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

At large separation, the finite-thickness van der Waals combination has the expansionThus the attraction governed by the Hamaker constant decays asand the screened electrostatic term decays exponentially on the Debye–Hückel screening length. The Helfrich repulsion, however, decays only asConsequently the total interaction approaches its unbound value zero from above as $d\to \infty$.

A finite bound minimum must have nonpositive energy to beat the state at infinity. Because the large-$d$ interaction is positive, such a minimum cannot move continuously to infinity while remaining globally stable. At the transition it instead becomes degenerate with the $d=\infty$ state at a finite spacing and then loses global stability. The equilibrium spacing therefore jumps from finite $d_{\ast }$ to infinity, making this a discontinuous, first-order unbinding transition.

For a dilute stack, the membrane number per unit normal length is $1/d=\varphi /\delta$. Dividing the fluctuation repulsion per area by the repeat distance gives its free-energy density

The short-range electrostatic repulsion and long-range van der Waals attraction enter at second-virial order. In general, for an effective pair energy $U_{A_H}(\Gamma )$ over relative configurations $\Gamma$, the second virial coefficient has the Mayer-integral formwith a fixed normalization by the microscopic membrane thickness making $B_2$ dimensionless here. The mean-field contribution is quadratic in membrane concentration. The required two-power free energy can therefore be writtenChanges of microscopic normalization merely rescale $B_2$ by a positive constant and do not affect the transition.

Write the pair energy asAt $A_H=0$ the interaction is repulsive, so the Mayer integrand and hence $B_2$ are positive. Increasing $A_H$ strengthens attraction andFor sufficiently strong attraction, negative configurations dominate and $B_2<0$. Continuity therefore gives a critical $A_H^{\ast }$ with $B_2(A_H^{\ast })=0$. The Taylor theorem giveswhere $r=-B_2^{\prime }(A_H^{\ast })>0$.

Write the free energy asWhen $A_H<A_H^{\ast }$, $B_2>0$ and the global minimum on $\varphi \ge 0$ is ${\varphi }_{\ast }=0$, corresponding to an unbound stack. When $A_H>A_H^{\ast }$, $B_2<0$, and minimization givesUsing $B_2\sim -r(A_H-A_H^{\ast })$ yieldsSince $d_{\ast }=\delta /{\varphi }_{\ast }$,so the continuous membrane unbinding transition has exponent