Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 344 (Theoretical Physics of Soft Condensed Matter)

Write $E(t)=\delta A/\delta x(t)$ for the functional derivative of the action. Under time reversal in classical mechanics, $\dot{x}$ changes sign whereas $E$ is even by assumption. Thus the forward and reversed Gaussian white noise histories associated with the same geometric path areThe Onsager--Machlup path probability therefore givesHere the path probabilities are densities conditional on their respective initial states; they are not separately normalized bridges conditioned on both endpoints. The noise-to-path Jacobian determinant must be treated with a consistent discretization: it cancels when it is invariant under time reversal in classical mechanics. This is the usual additive-noise Langevin dynamics convention, including a constant-mass Underdamped Langevin dynamics or the time-reversal invariance of a path Jacobian in midpoint overdamped Langevin dynamics. A completely arbitrary velocity-dependent Lagrangian would require specifying that measure rather than deducing its cancellation solely from the parity of $E$.

For an autonomous Lagrangian, the Euler-Lagrange equation usesApply the chain rule to the Hamiltonian along an arbitrary smooth path, without assuming the unforced Euler-Lagrange equation:Consequently the energy balance for an autonomous Lagrangian isWith explicit time dependence, an additional $-{\partial }_{t}L$ enters $dH/dt$. For ideal Gaussian white noise, the identity is understood through smooth-noise regularization or the Stratonovich chain rule. The original PDF correctly differentiates $L$ with respect to $\dot{x}$ in $H$; the supplied TeX's derivative with respect to $x$, its endpoint $t^2$, and its ordinary derivative of $A$ are transcription errors.

At thermal equilibrium, microscopic reversibility equates the probabilities of a path and its reversed path when both include their Boltzmann distribution initial weights. Denote their endpoint states by $z_1,z_2$, including velocity if needed. Since the Hamiltonian is even under time reversal in classical mechanics,This detailed balance condition and the energy balance for an autonomous Lagrangian yieldThis is the fluctuation-dissipation relation for a Langevin particle: the strength of Gaussian white noise is fixed by the damping and temperature, with $k_B$ the Boltzmann constant.

For an equilibrium coarse-grained variable, the unresolved microscopic states contribute entropy; their statistical weight is encoded in the Helmholtz free energy, rather than in a single microscopic energy. Relative to the same reference measure, ${\rho }_{\mathrm{eq}}(x)\propto e^{-\beta F(x)}$, so microscopic reversibility becomesThis extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.

The Neumann boundary conditions give Fourier modes $q=n\pi /L$, including the spatially uniform mode $q=0$. In the frame translating with the mean deposition height, that mode has no restoring force and obeysHere $W_t$ is Brownian motion, independent of the initial height. More generally $⟨h_0(t{)}^2⟩=⟨h_0(0{)}^2⟩+{\sigma }^{2}t$ when these second moments exist. Removing the mean deposition drift does not remove the Gaussian white noise.

The Brownian zero mode of a fluctuating interface has no stationary distribution on the real height axis for ${\sigma }^2>0$. This conclusion does not depend on assuming finite variance: a stationary characteristic function would satisfy $\hat{\mu }(k)=\hat{\mu }(k)e^{-{\sigma }^2{k}^{2}t/2}$, hence vanish for every $k\ne 0$, contradicting its continuity at zero and $\hat{\mu }(0)=1$. Any stationary joint distribution of the whole height would have a stationary zero-mode marginal, which is impossible. Thus the full unpinned height has no Boltzmann equilibrium. Pinning the mean height, or retaining only the nonzero Fourier modes, removes this obstruction. The PDF has ${\dot{h}}_q$; the TeX's ${\hslash }_q$ is a transcription error.

For $q\ne 0$ and $\alpha >0$, each Fourier mode is an Ornstein-Uhlenbeck process with decay rate $r_q=\alpha q^2$. Its explicit Ornstein-Uhlenbeck solution isso the Itô isometry givesTo identify the overdamped Langevin dynamics, choose a friction ${\zeta }_q>0$, a harmonic oscillator potential $V_q={\zeta }_q\alpha q^2{h}_q^2/2$, and force noise $f_q={\zeta }_q{\eta }_q$. The fluctuation-dissipation relation for a Langevin particle then defines $k_B{T}_{\mathrm{eff}}={\zeta }_q{\sigma }^2/2$. Its Boltzmann distribution isThis is the stationary spectrum of a linear fluctuating interface, also following from the equipartition theorem. The numerical prefactor uses exactly the mode-noise normalization given in the paper. The nonzero modes can therefore have a stationary Gaussian distribution even though the unpinned zero mode cannot.

Use the paper's velocity gradient convention $A_{ij}={\partial }_i{v}_j$, so $\Omega =(A-A^T)/2$ and $D=(A+A^T)/2$. In a rigid motion there is rotation but no rate-of-strain tensor: $D=0$. A polar order parameter attached to the material must rotate exactly with it. In this convention its material derivative is $-\Omega \mathbf{p}$, and its corotational derivative of a polar vector must vanish.

Thus the compensating term is $+\Omega \mathbf{p}$ with coefficient exactly one. This follows from an objective time derivative and is independent of material properties. By contrast, the response to $D\mathbf{p}$ describes deformation and flow alignment of a polar order parameter; its coefficient $\xi$ can depend on the material. The transposed velocity gradient convention is essential to the sign of $\Omega$.

Let $h_i=\delta F/\delta p_i$ be the polar molecular field, with the positive functional derivative convention used in the paper. During a pure advective displacement $u_j$, the polar order parameter changes byThe free energy change is $\delta F=\int h_i\delta p_i{d}^{d}r$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and incompressibility ${\partial }_j{u}_j=0$. Integration by parts in the advective term givesRelabelling the dummy indices in the rotational and flow alignment of a polar order parameter terms givesTherefore $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j{d}^{d}r$, with the reversible stress of a polar liquid crystalFor a local free-energy density $f(\mathbf{p},\nabla \mathbf{p})$, an explicit choice isIts divergence is exactly $-p_k{\partial }_j{h}_k$, by the Euler-Lagrange equation for $h_k$. A pressure term can be reassigned in an incompressible flow; the chosen representative realizes the printed divergence without that ambiguity. The force density ${\partial }_i{\Sigma }_{ij}^p$ has mechanical power $-\int {\Sigma }_{ij}^p{\partial }_i{v}_j$, the negative of the advective free energy rate, which checks the stress sign.

The polar molecular field isFor a spatially uniform polar order parameter, $\mathbf{h}=(a+b{p}^2)\mathbf{p}$, so it is colinear with $\mathbf{p}$ and its magnitude is $|\mathbf{h}|=|a+b{p}^2|p$. The scalar factor can be negative; colinearity does not necessarily mean parallel orientation.

For the simple shear flow, the paper's velocity gradient convention givesSubstitute these matrices into $\dot{\mathbf{p}}=-\Omega \mathbf{p}+\xi D\mathbf{p}-\Gamma \mathbf{h}$. Projection along and perpendicular to the polar order parameter yieldsAssume $g\ne 0$, $\Gamma >0$, $b>0$ and $p>0$. The flow-alignment angle in planar shear must satisfyFor $\xi =-1$, use the cosine equation: $\theta =\pi /2$ modulo $\pi$, and the displayed tangent is infinite. The radial equation giveswhere the sign distinguishes the angular branches, and only positive right-hand sides are ordered solutions. Linearizing the angular equation gives $\delta \dot{\theta }=-g\xi \sin 2\theta \delta \theta$. For $|\xi |>1$, the stable angular branch therefore hasIts radial relaxation eigenvalue is $-2\Gamma b{p}^2<0$. At $|\xi |=1$ the angular linearization is marginal and $p^2=-a/b$ requires $a<0$. If $g=0$, the shear restriction disappears and any orientation is allowed for $p^2=-a/b>0$. The printed request for dependence only on $a,b,\Gamma ,g$ omits $\xi$: in general the magnitude necessarily depends on it.

At $\xi =0$, the polar order parameter obeysFor $a<0$, $b>0$, the magnitude relaxes to $\sqrt{-a/b}$ while the direction rotates continuously. Thus the likely ordered behaviour is tumbling rather than a fixed alignment.

More generally, for $|\xi |<1$ and $g\ne 0$, $\xi \cos 2\theta -1$ never vanishes, so the tumbling of a polar order parameter persists. The polar order parameter completes a full $2\pi$ rotation inobtained by integrating $dt=2|d\theta |/(|g|(1-\xi \cos 2\theta ))$ over a full rotation. An unoriented nematic director repeats after half this period, whereas a polar order parameter distinguishes opposite directions. For $\xi \ne 0$ its magnitude can also oscillate. If the polar order parameter decays to zero, its orientation is no longer a physical observable; the tumbling conclusion concerns a persistent ordered state.

Take $b>0$ for a stable quartic Landau free energy. The cubic-term removal in a conserved quartic Landau free energy uses $\varphi =\psi -c/(3b)$ and gives, apart from a constant,For a closed binary fluid mixture, the compositional order parameter has fixed integral. Adding $d\int \varphi$ therefore adds the same constant to every allowed state. Equivalently, in the common-tangent construction for phase coexistence, a linear term changes the tangent slope, hence the chemical potential, by $d$, but leaves the coexistence compositions unchanged. This is why linear composition bias leaves coexistence compositions unchanged; it would not mean an unchanged composition at a fixed externally imposed chemical potential.

In the shifted variable, the binodal compositions are ${\psi }_{\pm }=\pm \sqrt{-a^{\prime }/b}$ for $a^{\prime }<0$. They merge atThe continuous phase transition therefore has the same scalar quartic Landau theory as the symmetric binary fluid mixture, with a shifted critical composition and control parameter. In the mean-field approximation, the composition difference vanishes with order-parameter critical exponent $1/2$. This argument presumes the critical composition lies in the physically accessible composition range; conservation affects the dynamics, not this static phase coexistence construction.

The nematic order parameter is a symmetric second-rank tensor and a traceless second-rank tensor describing orientational anisotropy. For molecular unit axes $\mathbf{s}$, $Q_{ij}$ is proportional to $⟨s_i{s}_j-{\delta }_{ij}/3⟩$. It vanishes in an isotropic phase and is unchanged by head-tail reversal $\mathbf{s}\mapsto -\mathbf{s}$, unlike a polar order parameter. For uniaxial nematic order, the nematic director is thus an unoriented axis.

One convention for the Landau-de Gennes free energy through fourth order, respecting rotational symmetry, isFor a traceless second-rank tensor in three dimensions, the Cayley-Hamilton theorem gives $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$, so there is only one independent quartic rotational invariant of a symmetric traceless tensor. The only possible linear scalar is $\mathrm{Tr}Q=0$; without an external anisotropy no linear term survives.

The cubic rotational invariant of a symmetric traceless tensor distinguishes prolate and oblate forms of uniaxial nematic order. Head-tail reversal leaves $Q$ unchanged and does not impose $Q\mapsto -Q$. Thus a cubic term is allowed in three dimensions and generically produces a first-order phase transition, rather than a symmetry-enforced continuous onset. In two dimensions a symmetric traceless second-rank tensor has eigenvalues $s,-s$, hence $\mathrm{Tr}{Q}^3=0$: the cubic invariant vanishes identically in two dimensions. This is a statement about the Landau free energy; fluctuations in two dimensions require a separate treatment.

For the stated uniaxial nematic order normalization, the nematic director is an eigenvector of $Q$ with eigenvalue $\lambda$, and the two perpendicular eigenvalues are $-\lambda /2$. ThereforeUsing precisely the Landau-de Gennes free energy convention in the preceding solution givesIf the quartic invariant were instead normalized as $b\mathrm{Tr}{Q}^4/4$, its coefficient would be $\bar{b}=9b/32$; the physical predictions are unchanged after redefining $b$.

For a nonzero global minimizer of the free energy, compare opposite values of the scalar nematic order parameter:The lower one has $\bar{c}\lambda <0$, soThe claim concerns the stable ordered phase, not every metastable stationary point. For $\bar{c}=0$ the two signs are degenerate. For $\bar{c}\ne 0$, equality of the ordered and isotropic free energies, together with stationarity, gives ${\lambda }_t=-\bar{c}/(2\bar{b})$ and ${\bar{a}}_t={\bar{c}}^2/(4\bar{b})>0$. The finite jump ${\lambda }_t$ exhibits the first-order phase transition driven by the cubic invariant.

A scalar coupling linear in the nematic order parameter requires a second-rank anisotropy. A single electric field supplies no such tensor, while $E_i{E}_j$ does. Its isotropic part gives $E^2\mathrm{Tr}Q=0$. Hence the lowest joint order in $E$ and $Q$ of the electric-field coupling to nematic order isTerms depending only on $E$ do not affect minimization over $Q$. With $\vartheta$ the angle between the nematic director and the electric field,For $\chi \lambda >0$, the lowest free energy occurs at ${\cos }^2\vartheta =1$, so the nematic director aligns along either $+\mathbf{E}$ or $-\mathbf{E}$. Then $E_i{Q}_{ij}{E}_j=\lambda E^2$, givingUnlike the linear composition term in a conserved binary fluid mixture, this term changes the equilibrium nematic order parameter: its spatial integral is not fixed.

Within this analytic quartic Landau theory, equilibrium of a nonconserved scalar order parameter requires $f^{\prime }=0$. At a continuous phase transition, its restoring curvature vanishes, so $f^{\prime \prime }=0$. A nonzero $f^{\prime \prime \prime }$ would leave a cubic leading term in the Taylor expansion about that state and prevent it from being a local minimum. Thus a stable critical endpoint of a quartic Landau free energy requires all three derivatives to vanish, with $f^{\prime \prime \prime \prime }>0$. The condition $f^{\prime \prime }=0$ alone can instead identify a spinodal point.

For $f=\bar{b}{\lambda }^4+\bar{c}{\lambda }^3+\bar{a}{\lambda }^2+\bar{d}\lambda$, $\bar{b}>0$, the successive equations areThey giveAt these values, the free energy is exactly $f({\lambda }_c)+\bar{b}(\lambda -{\lambda }_c{)}^4$, proving stability. With no electric field, $\bar{d}=0$, so necessarily $\bar{c}=0$, followed by ${\lambda }_c={\bar{a}}_c=0$.

For $\bar{c}<0$ and $\chi >0$, the electric-field coupling to nematic order supplies the required negative ${\bar{d}}_c$ atTuning the quadratic coefficient to ${\bar{a}}_c$ simultaneously reaches the field-induced critical endpoint of the isotropic-nematic transition. The PDF's $\tilde{c}$ in the last clause is undefined; it is interpreted here as the previously defined $\bar{c}$. The endpoint is a prediction of the stable quartic, small-field mean-field approximation; higher-order terms or a field beyond the expansion's validity can shift it.