It occurs to me (though I'm hardly the first) that the decay $$ \mu^- \to e^- + \bar \nu_e + \nu_\mu $$ should be forbidden in electron-degenerate matter, since there must be an empty state available to receive the electron. How would one take this fact to make an order-of-magnitude estimate of the equilibrium muon density at the core of a white dwarf star?

(Technical answers and illuminating references are welcome.)


2 Answers 2


Though I agree with the logic of MariusMatutiae, I find I cannot reproduce their quantitative answer.

I get an electron number density of $1.2 \times 10^{41}$ m$^{-3}$ (is it just a unit thing?) for an electron Fermi energy of 30MeV.

In a carbon white dwarf with 2 mass units per electron, the Fermi energy of the electrons reaches 30MeV at densities of $4\times 10^{14}$ kg/m$^{3}$ - i.e. at densities which still do exceed the maximum possible in a White Dwarf because of instabilities caused by inverse beta decay or General Relativity (the maximum density of a WD is more like a few $10^{13}$ kg/m$^3$). But not as high as densities in a neutron star.

I have produced an applet that allows you to explore the parameter space in detail.


OK, but even given that, where do the muons come from? Actually, you need to create them with an energy of 105.6MeV from electrons and anti-neutrinos. If they reach some kind of equilibrium then the chemical potential (the Fermi energy) of the electrons and muons should be equal. Thus the electron energy threshold for muon production is normally considered to be more like 105.6MeV and consequently a factor of $(100/30)^3$ higher electron number densities and mass densities are required.

A similar calculation shows that in neutron stars, muon production is not really viable until densities reach several $\times10^{17}$ kg/m$^3$. It is a much higher density here because (a) the electrons and muons have the same Fermi energies when they are at equilibrium (b) the number of mass units per electron is more like 60.

  • $\begingroup$ Your applet is great! I really feel like I understand the problem much better now. $\endgroup$
    – rob
    Commented Oct 18, 2014 at 16:47
  • $\begingroup$ Thanks for that - I produced a few such things for a course on WDs and NSs that I teach. $\endgroup$
    – ProfRob
    Commented Oct 18, 2014 at 16:50
  • $\begingroup$ By the way, the muons wouldn't have to be produced inside the star. If muons in the core of the star were stable, the could be produced e.g. by cosmic ray spallation at the surface, like the muons in my house which come from the top of the atmosphere; those few that migrated to the core and cooled by elastic scattering would get stuck. But since there's never enough electron density to block the muon decays, the point is moot. $\endgroup$
    – rob
    Commented Oct 18, 2014 at 16:50
  • $\begingroup$ But the muon lifetime is only 2.2 microsecond so they probably could not even traverse the outer non-degenerate layer (~60 km) before decaying. $\endgroup$
    – ProfRob
    Commented Oct 18, 2014 at 16:57

Fun question. The muon density inside a white dwarf is negligible, because Fermi suppression does not really apply.

Fermi suppression is the technical name of the effect you were describing: the decrease in the rate of a process due to the fact that there are no free states to accommodate one of the decaying particles (an electron, in this case).

The muon has a mass of $105.6 MeV$, about 210 times more than the electron. Thus the electron resulting from the decay is necessarily highly relativistic. We shall then have Fermi suppression if the Fermi energy $E_F$ is larger than the energy the electron gains in the muon decay.

Now, the Fermi Energy for dense matter in the highly relativistic limit (see here, for instance) is $$ E_F = hc \left(\frac{3 n_e}{8\pi}\right)^{1/3} \approx 6\times 10^{-7} n_e^{1/3} eV $$

In order to have $E_F = 30 MeV$ (let us imagine that the other two particles carry away as much energy as the electron), we need $$ n_e \approx 10^{41}\; cm^{-3}\;\;\;. $$ Since inside a white dwarf there are as many protons as there are electrons, this imples a local density of $$ \rho \approx 10^{17}\; g \; cm^{-3} $$ which is high even for neutron stars, let alone white dwarves.

Thus the Fermi energy is never this ($30 MeV$) large in white dwarfs, and muon decays proceed unimpeded.

  • $\begingroup$ Nice. Evidently I should have done a basic computation. $\endgroup$ Commented Sep 14, 2014 at 15:56
  • $\begingroup$ I think you made a units mistake. $\endgroup$
    – ProfRob
    Commented Oct 18, 2014 at 11:51
  • $\begingroup$ I agree with @RobJeffries: I get an electron number density $n_e \approx 10^{-4}\,\mathrm{fm^{-3}} = 10^{41}\,\mathrm{m^{-3}} = 10^{35}\,\mathrm{cm^{-3}}$ $\endgroup$
    – rob
    Commented Oct 18, 2014 at 16:16

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