After a star becomes a White dwarf, it resists gravitational collapse mainly due to the electron degeneracy pressure. If the mass of the white dwarf is greater than the Chandrasekhar limit, the degeneracy pressure cannot resist the collapse any longer and is doomed to become a neutron star or a black hole. Why can't the degeneracy pressure keep on self-adjusting itself to resist collapse forever?
[Physics] Why can’t the degeneracy pressure self-adjust itself to resist gravitational collapse
astrophysicsblack-holesgravitational-collapsequantum mechanics
Related Solutions
How does the Pauli Exclusion Principle actually create a force?
The Pauli exclusion principle doesn't really say that two fermions can't be in the same place. It's both stronger and weaker than that. It says that they can't be in the same state, i.e., if they're standing waves, two of them can't have the same standing wave pattern. But for bulk matter, for our purposes, it becomes a decent approximation to treat the exclusion principle as saying that if $n$ particles are confined to a volume $V$, they must each be confined to a space of about $V/n$. Since volume goes like length cubed, this means that their wavelengths must be $\lesssim (V/n)^{1/3}$. As $V$ shrinks, this maximum wavelength shrinks as well, and the de Broglie relation then tells us that the momentum goes up. The increased momentum shows up as a pressure, just as it would if you increased the momenta of all the molecules in a sample of air. A degenerate body like a neutron star or white dwarf is in a state where this pressure is in equilibrium with gravity.
We think that most neutron stars are produced in the cores of massive stars and result from the collapse of a core that is already at a mass of $\sim 1.1-1.2 M_{\odot}$ and so as a result there is a minimum observed mass for neutron stars of about $1.2M_{\odot}$ (see for example Ozel et al. 2012). Update - the smallest, precisely measured mass for a neutron star is now $1.174 \pm 0.004 M_{\odot}$ - Martinez et al. (2015).
The same paper also shows that there appears to be a gap between the maximum masses of neutron stars and the minimum mass of black holes.
You are correct that current thinking is that the lower limit on observed neutron star and black hole masses is as a result of the formation process rather than any physical limit (e.g. Belczynski et al. 2012 [thanks Kyle]).
Theoretically a stable neutron star could exist with a much lower mass, if one could work out a way of forming it (perhaps in a close binary neutron star where one component loses mass to the other prior to a merger?). If one just assumes that you could somehow evolve material at a gradually increasing density in some quasi-static way so that it reaches a nuclear statistical equilibrium at each point, then one can use the equation of state of such material to find the range of densities where $\partial M/\partial \rho$ is positive. This is a necessary (though not entirely sufficient) condition for stability and would be complicated by rotation, so let's ignore that.
The zero-temperature "Harrison-Wheeler" equation of state (ideal electron/neutron degeneracy pressure, plus nuclear statistical equilibrium) gives a minimum stable mass of 0.19$M_{\odot}$, a minimum central density of $2.5\times10^{16}$ kg/m$^3$ and a radius of 250 km. (Colpi et al. 1993). However, the same paper shows that this is dependent on the details of the adopted equation of state. The Baym-Pethick-Sutherland EOS gives them a minimum mass of 0.09$M_{\odot}$ and central density of $1.5\times10^{17}$ kg/m$^3$. Both of these calculations ignore General Relativity.
More modern calculations (incorporating GR, e.g. Bordbar & Hayti 2006) get a minimum mass of 0.1$M_{\odot}$ and claim this is insensitive to the particular EOS. This is supported by Potekhin et al. (2013), who find $0.087 < M_{\rm min}/M_{\odot} < 0.093$ for EOSs with a range of "hardness". On the other hand Belvedere et al. (2014) find $M_{\rm min}=0.18M_{\odot}$ with an even harder EOS.
A paper by Burgio & Schulze (2010) shows that the corresponding minimum mass for hot material with trapped neutrinos in the centre of a supernova is more like 1$M_{\odot}$. So this is the key point - although low mass neutron stars could exist, it is impossible to produce them in the cores of supernovae.
Edit: I thought I'd add a brief qualitative reason why lower mass neutron stars can't exist. The root cause is that for a star supported by a polytropic equation of state $P \propto \rho^{\alpha}$, it is well known that the binding energy is only negative, $\partial M/\partial \rho>0$ and the star stable, if $\alpha>4/3$. This is modified a bit for GR - very roughly $\alpha > 4/3 + 2.25GM/Rc^2$. At densities of $\sim 10^{17}$ kg/m$^3$ the star can be supported by non-relativistic neutron degeneracy pressure with $\alpha \sim 5/3$. Lower mass neutron stars will have larger radii ($R \propto M^{-1/3}$), but if densities drop too low, then it is energetically favorable for protons and neutrons to combine into neutron-rich nuclei; removing free neutrons, reducing $\alpha$ and producing relativistic free electrons through beta-decay. Eventually the equation of state becomes dominated by the free electrons with $\alpha=4/3$, further softened by inverse beta-decay, and stability becomes impossible.
Best Answer
The basic problem is that for a sufficiently massive star, the electrons become relativistic. The fine details of this calculation are rather complicated, but you can get a qualitative sense of the argument as follows:
For non-relativistic fermions at zero temperature, it is possible to show that the total energy of $N$ particles in a box of volume $V$ is proportional to $N^{5/3}/V^{2/3}$. This can be done via counting the density of states, and using the fact that the energy of a non-relativistic particle obeys $E \propto |\vec{p}|^{2}$. For a spherical volume of radius $R$, we have $R \propto V^{1/3}$, and the number of fermions present is proportional to the mass. This means that the total energy of the fermions is proportional to $M^{5/3}/R^2$. This energy is positive.
On the other hand, the gravitational energy of a solid sphere is negative and proportional to $M^2/R$. This means that the total energy is the sum of a negative $R^{-1}$ term and a positive $R^{-2}$ term, and such a function will have a minimum somewhere. This will be the equilibrium point. At smaller radii, the energy of the degeneracy grows faster than the binding energy decreases, pushing the radius back to larger values. At larger radii, the reverse occurs. This means that the star will be stable.
This argument doesn't hold up to arbitrarily large energies, though, because eventually the Fermi energy of the electrons exceeds the rest energy of the electron; in other words, the electrons become relativistic. This changes the relationship between energy and momentum of the electrons. For highly relativistic electrons, we have $E \propto |\vec{p}|$ instead; and going through the same calculations (neglecting the electron mass entirely), we find that the total energy of a relativistic fermion gas is proportional to $N^{4/3}/V^{1/3} \propto M^{4/3}/R$.
The gravitational binding energy, on the other hand, remains negative and proportional to $M^2/R$. This implies that the overall energy is itself proportional to $1/R$, and there is no extremum of the total energy of the system. Since the fermion energy and the binding energy always increase or decrease at exactly the same rate, there will be no stable equilibrium radius. The star will either blow itself apart or collapse in on itself, depending on whether the kinetic energy of the fermions or the gravitational binding energy wins out.