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White-dwarf stars

A main-sequence hydrogen-burning star, such as the Sun, is maintained in equilibrium via the balance of the gravitational attraction tending to make it collapse, and the thermal pressure tending to make it expand. Of course, the thermal energy of the star is generated by nuclear reactions occurring deep inside its core. Eventually, however, the star will run out of burnable fuel, and, therefore, start to collapse, as it radiates away its remaining thermal energy. What is the ultimate fate of such a star?

A burnt-out star is basically a gas of electrons and ions. As the star collapses, its density increases, so the mean separation between its constituent particles decreases. Eventually, the mean separation becomes of order the de Broglie wavelength of the electrons, and the electron gas becomes degenerate. Note, that the de Broglie wavelength of the ions is much smaller than that of the electrons, so the ion gas remains non-degenerate. Now, even at zero temperature, a degenerate electron gas exerts a substantial pressure, because the Pauli exclusion principle prevents the mean electron separation from becoming significantly smaller than the typical de Broglie wavelength (see the previous section). Thus, it is possible for a burnt-out star to maintain itself against complete collapse under gravity via the degeneracy pressure of its constituent electrons. Such stars are termed white-dwarfs. Let us investigate the physics of white-dwarfs in more detail.

The total energy of a white-dwarf star can be written

E = K + U,
\end{displaymath} (682)

where $K$ is the total kinetic energy of the degenerate electrons (the kinetic energy of the ion is negligible) and $U$ is the gravitational potential energy. Let us assume, for the sake of simplicity, that the density of the star is uniform. In this case, the gravitational potential energy takes the form
U = -\frac{3}{5}\frac{G\,M^2}{R},
\end{displaymath} (683)

where $G$ is the gravitational constant, $M$ is the stellar mass, and $R$ is the stellar radius.

Let us assume that the electron gas is highly degenerate, which is equivalent to taking the limit $T\rightarrow 0$. In this case, we know, from the previous section, that the Fermi momentum can be written

p_F = {\mit\Lambda}\left(\frac{N}{V}\right)^{1/3},
\end{displaymath} (684)

{\mit\Lambda} = (3\pi^2)^{1/3}\,\hbar.
\end{displaymath} (685)

V = \frac{4\pi}{3}\,R^3
\end{displaymath} (686)

is the stellar volume, and $N$ is the total number of electrons contained in the star. Furthermore, the number of electron states contained in an annular radius of ${\bf p}$-space lying between radii $p$ and $p+dp$ is
dN = \frac{3\,V}{{\mit\Lambda}^3}\, p^2\,dp.
\end{displaymath} (687)

Hence, the total kinetic energy of the electron gas can be written
K = \frac{3\,V}{{\mit\Lambda}^3}\int_0^{p_F} \frac{p^2}{2\,m...
\end{displaymath} (688)

where $m$ is the electron mass. It follows that
K = \frac{3}{5}\,N\,\frac{{\mit\Lambda}^2}{2\,m}\left(\frac{N}{V}\right)^{2/3}.
\end{displaymath} (689)

The interior of a white-dwarf star is composed of atoms like $C^{12}$ and $O^{16}$ which contain equal numbers of protons, neutrons, and electrons. Thus,

M = 2\,N\,m_p,
\end{displaymath} (690)

where $m_p$ is the proton mass.

Equations (682), (683), (685), (686), (689), and (690) can be combined to give

E = \frac{A}{R^2} - \frac{B}{R},
\end{displaymath} (691)

$\displaystyle A$ $\textstyle =$ $\displaystyle \frac{3}{20}\left(\frac{9\pi}{8}\right)^{2/3}\frac{\hbar^2}{m}
\left(\frac{M}{m_p}\right)^{5/3},$ (692)
$\displaystyle B$ $\textstyle =$ $\displaystyle \frac{3}{5}\,G\,M^2.$ (693)

The equilibrium radius of the star $R_\ast$ is that which minimizes the total energy $E$. In fact, it is easily demonstrated that
R_\ast = \frac{2\,A}{B},
\end{displaymath} (694)

which yields
R_\ast = \frac{(9\pi)^{2/3}}{8}\,\frac{\hbar^2}{m}\,\frac{1}{G\,m_p^{\,5/3}\,
\end{displaymath} (695)

The above formula can also be written
\frac{R_\ast}{R_\odot}= 0.010\left(\frac{M_\odot}{M}\right)^{1/3},
\end{displaymath} (696)

where $R_\odot= 7\times 10^5\,{\rm km}$ is the solar radius, and $M_\odot = 2\times 10^{30}\,{\rm kg}$ is the solar mass. It follows that the radius of a typical solar mass white-dwarf is about 7000km: i.e., about the same as the radius of the Earth. The first white-dwarf to be discovered (in 1862) was the companion of Sirius. Nowadays, thousands of white-dwarfs have been observed, all with properties similar to those described above.

next up previous
Next: The Chandrasekhar limit Up: Quantum statistics Previous: Conduction electrons in a
Richard Fitzpatrick 2006-02-02