Degeneracy Pressure in Stars

The pressure exerted by fermions squeezed into a small box is what keeps cold stars from collapsing. White Dwarfs are held up by electrons and Neutron Stars are held up by neutrons in a much smaller box.

We can compute the pressure from the dependence of the energy on the volume for a fixed number of fermions.

$$\begin{eqnarray*} dE&=&\vec{F}\cdot d\vec{s}=PAds=PdV \\ P&=&-{\partial E_{\mat... ...={\pi^3\hbar^2\over 15m}\left({3n\over\pi}\right)^{5\over 3} \\ \end{eqnarray*}$$

The last step verifies that the pressure only depends on the density, not the volume and the $N$ independently, as it should. We will use.

$$\bgroup\color{black} P={\pi^3\hbar^2\over 15m}\left({3N\over\... ...r 15m}\left({3N\over\pi}\right)^{5\over 3}V^{-5\over 3} \egroup$$

To understand the collapse of stars, we must compare this to the pressure of gravity. We compute this approximately, ignoring general relativity and, more significantly, the variation of gravitational pressure with radius.

$$\begin{eqnarray*} E&=&-\int\limits_0^R {GM_{\mathrm inside}4\pi r^2\rho\over r}d... ...dr \\ &=&-{(4\pi)^2\over 15} G \rho^2 R^5=-{3GM^2\over 5R} \\ \end{eqnarray*}$$

The mass of the star is dominated by nucleons.

$$\bgroup\color{black} M=NM_N \egroup$$

Putting this into our energy formula, we get.

$$\bgroup\color{black} E=-{3\over 5}G(NM_N)^2\left({4\pi\over 3}\right)^{1\over 3}V^{-{1\over 3}} \egroup$$

We can now compute the pressure.

$$\bgroup\color{black} P_g=-{\partial E\over\partial V} =-{1\ov... ...)^2\left({4\pi\over 3}\right)^{1\over 3}V^{-{4\over 3}} \egroup$$

The pressures must balance. For a white dwarf, the pressure from electrons is.

$$\bgroup\color{black} P_e={\pi^3\hbar^2\over 15m_e}\left({3N_e\over\pi}\right)^{5\over 3}V^{-{5\over 3}} \egroup$$

We can solve for the radius.

$$\bgroup\color{black} R=\left({3\over 4\pi}\right)^{2\over 3}{... ...{3\over \pi}\right)^{5\over 3}{N_e^{5\over 3}\over N^2} \egroup$$

There are about two nucleons per electron

$$\bgroup\color{black} N\approx 2N_e\egroup$$

so the radius becomes.

$$\bgroup\color{black} R=\left({81\pi^2\over 512}\right)^{1\over 3}{\hbar^2\over Gm_eM_N^2}N^{-{1\over 3}} \egroup$$

The radius decreases as we add mass. For one solar mass, $N=10^{57}$, we get a radius of 7200 km, the size of the earth. The Fermi energy is about 0.2 MeV.

A white dwarf is the remnant of a normal star. It has used up its nuclear fuel, fusing light elements into heavier ones, until most of what is left is Fe $^{56}$ which is the most tightly bound nucleus. Now the star begins to cool and to shrink. It is stopped by the pressure of electrons. Since the pressure from the electrons grows faster than the pressure of gravity, the star will stay at about earth size even when it cools.

If the star is more massive, the Fermi energy goes up and it becomes possible to absorb the electrons into the nucleons, converting protons into neutrons. The Fermi energy needs to be above 1 MeV. If the electrons disappear this way, the star collapses suddenly down to a size for which the Fermi pressure of the neutrons stops the collapse (with quite a shock). Actually some white dwarfs stay at earth size for a long time as they suck in mass from their surroundings. When they have just enough, they collapse forming a neutron star and making a supernova. The supernovae are all nearly identical since the dwarfs are gaining mass very slowly. The brightness of this type of supernova has been used to measure the accelerating expansion of the universe.

We can estimate the neutron star radius.

$$\bgroup\color{black} R\rightarrow R{M_N\over m_e}N^{1\over 3}2^{-{5\over 3}}=10 \egroup$$

Its about 10 kilometers. If the pressure at the center of a neutron star becomes too great, it collapses to become a black hole. This collapse is probably brought about by general relativistic effects, aided by strange quarks.