Stars in the early universe and stellar processes

[Teichii492]

Hey PF,

Since there are stars that can be powered predominantly (>50%) by the CNO cycle, which requires carbon as a catalyst, and i understand the core temperatures of these stars is about 10⁶ K. Does this mean that stars where the triple-alpha process is dominant (10⁸ K) had to exist and die previously for there to be enough carbon available to dominantly CNO power a star.
I'd say the bigger question attached to this is "Is there a limit to the earliest period where it is possible for stars to exist powered more than 50% by the CNO cycle?"

and maybe to go a step further
"alternatively, does this mean that regardless of the over density regions in the primordial times, where it would be more likely for a hot star to form, (correct me if I've misunderstood the consequences of over and underdensity regions) that there still couldn't be dominantly CNO powered stars (occupying a region around 10⁶K) until enough hotter stars existed to generate enough carbon for the cooler stars to exist?"

I feel like I'm missing something blatantly obvious here? Possibly, I've made the assumption that i am ignorant of either an early universe carbon nucleosynthesis event, or just that i am not appreciating the abundance of carbon generated by supernova nucleosynthesis and rare fusion events in some stars.

 

[mfb]

I'd say the bigger question attached to this is "Is there a limit to the earliest period where it is possible for stars to exist powered more than 50% by the CNO cycle?"

Sure.
The first generation of stars didn't have heavier elements than helium (at least not in relevant quantities). They also didn't have effective cooling mechanisms, so they were quite massive and short-living, their ash then added C(N)O to the universe.

 

[snorkack]

Hey PF,

Since there are stars that can be powered predominantly (>50%) by the CNO cycle, which requires carbon as a catalyst, and i understand the core temperatures of these stars is about 10⁶ K. Does this mean that stars where the triple-alpha process is dominant (10⁸ K) had to exist and die previously for there to be enough carbon available to dominantly CNO power a star.

Only sufficiently massive stars.
Massive stars with low metallicity are hotter than higher metallicity stars of similar mass. Because pp cycle is slower, and it takes higher temperature before the pp luminosity can stop contraction of the star.
Sufficiently hot and metal-poor stars can, while still fusing protium by pp process, heat up to the point where triple alpha process also operates, and produces the carbon necessary to shift the hydrogen fusion from pp to CNO cycle.
Of course, this process is self-limiting to a certain low but nonzero metallicity. And for sufficiently massive and metal-poor stars, it can happen already in the pre-main-sequence evolution, long before protium is exhausted.

 

[Teichii492]

Only sufficiently massive stars.
Massive stars with low metallicity are hotter than higher metallicity stars of similar mass. Because pp cycle is slower, and it takes higher temperature before the pp luminosity can stop contraction of the star.
Sufficiently hot and metal-poor stars can, while still fusing protium by pp process, heat up to the point where triple alpha process also operates, and produces the carbon necessary to shift the hydrogen fusion from pp to CNO cycle.
Of course, this process is self-limiting to a certain low but nonzero metallicity. And for sufficiently massive and metal-poor stars, it can happen already in the pre-main-sequence evolution, long before protium is exhausted.

Brilliant, i think both the answers here field enough of what i wanted to know. I'll look more into the details of stellar evolution i think. Thanks a million!

 

[phyzguy]

As snorkack said, stars don't have to live and die to produce the CNO needed to start the CNO cycle. A massive star that starts being composed only of H and He will build up its own CNO and switch over to the CNO cycle fairly quickly. I ran some simulations a while back with the Mesa stellar evolution code and found that it only takes about 10,000 years for a star to build up enough CNO to power the CNO cycle.

 

[Ken G]

And I'll make a prediction for you, if you still have access to those calculations: as the C builds up and CNO-cycle fusion of H kicks in, the nuclear cross sections, as a function of temperature, obviously increase markedly for the more massive stars. But the luminosity of the star will not increase much at all.

 

[snorkack]

And I'll make a prediction for you, if you still have access to those calculations: as the C builds up and CNO-cycle fusion of H kicks in, the nuclear cross sections, as a function of temperature, obviously increase markedly for the more massive stars. But the luminosity of the star will not increase much at all.

Go on with predictions:
Core temperature should fall;
The concentration of luminosity near the centre, with highest temperature and metallicity, should promote convection in the core of the star.

 

[Teichii492]

As snorkack said, stars don't have to live and die to produce the CNO needed to start the CNO cycle. A massive star that starts being composed only of H and He will build up its own CNO and switch over to the CNO cycle fairly quickly. I ran some simulations a while back with the Mesa stellar evolution code and found that it only takes about 10,000 years for a star to build up enough CNO to power the CNO cycle.

That sounds quite interesting. I'll assume that the carbon generation events are rare and that this is made negligible because of the sheer mass of the star and thus the temperature of the core? or is it more complicated than that?

Also i would like to refine an earlier question. Does the mass of the star formed from a gas cloud merely depend on the density fluctuations present in that space after the radiation era, if we are talking about the first stars? and thus a high region of overdensity would cause more matter from the relatively homogeneous gas cloud to collapse to a given point?

A little further, does this make it possible for black holes to form after 100 million years simply due to such regions of extreme overdensity creating very very short lived, massive stars in the early universe?

 

[Ken G]

Go on with predictions:
Core temperature should fall;

Certainly, because the star will self-regulate its core temperature to avoid over-producing light faster than it can diffuse out, which is the dominant physics that sets the luminosity. There will be some small feedback into the luminosity, when the structure adjusts, but the radiative diffusion rate is surprisingly insensitive to the temperature, so the luminosity should not change like you might expect, given how the fusion cross-section function is rising. I hope this experiment is done, I'd like to know exactly how it comes out-- but it seems clear that the luminosity will not increase as much as the fusion cross section function, in fact it might even drop. But if you really want luminosity to rise, reduce the radiative opacity! Now you will see a whopping effect, way more important than anything happening to fusion cross sections, because the dominant physics that sets luminosity in reasonably massive main-sequence stars is simply the rate of escape of light.

The concentration of luminosity near the centre, with highest temperature and metallicity, should promote convection in the core of the star.

Yet this rarely sets the timescale for heat escape, i.e., luminosity, until you get to extremely massive stars that are almost fully convective. In fact, I've always wondered about what is happening for the super-massive stars, because they are both highly convective, and very far from the Hayashi track, so there is something that requires explaining there. But if we steer clear of the very highest mass stars (where radiative diffusion may give way to convection), and to the very lowest mass stars (which get degenerate before reaching the main sequence), then we can use the simple luminosity arguments I'm giving, regardless of the fusion environment.

 

[Ken G]

A little further, does this make it possible for black holes to form after 100 million years simply due to such regions of extreme overdensity creating very very short lived, massive stars in the early universe?

You are actually asking a research-level question here. It is still debated if the seed black holes for what end up being the supermassive black holes in quasars are "quasistars" (very massive stars that just kind of fall into their own central black hole), of more normal versions of supernovae that leave black hole remnants that grow via mergers. What we desperately need are observations of the first stars! It was originally thought that "quasistars" would essentially eat too much of their own luminosity to be visible, but a recent paper (http://arxiv.org/abs/1509.07511) suggests otherwise.

 

[snorkack]

But if you really want luminosity to rise, reduce the radiative opacity! Now you will see a whopping effect, way more important than anything happening to fusion cross sections, because the dominant physics that sets luminosity in reasonably massive main-sequence stars is simply the rate of escape of light.

Then the next prediction:
Low mass, low metallicity stars should have higher luminosity on Henyey track than equally massive stars of higher metallicity.

 

[Ken G]

Then the next prediction:
Low mass, low metallicity stars should have higher luminosity on Henyey track than equally massive stars of higher metallicity.

Yes, as long as the mass does not go so low that the stars are going degenerate-- that messes up the connection between pressure and temperature that is assumed in Eddington-type models. If that happens,, I'd have to do a different estimate to see what would happen. But if we steer clear of red dwarfs, then yes, low metallicity should certainly lead to higher luminosity at the same mass. So now all that remains is, does someone want to run Mesa and check the predictions, or can phyzguy just look up the results already existing in those old calculations?

 

[snorkack]

Yes, as long as the mass does not go so low that the stars are going degenerate-- that messes up the connection between pressure and temperature that is assumed in Eddington-type models. If that happens,, I'd have to do a different estimate to see what would happen. But if we steer clear of red dwarfs,

Why is Hayashi track different from Henyey track - because of degeneracy, or just because of convection?

 

[Ken G]

Why is Hayashi track different from Henyey track - because of degeneracy, or just because of convection?

I think it's all pretty much about the convection. The Hayashi track is not a law that convective stars must follow, it is a limit that says there is simply no equilibrium solution for a star to the right of that track. The physical reason is essentially that if a star were to the right of it, it would go convective, which would bring more heat to the surface and raise the surface T. So when stars try to get to the right of that limit, they go fully convective instead.

This effect does depend on the opacity at the surface of the star, and first generation stars have surfaces without metals, even if their cores are doing some triple-alpha fusion, so they might not hit the Hayashi limit at all. But, it is often thought that such stars would be of high mass, and high-mass stars don't have much of a Hayashi phase anyway, they become radiative very quickly, and they also undergo fusion very quickly, so it might not matter much if they don't have a Hayashi limit.

Anyway, the important thing about the Hayashi limit is that it is one of the rare situations where stars essentially work "outside-in", instead of "inside-out." By that I mean, we usually imagine the star is its interior, and the surface just has to kind of "play ball" with whatever is going on inside. The classic example of this is the Henyey track and the main sequence, where the luminosity is primarily set by the rate that light leaks out, and the radius is set by the history of contraction, so the surface has to come to whatever temperature will allow that luminosity to be ejected into space.

But not the Hayashi track, because there, if the surface T gets too low, the T gradient gets too steep, and the star goes convective. So what this means is, the surface T cannot go below about 3000 K, so hits that limit and is typically in that vicinity (it depends a bit on mass). So if we have the radius of the star from its contraction history, knowing its surface T then gives its luminosity, in an outside-in fashion. Or, if it is a red giant, then we don't know the radius, we know the luminosity (it comes from radiative escape and shell fusion working together self-consistently) and the surface T, so it sets the radius-- that's why they puff out into giants.

So to answer your question, I believe what matters is the convection. The limit is derived from surface physics, and even red dwarfs shouldn't be degenerate at their surfaces.

 

[snorkack]

I think it's all pretty much about the convection. The Hayashi track is not a law that convective stars must follow, it is a limit that says there is simply no equilibrium solution for a star to the right of that track. The physical reason is essentially that if a star were to the right of it, it would go convective, which would bring more heat to the surface and raise the surface T. So when stars try to get to the right of that limit, they go fully convective instead.

This effect does depend on the opacity at the surface of the star, and first generation stars have surfaces without metals, even if their cores are doing some triple-alpha fusion, so they might not hit the Hayashi limit at all.

But not the Hayashi track, because there, if the surface T gets too low, the T gradient gets too steep, and the star goes convective. So what this means is, the surface T cannot go below about 3000 K, so hits that limit and is typically in that vicinity (it depends a bit on mass).

Yes, but all stars come from the right of Hayashi track!
If gas clouds are 2,7 K now, and maybe 30 K in early universe, what does a star look like while it is evolving from surface temperature 300 K to surface temperature 3000 K?

 

[phyzguy]

Yes, but all stars come from the right of Hayashi track!
If gas clouds are 2,7 K now, and maybe 30 K in early universe, what does a star look like while it is evolving from surface temperature 300 K to surface temperature 3000 K?

Figure 6 of this paper shows a simulation of the evolution of a collapsing protostar as it evolves from a few K up to ~100,000K central temperature. It best steadily hotter and denser as it collapses.

 

[snorkack]

Figure 6 of this paper shows a simulation of the evolution of a collapsing protostar as it evolves from a few K up to ~100,000K central temperature. It best steadily hotter and denser as it collapses.

Yes. That tracks the central, not "surface" temperature.
The compressibility of hydrogen gets very big around 2000 K, because of dihydrogen dissociation.
Does the location of star surface, and the density above surface, depend significantly on metallicity?

 

[Ken G]

Yes, but all stars come from the right of Hayashi track!

True, but one would not call that a star, and the reason is because they come from a different equilibrium solution, on the "other side" of the Jeans instability, if you will. There are two important physical differences: 1) the energy transport timescales are faster than the dynamical time, so the gas is treated as nearly isothermal instead of having a convective or radiative temperature gradient. 2) the force balance is one in which the internal gravity of the "object" is largely insignificant. That situation breaks completely from the kinds of simple assumptions that go into the Hayashi track.

 

[snorkack]

True, but one would not call that a star, and the reason is because they come from a different equilibrium solution, on the "other side" of the Jeans instability, if you will. There are two important physical differences: 1) the energy transport timescales are faster than the dynamical time, so the gas is treated as nearly isothermal instead of having a convective or radiative temperature gradient.

How so?
High metallicity gas, at least, becomes optically thick long before it reaches even central temperature of 2000 K, let alone Hayashi track.

 

[Ken G]

How so?
High metallicity gas, at least, becomes optically thick long before it reaches even central temperature of 2000 K, let alone Hayashi track.

I don't think that's true, If it's in hydrostatic equilibrium with an important self-gravity, the Hayashi limit should apply, and the star should be fairly hot. The usual picture is that stellar gas starts out so spread out it is essentially isothermal and has no important gravity, so it's not a star. Then it hits the Jeans instability, and goes out of equilibrium, until it finds a new hydrostatic solution where it's own gravity is very important, and the gas is way hotter. This is where the Hayashi limit becomes relevant, in my understanding of the situation. If there is a short phase where some of these concepts are ambiguous, so be it, but the Hayashi track is reached soon enough.

 

[snorkack]

I don't think that's true, If it's in hydrostatic equilibrium with an important self-gravity, the Hayashi limit should apply, and the star should be fairly hot.

In other words, before it reaches Hayashi limit, it is out of hydrostatic equilibrium.

The usual picture is that stellar gas starts out so spread out it is essentially isothermal and has no important gravity, so it's not a star. Then it hits the Jeans instability, and goes out of equilibrium, until it finds a new hydrostatic solution where it's own gravity is very important, and the gas is way hotter. This is where the Hayashi limit becomes relevant, in my understanding of the situation. If there is a short phase where some of these concepts are ambiguous, so be it, but the Hayashi track is reached soon enough.

The free fall period may be short in time, but a lot happens in that time.
So... any convective shells are, by definition, not in free fall.
What would you call "surface" of a star? The surface where the gas becomes optically thick and radiation escapes? Or the surface where infalling gas decelerates and transfers its momentum to hydrostatic pressure?

 

[Ken G]

In other words, before it reaches Hayashi limit, it is out of hydrostatic equilibrium.

Yes.

The free fall period may be short in time, but a lot happens in that time.
So... any convective shells are, by definition, not in free fall.

Yes.

What would you call "surface" of a star? The surface where the gas becomes optically thick and radiation escapes?

Yes.

Or the surface where infalling gas decelerates and transfers its momentum to hydrostatic pressure?

There may be some point in finding a surface like that at some stage in the star's evolution (ETA: as in the paper cited by phyzguy), but once the whole star is in equilibrium, that isn't happening any more. In other words, the details of the transition from free fall to hydrostatic equilibrium may be complicated, but the transition does occur, and the Hayashi limit holds afterward.

 

[snorkack]

Is Hayashi limit most connected to the maximum of compressibility, or maximum of opacity?

 

[Ken G]

It's the surface opacity, it falls off a cliff below 3000 K because metals aren't being ionized any more and there are not free electrons (molecular opacity kicks in at some point, but that's cooler still). When the surface opacity is plummeting, the star can't find a place to put the surface, because it would have to be at too high a pressure (the surface pressure is basically the weight of a mean-free-path of gas, so it is inversely proportional to the opacity). So it has to keep the surface up near 3000 to avoid that opacity cliff.

Now, since first stars don't have metals at their surface, it's not clear they would even have a Hayashi limit, but they're also said to be high-mass stars, so those stars don't have much of a Hayashi phase anyway.

 

[snorkack]

Remove metals, and you´ll still have H ionization providing some electrons for hydride ions. At somewhat higher temperature and lower opacity value.

 

[Ken G]

The opacity will certainly be a lot lower if you need to get the electrons from hydrogen! But it might be enough to support a Hayashi track at somewhat higher surface T, for all I know.

 

[snorkack]

Another important thing is the possibility or otherwise of hydrostatic balance.
As demonstrated, when V is decreasing with 3rd power of R, P is increasing with inverse 4th power of R.
This relies purely on laws of gravitation, assumption of hydrostatic balance and constant radial distribution of density. Notably, it does not include express assumptions as to opacity, or ideality of gas.
Ideal gas law is
PV=nRT
Now, the "virial theorem" does not only require the gas to be ideal (absence of degeneracy as well as light pressure), but also for n to be independent on T (and V). Then T increases with inverse of R.
But suppose n is NOT independent on T (nor V)? Like, dihydrogen molecules dissociate first to hydrogen atoms (eventually doubling n), and then to protons and electrons (doubling n again)?
At low densities, plasma can still have ideal gas pressure. It is still true that
PV=RnT
but note that now it is the expression nT that has to grow with inverse of radius - not T alone.
CAN nT grow with inverse of R?
With monoatomic gas - no problem. On adiabatic compression of monoatomic gas, P grows with 5/3 power of density, and T therefore with 2nd power of inverse R. As the star shrinks, P exceeds gravity, and the star stays in hydrostatic equilibrium unless and until some heat gets out of core.
Hydrogen atoms are monoatomic gas.
Protons and electrons also are monoatomic gas.
Molecular dihydrogen is not a monoatomic gas.
But a diatomic gas still has P grow with 7/5 power of density, and T therefore still with 6/5 power of inverse R. And the admixture of monoatomic He also increases the temperature.
But what if the hydrogen is dissociating to atoms, and then electrons? How fast is nT then growing with density?

 

[Ken G]

Another important thing is the possibility or otherwise of hydrostatic balance.
As demonstrated, when V is decreasing with 3rd power of R, P is increasing with inverse 4th power of R.
This relies purely on laws of gravitation, assumption of hydrostatic balance and constant radial distribution of density. Notably, it does not include express assumptions as to opacity, or ideality of gas.

Yes, one can also use the virial theorem to get the same connection between P and M,.R.

Now, the "virial theorem" does not only require the gas to be ideal (absence of degeneracy as well as light pressure),

Actually, the virial theorem only requires a negligible role of light pressure, it's just fine with degeneracy. This is a common misconception about degeneracy! Degeneracy has essentially no impact at all on what sets the pressure or kinetic energy of the gas, all it does is change the equation for temperature from the ideal gas law, T = PV/Nk, to something more complicated.

but also for n to be independent on T (and V). Then T increases with inverse of R.

Anyway, we can agree that T is inverse of R, under these conditions-- what I mean by the virial theorem isn't a constaint on T, it is a constraint on the kinetic energy (which, nonrelativistically, is half the magnitude of the potential energy). Let's not worry about the virial theorem, you are talking about T ~ 1/R, and that's fine for an ideal gas with negligible light pressure, i.e., stars that are similar to our Sun.

But suppose n is NOT independent on T (nor V)? Like, dihydrogen molecules dissociate first to hydrogen atoms (eventually doubling n), and then to protons and electrons (doubling n again)?
At low densities, plasma can still have ideal gas pressure. It is still true that
PV=RnT
but note that now it is the expression nT that has to grow with inverse of radius - not T alone.
CAN nT grow with inverse of R?

Sure, the T just does what it needs to do. In a star like the Sun, the ideal gas law is essentially an equation that sets T, all the other variables are set in other ways.

But a diatomic gas still has P grow with 7/5 power of density, and T therefore still with 6/5 power of inverse R. And the admixture of monoatomic He also increases the temperature.
But what if the hydrogen is dissociating to atoms, and then electrons? How fast is nT then growing with density?

There are certainly important affects on T when that's happening, I'm not sure exactly which ramification you are interested in. For example, it certainly changes the convection instability criterion.

 

[snorkack]

Let's not worry about the virial theorem, you are talking about T ~ 1/R, and that's fine for an ideal gas with negligible light pressure, i.e., stars that are similar to our Sun.

And n independent on T.

Sure, the T just does what it needs to do.

No, it does not.

There are certainly important affects on T when that's happening, I'm not sure exactly which ramification you are interested in. For example, it certainly changes the convection instability criterion.

What I care about is hydrostatic instability criterion.
Obviously, a body of gas which is freely radiating at timescales faster than free fall timescale cannot reach hydrostatic equilibrium. Self-gravity requires nT to grow with inverse R, and it´s not growing at all because T is constant and in equilibrium with external radiation field.
How about gas which is opaque, but dissociating?
nT obviously is growing, because n is growing, and T also is growing because that´s what is changing n.
But dissociation takes up a lot of energy. T grows slowly.
How fast is nT growing, in the end? Specifically, is it growing fast enough to keep up with inverse R?

 

[Ken G]

Obviously, a body of gas which is freely radiating at timescales faster than free fall timescale cannot reach hydrostatic equilibrium.

That is not only not obvious, it's wrong. The interstellar medium being an obvious counterexample. Indeed, the instabilities that the ISM is subject to tend to be thermal instabilities, not hydrodynamical instabilities. Normally, if the energy transport timescales are shorter than the free-fall time, we treat the gas as having a spatially constant temperature, which then serves as a constraint on the hydrostatic equilibrium. However, in stars, it is invariably the other way around, and the energy transport timescale is long-- so the hydrostatic equilibrium imposes a gross constraint on the temperature and density (via nT) that the energy transport physics simply has to deal with, and can only alter if structures develop on such small scales that energy transport timescales can compete with sound crossing times. That's essentially never true for the star as whole, though.

Self-gravity requires nT to grow with inverse R, and it´s not growing at all because T is constant and in equilibrium with external radiation field.

That is a very standard situation in the ISM, and is often in hydrostatic equilibrium-- whenever you are on the "other side" of the Jeans instability from where stars are (such as in a giant molecular cloud that is still forming and has not reached the Jeans limit). In stars, the big difference is that the energy transport timescales get much longer than the dynamical timescales, so you always end up finding hydrostatic equilibrium pretty quickly-- unless you have a supernova, or at least a pulsational instability.

How about gas which is opaque, but dissociating?

In an ideal gas, the structure will simply adjust so that nT supports the pressure that is being imposed. That can mean adjusting n, or adjusting T, they just figure out what they need to do. Since T is ruled by the energy transport, you are right that often n will be what is free to adjust, but if there's dissociation, that just means nT will need to take that into account. That's why it's better to think in terms of pressure anyway, the force balance gives you the pressure, and once you have that, the nT just finds the appropriate degree of ionization, given the T constraints, to give the necessary P.

nT obviously is growing, because n is growing, and T also is growing because that´s what is changing n.

This is what I'm trying to explain. The force balance just tells you that P is set by R. So if you imagine R as an external variable you are gradually reducing and wondering what happens to the stellar structure, that means you are given P. Now for an ideal gas, that just sets nT, where n must include the ionization that is consistent with T. So that's all the star does-- find the T and the ionization that is mutually consistent and satisfies the necessary nT. That's why I said the ideal gas physics is responsible for setting T. The energy transport then simply has to monkey with that T structure to make it all work out, but this basically means the energy transport is whatever it needs to be to support that T. So the basic logic of a contracting ideal-gas star is this: given M, we have that R sets nT. Here n is also set by R, combined with the ionization that works with T, so this is what determines T: we know nT, and we know the ionization physics. Then the energy transport takes that T, and figures out how fast it transports heat. But all that affects is the timescale for continuing on to the next lower R, when you look at much longer timescales than are of interest in the force balance.

But dissociation takes up a lot of energy. T grows slowly.

The timescales for change are a much different issue from the dynamical timescale on which force balance is attained.

How fast is nT growing, in the end? Specifically, is it growing fast enough to keep up with inverse R?

Yes, that is the requirement for force balance, it is the fastest timescale that must be satisfied. In a star, it's usually set up in only a few hours!

 

[snorkack]

However, in stars, it is invariably the other way around, and the energy transport timescale is long-- so the hydrostatic equilibrium imposes a gross constraint on the temperature and density (via nT)

It can only do so if the heat is available to maintain hydrostatic equilibrium.

In stars, the big difference is that the energy transport timescales get much longer than the dynamical timescales, so you always end up finding hydrostatic equilibrium pretty quickly-- unless you have a supernova, or at least a pulsational instability.

There are regions where hydrostatic equilibrium cannot and will not be found. Supernova is one of them, but not the only one.

In an ideal gas, the structure will simply adjust so that nT supports the pressure that is being imposed. That can mean adjusting n, or adjusting T, they just figure out what they need to do.

Only if they find the energy to change n and T enough.

Since T is ruled by the energy transport

Not on free fall timescales. On free fall timescales, it is ruled by adiabatic heat capacity.

, you are right that often n will be what is free to adjust, but if there's dissociation, that just means nT will need to take that into account. That's why it's better to think in terms of pressure anyway, the force balance gives you the pressure, and once you have that, the nT just finds the appropriate degree of ionization, given the T constraints, to give the necessary P.

No, it does not. nT is given by the adiabatic compression, and if the resulting P does not keep up with gravity, there is no force balance - unless you count into force balance the radial acceleration of free fall.

This is what I'm trying to explain. The force balance just tells you that P is set by R.

No, it tells that V is set by R. And it gives the P needed to balance gravity. It does not tell whether the force will be balanced.

So if you imagine R as an external variable you are gradually reducing and wondering what happens to the stellar structure, that means you are given P. Now for an ideal gas, that just sets nT, where n must include the ionization that is consistent with T.

No, we are given V. For an ideal gas, it is adiabatic compression that gives us nT.
If nT gives us P which is bigger than necessary, then contraction stops, and then continues gradually as T is decreased. If, however, nT gives less than necessary P, then force balance will give further contraction at free fall timescales.

Then the energy transport takes that T, and figures out how fast it transports heat. But all that affects is the timescale for continuing on to the next lower R, when you look at much longer timescales than are of interest in the force balance.

The timescales for change are a much different issue from the dynamical timescale on which force balance is attained. Yes, that is the requirement for force balance, it is the fastest timescale that must be satisfied. In a star, it's usually set up in only a few hours!

The timescales for change are only much longer than free fall timescales as long as hydrostatic balance can be reached. And in some conditions, it cannot be attained.

 

[Ken G]

It can only do so if the heat is available to maintain hydrostatic equilibrium.

The heat is available-- gravity makes sure of that.

There are regions where hydrostatic equilibrium cannot and will not be found. Supernova is one of them, but not the only one.

Supernovae occur only when the gas goes relativistic. That's not happening here. In general, nonrelativistic stars find it very easy to reach hydrostatic equilibrium. Even pulsating stars pulsate around a hydrostatic equilibrium state.

Only if they find the energy to change n and T enough.

Never a problem-- plenty of gravitational energy is available as the star moves to lower and lower R. The energy equation mostly only determines how long it takes to move to lower R, based on the rate that heat escapes the system.

Not on free fall timescales. On free fall timescales, it is ruled by adiabatic heat capacity.

And adiabatic heat capacity is dynamically stable, that's why stars are dynamically stable.

No, it does not. nT is given by the adiabatic compression, and if the resulting P does not keep up with gravity, there is no force balance - unless you count into force balance the radial acceleration of free fall.

Well, we've gotten far afield of the question now. It sounds like you are asking if a changing ionization balance can make it impossible for a star to find a hydrostatic equilibrium. No, it cannot. The nT will simply find the self-consistent value that supports the necessary P, while n is consistent with the ionization equation and T satisfies the energy transport equation. The latter is quite flexible for maintaining a wide array of T, whatever is necessary.

Even pulsational stars, like Cepheids, don't break this, for they oscillate around a stable force balance, and their oscillation is due to a thermal instability, not a dynamical instability (the ionizing layers receive heat when contracted, and lose heat when expanded, that's in the energy equation, so is not a dynamical instability). Supernovas are totally different-- they require the gas be relativistic, which we are not talking about here.

No, we are given V. For an ideal gas, it is adiabatic compression that gives us nT.

No.

If nT gives us P which is bigger than necessary, then contraction stops, and then continues gradually as T is decreased.

If nT gives a P that is bigger than necessary, there's no contraction in the first place. Since the force-balance timescale is so fast, we will never have nT be anything but what it needs to be for force balance, the evolution will occur from hydrostatic equilibrium to hydrostatic equilibrium. The reason there is evolution at all is heat escape, so heat escape sets the timetable for evolution through states that have no need for any adiabatic energy equation. All you need the adiabatic energy equation for is to establish dynamical stability, which fiercely holds in nonrelativistic stars-- even those undergoing ionization changes.

If, however, nT gives less than necessary P, then force balance will give further contraction at free fall timescales.

That is what does not happen in nonrelativistic stars, like the Sun. It only happens when the Jeans instability first sets in, during initial star formation, or if the star goes relativistic, which never happens during the main sequence phase.

The timescales for change are only much longer than free fall timescales as long as hydrostatic balance can be reached. And in some conditions, it cannot be attained.

Those situations were already described above. They do not happen on the main sequence. This was one of Eddington's first discoveries, even before fusion was discovered.

 

[snorkack]

Never a problem-- plenty of gravitational energy is available as the star moves to lower and lower R.

Gravity provides only the energy of adiabatic compression.

No.If nT gives a P that is bigger than necessary, there's no contraction in the first place.

If nT gives a P that is bigger than necessary, there is stable hydrostatic equilibrium, where a small loss of heat causes small contraction.

Since the force-balance timescale is so fast, we will never have nT be anything but what it needs to be for force balance, the evolution will occur from hydrostatic equilibrium to hydrostatic equilibrium. The reason there is evolution at all is heat escape, so heat escape sets the timetable for evolution through states that have no need for any adiabatic energy equation.

Adiabatic energy equation gives the amount of heat left over which has to escape. If no energy needs to escape, there is no hydrostatic equilibrium.

All you need the adiabatic energy equation for is to establish dynamical stability, which fiercely holds in nonrelativistic stars-- even those undergoing ionization changes.
That is what does not happen in nonrelativistic stars, like the Sun. It only happens when the Jeans instability first sets in, during initial star formation, or if the star goes relativistic, which never happens during the main sequence phase.
Those situations were already described above. They do not happen on the main sequence. This was one of Eddington's first discoveries, even before fusion was discovered.

Yes. And dissociation of hydrogen takes place at temperatures cooler than Hayashi limit and main sequence.

 

[Ken G]

Gravity provides only the energy of adiabatic compression.

Call it what you like, but gravity always provides twice the kinetic energy that pressure needs to support the star, whenever the gas is nonrelativistic.

If nT gives a P that is bigger than necessary, there is stable hydrostatic equilibrium, where a small loss of heat causes small contraction.

Again, the heat transport is way slower than the sound crossing time. The star is always in hydrostatic equilibrium as a very good approximation, as long as it stays nonrelativistic.

Adiabatic energy equation gives the amount of heat left over which has to escape. If no energy needs to escape, there is no hydrostatic equilibrium.

I have no idea what you are talking about. A star is in hydrostatic equilibrium. If it suffers a net loss of heat, it very gradually shifts through a sequence of hydrostatic equilibriums, generally of lower and lower R. If the ionization changes during that process, it merely changes the amount that R needs to change. The timescale for the change is controlled by the rate of heat loss. That's pretty much it.

Yes. And dissociation of hydrogen takes place at temperatures cooler than Hayashi limit and main sequence.

If it happens prior to the establishment of a hydrostatic equilibrium, we don't call it a star yet. After that equilibrium, we have the Hayashi limit.

 

[snorkack]

Call it what you like, but gravity always provides twice the kinetic energy that pressure needs to support the star,

Gravity provides a certain amount of energy without caring where it goes. If it goes to translational kinetic energy, it contributes to pressure. If it goes to something else, like internal rotational energy of molecules, it does not.

whenever the gas is nonrelativistic.
Again, the heat transport is way slower than the sound crossing time. The star is always in hydrostatic equilibrium as a very good approximation, as long as it stays nonrelativistic.
I have no idea what you are talking about. A star is in hydrostatic equilibrium. If it suffers a net loss of heat, it very gradually shifts through a sequence of hydrostatic equilibriums, generally of lower and lower R. If the ionization changes during that process, it merely changes the amount that R needs to change. The timescale for the change is controlled by the rate of heat loss. That's pretty much it.
If it happens prior to the establishment of a hydrostatic equilibrium, we don't call it a star yet. After that equilibrium, we have the Hayashi limit.

A body whose interior is under 2000 K, and supported by hydrostatic pressure of ideal gas diatomic molecular hydrogen, is in hydrostatic equilibrium. Is it a star?
And no, gravity does not provide 200 % kinetic energy that pressure needs to support it. It provides just 120 %. Gravity does provide 200 %, but 80 % go to molecular rotations, not translations.