I'm trying to get some specific numbers for electron degeneracy that I can understand, using a concrete example.

Take for example this portion of carbon crystal:

carbon crystal lattice

  1. Exactly how much energy would be required to overcome the electron degeneracy pressure of these 18 carbon atoms, in a theoretical laboratory capable of such things.
  2. I believe that neutrinos would be emitted as a byproduct of the collapse, is that correct? What other reactions if any would be needed to fully account for conservation of energy/momentum etc?
  • 2
    $\begingroup$ It sounds an interesting/practical question. Have you tried to compute the pressure $P$ based on the energy volume relation: $P \propto dE/dV = (dE(k)/dk)(dk/dV)$; I suppose Fermi energy/Fermi momentum $k_F$ would come into the formula expression. But for the carbon crystal, you may need to specify a orbital model (such as tight-binding model). And then perhaps specify the occupation number in $P \propto dE/dV|_{k=k_F}$. (though your carbon model is not metal.) That is my intuitive thought. $\endgroup$
    – wonderich
    Jan 5, 2014 at 21:00
  • $\begingroup$ Look at a derivation of the radius of a white dwarf with respect to mass, electron degeneracy pressure comes into play there. $\endgroup$ Jun 9, 2014 at 20:41
  • $\begingroup$ Degeneracy pressure isn't "overcome", it is bypassed. As soon as the electrons have around 13 MeV they neutronise with protons in the carbon nuclei. $\endgroup$
    – ProfRob
    Aug 31, 2017 at 6:52

3 Answers 3


I'm not sure of the exact answer to your first question, as the numerical answer will depend on the electron density, and I'm not sure what dimensions your structure would have.

To answer the second part of your question, if sufficient pressure was applied, then the most energetically favorable process would no longer to be to stack the electrons in higher energy states. Instead they would combine with protons to create neutrons and electron neutrinos in the electron capture process: $$e + p \rightarrow n + v$$

This, more or less, is how neutron stars come about.


I've wondered about this too, but never pursued it before. Here is an educated guess based on dimensional analysis.

A pressure has the same units as an energy density: $$ \mathrm{ \frac N{m^2} = \frac{N\,m}{m^3} = \frac J{m^3} } $$

Electron capture takes kind of 1 MeV of energy, so maybe the transition becomes allowed once the energy density is something like 1 MeV/atom. Let's test this idea out: I vaguely remember that coal (or something) turns to diamond somewhere above one gigapascal: \begin{align} \mathrm{ 1 \,GPa }&= \mathrm{10^9\frac J{m^3} \cdot \frac{1\,eV}{1.6\times10^{-19} \,J} \cdot \left(\frac{1\,m}{10^9\,nm}\right)^3 }\\ &\mathrm{= \frac 58\times10^{1}\,\frac{eV}{nm^3} }\\ &\mathrm{= 6 \,\frac{eV}{nm^3} }\end{align} Diamond is pure carbon, density 3.5 g/cm$^3$, so the volume of one atom is \begin{align} V &= \mathrm{ \frac{1\,cm^3}{3.5\,g} \cdot \frac{12\,g}{1\,mole} \cdot \frac{1 \,mole}{6\times10^{23}\,atoms}\cdot \left(\frac{10^9\,nm}{100\,cm}\right)^3} \\ &\mathrm{ = \frac 2{3.5} \times10^{-2} \frac{nm^3}{atom} }\\ &\mathrm{ = 0.006 \frac{nm^3}{atom} }\end{align} Graphite (the initial state) has a density closer to 2 g/cm$^3$ so the volume of each atom is about 0.0035 nm$^3$.

This suggests that the transition from carbon to diamond at constant pressure takes an energy $E = P\Delta V \approx 0.015$ eV. This corresponds pretty closely to a tabulation of formation enthalpies which includes diamond: creating diamond requires about 1.9 kJ/mole, or about 20 meV/atom. Shockingly close! I must have made a set of mistakes that cancel each other out :-)

Assuming this method is sound, the energy for electron capture (and other nuclear transitions) is generally about 1 MeV. So if your block of carbon kept the same density 0.006 nm$^3$/atom, and electron that disappears frees up 1/6 of the atom's volume, you'd need $$ P = \frac E{\Delta V} = \mathrm{{10^6\,eV} \cdot \frac{6\,GPa\,nm^3}{1\,eV} \cdot \frac{6\, electrons/atom}{0.006\,nm^3/atom} = 6\times10^{9}\,GPa }$$ Here's some company that thinks 15,000 bar = 1.5 GPa is a lot of pressure, so we have the sanity check that this estimate $P$ for the degeneracy pressure is "astronomical" (it's sixty trillion atmospheres).

I'm very interested to see if you get an answer from someone who isn't winging it.

Two years after I wrote this answer, Rob Jeffries adds in a comment

Neutronisation threshold in Carbon nuclei is 13.9 MeV. This is reached by degenerate electrons at pressure of about $10^{19}$ GPa.

So I guessed the neutronization energy to within a factor of ten, but my pressure is kind of embarrassingly off. But I explicitly made the assumption that the density of the diamond wouldn't change between ordinary conditions and the onset of neutron capture --- which of course isn't what happens. In nature, the "overcoming" of electron degeneracy pressure occurs only in white dwarfs and degenerate stellar cores, where the density is as high as $10^7\rm\,g/cm^3$.

If you take my estimate $10^{10}\rm\,GPa$ above, increase the energy by a factor of ten, and decrease the volume per "atom" by a factor of $10^7$, you get within a factor of ten of RobJeffries's statement of the upper limit of electron degeneracy pressure in white dwarfs.

  • $\begingroup$ Neutronisation threshold in Carbon nuclei is 13.9 MeV. This is reached by degenerate electrons at pressure of about $10^{28}$Pa. $\endgroup$
    – ProfRob
    Aug 31, 2017 at 7:00
  • $\begingroup$ @RobJeffries Very interesting. Apparently my educated guesses were on the right track; see edit. $\endgroup$
    – rob
    Aug 31, 2017 at 12:55
  • $\begingroup$ The density for neutronisation in Carbon is $3.9\times 10^{10}$ g/cc. $\endgroup$
    – ProfRob
    Sep 1, 2017 at 16:23
  • $\begingroup$ @RobJeffries Very interesting. Apparently my educated guesses were not on the right track. I'm curious about where I go off the rails (I have some guesses). I might put together a new question, or perhaps you could post a fuller answer to this one? $\endgroup$
    – rob
    Sep 1, 2017 at 17:11
  • $\begingroup$ The threshold density is simply that for which the Fermi energy of the electrons reaches the neutronisation energy. $\endgroup$
    – ProfRob
    Sep 1, 2017 at 22:36

Degeneracy pressure is not "overcome" it is simply bypassed by electron capture (a.k.a. neutronisation or inverse beta decay). This happens when the density-dependent Fermi energy of the electrons reaches the threshold energy for the endothermic electron capture process. $$ ^{12}_{6}\rm C + e \rightarrow ^{12}_{5} \rm B + \nu_e $$

The threshold energy is 13.9 MeV. If you set the electron Fermi energy to be 13.9 MeV, assume totally ionised carbon and ideal degeneracy pressure, you can work out that the electrons are highly relativistic and the density of the material must reach $3.9 \times 10^{13}$ kg/m$^{3}$.

In terms of "how much energy" must be supplied to cause this, well that isn't a very well posed question. A better one is what pressure must be applied to squeeze the material to this density. This will be given by the expression for relativistic electron degeneracy pressure and is about $10^{28}$ Pa.


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