Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 317 / 3 / i

Table of contents
- Solution i

Solution

Assume zero temperature, complete ionization, constant composition, noninteracting electrons, negligible ion thermal pressure, and Newtonian stellar gravity. Ions supply almost all the mass, while electron degeneracy pressure supplies support. By the Pauli exclusion principle, the two electron spin states fill a momentum sphere up to Fermi momentum

p_{F}

. Counting states gives

n_{e} = \frac{2}{( 2 π ℏ ) ^{3}} \frac{4 π p _{F}^{3}}{3} = \frac{p _{F}^{3}}{3 π ^{2} ℏ ^{3}}, ρ = μ_{e} m_{u} n_{e},

where

μ_{e}

is mean molecular weight per electron. The momentum flux of this isotropic Fermi gas is

P = \frac{2}{3 ( 2 π ℏ ) ^{3}} \int_{p \leq p_{F}} p v (p) d^{3} p = \frac{1}{3 π ^{2} ℏ ^{3}} \int_{0}^{p_{F}} \frac{p ^{4} c ^{2}}{m _{e}^{2} c ^{4} + p ^{2} c ^{2}} d p .

x = p_{F} / (m_{e} c)

and performing the integral gives the equation of state of a cold electron gas

P = \frac{m _{e}^{4} c ^{5}}{24 π ^{2} ℏ ^{3}} [x (2 x^{2} - 3) 1 + x^{2} + 3 arsinh x], x = \frac{ℏ}{m _{e} c} (\frac{3 π ^{2} ρ}{μ _{e} m _{u}})^{1/3} .

In the nonrelativistic limit

p_{F} ≪ m_{e} c

,

v (p) ≃ p / m_{e}

, so

P = K_{NR} ρ^{5/3}, K_{NR} = \frac{ℏ ^{2} ( 3 π ^{2} ) ^{2/3}}{5 m _{e} ( μ _{e} m _{u} ) ^{5/3}} .

In the ultrarelativistic limit

p_{F} ≫ m_{e} c

,

v (p) ≃ c

, so

P = K_{R} ρ^{4/3}, K_{R} = \frac{ℏ c ( 3 π ^{2} ) ^{1/3}}{4 ( μ _{e} m _{u} ) ^{4/3}} .

These are polytropic equations of state with indices

3/2

and

3

, respectively. The relativistic softening underlies the Chandrasekhar mass limit; Coulomb corrections, thermal effects, rotation, and general relativity are excluded from this idealized derivation.

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