 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-3/35d/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Below

T_{c}

the fugacity is

z = 1

in the continuum ideal-gas approximation. The zero-energy condensate contributes no internal energy, so the same Bose integral with one additional factor of energy gives

E = B Γ (\frac{11}{2}) ζ (\frac{11}{2}) (k_{B} T)^{11/2} .

Eliminate

B

using

N = B Γ (9/2) ζ (9/2) (k_{B} T_{c})^{9/2}

and

Γ (11/2) = (9/2) Γ (9/2)

E = \frac{9}{2} N k_{B} T \frac{ζ ( 11/2 )}{ζ ( 9/2 )} (\frac{T}{T _{c}})^{9/2} (T \leq T_{c}) .

The excited fraction is

(T / T_{c})^{9/2}

and the remaining particles occupy the Bose-Einstein condensate. For fixed

B

, the condensed-state energy at fixed

T

is independent of how many additional particles are placed in the zero-energy ground state.

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