Write the pair energy as
$$
U_{A_H}(\Gamma)=U_{\rm rep}(\Gamma)-A_HW(\Gamma),
\qquad W(\Gamma)>0.
$$
At $A_H=0$ the interaction is repulsive, so the Mayer integrand and hence $B_2$ are positive. Increasing $A_H$ strengthens attraction and
$$
\frac{dB_2}{dA_H}
=-\frac1{2k_BT}\int
W(\Gamma)e^{-U_{A_H}(\Gamma)/(k_BT)}\,d\Gamma<0.
$$
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
$$
\boxed{
B_2(A_H)
=B_2'(A_H^*)(A_H-A_H^*)+\cdots
\sim r(A_H^*-A_H)},
$$
where $r=-B_2'(A_H^*)>0$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-355/3/c/solution