Stars in the early universe and stellar processes

[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.

 

[Ken G]

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.

Why bother to bring this up? We know that the amount that goes into rotation does not refute what I said above-- gravity always provides plenty of energy to maintain pressure balance in nonrelativistic gas. If there is some other place that energy can go for awhile, it just means the contraction will proceed even further, until the energy must once again go into pressure and balance can be achieved again. Stars are very stable to finding hydrostatic equilibrium.

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?

I think not. Such objects cannot be in hydrostatic equilibrium, by the Hayashi theorem. They will continue to collapse until they get hotter.

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.

120% is enough.

 

[snorkack]

Why bother to bring this up? We know that the amount that goes into rotation does not refute what I said above-- gravity always provides plenty of energy to maintain pressure balance in nonrelativistic gas.

Only for a diatomic gas.

If there is some other place that energy can go for awhile, it just means the contraction will proceed even further, until the energy must once again go into pressure and balance can be achieved again. Stars are very stable to finding hydrostatic equilibrium.

Adding 2 degrees of freedom, for diatomic gas, decreases the stability fivefold. Add a third degree of freedom, for example of vibrations, and the stability vanishes and hydrostatic equilibrium is impossible - only free fall is allowed.

I think not. Such objects cannot be in hydrostatic equilibrium, by the Hayashi theorem. They will continue to collapse until they get hotter.120% is enough.

If 120 % is enough, then why is hydrostatic equilibrium impossible?

 

[Ken G]

Only for a diatomic gas.

And monatomic gas, and combinations of monatomic and diatomic gases-- in short, the ones we actually encounter. It is perhaps an interesting point that a star made of methane might not be able to avoid collapsing until the methane dissociates, but it would dissociate, so the point is somewhat moot.

If 120 % is enough, then why is hydrostatic equilibrium impossible?

It isn't. What is impossible is an equilibrium in both the force equation and the energy equation.

 

[snorkack]

And monatomic gas, and combinations of monatomic and diatomic gases-- in short, the ones we actually encounter. It is perhaps an interesting point that a star made of methane might not be able to avoid collapsing until the methane dissociates, but it would dissociate, so the point is somewhat moot.

It isn´t. Because if, at some point, rather than radiate away the gravitational energy gradually through contraction in hydrostatic equilibrium, the star undergoes a free fall collapse until it dissociates, the accumulated energy of free fall must be converted into heat at even faster than free fall timescales at the end of the fall.

It isn't. What is impossible is an equilibrium in both the force equation and the energy equation.

If force equation can remain in equilibrium but energy equation does not, what does a star look like while evolving towards Hayashi track from lower temperatures?

 

[Ken G]

It isn´t. Because if, at some point, rather than radiate away the gravitational energy gradually through contraction in hydrostatic equilibrium, the star undergoes a free fall collapse until it dissociates, the accumulated energy of free fall must be converted into heat at even faster than free fall timescales at the end of the fall.

I know all that, this is exactly what happens to gas that passes into the Jeans instability. It has to find a new hydrostatic equilibrium, after a period of free fall, and this is what raises the temperature enough to cause dissociation of the molecules. There are all kinds of interesting cases, but the standard case, such as our Sun, involves ideal gases, and the molecules dissociate, and even if they didn't, the primary molecule would be diatomic anyway. So yes, your point is indeed moot.

If force equation can remain in equilibrium but energy equation does not, what does a star look like while evolving towards Hayashi track from lower temperatures?

The problem is that the hydrostatic equilibrium is not stable for stars to the right of the Hayashi limit. So although there is plenty of kinetic energy in the star to prevent further gravitational contraction, it is like a bunch of pencils balancing on their points, if the T gradient results in too low a surface temperature. So each of the pencils fall over, though no contraction of the star as a whole is needed-- all that is needed is that the kinetic energy is reapportioned such that the T structure readjusts, and the surface T goes up to the Hayashi limit. That all happens on the energy transport timescale, so it's nothing like the hydrostatic equilibrium timescale that virializes the gas, but it is still very quick on an evolutionary timescale. So we don't find stars to the right of the Hayashi track.

 

[snorkack]

ideal gases, and the molecules dissociate, and even if they didn't, the primary molecule would be diatomic anyway. So yes, your point is indeed moot.

A diatomic molecule with a vibrational degree of freedom would also have 6.

The problem is that the hydrostatic equilibrium is not stable for stars to the right of the Hayashi limit. So although there is plenty of kinetic energy in the star to prevent further gravitational contraction,

Then is the hydrostatic equilibrium stable or not?

it is like a bunch of pencils balancing on their points, if the T gradient results in too low a surface temperature. So each of the pencils fall over, though no contraction of the star as a whole is needed-- all that is needed is that the kinetic energy is reapportioned such that the T structure readjusts, and the surface T goes up to the Hayashi limit. That all happens on the energy transport timescale, so it's nothing like the hydrostatic equilibrium timescale that virializes the gas, but it is still very quick on an evolutionary timescale. So we don't find stars to the right of the Hayashi track.

But energy can only be transported from a hotter body to cooler. What would a body look like whose centre is slightly cooler than Hayashi limit?

 

[Ken G]

A diatomic molecule with a vibrational degree of freedom would also have 6.

Another moot point-- the vibrational degree of freedom does not get excited until the molecule is being dissociated anyway. That's why it doesn't matter.

Then is the hydrostatic equilibrium stable or not?

It is stable once the surface temperature is raised to the Hayashi limit. That's what the Hayashi limit is.

But energy can only be transported from a hotter body to cooler. What would a body look like whose centre is slightly cooler than Hayashi limit?

As I said above, such a body is still in free fall, coming across the Jeans limit, and is not considered a star yet. A star is defined as being when the timescale for change of the object is much larger than its free-fall time, and is on the side of the Jeans limit where the self-gravity is important. So in effect, a star is an object that "wants" to collapse gravitationally, but doesn't.

 

[snorkack]

Another moot point-- the vibrational degree of freedom does not get excited until the molecule is being dissociated anyway.

Then what´s the effect of dissociation on heat capacity?

It is stable once the surface temperature is raised to the Hayashi limit. That's what the Hayashi limit is.
As I said above, such a body is still in free fall, coming across the Jeans limit, and is not considered a star yet.

If diatomic molecular hydrogen has sufficiently small heat capacity to reach hydrostatic equilibrium, then how is it in free fall?

A star is defined as being when the timescale for change of the object is much larger than its free-fall time, and is on the side of the Jeans limit where the self-gravity is important. So in effect, a star is an object that "wants" to collapse gravitationally, but doesn't.

And a body of molecular hydrogen supported by ideal gas pressure against self-gravity meets that definition.

 

[Ken G]

Then what´s the effect of dissociation on heat capacity?

No significant effect because it's only temporary, and H has dissociated before you have anything you could call a star, i.e., before hydrostatic equilibrium is achieved. It certainly has a destabilizing effect on the force balance, but again, the force balance is already destabilized, that's why stars form in the first place.

If diatomic molecular hydrogen has sufficiently small heat capacity to reach hydrostatic equilibrium, then how is it in free fall?

The gravitational instability plays out in several phases. The first phase is the Jeans instability, which is the phase when the energy transport timescale is shorter or comparable to the free-fall time. This is normally treated as an "isothermal" phase of free fall, so dissociation would not occur. Then as optical depth builds, the energy transport timescales get longer than the free-fall time, and the temperature starts to rise. That's when dissociation occurs, but the star has still not achieved hydrostatic equilibrium, in part due to the energy going into dissociation, and in part because the outer regions of the star still have a longer free-fall time than the age of the system. That's what can lead to complicated structure like infall shocks, it's a dynamical phase that is still considered to be the formation of the star. Eventually, a kind of global hydrostatic equilibrium sets in for the first time, the mass of the star is determined, and the temperature is high enough that dissociation of H is over with. This is the first time we would call it a star, or a protostar, rather than a dynamical object that is in the process of forming a star. And this is also the phase where the Hayashi limit applies, which determines how the kinetic energy is proportioned inside the star such that the hydrostatic equilibrium is locally stable. That also leads to a surface T in the general vicinity of 3000 K, due to the surface opacity effects (largely from H minus) that cause the Hayashi limit.

And a body of molecular hydrogen supported by ideal gas pressure against self-gravity meets that definition.

That phase of star formation doesn't exist. There are not stars whose primary constituent is molecular hydrogen, we just don't find objects like that because the molecular phase does not coincide with the stages we regard as stars. I don't claim to know all the details of the dynamical phase that is a star in formation, indeed it is very much on the research frontier. For example, we know angular momentum plays a key role, that's why forming stars have disks around them, and there could be a role for magnetic fields, though I haven't mentioned either rotation or magnetism in the above. But when we finally get a self-gravitating object with an age much longer than its free-fall time, that is in a stable hydrostatic equilibrium everywhere, we have what we call a star or at least a protostar, and it obeys the Hayashi limit, and is largely composed of monatomic hydrogen.

 

[snorkack]

H has dissociated before you have anything you could call a star, i.e., before hydrostatic equilibrium is achieved. It certainly has a destabilizing effect on the force balance, but again, the force balance is already destabilized, that's why stars form in the first place.
The gravitational instability plays out in several phases. The first phase is the Jeans instability, which is the phase when the energy transport timescale is shorter or comparable to the free-fall time. This is normally treated as an "isothermal" phase of free fall, so dissociation would not occur. Then as optical depth builds, the energy transport timescales get longer than the free-fall time, and the temperature starts to rise. That's when dissociation occurs

A recent gas cloud starts at 3 K and appreciable concentration of metals and dust, causing opacity at all wavelengths.
There is a wide range of temperatures between 3 K, where collapse starts, and 1500...2000 K, where dust evaporates, decreasing but not eliminating metal opacity.
So can a gas cloud become optically deep, and stop free fall, before heating to 1500 K?

Eventually, a kind of global hydrostatic equilibrium sets in for the first time, the mass of the star is determined, and the temperature is high enough that dissociation of H is over with. This is the first time we would call it a star, or a protostar, rather than a dynamical object that is in the process of forming a star. And this is also the phase where the Hayashi limit applies, which determines how the kinetic energy is proportioned inside the star such that the hydrostatic equilibrium is locally stable. That also leads to a surface T in the general vicinity of 3000 K, due to the surface opacity effects (largely from H minus) that cause the Hayashi limit.

So, what´s the stage immediately before reaching the Hayashi limit temperature? What determines the radius and luminosity where Hayashi limit is reached?

 

[Ken G]

A recent gas cloud starts at 3 K and appreciable concentration of metals and dust, causing opacity at all wavelengths.
There is a wide range of temperatures between 3 K, where collapse starts, and 1500...2000 K, where dust evaporates, decreasing but not eliminating metal opacity.
So can a gas cloud become optically deep, and stop free fall, before heating to 1500 K?

The formation of a star is a complex topic. Normally you have a "core" of a molecular cloud, where the density is high, that intially goes across the Jeans instability, though this picture already invokes idealizations. The free-fall time depends on density, so the regions that go into hydrostatic equilibrium (when the free-fall time drops below the age of the system) are in the highest density regions first. But more mass continues to infall, either by some kind of spherical Bondi accretion, or by an accretion disk if angular momentum is important. In either case, you can have an accretion shock, so part of the system can be in force balance, and part not. In any event, these complicated partially dynamical stages are not considered stars yet, and the assumptions in the Hayashi limit are not yet applicable.

So, what´s the stage immediately before reaching the Hayashi limit temperature? What determines the radius and luminosity where Hayashi limit is reached?

A locally stable self-gravitating solution that is everywhere in force balance-- a star model, rather than some stage in dynamical star formation.

 

[snorkack]

In any event, these complicated partially dynamical stages are not considered stars yet, and the assumptions in the Hayashi limit are not yet applicable.
A locally stable self-gravitating solution that is everywhere in force balance-- a star model, rather than some stage in dynamical star formation.

So what precisely are the assumptions of Hayashi limit?
And what determines the point where Hayashi limit is reached?

 

[Ken G]

So what precisely are the assumptions of Hayashi limit?

A self-gravitating sphere of stellar composition that not only contains enough internal kinetic energy to be globally virialized, but also has that kinetic energy apportioned to allow local hydrodynamical stability at every point.

And what determines the point where Hayashi limit is reached?

When the assumptions become valid, which is essentially when it becomes possible to have a configuration like that. It is the first long-lived phase in the life of what we call a star.

 

[snorkack]

What is the condition immediately preceding Hayashi track? Hydrostatically unstable free fall, or turbulent convection?

 

[Ken G]

I think the answer is that it is a state that is hard to characterize, because it does not submit to simple analysis. It may have different parts doing different things, some more dynamical than others. After all, "turbulence" is a kind of catch-all phrase. I suppose one would need to look at simulations of star formation, but still, you have to choose if you want to put in angular momentum, and magnetic fields, including interactions between the two like dynamos. I'd say it's a research frontier.

 

[snorkack]

What does a star passing through Hayashi forbidden zone look like?
There are 2 ways for a star to exist in Hayashi forbidden zone: either it is hydrostatically supported by degeneracy pressure, in which case it is small and of low luminosity, or else it is rapidly contracting.
And a star contracting through Hayashi forbidden zone should be big and bright (in near-infrared).
Since a lot of brown dwarfs are seen in near-IR, despite being small and dim, how about big and bright protostars in Hayashi forbidden zone?

 

[Ken G]

What does a star passing through Hayashi forbidden zone look like?

It isn't called a star at all, that's star formation. And as I said, one needs a different approach to answer that question-- rather than the approach we take for stars, which is asking "what set of simplifiying idealizations can help us understand the basic structure of stars in such a way that time becomes a kind of external driving parameter", one needs to follow the time evolution explicitly from stage to stage. That is much more difficult, because you cannot know what is happening in any stage without first knowing the situation in the previous stage.

There are 2 ways for a star to exist in Hayashi forbidden zone: either it is hydrostatically supported by degeneracy pressure, in which case it is small and of low luminosity, or else it is rapidly contracting.

As I said above, the Hayashi limit does not apply for degeneracy, it applies for ideal gases. Including degeneracy allows a continuous transition from star models to brown dwarf models to planet models, all the way down to the blueprints for your house. But the topic of the thread was defined early on.

Since a lot of brown dwarfs are seen in near-IR, despite being small and dim, how about big and bright protostars in Hayashi forbidden zone?

As I said, brown dwarfs can have long-lived phases in that region, because they can be in a stable hydrostatic equilibrium. Stars like the Sun, which are not degenerate until they become white dwarfs, cannot.