Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 3

The Young–Laplace equation gives a pressure excess $2\sigma /R$ inside a spherical droplet. Near coexistence, the common chemical-potential shift is $\mu \simeq \alpha \delta$, where $\alpha =f^{\prime \prime }({\varphi }_B)$, while changing phase changes the composition by approximately $2{\varphi }_B$. Balancing the capillary pressure against this thermodynamic shift gives the Gibbs--Thomson relationHence the compositions immediately outside and inside arewith

In the exterior write $\varphi =-{\varphi }_B+\tilde{\varphi }$. Linearization at the bulk minimum givesFor a large droplet, variations occur on scale $R$ much larger than the interfacial correlation length, so the second term is smaller by $O(\kappa /(\alpha R^2))$ and $\mu \simeq \alpha \tilde{\varphi }$. The conserved order-parameter dynamics then becomesDiffusion relaxes the exterior profile much faster than the droplet radius changes. In this quasistatic limit ${\partial }_t\tilde{\varphi }\simeq 0$, and therefore

The spherically symmetric Laplace equation with $\tilde{\varphi }(R)=\delta (R)$ and $\tilde{\varphi }(\infty )=\epsilon$ has solutionThusConservation at the moving interface, whose composition jump is $2{\varphi }_B$, gives $2{\varphi }_B\dot{R}=-J_r(R^{+})$ and hence

Since $\delta (R)=C/R$, the right side is proportional to $\epsilon /R-C/R^2$. It is negative for $R<R^{\ast }$, zero at $R^{\ast }=C/\epsilon$, positive for $R>R^{\ast }$, and approaches zero from above for large $R$. Thus $R^{\ast }$ is the unstable critical nucleus: smaller droplets dissolve, while larger droplets grow.

Using $\delta (R)=\sigma /(\alpha {\varphi }_{B}R)$, the growth law isForone has $F^{\prime }=8\pi \sigma R-4\pi \Delta R^2$. Therefore, with $\mathcal{M}(R)=M/(16\pi {\varphi }_B^2{R}^3)$,The first term in $F$ is the positive surface cost and the second is the negative bulk free-energy gain of converting a metastable volume. The nonzero stationary radius and barrier are the classical nucleation theory valuesSince $F^{\prime \prime }(R^{\ast })=-8\pi \sigma <0$, this stationary point is a maximum.

Thermal molecular motion causes the coarse droplet radius to fluctuate as well as drift down $F(R)$, so its effective equation must be an Overdamped Langevin dynamics. With $⟨\Lambda (t)\Lambda (t^{\prime })⟩=\delta (t-t^{\prime })$, detailed balance with equilibrium density proportional to $e^{-F/(k_{B}T)}$ imposes the Fluctuation-dissipation theorem. The Model A fluctuation-dissipation relation givesThe factor of mobility ensures that the diffusion in $R$ and the dissipative drift have the same equilibrium Gibbs distribution.

A subcritical droplet must make a rare thermal fluctuation from the metastable basin over the free-energy barrier $F^{\ast }$. Its probability carries the Boltzmann factor $e^{-F^{\ast }/(k_{B}T)}$, so the nucleation rate has the Arrhenius formFor a volume containing order one candidate subcritical droplet, the waiting time for a supercritical droplet and subsequent macroscopic growth is thereforeup to an algebraic kinetic prefactor. This is the Arrhenius nucleation time.