Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 35A / b / iii

Table of contents
- Solution iii

Solution

The zero-temperature internal energy is

E = \int_{0}^{E_{F}} ε g (ε) d ε = \frac{V E _{F}^{4}}{4 π ^{2} ℏ ^{3} c ^{3}} = \frac{3}{4} N E_{F} .

Using part (i),

p = \frac{N E _{F}}{4 V} = \frac{D}{4} N^{4/3} V^{- 4/3} .

Consequently the adiabatic equation of state is

p V^{a} = K, a = \frac{4}{3}, K = \frac{ℏ c ( 3 π ^{2} ) ^{1/3}}{4} N^{4/3} .

The pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys

p V = N k_{B} T

, so its corresponding volume exponent is one and its pressure vanishes as

T \to 0

.

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