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After H burning has finished, the he mass of the He core gradually increases, as does its density and temperature. Low-mass stars have denser cores when they reach a temperature at which He is ignited. The density is high enough that the electrons in the core are degenerate. In such conditions the heat from the nuclear reactions goes almost exclusively into ...

14

The scenario you describe may occur. On the other hand it may actually be that neutronisation in a white dwarf is the trigger for a thermonuclear type Ia supernova. You may be misunderstanding the Pauli Exclusion Principle (PEP).The PEP states that no two fermions can occupy the same quantum state, not that they cannot occupy the same space or be compressed ...

10

The density of white dwarfs is not hypothetical, it can be measured. The short answer is that the density is so high that a stable star can only be supported by electron degeneracy. Sirius B is an example. The radius can be estimated by combining the luminosity of the white dwarf with its temperature estimated from spectroscopy. The mass can then be ...

10

Stars are electrically neutral. In the optimal case for proton degeneracy, a star is made of hydrogen, maximising the proton number density. The number density of electrons is exactly the same. Since the Fermi momentum $p_F$ of a species is dependent only on number density$^1$, the Fermi momentum of the electrons and protons are the same. Pressure however$^... 10 Straightforwardly no. For a start there are almost no free protons inside a white dwarf. They are all safely locked away in the nuclei of carbon and oxygen nuclei (which are bosonic). There are a few protons near the surface, but not in sufficient numbers to be degenerate. Let us assume though that you were able to build a hydrogen white dwarf that had ... 9 In a white dwarf, the dense matter is not in its lowest energy configuration. Energy can still be extracted from the white dwarf material by fusion, provided it can be ignited. What exothermic nuclear reactions would there be that could take place in a neutron star? The bulk of the material is in the form of neutrons with a small number of protons and ... 9 The answer to whether a normal star can contain a core of degenerate neutrons is a yes. Thorne & Zytkow 1977 produced numerically models where a neutron star becomes embedded in the center of a massive giant or supergiant star, surrounded by a large gaseous envelope. In this scenario, the main source of energy becomes not nuclear fusion but instead ... 7 Z is the atomic number, A is the atomic mass. A white dwarf is typically made of carbon (A=12, Z=6) or Oxygen (A=16, Z=8). So Z/A = 0.5. 7 More massive stars indeed have higher pressures, but what's key is that they also have higher temperatures. After leaving the main sequence, they reach core temperatures of a few times$\sim10^8$Kelvin while their densities remain at something like$\sim10^4$g cm$^{-3}$- which, you can check, is within the nondegenerate regime. Therefore, helium fusion ... 7 Proton degeneracy is not important, because its effect is much smaller -- much like nuclear particles in theory also are dictated by gravity, but the electromagnetic and nuclear forces are dominating, since they are much stronger. Proton degeneracy is weaker than electron degeneracy due to the far greater mass of the proton compared to the electron. The ... 6 The test to see whether degeneracy pressure is going to be significant is to compare$kT$with the Fermi energy$E_F$The Fermi energy is the energy level up to which all energy states would be occupied in a completely degenerate fermion gas. It is given by (for non-relativistic conditions) $$E_F = \frac{h^2}{2m}\left(\frac{3}{8\pi}\right)^{2/3} n^{2/3},$$... 6 There are several reasons dark energy cannot be pressure due to the Pauli exclusion principle. First of all, pressure does not cause expansion of the universe, because pressure is not a force-- pressure gradients are a force (per unit volume), and the cosmological principle precludes them. Indeed, the pressure that exists everywhere only appears as a force ... 6 Gas giants like Jupiter consists mainly of Hydrogen and some Helium. If you gradually add mass to these planets then core temperatures will rise gradually and a stage will come where they will ignite like normal stars. Unlike white dwarfs where accretion of mass leads to type 1A supernova. 6 I am not sure what you mean by "thermal" pressure. Jupiter is supported by pressure, just like all objects that are in (approximate) hydrostatic equilibrium. That pressure is provided by your everyday, temperature-dependent Maxwell-Boltzmann ideal gas pressure in the outer parts, but the free electrons in the interior become degenerate and so in these ... 6 Massive stars do not undergo helium flash because they have core temperatures high enough to prevent the helium core from becoming electron-degenerate. Check here for some more information. Therefore, the star can burn helium in a smooth transition, instead of undergoing helium flash. In more details, the massive star's core heats up past the helium burning ... 5 No, hydrogen burning doesn't start until 82$m_{\rm Jup}$and deuterium burning starts at 13$m_{\rm Jup}$, so evidently 1.4$m_{\rm Jup}$is not enough for this experiment. Your misconception is that Jupiter's mass is about the maximum a planet can be before it starts to fuse hydrogen in its core and undergo a massive transformation. which is simply ... 4 @Knu8 was right that adding mass to a gas giant will turn it into a star long before the gas giant could become a white dwarf or neutron star. But that works if you add fusionable material such as hydrogen. If you add something that can't fuse into a heavier element, such as Iron or Tin, the matter just keeps accreting onto the gas giant until the planet ... 4 The limit to the neutron star mass is at least$2M_{\odot}\$, since at least two have measured masses as large as this (e.g. Demorest et al. 2010). The answer to your question is the same as the answer to what is the maximum mass of a neutron/quark star, since if you compress matter of any sort, this is ultimately what it will become. The answer to this ...

3

Basically, the Pauli exclusion principle says that two fermions (in this case, electrons) can't be in the same quantum state. To expand: No two electrons in an atom can share the same numbers for their four quantum numbers, properties that help describe the state of a particle. What are quantum numbers? The important consequence here is that no two electrons ...

3

So then why would gas giant cores be electron degenerate, but not become hot enough for nuclear fusion? Degeneracy isn't an on/off switch. It's a quantum mechanical aspect of pressure that is always present, just as is thermal pressure. A substance is highly degenerate if degeneracy pressure completely dominates over thermal pressure, non-degenerate if ...

3

Whether a white dwarf responds to the accretion of material by exploding or collapsing depends on the competition between energy being released in fusion reactions and energy being locked away by endothermic electron capture (neutronisation) reactions. It is thought that most white dwarfs of moderate mass have a C/O composition. They will need to accrete a ...

3

There are a variety of white dwarfs with various compositions, and analysing how they detonate in a supernova (or not) is an topic under investigation. A simple model, described in "How is the first detonation in Supernove type Ia triggered?" is of a helium shell initially igniting and that setting off the carbon in the core. In this type of Type 1a ...

3

There are problems with the scenario outlined above. Firstly, if the CNO cycle were to operate, it would have to operate for millions/billions of years to noticeably affect the relative abundances of CNO in the solar photsophere. That is because the core is embedded within a radiative zone where only comparatively slow mixing processes operate. A much more ...

3

The above answer handles the question as you posed it, but perhaps it should also be addressed the way you phrased the question in the title. Are you imagining that degeneracy pressure is associated with being a star? Degeneracy pressure is achieved simply by waiting long enough for a gas to lose enough heat. Jupiter is already pretty close to old enough ...

2

Another way to think of the answer is that the pressure of any nonrelativistic gas (degenerate or ideal) is always 2/3 the kinetic energy per unit volume. So we can say that when the electrons go degenerate, and contribute almost all of the pressure, it must be because they also contain almost all of the kinetic energy. This is true, the reason they ...

2

In Layman's terms, the Pauli exclusion principal wouldn't need to be overcome to form the black hole. A Neutron star of a certain size will shrink below it's Schwarzschild radius naturally. That's not hard to see. In fact, like white dwarfs, Neutron stars grow smaller in radius as they gain mass. The maximum mass wouldn't be much more than 2.5 or so ...

2

And here's another way to frame the situation. Degeneracy is not responsible for the pressure in the electrons in a white dwarf, they would have that same pressure once the star contracts to the same size even if it was all ideal gas (say, if the electrons were distinguishable). What is meant by "degeneracy pressure" is the perfectly mundane kinetic gas ...

2

The main method by which heat is transferred in a white dwarf is thermal conduction. Degenerate electrons have very long mean free paths and can conduct heat excellently. White dwarfs lose energy by emitting photons from their outer, non-degenerate surface layers. They may also emit neutrinos from their interiors when they are really young and hot. Gamma ...

2

If you have a gas of non-degenerate ions and degenerate electrons, then by definition you have an upper limit to the temperature, since if it were higher the electrons would not be degenerate. Below this temperature, the electron degeneracy pressure will always greatly exceed any perfect gas pressure from the ions. Since degeneracy pressure does not depend ...

1

This is fairly standard bookwork that I am not going to reproduce and you should understand that the lines on your diagram are "fuzzy" in the sense that they mark the loci where you can neither use one approximation or another. So first, the division between a classical gas and a quantum gas (note that all the gases in your plot are ideal - the term ideal ...

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