Is Pauli Exclusion Principle alone responsible for electron degeneracy pressure?
What happens to the electrons when gravity overcomes the degeneracy pressure? Are they no longer considered identical fermions?
Gravity does not overcome the degeneracy pressure. With more compressing force, object becomes smaller until degeneracy pressure rises enough to counteract that force.
Degeneracy pressure rises because more electrons are forced into higher-energy states due to Pauli Exclusion Principle (more elecrons are forced into the same volume, and all lower-energy states are already taken).
What does happen in accreting white dwarfs is that electrons eventually reach energies high enough for e + p -> n reaction.
Yet it is possible for gravity to overcome degeneracy pressure, for example if you merge two white dwarfs of a solar mass each. It doesn't mean the particles are no longer identical, it just means the system collapses due to the strong gravity. As the system collapses, the electrons gain energy, and so their deBroglie wavelengths get even smaller. Thus they always stay "out of each other's way", and the Pauli Exclusion Principle is never violated during the collapse (until they fall into a black hole or something else happens to them, as has been mentioned). Indeed, the whole reason the collapse can occur is that it never violates the Pauli exclusion principle, this is what prevents the Earth from collapsing under its own gravity-- to do so would indeed violate the PEP.
So in a core collapse supernova is the process that the electron degeneracy pressure is eliminated because the electrons are used up in the e + p -> n reaction causing the collapse or is it overcome by gravity first?
It's gravity first. There is a common misconception that degeneracy is a cause of pressure, such that if you waved a magic wand and put a label on every electron (making them distinguishable and hence free from the Pauli exclusion principle) that degeneracy pressure would disappear. That's not true, because degeneracy pressure is actually just normal gas pressure, in the sense that it stems from the motion of the particles (essentially their kinetic energy density), so if you eliminated the PEP, nothing at all would happen to the pressure, not right away anyway. Instead, the absence of degeneracy would allow the particles to lose heat, so the pressure would fall very slightly below gravity, and the object would very slowly contract. That contraction would eventually lead to higher temperatures and pressures, causing the electrons to go relativistic. That's what would cause gravity to win-- it's always the speed of light that is the limiting factor, that's what gives gravity the edge and causes catastrophic collapse.
In the case of core collapse without any magic wands, what actually happens is the electrons either go relativistic before they go degenerate (which gives them not only the energy to make neutrons, but also the energy to make photons that can break up the nuclei, eating up basically all the energy the star has released in its lifetime of fusing those nuclei, and that's what makes gravity win against pressure), or else they go degenerate first but still manage to go relativistic, leading to all the same results. An interesting technicality is that, although degeneracy can sometimes prevent core collapse by keeping the electrons from ever going relativistic (for core mass below 1.4 solar masses), there are situations where it actually makes the core collapse happen sooner (for core mass greater than 1.4 solar masses) by causing the electrons to go relativistic sooner (as degeneracy makes the electrons steal most of the ion kinetic energy).
Define "very slowly". It depends on temperature, since luminosity is proportional to T^4. At 10000K, the cooling may be slow, at one million kelvins, it's 100 million times faster.
I don't see how that is possible for degenerate electrons. Degenerate electrons (even relativistic ones) can't lose energy, and thus can't generate photons and can't "break up nuclei".
By "very slowly" I simply mean on heat transport timescales, rather than momentum transport timescales (the latter is generally way faster). In other words, on timescales long enough to assume hydrostatic equilibrium. In the case of stellar cores, heat transport times might be thousands to billions of years (except when neutrinos do it, so that's another key moment in the collapse of a core-- when it can make a significant neutrino flux), whereas momentum transport times might be seconds to hours.
Degeneracy is merely a useful approximation, the electrons in a collapsing core are not fully degenerate, so don't have zero temperature and can make photons. For example, in the case of a type Ia supernova, the temperature goes high enough to fuse carbon, even though the Chandrasekhar mass is derived by assuming complete degeneracy. In a core collapse, the temperature goes even higher. All degeneracy really means is that kT is way less than the kinetic energy per electron, but it can still be very high if the kinetic energy per electron goes highly relativistic. The terms "degenerate", "partially degenerate", and "fully degenerate", are not always carefully disambiguated, because complete degeneracy is such a useful reference point.
It's an interesting question how photons get made in highly degenerate plasma. For a given ratio E/kT, where E is the kinetic energy per electron (and that ratio is a measure of the degree of degeneracy), that ratio characterizes the ratio of the total kinetic energy in the electrons, to the total kinetic energy in the ions. So if that ratio is large, then the electrons have the kinetic energy. However, only a fraction of order (kT/E)^2 is available to create photons, i.e., that's what we might call the available "thermal energy" in the electrons, whereas all the kinetic energy in the ions is thermal. So the ratio of thermal energy in the electrons to the ions is kT/E and is small, perhaps suggesting that the ions will be responsible for making the photons, since they carry the heat content. However, to make light, you need an oscillating dipole moment, so you need the electrons and ions to work together. But the electrons have the larger acceleration, in inverse proportion to their mass, in that action/reaction. Since the Larmor formula says the radiated power scales as acceleration squared, putting this all together tells us the ratio of light made by electrons to light made by ions is kT/E times the square of their mass ratio. Hence, electrons make the photons, unless the degree of degeneracy is spectacularly high.
This zero temperature electron state confuses me. Does this mean that, while all of the lower electron energy states are taken, there is normally room at the higher energy states to change levels (partial degeneracy). So to keep the highest energy electrons from changing states the temperature must be zero (full degeneracy)? Does this really happen or is this an approximation used to make the calculations easier. I.e, the pressure contribution by electron thermal energy is so small we can ignore it?
Do you agree that air is degenerate?
I can only say that the statement "air is degenerate" would be highly misleading, to say the least. Most of the electrons in air are in their grounds states, but that just isn't what anybody would mean by "air is degenerate," they would mean air is a degenerate gas, which it isn't.
Yes, to keep them from being able to drop to any lower energy states.
It never happens, you never get to zero temperature.
Thank you Ken G. Nothing like the pleasure of mostly getting a concept!
Not only are most of the electrons in their ground states. Even if it were not the case, you could easily have a situation where most of the electrons are ionized - but the minority that are not, like the 2 innermost electrons of N5+ ions, being the innermost electrons, have such huge kinetic energy of their orbital movement that it outweighs the total kT energy elsewhere.
But what is the point of what you are saying? You also have mc^2 mass energy that vastly exceeds everything you are talking about, but that doesn't affect the ideal-gas nature of air either.
There still appears to be some confusion on this thread about what degeneracy is. Degeneracy is a quantum mechanical effect that alters the ground state of a system, that's it. But it is the thermodynamic nature of a ground state that is the crucial element. As a system approaches its ground state, its temperature drops way below what you think it should be had there not been a quantum mechanical ground state there, if you are tracking what is happening to the kinetic energy (and since energy is a useful conserved quantity, it is often a very good idea to track the kinetic energy). That is what I was saying above.
So this is the answer to the confusion snorkack is raising-- for example, if you have an ideal gas of neutral hydrogen atoms, that gas has a temperature that is related to the kinetic energy in the translational degrees of freedom of the particles, and it also has a temperature that is related to the excitation of the electrons in the atoms (and we normally take those two as the same for it to formally mean a temperature, but it's not important here). The fact that the translational degrees of freedom are nowhere near their ground state is why the temperature is just what you'd think if you look at the kinetic energy of motion of the ideal gas particles (i.e., 3/2 kT each), and the fact that the bound electrons are very close to their ground state is why the temperature is way lower than you would think if you look at the kinetic energy of the electrons within the atoms. All degeneracy does is alter the nature of the ground state, such that if we had neutral oxygen instead of neutral hydrogen, the ground state where the electrons reach zero temperature has less kinetic energy (due to the degeneracy) than it would if there was not degeneracy (i.e., if electrons were distinguishable). But the point is still that as the system approaches its ground state (which unfortunately in many contexts means we say the system is becoming "degenerate", an almost meaningless choice of words that causes no end of confusion), it means the temperature is forced to be way lower than you would think classically from the kinetic energy. That is the thermodynamic consequence of "degeneracy," and that's all it is, it isn't anything else. As such, that's all "degeneracy pressure" ever meant, though you will find all kinds of misconceptions around it being some kind of "force" or "source of pressure," neither of which is true.
A more useful way of stating the same is that a quantum system will have nonzero kinetic energy in ground state.
Not important. You could achieve a similar result with indistinguishable particles - they merely need to be bosons rather than fermions.
The description of a hydrogen-like atom, with a single particle orbiting the nucleus, would look much the same whether the nucleus is orbited by an electron or a negative pions - only parametres like mass would change.
2 electrons can fit in the same state (with 2 different spins that electron has).
But a lithium-like atom would be different - 3rd electron cannot fit into an occupied 1s orbital, but a 3rd negative pion could.
I guess if you think that isn't what I already said.
Your claim of "not importance" is logically incorrect. Your argument is like saying that it is "not important" that a broken leg can be healed by putting it in a cast because you could also remove the leg entirely.
I'm acting on the assumption that everyone reading this already knows that quite well.
Yes, that was my point above, with oxygen instead of lithium. The rest I assumed was already well known.
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