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A nucleus is very small and very dense. Its density is approximately $2.3 \times 10^{17}~\mathrm{kg/m^3}.$ So why can't a nucleus itself become a black hole?

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R. Chaudhuri
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    If that amount of mass was a Schwarzchild black hole, what would it's Schwarzchild radius be? Is that reasonable? – AHusain Jul 09 '16 at 05:19
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    Please correct the vastly incorrect number for the density of nuclear matter. – ProfRob Jul 09 '16 at 18:47
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    Btw, if you have a lot of nucleus, then their gravitational field essentially sums. The problem is that you need around 4 solar mass for that. – peterh Jul 10 '16 at 00:00

3 Answers3

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Let's take the carbon nucleus as a convenient example. Its mass is $1.99 \times 10^{-26}$ kg and its radius is about $2.7 \times 10^{-15}$ m, so the density is about $2.4 \times 10^{17}$ kg/m$^3$. Your density is ten orders of magnitude too high.

The Schwarzschild radius of a black hole is given by:

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

and for a mass of $1.99 \times 10^{-26}$ kg this gives us:

$$ r_s = 2.95 \times 10^{-53} \,\text{m} $$

This is far below the Planck length, so it is unlikely matter could be squeezed into a region that small i.e. a single carbon nucleus cannot form a black hole.

If we take the Planck length as $r_s$ and calculate the associated black hole mass the result is half the Planck mass $\tfrac{1}{2}\sqrt{hc/2\pi G}$, which is about $11\,\mu\text{g}$ or about $5.5 \times 10^{17}$ times larger than the mass of the carbon nucleus. This is the smallest mass that we expect could form a black hole.

John Rennie
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  • So, there is almost zero probability that any nucleus would ever become a tiny black hole, right ? – ABcDexter Jul 09 '16 at 06:11
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    @ABcDexter: it's meaningless to talk about the probability that any nucleus would ever become a tiny black hole unless you describe the specific process by which you think this might occur. However, on general grounds we believe it's impossible for anything lighter thn a Planck mass to form a black hole. – John Rennie Jul 09 '16 at 06:15
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    What's the relevance of the Planck mass? Are you (incorrectly) assuming that it represents the smallest possible size? – OrangeDog Jul 09 '16 at 18:57
  • @OrangeDog I would assume it's the mass that would have a Schwarzschild radius of Planck length. (Actually, it's the maximum allowed mass for point-masses) – Michael Jul 09 '16 at 19:55
  • @Michael I meant Planck length, but no, the Planck mass is not the maximum of anything. – OrangeDog Jul 10 '16 at 08:21
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    @OrangeDog: no, I'm not assuming the Planck length is the smallest possible length. My point is that any process intended to compress matter into a region smaller than a Planck length would require so much energy that a black hole (of the Planck mass) would form before you could complete the process. You would always end up with a Planck mass black hole. There isn't any way you can create a black hole with a lower mass than this, and therefore with a smaller Schwarzschild radius. Disclaimer: this assumes quantum gravity isn't even weirded than we think! – John Rennie Jul 10 '16 at 08:41
  • But you haven't actually given any reason for this – OrangeDog Jul 10 '16 at 09:14
  • @OrangeDog: I'm not sure if your point is merely editorial, i.e. if you're saying I should justify the claim, or if you're genuinely asking why it is so. If the latter then this is a well known argument - see for example this answer. I wouldn't say it's any more than an argument because we don't know enough about quantum gravity to be sure what goes on in those circumstances. – John Rennie Jul 10 '16 at 09:25
  • You should at least link that. At the moment you pull the Planck length out of thin air then conclude 11um is the smallest possible BH. – OrangeDog Jul 11 '16 at 17:46
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The minimum radius of a spherical object to be a black hole is given by :

r = 2Gm/ (c^2)

From this, I think we may be able to calculate the minimum density for the object to be a black hole, which is:

                      d = (21/704)((c^6)/((G^3)(m^2))     (Assuming pi = 22/7)

it is, d = (7.37 x 10^79 )/(m^2)

So, you can now guess how high it is. For URANIUM, it is 4.61 x 10^128 kg/metervolume

as you can see, it is in the order of 10 raised to power of 128 !!!

For hydrogen, it is 2.8 x 10^133 kg/metervolume

In the order of 10 raised to the power of 133 !!!

But, the average density of nucleus is much lower than the above values.

Thus, we can never expect nucleus to become a black hole.

Nature is always remarkable!!! It always helps us.....

Srikar Anand Yellapragada
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I haven't run the numbers, I assume JR is correct. I do know that the 'density' of a proton is much less than that required (isn't this obvious?) for gravitational collapse to occur (otherwise, it would). The only thing I want to add here is a cautionary note about our lack of understanding about quantum gravity. That is, as soon as you want to discuss gravity of sub-atomic particles we leave evidence based science behind and need to speculate (string theory, etc.). It has, for instance, yet to be established that the Schwarzchild Radius applies to quantum particles.

Li Zhi
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