup$
    – PM 2Ring
    Jan 17 at 4:03
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The mass of a black hole determines the radius of the horizon. There are proportionality constants involving Newton's constant G and the speed of light c. But, in units where those are 1, it's just mass=radius.

That means the volume of an object that is just barely outside its own horizon is proportional to the cube of the radius, and thus the cube of the mass.

Vol propotional to mass cubed
Density = mass / vol proportional to 1 /mass^2

That means the density of an object just outside its horizon radius decreases as the mass increases. So by the time you get a mass the size of our galaxy, the density when it is just outside its own horizon is roughly that of air.

But this does not prevent the formation of a horizon, nor a singularity. The singularity forms when the object goes below its horizon. This is because of the fact that, inside the horizon, all timelike paths lead closer to the center of the black hole. Lower density does not prevent that, since the density will rise without limit once the object goes inside its own horizon.

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