"Gravitational Compression in Neutron Stars"

[PeterDonis]

breaking a bunch of neutrons down into quark-degenerate matter ought to release at least some of the strong-interaction-binding energy that kept the quarks in a baryonic configuration

This reasoning would be valid if quark-degenerate matter were a possible state of matter at zero temperature, as baryonic configurations are. But it isn't; it can only exist to begin with at very high temperatures. Which means that the transition from baryonic configurations to quark-degenerate matter requires an input of energy; it is not a transition that will release energy, whether it's "binding energy" or anything else.

Note also that you can't use the kinetic energy gained in the collapse as the input of energy, at least not permanently, because that energy can be radiated away. In other words, if we took a bunch of neutrons and forced them to collapse somehow, and were able to control the process so that the kinetic energy of the collapse converted everything into quark-degenerate matter, the quark-degenerate matter could still emit radiation and convert right back into neutrons, since neutrons are a lower energy configuration.

why can't quarks released by the collapsing neutrons be bound in quark-degenerate or strange matter?

Because those configurations are higher energy configurations than neutrons. See above.

I'm assuming that 21 quarks bound together in quark-degenerate plasma is going to have a lower binding energy than 7 neutrons

And that is the incorrect assumption that is leading you astray. See above.

 

[PeterDonis]

Maybe core neutrons collapse into 1% quark matter with 99% energy that result in super intense X-rays which could result in intense positron/electron production.

As Jonathan Scott says, this would violate baryon number conservation. But we can look at it even without reference to baryons. Neutrons are composed of three quarks--one up quark and two down quarks. There are no antiquarks. So there is no way to convert 99% of the neutron's mass into energy; that would require matter-antimatter annihilation, i.e., quark-antiquark annihilation.

You could reduce a neutron's mass by a percent or two via weak interactions, converting it into a proton, if the neutron were in free space and not bound into a neutron star. But in neutron stars, neutrons are stable; weak interactions to convert them into protons would require an input of energy, and the resulting protons would just turn right back into neutrons, because the neutrons are a lower energy configuration. (As my response to sevenperforce should make clear, under the conditions inside a neutron star, neutrons are the lowest energy configuration possible, as far as I know.)

I have deleted all posts relative to ultra-relativistic jets; they are an interesting topic but not relevant to this discussion. Bernie G, please start a separate thread if you want to discuss the jets further, and please refrain from speculation.

 

[sevenperforce]

It might be helpful to take a step back and look at the starting premise of this thread:

We have to find a plausible scenario for a neutron star collapsing into a black hole. One such scenario would be a neutron star that is below the maximum mass limit, but not by much, accreting enough mass onto it to push it over the limit (for example, the neutron star could be in a binary system with a massive companion and material from the companion could fall onto the neutron star). If that scenario seems ok to everyone, then further discussion can be based on it.

Sounds good to me.

This reasoning would be valid if quark-degenerate matter were a possible state of matter at zero temperature, as baryonic configurations are. But it isn't; it can only exist to begin with at very high temperatures. Which means that the transition from baryonic configurations to quark-degenerate matter requires an input of energy; it is not a transition that will release energy, whether it's "binding energy" or anything else.

Ah. Quark-degenerate matter is a higher-energy state than baryonic matter. Got it!

No, this won't work. It is true that, if we look at the event horizon in a spacetime where an object like a neutron star (or an ordinary star) is collapsing to a black hole, the horizon forms at the center, $r = 0$, and moves outward until it reaches the Schwarzschild radius associated with the total mass of the object. But that does not mean the mass of the black hole starts at zero and slowly grows; what it means is that, until all of the matter in the object has collapsed below the event horizon, there is no clean way to separate the "black hole" from "the rest of the object".

What's more, you can't even have a static system with a radius just a little bit larger than the Schwarzschild radius associated with its mass. There is a theorem called Buchdahl's theorem which says that the minimum radius that any static system can have is 9/8 of the Schwarzschild radius associated with its mass. That means there is a finite "gap" between an object being stable in a static configuration and an object being a black hole; there is no continuous sequence of static configurations with gradually increasing mass that suddenly turns into black holes without any collapse in between.

Well, in this case, we'd obviously have a collapse happening, so it's definitely not a static system.

I'd also note that the Schwarzschild radius won't "form" at the center and grow outward, because a neutron star already has a pretty significant Schwarzschild radius. Heck, R_S for the Earth is already about half an inch; a two-solar-mass neutron star has a Schwarzschild radius at a little over half of its own radius. When such a neutron star collapses, the Schwarzschild radius will remain constant while the outer layers fall into it; once they've all fallen in, it's a black hole.

However, the core is a good deal denser than the average density of the neutron star, so in some instances the Schwarzschild radius of the core alone may be a greater proportion of the core radius than for the neutron star as a whole. Thus, if the core begins to collapse first (as would be expected), it's possible that the core will fall within its own Schwarzschild radius before the entire neutron star falls within its total Schwarzschild radius. At that point, you would have a core-mass black hole surrounded by a collapsing shell of neutron-star mantle-and-crust. That's the case where you would have a black hole which is "separate" from the rest of the object.

If the collapse of the core causes the density gradient to increase rapidly enough, then you could have the case where the inner core falls within its own Schwarzschild radius before the outer core does. Depending on the shape of that density gradient, the initial black hole could have an arbitrarily low mass, which leads to my suggestion.

I've also been discussing this in this thread that I actually started myself; I'd love to get your input over there.

When the event horizon forms at $r = 0$ and starts moving outward, it won't be producing Hawking radiation (at least according to our best understanding of Hawking radiation), for at least two reasons. First, the horizon is not in vacuum--it is embedded in the collapsing matter. The derivation of Hawking radiation being emitted from a horizon assumes vacuum. Second, the horizon is not a trapped surface--in other words, its area is not constant. The area of the horizon grows until all the collapsing matter has fallen inside it. The derivation of Hawking radiation, if you look at the details, assumes that the horizon is a trapped surface--that its area is not growing.

So this proposed mechanism for stopping a black hole from forming, at least if we use the current understanding of Hawking radiation, won't work. However, it should be noted that our current understanding of Hawking radiation and how it is produced might not be correct.

Indeed. I know that Hawking's equations were "set" using the model of a stable black hole in a vacuum, but there is no vacuum (since the CMBR is always causing SOMETHING to fall into the black hole) and there are no stable black holes (because, Hawking radiation).

Moreover, while the observed match between the spectrum of thermally-radiating objects and the predictions of Planck's Law are always statistical in nature, Hawking radiation is unique in that it is predicts an exact match to Planck's Law. The reason that Planck's Law matched statistical observations, of course, was due to the quantization of energy levels in thermal emission. So even though Hawking's predictions cannot be derived with the sort of micro-black-hole that we're talking about here, I'm still interested in whether applying Hawking's predictions to a Planck-mass model could yield a statistical match to Hawking's predictions for macroscopic black holes, and thus serve as a starting point for a model of quantum gravity, or at least black hole quantization.

 

[PeterDonis]

I'd also note that the Schwarzschild radius won't "form" at the center and grow outward

I didn't say "Schwarzschild radius". I said "event horizon". They're not the same. See below.

a neutron star already has a pretty significant Schwarzschild radius

But it doesn't have an event horizon at all.

When such a neutron star collapses, the Schwarzschild radius will remain constant while the outer layers fall into it

The Schwarzschild radius remaining constant is just another way of saying the externally measured mass remains constant. (This assumes that no radiation is emitted during the collapse process, which is highly unlikely in the real world, but we can assume it for this thought experiment.) It does not mean that there is an event horizon sitting there waiting for things to fall in.

What actually happens, as I said before, is that, as the star collapses, an event horizon forms at the very center, at $r = 0$, and moves outward (increases in radius) as the collapse continues. At the point where the surface of the star is just passing through the horizon, the horizon is just at the Schwarzschild radius corresponding to the mass of the system--and once there, it stays there.

Actually, though, this way of putting things can be misleading. A better way to put it starts with recognizing the definition of the event horizon: it is the boundary of the region of spacetime from which light signals cannot escape. So what is actually happening is that there is a particular event at $r = 0$ such that, if an outgoing light signal is emitted from that event, it will intersect the surface of the collapsing star just at the Schwarzschild radius corresponding to the mass of the system, and will then be trapped there forever, unable to move any further outward. The event horizon is just the set of all possible outgoing light rays emitted from that event at $r = 0$, in all possible directions; they will all intersect the star's surface at the same radius (because we are assuming spherically symmetric collapse), and will all be trapped there, forming a 2-sphere of outgoing light rays that stays at that same radius forever.

I would advise rethinking the rest of your proposed scenarios (with varying densities of core vs. outer parts, etc.) in the light of the above.

I'd love to get your input over there.

I'll take a look.

there is no vacuum (since the CMBR is always causing SOMETHING to fall into the black hole)

Yes. This is usually taken to mean that no black hole can actually emit Hawking radiation until the temperature of the CMBR falls below its Hawking temperature, which will take something like $10^{67}$ years for a one solar mass black hole (and far longer for the much larger holes at the centers of galaxies and quasars). But nobody has actually done a rigorous analysis of this, as far as I know; it's just the obvious heuristic guess based on what we currently know.

applying Hawking's predictions to a Planck-mass model

I'm not sure what you are thinking of here. Hawking's prediction for a Planck mass black hole is that it will evaporate immediately, with no time lapse--i.e., that such a hole can't really exist since it will evaporate as soon as it is formed. This is not something that can be usefully analyzed statistically, as far as I can see.

 

[sevenperforce]

The Schwarzschild radius remaining constant is just another way of saying the externally measured mass remains constant. (This assumes that no radiation is emitted during the collapse process, which is highly unlikely in the real world, but we can assume it for this thought experiment.) It does not mean that there is an event horizon sitting there waiting for things to fall in.

Sure, I get that.

Actually, though, this way of putting things can be misleading. A better way to put it starts with recognizing the definition of the event horizon: it is the boundary of the region of spacetime from which light signals cannot escape. So what is actually happening is that there is a particular event at $r = 0$ such that, if an outgoing light signal is emitted from that event, it will intersect the surface of the collapsing star just at the Schwarzschild radius corresponding to the mass of the system, and will then be trapped there forever, unable to move any further outward.

I would advise rethinking the rest of your proposed scenarios (with varying densities of core vs. outer parts, etc.) in the light of the above.

I'm not quite sure how this would change the scenario. If there is a particular event at $r = 0$ such that an outgoing light signal emitted from that event would intersect the surface of the collapsing core just at the Schwarzschild radius corresponding to the mass of the core, then you have a core-mass black hole already inside the collapsing neutron star. Similarly, if there is a particular event at $r = 0$ such that an outgoing light signal emitted from that event would intersect the inner-core/outer-core boundary just at the Schwarzschild radius corresponding to the mass of the inner core, then you have an inner-core-mass black hole at the center of the collapsing core.

Hawking's prediction for a Planck mass black hole is that it will evaporate immediately, with no time lapse--i.e., that such a hole can't really exist since it will evaporate as soon as it is formed. This is not something that can be usefully analyzed statistically, as far as I can see.

Well, any analysis might be completely pointless if Hawking radiation predictions break down at a larger scale, but if they don't, then there might be a useful statistical analysis of what would happen as the Planck scale is approached, even if we're not dealing specifically with the Planck mass. For instance, trying to derive the minimum mass by looking at where the math would no longer make sense, like when the peak wavelength of emitted radiation would correspond to a particle energy exceeding half the energy of the object.

 

[PeterDonis]

If there is a particular event at r=0r=0r = 0 such that an outgoing light signal emitted from that event would intersect the surface of the collapsing core just at the Schwarzschild radius corresponding to the mass of the core, then you have a core-mass black hole already inside the collapsing neutron star.

Yes, you do. But it wasn't sitting there all the time at the neutron star's Schwarzschild radius. It formed as a part of the collapse process, started at zero radius, and gradually increased to the Schwarzschild radius. And if the rest of the star is going to collapse as well, then the hole won't stay at the Schwarzschild radius of the core; it will keep expanding until all the matter has fallen inside the Schwarzschild radius for the whole star. In other words, yes, while the collapse is happening, you will be able to look at it as a black hole being inside a collapsing star; but it won't be a static black hole inside a collapsing star. So you can't use intuitions that are only valid for static holes, for example about "accretion of matter". Matter is collapsing, and during the collapse more and more matter is inside the growing horizon, but this process is different from the process of accretion of matter onto a hole that has been sitting there static for a long time, surrounded by vacuum, and then suddenly has a large amount of matter fall into it.

A key thing to keep in mind here is that the definition of the event horizon is inherently "teleological"; that is, it depends on what is going to happen in the future. In other words, there is no way to tell locally where the horizon is; to know where it is, you have to know the entire future of the spacetime. So your normal intuitions about objects don't work; you can't think of the horizon as something that is forming because of what already happened. It is forming because of what is going to happen--because all the matter is going to collapse inside the Schwarzschild radius corresponding to its mass. So you can't have a scenario where only part of the matter falls in and then you have a static hole, because if only part of the matter falls in, either the horizon won't form at all, or it won't stay static (as above).

trying to derive the minimum mass by looking at where the math would no longer make sense

We already know at least a heuristic answer to this: the minimum mass is the Planck mass. There is nothing in the math that shows any problem for any hole larger than that, at least as we understand it today.

 

[sevenperforce]

If the rest of the star is going to collapse as well, then the hole won't stay at the Schwarzschild radius of the core; it will keep expanding until all the matter has fallen inside the Schwarzschild radius for the whole star. In other words, yes, while the collapse is happening, you will be able to look at it as a black hole being inside a collapsing star; but it won't be a static black hole inside a collapsing star. So you can't use intuitions that are only valid for static holes, for example about "accretion of matter". Matter is collapsing, and during the collapse more and more matter is inside the growing horizon, but this process is different from the process of accretion of matter onto a hole that has been sitting there static for a long time, surrounded by vacuum, and then suddenly has a large amount of matter fall into it.

Just so I'm sure we're on the same page -- in the example case, if some event outside the event horizon suddenly arrested the collapse of the outer layers and blasted them away from the growing event horizon, the object left behind would be a static black hole, correct?

A key thing to keep in mind here is that the definition of the event horizon is inherently "teleological"; that is, it depends on what is going to happen in the future. In other words, there is no way to tell locally where the horizon is; to know where it is, you have to know the entire future of the spacetime. So your normal intuitions about objects don't work; you can't think of the horizon as something that is forming because of what already happened. It is forming because of what is going to happen--because all the matter is going to collapse inside the Schwarzschild radius corresponding to its mass. So you can't have a scenario where only part of the matter falls in and then you have a static hole, because if only part of the matter falls in, either the horizon won't form at all, or it won't stay static (as above).

As I understand it, this is the problem that Hawking himself runs into with the firewall problem. If Hawking radiation is being produced at or just above the event horizon of a black hole, then an infalling observer would definitely notice the event horizon locally, which isn't actually allowed.

If the core is dense enough, and collapses fast enough, then I would presume it is possible that it can be swallowed up by an event horizon corresponding to its own Schwarzschild radius apart from the separate, larger event horizon corresponding to the Schwarzschild radius of the object as a whole, no?

We already know at least a heuristic answer to this: the minimum mass is the Planck mass. There is nothing in the math that shows any problem for any hole larger than that, at least as we understand it today.

Correct me if I'm wrong, but I thought the Planck mass was the maximum mass for a point particle, since any point particle with a mass greater than the Planck mass will be a black hole.

It is narrowly possible to have a black hole which is smaller than the Planck mass but is still within the boundaries of other Planck-scale values. For example, as in the other thread, a black hole of 0.75 Planck masses will have a Schwarzschild radius of 1.5 Planck lengths and an evaporation lifetime which is safely above the Planck time.

 

[PeterDonis]

in the example case, if some event outside the event horizon suddenly arrested the collapse of the outer layers and blasted them away from the growing event horizon, the object left behind would be a static black hole, correct?

In other words, enough material collapses to form a black hole, just less than the original total mass? Yes, if we assume everything else got radiated away and didn't fall in, what was left behind would be a static black hole with the mass of whatever did fall in.

But a key assumption in your quote above is embodied in the words "the growing event horizon". By specifying that, you are specifying that enough matter is going to fall into make a black hole. So given that specification, it's impossible for a black hole not to form, because you already specified that it did.

To see why this matters, consider an alternate scenario: a large mass is imploding, say 5 solar masses; 1.5 solar masses worth has collapsed inside the Schwarzschild radius for 5 solar masses; but before the rest of the 5 solar masses can fall in, something arrests the collapse and blasts the rest of the mass away. What will be left behind will not be a black hole; it will be a 1.5 solar mass neutron star. And in this case, no horizon will ever form at the center at $r = 0$. Even if the density there gets higher, temporarily, than the central density of the final 1.5 solar mass neutron star, that won't be sufficient to form a black hole.

In other words, the formation of the event horizon is not dictated by the density at $r = 0$; the rule isn't that when that density reaches a particular value, the horizon forms. The rule is that the horizon forms if enough matter is going to collapse to make a black hole. Event horizon formation is not a local process, and you can't analyze it in terms of local variables like the density; that's not how it works.

I thought the Planck mass was the maximum mass for a point particle, since any point particle with a mass greater than the Planck mass will be a black hole.

I emphasize that we are talking heuristic speculation here; nobody has a firm theory for this. But yes, according to the speculative viewpoint you are describing, the Planck mass is the maximum possible mass for a point particle and the minimum possible mass for a black hole. The latter is what I said.

The problem with this, from a GR viewpoint, is that we don't have a firm model of what a point particle would be like in terms of classical spacetime curvature. Some physicists have suggested using the "super-extremal" Kerr-Newman spacetime geometries for this (these geometries describe rotating, charged black holes when they are "sub-extremal", i.e., when their mass is larger than their charge + spin, in geometric units; but "super-extremal" Kerr-Newman geometries describe naked singularities, with no event horizons, that have spin and charge). But there are a number of problems with this approach, one of which is that these geometries are unstable against small perturbations, so this model would predict that elementary particles could not be stable, but would be quickly destroyed by the smallest fluctuation in their environment.

It is narrowly possible to have a black hole which is smaller than the Planck mass but is still within the boundaries of other Planck-scale values. For example, as in the other thread, a black hole of 0.75 Planck masses will have a Schwarzschild radius of 1.5 Planck lengths and an evaporation lifetime which is safely above the Planck time.

Again, I emphasize that all this is heuristic speculation; we do not have a firm theory that says this is how Planck scale physics work. All of these calculations should be taken with a huge helping of salt. They don't really tell us anything except that the Planck scale appears to be the scale at which we expect new physics to emerge. But the Planck scale is twenty orders of magnitude smaller than the smallest scale we can probe with current experiments, so we're not likely to get any data on the matter any time soon.

 

[sevenperforce]

In other words, enough material collapses to form a black hole, just less than the original total mass? Yes, if we assume everything else got radiated away and didn't fall in, what was left behind would be a static black hole with the mass of whatever did fall in.

But a key assumption in your quote above is embodied in the words "the growing event horizon". By specifying that, you are specifying that enough matter is going to fall into make a black hole. So given that specification, it's impossible for a black hole not to form, because you already specified that it did.

My confusion/exception is over the highlighted bit above. Talking about "enough material" doesn't quite make sense because there isn't a minimum-mass black hole, at least not on these scales. There is a minimum mass for a static object to collapse into a black hole, but I'm talking about a collapse which has already physically begun.

To see why this matters, consider an alternate scenario: a large mass is imploding, say 5 solar masses; 1.5 solar masses worth has collapsed inside the Schwarzschild radius for 5 solar masses; but before the rest of the 5 solar masses can fall in, something arrests the collapse and blasts the rest of the mass away. What will be left behind will not be a black hole; it will be a 1.5 solar mass neutron star. And in this case, no horizon will ever form at the center at $r = 0$. Even if the density there gets higher, temporarily, than the central density of the final 1.5 solar mass neutron star, that won't be sufficient to form a black hole.

Slight modification of that scenario. A large mass is imploding, say 5 solar masses; 1.5 solar masses worth has collapsed inside the Schwarzschild radius for 1.5 solar masses before the rest of the 5 solar masses has collapsed inside the Schwarzschild radius for 5 solar masses, and before the rest of the 5 solar masses can fall in, something arrests the collapse. Is that a possible scenario? If so, that's precisely the scenario I've been interested in from the beginning.

Again, I emphasize that all this is heuristic speculation; we do not have a firm theory that says this is how Planck scale physics work. All of these calculations should be taken with a huge helping of salt. They don't really tell us anything except that the Planck scale appears to be the scale at which we expect new physics to emerge. But the Planck scale is twenty orders of magnitude smaller than the smallest scale we can probe with current experiments, so we're not likely to get any data on the matter any time soon.

Probably not. However, if quantization at the Planck scale produces a model which matches macroscopic behavior on measurable scales (for example, Hawking radiation as tunneling-decay of Planck-scale quantized black holes generating a statistical blackbody curve matching Hawking's predictions for macroscopic black hole thermal radiation spectra), that's useful.

 

[PeterDonis]

Talking about "enough material" doesn't quite make sense because there isn't a minimum-mass black hole, at least not on these scales.

There isn't a minimum mass black hole in the sense that there exists a stable solution to the Einstein Field Equation describing a black hole with a mass of, say, 1/10 the mass of the Sun, yes.

But there is a minimum mass black hole in the sense that there is no feasible way for an object under the minimum mass limit for a neutron star to collapse to a black hole. That's what I meant when I said "enough material"--there has to be enough mass to be over the maximum mass limit for a neutron star. (The limit for a neutron star is the important one because it is the largest maximum mass limit for any stable configuration that isn't a black hole.)

1.5 solar masses worth has collapsed inside the Schwarzschild radius for 1.5 solar masses before the rest of the 5 solar masses has collapsed inside the Schwarzschild radius for 5 solar masses, and before the rest of the 5 solar masses can fall in, something arrests the collapse. Is that a possible scenario?

Yes, but not quite as you describe it. Remember Buchdahl's theorem and the 9/8 factor? If the rest of the 5 solar masses is going to be stopped from collapsing, it has to be stopped before it falls inside 9/8 of the Schwarzschild radius for 5 solar masses. Otherwise what you have is not a 1.5 solar mass black hole with matter outside it, but a 5 solar mass black hole in the process of forming; there is no intermediate configuration which can be described as a stable 1.5 solar mass black hole.

This is the reason why I emphasized that there has to be a region of empty space between the core that collapses to a black hole and the rest of the matter that is somehow stopped from collapsing. The two regions of matter can't be continuous, because if they are, the outer one can't be outside the Buchdahl's theorem limit.

 

[PeterDonis]

Hawking radiation as tunneling-decay of Planck-scale quantized black holes generating a statistical blackbody curve matching Hawking's predictions for macroscopic black hole thermal radiation spectra

I've never seen anything like this described in a peer-reviewed paper. Have you? On its face it doesn't seem feasible; to get a prediction for macroscopic black holes, you have to look at macroscopic black holes, not Planck scale ones. A model of Planck scale black holes can only give you information about the spectrum of Planck scale holes. A macroscopic black hole is not just a huge number of Planck scale black holes mashed together.

"Planck scale physics" is not an all-purpose get out of jail free card that let's you say whatever you want. It's just a heuristic guess about the way a correct quantum gravity theory is going to look when we find it. Heuristic guesses like that have been wrong before.

 

[sevenperforce]

But there is a minimum mass black hole in the sense that there is no feasible way for an object under the minimum mass limit for a neutron star to collapse to a black hole. That's what I meant when I said "enough material"--there has to be enough mass to be over the maximum mass limit for a neutron star. (The limit for a neutron star is the important one because it is the largest maximum mass limit for any stable configuration that isn't a black hole.)

Presumably, the notable exception to this rule would be if some smaller amount of mass were imploded (perhaps by shockwaves in a collapsing star or supernova) at a great enough speed that it collapsed into its own Schwarzschild radius on its own (or at least substantially before the outer shell had done so).

I've never seen anything like this described in a peer-reviewed paper. Have you? On its face it doesn't seem feasible; to get a prediction for macroscopic black holes, you have to look at macroscopic black holes, not Planck scale ones. A model of Planck scale black holes can only give you information about the spectrum of Planck scale holes. A macroscopic black hole is not just a huge number of Planck scale black holes mashed together.

I haven't seen this described, no. But I figured it would be potentially useful to look at whether modeling a macroscopic black hole as a huge number of Planck scale black holes on the surface of their collective event horizon would match any of the predictions for the spectrum of a macroscopic black hole. Physics has a pretty good record of explaining otherwise-anomalous behavior by figuring out the right scale at which to quantize it (e.g., photoelectric effect). No promise of results, but worth casual investigation.

 

[PeterDonis]

Presumably, the notable exception to this rule would be if some smaller amount of mass were imploded (perhaps by shockwaves in a collapsing star or supernova) at a great enough speed that it collapsed into its own Schwarzschild radius on its own (or at least substantially before the outer shell had done so).

If the shock waves end up separating the matter into two disconnected regions, yes, you could look at this as forming a smaller black hole and then having a second shell of matter fall into it (or not, if the shock waves end up blasting the rest of the matter outward fast enough). But the regions have to be disconnected; otherwise, as I said before, you just have a larger black hole in the process of forming.

modeling a macroscopic black hole as a huge number of Planck scale black holes on the surface of their collective event horizon

This doesn't make sense as you state it. You might be thinking of a model in which the horizon of a large black hole is modeled as a configuration of a large number of area "quanta", where each quantum of area is 1/4 of the Planck area (the factor 1/4 comes from the calculations by Hawking and Bekenstein in the 1970s, which have been replicated by all other models since). But this doesn't mean the large hole is modeled as a large number of Planck scale holes. Heuristically, it means a Planck scale hole is a quantum state with "area number" (the number of area quanta) equal to one, while a macroscopic hole is a quantum state with an area number that is very, very large (on the order of $10^{76}$ for a solar mass hole).

As far as whether this model predicts anything useful for macroscopic black holes, AFAIK it is consistent with the previously known formula for Hawking radiation. But that formula itself has not been tested by observation, and there is no expectation that it will be so tested any time soon. So all we really have is two different proposals for quantum gravity that happen to agree on one theoretical prediction.

No promise of results, but worth casual investigation.

Proposals along these lines have already gotten a lot more than "casual investigation". There are a lot of theoretical physicists working on quantum gravity--string theory, loop quantum gravity, and other more esoteric proposals, and that has been the case for at least two decades and arguably three. None of the ideas you are suggesting are new. In fact that's the problem with quantum gravity research: all that effort and the basic heuristic model is still the same one that Hawking and Bekenstein came up with more than forty years ago. (AFAIK neither of them actually articulated the "area quantum" heuristic I described above, but it's obvious once you have Bekenstein's formula saying that a black hole's entropy is proportional to the horizon area.)

 

[sevenperforce]

If the shock waves end up separating the matter into two disconnected regions, yes, you could look at this as forming a smaller black hole and then having a second shell of matter fall into it (or not, if the shock waves end up blasting the rest of the matter outward fast enough). But the regions have to be disconnected; otherwise, as I said before, you just have a larger black hole in the process of forming.

Then...getting back to the OP in this thread...

...if a neutron star exceeding the ~2 M_☉ limit collapses, and this collapse takes place in such a way that the inner core is accelerated inward more rapidly than the rest of the neutron star, then a gap would form. If the imploding portion of the detached inner core was small enough, then it is possible that the output of its Hawking radiation could arrest of the collapse of the rest of the object and blast it away in a hypernova, then subsequently evaporate entirely.

This could explain the mass gap between neutron stars and black holes, as not even a neutron star which accretes to collapse would result in a black hole between 2 and 5 solar masses. It would probably also predict some pretty specific hypernova light curves that we could watch for.

 

[PeterDonis]

if a neutron star exceeding the ~2 M☉ limit collapses, and this collapse takes place in such a way that the inner core is accelerated inward more rapidly than the rest of the neutron star, then a gap would form

Possible, yes.

If the imploding portion of the detached inner core was small enough, then it is possible that the output of its Hawking radiation could arrest of the collapse of the rest of the object and blast it away in a hypernova, then subsequently evaporate entirely.

Nope. The inner core would still have to be larger than the maximum mass limit for a neutron star; otherwise it would just become a smaller neutron star, not a black hole. Hawking radiation from a black hole of that mass is completely negligible.

This could explain the mass gap between neutron stars and black holes

No, it couldn't. See above.

 

[sevenperforce]

You might be thinking of a model in which the horizon of a large black hole is modeled as a configuration of a large number of area "quanta", where each quantum of area is 1/4 of the Planck area (the factor 1/4 comes from the calculations by Hawking and Bekenstein in the 1970s, which have been replicated by all other models since).

Yeah, I'm familiar with that model, at least in passing. I was going in a slightly different direction, though.

If Hawking's predictions are correct all the way down to the Planck scale, then there would have to be a minimum-mass black hole not based on the Planck mass, but based on the point at which the Hawking radiation particles would have an energy equal to half the mass-energy of the black hole. As I stated above (or maybe it was in the other thread), Hawking's model predicts an exact thermal blackbody curve for black holes, but (as far as I can tell) doesn't really provide a mechanism. However, if such a minimum-mass black hole (and corresponding maximum-energy Hawking radiation wavelength peak) exists, then we could examine whether macroscopic black-hole Hawking radiation spectra could be produced by redshifted and diffracted quantized emission at that wavelength. It is (again, as far as I can tell) a possible way of quantizing black hole behavior that hasn't yet been proposed.

 

[sevenperforce]

If the imploding portion of the detached inner core was small enough, then it is possible that the output of its Hawking radiation could arrest of the collapse of the rest of the object and blast it away in a hypernova, then subsequently evaporate entirely.

Nope. The inner core would still have to be larger than the maximum mass limit for a neutron star; otherwise it would just become a smaller neutron star.

But not if the initial stage of collapse caused it to implode rapidly enough, right?

For example, if the inner core comprised a region of quark-gluon plasma on the other of a few thousand tonnes, very near the 9/8 limit of stability, and the shockwave when the mass limit was finally exceeded compressed this core at a speed far exceeding the speed of collapse for the rest of the core, then it would collapse into a very small black hole and produce enough radiation pressure from Hawking radiation to arrest and reverse the collapse of the rest of the neutron star.

 

[PeterDonis]

sevenperforce, at this point you are just speculating, and since you are giving no theory or math to back up your speculations, I can't really give any useful comments. If you are really interested in these topics, you need to spend some time going through the literature on quantum gravity and numerical simulations of gravitational collapse, to see if there are any actual mathematical models for the scenarios you are proposing. Otherwise we're just indulging in handwaving.

If Hawking's predictions are correct all the way down to the Planck scale, then there would have to be a minimum-mass black hole not based on the Planck mass, but based on the point at which the Hawking radiation particles would have an energy equal to half the mass-energy of the black hole.

I don't know where you are getting this from. If you have a peer-reviewed paper that shows how this result is derived, or if you can give the derivation yourself, then please show your work. Otherwise, as above, you're just handwaving.

not if the initial stage of collapse caused it to implode rapidly enough, right?

Have you tried to calculate how rapidly "rapidly enough" would be and how small the resulting black hole could be for a given speed of collapse? You might try looking into the literature on primordial black holes and what conditions would be required to form them, so you can come up with some actual calculations. Otherwise, again, you're just handwaving.

 

[PeterDonis]

Have you tried to calculate how rapidly "rapidly enough" would be and how small the resulting black hole could be for a given speed of collapse?

As an example of such a calculation, consider that the Hawking temperature of a black hole is about $6 \times 10^{-8} \times M_\text{S} / M$ K, where $M_\text{S}$ is one solar mass. If we suppose that a "hypernova" of some sort requires a temperature of, say, $6 \times 10^8$ K (probably a significant underestimate) in order to blast the upper layers of a collapsing star away and prevent them from falling in with the core, then a hole mass of about $10^{-16} M_\text{S}$ would be required, or about $10^{14}$ kg.

What density does this correspond to? We can't just calculate the Euclidean volume of a sphere with radius equal to the Schwarzschild radius for the above mass, because space in this case is not Euclidean. But we can estimate the density using the math of the Oppenheimer Snyder model, at the instant when collapsing matter of this mass would just be falling through its Schwarzschild radius. The key is that the matter region in this model looks like a portion of a closed universe that is collapsing toward a "Big Crunch", and we can calculate the 3-volume of such a closed universe if we know the scale factor and the range of spatial coordinates that it occupies.

The formulas for these at the instant when the collapse starts are given in the page on the O-S model that I linked to earlier:

$$
A_0 = \sqrt{\frac{R_0^3}{2M}}
$$
$$
\sin \chi_0 = \sqrt{\frac{2M}{R_0}}
$$

where $M$ is the mass of the collapsing matter and $R_0$ is the radial coordinate of its surface at the instant the collapse starts. Since we are talking about the collapse of a pre-existing neutron star, we should have $R_0$ as a fairly small multiple of the Schwarzschild radius $2M$; I will assume $R_0 = 6M$ here. Then we have $A_0 = 6 \sqrt{3} M$ and $\sin \chi_0 = \sqrt{1/3}$, which gives $\chi_0 = 0.616$.

Next, we compute what $A$ is when the surface of the matter reaches $R = 2M$. We have

$$
R = \frac{1}{2} R_0 \left( 1 + \cos t \right) = \frac{1}{3} R_0
$$

which gives $1 + \cos t = 2/3$, and

$$
A = \frac{1}{2} A_0 \left( 1 + \cos t \right) = \frac{1}{2} A_0 \frac{2}{3} = \frac{1}{3} A_0 = 2 \sqrt{3} M
$$

which is nice and neat. Then we just need to integrate over the spatial metric to get the 3-volume of a spatial slice:

$$
V = \int_0^{\chi_0} A^3 d\chi 4 \pi \sin^2 \chi = 4 \pi A^3 \frac{1}{2} \left( \chi_0 - \sin \chi_0 \right)
$$

which evalutes to

$$
V = 48 \pi \sqrt{3} M^3 \left( \chi_0 - \sin \chi_0 \right)
$$

This gives $V \approx 9.96 M^3$, but $M$ here is in geometric units, so we need to add a factor of $(G / c^2)^3$ to obtain $V = 4.07 \times 10^{-81} M^3$, where $M$ now is in kilograms. So the density will be $M / V = 10^{81} / 4.07 M^2 \approx 10^{52}$ kg per cubic meter.

I'll leave it to you to compare this to typical neutron star or even quark-gluon plasma densities to figure out how likely it is that any kind of shock wave could compress matter to that density without being successfully resisted by the pressure of the matter.

 

[PeterDonis]

we should have $R_0$ as a fairly small multiple of the Schwarzschild radius $2M$; I will assume $R_0 = 6M$ here

You might wonder what happens if we assume a much larger value for $R_0$, corresponding to the collapse of an ordinary star rather than a neutron star. The answer is that it makes the density we are looking for even higher!

Let's work this through for $R_0 / 2M = 10^6$, which is a typical value for an ordinary star. This gives $A_0 = 10^3 R_0$ and $\sin \chi_0 = 10^{-3}$, which gives $\chi_0 - \sin \chi_0 \approx 2 \times 10^{-10}$. Looking at the equation for $A$, we see that it can be expressed as

$$
A = \frac{A_0}{R_0} R
$$

which gives $A = 2 \times 10^3 M$ when $R = 2M$. So the proper volume for the mass $M$ is now

$$
V = 2 \pi A^3 \left( \chi_0 - \sin \chi_0 \right) \approx 2.51 M^3
$$

which is smaller by almost a factor of 4 than the value in my previous post, and therefore means a density almost a factor of 4 higher.

 

[sevenperforce]

If Hawking's predictions are correct all the way down to the Planck scale, then there would have to be a minimum-mass black hole not based on the Planck mass, but based on the point at which the Hawking radiation particles would have an energy equal to half the mass-energy of the black hole.

I don't know where you are getting this from. If you have a peer-reviewed paper that shows how this result is derived, or if you can give the derivation yourself, then please show your work. Otherwise, as above, you're just handwaving.

Certainly. I was speaking qualitatively for the sake of brevity up until now, but the math is definitely where things get interesting.

Extrapolating Hawking's predictions all the way down to the Planck scale for very, very small black holes will run you into nonsense pretty fast because the power output goes to infinity as the mass goes to zero. It may be, of course, that Hawking's equations simply no longer hold well before this point is reached. But if they do hold true all the way down, then the very last step of black hole evaporation must necessarily involve the entire remaining mass-energy of the black hole being released at once. However, Hawking's model requires that the radiation match a blackbody curve corresponding to the black hole temperature. Since the wavelength is also going to zero as the mass goes to zero, the particle energy is necessarily going to go to infinity, and so the minimum possible mass will correspond to the point where the peak wavelength of the Hawking radiation corresponds to a particle energy E_p = hc/λ(T_H) matching the total mass-energy of the black hole E = Mc²:

hc/λ(T_H) = Mc² or M = h/λc(T_H)

For the sake of momentum conservation, however, this has to be adjusted; you actually need two particles, each with half the mass-energy of the black hole. I did the math in the other thread here.

 

[sevenperforce]

As an example of such a calculation, consider that the Hawking temperature of a black hole is about $6 \times 10^{-8} \times M_\text{S} / M$ K, where $M_\text{S}$ is one solar mass. If we suppose that a "hypernova" of some sort requires a temperature of, say, $6 \times 10^8$ K (probably a significant underestimate) in order to blast the upper layers of a collapsing star away and prevent them from falling in with the core, then a hole mass of about $10^{-16} M_\text{S}$ would be required, or about $10^{14}$ kg.

What density does this correspond to? We can't just calculate the Euclidean volume of a sphere with radius equal to the Schwarzschild radius for the above mass, because space in this case is not Euclidean. But we can estimate the density using the math of the Oppenheimer Snyder model, at the instant when collapsing matter of this mass would just be falling through its Schwarzschild radius. The key is that the matter region in this model looks like a portion of a closed universe that is collapsing toward a "Big Crunch", and we can calculate the 3-volume of such a closed universe if we know the scale factor and the range of spatial coordinates that it occupies.

This gives $V \approx 9.96 M^3$, but $M$ here is in geometric units, so we need to add a factor of $(G / c^2)^3$ to obtain $V = 4.07 \times 10^{-81} M^3$, where $M$ now is in kilograms. So the density will be $M / V = 10^{81} / 4.07 M^2 \approx 10^{52}$ kg per cubic meter.

I'll leave it to you to compare this to typical neutron star or even quark-gluon plasma densities to figure out how likely it is that any kind of shock wave could compress matter to that density without being successfully resisted by the pressure of the matter.

Perfect! This is exactly what I was trying to get a feel for.

I would presume that neutron-degenerate matter (and, subsequently, quark-gluon plasma) will resist up to a certain failure pressure, and then they will essentially "collapse" and stop resisting altogether. Is that correct, or will degeneracy pressure continue to provide resistance even after they have been compressed past their failure point?

 

[PeterDonis]

Hawking's model requires that the radiation match a blackbody curve corresponding to the black hole temperature

More precisely, Hawking's model requires that the radiation match a black body curve for macroscopic black holes, i.e., holes for which the corresponding "quantum number" (the number of "area quanta" or whatever other microstate turns out to be the right one for doing statistical mechanics) is very large. For small quantum numbers, the concept of "black body curve" has no meaning. That's not just true of black holes; it's true of any system whatsoever. Go and look at how the black body formula is derived for the case of an ordinary object: you will see that it assumes that the number of atoms in the object (here atoms are the microstates on which statistical mechanics is based) is very large. If it isn't, the derivation doesn't work.

The rest of your reasoning is based on the mistake I just described, so it isn't valid.

 

[PeterDonis]

the very last step of black hole evaporation must necessarily involve the entire remaining mass-energy of the black hole being released at once.

Yes, because one Planck mass (more or less--actually, as I said in a previous post, the quantum is 1/4 of the Planck area, so the minimum mass for a hole would be the mass corresponding to that area) is the smallest possible mass, so when a one Planck mass black hole evaporates, it does it in one step, so to speak. This has nothing to do with black body radiation (as my previous post showed, the concept of "black body radiation" has no meaning for systems with small quantum numbers). It's just a simple consequence of the fact that black hole masses are quantized in units of (more or less) one Planck mass (if it is a fact--bear in mind, again, that all this is speculation based on a heuristic guess about quantum gravity, with no experimental evidence to back it up).

 

[sevenperforce]

Yes, because one Planck mass (more or less--actually, as I said in a previous post, the quantum is 1/4 of the Planck area, so the minimum mass for a hole would be the mass corresponding to that area) is the smallest possible mass, so when a one Planck mass black hole evaporates, it does it in one step, so to speak. It's just a simple consequence of the fact that black hole masses are quantized in units of (more or less) one Planck mass (if it is a fact--bear in mind, again, that all this is speculation based on a heuristic guess about quantum gravity, with no experimental evidence to back it up).

Right; I was supposing that perhaps this speculation and heuristic guess are incorrect, and wondering if there was another way to come up with a minimum mass that could provide predictions we could actually test.

More precisely, Hawking's model requires that the radiation match a black body curve for macroscopic black holes, i.e., holes for which the corresponding "quantum number" (the number of "area quanta" or whatever other microstate turns out to be the right one for doing statistical mechanics) is very large. For small quantum numbers, the concept of "black body curve" has no meaning. That's not just true of black holes; it's true of any system whatsoever. Go and look at how the black body formula is derived for the case of an ordinary object: you will see that it assumes that the number of atoms in the object (here atoms are the microstates on which statistical mechanics is based) is very large. If it isn't, the derivation doesn't work.

Does emission at the peak of the wavelength function satisfy the conditions of the blackbody curve, even if the derivation itself wouldn't be valid?

The way I see it, since we can't probe the Planck scale directly (at least not now), the only way to test any quantization models would be to relate some quantum-level phenomena (like tunneling decay of quanta) to larger, measurable things (like black hole blackbody spectra) in a predictable way. Einstein didn't have a way of showing anybody individual photons, but he proved that light was composed of photons by using quantization to explain the photoelectric effect.

 

[PeterDonis]

Does emission at the peak of the wavelength function satisfy the conditions of the blackbody curve, even if the derivation itself wouldn't be valid?

Of course not. Emission at one wavelength is not the same as emission over a whole range of wavelengths, which is what the black body curve describes.

The way I see it, since we can't probe the Planck scale directly (at least not now), the only way to test any quantization models would be to relate some quantum-level phenomena (like tunneling decay of quanta) to larger, measurable things (like black hole blackbody spectra) in a predictable way.

Yes. Now go read up on the quantum gravity literature to see all the proposals that have already been made for trying to do that. As I said before, this area of research is several decades old. You would be well advised to familiarize yourself with what has already been done.

 

[PeterDonis]

Thread closed for moderation.