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If a large star goes supernova, but not enough mass collapses to form a black hole, it often forms a neutron star. My understanding is that this is the densest object that can exist because of the Pauli exclusion principle: It's made entirely of degenerate matter, each particle of which cannot occupy the same quantum state of any other.

So these objects are so massive that they gravitationally lens light. If you make them more massive, they bend the light more. Keep going and going until they bend the light so much that light passing near the surface can barely escape. It's still a neutron star. Add a bit more mass, just enough that light passing just over the surface cannot escape. Now it's a black hole with an event horizon (I think?). Does this mean the neutron star has become a singularity? Isn't it still just a neutron star just beneath the event horizon?

Why are black holes treated as having a singularity instead of just an incredibly massive neutron star at its center? Does something happen when an event horizon is "created?"

Qmechanic
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Carson Myers
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    Strictly speaking, all objects gravitationally lens light. One of the main techniques we use to find extra-solar planets is micro-lensing. See http://en.wikipedia.org/wiki/Gravitational_microlensing#Detection_of_extrasolar_planets – Alasdair Allan Jun 02 '11 at 09:43
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    I remember reading somewhere that for ordinary degenerate matter, the event horizon is inside the object, so it is not a black-hole or a singularity. I think some basic calculations could help here. For instance, a neutron star is several kms in radius, while the event horizon will be a few hundred meters. Also, it is believed that there are more degenerate states, for instance quark matter. – Jus12 Jul 29 '11 at 16:40
  • @Jus12: Pedantically, the event horizon would form if the mass were compressed to a radius inside the current size of the ordinary or degenerate body. For the Earth the critical size in on order of 1 centimeter. – dmckee --- ex-moderator kitten Mar 09 '12 at 18:49
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    @Jus12: there is no event horizon anywhere in ordinary matter. Once an event horizon forms, all of the matter inside must collapse to a black hole. This is the key difference between a neutron star and a black hole--a neutron star is made up of (very dense) ordinary matter in a stable configuration. A black hole is essentially a pointlike or ringlike concentration of mass with infinite density. There is no way to perturb a neutron star and get a stable black-hole like configuration. – Zo the Relativist Aug 07 '12 at 14:44
  • @Jerry Schirmer. My error.. I meant "the event-horizon for a blackhole of equivlent mass", not the event horizon itself. – Jus12 Aug 11 '12 at 06:02
  • Depressing to see that the only correct answer (Victor's) is buried under 5 completely incorrect ones. The incorrect ones are incorrect because they argue from properties of the Schwarzschild metric. The Schwarzschild metric doesn't describe the process of gravitational collapse. –  May 05 '13 at 02:41

10 Answers10

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Short answer is yes.

But if you want to nit pick, I could argue that when a star collapses to form a BH, it first forms a horizon before the singularity forms (cannot form a "naked singularity"). And since time inside the horizon is essentially frozen with respect to that of an observer outside, the singularity NEVER forms. Yet from the point of view of the collapsing star, the singularity forms in about a millisecond after the horizon.

Jeremy
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In classical General Relativity, once an event horizon forms, every particle inside that event horizon will inevitably travel in the direction of the center of the black hole. This is what is meant by "gravitational collapse" and how matter comes to form a singularity in the center- no matter how small it is, or how close to the center it is, nothing can prevent it from approaching ever closer to the center. From the point of view of the object itself, it does reach the center in a finite time.

In some more exotic theories of physics, such as string theory or loop quantum gravity, the quantized nature of space and time comes to the rescue and prevents a singularity from forming, so a maximum, finite density is reached and an equilibrium is maintained in the center. This is similar conceptually to what you describe, but still a more exotic and much, much denser material than neutron star-stuff.

The density we're talking about here would be approximately one Planck Mass per cubic Planck Length, in other words 2.176 51 × 10^−8 kg / (1.616 199 × 10^−35 m)^3 ~= 5.15556^96 kg/m^3, where neutron star material is "only" (roughly) 10^18 kg/m^3.

In either case, however, outside the event horizon, the black hole can be treated mathematically and observationally as a simple singularity, so for observational calculations, there is no "value added" in worrying about the inner workings of the black hole. The theorem describing this is colloquially called "Black holes have no hair." This theorem was proven and coined by John Wheeler, the same physicist who coined the phrase "black hole" in the first place.

Andrew
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    This is only true for neutral uncharged black holes. Nothing massive can reach the center for rotating or charged black holes, only light. – Ron Maimon Aug 09 '12 at 08:09
  • The No Hair results were not arrived at by Wheeler. Israel did it first for spherical BHs, Carter later for axis yammered rotating BH's. Wheeler has been credited with the term Black Hole, he said it was Bekenstein. – Bob Bee Sep 04 '16 at 04:29
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If there is a trapped null surface, and one of the energy conditions like the null energy condition, or the weak energy condition is satisfied, and space outside is noncompact, there has to exist either a singularity or closed timelike curve inside the black hole.

See Penrose and Hawking.

Victor
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  • "Pensose and Hawking" --> "Hawking and Ellis"? Or http://en.wikipedia.org/wiki/Penrose%E2%80%93Hawking_singularity_theorems –  May 05 '13 at 02:34
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So, you are hoping a neutron star's edge could be just under the event horizon and be stable there - in other words, the neutron star would not collapse into the singularity that would form.

That is not possible. See my answer here: Why can't you escape a black hole? It has a nice picture that explains that once you pass the event horizon, the curvature of space-time essentially rotates the time direction to point into a spatial direction towards the center of the black hole. So just as you cannot resist moving forward in time, the edge of the neutron star cannot resist falling into the center of the black hole - that is it's future time direction.

Community
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FrankH
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Wikipedia seems to indicate they they all do:

At the center of a black hole as described by general relativity lies a gravitational singularity, a region where the spacetime curvature becomes infinite. For a non-rotating black hole this region takes the shape of a single point and for a rotating black hole it is smeared out to form a ring singularity lying in the plane of rotation. In both cases the singular region has zero volume. It can also be shown that the singular region contains all the mass of the black hole solution. The singular region can thus be thought of as having infinite density.

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    Sure, but is this just because we don't know more about them? Is it just for mathematical simplicity or does the matter of a neutron star actually collapse to zero volume when it becomes massive enough for an event horizon to form? – Carson Myers Jun 01 '11 at 20:12
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This is a question for the Physics forum.

The honest answer is that we don't know for sure. General relativity is a classic (non-quantum) theory, so it should fall apart at very small scales and very high energy densities - exactly what's supposed to happen in a singularity. We're still waiting for the quantum relativity theory to emerge; if and when that happens, we'll know a lot more about singularities.

WRT black holes, it is perhaps prudent to say that, the deeper in you go, the less we actually know what's going on.

We know a lot about the region surrounding the event horizon; we're pretty confident we got that right, and actually we have observations nowadays matching the theory.

We believe we know a little about stuff happening within the event horizon, but things are getting a bit foggy there.

We can't in all honesty say that we know much about what happens down at the bottom, in the very singularity; that's where general relativity divides by zero and goes belly up.

So, take everything with a grain of salt and keep an open mind.

Florin Andrei
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  • It is true that we do not know what happens at the singularity or very near the singularity, but for astrophysical black holes, classical General Relativity should hold well everywhere in the vicinity of the event horizon. – FrankH Mar 09 '12 at 13:33
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The maximum mass for a neutron star is the Tolman–Oppenheimer–Volkoff limit and is thought to be between 1.5 and 3 solar masses, the range being due to uncertainties of the equation of state of matter at these extreme densities. If the mass of a neutron star exceeds this limit then implosion to a black hole is assumed to be inevitable, there is no force that can repel the collapse according to general relativity.

So black holes are distinct from neutron stars and an event horizon only forms around the black hole.

The quantum degeneracy pressure of electrons that you mentioned occurs inside white dwarfs. In neutron stars the quantum degeneracy pressure of neutrons is responsible.

high_hopes
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All black holes contain singularities, however not all singularities involve black holes. A neutron star may be dense, matter the size of a pinhead can weigh as much as the earth, but there seems to be a mathematical cut-off point beyond which a black hole is formed. The first step for this is the formation of the even horizon, and everything within the event horizon is your singularity. If a neutron star's mass increases in relation to its radius to form its critical circumference (a star 10x heavier than our sun would have a critical circumference of around 110 miles, or a 20 mile radius), it undergoes gravitational collapse and you have your black hole. Beyond it, the matter is so infinitely dense that the gravitational pull sucks every photon of light into its centre. At this point you have your singularity, where the infinite density mean space and time as we know it, cease to exist. You find yourself in a constant state of chaotic equilibrium; like the ultimate drag car using up hundreds of kilos of fuel every second.

faz
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    What does it mean that "not all singularities involve black holes"? Have you disproved cosmic censorship? It is false that inside a black hole everything is sucked to the center-- it's just not true for rotating or charged solutions. There is no consensus on what happens in those. – Ron Maimon Aug 09 '12 at 08:08
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A black hole is a supper massive object with a very intense gravitational field, and so is a neutron star, but the difference between the two is, that, light can escape the gravitational field of a neutron star, but not that of a black hole, that is why it is called a black hole.

Some stars are massive enough to become black holes, but some are not, they either become dwarf stars (like our sun would), or neutron stars. Stars above the chandrashekhar limit (about 2 solar masses)would become black holes, have an escape velocity more than the speed of light, while those below it wont.

An event horizon is just light trying to escape a black hole, but stuck in an orbit around it.

Singularity is just a hypothesis, nobody knows if it really exists or not, but its not necessary for a black hole (according to me). I believe in the exclusion principle, but that would be at the string level (string theory) at a scale of $10^-33$ metres and that's how far a 'singularity' can go.

Shaun Fletcher
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  • I don't see how this addresses the question. The OP asks whether a black hole must necessarily have a singularity at the centre. The answer is basically yes or no (actually "yes" as the Hawking Penrose theorems assure us) but you commit to neither. – John Rennie Mar 19 '15 at 10:59
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    Also, an event horizon is not light stuck in an orbit. Light's only orbit around a Schwarzschild black hole is the photon sphere at 1.5 times the radius of the event horizon. Any lower than this and the light must escape or fall in completely, no orbiting allowed. – Jim Mar 20 '15 at 17:18
  • @Jim The horizon is lightlike, so photons tangential to it would neither fall nor escape. – safesphere May 21 '19 at 07:57
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    @safesphere that is not an accurate interpretation. A lightlike horizon means photons travelling radially outward would not make any headway. I advise reading up on the photon sphere first. I assure you that the math checks out – Jim May 21 '19 at 12:08
  • @Jim Lightlike means null, but not what you wrote. Please do tell what happens to the photons emitted at the horizon and tangential to it. Do they fall or escape in your opinion? – safesphere May 21 '19 at 16:51
  • @safesphere My opinion aside, the mathematics say that photons emitted tangential to and at the event horizon fall into it – Jim May 22 '19 at 11:37
  • @Jim Great, can you provide a reference to this calculation? – safesphere May 25 '19 at 01:59
  • @safesphere If you had checked wikipedia as I have been suggesting all along, you'd have seen exactly the calculation you are currently asking for. The photon sphere is the last stable orbit. It is where photons travel in a circular orbit. Anything lower than this, obviously, requires tangential photons to fall inward because of simple logic – Jim May 27 '19 at 14:08
  • @Jim You forget that the speed of light at the horizon is zero. Here is one relevant calculation: https://physics.stackexchange.com/questions/465313/are-the-horizon-generators-radial-null-geodesics-also/465461?noredirect=1 - Note there that the horizon generators are null geodesics that are neither ingoing nor outgoing. This means, light within the horizon is neither falling nor flying away. It is not circling either, so these geodesic are not orbits. – safesphere May 27 '19 at 16:50
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NO: all black-holes do NOT have a singularity. The following description of a “radiation-structured” BH is conjectural; but it answers your question about singularities, and it makes sense. I've omitted all but the key ideas in describing this concept, and there's a lot of detail that may, as yet, be unexplainable:

When particle momentum in a developing black-hole (not quite yet a BH) reaches a critical level, gravitational potential has produced a value of particle momentum that conflicts with the values allowed by quantum mechanics, and nature responds by transforming high-energy particle-dynamics into high energy radiation. This radiation is manifested at or near the BH boundary, rather than in the form of a singularity.

This critical event establishes a BH boundary 'density' corresponding to E/surface-area = energy-density. This energy density remains constant regardless of BH size and energy content ; a formative-BH-star’s particle-driven temperature also remains constant as it’s size and energy and entropy increase, following the transformation of a proto- blackhole into a BH. Additional energy, after BH formation, joins the BH boundary energy, rather than enters the BH interior. There is no singularity associated with this concept of a “radiation-structured” black-hole. Comments are appreciated…RobertO

RobertO
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    This is total nonsense. –  Jan 28 '19 at 04:17
  • Hi Ben: Thanks for your comment, and I would appreciate why you believe that a radiation structured BH does not make sense. Thanks, RobertO – RobertO Jan 29 '19 at 18:07