33
$\begingroup$

Why does there have to be a singularity in a black hole, and not just a very dense lump of matter of finite size? If there's any such thing as granularity of space, couldn't the "singularity" be just the smallest possible size?

$\endgroup$
  • 2
    $\begingroup$ The "singularity" in a black hole doesn't have any matter in it, necessarily. A "singularity" doesn't mean a point of infinite density, it means a place you can reach be geodesics where space-time is not a manifold. $\endgroup$ – Ron Maimon Jan 1 '12 at 17:40
  • 2
    $\begingroup$ duplicated or nearly duplicated by physics.stackexchange.com/q/75619 $\endgroup$ – Ben Crowell Aug 29 '13 at 22:30
  • $\begingroup$ related: physics.stackexchange.com/questions/144447/… A singularity doesn't have a well-defined volume (so it isn't necessarily zero), and it doesn't even have a well-defined dimensionality (so it isn't necessarily analogous to a point, or to a surface). $\endgroup$ – Ben Crowell May 16 '18 at 18:48
  • $\begingroup$ However you can understand why people go about talking about singularities as a ''point'' because media and some scientific articles have said so. Points of course, have no dimensions, no volume. The terminology ''point singularity'' is often used by layman. $\endgroup$ – Gareth Meredith Mar 25 at 4:29
22
$\begingroup$

It's important to understand the context in which statements like "there must be a singularity in a black hole" are made. This context is provided by the model used to derive the results. In this case, it was classical (meaning "non quantum") general relativity theory that was used to predict the existence of singularities in spacetime. Hawking and Penrose proved that, under certain reasonable assumptions, there would be curves in spacetime that represented the paths of bodies freely falling under gravity that just "came to an end". For these curves, spacetime behaved like it had a boundary or an "edge". This was the singularity the theory predicted. The results were proved rigorously mathematically, using certain properties of differential equations and topology.

Now in this framework, spacetime is assumed to be smooth - it's a manifold - it doesn't have any granularity or minimum length. As soon as you start to include the possibilities of granular spacetime, you've moved outside the framework for which the original Hawking Penrose theorems apply, and you have to come up with new proofs for or against the existence of singularities.

$\endgroup$
  • $\begingroup$ So, then it could be said that any way you look at it, GR breaks down (at the singularity) - singularity or not? $\endgroup$ – Per Jan 1 '12 at 16:12
  • 9
    $\begingroup$ Yes, I've often heard it phrased as "GR predicts its own downfall". This is maybe a bit harsh - it's more a case that GR is making a statement about its domain of applicability. $\endgroup$ – twistor59 Jan 1 '12 at 16:19
  • 1
    $\begingroup$ @Ron Maimon, well we don't know that yet, there may be naked singularities, the question is still open. But I agree with you strange properties of spacetime doesn't mean that somethng is wrong with the theory, it could be that space time has strange properties. It is not the first in physics. $\endgroup$ – MBN Jan 10 '12 at 6:30
  • 1
    $\begingroup$ @pabouk: Not clear. The singularity is timelike if the black hole is spinning or charged, and in this case, we don't have exact collapse solutions. It should be possible to figure out the behavior of exact charged collapsing dust, but it hasn't been done. The singularity theorem has been misinterpreted to mean that there is a spacelike singularity which swallows all matter, and this is only true for spherically symmetric collapse, it fails when the singularity is timelike. My own opinion about what happens in this case is that there is a partial explosion, but this is not what others think. $\endgroup$ – Ron Maimon Oct 9 '14 at 0:55
  • 1
    $\begingroup$ @pabouk: Regarding 'settled state', the end point for the black hole is always a Kerr-Neumann static solution, the major issue is what happens to the interior matter in the case that it misses the singularity, i.e. when you have a charged/rotating collapse. My feeling on this is that it is partially ejected, and the black hole is only produced to the extent that mass-energy is destroyed at the singularity, and this is only massless stuff, for highly charged or highly rotating black holes. I suspect the full answer requires string theory, the classical theory has nonsense extra universes. $\endgroup$ – Ron Maimon Oct 9 '14 at 0:58
2
$\begingroup$

See Carter 1968 for why rotating black holes that have incoming disturbances may not have a singularity at all.

A stationary non - rotating hole will have a singularity. But no one thinks that these exist in nature. But with rotation that singularity 'shrinks' to a ring. The set of paths that hit the singularity is shrunk to a mathematical 2D plane from 'all directions' with the Swarzschild Soln. Then with incoming 'noise' it may be that there are no paths - geodesics - that lead to a singularity.

http://luth.obspm.fr/~luthier/carter/trav/Carter68.pdf

All exact solutions of General Relativity are done with asymptotically flat space, which does not exist in the real world. So while the theory of GR admits singularities, in a real classical GR world they likely don't exist.

Carter actually always talks about a singularity, but one with no paths to it. No ouchy at the end of a path. With no paths to a singularity - is it really there? I would think not, and as Carter points out, others do too. (Lifshitz and Khalatnikov).

$\endgroup$
  • 1
    $\begingroup$ There must be null geodesics which hit the singularity by the Penrose theorem--- the outgoing null rays have to defocus somewhere, and they can't do it at any ordinary point. There is no requirement that any non-null geodesic must hit the singularity, and I don't think they do for rotating/charged black holes, although this is a minority position. $\endgroup$ – Ron Maimon Jan 2 '12 at 9:06
  • $\begingroup$ -1: I have to downvote, because you haven't fixed your wrong statement that Carter says that there is no singularity after perturbation. Carter is NOT saying thi: it contradicts the most basic implications of Penrose's theorem. He is only noting that a generic rotating+charged perturbation leads to a time-like singularity, so that only null rays hit the singularity. This is the content of the paper you linked, it doesn't say what you say here. $\endgroup$ – Ron Maimon Jan 3 '12 at 6:03
  • 2
    $\begingroup$ See the paper, in the 'Implications' section at the end. "Thus as the symmetry is progressively reduced... the extent of the class of geodesics reaching the singularity is steadily reduced likewise, until in the case with both charge and rotation there are almost none at all, which suggests that after further reduction of symmetry, incomplete geodesics might cease to exist altogether." As I said Carter thinks that real black holes in nature don't have any geodesics that end on the singularity. $\endgroup$ – Tom Andersen Jan 9 '12 at 21:10
  • $\begingroup$ Another quote, earlier in the paper: "Thus we conclude that when the solution is charged, no timelike geodesics can reach the singularity". I think I see what you are saying - that Carter says the singularity is there, its just that there are no paths to it. I don't see the difference myself - if there are no paths to an 'object' - does the object exist? $\endgroup$ – Tom Andersen Jan 9 '12 at 21:16
  • $\begingroup$ The key word is "almost", and I agree with Carter that nearly all paths miss the singularity. Only a measure zero set of all geodesics, null geodesics, hit the singularity and it doesn't have to be all null geodesics either. Carter is saying "the singularity is there, and there are very few paths to it", namely null geodesics directed at the ring. The idea that the singularity can be removed is speculation by Carter, which it is not really polite to recall, because it only shows he hadn't fully internalized the singularity theorems in 1968 (I am sure he did subsequently). $\endgroup$ – Ron Maimon Jan 9 '12 at 22:59
2
$\begingroup$

Because otherwise general relativity would contradict itself. The event horizon of a black hole is where not even light can escape. Below the horizon all photons must fall. In relativity theory all observers measure the speed of light the same, c; that's a postulate of the theory. Then all physical things (including observers) at and below the horizon must fall and keep falling, lest they measure the speed of light emitted upward to be something other than c. If you could stand on a very dense lump of matter of finite size at the center of a black hole, and pointed a flashlight upward, the photons would somehow have to fall to the ground (without moving upward at all) and you wouldn't measure the speed of light to be c in the upward direction. The theory would be broken. The singularity is the "can't fall further" point and the theory becomes inapplicable there.

$\endgroup$
2
$\begingroup$

The chosen answer is quite good. This is a general answer for an unsophisticated audience whose naive questions are sent here as a duplicate, as in this case.

Classical physics developed when calculus and differential equations entered the field and made possible the mathematical modeling of observations and data; before the times of Newton the models had not advanced further than using algebra and euclidean geometry.

The mathematical formulas appearing in classical physics are full of singularities. Take the 1/r potentials in electricity and gravity. The approach to r=0 predicts larger and larger fields, up to infinity. This is not a problem because classically any object has a volume, no matter how small, and it was understood that the infinities were theoretical extrapolations, for the classically non existent states of point particles. Any particles were presumed to have a mass that could not be compressed to a point, so these singularities were not a problem. When experiments started getting data below the nanometer level, Quantum Mechanics had to be invented in order to explain the data, and quantum mechanics comes with the Heisenberg uncertainty principle,HUP, which turns all singularities into a fuzzy region . The electron does not fall on the proton but is constrained to orbit around it. The same with the electron on the positron. Free electrons are posited to be zero point particles with mass, but there is no infinity in the field due to the HUP, active in any interaction that would define r.

General relativity is also a mathematical model for very large scales and energies and mass, for the gravitational observations. As the chosen answer shows a mathematical singularity is extrapolated in the mathematical description of classical black holes.

The same is true for the original mathematical model of the big bang cosmological model, where a different type of singularity was postulated using General Relativity, where all the presently seen energy of the universe appeared. Astrophysical observations forced the conclusion that at the very beginning quantum mechanics has to be used, so the beginning of the universe is a fuzzy region and the mathematical singularity non existent. ( we are still waiting for a definitive quantization of gravity though).

So the title:

Why singularity in a black hole, and not just “very dense”?

can be answered by: It is not a singularity but the concept of "dense" is quantum mechanical, "dense probability distributions for the energy content" generating a quantum mechanical mathematical fuzziness around the classical singularity point.

$\endgroup$
0
$\begingroup$

Actually there are no singularities inside the black hole. It is just a mathematical special point on the coordinate system, which does not correspond to any real-world singularities.

The event horizon of a black hole is just an impenetrable barrier on which the time is frosen so nothing can pass it. In another model the black hole as a whole behaves as a viscous liquid with quite limited density (the density decreases as the BH's mass rises).

$\endgroup$
0
$\begingroup$

For any experiments a spherical Black Hole behaves the same way as if its mass was uniformly distributed over its surface or uniformly distributed over its volume or concentrated in its center. These variants are indistinguishable.

It is impossible to find exact distribution of mass inside a black hole because it has no internal structure, due to holographic principle (if it had, it would be possible to transfer information out of black hole via gravitation waves).

$\endgroup$
0
$\begingroup$

My understanding is that the Uncertainty Principle forbids point masses, which would have 0 uncertainty in position, and thus total uncertainty in momentum. The well-known result is that no particle can be confined in a region smaller than its wavelength. Three or more solar masses in the space of one particle is indeed a very high density, but not infinite.

It is the same as the reason why electrons, with a much longer wavelength than protons and neutrons, cannot fall into the nucleus of an atom. They are already as close as they can get.

$\endgroup$
0
$\begingroup$

My two cents; singularities do not need to form. They are for all intents and purpose, a rough model that neglects quantum physics.

If you have the right kind of argument, you can have models that avoid singularities. The event horizon, is not a true singularity but is in fact a coordinate phenomenon (ie. space becomes timelike and time spacelike). There are today, models which attempt to explain the collapse of a star in such a way, that singularities do not form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.