Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 10E (Cosmology)

As $T\downarrow 0$, the Fermi-Dirac distribution tends to a step function: $\bar{n}(p)=1$ for $ϵ(p)<{\mu }_0$, and $0$ for $ϵ(p)>{\mu }_0$. The value at equality is immaterial to the integrals. The Fermi momentum is defined by $ϵ(p_F)={\mu }_0$: the Pauli exclusion principle fills the lowest available states and forbids multiple occupation of a state with a fixed spin. Thus the sketch is a horizontal line at one up to $p_F$, followed by a jump to zero.

In the ultrarelativistic, $ϵ(p)\simeq cp$, givingEliminate $p_F$ to obtainThe supplied expression for $ϵ$ includes rest energy; in this limit its leading term is also the leading kinetic energy.

Electrical neutrality gives $n_e\simeq n_p$ at the accuracy of the proposed estimate. With $V\sim R^3$ and $n_p\sim M/(m_p{R}^3)$, the degenerate electron gas result givesThe gravitational potential energy is $E_{\mathrm{grav}}\sim -G{M}^2/R$. Both scale as $R^{-1}$, so in the strictly ultrarelativistic approximationFor sufficiently large mass the bracket is negative, and contraction lowers the energy without bound in this model: electron degeneracy pressure cannot halt collapse. Equating the scales gives the Chandrasekhar massBelow this scale, the strictly ultrarelativistic expression alone has no finite-radius energy minimum; including the transition to nonrelativistic electrons produces the stable white dwarf branch. Composition and the stellar density profile supply the numerical factors omitted by this scaling estimate.