When will neutronization occur in the black hole formation process?

In summary: The second part of my question is outside the context of GR:If there were no spacetime for the mass to curve, would a black hole actually ever form? This is outside the context of GR because without spacetime, the theory of GR would not exist.
  • #36
PeterDonis said:
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.
 
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  • #37
Ken G said:
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.

Ken G said:
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.
 
  • #38
PeterDonis said:
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.
 
  • #39
Ken G said:
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 :oops: , 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.

Ken G said:
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.

Ken G said:
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).
 
  • #40
PeterDonis said:
I realize I'm the one who brought up supernovas earlier in the thread :oops: , 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.
 
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  • #41
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.
 
  • #42
PAllen said:
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.
 
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  • #43
Ken G said:
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.

Ken G said:
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.

Ken G said:
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.

Ken G said:
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.
 
  • #44
PeterDonis said:
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.
 
  • #45
Ken G said:
Degeneracy pressure is always the same thing as kinetic pressure, degeneracy pressure is kinetic pressure.

I disagree. See below.

Ken G said:
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.

Ken G said:
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.
 
  • #46
PeterDonis said:
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.
 
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  • #47
Ken G said:
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".

Ken G said:
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).
 
  • #48
PeterDonis said:
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.
 
  • #49
Ken G said:
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.

Ken G said:
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.

Ken G said:
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.

Ken G said:
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.)
 
  • #50
PeterDonis said:
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.
 
  • #51
To clarify this, a reference might help. For example, look at the second page in http://jila.colorado.edu/~pja/astr3730/lecture17.pdf, which derives degeneracy pressure by starting with the expression for kinetic pressure. The nonrelativistic substitution of mv where they have p shows by inspection that the pressure they derive is 2/3 of the kinetic energy density, which is a standard aspect of nonrelativistic kinetic pressure in three dimensions. This means that degeneracy pressure is not unusual in how it relates pressure to kinetic energy density, it is unusual only in how it connects the average kinetic energy per particle with a temperature. I say this means degeneracy pressure is a thermodynamic, not mechanical, effect, so its common label can be quite misleading.
 
  • #52
Ken G said:
As to how relativistic the electrons are, yes, they are very relativistic. All the same, they are not completely relativistic

Well, "completely relativistic" would mean "lightlike", correct?

Ken G said:
To clarify this, a reference might help.

Thanks, I'll take a look.
 
  • #53
This may just be an issue of semantics. For a non-relativistic ideal gas the energy of each mode is ##E_k = \frac{\hbar^2 k^2}{2m}## which is just the spectrum of the free particle Hamiltonian and as such constitutes the kinetic energy of each mode. The number density of the Fermi-Dirac distribution is then ##n = \frac{g}{6\pi^2}k_F^3## where ##g## is the degeneracy of states and ##k_F## is the Fermi momentum. From this one obtains the degeneracy pressure ##P_F = \frac{2}{5}nE_F## where as usual ##E_F \equiv \lim_{T\rightarrow 0} (\frac{\partial F}{\partial n})_{T,V}## in the canonical ensemble, with ##F## the Helmholtz free energy.

In this sense the degeneracy pressure of an ideal gas is certainly due to kinetic energy. I don't see where the existence of pressure due to kinetic energy in the degeneracy limit implies some kind of motion in this limit as this is a quantum mechanical effect after all and the concept of motion is moot in quantum mechanics, not to mention there always exists kinetic energy in quantum mechanical ground states.

However the statement
Ken G said:
I say this means degeneracy pressure is a thermodynamic, not mechanical, effect, so its common label can be quite misleading.

doesn't make sense to me as thermodynamic degrees of freedom such as pressure are obtained from averages over statistical ensembles of mechanical degrees of freedom. Again it may just be an issue of semantics.
 
  • #54
PeterDonis said:
Well, "completely relativistic" would mean "lightlike", correct?
Correct-- at which point the adiabatic index is 4/3, the value needed for only marginal stability against contraction.
Thanks, I'll take a look.
Sorry for not providing it sooner.
 
  • #55
WannabeNewton said:
In this sense the degeneracy pressure of an ideal gas is certainly due to kinetic energy. I don't see where the existence of pressure due to kinetic energy in the degeneracy limit implies some kind of motion in this limit as this is a quantum mechanical effect after all and the concept of motion is moot in quantum mechanics, not to mention there always exists kinetic energy in quantum mechanical ground states.
I agree that the meaning of "motion" is rather ambiguous in the quantum limit, but the meaning of "kinetic" is not, i.e., we still use the term "kinetic energy", and its informal meaning as "energy of motion." My point is simply that degeneracy pressure is a thermodynamic, not mechanical, effect-- it relates to the temperature, not the pressure, if the conserved quantities of mass and energy are being tracked in the evolution of some object.
However the statement [that I've repeated here] doesn't make sense to me as thermodynamic degrees of freedom such as pressure are obtained from averages over statistical ensembles of mechanical degrees of freedom. Again it may just be an issue of semantics.
What I mean is that it is something that affects the temperature we associate with the energy, not the pressure we associate with that same energy. Is that not the standard distinction between a "thermodynamic" vs. "mechanical" effect?
 
  • #56
On the question of when neutronization sets in, I've looked up some hard numbers. If you take the maximum energy an electron can get from a beta decay, and equate that to the Fermi energy of a fully degenerate electron gas, you find that beta decay is suppressed at a density of about 20 million g/cc (for example, in Kippenhahn and Wiegert, page 135). The density of a solar mass of material in the volume of the Earth is about 1/10 that amount, suggesting that if you compress such a radius by anything more than a factor of 2, you will begin neutronization. Yet a crucial question is still, what is the timescale? Since neutronization does not appear in the force equation, it appears in the composition and energy equations, if the neutronization timescale is way longer than the sound crossing time, you still have a quasi-steady force equilibrium, so you still have dynamical stability (for the reasons given above-- the highly relativistic electron adiabatic index is still a little above 4/3). Thus my point is, you must wait for something that eats up the kinetic energy, and that will be whichever of these eats up that energy the fastest: neutronization, the Urca process, or photodistintegration.

So collapse must occur as soon as one of these processes plays out on the sound crossing time, since then you will have an instability in the energy equation which runs away on timescales shorter than you can establish a force balance. It's not obvious how small the radius needs to get before this sets in, but it has to happen when the object is still much larger than a neutron star (so for a mass appreciably less than the Chandra mass), or there would not be enough gravitational energy left for the sudden release needed in a supernova. My point is, that release must happen, not by virtue of the force equation alone, but by virtue of something happening in the energy equation which eats up the kinetic energy that goes into the pressure needed in the force equation. That is not what would be called a dynamical instability, like a pencil on its point, even though core collapse is sometimes explained that way.
 

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