Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 344 (Theoretical Physics of Soft Condensed Matter)

For a binary fluid mixture, the compositional order parameter $\varphi (\mathbf{r})$ may be taken as the local concentration difference between the two species. In a closed systemis fixed by the total amount of each species. This is the sense in which $\varphi$ is conserved; locally it changes through a current and obeys a continuity equation.

For a symmetric mixture, the Landau-Ginzburg theory free energy isA term $h{\int }_V\varphi ,d\mathbf{r}$ is constant on the allowed configurations and therefore changes neither equilibrium probabilities nor the dynamics. Equivalently, it merely shifts the chemical potential by a spatial constant, whose gradient vanishes.

The coupling $\frac{c}{2}\int \varphi p^2$ is linear in $\varphi$, but its coefficient $p^2(\mathbf{r})$ is generally position dependent and dynamical. It therefore cannot be reduced to a constant times the conserved integral of $\varphi$; it changes both the local chemical potential of the mixture and the tendency of the polar order parameter to order.

Interchanging the labels of the two species sends $\varphi \mapsto -\varphi$ while leaving the even terms in $\varphi$ unchanged. It sends $c\mapsto -c$, so one may choose $c>0$ without loss of generality.

The isotropic disordered system is invariant under spatial inversion, which sends the polar order parameter $p\mapsto -p$ while leaving the scalar composition $\varphi$ unchanged. A term $\varphi p$ is odd under this symmetry and is forbidden; in more than one dimension it also fails to be a rotational scalar. Finally, if no higher even powers are retained, $b<0$ or $B<0$ makes the free energy unbounded below as the corresponding field grows. Thermodynamic stability therefore requires $b,B>0$.

At fixed uniform $\varphi$, the nonconserved variable $p$ can relax freely, so mean-field approximation minimizeswith respect to $p$. The stationary solutions obeyThus $p=0$ when $A+c\varphi \ge 0$, while $p^2=-(A+c\varphi )/B$ when $A+c\varphi <0$. Substitution givesWriting this in the requested form yieldsbecause $A+c\varphi =c(\varphi -{\varphi }_o)$. The value ${\varphi }_o$ is the composition at which the coefficient of $p^2$ changes sign, so it is the mean-field threshold for polar ordering.

With ${\varphi }_o=0$,For $C<a$, both branches are locally convex at the origin. Atthe negative-side curvature vanishes; for $C>a$ the interval near $0^{-}$ has $f^{\prime \prime }=a-C+3b{\varphi }^2<0$, so a homogeneous composition there is unstable and the equilibrium free energy is its convex envelope. The sketch therefore has an ordinary upward quartic on the positive side and, beyond $C_c$, a negative-curvature shoulder and a minimum on the negative side.

The two binodal compositions ${\varphi }_1<0<{\varphi }_2$ are the contact points of the common-tangent construction. Equivalently, they solveThe first equality is equality of chemical potential; the second is equality of pressure. Heretogether with the intercept equation determines the two densities. As $C\downarrow a$, both contact points approach zero continuously with scale $|{\varphi }_{1,2}|\propto \sqrt{C-a}$, so this mean-field onset of phase separation is continuous.

When $b={\kappa }_1=0$, stationarity in $\varphi$ givesSubstituting this value, or completing the square in the Gaussian integral over $\varphi$, givesWhen $A>0$ and the quartic term is negligible, each Fourier mode is Gaussian with the Ornstein--Zernike correlation functionup to the chosen Fourier normalization. Its correlation length is ${\xi }_p=\sqrt{\kappa /A}$.

The eliminated field obeys $\varphi =-c{p}^2/(2a)$, so its connected correlation function isFor the zero-mean Gaussian field of part (i), Wick theorem givesIt is therefore nonnegative and has the square of the Ornstein--Zernike spatial form. In particular, if $C_p(r)$ has exponential factor $e^{-r/{\xi }_p}$, then $C_{\varphi }(r)$ has $e^{-2r/{\xi }_p}$ and correlation length ${\xi }_p/2$, with the algebraic prefactor also squared.

TakeIts variational chemical potential isConservation gives $\dot{\varphi }=-\nabla \cdot \mathbf{J}$. Assuming an isotropic constant mobility $M>0$, local linear irreversible thermodynamics gives $\mathbf{J}=-M\nabla \mu$. Hence the noiseless conserved order-parameter dynamics isIt decreases the free energy because $\dot{F}=-\int M|\nabla \mu {|}^{2}d\mathbf{r}\le 0$ under closed or no-flux boundary conditions.

For $a<0$ and $b>0$, planar coexistence occurs at $\varphi =\pm {\varphi }_B$ with ${\varphi }_B=\sqrt{-a/b}$. Put $\alpha =f^{\prime \prime }({\varphi }_B)=a+3b{\varphi }_B^2=-2a>0$. A common small shift $\delta$ gives equal chemical potentials to first order,The thermodynamic pressure is $P=\varphi \mu -f$. Its difference between the positive interior and negative exterior is thereforeFor a sphere the Young–Laplace equation gives $P_{+}-P_{-}=2\sigma /R$. Consequently the Gibbs--Thomson relation is

The curved interface requires exterior composition $-{\varphi }_B+\delta$, but the far field supplies only $-{\varphi }_B$. Material therefore diffuses away from the droplet, which evaporates. Write $\varphi =-{\varphi }_B+g$ outside. Linearization gives $\mu =\alpha g$, and the quasistatic condition $\dot{\varphi }=M{\nabla }^2\mu \simeq 0$ givesSpherical symmetry yields $g(r)=\delta R/r$. Taking the normal outward from the droplet,The interface converts positive-phase material into negative-phase material. Integrating the continuity equation through the moving interface gives the Stefan conditionwhere $[\varphi ]=2{\varphi }_B$, and hence

For $h=\eta \sin (qx)$, the linearized mean curvature isThe boundary value in the upper phase is thereforeThe decaying harmonic extension into $z>0$ isWith the normal directed into the upper half-space,

Using only this upper-phase flux in $\dot{h}=v_n=-J_n/(2{\varphi }_B)$ givesPositive surface tension raises the chemical potential at a crest, drives material away from it, and therefore smooths the interface, which fixes the negative sign of the decay rate.

The dynamics is nonlocal because the conserved composition must diffuse through the bulk. Solving Laplace's equation maps a boundary Fourier amplitude to its normal derivative by multiplication by $|q|$. Combining this Dirichlet-to-Neumann factor with the curvature factor $q^2$ produces the nonanalytic $|q{|}^3$ rate described by diffusion-limited relaxation of an interface.

Both coexisting phases transport the conserved order parameter. The calculation above included only the harmonic field and normal current in the upper half-space. The lower half-space contributes an equal flux into the interface for equal mobilities and symmetric phases. The total interface speed is therefore twice the one-sided result, so the correct coefficient is

The antisymmetric velocity-gradient tensor $\mathbf{\Omega }$ is the local angular velocity of a rigid-body rotation. Under such a rotation a material vector must rotate at exactly the same angular velocity, independently of its molecular constitution. The coefficient of $\mathbf{\Omega }\cdot \mathbf{p}$ is therefore fixed to one by rotational covariance, also called material objectivity. In contrast, the symmetric strain-rate tensor $\mathbf{D}$ deforms rather than merely rotates the material, so its flow-alignment coefficient $\xi$ is material dependent.

During a small incompressible displacement $u_i=v_i\Delta t$, pure advection changes the polar field byThe free-energy change is $\delta F=\int h_i\delta p_i,d\mathbf{r}$, where $h_i=\delta F/\delta p_i$. Integrating the translational term by parts and separating the antisymmetric and symmetric parts of ${\nabla }_i{u}_j$ identifies the order-parameter stress. Up to an arbitrary isotropic pressure, its three contributions obeyThe first term is the distortion or Ericksen force density, the second transfers antisymmetric rotational torque, and the third is the symmetric flow alignment of a polar order parameter contribution.

For a uniform field, $\mathbf{v}\cdot \nabla \mathbf{p}=0$, and the stated flow has $\mathbf{\Omega }=0$ andThe equation $D\mathbf{p}/Dt=-\Gamma \mathbf{h}$ becomesThus

The shifted molecular field is the gradient ofThus $A=\mathrm{diag}(a+\alpha ,a-\alpha /2,a-\alpha /2)$.

If $\alpha <0$, the $x$ mode softens first atThe extensional flow already selects the $x$ axis, and ordering chooses one of its two polar directions. The transition is continuous and spontaneously breaks the remaining inversion symmetry $p_x\mapsto -p_x$.

If $\alpha >0$, the degenerate transverse modes soften first atThis transition is also continuous. The flow preserves rotations about the $x$ axis, while the ordered vector chooses an azimuthal direction in the $yz$ plane and spontaneously breaks that $SO(2)$ symmetry.

For $\alpha <0$, choose one ordered state $\mathbf{p}=(p,0,0)$ with $p^2=(-a-\alpha )/b$. Stationarity gives $\tilde{\mathbf{h}}=0$, henceUniformity makes ${\nabla }_i{\Sigma }_{ij}^{(1)}=0$, so its spatially constant part can be absorbed into the pressure. Since $\mathbf{h}$ is parallel to $\mathbf{p}$, the antisymmetric term ${\Sigma }^{(2)}$ vanishes. The symmetric term isup to isotropic pressure. Using $\alpha =-\xi \gamma /\Gamma$, its only deviatoric component is ${\Sigma }_{xx}^p=({\xi }^2\gamma /\Gamma )p^2$.