core28 min studyLesson 320 in Physics

Thermodynamic convexity and stability

Connect entropy maxima and energy curvature to measured response signs while respecting ensemble constraints, coupled variations, and phase separation.

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What you’ll understand

  • Distinguish a stationary state from a stable constrained equilibrium.
  • Test coupled second variations rather than diagonal signs alone.
  • Relate heat-capacity and compressibility signs to the correct potential curvature.

Equal temperatures and balanced pressures make the first-order variation of a suitable equilibrium quantity vanish. Stability asks a stronger question: does every allowed small redistribution move the system away from its preferred extremum? Curvature answers that question locally. The correct curvature depends on the constraints and on which variables are allowed to fluctuate. We will connect entropy concavity to positive response functions, explain why a free energy can be concave in temperature while still obeying a minimum principle, and test a coupled two-coordinate example.

The extremum must resist allowed variations

An isolated additive macroscopic system has maximum entropy at stable equilibrium for fixed total energy, volume, and amount. Imagine redistributing a little energy between two equal subsystems while preserving the total. The first entropy variation vanishes when their temperatures agree. The second must be nonpositive, so a further spontaneous redistribution cannot increase the already maximal entropy. This gives concavity of entropy in its extensive coordinates on the ordinary stable equilibrium domain.

The equivalent energy representation is convex in entropy, volume, and amount. Extensivity creates a zero-curvature scaling direction when all extensive coordinates scale together, so the general Hessian condition is semidefinite rather than automatically strictly positive. Restricting amount and considering independent fluctuations in a smooth stable single phase can give strict positivity in the remaining relevant directions. Constraints can remove some candidate variations, and nonadditive systems such as self-gravitating matter lie outside the simple bulk argument used here.

(∂2S∂U2)V,ν=−1T2CV≤0,(∂2U∂S2)V,ν=TCV≥0\left(\frac{\partial^2S}{\partial U^2}\right)_{V,\nu}=-\frac{1}{T^2C_V}\le0,\qquad \left(\frac{\partial^2U}{\partial S^2}\right)_{V,\nu}=\frac{T}{C_V}\ge0
These relations assume positive absolute temperature and a smooth ordinary additive equilibrium phase with finite responses. They show why positive fixed-volume heat capacity supports both entropy concavity and energy convexity.↔ Scroll wide equations horizontally.

A minimum principle does not make every curvature positive

At fixed bath temperature, volume, and amount, Helmholtz free energy is minimized over allowed internal configurations. Temperature itself is imposed by the bath, so this does not require F to have a minimum as a function of its temperature argument. In fact FTT=−CV/TF_{TT}=-C_V/T is negative for a stable positive-capacity phase. By contrast, its volume curvature at fixed temperature is −(∂p/∂V)T=1/(VκT)-(\partial p/\partial V)_T=1/(V\kappa_T) and is positive for positive compressibility.

Thus F has different curvature signs in its different natural variables. The Gibbs potential is similarly concave in its intensive temperature and pressure arguments on stable branches. These signs reflect a Legendre transform, not an instability. Diagnose stability using the appropriate constrained variations and responses instead of asserting that “free energy must be convex in everything.” When a homogeneous equation of state predicts pressure rising with volume at fixed temperature, however, the resulting negative compressibility does identify a mechanical instability in this ordinary setting.

(∂2F∂T2)V,ν=−CVT,(∂2F∂V2)T,ν=1VκT,K=(abbc)⪰0 ⟺ a≥0, c≥0, ac−b2≥0\left(\frac{\partial^2F}{\partial T^2}\right)_{V,\nu}=-\frac{C_V}{T},\qquad \left(\frac{\partial^2F}{\partial V^2}\right)_{T,\nu}=\frac1{V\kappa_T},\qquad K=\begin{pmatrix}a&b\\b&c\end{pmatrix}\succeq0\ \Longleftrightarrow\ a\ge0,\ c\ge0,\ ac-b^2\ge0
The final equivalence concerns a real symmetric two-coordinate Hessian K for dimensionally compatible normalized coordinates. Strict positive definiteness instead requires a positive leading diagonal and positive determinant.↔ Scroll wide equations horizontally.

Coupling and global alternatives matter

For two internal coordinates, positive individual diagonal curvatures are necessary but insufficient. A sufficiently large cross coupling can create a downhill combined direction even when varying either coordinate alone raises the potential. The Hessian eigenvalues or the two-by-two determinant reveal this possibility. A negative eigenvalue means local instability. A zero eigenvalue requires examination of higher orders or a symmetry or coexistence constraint; it is not by itself proof of either a strict minimum or an instability.

Local stability also differs from global equilibrium. A metastable free-energy minimum can persist even though a distant configuration has lower free energy, because a barrier separates them. In bulk phase separation, the stable free energy forms a convex envelope rather than retaining an unstable homogeneous segment. A spinodal marks loss of local stability, while coexistence is set by equality of phase potentials. These distinctions connect curvature measurements to observable responses without pretending that curvature alone determines nucleation barriers or transition rates.

Positive diagonal curvatures can hide a saddle

At fixed temperature and volume, a model’s internal normalized coordinates x and y have local free-energy change ΔF=(100 J)(2x2+6xy+2y2)/2\Delta F=(100\,\mathrm J)(2x^2+6xy+2y^2)/2. Test the origin’s stability and evaluate the change for x=0.10,y=−0.10x=0.10,y=-0.10.

  1. The dimensionless Hessian factor is K=(2332)K=\left(\begin{smallmatrix}2&3\\3&2\end{smallmatrix}\right). Each diagonal is positive, so varying only x or only y increases the quadratic free energy.

  2. Its determinant is 4−9=−54-9=-5, and its eigenvalues are 5 and −1. The negative eigenvalue corresponds to an opposite-sign coordinate change, so the origin is a saddle rather than a local minimum.

  3. Substitution gives 2(0.01)+6(−0.01)+2(0.01)=−0.022(0.01)+6(-0.01)+2(0.01)=-0.02. Multiplying by 100/2100/2 yields ΔF=−1 J\Delta F=-1\,\mathrm J, an explicit allowed downhill fluctuation.

The result

The origin is locally unstable despite positive diagonal curvatures. The coupled variation lowers free energy by 1 J in the stated quadratic model.

Explore the model

Interactive lab

Interactive lab · function

Stationary points need curvature information

Reset, set Coefficient a = 1 and Position x = 0, and predict zero slope with upward curvature. For a fresh comparison Reset, set a = −1 and x = 0, and predict the same zero slope with downward curvature. Move x to 0.5 within each selected curve to inspect the opposite tangent signs.

Stationary points need curvature informationThis 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 forThe instrument tests only a one-coordinate parabola. It cannot display coupled Hessian eigenvalues, phase separation, ensemble constraints, or a material’s compressibility; those require the lesson’s separate calculations.

Check your understanding

Why is a negative F temperature-temperature derivative compatible with stable equilibrium at fixed bath temperature?

Try it yourself

A smooth single-phase sample near volume 2.0 L has pressure fall from 200 kPa to 195 kPa when volume increases by 0.10 L at fixed temperature. Estimate its isothermal compressibility. Separately, find FTTF_{TT} at 300 K if CV=12 J K−1C_V=12\,\mathrm{J\,K^{-1}}.

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

Give me a hint

Convert litres to cubic metres, use minus the reciprocal volume times the local volume-pressure slope, and retain the thermodynamic curvature sign.

Compare with the solution

The estimated pressure-volume slope is −5000/0.00010=−5.0×107 Pa m−3-5000/0.00010=-5.0\times10^7\,\mathrm{Pa\,m^{-3}}. Thus κT=−1/[V(∂p/∂V)T]=1/[0.002(5.0×107)]=10−5 Pa−1\kappa_T=-1/[V(\partial p/\partial V)_T]=1/[0.002(5.0\times10^7)]=10^{-5}\,\mathrm{Pa^{-1}}, a positive mechanically stable response locally. The temperature curvature is FTT=−12/300=−0.040 J K−2F_{TT}=-12/300=-0.040\,\mathrm{J\,K^{-2}}. Its negative sign is consistent with positive heat capacity and does not contradict the constrained free-energy minimum principle.

Take it with you

  • Stability concerns second variations under specified conservation and reservoir constraints.
  • Positive diagonal responses do not exclude a coupled unstable direction.
  • Local stability, metastability, coexistence, and spinodal loss describe different features of a free-energy landscape.

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