Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 (Theoretical Physics of Soft Condensed Matter)

Because $\varphi$ is a conserved composition density, an infinitesimal material displacement $\mathbf{r}\mapsto \mathbf{r}+\mathbf{u}$ changes it at fixed position byThe second term is essential when the deformation is compressible. By the definition of the chemical potential, and taking $\mathbf{u}$ to vanish on the boundary,The same free-energy change written in terms of the stress tensor isSince $\mathbf{u}$ is arbitrary,This is the Korteweg force density of a diffuse-interface mixture.

Functional differentiation givesFor $\Pi =\mu \varphi -\mathbb{F}$ and ${\Sigma }_{ij}=-\Pi {\delta }_{ij}-\kappa ({\nabla }_i\varphi )({\nabla }_j\varphi )$,The bracket vanishes by the expression for $\mu$, while the two mixed second-derivative terms cancel. Therefore

Equilibrium minimizes $F$ at fixed total composition $\int \varphi$, so a Lagrange multiplier gives $\delta F/\delta \varphi ={\mu }_0$, independent of position. In the two homogeneous phases,The symmetric coexistence pair therefore has ${\mu }_0=0$. For a planar profile depending only on the normal coordinate $x$,orup to reversal of the two phases.

Put $\varphi ={\varphi }_{B}g(u)$, $u=(x-x_0)/{\xi }_0$, and ${\xi }_0^2=-2\kappa /a$. Since $b{\varphi }_B^2=-a$, the interface equation reduces toMultiplication by $g^{\prime }$ and use of $g(\pm \infty )=\pm 1$, $g^{\prime }(\pm \infty )=0$ gives the first integralFor the increasing profile, $g^{\prime }=1-g^2$, so $\mathrm{artanh}g=u$ after shifting $x_0$. Hence the phi-four diffuse interface isThe sign chooses the orientation and the translation zero mode $x_0$ sets the interface position.

Far from a planar interface, the equilibrium profile equals a bulk minimum and contributes the uniform density $f({\varphi }_B)$. The localized transition layer contributes an additional free energy proportional to its area $A$, because translation invariance makes the excess per unit area independent of position. By the definition of surface tension as interfacial excess free energy per area,

For a profile depending only on $x$, equation (2) gives ${\Sigma }_{yy}=-\Pi$ and ${\Sigma }_{xx}=-\Pi -\kappa ({\varphi }^{\prime }{)}^2$. ThusFor ${\varphi }_E={\varphi }_B\tanh [(x-x_0)/{\xi }_0]$,Changing variable to $u=(x-x_0)/{\xi }_0$ yieldsUsing ${\varphi }_B^2=-a/b$ and ${\xi }_0^2=-2\kappa /a$ gives the positive interfacial tension of a phi-four diffuse interface

A polar liquid crystal distinguishes the two ends of each constituent. Its orientational order is described by a vector $\mathbf{p}$, and $\mathbf{p}$ and $-\mathbf{p}$ represent different states. A nematic liquid crystal has head-tail symmetry, so its director obeys $\mathbf{n}\sim -\mathbf{n}$. Its lowest-rank faithful nematic order parameter is the symmetric traceless tensorwhich is unchanged by $\mathbf{n}\mapsto -\mathbf{n}$.

The Fourier transform sends each spatial derivative to $i{q}_i$, so at Gaussian levelFor $\kappa <0<\gamma$, minimizing over $s=q^2\ge 0$ givesThe minimum kernel is $G(q_0)=a-{\kappa }^2/(4\gamma )$. Gaussian fluctuations first diverge when this vanishes, so the nonzero-wavevector soft-mode sphere becomes unstable at

For candidate (i), $\mathbf{p}=p_0\hat{\mathbf{p}}\cos q_{0}x$. Since $G(q_0)=\hat{a}:=a-a_c$, $⟨{\cos }^2{q}_{0}x⟩=1/2$, and $⟨{\cos }^4{q}_{0}x⟩=3/8$, its mean-field free-energy density isFor $\hat{a}<0$, stationarity givesand substitution yieldsFor $\hat{a}\ge 0$, the minimum is $p_0=0$. The amplitude therefore vanishes continuously as $p_0\propto (a_c-a{)}^{1/2}$ on approaching $a_c$ from below, which is a continuous mean-field transition.

For candidate (ii), $|\mathbf{p}|=p_0$ everywhere and each component has wavevector magnitude $q_0$. The quadratic density is therefore $\hat{a}{p}_0^2/2$, while the quartic density is $b{p}_0^4/4$ without trigonometric averaging:For $\hat{a}<0$,Since $(-1/4)/(-1/6)=3/2$, the helical structure's free energy is $50\%$ more negative than that of candidate (i).

Every term in the free energy contracts the vector components with the Euclidean inner product: it depends only on $|\mathbf{p}{|}^2$, $|\mathbf{p}{|}^4$, $({\nabla }_i{p}_j)({\nabla }_i{p}_j)$, and $|{\nabla }^2\mathbf{p}{|}^2$. A constant $\mathcal{R}\in SO(3)$ preserves all these contractions and commutes with spatial differentiation. Thereforeso every constant rotation of candidate (ii) has the same free energy. The orientation of the rotation plane of the polar helical smectic is thus continuously degenerate.

Let ${\mathbf{e}}_1,{\mathbf{e}}_2$ span the rotating plane and let $\mathbf{n}={\mathbf{e}}_1\times {\mathbf{e}}_2$ be its normal. For modulation along $\hat{\mathbf{x}}$,and period averaging givesFor $\lambda >0$, this is minimized by $\mathbf{n}\parallel \hat{\mathbf{x}}$, so the rotation plane is perpendicular to the modulation direction and the helix is transverse. For $\lambda <0$, it is minimized energetically by maximizing the bracket: $\mathbf{n}\perp \hat{\mathbf{x}}$, so the modulation direction lies in the rotation plane. Thus either sign lifts the full rotational degeneracy, leaving only the rotations consistent with its selected relative orientation. A sufficiently small negative $\lambda$ does not overcome the stabilizing higher-gradient terms.

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.