Why are black hole singularities posited when we now believe singularities no longer exist? I understand that once we pass the event horizon we fall into what’s been referred to as a singularity, and I know that hypothesis is popular. How do we know that all the matter/light that has been pulled in doesn’t just get pushed out into a different universe/uninhabited area of dark matter?

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    $\begingroup$ Present day physics is not about opinions and beliefs and popularity. It is about observations and measurements of nature modeled mathematically very strictly , with models that are predictive. Our cosmological models depend on general relativity, which has mathematical singularities. There is continuous effort to quantize gravity definitively , if/when it is successful it will give a different form to what is now mathematically modeled as a singularity, $\endgroup$
    – anna v
    Feb 10, 2022 at 20:02

4 Answers 4


16 February 22 update I have now revised this explanation and included a lot more detail because I am finding that a lot of people who are really interested in black holes and what might be inside them are not well aware of vital background astrophysics and do not fully understand what I am saying so I am including a lit more background information in what is quite a complex explanation.

Consider the "centre of gravity" of any object. This is the point of balance it is a mathematical singular point and exists for any body whatever it is.

Now consider the centre of gravity of a planet or star. This is the effective point where all the gravitational attraction of the earth is directed the surface of the earth happens to stop us getting there but if we was a hole right through the middle of the earth is is where you would end up if you fell down it this is a singular point or a singularity all gravitating bodies have them, even black holes.

The singularity is a mathematical concept not a physical concept. A lot of the misunderstandings in physical science are caused by students taking the mathematical concepts too literally and not tempering it with true physical understanding.

Think of the singularity like a target towards which all the matter that is collapsing is trying to get to. Clearly it cannot all be in the same place physically so it means that we know for any object something is going to prevent this. But the big question is what will happen? so it might be a good idea to have a go at answering this question in a simple way that anyone can understand.

Next let me debunk a lot of myths about an event horizon. The event horizon of a black hole is simply just like the horizon on the earth's surface. We see a ship disappear over the horizon as it sails away but we know that it has not been destroyed we just cannot see it because the earth is in the way. The event horizon of a black hole is just like this, we cannot see any more of what is going on because the force of gravity and the finite fixed speed of light prevent us from seeing what is going on from a distance. Consequently we must think of ourselves as joining the particles involved in the process and look at things physically from the point of view of the matter taking part in the collapse and not from looking at a distance.

Even the extreme conditions inside a collapsing neutron Star are mild compared with the energies involved with the LHC particle collisions, where they claim they are showing that the laws of physics we are familiar with still apply back to the first picoseconds or nanoseconds of the Big Bang. It is useful to note that light travels about 30cm in one nanosecond

So it is clear that in the first instants of the collapse inside the event horizon we CAN model this collapse. It is important to note that the collapse will start to take place some time before any event horizon has formed

Let us now look at this using the fundamental classical equations of a non rotating black hole, A very useful applet reference is:-


This will be very important later in this presentation.

Now it is a simple calculation to show that a thin massive shell collapsing from a given radius to half this radius as part of a massive body collapsing under its self gravity gains a constant amount of energy. This is one of the fundamental equations of star formation in astrophysics.

Firstly this tells us that particles in any mass collapsing along their "proper time" geodesics towards the body's mathematical singularity (the centre of gravity of any object) under inverse square law gravity will release an infinite quantity of energy unless this collapse is prevented in some way.

For cold interstellar gas and dust this state happens with the formation of a star, planet or asteroid. asreroids just aggregate. planets get quite hot during the process and can melt and differentiate chemically. Larger objects get hotter as they collapse under gravity and stars are so hot that the atoms are dissociated into a plasma and it is the simple gas pressure of the heat that stops the collapse

For a star, initially this gravitational collapse produces enough heat to make it shine although collapse driven young stars are frequently hidden deep inside dust clouds that form them but can be seen using infra red astronomy. If nothing further happened the star would cool and fade in a few million years. This caused a great deal of worry for astrophysicists during the 19th Century because it was becoming clear that the geological record showed that the earth was much older than this!

Eventually if the star is big enough the core of the star will get hot enough to start hydrogen to helium fusion in its core. This is helped by the presence trace quantities of "metals" or nuclei above the "big bang" elements of hydrogen helium and a trace lithium in its core. This produces energy at a reasonably stable rate dependant on the mass of the star and keeps the star in the "main sequence" for much longer than the simple driver of gravitational collapse energy would do and solves the problerm od the old earth!

For a small to modest sun sized star this is almost the end of the story because when it runs out of hydrogen in its core. The volume round the core collapsed under its self gravity to generate more heat it collapses a bit further under gravitation to make the helium core hotter and start "burning" hydrogen in a shell around its core. This makes the star "brighter" by producing more energy than the core did. To release more energy from its surface the star swells up to increase its surface area but at the same time the pressure is less at its surface and the star cools to become a red giant as the shell gets bigger internal collapse gets greater and the star continues to become an "asymptotic giant" and a "planetary nebula" with, at its centre, a planet sized helium white dwarf star of degenerate gas slowly cooling down over a vast length of time.

Now we must move on to a large star at the end of its life larger stars can get hotter because they have stronger gravity and need higher gas pressures to stop the collapse so when the hydrogen runs out in the core the collapse to form a shell allows helium burning to form carbon nitrogen and oxygen the vital elements of life there is a set of asymptotic giant arars producing white dwarf stars with carbon and neon silicon and higher nuclei in them the final state is iron becauase beyond iron furthe nucleosynthesis absorbs energy rather than generates it and further core collapse must happen to keep up the energy flow to maintain the star.

This pushes the core "white dwarf" temperature above its Chandrasekhar limit which occurs with a core mass of about 1.4 solar masses. This is the limit beyond which electron degeneracy pressure cannot support the star and it starts to collapse from the centre of the degenerate material to form a neutron star which is supported by neutron Degeneracy pressure the forcing of the electrons to combine with the protons to form Neutrons absorbs energy and cools the core driving the collapse. It also produces a vast flux of antineutrinos which effectively blows the outer layers of the star to pieces which after a period of expansion appears in the sky as a supernova of type II

The situation here is quite complex because there are three routes depending on the mass of the large star.

1 For the lowest mass stars in this state the neutron star appears as the residue of the supernova

2 For the highest mass stars the core collapse does not produce a neutron star but matter antimatter pair production happens this also cools the core because of the energy loss into matter but the supernova explosion produces no residue and disrupts the star completely.

3 For the intermediate mass case, after the initial formation of the neutron star the some of the material falls back onto the neutron star and this extra mass takes the star above its "Tolman-Oppenheimer-Volkoff limit of around 2.15 solar masses (this is the most recent figure) at this point the core of the star starts to collapse under its self gravity. this has not formed an event horizon or even a photon sphere yet it is just collapsing under its own gravity which is relatively low because it is only fraction of the mass of the whole neutron star

It is heading towards its Singular point so what is going to stop it so look at the hawking radiation calculator.

The particles are all interacting with each other (the mean free time between particle interactions in a neutron star is around 10e-25seconds) and getting hotter and creating particle antiparticle pairs. this will increase the gravity locally and drive the collapse further eventually a small black hole will form but there is no change in the laws of physics from the point of view of the particles and the collapse will continue inside the horizon there will also be very little angular momentum

The radiation and particles can escape a short distance from the main collapsing mass but they will always fall back into the mass. This is the inner event horizon of a simple black hole This is well known from the rotating Kerr caseand Roy Kerr himself if asked will confirm this as he did in a private conversation.

Now this second event horizon will be emitting hawking radiation into its own event horizon because of the greater curvature of the gravitational field. As the collapse towards the singularity progresses this Hawking radiation will increase and looking at the Hawking calculator shows that it increases faster than the energy generated by the collapse creates energy so eventually the flow outwards will overtake the new energy created by the collapse and finally the flow outwards will equal the flow coming back from the reflected energy returning into the collapsed hole insides it now distant event horizon to infinity.

It is only after this has happened that the event horizon will eat its way out ti the surface of the neutron star while the inner event horizon will collapse and turn into a kerr ring singularity that will stabilise the hole

This is the firewall that is deep inside the black hole and is probably a sub plank dimensions. The calculations of the dimensions and energy levels for a given mass of black hole are very difficult and I have not done them yet but the way the equations work with each other is perfectly clear.

So inside every black hole is a white hole hidden by the event horizon

It is quite possible that enough energy will be released to create a new big bang universe just like our own but that needs the same concepts used with a Kerr black hole. Looking at this gives even more insights into quantum string, brane, gravitational and cosmological theories but that is beyond the limits of answering this simple question and debunking a lot of the spurious black hole mysteries that popular science pundits like to promulgate.

The point I am making is that it is quite possible to analyse what is going on using our existing physics.

  • $\begingroup$ Welcome to PSE Sir! I admire the mission you state in your profile, but I'd like to point out two inconsistencies in your answer. "collapsing to a mathematical singularity under inverse square law gravity will release an infinite quantity of energy" - This relates only to Newtonian gravity, which is a force field and for this reason possesses potential energy. This energy diverges at the center of a mass point (Newtonian singularity), as your statement describes. However, in GR, gravity is not a force, so no potential energy exists in principle and thus your quoted statement does not apply. $\endgroup$
    – safesphere
    Feb 12, 2022 at 6:24
  • $\begingroup$ In the Schwarzschild (or Kerr) spacetime, energy is conserved. If two objects with the masses $m$ and $M$ at infinity fall to each other and no energy is dissipated away in the collision (no radiation), then the total mass/energy of the combined object in GR is $m+M$ regardless of how small the resulting radius is (ignoring any quantum effects). So conceptually, a free fall of a particle to a (hypothetical) singularity does not release an infinite energy and doesn't change the energy of the system at all. Please see this question for details: physics.stackexchange.com/questions/356029 $\endgroup$
    – safesphere
    Feb 12, 2022 at 6:31
  • $\begingroup$ The second inconsistency is that you seem to be thinking of the singularity as a point (and equivalently of the second horizon as approximately a sphere). however, the Schwarzschild singularity is an infinitely long line existing in the future. See: math.stackexchange.com/questions/2929400 - And equivalently the inner Kerr horizon is a spherinder (the hypersurface of a 4D cylinder of the spherical type): hi.gher.space/wiki/Spherinder - that exists as a future moment of time for those outside of it - it cannot radiate anything to the past that you refer to as "outwards". $\endgroup$
    – safesphere
    Feb 12, 2022 at 6:49
  • $\begingroup$ In summary, your answer refers to two competing flows of energy inside a black hole, "inward" and "outward": "the flow outwards will overtake the new energy created by the collapse" - However, as I elaborated above, neither flow actually exists and neither does the very problem that you are trying to solve ("infinite energy"). Interesting answer anyway, so +1. $\endgroup$
    – safesphere
    Feb 12, 2022 at 6:58
  • $\begingroup$ Thank you for your reply. I appreciate what you are saying and it follows the general approach driven by the simple maths but it is a big one and Tony need some real physics to sort this out not just the maths. as comments are limited in length I will now start a new comment like you. $\endgroup$
    – Ian Kimber
    Feb 13, 2022 at 16:08

Depends on what you mean by "posit".

Our best description of gravity at the moment is General Relativity. The singularity is a direct consequence of the equations. It's not a hypothesis that is assumed.

Even when you assume that GR is incomplete and that the singularity is almost certainly not a complete description, that doesn't give you much insight into alternatives.

How do we know that all the matter/light that has been pulled in doesn’t just get pushed out into a different universe/uninhabited area of dark matter?

We don't. But at the same time, we have no evidence that it does. It's certainly not the only other option. Would assuming such a thing occurs give us insights into how other astronomical objects behave or how compact particles interact? If it doesn't, then there's no reason yet to replace the singularity in the model.

All models are incomplete. But we can still use them with that knowledge.

  • $\begingroup$ "The singularity is a direct consequence of the equations. It's not a hypothesis that is assumed." - Sure it is. The existence of spacetime passed the horizon is unfalsifiable and therefore is an assumption, not a fact, including the singularity. $\endgroup$
    – safesphere
    Feb 11, 2022 at 7:43
  • $\begingroup$ The GR model solution is a singularity. Even though this probably represents a limitation of the model rather than an accurate description of a physical black hole, it's the best we have until we can extend GR. Do you find texts claiming that such a solution is an accurate description of real black holes (rather than mathematical objects like a Schwarzschild black hole)? $\endgroup$
    – BowlOfRed
    Feb 11, 2022 at 7:54

A singularity like in general relativity will still have the properties of the exterior of a black hole if there isn't a point-like accumulation of matter in the center of the hole. If the particles are not truly point-like (like strings or alternatives), the singularity will be smoothened somehow, and there will not form an infinitesimal hole at infinity. As long as the mass can reside within the Schwarzschild radius, a black hole can exist. But without a hole in space at infinity. A GR relativity is incompatible with a beginning in time. If the initial state of the universe was truly point-like, as can be in GR, a beginning in time would have been impossible, as motion could never have started. Motion can't start without preceding motion, and exactly this would happen if the initial state is a classical GR singularity. It would be an eternally frozen state.A strong argument that new physics is present at high energy (or small distance) scales. It's therefore that the center of a black hole is not a point-like collection of mass.


As a rather speculative answer to your question, I will offer up this:

A black hole initiates when a star collapses into a singularity. Understanding the Pauli Exclusion Principle, we realize that two or more fermions (1/2 spin particles such as quarks or electrons) cannot simultaneously occupy the same quantum state within a quantum system. Thus, as a star collapses to a singularity, these particles must transform into energy that exits the rapidly forming black hole. This agrees with physicist John Wheeler (who coined the term Black Hole) who believed an imploding star converts its nucleons (protons and neutrons) into energy during black hole formation. Likewise, nothing with mass can reach c, so it would be impossible for a quark, with mass, to cross the distance between the even horizon and the singularity at the centre of a black hole.

However, bosons (spin 1 particles such as gluons) can occupy the same quantum state without limit. Therefore, when a star collapses all that remains of the original star are the gluons, compacted down into the singularity. Gluons are neither added nor subtracted during the collapse. We can arrive at a new definition of a black hole singularity – a star’s worth of gluons captured in a single point.

  • $\begingroup$ How can all particles be turned into energy if there is matter only. I mean matter and no antimatter. $\endgroup$ Feb 10, 2022 at 20:42
  • $\begingroup$ @Felicia You can see Wheelers ideas in the book Black Holes and Time Warps academia.edu/30472906/… on pages 244-253. It says: "Wheeler was fixated on his idea that quantum gravity must make nucleons in an imploding star dissolve away into radiation and escape the implosion. " Take a look at that chapter and you can see what he is talking about. $\endgroup$ Feb 10, 2022 at 20:58
  • $\begingroup$ What has the PEP got to do with anything? There are infinitely many quantum states available at a fixed position. $\endgroup$
    – ProfRob
    Feb 10, 2022 at 21:29
  • $\begingroup$ @ProfRob The original question is about the possibility of a singularity at the centre of a black hole. I'm offering a possible explanation for how a black hole can retain the mass and gravity of the originating star, but all held at a singularity. Since gluons are responsible for 99% of the mass of a nucleon, but are also immune from the PEP, this offers a possible answer that seems to align with Wheeler's ideas. $\endgroup$ Feb 10, 2022 at 21:36
  • $\begingroup$ @ProfRob "There are infinitely many quantum states available at a fixed position" - This is not true. A fixed position implies a total uncertainty of momentum, which implies motion, which implies changing positions. Thus your statement would apply at a fixed position, only if other positions existed to where the motion could be directed. When no other positions exist, no motion is possible, so your statement fails. I know this is a rough argument, cause a singularity is not even a "position", but still. Also, the Fermi energy needed would violate energy conservation. Thus you have no claim. $\endgroup$
    – safesphere
    Feb 11, 2022 at 7:18

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