Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 344 (Theoretical Physics of Soft Condensed Matter)

Take $\gamma >0$ so that the short-wavelength quadratic energy is stable. For $B=0$, a Fourier mode with wavevector magnitude $q$ has inverse Gaussian field theory covarianceFor $\kappa <0$, its minimum is atThe zero-average constraint excludes the uniform mode but allows $\varphi =A\cos (q_{0}z)$. At $a=a_c$ the modes on the nonzero-wavevector soft-mode sphere become soft; below it they have negative quadratic energy, so the homogeneous state loses stability. A modulation along any chosen direction is the simplest smectic phase pattern. The quadratic instability selects its wavelength; with $B=0$, a nonlinear stabilizing interaction would still be needed to determine its amplitude.

Average the trial field over an integer number of periods. The squared-field, squared-gradient and squared-Laplacian averages are $A^2/2$, $q_0^2{A}^2/2$ and $q_0^4{A}^2/2$, while $⟨{\varphi }^2|\nabla \varphi {|}^2⟩=A^4{q}_0^2/8$. Hence the single-mode smectic mean-field free energy isFor the physically stabilizing case $B>0$, minimization givesThe amplitude tends continuously to zero at $a_c$, so this mean-field approximation predicts a continuous phase transition. The ordered minimum has free-energy density $-(a_c-a{)}^2/(8B{q}_0^2)$.

A nonzero $B$ alone is not sufficient for this conclusion: $B<0$ makes the displayed quartic amplitude energy unbounded below and requires higher-order stabilization. The continuous-transition statement uses $B>0$ and the stipulated fixed wavevector $q_0$.

Apply the Jensen inequality to the convex exponential function under the trial equilibrium measure:The given identity therefore implies $e^{-F}\ge e^{-F_0-⟨H-H_0{⟩}_0}$. Taking minus the logarithm reverses the inequality and proves the Feynman-Bogoliubov inequalityThe units $\beta =1$ make $H$ and $F$ dimensionless here. Equality holds when $H-H_0$ is constant on the support of the trial measure. Optimizing the trial parameters tightens an upper bound, which is the basis of the Gaussian variational approximation.

Choose an even positive $J(q)$. Each independent complex Fourier mode of the real field has density $(J(q)/\pi )e^{-J(q)|{\varphi }_q{|}^2}$, soThe quadratic part therefore has expectation ${\sum }_q^{+}G(q)/J(q)$. For the quartic part, apply Isserlis theorem to its four jointly centered Gaussian random variables, or to their real and imaginary components. Pairing the two differentiated fields together imposes $q_2=-q_1$, and its contribution isEach of the two other pairings instead gives $-(B/V)|{\sum }_{q}q/J(q){|}^2=0$, because the summand is odd under $q\mapsto -q$. Thus only the displayed contribution remains. Each full sum is twice its positive-wavevector sum for a real field; putting $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ gives $⟨H_4{⟩}_0=4B{S}_2{S}_0/V$.

Substitute these expectations, $F_0$ and $⟨H_0{⟩}_0$ into the Feynman-Bogoliubov inequality:The zero mode is absent by the composition constraint. The counting uses one representative of every nonzero $\{q,-q\}$ pair, so no extra factor of two is assigned to an independent complex amplitude.

Differentiate the bound with respect to one $J(q)$, using $\partial S_0/\partial J(q)=-J(q{)}^{-2}$ and $\partial S_2/\partial J(q)=-q^{2}J(q{)}^{-2}$:Thus an interior minimizing Gaussian variational kernel for a gradient quartic interaction hasUse the half-space mode density given in the question. In the thermodynamic limit, the coupled self-consistency equations areThe admissible solution must have $J(q)>0$. The coefficient $\gamma$ is unchanged because the interaction's derivative contributes only a constant and a $q^2$ term.

The formal continuum integrals require the coarse-graining ultraviolet cutoff $\Lambda$ in three dimensions: the first radial integrand tends to a constant at large $k$, so taking $\Lambda =\infty$ would leave $\bar{a}$ divergent. The low-wavevector fluctuation argument is independent of this short-distance regularization.

In the usual three-dimensional isotropic setting, the transition is expected to be discontinuous, by the Brazovskii fluctuation-induced first-order transition mechanism.

If $\bar{\kappa }<0$, write $q_{\ast }^2=-\bar{\kappa }/(2\gamma )$ and $r=\bar{a}-{\bar{\kappa }}^2/(4\gamma )$. Then $J(q)=r+\gamma (q^2-q_{\ast }^2{)}^2$. A continuous finite-wavevector transition would require $r\to 0$ with $q_{\ast }>0$. But the soft-mode shell integral behaves asThe positive fluctuation corrections in the self-consistency equations diverge in that limit. In particular $\bar{\kappa }=\kappa +2B{I}_0$ cannot stay at a finite negative value while $I_0$ diverges. Thus the isotropic disordered state cannot lose stability through a zero-mass mode on a fixed nonzero shell in this approximation.

Ordering is instead expected through a free-energy crossing with a finite-amplitude smectic phase, giving a first-order phase transition. This is a reasoned fluctuation prediction for the finite-wavevector transition; the disordered Gaussian variational approximation alone does not locate the coexistence point or calculate the amplitude jump. A zero-wavevector instability or a lower-dimensional setting would require a separate analysis.

Write ${\mu }_j=\delta F/\delta p_j$ and $\int ={\int }_{t_1}^{t_2}dt\int d\mathbf{r}$. Eliminating the noise gives $f=\dot{p}+\Gamma \mu$. The Onsager--Machlup path probability is thereforeThese are conditional path densities given the corresponding starting configuration. For a time-even vector order parameter, reversal reads the configurations backward, changes $\dot{p}$ to $-\dot{p}$, and keeps $\mu$ unchanged on each corresponding configuration.

The two Gaussian white noise measures have the same normalization. With a common midpoint discretization, the noise-to-path Jacobian also agrees under reversal by time-reversal invariance of a path Jacobian. The common factor $\mathcal{K}[p]$ may include that path-dependent Jacobian; it need not be a universal constant. This makes the shared-factor assertion precise while leaving the requested ratio unaffected.

The difference between the two squared actions is $|\dot{p}+\Gamma \mu {|}^2-|-\dot{p}+\Gamma \mu {|}^2=4\Gamma \dot{p}\cdot \mu$. The common factor cancels, givingThe last equality uses the functional chain rule with the same midpoint convention as the Onsager–Machlup functional.

At equilibrium, microscopic reversibility requires $e^{-F_1/(k_{B}T)}{P}_F=e^{-F_2/(k_{B}T)}{P}_B$, equivalently $P_F/P_B=e^{-(F_2-F_1)/(k_{B}T)}$. Comparing the coefficients of an arbitrary free-energy change gives the model A fluctuation-dissipation relation

The Fourier transform turns ${\nabla }^2$ into multiplication by $-q^2$. Each Cartesian component is an Ornstein-Uhlenbeck process:Since $r(q)>0$, the transient from the remote past vanishes, and the integrating-factor solution isUsing the stated Gaussian white noise covariance and assuming $t_1\ge t_2$,Interchanging the times gives the stationary two-time correlation functionAt equal times this is the equilibrium Gaussian field theory covariance, and the decay time is $r(q{)}^{-1}$.

Put $u_j={\dot{p}}_j+{\partial }_i{W}_{ij}$. The noises required by a joint trajectory are $f_j=u_j+\Gamma {\mu }_j$ and $N_{ij}=W_{ij}+M{\partial }_i{\mu }_j$. Their independent Gaussian white noise weights give the joint path probability of an order parameter and its flux:For the physically reversed path, $p$ is time-even and the transport flux $W$ is time-odd. Therefore $u\mapsto -u$, $W\mapsto -W$, andThe common factor includes the two noise normalizations and the midpoint Jacobian determinant. Its calculation must include both relaxation channels. On a finite spatial grid, let $\mathsf{D}$ and $\mathsf{G}$ represent the divergence and gradient, with $\Delta =\mathsf{DG}$ the discrete Laplacian, and let ${\mathcal{H}}_n=\partial \mu /\partial p$ be the Hessian matrix of $F$ at the midpoint of time step $n$. Differentiating the two required noises with respect to the next configuration and the interval flux gives the Jacobian matrixUsing the Schur complement, its Jacobian determinant, apart from the path-independent power of $\Delta t$, isReversal visits the same midpoint configurations in reverse order, so the product of these factors is unchanged. The joint time-reversal invariance of a path Jacobian therefore involves the complete relaxation operator $\Gamma I-M\Delta$.

Taking the action difference givesRepeated component indices are summed. The second squared norm sums over both flux indices. This is mixed conserved and nonconserved order-parameter dynamics; reversing the order parameter history while leaving the flux unreversed would give the wrong probability ratio.

Set the noise strengths to the model A fluctuation-dissipation relation and the fluctuation-dissipation relation for a conserved flux:The preceding logarithmic ratio becomesThe last two terms combine as ${\partial }_i({\mu }_j{W}_{ij})$. They integrate to zero under the periodic boundary conditions, by integration by parts. The remaining term is $F_2-F_1$ by the functional chain rule, henceAfter including the equilibrium initial weights, the forward and reversed histories have equal probability. Thus microscopic reversibility is satisfied by the two channels together, with each channel's own thermal noise strength.

Use the stable parameters $b,\kappa >0$. The bulk stationary points satisfy $f^{\prime }(\varphi )=a\varphi +b{\varphi }^3=0$. For $a<0$, $\varphi =0$ has negative curvature, while the two minima areBoth have $f(\pm {\varphi }_B)=-a^2/(4b)$ and $f^{\prime \prime }(\pm {\varphi }_B)=-2a>0$. To establish minimization at fixed total composition, writeFor mean composition $\bar{\varphi }$ with $|\bar{\varphi }|<{\varphi }_B$, a mixture of the two minima with volume fraction $\lambda =(\bar{\varphi }+{\varphi }_B)/(2{\varphi }_B)$ in the positive phase attains this lower bound in the bulk. Macroscopic slabs have only a subextensive interfacial excess free energy, so the bound is attained per unit volume in the thermodynamic limit. Thus constrained minimization gives phase coexistence in the mean-field approximation; at the endpoints only one phase is needed.

The two equilibrium conditions also follow directly by constrained optimization. Introduce a Lagrange multiplier $\mu$ for $\lambda {\varphi }_{+}+(1-\lambda ){\varphi }_{-}=\bar{\varphi }$ in the mixture's bulk free-energy density. Varying ${\varphi }_{+}$ and ${\varphi }_{-}$ gives $f^{\prime }({\varphi }_{+})=f^{\prime }({\varphi }_{-})=\mu$; varying $\lambda$ gives $f({\varphi }_{+})-\mu {\varphi }_{+}=f({\varphi }_{-})-\mu {\varphi }_{-}$. These equate the chemical potentials and pressures, respectively.

Equivalently, their bulk chemical potentials are both zero and their pressures are equal:These are precisely the conditions of the common-tangent construction for phase coexistence: the horizontal supporting tangent touches both minima. Equal chemical potentials prevent net composition exchange, and equal pressure gives mechanical equilibrium at a flat interface.

Apply force balance to one hemisphere. The pressure force is $\pi R^2({\Pi }_{\mathrm{in}}-{\Pi }_{\mathrm{out}})$, while the interfacial tension around its equator gives $2\pi R\gamma$. The Young–Laplace equation therefore yieldsTo leading order in curvature, let the two bulk compositions be ${\varphi }_{+}={\varphi }_B+{\delta }_{+}$ and ${\varphi }_{-}=-{\varphi }_B+{\delta }_{-}$. Their curvatures agree, $\alpha =f^{\prime \prime }(\pm {\varphi }_B)=-2a$, so equal chemical potentials require $\alpha {\delta }_{+}=\alpha {\delta }_{-}$ and hence a common shift $\delta$.

Since $d\Pi /d\varphi =\varphi f^{\prime \prime }(\varphi )$, expansion about the flat coexistence compositions gives ${\Pi }_{+}-{\Pi }_{-}=2{\varphi }_B\alpha \delta +O({\delta }^2)$. Matching the capillary pressure proves the Gibbs--Thomson relation in these variables:The pressure jump balances interfacial forces; it does not permit unequal chemical potentials. A common shift in both compositions supplies a common interfacial chemical potential while their pressures differ. The expansion assumes $R$ large compared with the phi-four diffuse interface width.

The far-field composition $-{\varphi }_B$ has chemical potential zero. Local curved-interface coexistence imposes $\mu (R)={\mu }_s=\gamma /({\varphi }_{B}R)$. The decaying spherically symmetric solution of the exterior Laplace equation isWith constant mobility, its outward current isThe linearized exterior composition is $\varphi (r)=-{\varphi }_B+\mu (r)/\alpha$, consistent with the surface shift in the previous part.

For diffusion-controlled droplet growth, the interior is uniform to leading order and has no diffusive current. Replacing exterior material by interior material requires the composition jump $2{\varphi }_B$, so conservation gives $2{\varphi }_B\dot{R}=-J_r(R)$. HenceThe negative sign represents evaporation. The total outward current is $4\pi R^2{J}_r(R)=4\pi M\gamma /{\varphi }_B$, independent of $R$ in this leading quasi-static capillary approximation.

Integrate the diffusion-controlled droplet growth equation: $d(R^3)/dt=-3M\gamma /(2{\varphi }_B^2)$. ThusSetting the radius to zero givesThis is the lifetime predicted by the sharp-interface quasi-static model. The final stage at a radius comparable to the interface width falls outside that approximation, while the leading lifetime for a large initial droplet is dominated by the larger radii.

The flat interface is an equilibrium reservoir with $\mu =0$, while the droplet has the constant surface value ${\mu }_s=\gamma /({\varphi }_{B}R)$. The exterior chemical potential satisfies the Laplace equation, tends to zero far away, and obeys these two surface values. These are exactly the electrostatic boundary conditions at a conductor for a sphere held at potential ${\mu }_s$ above a grounded infinite plane.

Introduce an arbitrary permittivity $ϵ$. In the electrostatic problem, $Q=-ϵ{\int }_{S_R}{\partial }_n\mu dS=C(R,h){\mu }_s$. Therefore the diffusion-capacitance analogy gives the outward currentFor an isolated sphere, its capacitance is $C(R,\infty )=4\pi ϵR$, so $I(R,\infty )=4\pi MR{\mu }_s=4\pi M\gamma /{\varphi }_B$. Dividing givesThus sphere-plane capacitance determines the evaporation enhancement. The assumed spherical shape gives the corresponding law $\dot{R}=-M\gamma c(R/h)/(2{\varphi }_B^2{R}^2)$ for droplet evaporation near a planar reservoir. The local surface current is generally nonuniform, even though its total is described by a single capacitance.

For a fixed center height $h$, the radius decreases according to the preceding current enhancement. ThusWithin the stipulated logarithmic approximation, $c(y)=1+y/2+O(y^2)$, so $1/c(y)=1-y/2+O(y^2)$. Expanding uniformly over $0\le R\le R_0\ll h$ and integrating givesTherefore the leading correction to the isolated-droplet lifetime isThe negative correction reflects faster evaporation into the nearby reservoir. This leading coefficient also agrees with the small-$R/h$ expansion of the exact sphere-plane capacitance; the stipulated logarithmic approximation need not match higher orders.