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Is Pauli Exclusion Principle alone responsible for electron degeneracy pressure?
BenAS said:What happens to the electrons when gravity overcomes the degeneracy pressure? Are they no longer considered identical fermions?
nikkkom said:What does happen in accreting white dwarfs is that electrons eventually reach energies high enough for e + p -> n reaction
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.Pete Muller said: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?
Ken G said: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.
Ken G said: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
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.nikkkom said: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.
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.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".
Ken G said: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.
Do you agree that air is degenerate?Ken G said: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.
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.snorkack said:Do you agree that air is degenerate?
Yes.Pete Muller said: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).
Yes, to keep them from being able to drop to any lower energy states.So to keep the highest energy electrons from changing states the temperature must be zero (full degeneracy)?
It never happens, you never get to zero temperature.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?
Ken G said: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,"
A more useful way of stating the same is that a quantum system will have nonzero kinetic energy in ground state.Ken G said: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.
Not important. You could achieve a similar result with indistinguishable particles - they merely need to be bosons rather than fermions.Ken G said: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).
I guess if you think that isn't what I already said.snorkack said:A more useful way of stating the same is that a quantum system will have nonzero kinetic energy in ground state.
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.Not important. You could achieve a similar result with indistinguishable particles - they merely need to be bosons rather than fermions.
Obviously.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.
I'm acting on the assumption that everyone reading this already knows that quite well.2 electrons can fit in the same state (with 2 different spins that electron has).
Yes, that was my point above, with oxygen instead of lithium. The rest I assumed was already well known.But a lithium-like atom would be different - 3rd electron cannot fit into an occupied 1s orbital, but a 3rd negative pion could.
Electron degeneracy pressure is a quantum mechanical effect that occurs when a large number of electrons are squeezed into a small volume. It is the force that keeps white dwarf stars from collapsing under their own gravity.
According to the Pauli exclusion principle, no two electrons can occupy the same quantum state. When electrons are squeezed into a small volume, they are forced into higher energy states and create a pressure that counteracts the force of gravity.
Electron degeneracy pressure is commonly observed in white dwarf stars, which are the remnants of low to medium mass stars. It is also present in neutron stars, which are the remnants of high mass stars.
Electron degeneracy pressure is a quantum mechanical effect that arises from the behavior of electrons. It is distinct from thermal pressure, which is caused by the random motion of particles, and radiation pressure, which is caused by the transfer of energy through photons.
If the force of gravity is strong enough to overcome electron degeneracy pressure, the object will continue to collapse. This can result in the formation of a black hole or a supernova explosion.