Why does the core of a low mass star contract after reaching electron degeneracy?

Question

I am learning about the lives and deaths of stars with a solar mass of 0.4-2. What I understand is that once the star stops hydrogen fusion after using up all its hydrogen, the star leaves the main sequence and starts becoming a red giant. The pressure of gravity causes the core to contract (no outwards pressure anymore because there is no more hydrogen fusion) and thus become hotter and hotter. At a certain point, the core stops contracting because the electrons are packed as densely as nature allows them to be (electron degeneracy).

At the same time, hydrogen fusion is taking place in the shell material of the star, dumping 'helium ash' on the core, which causes more mass to press upon the core, causing the core to contract even more. And that is my question: how can the core contract more after having reached electron degeneracy? I thought they could not be packed more closely?

Answer 1

Electron degeneracy does not lead to an infinitely hard equation of state.

The Pauli exclusion principle does not say that two fermions cannot occupy the same space; it says they cannot occupy the same quantum state. What this means is that as you squish the electrons together they have to occupy higher and higher momentum states. It is this non-zero momentum (even if cold) that leads to degeneracy pressure.

In the core of a star this pressure is proportional to number density of electrons $n_e^{5/3}$, which could be written in terms of mass density as $(\rho/\mu_e m_u)^{5/3}$, where $\mu_e$ is the number of mass units per electron.

The equation of state $$ P \propto \left(\frac{\rho}{\mu_e m_u}\right)^{5/3}$$ is certainly compressible. If you do the sums with this equation of state, then you can do this in two ways. The mass density can be increased by adding mass to the core; an approximate calculation can show that $R \propto \mu_e^{-5/3} M^{-1/3}$ and hence $\rho \propto M^2$, where $M$ would be the core mass. Alternatively one could increase the mass per electron by changing the core composition.

In the core of a low mass star ascending the giant branch, there is mainly the first and a bit of the second. Helium ash gets dumped onto the core from the hydrogen burning shell thus increasing the degenerate core mass, and helium has twice the mass units per electron of hydrogen. Thus core compression continues until He is ignited.

The degenerate core of a low mass star is still not as dense as a degenerate white dwarf like Sirius B...

Answer 2

If you read further on stellar evolution, if the mass of the original star is large enough electron degeneracy is overcome and the star becomes a neutron star, stable because of neutron degeneracy.

This degeneracy is due to the Pauli exclusion principle which does not allow fermions of the same mass and charge to be in a single quantum state. What happens with neutron stars is that the pressures , and thus the kinetic energies of these electrons, become so high that the energy difference between protons and neutron is overcome, and an electron hitting a proton turns it into a neutron. The electron disappears, and in neutron stars all baryons turn to neutrons and neutrinos ( inverse to beta decay) which run away to infinity.

This process can always happen thermodynamically, as there are always high kinetic energy terms in all distributions. In your white dwarf example, some electrons from the high kinetic energy tail of the distribution, hitting protons will fuse into neutrons. The mass in your example is not high enough to turn into a neutron star, but excess mass falling can contract the core from the new neutrons in the process. Thus the electron degeneracy is not the whole story.