advanced32 min studyLesson 332 in Physics

Landau free energy and first-order transitions

Use symmetry-constrained local free-energy expansions to distinguish equilibrium coexistence, metastable branches, and mean-field spinodals.

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What you’ll understand

  • State the analyticity and small-order-parameter assumptions in Landau theory.
  • Find coexistence by comparing free-energy values rather than curvatures alone.
  • Locate local spinodals and explain their relation to nucleation and hysteresis.

A local free-energy landscape organizes several different questions: which uniform states are stationary, which are locally stable, and which have the lowest free energy? A first-order transition is decided by the last question. A metastable state may persist after it ceases to be the equilibrium minimum, while a spinodal marks the disappearance of a local minimum within the chosen mean-field landscape. These are distinct locations, not interchangeable definitions of the transition temperature.

Symmetry restricts the expansion but does not choose coefficients

For a scalar order parameter with reversal symmetry and no conjugate field, the local expansion contains only even powers. A field adds a linear term that explicitly breaks that symmetry. Analyticity of this coarse-grained local function near small order parameter is an assumption; the final thermodynamic free energy can become nonanalytic after minimization and the infinite-system limit. One must not assume the exact critical free energy itself has an ordinary Taylor series everywhere.

A positive quartic coefficient gives the familiar continuous mean-field onset when the quadratic coefficient changes sign. A negative quartic term requires a positive higher-order stabilizer, for which the sextic term is the simplest choice. This truncated polynomial is useful only over amplitudes where omitted terms remain small or where it is explicitly adopted as an illustrative model. Symmetry alone does not guarantee a continuous transition, fix a critical temperature, or determine the numerical coefficients.

f(m)=r2m2+u4m4+v6m6−hm,f′(m)=rm+um3+vm5−hf(m)=\frac r2m^2+\frac u4m^4+\frac v6m^6-hm,\qquad f'(m)=rm+um^3+vm^5-h
m is dimensionless and f is a uniform free energy per cell measured in a fixed energy unit E0; r,u,v,h use that same unit. The sextic model assumes v>0, a reversal-symmetric zero-field system, and controlled neglect of higher powers or an explicitly illustrative polynomial.↔ Scroll wide equations horizontally.

Equal minima determine coexistence

At zero field, the disordered stationary point is m=0m=0. Nonzero stationary points obey a quadratic equation for q=m2q=m^2. For negative u, a pair of ordered minima can coexist locally with the central minimum while the quadratic coefficient is still positive. The first-order coexistence condition requires the nonzero stationary point to have the same free energy as the origin, not merely that the two states both exist.

Combining stationarity with equality of free energies gives a finite jump in squared order parameter and a positive coexistence value of r. In a sample with a conserved spatial average, the equilibrium state between coexisting values may separate into domains, adding interfacial costs at finite size. The resulting equilibrium free energy respects convexity in its proper constrained variable. The nonconvex local polynomial should therefore be read as a coarse-grained landscape rather than the fully convex thermodynamic answer for arbitrary constrained mixtures.

qcoex=−3u4v,rcoex=3u216v,rsp,ord=u24v,rsp,0=0q_{\mathrm{coex}}=-\frac{3u}{4v},\qquad r_{\mathrm{coex}}=\frac{3u^2}{16v},\qquad r_{\mathrm{sp,ord}}=\frac{u^2}{4v},\qquad r_{\mathrm{sp,0}}=0
These zero-field sextic-model results assume u<0 and v>0. q=m² is dimensionless; r values are in the same energy-per-cell unit as u and v. The two spinodals are local mean-field stability endpoints, not additional equilibrium coexistence points.↔ Scroll wide equations horizontally.

A barrier makes kinetics different from equilibrium

A local minimum with higher free energy is metastable. Reaching the equilibrium phase requires an appropriately sized fluctuation that creates a nucleus and overcomes an interfacial barrier. Its lifetime depends on temperature, sample volume, defects, and dynamics, none of which is determined by uniform minimization alone. Sweeping a control parameter too rapidly can therefore follow different metastable branches and produce hysteresis even though equilibrium comparison selects a unique coexistence condition.

A spinodal solves stationarity and vanishing curvature together: a minimum merges with an intervening maximum. The central minimum loses curvature at r=0, while the ordered extrema first appear at the discriminant threshold above coexistence. In short-range finite-dimensional systems, nucleation can end metastable persistence before this ideal endpoint, and a sharp thermodynamic spinodal need not exist. A smooth experimental fit with a negative quartic coefficient is consequently a model inference requiring checks, not a stand-alone proof of a phase boundary.

A sextic model with simple coexistence values

Take u=−4E0u=-4E_0, v=3E0v=3E_0, and zero field in the sextic free energy. Find the coexistence r, the magnitude of the order-parameter jump from the central state, and both local spinodal values. Compare their order along increasing r.

  1. The equal-depth condition gives qcoex=−3(−4)/(4×3)=1q_{\mathrm{coex}}=-3(-4)/(4\times3)=1. Therefore the ordered minima at coexistence have ∣m∣=1|m|=1, separated by a finite jump from the central value zero.

  2. Insert the same coefficients into rcoex=3u2/(16v)r_{\mathrm{coex}}=3u^2/(16v) to obtain rcoex=1E0r_{\mathrm{coex}}=1E_0. At this value f/E0=12m2(1−m2)2f/E_0=\tfrac12m^2(1-m^2)^2, making the equal values at zero and plus or minus one explicit.

  3. The central spinodal is zero, and the ordered spinodal is u2/(4v)=16/12=4/3u^2/(4v)=16/12=4/3 in units E0. Thus coexistence lies strictly between the two endpoints, and neither endpoint equals the equilibrium crossing.

The result

Coexistence occurs at r=E0 with |m| jumping from zero to one; local spinodals are r=0 and r=4E0/3.

Explore the model

Interactive lab

Interactive lab · function

Curvature is a local test

For each fresh run, Reset and set BOTH Coefficient a and Position x. At x=0 compare a=1 and a=−1, predicting a local minimum and maximum of y=ax² with zero tangent slope in both cases. This quadratic instrument cannot display the sextic landscape, two minima, nucleation, or coexistence.

Curvature is a local testThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forStationarity alone does not determine stability. The lab isolates a local curvature distinction and cannot locate a first-order transition.

Check your understanding

For the zero-field sextic Landau model, which criterion locates equilibrium coexistence between m=0 and a nonzero uniform minimum?

Try it yourself

Use the same model with u=−4E0u=-4E_0, v=3E0v=3E_0, but set r=5E0/4r=5E_0/4. Its nonzero stationary squared amplitudes are q=1/2q=1/2 and q=5/6q=5/6. Classify them and decide whether the ordered local minimum or the origin is the equilibrium uniform state.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Evaluate the curvature at a stationary point using f′′=2q(u+2vq)f''=2q(u+2vq), then compare its free energy with f(0)=0.

Compare with the solution

At q=1/2q=1/2, the curvature is 2(1/2)(−4+6/2)E0=−E02(1/2)(-4+6/2)E_0=-E_0, so these are maxima. At q=5/6q=5/6, it is 2(5/6)(−4+5)E0=5E0/32(5/6)(-4+5)E_0=5E_0/3, so these are local minima. At a nonzero stationary point the free energy simplifies to f/E0=q2−q3f/E_0=q^2-q^3, giving 25/216>025/216>0 at q=5/6. The origin has positive curvature r and free energy zero, so it is the equilibrium uniform state. The ordered local minima are metastable, consistent with r lying above coexistence but below the ordered spinodal.

Take it with you

  • An analytic local Landau expansion is an assumption about coarse-grained variables.
  • Coexistence compares minimum values; spinodals test loss of local stability.
  • Hysteresis and metastable lifetimes require kinetic information beyond the uniform polynomial.

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