When will neutronization occur in the black hole formation process?

[bcrelling]

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

 

[Matterwave]

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.

 

[Doug Huffman]

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.

 

[PeterDonis]

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.

 

[tionis]

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?

 

[PeterDonis]

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.

 

[bcrowell]

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.

 

[tionis]

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?

 

[PeterDonis]

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.

 

[PAllen]

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.]

 

[bahamagreen]

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.

 

[PAllen]

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.

 

[bahamagreen]

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"?

 

[PAllen]

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.

 

[tionis]

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?

 

[martinbn]

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.

 

[PAllen]

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.

 

[PAllen]

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.]

 

[PeterDonis]

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).

 

[PAllen]

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.

 

[tionis]

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

 

[mfb]

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

No, the mass stays the same.

 

[tionis]

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?

 

[PAllen]

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.

 

[PeterDonis]

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).

 

[stevebd1]

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).

 

[PeterDonis]

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).

 

[Ken G]

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.

 

[PeterDonis]

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.

 

[PAllen]

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.

 

[Ken G]

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, more so than why you get a huge energy release from core collapse).

 

[Ken G]

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.

 

[PeterDonis]

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.

 

[Ken G]

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 separate 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.

 

[PeterDonis]

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.

 

[Ken G]

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?

That is an important issue because it properly focuses our attention where it needs to be: what is happening to the heat. There are really two separate considerations that are required to have core collapse. The first is there even at zero temperature, which asks, is there a hydrostatic equilibrium at all for the electrons, or is gravity too strong? That's the Chandra mass. The second is, when contraction starts, will there be enough gravitational energy released to renew the pressure support? That's a rather different question, because it does not assume the temperature is zero, and if the temperature is not zero, adiabatic gravitational contraction will increase the temperature. Since the contraction is neither adiabatic nor at zero temperature, attention shifts to what happens to the heat. So that's where we must look for thermal instability as the necessary cause of core collapse.

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.

What I was saying was going back to the OP where we add mass very slowly, say a particle at a time. If we make the simplifying assumption used in the derivation of the Chandra mass, that the temperature is zero, what we are saying is that the temperature constantly stays at zero as we add mass, particle by particle. There is obviously always a force balance in that situation, right up until you have a black hole (or at least GR-type instabilities very near to the creation of a black hole). So there's no core collapse there because the processes that lead to thermal instability, like photodisintegration and the Urca process, don't happen at zero temperature. This is why I'm saying that the loss of force balance has to do with thermal processes that remove heat, and do so in a runaway kind of way that ultimately leads to thermal timescales shorter than the force-balance timescale, and that's what produces the free fall. The basic assumptions that go into the Chandrasekhar mass will not give a loss of force balance when the mass is added slowly enough, so that's why the Chandra mass calculation is just a benchmark for the mass scale of interest, not a physical description of the core collapse process.

So what, the equation of state just magically agrees to not change until the conversion into neutrons is complete?

None of my argument relies on that. The equation of state of the neutrons is of no importance, because the equivalent to the "Chandra mass" when dealing with neutrons is higher than for electrons. Hence, the EOS for neutrons is never part of the explanation of core collapse, but it is part of the explanation of the completely separate process of core bounce.

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..

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. ;)

Again, since the Chandra mass for neutrons is higher than for electrons, the neutron EOS never plays any role in the reasons we have a core collapse.

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?

I can only repeat, those "other forces" clearly have nothing to do with why we get a core collapse, since they oppose collapse. The effects of beta decay that matter are how they remove heat when the neutrinos escape. That is not the neutron EOS, that is the thermal instabiilty that contributes to the core collapse.

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.

That is what I have been saying-- you are talking about a contributing factor to thermal instability that is the real reason for core collapse. The existence of a Chandra mass just tells you that you will have a compact object, it does not tell you that you will lose force balance along the path to reaching that compact object. So again, the Chandra mass is a benchmark for when you might get a thermal instability and a core collapse, but the actual existence of a core collapse requires analysis of the thermal instability. That's the part that answers the OP question, of why you don't just smoothly contract when you add mass slowly-- as you approach the Chandra mass, a thermal instability sets in, and at that point you don't even need to add any more mass at all.

 

[PeterDonis]

If we make the simplifying assumption used in the derivation of the Chandra mass, that the temperature is zero, what we are saying is that the temperature constantly stays at zero as we add mass, particle by particle. There is obviously always a force balance in that situation, right up until you have a black hole (or at least GR-type instabilities very near to the creation of a black hole).

As I said in an earlier post, GR says that hydrostatic equilibrium is impossible for an object with a radius less than 9/8 of the Schwarzschild radius for its mass. The "redshift factor" for that radius is $\sqrt{1 - 2M / r} = \sqrt{1 - 8/9} = 1/3$, which is still quite a bit different from zero (the limiting value at the horizon of a black hole with that mass). So when an object at zero temperature contracts to this radius (such an object will be a neutron star, not a white dwarf--a white dwarf will have to convert to a neutron star well before it can contract to this point; see further comments below), force balance is no longer possible, even though the object is still well short of forming a black hole. The rest of the process of forming a black hole cannot happen gradually.

I'm not sure if the above is what you mean by "GR-type instabilities very near to the creation of a black hole", but I would not call the above radius "very near", because, as noted, the redshift factor is still well away from zero, so the object will not have "disappeared from view"; it will still be easily visible. So you won't see the object slowly getting redder and redder right up to the zero point; you will see it slowly getting redder and redder up to the 1/3 point, and then it will collapse, and the rate at which it gets redder and disappear will increase a lot.

The equation of state of the neutrons is of no importance, because the equivalent to the "Chandra mass" when dealing with neutrons is higher than for electrons.

The neutron star mass limit is higher, yes, but that just means that, for the process described in the OP, slowly adding mass to an object, there will be two stages on the way from white dwarf to black hole. The first stage will be when the white dwarf reaches a point where it starts to convert into a neutron star; at that point the process won't be gradual because of some combination of thermal instability (neutrinos and other processes carrying away heat very fast) and force instability (the equation of state changing because the chemical composition is changing). If we assume that this process ends in a neutron star, not a black hole, then there will be a second stage when the neutron star reaches its mass limit and starts to contract; that process must become unstable when the limiting radius given above is reached, at which point it will no longer be gradual, as above, and that will be due to a lack of any possible hydrostatic equilibrium inside the limiting radius (note, as I said before, that this is true regardless of temperature). So this stage of collapse is driven primarily by lack of force balance.

If you're just looking at the white dwarf to neutron star stage, and assuming that the process was gradual up to that point, then I can see how the neutron star EOS would not have a significant effect, because, as you say, the mass limit is higher so there is guaranteed to be a stable neutron star configuration available, it's just a matter of how it is reached. But the same will not be true for the neutron star to black hole stage, for the reasons given above.

 

[Ken G]

As I said in an earlier post, GR says that hydrostatic equilibrium is impossible for an object with a radius less than 9/8 of the Schwarzschild radius for its mass.

Yes, I was referring to your earlier point on that when I mentioned instability very near creation of a black hole. That's way too late to have a "core collapse"-- a core collapse takes an object the size of a small planet and turns it into the size of a small city, and does it on the free-fall time. So that process is the thermal instability process I'm talking about. The final throes of the process would indeed have to be out of force balance, but if that were the only time it was out of force balance, it's not much of a core collapse and won't make a supernova.

I'm not sure if the above is what you mean by "GR-type instabilities very near to the creation of a black hole", but I would not call the above radius "very near", because, as noted, the redshift factor is still well away from zero, so the object will not have "disappeared from view"; it will still be easily visible.

That is indeed what I meant by "very near", in the sense that 9/8 is close to 1. But more importantly, it's not what is meant by a core collapse.

The neutron star mass limit is higher, yes, but that just means that, for the process described in the OP, slowly adding mass to an object, there will be two stages on the way from white dwarf to black hole.

Right-- but neither is a core collapse, without the thermal instability.

The first stage will be when the white dwarf reaches a point where it starts to convert into a neutron star; at that point the process won't be gradual because of some combination of thermal instability (neutrinos and other processes carrying away heat very fast) and force instability (the equation of state changing because the chemical composition is changing).

It's not a combination of those, because the force effects don't matter-- the object is in free fall due to the thermal instability, the forces are not doing anything until you get all the way down to the core bounce, after the core collapse has already occurred. All the force effects due is allow the conditions to be ripe for the thermal instability-- but after that, the core collapse is caused by the thermal instability. Sure, we have a bunch of equations in play, and changing any of them will change what happens. My point is that the usual explanation for core collapse misses something quite important. The usual explanation is that no force balance exists above the Chandra mass. This leads the OP to ask, why can't you just get a gradual contraction if you add mass gradually, such that you arrive at a compact object by the time you arrive at the Chandra mass? That's a very good question, and it emerges because the usual explanation is lacking-- it fails to identify the key role of thermal instability that happens at a nonzero temperature, even though the Chandra mass is derived as a benchmark in the idealized limit of zero temperature.

If you're just looking at the white dwarf to neutron star stage, and assuming that the process was gradual up to that point, then I can see how the neutron star EOS would not have a significant effect, because, as you say, the mass limit is higher so there is guaranteed to be a stable neutron star configuration available, it's just a matter of how it is reached. But the same will not be true for the neutron star to black hole stage, for the reasons given above.

Yes I agree, I'm interested in the causes of what would normally be called "core collapse," a free fall of a small planet-sized core into a small city-sized compact object. I can't speak in detail about that process without a detailed model that solves all the equations, but I'm saying that it seems the crucial piece in explaining why that collapse is not in force balance is the presence of thermal instability that operates at nonzero temperature.

 

[PeterDonis]

The final throes of the process would indeed have to be out of force balance, but if that were the only time it was out of force balance, it's not much of a core collapse and won't make a supernova.

I realize I'm the one who brought up supernovas earlier in the thread , but a supernova is a very different process from the gradual adding of small amounts of mass to a compact object until a maximum mass limit for a particular type of object (white dwarf or neutron star) is reached. As I understand it, much of the energy released in a supernova is not because you're taking an object the size of a small planet (the white dwarf core) and contracting it to the size of a small city (the neutron star, assuming it ends up as one and not a black hole); it's because you're taking an object the size of a large star and contracting it to the size of a small city. The amount of energy available by collapsing the star is much larger (much more mass and much larger radius) than the amount of energy available by just collapsing the core.

That is indeed what I meant by "very near", in the sense that 9/8 is close to 1

But a redshift factor of 1/3 is a lot different from a redshift factor of zero. Or, if you invert it and talk about wavelength instead of frequency (which is more common in astrophysics), a redshift of 3 is a lot different from a redshift of infinity. That seems to me to be the more appropriate thing to focus on.

The usual explanation is that no force balance exists above the Chandra mass. This leads the OP to ask, why can't you just get a gradual contraction if you add mass gradually, such that you arrive at a compact object by the time you arrive at the Chandra mass? That's a very good question, and it emerges because the usual explanation is lacking-- it fails to identify the key role of thermal instability that happens at a nonzero temperature

Well, the thermal instability explanation also fails to identify a key factor: the fact that the equilibrium configuration of a given mass if it's a white dwarf is quite a bit larger (small planet-sized vs. city-sized, to use your own terms) than if it's a neutron star. Even if the neutron star is at high temperature, that won't change the radius of the equilibrium configuration appreciably: fusion temperatures are tens to hundreds of keV, whereas the neutron rest mass is nearly 1 GeV, five orders of magnitude higher, so thermal energy is insignificant compared to rest energy. So there is still a huge gap in size between white dwarf and neutron star where no stable configuration exists. That huge gap is what makes available a large amount of thermal energy from the conversion of a white dwarf to a neutron star of the same mass (just above the Chandrasekhar limit).

 

[Ken G]

I realize I'm the one who brought up supernovas earlier in the thread , but a supernova is a very different process from the gradual adding of small amounts of mass to a compact object until a maximum mass limit for a particular type of object (white dwarf or neutron star) is reached.

Yes, I admit I have interpreted the OP along the lines of "what happens when a black hole is formed", more so than hypothetical pathways that could involve maintaining zero temperature the whole time. If we only look at the force balance in degenerate electrons, a la the Chandra mass derivation, we would think that the radius would just keep shrinking if we add mass slowly enough. But that's only true if the temperature starts out zero and stays zero-- in a real star, it isn't zero. The zero temperature assumption is just an idealization to get the benchmark mass where the thermal instability kicks in, when we account for the nonzero temperature actually present in stars undergoing core collapse, we see that you can initiate core collapse by adding mass slowly, and at some point it no longer cares that you are adding mass at all, because the thermal instability has set in. You have to be close to the Chandra mass for this to happen, but the loss of force balance happens because the star was using the nonzero temperature as part of its pressure support, and that is the support that the thermal instability takes away. Because the equation of state of relativistic electrons is so soft, a small loss of support involves a substantial infall, which only increases the force imbalance by ramping up the thermal runaway. Eventually there is substantial inertia in the infall, so it no longer matters if a hydrostatic force balance would be possible at zero temperature, the inertia of the infall can only be stopped by hardening the equation of state (which happens for the various reasons you point out when neutronization occurs, but that is the explanation of the core bounce, not the core collapse that has already happened).

As I understand it, much of the energy released in a supernova is not because you're taking an object the size of a small planet (the white dwarf core) and contracting it to the size of a small city (the neutron star, assuming it ends up as one and not a black hole); it's because you're taking an object the size of a large star and contracting it to the size of a small city.

Some of the infalling mass was already in the core, some was in the envelope, and only the mass that ends up below the point of the core bounce can contribute gravitational energy to the supernova, but exactly how much is above or below the eventual core bounce is a detail of no great significance. My point was only that the initial radius is way larger than the kinds of distances where GR effects become important, so the free fall of core collapse starts from a rather Newtonian gravity environment that cannot have much to do with GR. There are also not free neutrons yet. So can look neither to GR, nor to the neutron equation of state, to understand why there is a free falling core collapse in a supernova, a free-fall that happens as you add the mass. As the Chandra mass is approached, the nonzero temperature in real stars becomes very important in the support of that object, and that temperature tends to rise as contraction occurs, until a thermal instability sets in that leads to core collapse before you get quite to the Chandra mass, even if you stop adding mass. Once this initiates, it will play out no matter how slowly you continue to add mass, and indeed you can stop adding mass altogether for all the difference it will make.

But a redshift factor of 1/3 is a lot different from a redshift factor of zero. Or, if you invert it and talk about wavelength instead of frequency (which is more common in astrophysics), a redshift of 3 is a lot different from a redshift of infinity. That seems to me to be the more appropriate thing to focus on.

You are talking about what a near-black-hole would look like if you added mass slowly and just let it constantly reach a new equilibrium at zero temperature. The OP does talk about that issue, but I'm imagining that the OP is asking about what happens in real astrophysical environments. I'm interpreting the OP question as basically asking "do black holes form suddenly or gradually, and is the difference how fast you add mass to them?" To that I would answer, when we look at real situations where black holes form, we find it happens suddenly, due to a thermal instability that kicks in as the Chandra mass is approached, independently of how fast the mass is added in real applications. But you could also say that if we imagine a hypothetical where the mass is added slowly, and the temperature is maintained at zero all the time (so no Urca process, no photodissociation), then the near-black-hole would still look significantly different from a black hole at the moment when gravitational instabilities cause it to collapse from 9/8 of the Schwarzschild radius to a black hole.

Well, the thermal instability explanation also fails to identify a key factor: the fact that the equilibrium configuration of a given mass if it's a white dwarf is quite a bit larger (small planet-sized vs. city-sized, to use your own terms) than if it's a neutron star. Even if the neutron star is at high temperature, that won't change the radius of the equilibrium configuration appreciably: fusion temperatures are tens to hundreds of keV, whereas the neutron rest mass is nearly 1 GeV, five orders of magnitude higher, so thermal energy is insignificant compared to rest energy.

Ah, but that's just the point-- even a small thermal energy is very important when the gas is relativistic. The reason stars undergo core collapse is that at some point, the loss of even that tiny fraction of the pressure support, due to the thermal instability that eats up that excess thermal energy, causes a runaway that causes even more of that extra support to be lost, and the equation of state of relativistic electrons is so soft that once contraction initiates, the force imbalance grows until it leads to free fall.

Basically, there are two ways that an approximately free-fall collapse could occur. It could start at zero temperature, with mass added with no additional heat, so zero temperature would always be maintained. The force-balance solution would say that as mass increases, the radius must shrink, and the inertial term required to have a shrinking radius will act as a kind of pressure support that will serve to make sure the collapse happens no faster than free-fall. So that works, if the mass is added fast enough, there can be a free-fall collapse. Or, the other way it could happen, is a small part of the support could be coming from a nonzero temperature, and as contraction occurs when mass is added, that temperature will rise in the absence of a thermal instability that removes that excess heat. The rising temperature will cause things like photodisintegration of iron and the Urca process, real things that actually happen in stellar core collapses. So we know that the latter model is what is actually happening, we need only to ask, is it important to recognize the thermal instability elements?

That's where your numbers come into play, but they need adjustment. Silicon fusion temperature is about 200-300 keV, and the characteristic kinetic energy of the core collapse ends at up to about 1 GeV, but it starts well below that. So the thermal excess is only about 2 orders of magnitude below the total kinetic energy when the collapse begins, it's an effect on the order of 1% roughly. That is still small, but the whole reason there is core collapse is that the relativistic electron equation of state is so soft that even a 1% change in the support results in a significant contraction before that 1% can be replaced by excess gravitational energy. In fact, as the contraction occurs, the gas gets even more relativistic, so the falling thermal excess becomes even more significant-- if there were no thermal instability, even a tiny thermal excess could prevent further collapse. You have to get that thermal energy out, it's anathema to core collapse. As such, the timescale for core collapse must be the timescale for the heat to be removed, and the timescale on which mass is being added becomes irrelevant, unlike in the zero-temperature version. Perhaps it's a minor point, as either process could give a core collapse. Perhaps the best way to sum this all up is as follows:

A core collapse happens when the timescale for the radius to change, either due to the removal of thermal excess energy due to a thermal instability or the addition of mass at zero temperature, is of order the free-fall time. Which view is more central to what is really happening probably requires a more detailed analysis. We can note that for the very massive stars, they have cores that are never highly degenerate and maintain force balance as they lose heat, but will undergo a core collapse when the thermal instability sets in. Stars with core masses closer to the Chandra limit do develop highly degenerate cores, but not zero temperature cores, so it's still not clear if the thermal instability is not what really happens in them as well.

So there is still a huge gap in size between white dwarf and neutron star where no stable configuration exists.

That isn't true, zero temperature solutions exist in that regime, in the idealization of an ideal gas with the masses of protons and electrons. They are not realized in actual stars, because of the history of how they came about. As they are crossing that regime, the thermal instability from photodisintegration and the Urca process rips the thermal support out from under the star, so it free-falls through that regime. Or, if we take the zero temperature idealization instead, then it's because there's a history of adding mass that has created an inertial term in the force balance that forces the star to cross that regime without finding the hydrodynamically stable solution that exists there. This is the crux of my point-- there are stable hydrostatic solutions that are possible in the core-collapse regime, they are not realized in practice because of the short timescales encountered (be they the short thermal timescale, or the short timescale for adding zero-temperature degenerate mass). The Chandra mass is a kind of landmark looming in the distance, but the actual situation does have possible hydrostatic solutions, one needs to put the time-dependent behavior in there to see why those do not occur.

 

[PAllen]

I would like to argue again that force imbalance is a perfectly reasonable description of what happens with collapse to a neutron star (or a BH). Further, I am quite skeptical that any temporary equilibrium can exist above the Chandra mass with a radius between dwarf and neutron star.

First, I'll go back to my gedanken experiment of adding iron filings slowly to an iron ball (slowly enough to stay pretty near absolute zero; this means waiting a while between infall of grains due to high KE contributed as the ball gets more massive). I would like to focus on the stage of adding mass to a black dwarf (I know much less about the physics of a supergiant planet slowly growing and then collapsing to a black dwarf; I suspect very little analysis relevant to this has been done, since it would never occur in the our universe, especially with all iron).

I claim there would be a point where a tiny addition of mass would lead to a near free fall collapse from black dwarf to neutron star, and that large energy release (primarily neutrinos) would accompany this. The trigger is that there is a fairly sudden change in the equilibrium rate of reverse beta decay versus beta decay. This leads to decrease of pressure (since neutrinos are so ineffective at providing pressure), rapid collapse, with each further amount of collapse further shifting the equilibrium in favor of reverse beta decay (and more electrons disappearing in favor of neutrinos). The result would be a mini-supernova. To me, rapid disappearance of pressure is perfectly reasonably described as a force imbalance.

Now, let's ask if we can create an interim temporary equilibrium between dwarf and neutron star by varying the conditions. Suppose we aim to do this by making the temperature rise dramatically as we add matter, to compensate for loss of electron pressure. For example, we can add a proportion of antimatter to generate extreme heat. The problem I see is that this creates many additional instabilities:

1) If thermal radiation exceeds 1 Mev, pair production is favored over annihilation, so your high temp photon pressure vanishes
2) Iron is very good at absorbing high energy photons (photo-disintegration).

These instabilities again lead to sudden pressure drops, which certainly qualify as force imbalance IMO.

Of course, iron black dwarfs don't exist (and most likely, no black dwarfs exist yet). However, if one imagined adding mass realistically to a white dwarf, the above factors still suggest that it would undergo a quite sudden collapse at a critical point.

So I would ask for at least a reasonable heuristic justification to the claim that there can exist a hydrostatic equilibrium between dwarf and neutron star.

 

[Ken G]

I would like to argue again that force imbalance is a perfectly reasonable description of what happens with collapse to a neutron star (or a BH). Further, I am quite skeptical that any temporary equilibrium can exist above the Chandra mass with a radius between dwarf and neutron star.

We all agree that force imbalance appears, it's nearly a free fall after all. I'm saying there exists a force balance, with a mass just below the Chandra mass and a radius between a white dwarf and a neutron star, but that force balance is simply not reached because of the thermal history of the system that creates a rather significant inertial term when those radii are encountered. And I'm also saying that were it not for the thermal instability that is the key here, you could indeed have a mass above the Chandra mass, and have a radius between dwarf and neutron star, and have a perfectly stable force balance. You simply don't have a zero temperature, so the nonrelativistic ions (which don't even appear in the derivation of the Chandra mass) are doing part of the heavy lifting. More on that at the end.

First, I'll go back to my gedanken experiment of adding iron filings slowly to an iron ball (slowly enough to stay pretty near absolute zero; this means waiting a while between infall of grains due to high KE contributed as the ball gets more massive). I would like to focus on the stage of adding mass to a black dwarf (I know much less about the physics of a supergiant planet slowly growing and then collapsing to a black dwarf; I suspect very little analysis relevant to this has been done, since it would never occur in the our universe, especially with all iron).

There is no dispute that if you keep the system at zero temperature, you cannot have a stable force balance above the Chandra mass, that's what the derivation of the Chandra mass shows. My point all along has been that the zero temperature idealization is just a means of getting a mass benchmark-- real core collapses do not occur at zero temperature. The temperature is small enough that the excess thermal energy it represents seems rather small, perhaps 1% or less of the total kinetic energy in there, but this support term is not negligible because it is its elimination that creates the fundamental instability that leads to core collapse in real supernovae (or so it seems to me, in the absence of real simulation data to pore over). But I do not dispute that there is a second path to core collapse-- you keep the temperature always at zero (essentially by imagining a system that starts out at zero temperature), and just add mass quickly enough that the smoothly dropping radius of the hydrostatic solution will show a time derivative that rivals the free-fall rate. If you add mass as quickly as that, you'll get a zero temperature core collapse. But if you add mass more slowly than that (say, "particle by particle"), then you will never get a core collapse at zero temperature-- you will have a force balance at all radii right down to near the Schwarzschild radius (where the GR instabilities set in that PeterDonis mentioned, or where neutronization sets in as you point out). But real core collapses set in sooner than that, and they happen at realistic fusion temperatures, because then the thermal instability will pull the rug out from under the hydrostatic solution at some point, no matter how slowly you add the mass (even if it is "particle by particle").

I claim there would be a point where a tiny addition of mass would lead to a near free fall collapse from black dwarf to neutron star, and that large energy release (primarily neutrinos) would accompany this. The trigger is that there is a fairly sudden change in the equilibrium rate of reverse beta decay versus beta decay. This leads to decrease of pressure (since neutrinos are so ineffective at providing pressure), rapid collapse, with each further amount of collapse further shifting the equilibrium in favor of reverse beta decay (and more electrons disappearing in favor of neutrinos). The result would be a mini-supernova. To me, rapid disappearance of pressure is perfectly reasonably described as a force imbalance.

Yet note that no such process appears anywhere in the derivation of the Chandra mass, so you are still going against the standard argument that core collapse occurs when you reach the Chandra mass because that's where the force balance disappears. What you are saying here is quite similar to what I have been saying all along-- that the core collapse is not caused by the loss of the existence of a hydrostatic solution at zero temperature when the Chandra mass is reached, it is caused by an instability that runs away before the Chandra mass is reached. That instability removes kinetic energy from the gas, removing pressure support. There are many types of processes that do this, there are Urca type processes that involve neutrino escape, and there is photodisintegration of the iron. Note these only happen at high temperature, so are fundamentally not zero temperature processes. For example, the Urca process normally requires the particles to convect across a temperature gradient (so that the nucleon captures a hot electron, emits a neutrino, convects up to lower temperatures, and then can beta decay and emit a neutrino again). At zero temperature, this doesn't work, because you cannot beta decay into a lower energy electron than the one that was originally inverse beta decayed away, as no such lower energy electron state is open. So the Urca process is only a process for removing excess thermal energy, which distinguishes it from the neutronization process you are talking about.

Now, I admit I have been wondering about the neutronization process itself (so just the original inverse beta decay), that may be able to occur at zero temperature and will result in energy loss via neutrino emission as you say, though it may also come rather late in the core collapse process. In any event, it is a third process that can remove energy, in addition to Urca and photodisintegration, and it isn't restricted to removing excess thermal energy, it might work even at zero temperature as you are arguing. If so, we still have the question of which of these three processes dominate, and should we invoke new language beyond "thermal instability" if we are not limited to thermal energy-- we could call it "energy loss instability" to be more generally inclusive.

Even so, the point is that we are not seeing the loss of a force balance solution at zero temperature, that solution is still there instantaneously at any stage of this process. So the problem is not in the force equation, the problem is with the sink term in the energy equation. That has really been my main point-- core collapse represents a loss of force balance that does not occur because no hydrostatic version of the force equation exists (as happens above the Chandra mass at zero temperature), it occurs because you have a sink term in the energy equation, i.e., there's no energy-static solution, even though there is a force-static solution, and the timescale of the former at some point goes faster than the timescale of the latter. So it's fundamentally not the complete absence of a possible balance in the force equation as occurs when the Chandra mass is reached, it's the presence of a sink in the energy that appears as the Chandra mass is neared. Whether or not that constitutes a "thermal instability" depends on which of these energy hogs is the dominant one in detailed solutions, because neutronization is not really a thermal process, I must agree. Also, neutronization would seem to completely slaughter the kinetic energy of the electron, once the neutrino escapes. So that would seem to favor neutronization as a key process, but neutronization happens rather late in the core collapse, so to really understand why core collapse happens in the first place, you may need to look at thermal mechanisms like the Urca process or photodisintegration-- processes that kick in before there is substantial neutronization. By the time you get neutronization, the core collapse may already be a fait accompli by virtue of having developed a large inertia. So again this comes under the heading of whether we are answering "how do black holes form", versus answering "what can happen in a kind of hypothetical formation process that maintains zero temperature." Those have rather different answers.

These instabilities again lead to sudden pressure drops, which certainly qualify as force imbalance IMO.

Yes, those are the kinds of "thermal instabilities" I have been talking about. The problem here is ambiguity in the phrase "loss of force balance", and whether it means "a solution that is not in force balance due to its history of interaction with an energy equation that has a very rapid timescale for change" versus "absence of a hydrostatic solution in the instantaneous force balance that makes no reference to energy losses." The latter is all you see in the usual descriptions that involve the Chandrasekhar mass, because that mass emerges entirely from the force equation with no reference to energy losses. I'm saying that the former meaning is the actual explanation for how core collapses occur.

So I would ask for at least a reasonable heuristic justification to the claim that there can exist a hydrostatic equilibrium between dwarf and neutron star.

There is a wide regime in there where neutronization would be slow, by which I mean on a much longer timescale than the sound crossing time, and so we only need to go into that realm and find the force balance at zero temperature that exists there. The only reasons we don't find stellar cores in that regime in real life is either because mass is not added to them as slowly as necessary to ignore the inertial term in the force balance as the hydrostatic radius changes, or because they don't start out at zero temperature, so they are subject to thermal instabilities in any regime where photodisintegration and/or the Urca process are active. I don't know which of those reasons is the more important one in real simulations of core collapse, but I would tend to think the latter, because the sound crossing time is so short that it might be hard to add mass that quickly. But I have heard it said that the final stages of fusion into iron goes pretty fast, so that rate might be what breaks the hydrostatic equilibrium. But in either case, it happens before you reach the Chandra mass-- so core collapse is not caused by the absence of a force balance when you set all the time derivatives to zero, like Chandrasekhar did-- that only serves to benchmark the important mass when core collapse happens, not to explain the core collapse process.

 

[PeterDonis]

My point was only that the initial radius is way larger than the kinds of distances where GR effects become important

Ok, that makes it clearer. I agree this is true for a normal star, and even for a white dwarf; but it's not true for a neutron star.

I'm imagining that the OP is asking about what happens in real astrophysical environments.

I'm not sure how realistic the "add mass one atom at a time" idealization is, astrophysically speaking, but I agree there are a wide range of possible situations one could consider, in terms of how fast mass gets added.

even a small thermal energy is very important when the gas is relativistic

For an electron gas, yes, because at fusion temperatures the electron kinetic pressure is of the same order of magnitude as the electron degeneracy pressure. For a neutron gas, no, because the kinetic pressure even at fusion temperature is much, much smaller than the neutron degeneracy pressure.

That isn't true, zero temperature solutions exist in that regime, in the idealization of an ideal gas with the masses of protons and electrons.

It is true that there are technically "solutions" in this regime, but they are unstable, like a pencil balanced on its point, so I'm assuming we are not considering them. There are no stable solutions between the white dwarf regime and the neutron star regime.

 

[Ken G]

Ok, that makes it clearer. I agree this is true for a normal star, and even for a white dwarf; but it's not true for a neutron star.

Yes, I keep interpreting the OP question around the causes of core collapse, but it could also be interpreted in terms of starting with a neutron star and asking what happens if you add mass to that, and does it gradually change appearance into a black hole. That's what you are answering, when you point out that it goes unstable at 9/8 the Schwarzschild radius, so won't finish a gradual transition into the appearance of a black hole. We are just talking about different phenomena.

I'm not sure how realistic the "add mass one atom at a time" idealization is, astrophysically speaking, but I agree there are a wide range of possible situations one could consider, in terms of how fast mass gets added.

Yes, so it's a good question as to how fast that really happens, and do we find two different phenomena. One is where we assume the core always maintains the temperature of the fusing shell that is adding that mass, and then we are seeking the smoothly updated hydrostatic equilibrium at that temperature, and at some point the change in the radius we get is happening so fast it rival the free-fall time, at which point we know we have lost force balance and have a core collapse. But this requires that we are not bottling up the heat that is in the smoothly contracting core, so alternatively, it is possible that before we get to the stage where the radius is changing at free-fall rates, we will find that it is changing rapidly enough that the heat cannot escape fast enough to maintain the fusion temperature. Then the core temperature grows enough to have photodisintegration of iron, and/or the Urca process, leading to a thermal instability that gets rid of the heat much faster and brings us back to the first situation. Since we know these thermal processes do indeed occur, the only question is, would that thermal energy prevent the core collapse in the absence of this thermal instability? I think it would, even though it is small, because it is locked up in the nonrelativistic protons, so the temperature will grow quite rapidly as it contracts-- absent the thermal instability. But as you point out, this could only forestall the collapse-- eventually neutronization will occur, and there are GR instabilities as well, so the process will always end catastrophically. My question is, which is the first thing that happens that starts the catastrophe? I believe the answer to that is the thermal instability.

For an electron gas, yes, because at fusion temperatures the electron kinetic pressure is of the same order of magnitude as the electron degeneracy pressure.

Degeneracy pressure is always the same thing as kinetic pressure, degeneracy pressure is kinetic pressure. For some reason, it is easy to find language that degeneracy pressure is some kind of new quantum mechanical type of pressure, but that's incorrect, it's just mundane kinetic pressure. The quantum mechanical effects are seen in the temperature of the gas, not its pressure, if one is fixing the kinetic energy content (which is a perfectly reasonable thing to fix, given its known energy history).

For a neutron gas, no, because the kinetic pressure even at fusion temperature is much, much smaller than the neutron degeneracy pressure.

That's what isn't true-- even for neutrons, kinetic pressure is the same as degeneracy pressure. It's just that for neutrons, you also have strong forces, altering the total pressure, and you have GR effects, so the equation of state is different from simple degeneracy pressure. But if we are concerned about the core collapse, we don't need to worry about neutrons, since the core collapse initiates when the gas is still electronic.

It is true that there are technically "solutions" in this regime, but they are unstable, like a pencil balanced on its point, so I'm assuming we are not considering them.

No, they are not unstable in the force balance. They are only unstable in the energy equation. If you set the time derivative of the energy to zero, you would have no trouble finding stable force balance in the regime between a white dwarf radius and a neutron star radius, that's what I'm saying.

 

[PeterDonis]

Degeneracy pressure is always the same thing as kinetic pressure, degeneracy pressure is kinetic pressure.

I disagree. See below.

The quantum mechanical effects are seen in the temperature of the gas, not its pressure

No, they aren't. Degeneracy pressure is present even at zero temperature, showing that quantum effects manifest directly in the pressure, with no "kinetic" (thermal) component at all. I realize that some use the terminology of "degeneracy motions" to describe how degeneracy pressure is produced, but this seems like a serious misnomer to me.

If you set the time derivative of the energy to zero, you would have no trouble finding stable force balance in the regime between a white dwarf radius and a neutron star radius, that's what I'm saying.

This amounts to agreeing with me that the solution is unstable against small perturbations (which will prevent the time derivative of energy from ever being exactly zero), which was my point. Yes, you can have a force balance if everything remains exactly the same forever; but that never actually happens. And as soon as you perturb the state slightly from that balanced state, there is no longer a force balance; the object must either expand back to white dwarf size and density or contract down to neutron star size and density. That is, once the state is perturbed, the forces that arise do not tend to restore the balanced state; they tend to increase the perturbation.

 

[Ken G]

Degeneracy pressure is present even at zero temperature, showing that quantum effects manifest directly in the pressure, with no "kinetic" (thermal) component at all.

Not so. You are equating a "kinetic component" to pressure with a "thermal component", but that's not correct. Thermal means that it connects to temperature, but kinetic just means it connects to motion. Degenerate electrons are free particles that have a pressure because they carry a momentum flux, that's kinetic pressure. What's more, if they are nonrelativistic, they obey the standard relation for kinetic pressure, P = 2/3 U, where U is the kinetic energy density. It's not a coincidence that the term "kinetic" appears twice there! That's why I find it so counterproductive for people to characterize degeneracy pressure as some special quantum mechanical non-kinetic pressure, because if I know the kinetic energy density, I know the pressure-- I don't even care if the particles are degenerate or not. Degeneracy is a thermodynamic effect, it influences the temperature that should be associated with a given energy and density, but it does not alter the pressure that should be associated with a given energy and density, that is the pressure that is simply associated with particle motion.

This amounts to agreeing with me that the solution is unstable against small perturbations (which will prevent the time derivative of energy from ever being exactly zero), which was my point. Yes, you can have a force balance if everything remains exactly the same forever; but that never actually happens.

All the same, there is an important difference between an energy-driven instability, and a momentum-driven instability. The issue is whether you find the instability when you allow time-dependent perturbations in the force equation, or in the energy equation. Physically, the distinction is, you have a momentum-driven instability if you "kick" the system (introduce a small force perturbation), and the force equation runs away (subject to energy conservation, typically adiabatic), and you have an energy-driven instability if you remove some kinetic energy and the energy equation runs away (subject to the force balance).

The reason this distinction is important is that the force equation, and the energy equation, often respond on very different timescales, so their associated instabilities will also play out on very different timescales. The timescale of the force equation is typically the sound crossing time (the time for the internal momentum to be appreciably altered by momentum transport), and the timescale of the energy equation is typically the Kelvin-Helmholtz time (the time for the total internal energy to change appreciably by energy transport). Often the latter is fantastically longer than the former, which allows for the whole concept of hydrostatic equilibrium in stars. However, in a core collapse, the timescales of the two become similar, and their distinction gets blurred as a result. All the same, it is a useful distinction to make, which is at the heart of what I've been saying above. In a nutshell, I've been saying that core collapse is the phenomenon you get when the energy equation is able to respond on the same timescale as the force equation, and that only happens because of processes that are very good at removing kinetic energy from the gas-- processes that do not even appear in the derivation of the Chandrasekhar mass, and processes that are never mentioned if one asserts that core collapse happens when the mass gets so high that no hydrostatic solution at zero temperature exists any more (the usual explanation).

And as soon as you perturb the state slightly from that balanced state, there is no longer a force balance; the object must either expand back to white dwarf size and density or contract down to neutron star size and density. That is, once the state is perturbed, the forces that arise do not tend to restore the balanced state; they tend to increase the perturbation.

If you just look at the momentum fluxes, you find that the forces that arise are perfectly stable. You have to look at the energy sinks, and only then do you see the instability. That's the summary of what I'm saying. This means core collapse is not a dynamical instability like a pencil on its point, which runs away with any added momentum. If you only kick the momentum, you don't get runaway of a stellar core, you need to do something to the internal energy that causes kinetic energy to be rapidly lost.

 

[PeterDonis]

You are equating a "kinetic component" to pressure with a "thermal component", but that's not correct.

You're quibbling over words. I was responding to your statement that "quantum mechanical effects are seen in the temperature of the gas, not its pressure" by pointing out that degeneracy pressure is present at zero temperature, so clearly that quantum effect is not seen in the temperature of the gas. It's no answer to that to say that the word "kinetic" doesn't have to mean "thermal".

If you just look at the momentum fluxes, you find that the forces that arise are perfectly stable.

Can you give a reference that discusses this? What you're saying is not my understanding of how the force balance works in this regime (meaning the regime between white dwarf size and density, and neutron star size and density).

 

[Ken G]

You're quibbling over words. I was responding to your statement that "quantum mechanical effects are seen in the temperature of the gas, not its pressure" by pointing out that degeneracy pressure is present at zero temperature, so clearly that quantum effect is not seen in the temperature of the gas. It's no answer to that to say that the word "kinetic" doesn't have to mean "thermal".

It's not a quibble, it's essential to understanding degeneracy pressure. I am well aware that degeneracy pressure exists at zero temperature, my point is that the pressure is kinetic, and it is there because of the presence of kinetic energy. The fact that it exists at zero temperature is because degeneracy drives the temperature to zero, it doesn't do anything to the pressure. Degeneracy causes the kinetic energy that is present because it is a conserved quantity that has a history in that gas to be partitioned among the particles in such a way that has no effect at all on the pressure (if nonrelativistic), but has a very significant effect on the temperature. Hence, the quoted statement is not only correct, it is the very guts of degeneracy. To say that it causes the temperature to be zero certainly does not contradict the claim that its effects are seen in the temperature!

Can you give a reference that discusses this? What you're saying is not my understanding of how the force balance works in this regime (meaning the regime between white dwarf size and density, and neutron star size and density).

No reference is needed, the issue is one of basic physics so I think the real problem is in communicating what I'm saying. Let's look at the force equation as an equation about how momentum gets transported and what it does when ti gets there, and the energy equation as an equation about how energy is transported and what it does when it gets there. I'm saying that if we ignore the energy equation and just focus on the force equation, we see no instability in the domain between a white dwarf and a neutron star. By ignoring the energy equation, I mean we allow no heat transport, and we ignore how changing the kinetic energy might lead to changes in the composition of the gas on the timescale of the sound crossing time. Those are not issues of "dynamical stability", like the stability of an orbit or whether a pencil can stand on its point. Dynamical stability is just about taking the force equation, and sticking some momentum in somewhere, and seeing if it recovers on the sound crossing time. An easier way to test that is to imagine simply squeezing the whole business uniformly, so we reduce the radius, and watch if it bounces back or falls in-- all on the sound crossing time, and ignoring the energy equation except that we are accounting for PdV work (because by the work-energy theorem, that's the energy we put in ourselves as part of our dynamical perturbation, it's not something that requires the energy equation to figure out and it doesn't happen on the timescale of the equilibration of the temperature or composition). So what I'm saying is, if we take a white dwarf and give it a mass that would have its radius be something like 1/10 the radius of a typical white dwarf, say 1/10 the radius of Sirius B, and ignore the energy equation by taking the temperature to be zero (so it's a black dwarf) and considering timescales too short for neutronization (I'm suggesting that the neutronization timescale would still be quite long at that radius, and might even be able to reach an equilibrium anyway, but in either case is an issue for the longer timescales of energy equilibration not the sound crossing-time scale of simple dynamical stability), and you give that object a little squeeze, it will bounce right back.

This claim can be demonstrated easily, because such an object would be below the Chandra mass, so its electrons are not fully relativistic and so the adiabatic index in the force balance is above 4/3. That means it bounces back. If the temperature is above zero, as in a real star, it would only bounce back even more, except that this now brings in additional terms in the energy equation that may lead to rapid instability and are what I have been talking about above.

 

[PeterDonis]

I am well aware that degeneracy pressure exists at zero temperature, my point is that the pressure is kinetic, and it is there because of the presence of kinetic energy.

Once again, reference please? This is not a use of the term "kinetic energy" that I'm familiar with, and I've already pointed out that viewing particles in degenerate states as having "degeneracy motion" is problematic. I would like something more than just your assertion that this viewpoint makes sense.

No reference is needed, the issue is one of basic physics so I think the real problem is in communicating what I'm saying.

Even if it is basic physics, a reference might do a better job of communicating it.

I'm saying that if we ignore the energy equation and just focus on the force equation, we see no instability in the domain between a white dwarf and a neutron star.

I still disagree. See below.

This claim can be demonstrated easily, because such an object would be below the Chandra mass, so its electrons are not fully relativistic and so the adiabatic index in the force balance is above 4/3. That means it bounces back.

Whether or not the electrons are relativistic is not a function of mass, it's a function of density. An object 1/10 the size of a normal white dwarf is 1000 times denser than a normal white dwarf. That makes its electrons extremely relativistically degenerate. (It also makes the neutronization time scale much shorter, btw.)

 

[Ken G]

Once again, reference please? This is not a use of the term "kinetic energy" that I'm familiar with, and I've already pointed out that viewing particles in degenerate states as having "degeneracy motion" is problematic. I would like something more than just your assertion that this viewpoint makes sense.

I don't understand what you are asking, are you asking for a reference that the equation pressure equals 2/3 of kinetic energy density applies to nonrelativistic degeneracy pressure? You will find that in any derivation of degeneracy pressure, it's where the concept comes from. The usual derivation is to solve for the kinetic energy density at which degeneracy produces zero temperature (a single accessible state), and multiply it by 2/3 to get the degeneracy pressure. Alternatively, you can solve for the pressure by equating it to the derivative of the kinetic energy with respect to volume, but you'll get the same answer-- and again it all traces to the kinetic energy. Degeneracy pressure is an absolutely mundane example of kinetic pressure, i.e., pressure that derives from kinetic energy of the particles.

Whether or not the electrons are relativistic is not a function of mass, it's a function of density.

Again the important focus needs to be on kinetic energy. The electrons go degenerate when their kinetic energy density corresponds to what you get when you have removed all the heat that is possible to remove. In a self-gravitating mixture of nonrelativistic protons and electrons, that happens at a kinetic energy density that is proportional to the density to the 5/3 power, divided by the electron mass. If you also know the electron density, you can then find the kinetic energy per electron, and compare that to the rest energy of the electron, to test if you were correct to use the nonrelativistic calculation. So yes, the density appears, but how relativistic something is is still an issue of its kinetic energy, there's nothing magical about including the quantum mechanics here.

An object 1/10 the size of a normal white dwarf is 1000 times denser than a normal white dwarf. That makes its electrons extremely relativistically degenerate. (It also makes the neutronization time scale much shorter, btw.)

But does it make the neutronization time compete with the sound crossing time? I'll bet you it does not. As to how relativistic the electrons are, yes, they are very relativistic. All the same, they are not completely relativistic, so as I said, the adiabatic exponent will be above 4/3. That means it will bounce back, it is dynamically stable-- even if it is very close to marginal stability.