6.11 Degeneracy Pressure

According to the previous sections, electrons, being fermions, behave in a way very differently from bosons. A system of bosons has very little energy in its ground state, as all bosons collect in the spatial state of lowest energy. Electrons cannot do so. At most two electrons can go into a single spatial state. A macroscopic system of electrons must occupy a gigantic number of states, ranging from the lowest energy state to states with many orders of magnitude more energy.

As a result, a free-electron gas of $I$ noninteracting electrons ends up with an average energy per electron that is larger than of a corresponding system of bosons by a gigantic factor of order $I^{2/3}$. That is all kinetic energy; all forces on the electrons are ignored in the interior of a free-electron gas, so the potential energy can be taken to be zero.

Having so much kinetic energy, the electrons exert a tremendous pressure on the walls of the container that holds them. This pressure is called degeneracy pressure. It explains qualitatively why the volume of a solid or liquid does not collapse under normally applied pressures.

Of course, degeneracy pressure is a poorly chosen name. It is really due to the fact that the energy distribution of electrons is not degenerate, unlike that of bosons. Terms like exclusion-principle pressure or “Pauli pressure” would capture the essence of the idea. So they are not acceptable.

The magnitude of the degeneracy pressure for a free-electron gas is

\begin{displaymath}
P_{\rm {d}} = {\textstyle\frac{2}{5}} \left(3\pi^2\right)^...
... \frac{\hbar^2}{2m_e}
\left(\frac{I}{{\cal V}}\right)^{5/3}
\end{displaymath} (6.18)

This may be verified by equating the work $-P_{\rm {d}}{ \rm d}{\cal V}$ done when compressing the volume a bit to the increase in the total kinetic energy ${\vphantom' E}^{\rm S}$ of the electrons:

\begin{displaymath}
- P_{\rm {d}} { \rm d}{\cal V}= {\rm d}{\vphantom' E}^{\rm S}
\end{displaymath}

The energy ${\vphantom' E}^{\rm S}$ is $I$ times the average energy per electron. According to section 6.10, that is $\frac35I$ times the Fermi energy (6.16).

A ballpark number for the degeneracy pressure is very instructive. Consider once again the example of a block of copper, with its valence electrons modeled as a free-electron gas, Using the same numbers as in the previous section, the degeneracy pressure exerted by these valence electrons is found to be 40 10$\POW9,{9}$ Pa, or 40 GPa.

This tremendous outward pressure is balanced by the nuclei that pull on electrons that try to leave the block. The details are not that simple, but electrons that try to escape repel other, easily displaced, electrons that might aid in their escape, leaving the nuclei unopposed to pull them back. Obviously, electrons are not very smart.

It should be emphasized that it is not mutual repulsion of the electrons that causes the degeneracy pressure; all forces on the electrons are ignored in the interior of the block. It is the uncertainty relationship that requires spatially confined electrons to have momentum, and the exclusion principle that explodes the resulting amount of kinetic energy, creating fast electrons that are as hard to contain as students on the day before Thanksgiving.

Compared to a 10$\POW9,{10}$ Pa degeneracy pressure, the normal atmospheric pressure of 10$\POW9,{5}$ Pa cannot add any noticeable further compression. Pauli’s exclusion principle makes liquids and solids quite incompressible under normal pressures.

However, under extremely high pressures, the electron pressure can lose out. In particular, for neutron stars the spatial electron states collapse under the very weight of the massive star. This is related to the fact that the degeneracy pressure grows less quickly with compression when the velocity of the electrons becomes relativistic. (For very highly relativistic particles, the kinetic energy is not given in terms of the momentum $p$ by the Newtonian value ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $p^2$$\raisebox{.5pt}{$/$}$$2m$, but by the Planck-Einstein relationship ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $pc$ like for photons.) That makes a difference since gravity too increases with compression. If gravity increases more quickly, all is lost for the electrons. For neutron stars, the collapsed electrons combine with the protons in the star to form neutrons. It is the degeneracy pressure of the neutrons, also spin $\leavevmode\kern.03em
\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ fermions but 2,000 times heavier, that carries the weight of a neutron star.


Key Points
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Because typical confined electrons have so much kinetic energy, they exert a great degeneracy pressure on what is holding them.

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This pressure makes it very hard to compress liquids and solids significantly in volume.

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\put(2,0){\makebox(0,0){\scriptsize\bf0}}
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...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
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Differently put, liquids and solids are almost incompressible under typical conditions.