Black hole creation process

  • Thread starter bcrelling
  • Start date
  • #1
69
2
We have a large mass, and we increase it slowly- dropping in one atom at a time. Will a black hole form suddenly, or will it gradually become blacker with the addition if each atom?

I assume that a mass marginally below the threshold must at least partially have the properties of a black hole?

Thanks
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,966
327
Creation of a black hole is quite catastrophic. It is not a gradual process. Degeneracy pressure can hold up the mass until it can't, at which point the mass basically crumples and catastrophically collapses into a black hole. Before the mass collapses, it can be quite dense, like a neutron star, but it is not dense enough to capture light and so is not "black", but it will become dense enough to redshift light emerging from the surface, so it does get redder.
 
  • #3
Doug Huffman
Gold Member
805
111
Your 'marginally' embraces the Sorites paradox, less one is it still marginally, less two, et cetera. I believe that your hypothetical would cause a catastrophe.

LOL quite a coincidence of both using catastrophe.
 
  • #4
38,063
15,845
We have a large mass, and we increase it slowly- dropping in one atom at a time. Will a black hole form suddenly, or will it gradually become blacker with the addition if each atom?

As Matterwave said, the process of forming a black hole is not gradual; there are abrupt transitions in the state of the matter. If we assume that no nuclear, chemical, or other reactions occur, and there are no thermal fluctuations (for example, we drop iron atoms at absolute zero onto a mass composed, at least at low masses, of other iron atoms at absolute zero), there are still at least three abrupt transitions we will observe, and a fourth that also qualifies as a transition even if it isn't necessarily abrupt:

(0) The transition from an object whose mass is too low for its own self-gravity to be significant, like a rock, to an object whose mass is large enough for self-gravity to be a significant factor in its structure, like a planet. This transition may not be abrupt; there are examples in nature of objects at various points along this spectrum (from rocks to asteroids to "dwarf planets" to planets).

(1) The transition from normal matter, made of atoms and held up against gravity by ordinary inter-atomic forces, to white dwarf matter, made up of electrons and nuclei not organized into atoms and held up against gravity by electron degeneracy pressure. This is an abrupt transition; there is no stable sequence of intermediate states in between ordinary matter and white dwarfs.

(2) The transition from white dwarf matter to neutron star matter, which is made of neutrons (electrons and protons are collapsed into neutrons during the process of neutron star formation) and held up against gravity by neutron degeneracy pressure. This is also an abrupt transition.

(3) The transition from neutron star matter to a black hole, when the mass of a neutron star exceeds the maximum possible mass at which the star can hold itself up against gravity. The basic reason there is such a maximum mass is that, as the mass increases, the neutrons get squeezed closer together and become relativistic, and relativistically degenerate matter has a lower adiabatic index (the exponent in the equation of state relating pressure and density) than non-relativistically degenerate matter. So as the density continues to go up as the mass increases, the pressure can no longer increase fast enough to keep up, and a point is reached where the star is no longer stable and collapses. This is an abrupt transition.

I assume that a mass marginally below the threshold must at least partially have the properties of a black hole?

A neutron star near the maximum possible mass (and therefore the minimum possible size) will have some gravitational redshift, but it will be fairly small by black hole standards; there is a large gap between the properties of such an object and the properties of a black hole with a mass just above the maximum possible neutron star mass.
 
  • Like
Likes John M. Carr and bcrelling
  • #5
tionis
Gold Member
458
67
One thing I don't understand is: Is it the imploding mass that causes a black hole to form, or is it the huge spacetime curvature formed around the imploding mass that causes the BH to form? In other words, if there were no spacetime for the mass to curve, would a BH actually ever form?
 
  • #6
38,063
15,845
if there were no spacetime for the mass to curve

Then we wouldn't be discussing General Relativity, we'd be discussing some other theory. Do you have one? If not, your question doesn't really have any point, because if we don't have a theory to constrain our speculations, we can say anything we want.
 
  • #7
bcrowell
Staff Emeritus
Science Advisor
Insights Author
Gold Member
6,723
426
The OP asked about a hypothetical process, but the replies talk about gravitational collapse of a star. Gravitational collapse of a star doesn't have to be the way a black hole forms. For example, there could be primordial black holes that were formed shortly after the big bang, long before stars formed.

We have a large mass, and we increase it slowly- dropping in one atom at a time. Will a black hole form suddenly, or will it gradually become blacker with the addition if each atom?

I assume that a mass marginally below the threshold must at least partially have the properties of a black hole?

There is no threshold, no minimum mass for a black hole. General relativity allows black holes to exist with any mass whatsoever. However, the known pathways to formation of a black hole from a dying star are pathways that only work if the star is fairly massive.

Hypothetically, one could have a black hole of any tiny size, and then any matter that you trickled into it would simply increase its mass from there.
 
  • #8
tionis
Gold Member
458
67
Then we wouldn't be discussing General Relativity, we'd be discussing some other theory. Do you have one? If not, your question doesn't really have any point, because if we don't have a theory to constrain our speculations, we can say anything we want.

The first part of my question is within the context of GR:

One thing I don't understand is: Is it the imploding mass that causes a black hole to form, or is it the huge spacetime curvature formed around the imploding mass that causes the BH to form?

The second part:

In other words, if there were no spacetime for the mass to curve, would a BH actually ever form?


.. is speculation on my part based on incomplete or misunderstood knowledge of the first part of the question, which is why I asked in the first place. Do you have an answer for the first part of my question?
 
  • #9
38,063
15,845
Do you have an answer for the first part of my question?

The black hole is the spacetime curvature that is formed by the imploding mass.
 
  • #10
PAllen
Science Advisor
8,861
2,067
To add to bcrowell's point, there is no minimum size to the Oppenheimer-Snyder collapse. In lay terms, that means if you posit matter that is incapable of pressure, then any amount of it will eventually (smoothly) collapse to a BH. General relativity does not include any theory of matter. Classical use of GR typically assumes that matter must locally behave as expected by SR (e.g. the dominant energy condition). However, this places no requirement on pressure, so the perfectly smooth (non-catastrophic) collapse of 1 gram of pressure-less mathematical dust to a BH is allowed in classical GR.

Peter's comments apply to matter as we know it, governed by equations of state based on the standard model of particle physics.

[edit: The closest real world analog I can think of the pressure-less dust is iron filings. Iron is incapable of either fission or fusion. If you add iron to an iron ball slowly enough to dissipate heat (which toward the end will be enormous ), the first catastrophe will be collapse from ordinary matter to neutron star with release of a flood of neutrinos. This would constitute some weak form of supernova (which obviously has never been observed). Then, as you kept adding iron, at somewhere above 3 solar masses, you would get catastrophic collapse to a BH. These are best guesses, as there is no process in the universe that slowly accumulates iron.]
 
Last edited:
  • Like
Likes martinbn and bcrowell
  • #11
993
48
The question was about dropping in one atom at a time... and the nature of the beginning of the thing becoming a BH.

It seems to me that the OP might be wondering:

Will the transition to BH happen after the addition of one particular atom in the series of atoms added?
(assuming that each dropping of an atom is spaced apart in time sufficient for its effect to be fully made manifest.)

Or better yet, let the things being dropped in be protons or hydrogen atoms... so the increment is small.

I think the OP is wondering if the pre-BH object is subject to HUP fluctuations and wondering how the addition of a single increment compares to the magnitude of these fluctuations... whether a fluctuation might be sufficient to put the object over the critical mass to become a BH when the BH is in the state where the addition of one more increment of dropped matter would otherwise do so...

Since the size at which an incrementally built BH is known, what orders of incremental mass compare to the fluctuation magnitude?

Could an object that is just short of becoming a BH make the transition through a HUP fluctuation?

Also, as to micro BHs, I don't know if it is thought that there are micro-versions of the above - micro objects just short of the density to become a BH. If so, would their HUP fluctuations with respect to their size be bigger than the large BHs? If so, it seems they would very subject to making the transition, and so quite rare in their pre-BH state.
 
  • #12
PAllen
Science Advisor
8,861
2,067
The question was about dropping in one atom at a time... and the nature of the beginning of the thing becoming a BH.

It seems to me that the OP might be wondering:

Will the transition to BH happen after the addition of one particular atom in the series of atoms added?
(assuming that each dropping of an atom is spaced apart in time sufficient for its effect to be fully made manifest.)

Or better yet, let the things being dropped in be protons or hydrogen atoms... so the increment is small.

I think the OP is wondering if the pre-BH object is subject to HUP fluctuations and wondering how the addition of a single increment compares to the magnitude of these fluctuations... whether a fluctuation might be sufficient to put the object over the critical mass to become a BH when the BH is in the state where the addition of one more increment of dropped matter would otherwise do so...

Since the size at which an incrementally built BH is known, what orders of incremental mass compare to the fluctuation magnitude?

Could an object that is just short of becoming a BH make the transition through a HUP fluctuation?

Also, as to micro BHs, I don't know if it is thought that there are micro-versions of the above - micro objects just short of the density to become a BH. If so, would their HUP fluctuations with respect to their size be bigger than the large BHs? If so, it seems they would very subject to making the transition, and so quite rare in their pre-BH state.
I would say much of this is unknown. Even the minimum mass of BH you would get by incrementally growing a neutron star is unknown (3 to 5 solar masses is an estimate, but that is an error bar of a whole sun). Qualititatively, all agree (including my hypothetical adding iron to a neutron star), that the final collapse to a BH would be sudden.
 
  • #13
993
48
By "sudden", does that mean at the speed of sound through the material of the object, or more like the speed of light around the surface... or does the geometry of the BH make the usual space and time measures confounded by the variations of an observer's position and motion? Maybe the transition must always appear "instantaneous"?
 
  • #14
PAllen
Science Advisor
8,861
2,067
By "sudden", does that mean at the speed of sound through the material of the object, or more like the speed of light around the surface... or does the geometry of the BH make the usual space and time measures confounded by the variations of an observer's position and motion? Maybe the transition must always appear "instantaneous"?
I don't know of any real analysis of a model of incremental BH formation from a neutron star (it would be a lot of work, and since it isn't known to occur, maybe no one has been motivated). Or maybe I've missed it in the literature. So I really can't answer the 'how fast' question with any confidence. It may not make much difference because the speed of sound would be a significant fraction of the speed of light for a neutron star (e.g. > .3 c).

On the other hand, viewed from afar, the process would be slower, but not really that slow - the object would become blacker than CMB filled empty space relatively fast, but much slower than as experienced by an particle of the neutron star.
 
  • #15
tionis
Gold Member
458
67
so, are there any spacetime-free solutions for mass/energy?
And if there are, what woud it say about an imploding mass?
Would we get to observe a mass collapse without becoming a black hole?
 
  • #16
martinbn
Science Advisor
2,764
1,124
Just one more comment. If the characteristic feature of a black hole is the event horizon, then you can get a black hole without anything unusual happening, at least for some time. Take a spherical configuration of stars that fall towards the centre. If you choose the masses well, an event horizon will form at the centre and grow outward i.e. you have a black hole. If you are somewhere there you will not see anything unusual. In fact a horizon may be growing and passing through you room right now.
 
  • #17
PAllen
Science Advisor
8,861
2,067
so, are there any spacetime-free solutions for mass/energy?
And if there are, what woud it say about an imploding mass?
Would we get to observe a mass collapse without becoming a black hole?
What do you mean by spacetime free solutions? Since SR and GR are built on spacetime, that seems to be asking what theory would be true if the currently best verified theories were wrong? There is no meaningful way to answer that. It is not like picking matter without electric charge, which is possible. There is just no theory left if you remove spacetime.
 
  • #18
PAllen
Science Advisor
8,861
2,067
Just one more comment. If the characteristic feature of a black hole is the event horizon, then you can get a black hole without anything unusual happening, at least for some time. Take a spherical configuration of stars that fall towards the centre. If you choose the masses well, an event horizon will form at the centre and grow outward i.e. you have a black hole. If you are somewhere there you will not see anything unusual. In fact a horizon may be growing and passing through you room right now.
Yes, I've used this example many times (stars as dust; implements hypothetical smooth Oppenheimer-Snyder type collapse), as a way of talking about what you would see throughout a collapsing volume, and also to discuss how radical a change to classical GR is implied by 'active horizon' hypotheses (e.g. firewalls).

[edit: similarly, if allowed to pose implausible initial conditions, if you had a sufficiently large sparse dust cloud, with a uniform density of e.g. air, but of a size just short its being in its own SC radius, then it would smoothly collapse further, with growth of event horizon from center, without any happening to the dust until well after it was already inside its event horizon.]
 
Last edited:
  • Like
Likes martinbn
  • #19
38,063
15,845
If you add iron to an iron ball slowly enough to dissipate heat (which toward the end will be enormous ), the first catastrophe will be collapse from ordinary matter to neutron star with release of a flood of neutrinos.

Wouldn't you pass through a white dwarf stage first? At that stage, the iron nuclei would still be separate entities, but the electrons would be degenerate. At the neutron star stage (the next stage), there would no longer be iron nuclei or electrons, just a big blob of neutrons. Ordinary iron ball -> white dwarf -> neutron star is the sequence according to the Harrison-Wakano-Wheeler equation of state (which is basically what we're talking about here--the discussion in Thorne's Black Holes and Time Warps is what I've been basing my comments on).
 
  • #20
PAllen
Science Advisor
8,861
2,067
Wouldn't you pass through a white dwarf stage first? At that stage, the iron nuclei would still be separate entities, but the electrons would be degenerate. At the neutron star stage (the next stage), there would no longer be iron nuclei or electrons, just a big blob of neutrons. Ordinary iron ball -> white dwarf -> neutron star is the sequence according to the Harrison-Wakano-Wheeler equation of state (which is basically what we're talking about here--the discussion in Thorne's Black Holes and Time Warps is what I've been basing my comments on).
More like a black dwarf, with different composition than any that form naturally (none are iron). But, yes, I forgot the stage of electron degeneracy. The only stage that would release much energy is the neutron star formation, when an enormous surge of neutrinos are released.
 
  • #21
tionis
Gold Member
458
67
What do you mean by spacetime free solutions? Since SR and GR are built on spacetime, that seems to be asking what theory would be true if the currently best verified theories were wrong? There is no meaningful way to answer that. It is not like picking matter without electric charge, which is possible. There is just no theory left if you remove spacetime.

Since spacetime couples to mass, I thought you could get rid of the spacetime part, just like you can have a matter-free spacetime and still have black holes form. I always thought black holes formed because some intrinsic property of matter, but now it's dawning on me that mass got nothing to do with it, that spacetime is actually responsible for it. For all we know, the collapsing mass is still there shrouded behind the horizon of spacetime. So the vacuum solutions that go to infinity and all that are for the collapsing spacetime. I think I finally got it. The star is still there, it's just that the spacetime around it is behaving as though the mass has gone to
infty.latex.gif
 
  • #22
35,976
12,840
Mass is always somewhere in spacetime. Spacetime is the framework for everything.
The star is still there
No, whatever remains has nothing to do with a star.
it is behaving as though the mass has gone to
proxy.php?image=http%3A%2F%2Fw2.syronex.com%2Fjmr%2Ftex%2Fimg%2Finfty.latex.gif
No, the mass stays the same.
 
  • #23
tionis
Gold Member
458
67
mfb -- yes, the hole's mass is the same as that of the collapsing object, but why does spacetime behaves as if the mass is infinite?
 
  • #24
PAllen
Science Advisor
8,861
2,067
mfb -- yes, the hole's mass is the same as that of the collapsing object, but why does spacetime behaves as if the mass is infinite?
It doesn't. It behaves as if the mass is M, the mass while the star was 'normal'. Inside the horizon, you have a curvature singularity, but this does not represent mass.
 
  • #25
38,063
15,845
More like a black dwarf, with different composition than any that form naturally (none are iron).

Yes, good point. This also affects the maximum mass limit; for a body composed of iron nuclei, it's around 1.2 solar masses, but if it's mostly hydrogen and helium nuclei, it's 1.44 solar masses (this is the limit that Chadrasekhar derived in the 1930's).
 
  • #26
749
41
mfb -- yes, the hole's mass is the same as that of the collapsing object, but why does spacetime behaves as if the mass is infinite?

On the outside of the event horizon (r>2M), spacetime is time-like according to Schwarzschild metric, this means that time (t) is temporal. Inside the event horizon (r>2M) spacetime is space-like which means space (r) is temporal, this means there is no stable radius, matter couldn't support itself at a constant r, regardless of pressure, within space-like spacetime and all matter collapses towards r=0 which is where the notion of the singularity and 'infinite' density comes from.

We have a large mass, and we increase it slowly- dropping in one atom at a time. Will a black hole form suddenly, or will it gradually become blacker with the addition if each atom?

I assume that a mass marginally below the threshold must at least partially have the properties of a black hole?

Thanks

To add to the other posts, according to the Schwarzschild interior metric, the point of no return is 9/4M (or 2.25M), this is the point when the time dilation at the centre of mass reaches zero and as the star collapsed further, the event horizon would begin to move outwards towards the surface as in this diagram (the pink line represents the event horizon, the blue lines the collapsing star).
 
  • #27
38,063
15,845
On the outside of the event horizon (r>2M), spacetime is time-like according to Schwarzschild metric, this means that time (t) is temporal. Inside the event horizon (r>2M) spacetime is space-like which means space (r) is temporal

You are not stating this correctly. Spacetime is not "timelike" or "spacelike"; there are always timelike, spacelike, and null vectors everywhere in spacetime. Coordinates like ##t## and ##r## can be timelike or spacelike (or null), but whether or not they are depends on the coordinate chart you choose. There are charts covering Schwarzschild spacetime in which the "time" ##t## is timelike everywhere, and there are charts in which ##r## is spacelike everywhere.

The correct way to state what you are trying to say here is this: Schwarzschild spacetime contains a set of curves along which the spacetime geometry is constant. In standard Schwarzschild coordinates, these are curves of constant ##r##, ##\theta##, ##\phi##. A more technical way of describing these curves is that they are integral curves of a Killing vector field; in standard Schwarzschild coordinates, this vector field is just the coordinate basis vector field ##\partial / \partial t## (technically this doesn't work at the horizon in these coordinates, because they are singular there, but there are ways to finesse that).

The key fact about this family of curves is this: outside the horizon, they are timelike; inside the horizon, they are spacelike; and on the horizon itself, they are null. This is an invariant fact about the spacetime geometry, independent of coordinates. It's unfortunate that many sources aren't careful enough about how they state this.

this means there is no stable radius, matter couldn't support itself at a constant r, regardless of pressure, within space-like spacetime and all matter collapses towards r=0 which is where the notion of the singularity and 'infinite' density comes from.

Just to clarify based on my comments above, since a curve of constant ##r## is spacelike inside the horizon, and since no particle of matter can travel along a spacelike worldline, it is impossible for matter to be static at constant ##r## inside the horizon. This in itself, however, does not tell you that the matter must be collapsing towards ##r = 0##; to show that, you need to show that all timelike curves inside the horizon have decreasing ##r## (which is easy to show).
 
  • #28
Ken G
Gold Member
4,462
333
I actually don't think the formation of a neutron star or a black hole involves a catastrophe in the force balance, though it is popular to explain it that way. To me, a catastrophe in the force balance would require that there be no hydrostatic solution, say via a bifurcation, even when we treat the system as not undergoing any thermal changes, i.e., not losing heat to its surroundings and not undergoing internal changes in composition. Creation of a neutron star is often characterized as what you get when you have a white dwarf at zero temperature whose mass exceeds the Chandrasekhar mass-- there is indeed no hydrostatic solution for such an object, so we might characterize its collapse as a catastrophe. However, this is an artifact of the questionable assumptions, because there is no reason to assume you have an object that maintains zero temperature in a real core-collapse scenario. Instead, you have a mass, which can exceed the Chandrasekhar mass, and you have an internal energy density, which because of the history of the object can certainly yield a nonzero temperature. That object can have a hydrostatic solution if the energy density is high enough, and indeed it seems to me it normally will be, because objects like this are generally created by dropping in mass that either already has a high energy content, or falls in from a far enough distance to acquire a high energy content. So I think it would be pretty unexpected to ever arrive at a state that does not have a hydrostatic solution, such that it could be said to collapse as a true catastrophe.

If that is indeed true (and an alternative possibility is that the general relativity of the situation leads to unstable orbits such that hydrostatic equilibrium is indeed impossible), then the core collapse could involve an object that has a hydrostatic solution that is possible, for its given energy content, if that object could just be left alone long enough to find that solution. But the object is not left alone, it is undergoing change that comes either from loss of heat via neutrino losses (perhaps the URCA process), or from photodisintegration of its iron that is changing its internal composition. When the gas is highly relativistic, even if it does have a hydrostatic solution if none of these changes were occurring, the rate at which the hydrostatic structure would need to reconfigure to adapt to these changes is strongly leveraged by the fact that relativistic gas is very soft-- by which I mean simply that large reconfigurations are necessary given even small internal changes in composition or heat content. At some point, these reconfigurations are so leveraged that they take much longer than just the sound crossing time, and meanwhile the thermal timescale is getting shorter and shorter because it could in principle get as short as the light crossing time. When the thermal timescale gets shorter than the force-balance timescale, you get free fall, but not because there exists no instantaneous hydrostatic solution, it is because the hydrostatic solution takes too long to achieve. So the catastrophe is a thermal instability, not a catastrophe in the force balance. The force balance contributes to the problem only by how soft it is-- small thermal changes lead to large reconfigurations which drive further thermal changes, until the thermal physics runs away. Take away the thermal changes and the force balance might have no problem finding a continuously changing configuration as you add mass particle by particle, you would just drive a steady and continuous transition to a neutron star that could be made as slow as you like by adding particles as slowly as you like. The thermal runaway precludes that.

So what I'm saying is, I suspect that if you added mass gradually to a white dwarf, then before you even quite get to the Chandra mass you will start to see the thermal adjustment timescale becoming comparable to the force-balance establishment timescale, and when the two are about equal, you will lose control of the process-- it will take on a life of its own that no longer cares how fast you are adding particles, and will lead to a collapse, not because there is no force balance (as is true above the Chandra mass), but because there is a thermal instability that operates faster than the forces can equilibrate. That's not a formal catastrophe in the sense of a bifurcation in which the force balance is lost, but it is a catastrophe in the sense that it is a thermal runaway.

Of course, in real stars, we don't know how fast the mass is being added-- it might be added abruptly, like in a white-dwarf merger. But in that case, as I said above, there's still no reason to assume the temperature will be zero, so there could still be a force balance that is possible at the higher mass. You'll still have to wait for the heat to get out, so you will still be controlled by the thermal timescale, and the catastrophe will still happen when the thermal timescale gets short enough to rival the force-balance timescale. So I think that will be true whether you add mass slowly or abruptly-- the core collapse is a thermal instability that plays out in situations where there generally would be a hydrostatic solution, but it simply does not have time to set up given the rapidly changing thermal environment. The criterion for the catastrophe is equating the thermal adjustment rate to the relativistically softened force-balance rate, which happens before you get to the Chandra mass and does not matter how fast you add the particles once you get to that critical limit.
 
  • #29
38,063
15,845
That object can have a hydrostatic solution if the energy density is high enough

Of course this is true--that's what stars are, and a star can be in hydrostatic equilibrium with a mass much larger than the Chandrasekhar limit (or its analogue for neutron stars). But a star has an internal energy generation mechanism that can balance its rate of heat loss to space so that its temperature remains constant and its structure remains static.

A white dwarf with mass above the Chandrasekhar limit has no such mechanism. It can gradually contract and generate heat from the contraction, but this is not a hydrostatic equilibrium. In principle it could still take a long time to contract to the point where a serious instability set in; but since it can't maintain a static structure while losing energy to space (which it has to since space is effectively at zero temperature), the way a star can, it will inevitably reach a point where it is unstable at its current temperature (which may be close to zero temperature) and collapses.

Of course a white dwarf above the limit could collapse to a neutron star instead of a black hole; but the same logic applies to neutron stars, since there is also a maximum mass limit for those. Any object above that maximum mass limit will eventually collapse to a black hole, though it might take time if the object stars out at high enough temperature for kinetic pressure to be significant in determining its structure (and if its mass is not so large that even the extra kinetic pressure doesn't help).

objects like this are generally created by dropping in mass that either already has a high energy content, or falls in from a far enough distance to acquire a high energy content

I don't think this is really true; supernova explosions are currently thought to be a major source (if not the main source) of stellar mass black holes, and those don't require any external object to fall into the star that goes supernova.

an alternative possibility is that the general relativity of the situation leads to unstable orbits such that hydrostatic equilibrium is indeed impossible

GR does impose such a limit: it says that a spherically symmetric object with a radius less than 9/8 of the Schwarzschild radius for its mass cannot be in hydrostatic equilibrium. AFAIK no white dwarf comes anywhere near this limit, but some neutron stars come fairly close to it. Note that this result does not depend on the temperature of the object.

the core collapse could involve an object that has a hydrostatic solution that is possible, for its given energy content, if that object could just be left alone long enough to find that solution

I agree that this is possible, and that something like a supernova explosion could cause an object to collapse into a black hole even though, strictly speaking, the mass left behind would have enabled a hydrostatic equilibrium as a neutron star (or even possibly a white dwarf, though I think that would be unlikely since a star small enough to reach that mass after the explosion would not go supernova in the first place), if the process had happened slowly.

How common this is is a different question, which we can't really assess in too much detail at this point, since we don't know the exact maximum mass limit for neutron stars, we only have a range (IIRC about 1.5 to 3 solar masses), and we don't have very good mass values for many black hole candidates, so we can't assess what percentage of actual black holes have masses that are below the neutron star maximum mass limit.
 
  • #30
PAllen
Science Advisor
8,861
2,067
Those who do numerical simulations of these things claim that collapse to a neutron star has matter going an appreciable fraction of the speed of light, and that this occurs quickly. For a BH, matter crosses a near horizon 'observer' at near c. In the case of a neutron star, it is debatable 'how catastrophic' it is, because for a neutron star, speed of sound is over .3c, so the collapse is not 'dust like'. However for BH formation, different bits of matter are essentially independent of each other because mutual influence cannot propagate fast enough.
 
  • #31
Ken G
Gold Member
4,462
333
Of course this is true--that's what stars are, and a star can be in hydrostatic equilibrium with a mass much larger than the Chandrasekhar limit (or its analogue for neutron stars). But a star has an internal energy generation mechanism that can balance its rate of heat loss to space so that its temperature remains constant and its structure remains static.
That just says it has a history that has led to its internal energy, which is a crucial point that for some reason often gets overlooked when people use language that suggests this history, and the associated energy density, just vanish once the energy generation turns off. Of course that's not correct, the energy is still there, and so is the hydrostatic equilibrium. So the issue is always about tracking what happens to the heat-- nothing would ever happen if heat were not being lost, so the issue is, how fast is it being lost, and can force balance be maintained on that timescale. So it's always about the thermal timescale, not the absence of a possible force balance.
A white dwarf with mass above the Chandrasekhar limit has no such mechanism.
Not so, this is only if you assume the white dwarf is at zero temperature, which is the common assumption. That requires that the white dwarf has a substantial history of cooling, and that's exactly what it will not have if you are adding mass to it-- you have to track the timescales for cooling, compared to the timescale of the added mass. Sure, if you add mass slowly enough that the white dwarf always cools to zero temperature, then the Chandra mass is relevant-- but that's also the case that will not lead to any core collapse, just a gradual contraction in hydrostatic equilibrium right up to the formation of a neutron star. So what actually causes the collapse is the thermal instability that kicks in when you get close to the Chandra mass, and that doesn't care how slowly you add mass because it is an instability, not a bifurcation. The term "catastrophe" is a bit ambiguous about that distinction.
It can gradually contract and generate heat from the contraction, but this is not a hydrostatic equilibrium.
I don't know what you mean by that, it is perfectly standard for a protostar to be doing that, and be in hydrostatic equilibrium (it's just a quasi-steady equilibrium, but so are all equilibria, there is no such thing as any other type.)
In principle it could still take a long time to contract to the point where a serious instability set in; but since it can't maintain a static structure while losing energy to space (which it has to since space is effectively at zero temperature), the way a star can, it will inevitably reach a point where it is unstable at its current temperature (which may be close to zero temperature) and collapses.
What I'm saying is that there would never be any collapse if the temperature really stayed near zero. In other words, if we use the usual white dwarf idealization, which is a zero temperature, we are saying that the process is happening slowly enough for the heat to keep leaking out and maintaining zero temperature-- and you'd never get a core collapse in that case, so that cannot be the correct description, even though it is formally the meaning of a Chandrasekhar mass. This doesn't mean the Chandra mass is not relevant, or that Chandrasekhar was wrong, it just means that the Chandra mass is simply a benchmark-- the way it is derived cannot be taken as the physical process that actually produces core collapse, that would miss what is actually happening there.
I don't think this is really true; supernova explosions are currently thought to be a major source (if not the main source) of stellar mass black holes, and those don't require any external object to fall into the star that goes supernova.
That's why I also mentioned that you could have mass falling into the core that already has a high internal energy, as when you build an iron degenerate core that is built from the high temperature of shell fusion of silicon or some such thing. That it is at such a high temperature, and not zero temperature, is the reason it has access to thermal instabilities like photodisintegration and the Urca process. Again the key point is that the actual physics of core collapse must be a thermal instability, not a hydrodynamic instability, because the energy is there to create hydrostatic equilibrium-- the zero temperature assumption is not formally correct, it only works to get the benchmark mass where the thermal instability is going to rear up.
GR does impose such a limit: it says that a spherically symmetric object with a radius less than 9/8 of the Schwarzschild radius for its mass cannot be in hydrostatic equilibrium. AFAIK no white dwarf comes anywhere near this limit, but some neutron stars come fairly close to it. Note that this result does not depend on the temperature of the object.
Yes, I agree that GR could throw a monkey wrench into the picture, such that it would not need to be a thermal instability, it could be a mechanical instability related to orbital instability. I'm not sure how important this is for the whole process of core collapse, as it seems like a rather late-stage event in the whole core collapse process (and may relate more to whether you get a neutron star or a black hole, moreso than why you get a huge energy release from core collapse).
 
  • #32
Ken G
Gold Member
4,462
333
Those who do numerical simulations of these things claim that collapse to a neutron star has matter going an appreciable fraction of the speed of light, and that this occurs quickly. For a BH, matter crosses a near horizon 'observer' at near c. In the case of a neutron star, it is debatable 'how catastrophic' it is, because for a neutron star, speed of sound is over .3c, so the collapse is not 'dust like'. However for BH formation, different bits of matter are essentially independent of each other because mutual influence cannot propagate fast enough.
Yes, any physical explanation is going to have to idealize the solutions, and so the two limits often invoked are the quasi-static limit of force balance, and the opposite limit of free-fall. Core collapse is generally regarded as a decent example of the latter, but my point is, this is often incorrectly attributed to the absence of a possible force balance when the mass gets high, independent of the other necessary assumptions that get swept under the rug. I'm saying that given the kinetic energy content that is in there, there is a possible force balance-- so what causes the core collapse must be the thermal instability that rapidly removes that kinetic energy, either via neutrino escape or photodisintegration of iron. Those processes are good at removing kinetic energy but require that the temperature remains high, so they self-regulate the thermal environment in a way that doesn't care how quickly mass is being added once they take on a life of their own. The hydrostatic analysis that gives us the Chandra mass tells us the mass benchmark where this will happen, but it still wouldn't happen without the thermal instability that siphons off the gravitational energy and converts contraction into collapse.
 
Last edited:
  • #33
38,063
15,845
So it's always about the thermal timescale, not the absence of a possible force balance.

Type II supernovas happen on a time scale of seconds, so the thermal timescale can evidently be pretty fast, fast enough to make any possible force balance irrelevant, by your own reasoning.

What I'm saying is that there would never be any collapse if the temperature really stayed near zero.

And if you mean this to apply to a white dwarf above the Chandrasekhar limit, this is simply wrong. Such an object cannot support itself against gravity with electron degeneracy pressure, so at zero temperature (i.e., zero kinetic pressure) it must collapse. That's what the Chandrasekhar limit means. The analogous limit for a neutron star is the largest mass at which an object can support itself against gravity with neutron degeneracy pressure; so a neutron star larger than that limit must collapse at zero temperature.

You seem to have in mind a model in which a white dwarf above the Chandrasekhar limit, but not above the analogous limit for a neutron star, slowly contracts from white dwarf size to neutron star size. Unfortunately, this is not possible either, because there are no quasi-stable states in that size range, so there is no way for a white dwarf to slowly move through states that are "close to equilibrium" on its way to neutron star size. This is because there is a sharp discontinuity in the equation of state for matter between the white dwarf regime and the neutron star regime; the transfer from being supported against gravity by electron degeneracy pressure to being supported by neutron degeneracy pressure is not gradual and cannot be undergone a little bit at a time as the star contracts.

The hydrostatic analysis that gives us the Chandra mass tells us the mass benchmark where this will happen, but it still wouldn't happen without the thermal instability that siphons off the gravitational energy and converts contraction into collapse.

It's worth noting that the thermal instability you mention (mainly neutrino escape as electrons and protons in the contracting white dwarf undergo inverse beta decay) is the reason for the sharp discontinuity in the equation of state that I mentioned above. But that discontinuity doesn't just affect the thermal environment (which affects the force balance indirectly, by removing kinetic pressure); it also affects the force balance directly, by changing the adiabatic index for the degeneracy pressure. For a white dwarf near the Chandrasekhar limit, the adiabatic index approaches 4/3 (the relativistic degeneracy value); but for a white dwarf over the limit that is contracting and starting to undergo conversion to a neutron star, the adiabatic index drops precipitously, to well under 1, and doesn't come back up again until the neutron star regime is reached (where it is 5/3, the non-relativistic degeneracy value, for a less massive neutron star and gradually decreases towards 4/3 as the star becomes more relativistically degenerate). So once the white dwarf starts conversion to a neutron star, its degeneracy pressure drops off and it collapses.
 
  • #34
Ken G
Gold Member
4,462
333
Type II supernovas happen on a time scale of seconds, so the thermal timescale can evidently be pretty fast, fast enough to make any possible force balance irrelevant, by your own reasoning.
Yes that's the point-- the supernova occurs because the thermal timescale can be that short. So the problem is not the absence of a possible force balance, it is the presence of a really short thermal timescale.
And if you mean this to apply to a white dwarf above the Chandrasekhar limit, this is simply wrong. Such an object cannot support itself against gravity with electron degeneracy pressure, so at zero temperature (i.e., zero kinetic pressure) it must collapse.
Again, it can support itself if it is not at zero temperature. And if it gradually approaches zero temperature, it will be in force balance all the while, and just become a black hole, with no core collapse ever, before the temperature ever reaches zero. That this is not precisely what happens is simply a matter of the very short thermal timescales owing to the processes that remove kinetic energy.
That's what the Chandrasekhar limit means.
And the Chandrasekhar mass is a zero-temperature limit, so again the issue is always, how is that limit approached-- in short, what is the thermal timescale.
You seem to have in mind a model in which a white dwarf above the Chandrasekhar limit, but not above the analogous limit for a neutron star, slowly contracts from white dwarf size to neutron star size. Unfortunately, this is not possible either, because there are no quasi-stable states in that size range, so there is no way for a white dwarf to slowly move through states that are "close to equilibrium" on its way to neutron star size.
The point is, there are quasi-stable states available there, they simply involve a lot of kinetic energy that prevents complete degeneracy from appearing. If we turn off the thermal processes that remove kinetic energy, such as the Urca process and photodissociation (neither of which appear in the derivation of the Chandra mass), then we never get a core collapse. What Chandrasekhar showed with his celebrated mass limit is that a white dwarf of that mass must contract. That it contracts in a free fall core collapse is all about the thermal processes, processes that do not even appear in the derivation of the Chandra mass. In short, people miss a big part of the story of core collapse when they go straight to the Chandra mass and assert things like "no force balance is possible so you must have a free fall", that's what is wrong. What is right is that no force balance is possible at the end of a process of releasing or locking up heat, but whether or not that leads to a collapse requires analysis of the thermal processes and their timescales because core collapse is a thermal instability, not a hydrodynamic one.
This is because there is a sharp discontinuity in the equation of state for matter between the white dwarf regime and the neutron star regime; the transfer from being supported against gravity by electron degeneracy pressure to being supported by neutron degeneracy pressure is not gradual and cannot be undergone a little bit at a time as the star contracts.
The difference in those equations of state brings in a new issue, which is whether or not you get a core bounce and an explosion. That's seperate from the issue of whether or not you get a free-fall core collapse. If neutrons had exactly the same equation of state as electrons, we'd still have core collapse in much the same way we have now, because the thermal instability has nothing to do with that equation of state. We just wouldn't have core bounce, and we wouldn't have neutron stars, we'd just have the rapid and relatively quiet collapse into black holes that we think we now get with very high mass stars.
It's worth noting that the thermal instability you mention (mainly neutrino escape as electrons and protons in the contracting white dwarf undergo inverse beta decay) is the reason for the sharp discontinuity in the equation of state that I mentioned above.
No, the equation of state is not about what happens while you are turning electrons and protons into neutrons, it is about what you have after that is all over with (higher mass particles with strong forces and strong gravity). That's all relevant to the core bounce as the equation of state stiffens. The thermal instability simply involves the removal of heat as neutronization occurs, which is different from the forces that neutrons experience after they form. In other words, one shows up in the energy equation, and the other in the force equation. The neutron equation of state involves effects that reduce how relativistic the particles are because there are no more low-mass electrons, and there are repulsive forces between the neutrons (not degeneracy, that isn't a repulsive force it's just the kinetic energy content there), so that stiffens the pressure and generates the core bounce. But the core collapse is a thermal instability, it shows up in the energy equation when we see that lots of kinetic energy is being locked up or lost.
For a white dwarf near the Chandrasekhar limit, the adiabatic index approaches 4/3 (the relativistic degeneracy value); but for a white dwarf over the limit that is contracting and starting to undergo conversion to a neutron star, the adiabatic index drops precipitously, to well under 1, and doesn't come back up again until the neutron star regime is reached (where it is 5/3, the non-relativistic degeneracy value, for a less massive neutron star and gradually decreases towards 4/3 as the star becomes more relativistically degenerate). So once the white dwarf starts conversion to a neutron star, its degeneracy pressure drops off and it collapses.
There is a lot of confusion about the adiabatic index in an equation of state. If the process were really adiabatic, the connection between pressure and density always looks like a 5/3 power law for nonrelativistic gas, and 4/3 for relativistic, regardless of degeneracy (though since neutrons experience other forces, you can get below 1 for them, if the process is adiabatic-- like a zero temperature condition). But the process is not adiabatic, and the temperature is not zero, there is important heat content because the gas is at some kind of fusion temperature initially. So we must track the thermal timescales, we must follow the heat. If we could really do that by simply asserting the zero-temperature equations of state for electrons and/or neutrons, we'd never get a core collapse, because if the temperature stays zero, the radius is always a continuous smooth function of the mass and the composition, both of which vary continuously as we add mass to create the contraction. That smooth solution assumes a quasi-steady force balance at every stage of the way, so is appropriate for a slow enough rate of adding mass (the "particle by particle" approach mentioned in the OP). But that is not what would really happen, because the temperature is not zero, there is heat locked up in the star that prevents contraction until thermal processes allow that heat to be removed. If there are thermal processes that remove that heat in an unstable way, you get the thermal instability that produces the core collapse. If there are not thermal processes that remove heat in an unstable way (i.e., that don't runaway until they happen faster than the sound crossing time), then you can keep all the same equations of state, and you'll still never get a core collapse if the rate you are adding mass is slow enough-- which is what I believe the OP was asking about.
 
Last edited:
  • #35
38,063
15,845
it can support itself if it is not at zero temperature.

It can support itself if kinetic pressure (and therefore temperature) is sufficiently high. But that means fusion temperature, not just "not zero" temperature. What happens when the temperature is well below fusion temperature but well above zero temperature?

if it gradually approaches zero temperature, it will be in force balance all the while

I disagree, but I think the disagreement is more about the equation of state than about the thermal issue (see further comments below). I agree that the thermal timescale is an important factor if a hydrostatic equilibrium is possible; but I think we disagree about when a hydrostatic equilibrium is even possible.

the equation of state is not about what happens while you are turning electrons and protons into neutrons, it is about what you have after that is all over with (higher mass particles with strong forces and strong gravity).

So what, the equation of state just magically agrees to not change until the conversion into neutrons is complete? The object still has a structure while that process is going on, and that structure is still affected by degeneracy pressure; so the sharp drop in degeneracy pressure during the conversion process is going to affect stability. Whether you are willing to call that a "change in the equation of state" is a matter of words, not physics. How much it affects it will depend on how much degeneracy pressure contributes to hydrostatic equilibrium, but as above, there is a wide range of temperatures between "zero temperature" and "fusion temperature" where degeneracy pressure still contributes significantly to hydrostatic equilibrium..

If the process were really adiabatic, the connection between pressure and density always looks like a 5/3 power law for nonrelativistic gas, and 4/3 for relativistic, regardless of degeneracy (though since neutrons experience other forces, you can get below 1 for them, if the process is adiabatic-- like a zero temperature condition).

In other words, "the index is 5/3 for nonrelativistic, and 4/3 for relativistic, except when it isn't". While this is true, it doesn't seem to be very helpful. ;)

As for the "other forces", which is the point, neutrons experience those regardless of temperature, and the effects on degeneracy pressure of inverse beta decay happen regardless of temperature, correct? If a white dwarf contracts enough for inverse beta decay to start, it will experience a sharp drop in degeneracy pressure regardless of its temperature, so hydrostatic equilibrium will be affected even if it is hot.
 

Related Threads on Black hole creation process

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
11
Views
961
  • Last Post
Replies
17
Views
6K
  • Last Post
Replies
4
Views
1K
  • Last Post
2
Replies
45
Views
5K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
1
Views
3K
Replies
6
Views
3K
Top