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When something is compressed to within its Schwarzschild radius, it will compress to a singularity within finite time regardless of what forces are involved. But is there some lesser point at which neutron degenerate matter won't be able to support itself? It takes a lot of mass to make a Schwarzschild radius larger than an atom, but is that actually necessary to make a black hole?

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  • $\begingroup$ We don't know. The only black holes we know about are stellar size or larger. If smaller chunks of matter can be compressed to a black hole, at all, is experimentally untested. You can, of course, extrapolate general relativity and deal with the question as a matter of faith in a theory that hasn't failed, yet. $\endgroup$ – CuriousOne Jun 10 '15 at 22:02
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The answer you your question is no. There is a radius, larger than the Schwarzschild radius at which a neutron star, quark matter, whatever its equation of state, cannot be supported against collapse.

There are limits imposed by causality and General Relativity on the structure of compact stars. In "Black Holes, White Dwarfs and Neutron Stars" by Shapiro & Teukolsky, pp.260-261, it is shown, approximately, that even if the equation of state hardens to the point where the speed of sound equals the speed of light, that $(GM/Rc^2)<0.405$.

The Schwarzschild radius is $R_s=2GM/c^2$ and therefore $R > 1.23 R_s$ for stability. This limit is reached for a neutron star with $M \simeq 3.5 M_{\odot}$. A more accurate treatment in Lattimer (2013) suggests that a maximally compact neutron star has $R\geq 1.41R_s$.

If the equation of state is softer, then collapse will occur at smaller masses, and higher densities but at a similar multiple of $R_s$.

Thus it is not necessary to compress matter within $R_s$ to form a black hole.

The picture below (from Demorest et al. 2010) shows the mass-radius relations for a wide variety of equations of state. The limits in the top-left of the diagram indicate the limits imposed by (most stringently) the speed of sound being the speed of light (labelled "causality" and which gives radii slightly larger than Shapiro & Teukolsky's approximate result) and then in the very top left, the border marked by "GR" coincides with the Schwarzschild radius. Neutron stars become unstable where their mass-radius curves peak, so stable neutron stars are always significantly larger than $R_s$ at all masses.

Neutron star mass-radius relations

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  • $\begingroup$ That's a very nice diagram. We still have to extrapolate down from neutron star mass to an arbitrary mass to answer the question, though. Do we have any observation suggesting that smaller bodies can become black holes? $\endgroup$ – CuriousOne Jun 10 '15 at 22:49
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    $\begingroup$ @CuriousOne My answer does answer the question posed in the title: i.e. No. Observationally it is found that all black holes are more massive than $4M_{\odot}$. There is a clear gap between the most massive neutron stars and the least massive black holes, that is not decisively explained yet. There are no stable, self-gravitating solutions for neutron matter below about $0.15M_{\odot}$. $\endgroup$ – Rob Jeffries Jun 10 '15 at 22:59
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If the supernova remnant is less than about 3 solar masses, a neutron star will be the result. Neutron degeneracy will hold up the star, just as electron degeneracy holds up a white dwarf. Gravity wants to crush the neutrons out of existence, but neutrons are fairly solid, and push back. However, if the mass of the supernova remnant is greater than about 3 solar masses, the greater gravity will crush even the neutrons out of of existence. If this were to happen, the star would collapse further. In fact, if neutrons are crushed out of existence, then we know of no other force that can resist the crush of gravity. The star will, in theory, collapse all the way down to a black hole

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  • $\begingroup$ The maximum mass supportable by ideal neutron degeneracy pressure is $0.7M_{\odot}$. More massive neutron stars exist because of the strong repulsion between nucleons in asymmetric nuclear matter. The upper limit for a stable neutron star is at least $2M_{\odot}$ and less than $3.5M_{\odot}$, but where between these limits is currently unknown. $\endgroup$ – Rob Jeffries Jun 10 '15 at 22:25
  • $\begingroup$ A supernova remnant is actually the ejecta from the supernova event. You want compact object here (sometimes stellar remnant is used). $\endgroup$ – Kyle Kanos Jun 10 '15 at 22:29
  • $\begingroup$ And how close does it have to get to the Schwarzschild radius to get that to happen? $\endgroup$ – DanielLC Jun 10 '15 at 23:39

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