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In the news report 'A White Dwarf Living On The Edge' (Keck Observatory), Ilaria Caiazzo writes about a $1.35 M_\odot$ white dwarf which was formed by the merging of two less massive white dwarfs:

“This is highly speculative, but it’s possible that the white dwarf is massive enough to further collapse into a neutron star [...] It is so massive and dense that, in its core, electrons are being captured by protons in nuclei to form neutrons. Because the pressure from electrons pushes against the force of gravity, keeping the star intact, the core collapses when a large enough number of electrons are removed.”

The result would be a neutron star, the two original white dwarfs escaping the fate of a Type Ia supernova. This process differs from the usual story, in which a neutron star can only remain after the core collapse (non-Type Ia) supernova of a single, more massive star without a white dwarf being formed.

Is it through 'inverse beta decay' (neutronisation) that electrons would be removed in Caiazzo's hypothesis? Is this one of the reasons why a white dwarf cannot quite reach the Chandrasekar limit (the other reasons being the effect of general relativity and Coulomb interactions of electrons and nuclei)?

Related questions: White dwarfs: limits to stability and neutronisation in nuclei

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Yes, inverse beta decay results in the removal of electrons from the degenerate electron gas. At fixed volume, this would lower the electron number density and hence lower the degeneracy pressure. It is possible that the star would then contract, the electron number density would be raised and the increased pressure would still support the star. However, there is a reasonably straightforward bit of analysis that shows that if the adiabatic index $\alpha$, where the relationship between pressure and density is $P \propto \rho^{\alpha}$, is lower than 4/3 then no stable equilibrium is attainable.

Let's look at a massive white dwarf supported by electron degeneracy. Firstly, why are we looking at massive white dwarfs? Because inverse beta decay is endothermic and requires the captured electrons to have relativistic energies $\sim 10$ MeV. This occurs at the high densities inside massive white dwarfs since the degenerate electron Fermi energy also increases with electron number density. When the Fermi energy reaches the inverse beta decay energy threshold, then the process can start to occur.

At these Fermi energies, the relativistic electron gas has a pressure proportional to electron number density, $n_e$, to the power of a little more than 4/3. If the composition of the gas is fixed, this also means $\alpha$ is just a little bit larger than 4/3 and stability can be found, because the density $\rho = \mu_e n_e m_u$, where $m_u$ is an atomic mass unit and $\mu_e$ is the number of mass units per electron in the gas.

However, if inverse beta decay starts then electrons will be removed, $\mu_e$ will increase and $n_e$ decreases. This has the effect of tipping $\alpha$ below 4/3 and triggering the collapse of the star. In the absence of other effects, this would mean the white dwarf becomes unstable at some finite density (the density at which the Fermi energy reaches the inverse beta decay energy threshold) and hence at a mass that is lower than the canonical Chandrasekhar limit, that assumes the star can reach infinite density.

Whether the collapse could result in a neutron star is critically dependent on the exact composition. If the electron energy threshold for inverse beta decay occurs at densities insufficient to ignite thermonuclear reactions, then it's possible the collapsing white dwarf might escape a supernova. The density thresholds for inverse beta decay, for the onset of "pycnonuclear" ignition and of the instability caused by General Relativity (the fact that pressure appears on the RHS of the hydrostatic equilibrium equation) are very similar for a carbon white dwarf.

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    $\begingroup$ Thank you for another clear and complete answer. Congratulations for passing the 100k reputation mark, showing an immense effort to answer a variety of questions. $\endgroup$
    – gamma1954
    Jul 16 at 8:38
  • $\begingroup$ @gamma1954 no, I think that is correct. nyuscholars.nyu.edu/en/publications/pyconuclearfusionrates Actually, it seems both are used. Maybe others are making a similar mistake to me, I'm not sure. A brief research suggests you are right and that I and many others have made the same mistake. $\endgroup$
    – ProfRob
    Jul 16 at 11:27

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