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Two examples are sodium dodecyl sulfate, with a negatively charged sulfate head and a hydrocarbon tail, and cetrimonium bromide, with a positively charged ammonium head and a hydrocarbon tail. Both are amphiphiles: one region is hydrophilic and another is hydrophobic.

In water, the hydrophobic effect favors association of the tails into a core, reducing their exposed area while the hydrated heads remain outside. This can lower the total Gibbs free energy despite loss of translational entropy. Head-group repulsion, interfacial cost and chain-packing constraints oppose unlimited growth, so finite micelles are favored under suitable conditions above the critical micelle concentration.

The appropriate geometric measure is the amphiphile packing parameter:For a spherical micelle with $n$ tails and core radius $r_c$, the volume and area relations are $nv=4\pi r_c^3/3$ and $n{a}_0=4\pi r_c^2$. Dividing gives $r_c=3v/a_0$, so a sphere fits without overextending a tail if $P\le 1/3$. Here $P<1/3$ andThus the packing model predicts spherical micelles. This is a preferred aggregate geometry, not a claim that concentration, temperature or ionic conditions can never alter the morphology.

The Flory–Huggins solution theory treats a polymer solution as an incompressible lattice mixture, with one solvent molecule occupying one site and one chain occupying $N$ connected sites. Thus the supplied $f$ is naturally the mixing free energy per lattice site; if a site has volume $v_0$, the free-energy density is $f/v_0$.

The term $k_{B}T(\varphi /N)\ln \varphi$ is the polymer translational entropy of mixing contribution. There are only $1/N$ as many independently translating chains as monomers, which explains the factor $1/N$. The term $k_{B}T(1-\varphi )\ln (1-\varphi )$ is the solvent entropy of mixing contribution. Both lower the mixing free energy for $0<\varphi <1$.

The term $k_{B}T\chi \varphi (1-\varphi )$ is the mean-field excess interaction contribution from replacing like contacts with polymer–solvent contacts. In the simplest energetic lattice model,where $z$ counts neighboring contacts and the ${ϵ}_{ij}$ are pair contact energies. Positive $\chi$ penalizes unlike contacts and can favor demixing. Because chain connectivity suppresses the polymer's translational mixing entropy, relatively modest unfavorable interactions can cause separation of long chains. More general effective $\chi$ can also include entropic effects.

At the thermodynamic critical point, the two spinodal branches merge: $f^{\prime \prime }=0$ and $f^{(3)}=0$. The third derivative isFor $0<\varphi <1$, its zero gives $1-{\varphi }_c=\sqrt{N}{\varphi }_c$. Substitution in the spinodal relation yieldsAn absolute critical temperature requires an interaction–temperature relation. Under the usual energetic approximation $\chi =C/T$, define the theta temperature $\Theta =2C$, at which an infinite chain has $\chi =1/2$. ThenThus $T_c$ tends to a nonzero constant, $T_c\sim N^0$, while $\Theta -T_c\sim N^{-1/2}$. It is the temperature deficit, not $T_c$ itself, that scales as $N^{-1/2}$.

More generally, if a smooth $\chi (T)$ obeys $\chi (\Theta )=1/2$ and ${\chi }^{\prime }(\Theta )\ne 0$, thenThis is the critical point of a Flory–Huggins polymer solution. The same leading chain-length exponent follows for a smooth temperature law, while its coefficient depends on the omitted solvent thermodynamics. The lattice interpretation connects the chain-length dependence of mixing entropy to the theta condition $\chi =1/2$.^[ref].

Here $T$ is the absolute temperature and $k_B$ the Boltzmann constant. An adsorption blob contains $g\sim (\xi /b{)}^2$ monomers, so there are $N/g\sim N(b/\xi {)}^2$ such blobs. Confining a blob removes an amount of conformational entropy of order $k_B$; thus the positive confinement contribution to the Helmholtz free energy isThe statement about one blob's entropy is used as an entropy loss relative to the free chain, rather than as an absolute entropy.

In the assumed slab model, contacts occur within a monomer-scale distance $b$ of the substrate. The contact fraction is approximately $b/\xi$, giving $N_c\simeq Nb/\xi$. Multiplying this by the negative contact energy yieldsThe two contributions express the competition between loss of conformational entropy and gain of adsorption energy. The numerical coefficients are the ones stipulated by this blob approximation.

The stretched segment is a Gaussian chain of $n$ Kuhn length steps, with unperturbed root-mean-square end-to-end distance $l=b\sqrt{n}$. Its end-to-end probability density is proportional to $\exp [-3z^2/(2n{b}^2)]$, so its leading elastic Helmholtz free energy is $3k_{B}T{z}^2/(2n{b}^2)$. Adding the remaining segment's adsorption free energy givesAn additive constant independent of $n,z$ is irrelevant. At the optimal adsorption height this becomesThe original PDF's hint labels the positive quantity proportional to $k_{B}T{R}^2/l^2$ an entropy. It has units of energy and is the stretching free energy. The corresponding change of conformational entropy is $\Delta S=-3k_B{R}^2/(2l^2)$. This dimensional and sign correction is essential to the minimization.

This is the leading extensive Gaussian chain approximation: endpoint normalization and junction corrections give subleading terms. It requires many stretched monomers and $z\ll nb$, rather than extension near the full contour length.

For dissolved, nonadsorbing polyethylene glycol, the usual new interaction is a depletion force. Polymer centres are excluded from a layer of thickness of order their radius of gyration $R_g$ around each particle. When the particles approach, their exclusion layers overlap. This increases the volume accessible to polymer and its translational entropy, lowering the colloid pair free energy. Equivalently the surrounding polymer osmotic pressure pushes the particles together.

In the dilute-polymer Asakura-Oosawa model, let $h=R-2a$ be the surface gap and $R_e=a+R_g$. Thenfor $2a\le R<2R_e$, and $U_{\mathrm{dep}}=0$ for $R\ge 2R_e$. Here ${\Pi }_p\simeq n_p^{(\mathrm{polymer})}{k}_{B}T$ is the reservoir polymer osmotic pressure. The overlap-volume formula follows by adding two equal spherical cap volumes with height $R_e-R/2$. Thus the added attraction has range about $2R_g$ and grows with polymer concentration. The total interaction is approximately hard exclusion plus screened electrostatic repulsion plus Van der Waals force and depletion attractions.

Increasing polymer concentration therefore creates a family of phase diagrams, rather than a unique shift determined by polymer size alone. Where the attraction overcomes repulsion, colloid-poor fluid can coexist with a colloid-rich phase, and fluid–crystal coexistence generally broadens. For sufficiently long-ranged attraction, stable fluid–fluid separation can occur. For the stipulated small polymer, the attraction is short-ranged: fluid–fluid separation can be metastable to crystallization, and a rapid attraction increase can instead produce an arrested colloidal gel. Residual long-ranged repulsion can oppose bulk separation. Simulations and free-volume calculations confirm that longer-ranged electrostatic repulsion can counteract depletion-driven separation.^[ref]

The nonadsorption assumption matters. If the polyethylene glycol adsorbs to the particles, a dense attached layer can give steric stabilization, whereas chains attaching to two particles can produce polymer bridging. The paper does not specify surface affinity or polymer concentration, so depletion is the standard conditional answer, not a prediction of an inevitable particular final phase.

Let the channel length be ${\ell }_c\gg R$ and assume uniform solution conductivity. Its Ohmic resistance scales as ${\ell }_c/(\sigma R^2)$, whereas access resistance of a reservoir mouth scales as $1/(\sigma R)$. The fraction of voltage in the entrance regions is therefore of order $R/{\ell }_c$. Most of the potential drop lies inside the long channel; the small entrance field region is neglected in the model, not asserted identically zero.

Write $x$ for inserted contour length and choose $qE>0$ to pull the chain inward. The charge already inside feels force $qEx$, givingConfinement removes polymer configurations. In the stiff-chain regime, the confinement deflection length obeys $R^2\sim {\lambda }^3/{\ell }_p$, hence $\lambda \sim (R^2{\ell }_p{)}^{1/3}$. A free-energy cost of order $k_{B}T$ per deflection gives a roughly constant resisting entropic force $f\sim k_{B}T/\lambda$. The insertion barrier of a charged polymer follows from $F(x)\simeq fx-qE{x}^2/2$, with an entry maximum at $x_{\ast }=f/(qE)$ when that point lies within the insertion interval. Its height is $f^2/(2qE)$. The field supplies work against this confinement entropy.

Put ${\zeta }_{\ell }=2\pi \eta /\ln (R/a)$, so $\gamma (x)={\zeta }_{\ell }x$. After capture, neglect the entry barrier, exterior-chain drag and thermal corrections to strong directed motion. ThenThe confined-polymer entry and recoil times have different length dependence; this entry graph is linear through the origin. Charge and drag both grow with inserted length, leaving a constant velocity. The same cancellation explains length-independent mobility in an ideal free-draining electrophoresis model. It does not include a capture waiting time or claim that every free-solution chain has identical hydrodynamic drag.

A measured superlinear entry time can arise when exterior-coil drag, tension propagation, wall adsorption or nonuniform friction adds length dependence absent from this model. For example $\gamma (x)\propto x^p$ with the same $qEx$ drive gives $\tau \propto L_0^p$ for $p>0$. Under the assumptions of uniform wall charge and polymer entry scaling, a homogeneous thin-double-layer electro-osmotic model adds a length-independent fluid velocity $u_{\mathrm{eo}}=-ϵ{\zeta }_{w}E/\eta$. It changes the drift coefficient, not its length exponent. This does not rule out charge heterogeneity, adsorption or near-cancellation followed by diffusion in a real channel; those are additional mechanisms.

With the field off, the confinement entropy drives outward motion. Assuming the constant force $f$ and the same local friction,The small final end region lies outside the long-confined-segment approximation but gives only a small correction. Thus ideal insertion scales as $L_0$, while recoil scales as $L_0^2$.

TODO: user-deferred DNA-specific numerical stopping-voltage estimate and its biological parameter-based inference. No biological quantities are evaluated.

For entropic-trap length separation, a reservoir/channel array can repeat length-dependent capture, confinement and relaxation delays, turning them into different mean residence times. The uniform inside-channel drift speed derived above is itself length independent, so that drift alone cannot separate lengths. Separation must retain length dependence of the barrier or of the capture/recoil dynamics and can be tuned with a forward-biased time-dependent field. For large bulk throughput, pressure-driven advection is preferable to diffusion alone; a pressure gradient can be combined with electric capture control. In the unobstructed continuum limit $Q=\pi R^4\Delta P/(8\eta {\ell }_c)$. Excessive driving erases trapping differences, so throughput and resolution compete; the question supplies no geometry or adsorption law selecting a unique optimal drive.

Write $\epsilon ={\epsilon }_0{\epsilon }_r$ and take $n_0$ to be the reservoir density of each monovalent ion species. Equal electrochemical potentials in equilibrium give the Boltzmann ion densities in a symmetric electrolyte,Combining this with the electrostatic Poisson equation gives the planar Poisson-Boltzmann equation,The Debye length is ${\kappa }^{-1}$. If the symbol $n_0$ instead denotes the combined density of both species, the factor two is absent; the per-species convention above makes the normalization explicit. At $|e\psi |\ll k_{B}T$, the linearized planar screening solution obeys ${\psi }^{\prime \prime }={\kappa }^2\psi$. With $\psi (0)=\zeta <0$ and $\psi (\infty )=0$,

The positive ions are counterions and the negative ions are co-ions. At weak potential,They approach $n_0$ from above and below respectively, with equal first-order excess and deficit. For stronger negative potential, interpret the source's strong-potential comparison as a statement about $|e\psi |$, since $\psi$ itself is negative. Counterions become strongly concentrated near the wall, while co-ions are exponentially depleted and remain nonnegative. Linearization misses this asymmetry and would eventually predict unphysical negative co-ion density. Point-ion mean-field theory itself eventually needs finite-size and correlation corrections; it provides no steric cap at arbitrarily large counterion concentration.

For steady tangential electro-osmosis, with no imposed pressure gradient and negligible inertia, the Stokes equation is $\eta v^{\prime \prime }+{\rho }_{e}E=0$. Since ${\rho }_e=-\epsilon {\psi }^{\prime \prime }$,The conditions $v(0)=0$ and $v^{\prime }(\infty )=0$ giveThe printed constant speed is the bulk velocity outside the double layer, not the velocity at the no-slip wall. With negative $\zeta$ the positive counterion excess drives fluid along the applied field. The far-field value is the Helmholtz-Smoluchowski electro-osmotic velocity.^[ref]

For the finite channel, first model the added wall as an ideally uncharged insulating boundary with no normal field on its outer side: ${\psi }^{\prime }(d)=0$, while $\psi (0)=\zeta$. The one-charged-wall electro-osmotic channel then hasIntegrating the same force balance and enforcing $v(0)=v(d)=0$ givesFor $U>0$, the maximum occurs atIn that limit $v/U\simeq 1-x/d-e^{-\kappa x}$: the velocity rises rapidly from zero at the charged wall, reaches a maximum a few screening lengths away, then falls approximately linearly to zero at the neutral wall. One charged wall does not give a symmetric plug across a channel with no-slip at the other wall.

An alternative electrode condition $\psi (d)=0$ means a wall held at reservoir potential; it is not the same as zero surface charge. Under that convention,with $\cosh [\kappa (d-x_{\max })]=\sinh (\kappa d)/(\kappa d)$. Both boundary conventions have the same leading thin-double-layer profile above. An externally imposed pressure gradient would add a pressure-driven profile; none is specified here.

For dilute nonadsorbing uncharged polymers in equilibrium or sufficiently weak flow, the entropic polymer depletion near nonadsorbing walls suppresses chain configurations near each wall. A depletion layer of order the coil radius of gyration is expected at both walls, with a bulk-like central region when $d$ is much larger than the coil size. Strong confinement rounds this into a central enrichment. Neutrality removes a direct electrophoretic force, not excluded-volume entropy. Adsorption, concentration and strong-flow migration were not specified, so this is a conditional equilibrium sketch rather than a universal flowing-polymer profile.

Fluctuation-induced forces arise because separation changes the fluctuations available to a system and therefore changes its free energy. The force is $F=-\partial \mathcal{F}/\partial d$ at fixed relevant control variables, not necessarily a force from a permanent charge or dipole.

For two neutral polarizable particles, correlated electromagnetic fluctuations give a short-distance dispersion attraction of the form $-C_6/r^6$. Retardation changes the asymptotic distance dependence. Boundaries produce a related Casimir force; for parallel ideal conducting plates at zero temperature,The negative pressure denotes attraction. Material response, nonplanar geometry and finite temperature alter the ideal result. The effect can be described through the separation-dependent electromagnetic mode spectrum rather than by treating a divergent absolute vacuum energy as observable.^[ref]

In a classical depletion force, smaller particles or polymer coils are excluded from a region near larger surfaces. Overlap of excluded regions increases their available volume. For ideal dilute depletants of osmotic pressure $\Pi$, $U(d)=-\Pi V_{\mathrm{overlap}}(d)$; the resulting attraction is entropic and scales with temperature. Confined interface fluctuations can instead generate repulsion when close approach removes entropy. Near a continuous phase transition, a large correlation length makes boundary-conditioned order-parameter fluctuations produce a critical Casimir force. Its sign depends on the boundary preferences, so fluctuation forces are not universally attractive.

The useful common idea is a free-energy difference with explicit boundary conditions; quantum vacuum, thermal field and excluded-volume mechanisms have different physical fluctuations and different ranges.

Work in the rest frame of the planar shock, with nonzero mass flux $j=\rho u$. Conservation of mass gives $[j]=0$; the tangential and normal components of balance of linear momentum give $[jv]=0$ and $[p+ju]=0$. Thus $[v]=0$. For an ideal gas, specific internal energy is $e=c_{v}T=p/[(\gamma -1)\rho ]$, and specific enthalpy is $e+p/\rho =\gamma p/[(\gamma -1)\rho ]$. Total energy flux is $j[(u^2+v^2)/2+e+p/\rho ]$. Its continuity, with $v$ already continuous, givesPut $c^2=\gamma p/\rho$ and $M=u/c$. Since $p=ju/(\gamma M^2)$, the normal component of balance of linear momentum says $u[1+1/(\gamma M^2)]$ is constant. HenceThe energy equation says $u^2[1/2+1/((\gamma -1)M^2)]$ is constant, givingSet $a=M_{-}^2$, $b=M_{+}^2$, square the first relation and equate to the second. After clearing positive denominators the equation factors asThe first factor is the unchanged state. For a genuine shock the second factor vanishes, yieldingThe physical selection principle is entropy increase in the direction of flow. For $u_{\pm }>0$, the upstream side is $x<0$ and the downstream side is $x>0$. A compressive entropy-admissible shock has $\boxed{M_{-}>1,M_{+}<1}$; interpreting the printed “ahead” and “behind” spatial labels must respect these explicit side assignments. The expansion-shock branch would violate the second law.

Substitution gives the velocity ratioSince $v$ is unchanged while the normal component decreases, $|v|/u$ increases. Thus an oblique flow bends towards the shock plane, equivalently away from its normal. If $v=0$ there is no angular bending; the flow stays normal and simply slows. The tangential continuity, not a change in the tangential speed, causes the deflection.

The first energy term is the rate of change of internal plus kinetic energy per unit volume; the second transports that energy with the moving gas; the pressure term is the mechanical work flux across the boundary. There is no heat flux or viscous work in the stated inviscid adiabatic model.

Use mass conservation to replace a conservative derivative of $\rho e$ by $\rho De/Dt$. Dotting the momentum equation with $\mathbf{u}$ gives $\rho D(|\mathbf{u}{|}^2/2)/Dt=-\mathbf{u}\cdot \nabla p$. Subtraction from total energy leavesNow $p=\rho RT$ and $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$ implyThus $\boxed{DS/Dt=0}$: smooth adiabatic inviscid trajectories are isentropic. Shocks need an entropy inequality rather than this smooth equality.

Adding $\nabla \cdot (k\nabla T)$ models Fourier heat conduction, with heat flux $-k\nabla T$. The same subtraction gives $\rho c_{v}DT/Dt+p\nabla \cdot \mathbf{u}=\nabla \cdot (k\nabla T)$, henceFor a fixed container use impermeability $\mathbf{u}\cdot \mathbf{n}=0$, insulation $k\nabla T\cdot \mathbf{n}=0$, and mass conservation. ThenThe boundary term vanishes and $\nabla (1/T)=-\nabla T/T^2$ supplies the positive sign. For $T>0$, $k\ge 0$, total entropy cannot decrease: this is entropy production by Fourier heat conduction.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.

For $U<0$ the same positive algebraic root remains real and satisfies $0<V<c_0$, because $V^2-c_0^2=(\gamma +1)UV/2<0$. The jump relations then give a density decrease, ${\rho }_1/{\rho }_0=V/(V-U)<1$, and a pressure decrease. Where both states remain thermodynamically positive, this is an expansion shock: its upstream shock-frame flow is subsonic and its entropy change has the forbidden sign. At sufficiently large negative $U$ the formula even predicts nonpositive pressure, an additional reason it cannot represent a physical gas state.

The observed outgoing wave is instead a continuous isentropic rarefaction wave. Its plus-family rays spread from the initial piston corner rather than coalescing. The fan head enters the undisturbed gas at $x=c_{0}t$; for moderate recession the tail is $x=(U+c_1)t$, with the uniform gas next to the piston moving at $U$ and $c_1=c_0+(\gamma -1)U/2>0$. The appropriate Riemann invariant is $u-2c/(\gamma -1)=-2c_0/(\gamma -1)$, so inside the fan, with $\xi =x/t$,These expressions show the spreading characteristics in the right panel of the preceding figure. If $U\le -2c_0/(\gamma -1)$, the gas cannot follow the piston with positive sound speed; a vacuum gap forms, bounded by the limiting rarefaction wave edge. Thus the fan description also identifies the strong-recession limit.

Conservation of the parcel's entropy makes $p/{\rho }^{\gamma }$ constant along its trajectory. Its initial value is $p_0/{\rho }_0^{\gamma }$, soThis is a relation following the specified parcel; parcels with different initial entropy can have different constants.

Continuity gives $D\log \rho /Dt=-\nabla \cdot \mathbf{u}$. Using $p=\rho RT$, the temperature equation gives $D\log T/Dt=-(R/c_v)\nabla \cdot \mathbf{u}$. ThereforeConsequentlyEach smooth parcel evolves isentropically under these adiabatic inviscid equations; the argument does not cover entropy-producing shocks.

Assume $b>0$ and temperature-independent $d$ for the local model. The zero-field stationary points obeyFor $d\ne 0$, the favoured ordered sign is opposite to $d$. The nonzero extrema first appear at $a=d^2/(4b)$; their existence is not yet a coexistence condition. To find the equilibrium transition, require both $F^{\prime }(m_{\ast })=0$ and $F(m_{\ast })=F(0)$. Substituting $a=-d{m}_{\ast }-b{m}_{\ast }^2$ into the free energy givesHenceAt coexistence the entire potential factorizes:It has equal minima at zero and $m_{\ast }$ separated by a barrier. Every fixed nonzero cubic coefficient produces a first-order transition in this quartic model, with a finite magnetization jump before the quadratic coefficient reaches zero.

For $d^2/(4b)>a>a_{\ast }$, the ordered minimum is metastable and zero is globally stable. For $a_{\ast }>a>0$, the ordered state is globally stable and zero is metastable. At $a=0$ the zero minimum loses local stability. Thus the spinodals are $a=d^2/(4b)$ and $a=0$, distinct from $a_{\ast }$. For $a<0$, the opposite-sign nonzero state can also be a local minimum, but the cubic term favours the state of sign $-d$.

If $a=a_0(T-T_c)/T_c$, the transition is shifted to $T_{\ast }=T_c[1+2d^2/(9b{a}_0)]$. With $b,d$ constant and $a^{\prime }(T)>0$, the entropy jump from the disordered to ordered phase is $-a^{\prime }(T_{\ast })m_{\ast }^2/2$, giving a latent heat of magnitude $T_{\ast }{a}^{\prime }(T_{\ast })m_{\ast }^2/2$. As $d\to 0$, the jump and shift vanish and the original continuous transition is recovered.

A genuine zero-field uniaxial ferromagnet with exact spin-reversal Z2 symmetry forbids this cubic term. The modified model therefore presumes that symmetry has been changed; it must not be presented as a symmetry-allowed correction for the original ferromagnet.

For the isotropic zero-field classical Heisenberg model of a ferromagnet, the analytic vector Landau expansion uses the Cartesian components of the vector order parameter near $\mathbf{m}=0$. rotational symmetry makes its scalar terms functions of $\mathbf{m}\cdot \mathbf{m}$. Global spin inversion symmetry also requires $f(\mathbf{m})=f(-\mathbf{m})$. Through fourth order this givesA positive quartic coefficient stabilizes the unconstrained truncated free energy. The disordered phase has $b>0$ and its minimum at $m=0$; the aligned phase has $b<0$ and a minimum at nonzero $m$. At the continuous transition $b=0$. The attractive alignment interaction supplies an energetic tendency toward negative quadratic curvature; orientational entropy favors the disordered phase at high temperature, so the sign of $b$ is a free-energy question rather than a consequence of the pair energy alone.

The analyticity assumption matters: the variable printed as $m$ is a nonnegative magnitude, so reversing $\mathbf{m}$ does not send that scalar to $-m$. Terms proportional to $|\mathbf{m}|$ or $|\mathbf{m}{|}^3$ would be rotationally invariant but nonanalytic at the vector origin. Standard analytic Landau theory excludes them. Without that assumption, the stated pair symmetry alone would not exclude odd powers of the magnitude.^[ref]

The original Von Neumann entropy depends on its eigenvalues, with $0\log 0$ defined by continuity:After rank-one dephasing, the eigenvalues are the retained diagonal populations $1/2,1/4,1/4$, givingdephasing removes coherences in the specified basis, increasing the system's mixedness while preserving those measurement probabilities. It cannot be a closed-system unitary transformation, since the eigenvalues change. The subsystem entropy increase does not by itself assert an increase in the entropy of a closed system plus its environment.

For low temperature, factor out the largest Boltzmann factor from the dominant eigenvalue:Thus the Helmholtz free energy per site has the expansionDifferentiate at fixed $J$, remembering that $K=J/(k_{B}T)$. The entropy per site isThe nonzero residual entropy comes from an exponentially large ground-state degeneracy, generated by geometrical frustration. A triangle with antiferromagnetic energy $J[st+(s+t)\tau ]$ has minimum energy $-J$: if $s=t$, the apex must have the opposite sign; if $s=-t$, either apex sign is permitted. The two all-equal configurations instead have energy $3J$. Thus exactly one of a minimum-energy triangle's bonds is frustrated. The bridge can simultaneously minimize its energy at $-J$ by requiring $t=-s^{\prime }$, where $s^{\prime }$ is the next cell's first base spin.

For fixed $s$ and $s^{\prime }$, the number of allowed apex choices is therefore two when $s=s^{\prime }$ and one when $s=-s^{\prime }$. The zero-temperature counting transfer matrix on successive first base spins isThe eigenvalues of $G$ are $3$ and $1$. Dividing $k_B\ln g_M$ by the $3M$ sites and taking the thermodynamic limit gives $k_B\ln 3/3$, independently confirming the derivative calculation. This does not assert a unique ordered ground state: the extensive choice of frustrated bonds is precisely what leaves entropy at zero temperature. At high temperature, all Ising spins become independent and $s\to k_B\ln 2$ per site.

Choose a coarse-grained magnetization $m(\mathbf{x})$ as the order parameter: it vanishes in the paramagnetism phase and develops a spontaneous nonzero value in the ferromagnetism phase. In zero applied magnetic field, spin reversal sends $m\mapsto -m$, so the local Landau free energy contains only even powers. A minimal spatially inhomogeneous free-energy functional isThe positive gradient coefficient penalizes rapid spatial variations. Near an ordinary transition take $a(T)=a_0(T-T_0)$, $a_0>0$, and positive $b$; retain the sextic term when $b$ can approach or cross zero. The source $h$ is conjugate to the magnetization and explicitly breaks spin reversal.

Determine equilibrium by minimizing this free-energy functional, imposing the appropriate boundary conditions. Its Euler-Lagrange equation isFor a uniform system at $h=0$, compare the values of all stable stationary solutions, rather than accepting every solution of this equation. The selected global minimum gives the equilibrium magnetization; derivatives of its free energy give entropy, heat capacity and magnetic susceptibility. At an ordinary transition the two ordered minima emerge continuously from zero. Spatial dependence also permits domain walls and an estimate of the correlation length.

This is a mean-field approximation: it selects a minimizing profile and neglects statistical fluctuations around it. The symmetry constrains the expansion, but its numerical coefficients are phenomenological. Close enough to a transition, thermal fluctuations near a critical point can invalidate the resulting mean-field critical exponents.^[ref]