# Could the singularity of a black hole just be an iron / dark matter sphere? [duplicate]

A singularity would break physics wouldn't it? How does a massive star collapse and suddenly fall into a point in space with no dimensions what so ever when it had a core and perfect balance between gravity and mass before its death?

This would certainly make no sense in the process right. Is there a simulation where this kind of behaviour is illustrated?

The only way the singularity of a black holes could make sense is when for example a massive star collapses due to gravity and its core being unable to fuse iron into a heavier element, the only thing left is pretty much the core being either iron or a new unknown element heavier than iron cough dark matter? cough .

This core's gravitational force is so strong that it bends light though if we were remove the force and just look under the hood of the object, we would simply just see a spherical object made out of iron or something heavier / unknown like perhaps dark matter.

TL;DR the core remains but we can't see it due to its gravitational force / strength thus a singularity would not exist.

How likely is that kind of concept? (I'm definitely not an expert in this field but I want to understand this concept better and see whether my ideas could possibly make sense)

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• why don't you read this hyperphysics.phy-astr.gsu.edu/hbase/Astro/blkhol.html , which is the main stream physics model for black holes. You seem to misunderstand the process. After the neutron star stage there are no nuclei in stars that will become black holes.The singularity is a mathematical singularity and it is expected that a quantized gravitational theory will make a fuzzy point of it as happens with the big bang en.wikipedia.org/wiki/Big_Bang , but we have to wait for a definite theory. – anna v Feb 3 at 14:46
• Ignore black holes for now and look at neutron stars. These things exist, they're observable, and they're much, much denser than iron. So no, your conception of an iron sphere isn't plausible. As for dark matter spheres, mathematically, the density at a point goes to infinity, so that's not plausible either; however it is possible vast amounts of black holes can be dark matter. – Allure Feb 3 at 14:48
• thanks for the link! Yes of course, neutron stars! How could I've forgotten those and iron. Thanks for the hint, I hope in the future we will solve this mystery once and for all. – StellarEquilibrium Feb 3 at 15:04
• "A singularity would break physics wouldn't it?" No, it would not. It would break a particular physical theory, in a regime in which we have no expectation that this theory should still be valid. – Rococo Feb 3 at 19:38

Iron can't exist in hydrodynamic equilibrium under the relevant conditions. We also observe at least some black holes whose masses are above the Tolman-Oppenheimer-Volkoff limit, so they can't be degenerate electron or neutron matter.

The best candidates for dark matter are WIMPs, and the best candidates among WIMPs are fermions. Under this suggestion, dark matter undergoing runaway gravitational collapse would be expected to form degenerate matter, much like a white dwarf or neutron star.

The Chandrasekhar limit is proportional to $$m^{-2}$$, where $$m$$ is the mass of the electron. Since dark matter particles have a mass greater than the mass of the electron, the Chandrasekhar limit for dark matter would be much lower, and therefore you're not going to get solar-mass degenerate-matter stars made of dark matter. (It would also be difficult to get such a thing to form by gravitational collapse.) I've spelled out the details of the estimate in more detail at the end of this answer.

Another popular dark matter candidate is the axion, which is a boson. Most models limit the mass of stable stars made out of axios to extremely small values. It does not appear possible to get these objects with the kinds of masses we see, for example, in Sag A*.

A singularity would break physics wouldn't it?

Details of the estimate for fermionic DM

Ignoring factors of $$\pi$$, etc., the pressure of a relativistic, degenerate gas is $$P\sim hc(n/V)^{4/3}$$. The pressure at the core of the star is $$P\sim GM^2/r^4$$, where $$M$$ is the total mass of the star. The star contains roughly equal numbers of neutrons, protons, and electrons, so $$M=Knm$$, where $$m$$ is the mass of the electron, $$n$$ is the number of electrons, and $$K\approx 4000$$. Setting the pressure at the core equal to the degeneracy pressure of a relativistic gas, we find that the Chandrasekhar limit is $$\sim (hc/G)^{3/2}(Km)^{-2}=6M_\odot$$. This is ignoring factors of $$\pi$$, etc., but is still on the right order of magnitude, and I expect the dependence on the variables to be right. For a star made out of dark matter, we have $$K=1$$, but current estimates of the mass of WIMPs are in the range near about 10 GeV, or $$2\times10^4 m_e$$. This makes the factor $$Km$$ greater by something like a factor of 5, so that the Chandrasekhar limit should be smaller by about a factor of 25. Of course there is considerable uncertainty here, but it does seem sufficient to rule out the existence of stars made out of degenerate WIMP matter with masses greater than the white-dwarf version of the Chandrasekhar limit.

• @A.V.S.: If I'm wrong, I'd like to understand what my mistake is. Please let me know. I'll spell out my reasoning in more detail in the question so that we have something more specific to discuss. – Ben Crowell Feb 3 at 17:07
• Two downvotes, neither with any specific explanation. I don't care about the gamification, but if I'm wrong, downvotes without explanation aren't helping me to learn from my mistake. – Ben Crowell Feb 3 at 17:29
• degeneracy pressure of a relativistic gas this is relevant for Fermi gas. As long as we are speculating, there is no reason that DM would be fermions, that it would be degenerate, or that it would be 'gas' (as in weakly interacting). And so, my overall issue is that you jump into speculation without explicitly (or implicitly) noting what is known, what you infer from OP and what are your own assumptions and what are existing speculative constructs. – A.V.S. Feb 3 at 17:32
• @A.V.S.: I see. Thanks for the explanation. My answer does assume WIMPs as dark matter candidates. If they are the lightest supersymmetric particle (LSP), then the main candidate is the neutralino, which is a fermion. The sneutrino is possible but not in the MSSM. So it seems quite plausible to me to take fermions as the likely DM particle, but you're right, I didn't highlight that assumption. Thanks for pointing that out, which prompted me to learn more about DM candidates. [...] – Ben Crowell Feb 3 at 18:06
• [...] Having said that, I don't agree with the further criticisms you list (assuming fermionic DM). Degeneracy is not an assumption, it's a logical conclusion because runaway gravitational collapse is guaranteed to increase the concentration beyond the quantum concentration. It seems natural to assume that it's weakly interacting, since that's a defining characteristic of DM. – Ben Crowell Feb 3 at 18:09