Do Black Holes Exist? Maths & Singularity Explained

[wolram]

Do Black Holes exist? after all the maths breaks down at the singularity Like the maths of the singularity breaks down at the beginning of the universe, what if there is no need for the maths to break down and these singularities do not exist?

 

[FactChecker]

A black hole begins long before the math of a singularity breaks down. Many have been "seen" by observing the behavior around them. Although the math at the singularity in the middle of a black hole might be a problem, that is only at the very center.

 

[wolram]

A black hole begins long before the math of a singularity breaks down. Many have been "seen" by observing the behavior around them. Although the math at the singularity in the middle of a black hole might be a problem, that is only at the very center.

It is also a problem at the universe singularity, why should we belie they exist, may be Black holes exist but they do not have the properties we attributes give to them

 

[PeterDonis]

Do Black Holes exist?

This is a vague question. Let me try to answer some more precise versions of it.

(1) Do the singularities that show up in idealized models in GR (of black holes and of the universe) describe something that actually exists in our universe? Answer: almost certainly not. Physicists regard these features of the models in GR as an indication that GR breaks down in this regime and needs to be replaced with a more comprehensive theory

(2) Do objects exist in our actual universe that, in practical terms, are equivalent to the idealized black holes that show up in the models, even if there aren't actual singularities inside them? Almost certainly yes. In other words, there are almost certainly regions of spacetime into which objects can fall and never come out, at least not in any practical sense. (Coming out in $10^{67}$ years as Hawking radiation doesn't count.)

 

[PeterDonis]

may be Black holes exist but they do not have the properties we attributes give to them

See my post #4.

It is also a problem at the universe singularity

The same answer I gave in post #4 applies to this: there is almost certainly not an "initial singularity" in our universe like the one that appears in the idealized models, but that does not mean the early universe was not very hot, very dense, and with very strong spacetime curvature. So the idealized model is still a very good approximation back to a very early time, which is all we need it for in practical terms anyway.

 

[wolram]

See my post #4.
The same answer I gave in post #4 applies to this: there is almost certainly not an "initial singularity" in our universe like the one that appears in the idealized models, but that does not mean the early universe was not very hot, very dense, and with very strong spacetime curvature. So the idealized model is still a very good approximation back to a very early time, which is all we need it for in practical terms anyway.

What can replace the idealised model, how can we do away with the singularity.

 

[PeterDonis]

What can replace the idealised model, how can we do away with the singularity.

By finding a more comprehensive theory that covers the regime where GR appears to break down. In the meantime, as I said, we can still use the GR models in the regime where they don't break down, so in practical terms not having a solution (yet) to this issue has no real impact on the work scientists are doing. It's just one of many open questions for research.

 

[newjerseyrunner]

What can replace the idealised model, how can we do away with the singularity.

There are lots of ways to deal with singularities. If scientists knew how to do it for this case, they would, they just haven't figured it out yet though.

Aren't there singularities in Standard Model too? I was under the impression that QM had point interactions, which happen at a single point in space and time, which causes mathematical problems. String theory doesn't have the same problem because the extra dimension removes any points and allows you to do time slicing in a relative way.

I'm pretty sure neither String theory or LQG have singularities at the hearts of black holes, so there are ways around it. The theories will approximate to the standard model though for any energy level that we are able to test with, so I don't think we'll have any idea for a while.

 

[PeterDonis]

I'm pretty sure neither String theory or LQG have singularities at the hearts of black holes

Nobody knows how to construct a string theory or LQG model of a black hole (yet), so we don't know if this is true (although I know most workers in the field expect it to be true).

 

[timmdeeg]

(2) Do objects exist in our actual universe that, in practical terms, are equivalent to the idealized black holes that show up in the models, even if there aren't actual singularities inside them? Almost certainly yes. In other words, there are almost certainly regions of spacetime into which objects can fall and never come out, at least not in any practical sense.

Am I right that nevertheless geodesics are incomplete in this case, even if there are no singularities in the sense you mentioned ?

 

[PeterDonis]

Am I right that nevertheless geodesics are incomplete in this case, even if there are no singularities in the sense you mentioned ?

No, because geodesic incompleteness and the presence of singularities in the sense I mentioned are equivalent, since the definition of "singularity" in the sense I mentioned is geodesic incompleteness.

 

[timmdeeg]

No, because geodesic incompleteness and the presence of singularities in the sense I mentioned are equivalent, since the definition of "singularity" in the sense I mentioned is geodesic incompleteness.

Ah, yes. So the timelike geodesic of a particle which falls into such a (real) black hole is complete, because it remains a part of spactime. Kindly correct if wrong.

 

[PeterDonis]

the timelike geodesic of a particle which falls into such a (real) black hole is complete, because it remains a part of spactime.

"Remains a part of spacetime" is not the definition of geodesic completeness. An incomplete geodesic is still "part of spacetime".

The definition of geodesic completeness is that the geodesic can be extended to any value of its affine parameter. In the case of a timelike geodesic, that means that every possible value of proper time, from $- \infty$ to $+ \infty$, corresponds to some point on the geodesic. An incomplete geodesic is one for which that is not true--for example, a timelike geodesic that falls into an idealized black hole with a singularity at the center, for which values of proper time greater than or equal to some finite number do not have any corresponding point on the geodesic.

 

[timmdeeg]

"Remains a part of spacetime" is not the definition of geodesic completeness. An incomplete geodesic is still "part of spacetime".

Saying "Remains a part of spacetime" I meant the particle, not the geodesic it describes. Would this make sense?

EDIT after overthinking I doubt it. The definition for geodesic completeness you gave is quite clear and particles are not included.

 

[PeterDonis]

after overthinking I doubt it.

You doubt correctly.

 

[Chronos]

Modern arguments over the existence of black holes generally center on whether or not a singularity is necessary. Most theoreticians question the existence of singularities, while perfectly willing to concede the existence of event horizons. The math behind an event horizon without a singularity is not pretty, nor is that of a singularity. A change in the way we think about the fundamentals of geometry is probably necessary to resolve both issues.

 

[PeterDonis]

Most theoreticians question the existence of singularities, while perfectly willing to concede the existence of event horizons.

I would not put it quite that way. I think most theoreticians are perfectly willing to concede the existence of apparent horizons, i.e., surfaces where radially outgoing light, locally, does not move outward. But I don't think most theoreticians are willing to concede that the presence of apparent horizons necessarily entails the presence of event horizons. Many of the hypotheses about how quantum gravity effects change things end up amounting to: quantum gravity effects prevent the formation of actual event horizons, while still allowing the formation of apparent horizons.

Since, experimentally, the only way to tell the difference between an apparent horizon and an event horizon is to wait a very, very long time (on the order of the Hawking evaporation time, i.e., about $10^{67}$ years for a solar mass black hole, and the time goes up as the cube of the mass), practically speaking there is no way to know if the various black hole candidates we observe really have event horizons, or only apparent horizons.

 

[PeterDonis]

The math behind an event horizon without a singularity is not pretty

That's quite true, and it's one of the reasons why many theoreticians appear to be hoping that, when quantum effects are taken into account, all of those horizons will turn out to only be apparent horizons, not event horizons. The difficulties you refer to all go away if the horizons are only apparent horizons.

 

[zonde]

There is chain of theoretical arguments that lead to black hole model of GR with some assumptions here and there. There can be flaws at different stages of the argument.
The place where I see the flaw in that chain of arguments is treating Pauli exclusion principle (degeneracy "pressure") as a source of additional force.

 

[PeterDonis]

The place where I see the flaw in that chain of arguments is treating Pauli exclusion principle (degeneracy "pressure") as a source of additional force.

Why do you think this is a flaw? Do you have any references to back up this claim?

 

[zonde]

Why do you think this is a flaw? Do you have any references to back up this claim?

I don't have ready textbook references so I went over couple of threads about PEP not being a force.
From this thread Why is the Pauli Exclusion Principle not a force? posts #6 by tom.stoer, #20 by Vanadium 50 and #22 by vanhees71 seem relevant.
Vanadium 50 post seems to support the idea that PEP produces a force. But "particle in a box" example he is using generally refers to square potential well and not a physical box. So it does not really make sense to say that there is pressure and force against the "walls" of potential well.
In the thread Pauli exclusion principle: a Force or not? in post #4 tom.stoer makes the argument why PEP is not a force.
Based on the arguments made in these threads I could try to find some textbook reference if necessary.

Another side of my claim that might require some backing up is where the PEP as a force comes up in BH model. Popularizations are generally relying on PEP as a force in their explanations but that might be different from original rigorous argument. I tried to look through original papers but the argument is built on considerable tree of references so it is hard to get to the bottom of it.

 

[PeterDonis]

In the thread Pauli exclusion principle: a Force or not? in post #4 tom.stoer makes the argument why PEP is not a force.

No, he doesn't, at least not the way you are using the term "force".

The question you are asking is whether degeneracy pressure (i.e., pressure that is due to the Pauli exclusion principle rather than to kinetics) makes an additional contribution ("additional" meaning "in addition to kinetic pressure") to the pressure that determines the possible equilibrium states of a static object like a white dwarf or neutron star. The answer to that question is yes. We know this because we have done detailed numerical models that only match observed data if degeneracy pressure is included as an additional contribution to kinetic pressure.

The question that was being asked in the threads you linked to was whether the Pauli exclusion principle is a "force" in the same sense that, say, electromagnetism is a "force", i.e., whether the PEP is modeled as an "interaction" the same way EM is. The answer to that question is no. But that's a different question from the one being asked in this thread.

Another side of my claim that might require some backing up is where the PEP as a force comes up in BH model.

It shows up in the equations that determine the maximum possible mass of a white dwarf or a neutron star. Observations indicate that there is indeed a maximum possible mass for both types of objects; for white dwarfs the observed maximum matches the theoretical one pretty closely, for neutron stars I believe there is still about a factor of 2 uncertainty, but the existence of a maximum possible mass (somewhere between about 1.5 and 3 solar masses) is not disputed.

It should be noted that the contribution of the PEP to pressure for these objects is not "additional" in any practical sense: the degeneracy pressure due to the PEP is the only significant pressure that is counterbalancing gravity in these objects. So if PEP degeneracy pressure did not make a separate contribution from kinetic pressure, these objects would not exist; they would have collapsed immediately to black holes. In other words, the presence of PEP degeneracy pressure makes it harder to form a black hole, not easier.

 

[zonde]

No, he doesn't, at least not the way you are using the term "force".

I am trying to avoid nonstandard meaning of the term "force". So no, I hold on to the statement that "PEP is not a force" and instead I try to adjust my own understanding of the white dwarf and neutron star model with different terms and concepts.

The question you are asking is whether degeneracy pressure (i.e., pressure that is due to the Pauli exclusion principle rather than to kinetics) makes an additional contribution ("additional" meaning "in addition to kinetic pressure") to the pressure that determines the possible equilibrium states of a static object like a white dwarf or neutron star. The answer to that question is yes. We know this because we have done detailed numerical models that only match observed data if degeneracy pressure is included as an additional contribution to kinetic pressure.

Redefining "pressure" as kinetic energy per volume instead of force per area we can say that PEP determines minimum "kinetic pressure" for population of identical fermions within potential well. This seems uncontroversial and valid statement.

 

[PeterDonis]

Redefining "pressure" as kinetic energy per volume instead of force per area

This only works for kinetic pressure, i.e., pressure due to temperature. It doesn't work for degeneracy pressure, which is present at zero temperature and does not depend on temperature.

It is true that adding fermions to a population of identical fermions in a potential well requires adding energy to the system, but I don't think this energy can be usefully viewed as "kinetic" energy.

 

[zonde]

It is true that adding fermions to a population of identical fermions in a potential well requires adding energy to the system, but I don't think this energy can be usefully viewed as "kinetic" energy.

Actually adding a fermion does not require energy. You see the particle added has to come from outside of potential well so it has at least the level of energy required to be outside of potential well. So when you add fermion to population of identical fermions in a potential well we simply can't remove as much energy as we would be able to remove if it would be to only fermion in potential well.

 

[PeterDonis]

when you add fermion to population of identical fermions in a potential well we simply can't remove as much energy as we would be able to remove if it would be to only fermion in potential well.

This is a reasonable restatement, but it doesn't change the main point of what I said, the the energy in question is not usefully viewed as "kinetic" energy.

 

[zonde]

the the energy in question is not usefully viewed as "kinetic" energy.

But how do you arrived at that conclusion? Because it does not seem reasonable to me. Say you have five excited electrons in potential well. So all of them have some energy. One of them can fall to lowest energy level and give up it's energy. This energy should be considered as "kinetic energy". But other four electrons can't fall to lowest energy level any more so they should have some of their energy considered as 'kinetic energy" and some as other type of energy just because they were not the ones that have fallen to lowest energy level. And if no electron falls to lowest energy level? How do we split between "kinetic energy" and the other energy? This does not make sense to me.

 

[PeterDonis]

Say you have five excited electrons in potential well. So all of them have some energy. One of them can fall to lowest energy level and give up it's energy. This energy should be considered as "kinetic energy".

Why? If your answer involves the usual meaning of the word "fall", which you used, please think again.

 

[PeterDonis]

How do we split between "kinetic energy" and the other energy?

You can't. That's my point. None of the energy you are talking about is usefully thought of as "kinetic" energy, because it has nothing to do with the temperature of the electrons--they can all be at absolute zero and it would still be true that only one (or two if we consider their spin) can occupy the lowest energy level. So whatever it is that is keeping the other electrons from occupying the lowest energy level, it has nothing to do with "motion" of the electrons (they're all at absolute zero) and nothing to do with "kinetic" anything (the temperature is absolute zero).

 

[zonde]

You can't. That's my point. None of the energy you are talking about is usefully thought of as "kinetic" energy, because it has nothing to do with the temperature of the electrons--they can all be at absolute zero and it would still be true that only one (or two if we consider their spin) can occupy the lowest energy level. So whatever it is that is keeping the other electrons from occupying the lowest energy level, it has nothing to do with "motion" of the electrons (they're all at absolute zero) and nothing to do with "kinetic" anything (the temperature is absolute zero).

But zero temperature being equivalent to zero kinetic energy works only in classical physics and not when you bring QM into the picture. In QM even at zero temperature particles have non-zero momentum.