Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 355 / 3 / c

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- Solution c

Solution

Write the pair energy as

U_{A_{H}} (Γ) = U_{rep} (Γ) - A_{H} W (Γ), W (Γ) > 0.

At

A_{H} = 0

the interaction is repulsive, so the Mayer integrand and hence

B_{2}

are positive. Increasing

A_{H}

strengthens attraction and

\frac{d B _{2}}{d A _{H}} = - \frac{1}{2 k _{B} T} \int W (Γ) e^{- U_{A_{H}} (Γ) / (k_{B} T)} d Γ < 0.

For sufficiently strong attraction, negative configurations dominate and

B_{2} < 0

. Continuity therefore gives a critical

A_{H}^{*}

with

B_{2} (A_{H}^{*}) = 0

. The Taylor theorem gives

B_{2} (A_{H}) = B_{2}^{'} (A_{H}^{*}) (A_{H} - A_{H}^{*}) + \dots \sim r (A_{H}^{*} - A_{H}),

where

r = - B_{2}^{'} (A_{H}^{*}) > 0

.

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