How Are Electron Clouds Affected in Neutron Stars?

[imsmooth]

When matter is crushed in a neutron star are the atoms of molecules closer and the electrons are the same distance from the nucleus, or is the electron cloud closer to the nucleus? If the latter, how is this possible if the orbits have to have fixed orbitals due to the standing waves of the electron?

 

[Orodruin]

A neutron star contains no electrons or protons, and definitely no atoms or molecules.

Edit: Even a normal star is a plasma and does not contain atoms or molecules.

 

[Doug Huffman]

Atomic structure is obliterated in the crushing. https://en.wikipedia.org/wiki/Neutron_star#Structure

 

[mfb]

A neutron star contains no electrons or protons, and definitely no atoms or molecules.

In the inner region, the fraction of protons and electrons is negligible, right. Outside, you have protons and electrons as well, and close to the surface you even have a few neutral atoms if the neutron star is cold enough.

Edit: Even a normal star is a plasma and does not contain atoms or molecules.

Plasma just means a significant fraction of atoms are ionized - not all, and not necessarily fully ionized. At the surface of the sun, for example, at a temperature of ~1/2 eV, most hydrogen and helium atoms should be neutral.

If the latter, how is this possible if the orbits have to have fixed orbitals due to the standing waves of the electron?

The model of standing waves (as Bohr model) is outdated for nearly 100 years now.
Nearby particles can influence the energy levels - and in neutron stars you always have other things nearby. Even with pressures achievable on Earth you can compress solid matter (and therefore the distance between atoms) a bit.

 

[Orodruin]

In the inner region, the fraction of protons and electrons is negligible, right. Outside, you have protons and electrons as well, and close to the surface you even have a few neutral atoms if the neutron star is cold enough.
Plasma just means a significant fraction of atoms are ionized - not all, and not necessarily fully ionized. At the surface of the sun, for example, at a temperature of ~1/2 eV, most hydrogen and helium atoms should be neutral.

I stand corrected and confess being a naive and tired theorist at the time of posting.

 

[snorkack]

The model of standing waves (as Bohr model) is outdated for nearly 100 years now.

Orbitals are still standing waves, just that they are not lines like in Bohr model, but space filling waves.

 

[mfb]

Okay, but then there is no reason to assume that they cannot be influenced by the environment.

 

[Ken G]

The place where the electron "orbitals" are more or less packed as tightly as they can be, given the mass and associated gravity of the object (say about a solar mass), is when the radius of the object is about the size of Earth. That's a white dwarf, essentially (and note that the analogy to an atom is only in passing-- the binding force of the white dwarf or neutron star is gravity, not electric forces, so that's why the density is so much larger than in solid objects where the binding is electromagnetic). So to ever get something that is only a few solar masses to be as small as a neutron star (about the size of a city), you definitely need to get rid of almost all the electrons, which is what I believe the OP is asking. In essence, the electrons got swallowed up by protons, which is where all those extra neutrons came from. So you could say that was the effect of the environment-- to get rid of the electrons altogether, more so than squeezing them closer to the ions than you find in a white dwarf.

But the OP question can be recast into a form where it still has great significance-- instead of a neutron star, ask how the electrons can be that close to the protons in a white dwarf! The answer is that the energy the electrons have does not come from the electric attraction to the nearest ion, as in a Bohr atom, it comes from the gravity of all the ions! This changes everything, because even though gravity is much weaker than electric forces, there are a heck of a lot of ions and all their gravity adds up. So it is like a "Bohr atom" but with the energy given to the electron by the gravity of a huge number of ions, not the electric attraction of a single ion. That difference is all you need to understand the much higher densities of a white dwarf, and how the electrons can cozy up so much closer to the ions without disobeying quantum mechanics. Indeed, the electromagnetic interactions between the electrons and ions in white dwarfs is rather negligible! (The gravity doesn't act specifically on the electrons, it acts on the electron-ion fluid, which is kept together as a neutral fluid by electric forces. So it's not that the electric forces aren't there, it's that we don't need to include them explicitly in our understanding of where the electrons got their energy, we can just say the electric forces acted like a conduit to shunt gravitational energy into the electrons.)

 

[Chronos]

Elemental emission lines are difficult to explain in the absence of atoms.

 

[virgil1612]

The place where the electron "orbitals" are more or less packed as tightly as they can be, given the mass and associated gravity of the object (say about a solar mass), is when the radius of the object is about the size of Earth. That's a white dwarf, essentially (and note that the analogy to an atom is only in passing-- the binding force of the white dwarf or neutron star is gravity, not electric forces, so that's why the density is so much larger than in solid objects where the binding is electromagnetic). So to ever get something that is only a few solar masses to be as small as a neutron star (about the size of a city), you definitely need to get rid of almost all the electrons, which is what I believe the OP is asking. In essence, the electrons got swallowed up by protons, which is where all those extra neutrons came from. So you could say that was the effect of the environment-- to get rid of the electrons altogether, more so than squeezing them closer to the ions than you find in a white dwarf.

And if we talk in terms of pressure, it's useful to recall that the pressure that fights gravity in a white dwarf is the electron degeneracy pressure, while in a neutron star it is the neutron degeneracy pressure. Degeneracy pressure is a specifically quantum pressure (much bigger then the pressure of the electron (neutron) classical gas, that appears when the concentration of electrons (neutrons) is so high that the average distance between particles becomes close to the de Broglie associated wavelength.

 

[Ken G]

And if we talk in terms of pressure, it's useful to recall that the pressure that fights gravity in a white dwarf is the electron degeneracy pressure, while in a neutron star it is the neutron degeneracy pressure. Degeneracy pressure is a specifically quantum pressure (much bigger then the pressure of the electron (neutron) classical gas, that appears when the concentration of electrons (neutrons) is so high that the average distance between particles becomes close to the de Broglie associated wavelength.

Though note it depends what we are regarding as fixed when we make the comparison of degeneracy pressure to classical pressure. If we take the temperature as given, then the pressure is indeed much higher when the particles are identical fermions under degenerate conditions. But if we take the energy density as given, perhaps by the history of energization of the object, then the pressures are just the same-- degeneracy pressure is classical pressure, when the energy is treated as the given. This is a point that I feel receives far too little attention in the rush to imagine that degeneracy pressure is some kind of weird quantum mechanical form of pressure!

 

[virgil1612]

Though note it depends what we are regarding as fixed when we make the comparison of degeneracy pressure to classical pressure. If we take the temperature as given, then the pressure is indeed much higher when the particles are identical fermions under degenerate conditions. But if we take the energy density as given, perhaps by the history of energization of the object, then the pressures are just the same-- degeneracy pressure is classical pressure, when the energy is treated as the given. This is a point that I feel receives far too little attention in the rush to imagine that degeneracy pressure is some kind of weird quantum mechanical form of pressure!

I was thinking about pressure values during the evolution of a single star, for example comparing the present pressures in the Sun with the degeneracy pressure when the Sun will have become a white dwarf.

 

[Ken G]

I was thinking about pressure values during the evolution of a single star, for example comparing the present pressures in the Sun with the degeneracy pressure when the Sun will have become a white dwarf.

Yes, we can make that comparison, and the Sun will have way higher pressure when it is degenerate. But we can equally compare the Sun's pressure now to when it was first forming, and arrive at the same conclusion. Certainly as a star loses heat, it is perfectly natural for the star to contract to higher pressure. That process continues whether the star is far from degeneracy, or approaching degeneracy. Indeed, all degeneracy really does, when seen in this light, is stop the contraction and stop the rising pressure-- it doesn't have any effect on the pressure except to say when it cannot rise any more because there is no more heat loss and no more contraction.

 

[virgil1612]

Yes, we can make that comparison, and the Sun will have way higher pressure when it is degenerate. But we can equally compare the Sun's pressure now to when it was first forming, and arrive at the same conclusion. Certainly as a star loses heat, it is perfectly natural for the star to contract to higher pressure. That process continues whether the star is far from degeneracy, or approaching degeneracy. Indeed, all degeneracy really does, when seen in this light, is stop the contraction and stop the rising pressure-- it doesn't have any effect on the pressure except to say when it cannot rise any more because there is no more heat loss and no more contraction.

Ken, I would like to ask you something. It's about the thread where you wonder how come, in various places, there is the wrong information that the inside pressure and density would be greater for heavier stars, when in fact it's the opposite, and the interesting discussion about the nuclear reactions - luminosity causality relation. For some reason I cannot reply in that thread.

My question is, when we have a massive star that reaches the Eddington limit and basically disintegrates, I understand that it's the radiation pressure that balances gas pressure. Does this happen roughly in the whole volume of the star or just in the outer layers?

Thanks, Virgil.

 

[Ken G]

Ken, I would like to ask you something. It's about the thread where you wonder how come, in various places, there is the wrong information that the inside pressure and density would be greater for heavier stars, when in fact it's the opposite, and the interesting discussion about the nuclear reactions - luminosity causality relation. For some reason I cannot reply in that thread.

My question is, when we have a massive star that reaches the Eddington limit and basically disintegrates, I understand that it's the radiation pressure that balances gas pressure. Does this happen roughly in the whole volume of the star or just in the outer layers?

Thanks, Virgil.

The Eddington limit applies everywhere in the star, but note a couple important points. The limit is a limit on the luminosity, given the mass, and the luminosity means just the energy that is carried radiatively (it doesn't count convection), and the mass is just the mass interior to the point in question. Also, the limit includes a reference to the opacity, which often can be a known value (when it's from free electrons), but other times can be significantly enhanced by other types of opacity. Finally, it is not completely obvious that this limit is fundamental, because a star can have various ways of exceeding the Eddington limit, at least temporarily.

So having said all that, just what the "Eddington limit" is can be a bit subtle when applied to the interior of the star! The simplest version of it is when it is applied at the surface, because then the energy flux has to be all radiative, the mass has to be all the mass of the star, and most stars that approach this limit have all their surface opacity being from free electrons. But it is still important that the luminosity limit doesn't only apply at the surface, because we might imagine a star that can tolerate being out of hydrostatic equilibrium at its surface only, but it's a much tougher state of affairs if it is out of equilibrium throughout the star. So because it applies, in some sense, over the whole star, we tend to think of it as a limit on the maximum luminosity that the star can support, given its mass. (Note that it is sometimes turned around and described as a limit on the maximum mass of the star, given its mass-luminosity relationship, but that's wrong, because the mass-luminosity relationship would simply adjust such that any mass would be possible-- it's not the Eddington limit, but rather stability considerations that are not well known, that set the limit on how much mass a star can have.)

ETA: Second thoughts: The above subtleties are the reason we cannot be sure from theoretical grounds that stars cannot have a luminosity above the Eddington limit. If it happens in the deep interior, the star will just go convective and solve its problem, and indeed high-mass stars are highly convective. If it happens at the surface, the star can just go out of hydrostatic equilibrium, and have a strong wind, and indeed high-mass stars do have strong winds (though the energy losses to the wind can cause the luminosity seen at large distances to be reduced compared to the luminosity at the surface). Finally, the limit is derived in spherical symmetry, and stars can find various ways of breaking spherical symmetry, and indeed mass loss from very high-mass stars can often look non-spherically-symmetric. So on balance, I would say the Eddington limit is more of a benchmark than it is an absolute physical limit!

 

[virgil1612]

The Eddington limit applies everywhere in the star, but note a couple important points. The limit is a limit on the luminosity, given the mass, and the luminosity means just the energy that is carried radiatively (it doesn't count convection), and the mass is just the mass interior to the point in question. Also, the limit includes a reference to the opacity, which often can be a known value (when it's from free electrons), but other times can be significantly enhanced by other types of opacity.

So having said all that, just what the "Eddington limit" is can be a bit subtle when applied to the interior of the star! The simplest version of it is when it is applied at the surface, because then the energy flux has to be all radiative, the mass has to be all the mass of the star, and most stars that approach this limit have all their surface opacity being from free electrons. But it is still important that the luminosity limit doesn't only apply at the surface, because we might imagine a star that can tolerate being out of hydrostatic equilibrium at its surface only, but it's a much tougher state of affairs if it is out of equilibrium throughout the star. So because it applies, in some sense, over the whole star, we tend to think of it as a limit on the maximum luminosity that the star can support, given its mass. (Note that it is sometimes turned around and described as a limit on the maximum mass of the star, given its mass-luminosity relationship, but that's wrong, because the mass-luminosity relationship would simply adjust such that any mass would be possible-- it's not the Eddington limit, but rather stability considerations that are not well known, that set the limit on how much mass a star can have.)

Thank you, great.

 

[Ken G]

Thank you, great.

By the way, I added another paragraph at the end as a "second thought," sorry for not including that the first time.

 

[virgil1612]

Thanks. Concerning the question about why so many sources (good sources) keep repeating that "massive star" -> "bigger central pressure" than a smaller star, I think it's just human nature. One gets so used to the idea that everything about massive stars should be "massive"... And then it gets repeated because it's so "logical". I remember I was myself surprised when I understood it's the opposite.

 

[Ken G]

Thanks. Concerning the question about why so many sources (good sources) keep repeating that "massive star" -> "bigger central pressure" than a smaller star, I think it's just human nature. One gets so used to the idea that everything about massive stars should be "massive"... And then it gets repeated because it's so "logical". I remember I was myself surprised when I understood it's the opposite.

Yes I think you're exactly right-- some explanations sound so "truthy" that we forget to check them. But the right answer is always way more interesting than the truthy one, so the extra effort can really pay off!