ε = p c

g (ε) = \frac{V ε ^{2}}{π ^{2} ℏ ^{3} c ^{3}} .

lo g Z = \int_{0}^{\infty} g (ε) lo g (1 + e^{- β (ε - μ)}) d ε .

Since the grand potential satisfies

p V = k_{B} T lo g Z

p V = \frac{1}{3} \int_{0}^{\infty} ε g (ε) \frac{d ε}{e ^{β (ε - μ)} + 1} = \frac{E}{3} .

Thus

p V = A E, A = \frac{1}{3} .

ii

Solution

0 \leq ε < E_{F}

. Hence

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

Solving for the Fermi energy gives

E_{F} = D (\frac{N}{V})^{1/3}, D = ℏ c (3 π^{2})^{1/3} .

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} .

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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