Questions about the lifecycle of stars

[stefan r]

The part of my post that you highlighted refers to the non-degenerate shell, not the degenerate core.

You are right, that was the wrong sentence. Sorry. You probably understood it fine but I am not convinced someone reading it will.

This contraction increases the gravitational pull on the shell of hydrogen just outside the core.

Anything at a given radius has the same gravitational pull from a same mass object regardless of the object's density. The core contraction allows plasma from the shell to fall in. At the lower radius the plasma has a higher temperature.

If I understood it correct then there is no difference between the fusion in the non-degenerate shell and main-sequence stars. Identical temperature and pressure would spawn identical fusion rates.

High metal stars are more compact when they form and burn hydrogen faster. The core burns hydrogen faster as helium increases (assuming total mass stays the same). The older core would continue ramping the burn rate exponentially except that degeneracy pressure delays or stops the collapse.

Hypothetically, if degeneracy pressure did not exist would the sun burn straight to iron and a planetary nebula or would it form a micro-black hole?

 

[Ken G]

If I understood it correct then there is no difference between the fusion in the non-degenerate shell and main-sequence stars. Identical temperature and pressure would spawn identical fusion rates.

That would be true if the temperatures were the same, but Drakkith was pointing out that the temperature in the shell is much higher than typical in the core of a main sequence star, and that's the fundamental reason that red giants are so much more luminous than the main-sequence stars that give rise to them.

Hypothetically, if degeneracy pressure did not exist would the sun burn straight to iron and a planetary nebula or would it form a micro-black hole?

That's an important question, and a good way to understand what degeneracy actually does. If there was not degeneracy (say, if every electron was distinguishable somehow), then there would be nothing to stop the solar core from continuing to lose heat and continue to contract, raising its temperature without limit. When the electrons go ultra relativistic, the core would become vulnerable to catastrophic collapse. So it would burn to iron, and continue to contract after that, ultimately collapsing into a mini black hole (if there was not degeneracy in neutrons either).

 

[snorkack]

If there was not degeneracy (say, if every electron was distinguishable somehow),

Or if every electron were indistinguishable boson.

Come to think of - protons are fermions. Alpha particles are not.
Do nuclei repel each other in purely electrostatic manner, or also as indistinguishable fermions?
What allows two hydrogen atoms to repel one another? Their electrons are fermions, and so are their protons, but the atoms in total are bosons.

 

[Ken G]

Or if every electron were indistinguishable boson.

Come to think of - protons are fermions. Alpha particles are not.
Do nuclei repel each other in purely electrostatic manner, or also as indistinguishable fermions?

Electrostatic forces are not so important within nuclei. Instead, one typically just imagines the nucleons are physically pressed together with more or less constant size per nucleon, by the strong force. Degeneracy doesn't matter much either, because the protons and neutrons also press together, and they're distinguishable.

What allows two hydrogen atoms to repel one another? Their electrons are fermions, and so are their protons, but the atoms in total are bosons.

I would have expected hydrogen atoms to experience van der Waals attraction, rather than repulsion, but either way it's electrostatic, not fermionic. Where you see fermionic effects are in the multielectron atoms, where you have to open up the atoms and look at their guts. But what you are saying is, you can't make a cool degenerate gas out of neutral hydrogen atoms, and that's true-- what happens instead is the electrostatic forces create a crystal. But the electrons wouldn't stay in the neutral hydrogen, they'd tunnel out and become like a gas of free particles in a "box" made by the proton lattice, I presume that's what will eventually happen deep in the core of Jupiter as it cools, but I don't know if the repulsion then will be more from the crystal lattice or the degenerate electrons. It's the same issue for cooling white dwarfs, the ions eventually crystallize but I believe the repulsion is still mostly from the degenerate electron pressure.

 

[snorkack]

I would have expected hydrogen atoms to experience van der Waals attraction, rather than repulsion, but either way it's electrostatic, not fermionic. Where you see fermionic effects are in the multielectron atoms, where you have to open up the atoms and look at their guts. But what you are saying is, you can't make a cool degenerate gas out of neutral hydrogen atoms, and that's true-- what happens instead is the electrostatic forces create a crystal. But the electrons wouldn't stay in the neutral hydrogen, they'd tunnel out and become like a gas of free particles in a "box" made by the proton lattice,

But they don´t. While in solid diprotium, the molecules do tunnel to rotate around in their positions, the electrons are trapped in molecules. Solid hydrogen is an insulator.
For comparison: He-4 atoms are bosons. But while all He-4 atoms do occupy the same state, they do not occupy the same volume - despite consisting of indistinguishable bosons, ground state helium has a finite density. They still repel.

 

[Drakkith]

But they don´t. While in solid diprotium, the molecules do tunnel to rotate around in their positions, the electrons are trapped in molecules. Solid hydrogen is an insulator.

Are you talking about solid hydrogen as in metallic hydrogen or something else?

 

[Ken G]

But they don´t. While in solid diprotium, the molecules do tunnel to rotate around in their positions, the electrons are trapped in molecules. Solid hydrogen is an insulator.
For comparison: He-4 atoms are bosons. But while all He-4 atoms do occupy the same state, they do not occupy the same volume - despite consisting of indistinguishable bosons, ground state helium has a finite density. They still repel.

Good point, I forgot the Lennard-Jones potential (often used as a proxy for these kinds of forces) goes repulsive at close distance. It's electrostatic, the fact that the particles are bosons merely means that their ground state is the zero point kinetic energy. Perhaps we should say that the ground state of bosons acts like they have springs between nearest neighbors, whereas the ground state of fermions acts like a Fermi sea and is generally reached while the interparticle forces are still attractive.

 

[JohnnyGui]

Compressibility is more of a local property of matter. If the density is independent on pressure, so pressure increases suddenly for practically no change of density, then the density is constant. If density increases with square of pressure then the planet does not change in radius with added mass.
And if density increases proportional to pressure, as is the case for isothermal ideal gas, then the planet will shrink by itself, without any added mass.

So it's the compressibility of a material that determines whether the relationship between pressure and density is proportional or something else? If so, is there a way to calculate the amount of compressibility that would give a proportional relationship?