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Disclaimer: when talking about stellar evolution, physicists tend to talk about the core of the star. We go on about core collapse, and the effects thereof. This may clash with what you've heard about the sun's size expanding over the coming billion years to be larger than the orbit of Earth. The outer layers of a star expand as it ages, but the core collapses and becomes more dense as more "heavy" elements are fused.
It's something to bear in mind.
Also, I spent days creating this post, and I'm not sure where it's too heady and long-winded and where it's hard to follow and obtuse at this point.
Please follow up as needed.
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lol... "Post Quick Reply" couldn't be less true.
Originally Posted by CoccoBill
Alrighty then, not as simple as I expected, not surprisingly. In the visualizations it's always stressed how far the electrons actually are from the nucleus and how much empty space there is in between.
Atoms are crazy little things [citation needed].
Just so that you don't think I'm saying that it's common for an electron to be inside a nucleus:
It is the case that electrons spend some time in the nucleus... that is... there is a non-0 chance of occasionally detecting an electron inside a nucleus.
It's not a very probable situation, but it is definitely part of the description of atoms.
The most probable place to find an atomic electron is definitely outside a nucleus.
A nucleus is generally ~(10)^-15 m in diameter.
An atomic diameter is generally ~(10)^-10 m.
So the atom (rather, its electron cloud) is 10,000 times as wide as its nucleus.
Like I mentioned before, the (10)^-15 m is misleading. It represents some probability of finding a nucleon inside that diameter. Just as electrons are sometimes inside of nuclei, nucleons are sometimes outside this nuclear size (it's ridiculously more rare, though). The forces acting on nucleons to hold them together are different than gravitational or electromagnetic forces, and nuclei therefore have a "hard shell." Meaning that finding a nucleon much beyond its parent nucleus means that it is no longer a part of that nucleus. I.e. you didn't find one of "that" atom's nucleons outside it's nucleus, you found a free nucleon, not bound to any nucleus.
Originally Posted by CoccoBill
If matter is mostly nothing, it just led me to think what would happen if you squeeze matter enough, and just picturing applying enough force to pack all of the atoms (or nuclei) together, with no empty space in between. I do realize it's not like electrons are just there spinning on perfect orbits around the nucleus like earth around the sun, but they're all over the place, even at the same time. That would have just been such a satisfactory intuitive explanation. So what does exactly happen when the force applied is nearing the Chandrasekhar limit, is the distance between the electrons and the nuclei being limited or how do they end up butting heads?
Keep in mind that what we're talking about now is not atoms, but plasma. Plasma is not made of atoms, per se.
An atom is some number of electrons bound to a nucleus with the same number of protons and any number of neutrons. If protons and electrons are not in equal numbers, it is not an atom, strictly speaking, but an ion.
Plasma consists of fully ionized nuclei (the nuclei have no electrons bound to them) going crazy, mad bonkers in a sea of electrons, also going crazy, mad bonkers.
The plasma forms from pre-stellar atoms when the thermal energy (created by increasing pressure due to gravitational collapse) overcomes the electron binding energy. In any collision (particle interaction), now, the average thermal energy delivers more energy to the electron (or nucleus, doesn't matter) than the energy binding it to the atom, and so the nucleus / electron(s) become(s) unbound.
Now what we have is a bunch of nucleons bound gravitationally and a bunch of electrons bound gravitationally. These 2 clouds are also coupled to each other. Each cloud acts to disperse itself, being composed entirely of like charges. However, the clouds are of opposite charge to each other, which means they attract each other. The net effect is that the 2 clouds remain interspersed through each other, shielding each other from the extreme dispersing forces of like-charge repulsion.
So what we have is a state of plasma. The core of collapsing star is basically a single quantum object, composed of unfathomably many entangled particles. We're talking about electron degeneracy pressure, so let's ignore the nucleons for a minute. Remember before when the thermal energy went past some threshold, it turned the atoms of the proto-star in to a plasma. This change happened because it was energetically favorable for the electrons to be unbound from their nuclei. I.e. it would have taken MORE energy to hold the atoms together than it did to create the plasma.
A similar thing is going to happen with the creation of a white dwarf star. The core of the star started to fuse Iron, which is always the end of that star. Iron fusion is endothermic, not exothermic. The outward pressure holding off the full gravitational collapse of the star is due to exothermic fusion reactions of elements lighter than Iron. Those exothermic reactions provide an outward force to balance the gravitational force. With Iron, the reaction not only doesn't provide any outward force, it contributes to the inward pull. For all stars, Iron fusion means supernova is imminent. (The iron in your blood was literally made by the death of a star.)
What I'm describing is a runaway reaction in which the pressure in the core of the star is increasing as the volume (of the core) decreases. That volume contains a mostly constant number of electrons. (There are nuclear decays going on which cause it to fluctuate, but it's fairly constant.)
Now it's gonna get ugly. There's no way to summarize what's going on any better than you understand without getting into quantized wave mechanics.
OK, oscillators.
If I have 2 masses, connected by springs, then they can move in 2 basic ways. Either they move together as a unit, where the spring does nothing, or they bounce back and forth, always moving in opposite directions to each other in the center of mass rest frame. No matter what movement you see them exhibit, it is always some additive combination of those 2 states.
If I have 3 masses, connected by springs, then they can move in 3 basic ways. As always, they can move together as one unit, where the springs do nothing. This is always the lowest energy state. They have an intermediate state, where the 2 "outer" masses move in opposite directions of each other, but the "middle" one sits still. The highest energy state is always one in which every mass is moving in the opposite direction as its adjacent neighbors.
With quantized systems, here, the number of masses is quantized, there is always a highest energy state... A maximum amount of energy that can be (nearly losslessly) stored in that system.
OK, it's well different with electron states, and we're talking about the state of a ridiculous number of particles in an ensemble, but the fact that there is a maximum energy that can be stored in the system still holds. As the volume decreases, the number of available states decreases. When the number of available states is equal to the number of fermions (particles that don't share states) in those states, then extreme physics is going on.
The increasing pressure can no longer compress the core of the star, because the core is effectively incompressable under the applied pressure. It can't reduce its volume with that number of electrons in there, and the electrons have nowhere to go. It will take a LOT more energy to make it energetically favorable for the electrons to undergo a process called electron capture. In electron capture, a proton and an electron interact, and a neutron and electron neutrino are left.
In fact, no free proton has ever been witnessed to spontaneously decay. Free neutrons do spontaneously decay into a proton, electron and anti-electron neutrino. This is because the combined rest-mass (energy) of the resulting particles is less than the rest mass of the neutron (and other conservation laws are followed).
In general, for small atoms, the lowest energy state for the nucleus is to have an equal number of protons and neutrons. As we look at larger atoms, we see more and more neutrons than protons. This is because protons have the same sign of electric charge and they're stupid close together. The repulsive forces they express on each other are stupendous. So protons in a nucleus have more energy than neutrons in the nucleus, because they experience a force that the neutrons (not having electric charge) do not.
Nonetheless, if you put a few neutrons together in a nucleus with no protons, at least one of them kinda immediately decays into a proton, because neutrons have the same wave mechanics constricting them in a specific volume as we talked about before. (protons, too). So if you got a bunch of neutrons, and no protons, then it's definitely lower in energy to take one of the highest energy neutrons in that volume and move it to the lowest energy available proton spot. That doesn't even count the bonus energy of a spontaneous decay.
The practical upshot is that in order to compress the electron degenerate matter to neutron degenerate matter, there is a steep cost in energy to pay. Those neutrons are higher in energy than protons, and still confined to a finite volume. If the infalling matter of the collapsing stellar core is does not create enough pressure to pay this cost, then it basically hits an immovable wall and rebounds in a supernova.
The same rebound happens when it collapses to a neutron star, and the neutron degeneracy pressure cannot be overcome.
When the neutron degeneracy pressure is overcome, it collapses into a black hole (or some theorized objects, not yet observed).
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