Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 (Theoretical Physics of Soft Condensed Matter)

Impose the fixed conserved order parameterwith a Lagrange multiplier ${\mu }_0$. Stationarity ofunder every local variation givesThus the chemical potential is spatially constant at constrained equilibrium.

For constant $\kappa$ and a planar profile,Far inside either bulk phase, derivatives vanish and $\varphi =\pm {\varphi }_B$, where ${\varphi }_B^2=-a/b$. ThereforeSince part a makes $\mu$ constant, it vanishes throughout the interface.

The Euler–Lagrange equation is $\kappa {\varphi }^{\prime \prime }=a\varphi +b{\varphi }^3$. Multiplication by ${\varphi }^{\prime }$ and integration gives the first integralWith $\varphi ={\varphi }_{B}g$, $u=x/{\xi }_0$, and ${\xi }_0^2=-2\kappa /a$, this becomesThe monotone heteroclinic solutions areTranslation invariance leaves $x_0$ undetermined in an infinite system. A fixed global composition, boundary condition, or pinning field fixes this midpoint position.

Forthe functional derivative isThe constrained minimum again makes $\mu$ constant. Evaluating it in either uniform bulk phase gives $\mu =f^{\prime }(\pm {\varphi }_B)=0$, so $\mu =0$ everywhere.

For a planar equilibrium, multiply $\mu =0$ by ${\varphi }^{\prime }$ and use the chain rule. The result isMatching to either bulk phase fixes the constant:Consequentlyand the inverse profile is

The interfacial tension is the excess free energy per unit interfacial area:The first integral from part e equates the first two terms inside the brackets to $\kappa (\varphi )({\varphi }^{\prime }{)}^2/2$. Hence

For the quartic free energy,With constant $\kappa$, part e gives, between the two bulk values,Integration giveswhose inversion is exactly the hyperbolic-tangent profile in part c.

For a uniform field the gradient term vanishes and minimization over $p=|\mathbf{p}|$ givesWhen $a<0$, the nonzero minimum haswhile its direction is arbitrary by rotational symmetry. A sudden quench creates many independently oriented domains. Their mismatches contain gradients and topological defects, whose motion and pair annihilation are slow, so the uniform state is not reached promptly.

Outside a defect core writeAround any closed curve on which $\mathbf{p}\ne 0$, its topological charge is the winding numberSingle-valued polar order requires $q\in \mathbb{Z}$. A nonzero value prevents the field inside the loop from being continuously deformed to a uniform nonvanishing field, so the core must contain a zero or singularity.

In polar coordinates $(r,\phi )$, a radial asterand a circulating vortexboth have $q=+1$. The continuous family $\theta =\phi +\chi$, $0\le \chi \le \pi /2$, deforms one into the other without making $\mathbf{p}$ vanish away from the core, proving their topological equivalence. The hyperbolic configuration $\theta =-\phi$ has $q=-1$.

The field $\mathbf{p}=-p(r)\hat{\mathbf{r}}$ has angle $\theta =\phi +\pi$. Its defect is at $r=0$ and has $q=+1$. Sinceone hasThe local free-energy density is therefore

Outside the core, put $p=p_0$. The angular-gradient contribution isThe omitted core contribution is finite and independent of $L$.

A defect of charge $q$ has far-field elastic energy proportional to $q^2\log (L/r_0)$. Splitting $q$ into allowed charges $q_i$ with $\sum q_i=q$ lowers $\sum q_i^2$ unless the pieces already have the smallest permitted magnitude. Polar order distinguishes $\mathbf{p}$ from $-\mathbf{p}$, so its smallest nonzero charge is $|q|=1$. A nematic liquid crystal identifies $\hat{\mathbf{n}}\sim -\hat{\mathbf{n}}$; a rotation by only $\pi$ closes a loop in its order-parameter space, permitting $q=\pm 1/2$.

Let $\chi$ be the angle between $\hat{\mathbf{p}}$ and $\hat{\mathbf{n}}$. ThenFor $\zeta ,\lambda >0$, the coupling $-\zeta \lambda p^2\cos 2\chi /2$ is minimized by $\cos 2\chi =1$, soThe uniform terms depending on $p$ reduce toTheir nonzero minimum is

An isolated nematic half-defect reverses the representative director after one circuit. Alignment would then reverse the polar vector, which is not the same physical state. Half-integer defects can therefore occur only if accompanied by a branch wall on which polar order rotates sharply or vanishes. The wall has a tension proportional to its length, so half-defects are confined in pairs in the uniformly polar phase; they can deconfine where polar order disappears.

Under the Euclidean isometry $\tilde{\mathbf{r}}=\mathbf{a}+R\mathbf{r}$, where $R^{T}R=I$,ThereforeSince the integration parameter is unchanged,Both the induced metric and arc length are invariant.

For an oriented planar curve, the Frenet-Serret formulas areThe frame is orthonormal, so differentiating each scalar product givesIn matrix form this says that the connection matrix plus its transpose is zero; it must be antisymmetric.

The signed curvature isAn orientation-preserving isometry sends $\mathbf{t},\mathbf{n}$ to $R\mathbf{t},R\mathbf{n}$ and leaves $s$ unchanged, soAn orientation-reversing isometry reverses the chosen normal and hence the sign convention for signed curvature, while the geometric curvature $|k|$ remains invariant.

Because ${\partial }_u=\sqrt{g}{\partial }_s$ and ${\mathbf{r}}_t=U\mathbf{n}+W\mathbf{t}$, commuting ${\partial }_t$ and ${\partial }_u$ givesPreservation of orthonormality then givesThe tangential derivative of the velocity issoFinally commute the $s$ and $t$ derivatives in ${\partial }_s\mathbf{t}=k\mathbf{n}$, accounting for the evolving metric throughThe normal component gives

Since ${\partial }_t\sqrt{g}=\sqrt{g}({\partial }_{s}W-kU)$,Thus

Local force balance and moment balance on a slender rod areIn a planar rod, $\mathbf{M}$ and $\mathbf{m}$ point perpendicular to the plane.

Substituteinto force balance. With $\mathbf{\mu }={\mathbf{\gamma }}^{-1}$,Projection onto the local frame gives

ResolveThe Frenet-Serret formulas giveUsingtherefore yields

For $\alpha >0$, variation of constants gives the stationary Ornstein-Uhlenbeck processFor $t\ge t^{\prime }$ its covariance isSymmetry in $t,t^{\prime }$ therefore gives

The Fluctuation-dissipation theorem for mobility $\tilde{M}$ requiresComparison with the stated noise amplitude givesThis choice makes the stationary Fokker-Planck equation have the Boltzmann distribution proportional to $e^{-V/(k_{B}T)}$.

Let $\tau =|t-t^{\prime }|$. Integrating the equation for $x$ and using independence of the two noises givesWith the supplied integral,For $\alpha \tau \ll 1$, the active contribution is ballistic, $f_0^2{\tau }^2+O({\tau }^3)$, in addition to the Brownian term. For $\alpha \tau \gg 1$,so the long-time motion is diffusive with an enhanced diffusion coefficient.

Conditioned on $f$, the additive-noise trajectory has the Onsager–Machlup functionalUnder time reversal, $\dot{x}$ changes sign while $x$ and the active force $f$ are even. Subtracting the forward and backward conditional actions givesThe corresponding ratio for an Ornstein–Uhlenbeck path conditioned on its initial endpoint isConsequentlywhereIf stationary endpoint densities are included in the path measures, their ratio cancels the Ornstein–Uhlenbeck boundary term; the displayed convention is the endpoint-conditioned path probability used in the calculation.

The first part of $\Delta U$ is the change of the quadratic energy associated with the active Ornstein–Uhlenbeck force. The second is minus the change in the particle's potential energy, measured in thermal-noise units. The time integral is the work done by active propulsion, again divided by its noise scale; the full log ratio is the trajectory's time-reversal asymmetry or entropy production.

In a stationary confining state the two endpoint terms remain $O(1)$ as $T\to \infty$ and have zero mean, while the mean active work and the mean log ratio grow proportionally to $T$. Endpoint-term distributions approach time-independent distributions with positive and negative fluctuations. Under mixing assumptions, the time-integrated work has a large-deviation distribution: its central part becomes approximately Gaussian with mean and variance proportional to $T$, while its far tails scale exponentially in $T$. The log-ratio distribution obeys the corresponding fluctuation theorem symmetry.