What prevents a star from collapsing after stellar death?

[Ken G]

Note also that it is not just an issue of the age of the Sun, the Sun will never produce any iron. That only happens in much more massive stars. You don't necessarily have to be massive enough to go supernova, the highest mass stars that don't supernova can make white dwarfs with iron in them, but that's still much more massive than the Sun. The Sun will never get hot enough to fuse anything past carbon and oxygen in any significant quantity.

 

[Monsterboy]

If nickel and iron are one of the last elements to be formed in the stars then how did we come across elements as heavy as uranium etc ?

 

[Ken G]

A common and good question. The difference is in what types of nucleosynthesis release net heat, so play a key role in stellar evolution and can delay the ultimate fate of the star, and what types simply are caused by a hot environment, as stars often encounter during late stages. So when the focus is only on fusion up to iron, it means the focus is on powering the star and affecting the evolutionary timescales. If core collapse occurs, all bets are off on past nucleosynthesis, as a huge amount of gravitational energy is released, and a super bright radiation field appears that can disrupt nuclei and undo essentially all the products of past nucleosynthesis in the central regions of the star. These new nuclear effects can result in all kinds of nuclei, much of which can be spewed out in the supernova bounce. But there's no concern that they do not release heat, because there is plenty of heat to go around.

 

[Drakkith]

Look up the R-process on wikipedia. I'm on my phone right now, otherwise I'd post the link myself.

 

[Matterwave]

If nickel and iron are one of the last elements to be formed in the stars then how did we come across elements as heavy as uranium etc ?

There are 2 processes (and a third, tentative process) generally thought to be attributable for the formation of matter heavier than iron and nickel. They are the R-process, the S-process, and the tentative P-process.

The others have mentioned the R-process, which is thought to occur in supernovae environments. It stands for the rapid capture of neutrons onto seed nuclei. The neutrons are captured "rapidly" meaning they are captured far faster than beta decay is allowed to happen. As such, you build up very very neutron rich, unstable, nuclei which then decay via many beta-decays into stable nuclei. The S-process, which is thought to occur in a variety of astrophysical environments, chief among them being in asymptotic giant branch stars, is the slow capture of neutrons by seed nuclei (usually starting from something much lighter than iron, since AGB stars were not massive enough to create iron in the first place). The nuclei capture neutrons very slowly, much slower than the beta decay process, and the nuclei basically move up the valley of stability 1 nucleon at a time, always staying in the valley of stability.

These two processes, which although not entirely understood, are quite well studied, but are not able to explain every nuclei that we see (they do explain many/most of the heavier elements). There are some proton rich nuclei which should not be able to form from either the R-process or the S-process, because, essentially to get to those nuclei via the S or R process, you must pass by stable nuclei which won't decay in the right way to those nuclei. These are the so called "p-process nuclei". These nuclei were first conjectured to be created by a "proton-capture process" (hence the name p-process), but the viability of such a process is...tenuous. There is current active research into how these "p-process nuclei" are formed (probably by some modified version of proton capture), and this is the least understood of the processes.

 

[Ken G]

Thank you for that extremely cogent summary, I had never even heard of the p-process though I knew that the R-process is hard or impossible to do in the laboratory and a lot about it is not yet understood.

 

[BrandonP]

"What prevents a star from collapsing after stellar death?"

If an object becomes smaller than its Schwarzschild radius,
then it will collapse into a black hole.

If not, then it won't.

That's really all there is to it.

 

[Murtuza Tipu]

Your first paragraph is not quite right. Gas pressure does not "stop" upon formation of an iron core, it is merely that the star cannot generate further heat from nuclear reactions and becomes unstable to collapse. i.e. The star does collapse! Perhaps what you mean is what halts the collapse (sometimes) before the star disappears inside its own event horizon and becomes a black hole? The answer is the degeneracy pressure of neutrons that are formed (endothermically) in electron capture events as the star collapses.

The analogy of filled "shells" is not too bad. In quantum mechanics we find that there are a finite number of possible quantum states per unit momentum per unit volume (often called "phase space"). In a "normal" gas, the occupation of these quantum states is governed by Maxwell-Boltzmann statistics - progressively fewer of these states are filled, according to .

In a Fermi gas at very high density or very low temperature, we reach a situation where the Pauli exclusion principle limits the occupation of these states to 2 particles per energy/momentum state (one for each spin); particles that might otherwise have occupied very low energy states are forced to occupy states of higher energy and momentum. In a "completely degenerate" gas, which is a good approximation for the electrons in a White Dwarf star or the neutrons in a Neutron star (the relevant case here), all the energy states are filled up to something termed the Fermi energy, with zero occupation at even higher energies.

Pressure is caused by particles having momentum (this is just basic kinetic theory). The large number of fermions with non-zero (even relativistic in some cases) momentum is the reason that a degenerate gas exerts a pressure, even if its temperature is reduced to close-to-zero. In fact, once a gas of fermions approaches complete degeneracy, a change in temperature has almost no effect on its pressure.

One point I will take issue with in your question, is the statement that "such particles cannot occupy the same small volume of space". In fact, the restriction is on the occupation of phase space. In a neutron star, the neutrons are almost touching each other, with separations of m. You can cram lots of particles into a small volume, but only at the expense of giving them large momenta.

 

[Drakkith]

Pressure is caused by particles having momentum (this is just basic kinetic theory). The large number of fermions with non-zero (even relativistic in some cases) momentum is the reason that a degenerate gas exerts a pressure, even if its temperature is reduced to close-to-zero. In fact, once a gas of fermions approaches complete degeneracy, a change in temperature has almost no effect on its pressure.

Question. If a group of fermions is degenerate with many of them having relativistic momentum, and a small amount of energy is added to raise the temperature, what happens to the momentum of the fermions?

 

[avito009]

If an object becomes smaller than its Schwarzschild radius,
then it will collapse into a black hole.

But I thought the maximum size of a stable white dwarf, approximately 3 × 1030 kg (about 1.4 times the mass of the Sun). Stars with mass higher than the Chandrasekhar limit ultimately collapse under their own weight and become neutron stars or black holes. Stars with a mass below this limit are prevented from collapsing by the degeneracy pressure of their electrons.

So who is right Brandon or Me?

 

[Drakkith]

But I thought the maximum size of a stable white dwarf, approximately 3 × 1030 kg (about 1.4 times the mass of the Sun). Stars with mass higher than the Chandrasekhar limit ultimately collapse under their own weight and become neutron stars or black holes. Stars with a mass below this limit are prevented from collapsing by the degeneracy pressure of their electrons.

So who is right Brandon or Me?

A white dwarf is not smaller than its Shwarzchild radius so it doesn't collapse.

 

[Ken G]

So you are both right, though your answer is more useful because it can help us understand why and when stars can go below their Schwarzschild radii.

 

[avito009]

A white dwarf is not smaller than its Shwarzchild radius so it doesn't collapse.

So correct me if I am wrong. If a star has mass greater than the Chandrasekhar Limit, due to this greater mass the gravitational force is more (Moon has lesser gravity than Earth because mass of Moon is lesser than mass of Earth. So grater the mass greater the gravitational pull). So degeneracy pressure provided by the electrons is weaker than the inward pull of gravity. So when this mass is very less the schwarzschild radius is also less. So when the radius of the star falls below the schwarzschild radius then we get a black hole.

Am I right?

 

[Drakkith]

So when this mass is very less the schwarzschild radius is also less.

Exactly. You'd need to compress the Moon to a much smaller volume than the Sun in order for it to form a black hole.

 

[Murtuza Tipu]

Have a look at this applet (it is designed for electrons in WDs, but will do here). It shows occupation index vs energy for fermions. To get the momentum distribution, you multiply this by the density of momentum states g(p) = 8pi p^2/h^3. Look at what happens when you increase the temperature whilst keeping the density constant. You get a partial degeneracy and a tail of particles with energies higher than the Fermi energy.geogebratube.org/student/b87651#material/28528

 

[D H]

So correct me if I am wrong. If a star has mass greater than the Chandrasekhar Limit, due to this greater mass the gravitational force is more (Moon has lesser gravity than Earth because mass of Moon is lesser than mass of Earth. So grater the mass greater the gravitational pull). So degeneracy pressure provided by the electrons is weaker than the inward pull of gravity. So when this mass is very less the schwarzschild radius is also less. So when the radius of the star falls below the schwarzschild radius then we get a black hole.

Am I right?

Not quite.

The Chandrasekhar limit is not the boundary between a star and a black hole. This limit pertains to white dwarfs, burnt-out stars that are mostly carbon and oxygen and in which electron degeneracy pressure sustains the balance between pressure and gravitation. There are many stars that are significantly more massive than the Chandrasekhar limit. In fact, a star whose initial mass is considerably larger than the Chandrasekhar limit is needed to produce a white dwarf that is close to the Chandrasekhar limit. Intermediate mass stars become rather leaky once they reach the red giant phase. They expel the majority of their mass into space as they burn helium.

The Chandrasekhar limit is instead the boundary between a white dwarf and neutron star. It is also very close to the point at which a growing white dwarf forms a type IA supernova. A white dwarf that has a binary pair can steal mass from its partner. A type IA supernova occurs when the mass of the dwarf gets close to the Chandrasekhar limit. The temperature builds and builds as the the white dwarf ever close to the limit, eventually reaching the point where carbon fusion starts. This triggers carbon fusion throughout the star, and that in turn triggers oxygen fusion. This all happens very quickly. Temperature rises to the point where temperature pressure dominates over radiation pressure and then over gravity. At this point, the entire star blows up.

 

[UncertaintyAjay]

A quantum state for an elecrron is defined by 4 quantum numbers. The first- the principle quantum number- tells you in which shell, the electron can be found.The second - the azimuthal quantum number tells you the orbital in which it is located. There are 5 orbitals- s,p,d and f. The fifth- the g orbital is hypothetical. The azimuthal quantum number is 0, for s; 1 for p; 2 for d; 3 for f. An orbital can only hold two orbitals. So there are 3 degenerate p orbitals( orbitals with the same energy) , each containing 2 electrons, 5 degenerate d orbitals and 7 degenerate f orbitals. The magnetic quantum number tells you which of these degebrate orbitals the electron can be found in. The last number the spin quantum number doesn't have a classical analog. There is a spin up(+1/2) and a spin down(-1/2). Two electrons are in the same state if both of them have all four quantum numbers identical. This is forbidden by paukis excluskon principle.