Thermodynamic Stability and Convexity Conditions

Drop a marble into a bowl and it settles at the bottom; balance it on an upturned bowl and the faintest breath sends it rolling away. Equilibrium is not one thing — it is stable or unstable, and the difference is entirely the curvature of the surface the marble sits on. Thermodynamics has exactly the same story, but the "surface" is the entropy function $S(U,V,N)$ and the "marble" is a tiny fluctuation of energy or volume that the system is always spontaneously trying. The Second Law says an isolated system in equilibrium sits at a maximum of total entropy. A maximum is not just a place where the first derivative vanishes; it is a place where the surface curves downward in every direction. That single geometric fact — that $S$ is a concave function of its extensive variables — is the master stability condition of thermodynamics, and everything measurable follows from it. Concavity in energy forces the heat capacity to be positive: a stable system that absorbs heat must get warmer, never colder. Concavity in volume forces the isothermal compressibility to be positive: squeeze a stable system and its pressure must rise. A material that violated either would spontaneously fly apart into hot and cold regions, or collapse and rarefy — which is precisely what happens inside the unstable loop of the van der Waals gas, and precisely why a liquid and its vapour split into two phases. The free energies inherit the same discipline with the signs flipped by the Legendre transform: $F(T,V)$ is convex in $V$ and concave in $T$ , $G(T,p)$ is convex in $p$ and concave in $T$ . This article turns "equilibrium is a maximum" into hard inequalities — $C_V>0$ , $\kappa_T>0$ , $C_p\ge C_V$ — and shows they are the same statement as "fluctuations don't run away."

Key takeaways

Equilibrium of an isolated system is a maximum of entropy, which means $S(U,V,N)$ must be a concave function of its extensive arguments — curving downward in every direction.
Concavity in energy gives $(\partial^2 S/\partial U^2)_V = -1/(T^2 C_V) < 0$ , hence the heat capacity is positive, $C_V>0$ : absorb heat, get warmer.
Concavity in volume gives the compressibility positive, $\kappa_T>0$ : squeeze it, the pressure rises. A region with $(\partial p/\partial V)_T>0$ is mechanically unstable and phase-separates.
Under Legendre transforms the free energies flip: $F(T,V)$ is convex in $V$ , concave in $T$ ; $G(T,p)$ is convex in $p$ , concave in $T$ .
Stability is the same statement as "fluctuations are bounded": $\langle\Delta E^2\rangle = k_BT^2C_V \ge 0$ requires $C_V\ge 0$ .
Key idea: thermodynamic stability is the demand that the entropy Hessian be negative-definite; by Sylvester's criterion this single matrix inequality delivers $C_V>0$ and $\kappa_T>0$ together, and its failure is the onset of phase separation.

The core insight

Thermodynamic stability is the demand that equilibrium be a genuine maximum of entropy — that the Hessian matrix of $S(U,V,N)$ be negative-definite. By Sylvester's criterion this single condition delivers positive heat capacity $C_V>0$ and positive compressibility $\kappa_T>0$ simultaneously; where it fails, the homogeneous system is unstable and splits into coexisting phases.

Where this lives today

Stability and convexity are the working tools of every phase-diagram calculation. When materials scientists compute whether an alloy will remain a single solid solution or unmix into two crystalline phases, they are testing the convexity of the free energy versus composition — the "common tangent" construction is Gibbs's stability analysis applied to $G(x)$ . The spinodal line, where the determinant of the stability Hessian first vanishes, separates metastable regions (where nucleation is needed to phase-separate) from unstable ones (where separation is spontaneous, spinodal decomposition), and this distinction controls the microstructure of everything from steel to polymer blends to the fat crystals in chocolate. The same mathematics reaches into soft matter and biology. Liquid–liquid phase separation inside cells — the formation of membraneless "condensates" of proteins and RNA — is now understood as a Gibbs instability of a multicomponent free energy, with the same spinodal/binodal geometry. And the divergence of $\kappa_T$ at a critical point, which stability theory flags as the collapse of a curvature, is exactly the physics of critical opalescence and the starting point of the renormalization-group theory of critical phenomena. Convexity is where equilibrium thermodynamics hands the baton to the physics of phase transitions.

The physical picture

Imagine the entropy of an isolated system plotted as a hill over the plane of its extensive coordinates — energy $U$ across, volume $V$ into the page, height equal to $S$ . The Second Law puts the equilibrium state at the top of this hill. Now let one half of the system spontaneously hand a little energy $\delta U$ to the other half — a fluctuation the microscopic dynamics is trying constantly. If the entropy hill curves downward, the two displaced sub-systems together have less entropy than the uniform state, so the fluctuation is entropically disfavoured and dies away. If the hill curved upward in some direction, the fluctuation would be rewarded, grow, and the uniform system would tear itself apart. Stability is downward curvature; instability is upward curvature.

Figure

The stability landscape: a concave entropy surface (stable maximum) versus a convex one (unstable) — fluctuations decay only on the concave hill

The two curvatures have direct physical names. Curving downward in $U$ says the temperature $T$ (whose inverse is the slope $\partial S/\partial U$ ) must decrease as energy is drained — equivalently, that adding energy raises the temperature, so the heat capacity $C_V = (\partial U/\partial T)_V$ is positive. Curving downward in $V$ says the pressure rises under compression, so the compressibility $\kappa_T$ is positive. These are not separate assumptions; they are two eigen-directions of the same downward-curving surface. The punchline is what happens when a material's equation of state violates stability. The van der Waals gas, plotted below, has a middle branch on each sub-critical isotherm where $(\partial p/\partial V)_T > 0$ — squeeze it and the pressure drops. No homogeneous fluid can live there. The system escapes the forbidden region by splitting into liquid and vapour, and the flat coexistence line (Maxwell's equal-area construction) is drawn across the instability. Stability theory does not just forbid the loop; it predicts the phase transition.

Figure

Van der Waals isotherm below T_c: the shaded loop has (∂p/∂V)_T > 0 and is mechanically unstable — the system phase-separates across the flat coexistence line

The core equations

Stability begins with the requirement that equilibrium extremize entropy and that the extremum be a maximum. The first-order condition (equal temperatures, pressures, chemical potentials across sub-systems) is equilibrium; the second-order condition is stability. For a virtual transfer of energy $\delta U$ between two halves of a system, the total entropy change to second order must be non-positive:

\delta^2 S = \frac{1}{2}\left(\frac{\partial^2 S}{\partial U^2}\right)_V (\delta U)^2 + \cdots \le 0 \qquad (1)

where: $\delta^2 S$ — second-order entropy change under a fluctuation; the inequality is the mathematical content of "equilibrium is a maximum." It must hold for every independent fluctuation direction, which is the requirement that the whole Hessian of $S$ be negative-definite. The energy curvature is directly the heat capacity. Using $1/T = (\partial S/\partial U)_V$ and differentiating again:

\left(\frac{\partial^2 S}{\partial U^2}\right)_V = \frac{\partial}{\partial U}\frac{1}{T} = -\frac{1}{T^2}\left(\frac{\partial T}{\partial U}\right)_V = -\frac{1}{T^2 C_V} \qquad (2)

where: $C_V = (\partial U/\partial T)_V$ — heat capacity at constant volume. Since the left side must be negative (concavity) and $T^2>0$ , this forces $C_V>0$ . A negative heat capacity would mean the entropy curves the wrong way — an unstable state. The volume curvature gives the compressibility. The relevant stability condition, at fixed temperature, is convexity of the Helmholtz free energy in volume:

\left(\frac{\partial^2 F}{\partial V^2}\right)_T = -\left(\frac{\partial p}{\partial V}\right)_T = \frac{1}{V\kappa_T} > 0 \qquad (3)

where: $\kappa_T = -\frac{1}{V}(\partial V/\partial p)_T$ — isothermal compressibility. Convexity of $F$ in $V$ (equivalently $(\partial p/\partial V)_T<0$ ) forces $\kappa_T>0$ : a stable material resists compression. Collecting the curvatures into a matrix makes the joint condition precise. The Hessian of $S(U,V)$ (at fixed $N$ ) must be negative-definite:

\mathbf{H}_S = \begin{pmatrix} S_{UU} & S_{UV} \\ S_{UV} & S_{VV} \end{pmatrix} \prec 0 \quad\Longleftrightarrow\quad S_{UU}<0,\ \ S_{UU}S_{VV}-S_{UV}^2>0 \qquad (4)

where: the subscripts denote second partial derivatives; the right-hand pair is Sylvester's criterion for negative-definiteness. The two inequalities together are stronger than $C_V>0$ and $\kappa_T>0$ separately — they also guarantee stability against combined energy-and-volume fluctuations. Finally, the Legendre transforms fix the convexity/concavity signs of every potential. Because a Legendre transform trades a variable for its conjugate slope, it flips curvature in the transformed variable:

F=U-TS:\ \ F \text{ concave in } T,\ \text{convex in } V; \qquad G=F+pV:\ \ G \text{ concave in } T,\text{ and } p \qquad (5)

where: the master rule is potentials are convex in their extensive variables and concave in their intensive variables. Concavity in $T$ is the statement $\partial^2 F/\partial T^2 = -C_V/T < 0$ , the same $C_V>0$ viewed through the free energy.

Derivation AConcavity of S(U,V,N) forces C_V > 0 and κ_T > 0

The goal is to show that the single geometric requirement — equilibrium is a maximum of total entropy, so $S(U,V,N)$ is concave — translates into the two most-used stability inequalities of thermodynamics: positive heat capacity and positive compressibility. The strategy is to consider an isolated system split into two identical halves, let a small amount of energy (then volume) flow between them, and demand that the total entropy not increase. Because $S$ is extensive and the split is symmetric, the first-order change cancels and the second-order change is controlled by the curvature of $S$ .

Step 1. Split the system into halves A and B, each initially in the equilibrium state. Move energy $\delta U$ from B to A at fixed total energy: A gains $+\delta U$ , B loses $-\delta U$ . The total entropy change to second order is $\Delta S = \tfrac12 S_{UU}(\delta U)^2 + \tfrac12 S_{UU}(-\delta U)^2 = S_{UU}(\delta U)^2$ . Why: the first-order terms $S_U(+\delta U) + S_U(-\delta U)$ cancel exactly because both halves start at the same slope $1/T$ — that cancellation is the equilibrium condition; only the curvature survives.

Step 2. Stability demands $\Delta S \le 0$ for this spontaneous fluctuation, so $S_{UU} = (\partial^2 S/\partial U^2)_V \le 0$ : the entropy is concave in energy. Why: if $S_{UU}$ were positive the fluctuation would raise total entropy and grow without bound, splitting the system into a hot half and a cold half — a thermal runaway.

Step 3. Convert the curvature to a measurable. Since $(\partial S/\partial U)_V = 1/T$ , we have $S_{UU} = \partial(1/T)/\partial U = -T^{-2}(\partial T/\partial U)_V = -1/(T^2 C_V)$ . Concavity $S_{UU}<0$ therefore requires $C_V>0$ . Why: the heat capacity is exactly the reciprocal slope $(\partial U/\partial T)_V$ , and a negative one would make $S_{UU}$ positive — forbidden by Step 2.

Step 4. Repeat with a volume fluctuation at fixed energy: $S_{VV}\le 0$ . Working at fixed temperature instead (the physically natural constraint for compression) this is equivalent to convexity of $F$ in $V$ , i.e. $(\partial^2 F/\partial V^2)_T = -(\partial p/\partial V)_T = 1/(V\kappa_T)>0$ , so $\kappa_T>0$ . Why: squeezing a stable material must raise its pressure; $(\partial p/\partial V)_T>0$ would let a density fluctuation feed on itself, and the homogeneous phase collapses into coexisting phases. *Verified: $S_{UU}=-1/(T^2C_V)<0 \Rightarrow C_V>0$ and $(\partial p/\partial V)_T<0 \Rightarrow \kappa_T>0$ checked symbolically; the van der Waals equation of state is shown to develop an unstable branch $(\partial p/\partial V)_T>0$ below $T_c=8a/27Rb$ and none above, in verify_aq-sm-thermodynamic-stability-convexity.py.*

Go deeper · Derivation B · 2 minThe negative-definite entropy Hessian (Sylvester's criterion) — the owned joint condition

Derivation BThe negative-definite entropy Hessian (Sylvester's criterion) — the owned joint condition

This is the article's owned structural result: the two separate inequalities $C_V>0$ and $\kappa_T>0$ are corollaries of one matrix statement — that the $2\times2$ Hessian of $S(U,V)$ is negative-definite — and the matrix condition is strictly stronger because it also rules out instabilities that appear only under combined energy-and-volume fluctuations. The goal is to state the joint condition via Sylvester's criterion and verify it on a concrete stable system. The strategy is to compute the Hessian of the Sackur–Tetrode entropy and check its two principal minors.

Step 1. For a general fluctuation moving both energy and volume, $\delta^2 S = \tfrac12[S_{UU}(\delta U)^2 + 2S_{UV}\,\delta U\,\delta V + S_{VV}(\delta V)^2]$ . Stability requires this quadratic form to be $\le 0$ for all $(\delta U,\delta V)$ — i.e. the Hessian $\mathbf H_S \prec 0$ (negative-definite). Why: an isolated system explores energy and volume exchange simultaneously; stability must survive the worst-case mixed fluctuation, not just the two axis-aligned ones.

Step 2. By Sylvester's criterion, $\mathbf H_S\prec 0$ iff the leading minor is negative and the determinant positive: $S_{UU}<0$ and $S_{UU}S_{VV}-S_{UV}^2>0$ . The determinant condition is the extra content: it can fail even when both $S_{UU}<0$ and $S_{VV}<0$ hold, signalling an instability in a diagonal fluctuation direction. Why: a $2\times2$ symmetric matrix with two negative diagonal entries is still indefinite if the off-diagonal coupling is large enough — Sylvester's determinant test catches exactly that.

Step 3. Test on the ideal gas. With $S(U,V,N)=Nk_B[\ln(V/N)+\tfrac32\ln(U/N)]+\text{const}$ , compute $S_{UU}=-\tfrac{3Nk_B}{2U^2}<0$ , $S_{VV}=-\tfrac{Nk_B}{V^2}<0$ , and $S_{UV}=0$ . The leading minor is negative and the determinant $S_{UU}S_{VV}-0>0$ : both Sylvester conditions hold, so the ideal gas is unconditionally stable. Why: an ideal gas has no attractive interactions to drive condensation, so no instability appears at any $(U,V)$ — the Hessian is negative-definite everywhere.

Step 4. Contrast with an interacting fluid. When attractions are strong enough (van der Waals below $T_c$ ), the determinant condition fails on the middle branch and $\mathbf H_S$ ceases to be negative-definite — the analytic signature of the impending liquid–vapour split. Why: the unstable branch is where a small density fluctuation lowers the free energy; the homogeneous phase is no longer a maximum of $S$ , and coexistence takes over. *Verified: the Sackur–Tetrode Hessian has $S_{UU}=-3Nk/2U^2<0$ , $S_{VV}=-Nk/V^2<0$ , $S_{UV}=0$ , determinant $>0$ , and both numerical eigenvalues negative — negative-definite; the van der Waals determinant fails below $T_c$ . Checked in the verify script.*

Limiting cases

The stability conditions govern several boundary regimes:

Case	Result	Meaning
Marginal stability	$\det \mathbf H_S \to 0$	A stability limit (spinodal): one curvature direction goes flat, $\kappa_T\to\infty$ or $C_V\to\infty$ ; onset of phase separation.
Critical point	$S_{VV}=0$ and higher derivatives vanish	The spinodal and coexistence curves meet; $\kappa_T$ diverges, giving critical opalescence and large density fluctuations.
Ideal gas	$\mathbf H_S \prec 0$ everywhere	Unconditionally stable — no attractions, no instability, no phase transition.
Negative heat capacity	$C_V<0$	Impossible for a stable isolated system in equilibrium; occurs for self-gravitating systems, which are never in true thermodynamic equilibrium.
Le Chatelier's principle	any single response $>0$	A stable system responds to a perturbation so as to oppose it — the physical face of positive $C_V$ , $\kappa_T$ .

Common trap“A negative heat capacity is always unphysical”

Not quite — the correct statement is that a negative heat capacity is impossible for a stable system in the canonical setting (fixed $T$ ), where $C_V=\langle\Delta E^2\rangle/k_BT^2\ge 0$ is manifestly non-negative. But isolated self-gravitating systems (star clusters, a gas cloud collapsing under its own gravity) genuinely exhibit $C_V<0$ : add energy and they cool (a star that gains energy expands and its surface cools). This is not a violation of thermodynamics; it is a symptom that gravitating systems are non-extensive and their microcanonical and canonical descriptions are inequivalent. The stability theorem here assumes short-range forces and extensivity; drop those and negative heat capacity is real. Two more traps: (1) $\kappa_T>0$ is required but $\kappa_S$ (adiabatic) is a smaller positive number, not the same one; (2) the determinant condition is not implied by the two diagonal conditions.

Worked examples

1 · Where does the van der Waals gas become unstable? (a real onset)

For a van der Waals gas $p = \frac{RT}{v-b}-\frac{a}{v^2}$ (per mole, $v=V/n$ ), find the critical temperature and show explicitly that an isotherm just below it has a region where $(\partial p/\partial v)_T>0$ , i.e. mechanically unstable. Use $a=1,\ b=0.1,\ R=1$ in reduced units.

Step 1. Critical temperature. The critical point is the inflection where $(\partial p/\partial v)_T=(\partial^2 p/\partial v^2)_T=0$ ; solving gives the standard result $T_c = \frac{8a}{27Rb}$ . Plugging in: $T_c = \frac{8(1)}{27(1)(0.1)} = \frac{8}{2.7} = \mathbf{2.963}$ .

Step 2. Pick a sub-critical isotherm, $T = 0.9\,T_c = 2.667$ . Compute the slope $(\partial p/\partial v)_T = -\frac{RT}{(v-b)^2}+\frac{2a}{v^3}$ and scan $v$ . At $v=0.3$ : $-\frac{2.667}{(0.2)^2}+\frac{2}{0.027} = -66.7 + 74.1 = \mathbf{+7.4>0}$ — positive slope, mechanically unstable.

Step 3. Confirm the boundaries. At small $v$ (dense, $v=0.15$ ) and large $v$ (dilute, $v=1.5$ ) the slope is negative (stable). The unstable window lies between the two spinodal points where $(\partial p/\partial v)_T=0$ . Between them, a homogeneous fluid cannot exist.

Step 4. Above $T_c$ ( $T=1.5\,T_c=4.44$ ) the slope is negative for all $v$ : no unstable branch, no phase transition — the supercritical fluid. The stability condition alone locates the phase transition without any further input.

2 · Stability inequalities as measurable numbers (exam level)

(a) A monatomic ideal gas has $C_V=\tfrac32 nR$ ; verify it is positive and state what instability its negativity would imply. (b) Water at $25^\circ$ C has $\kappa_T=4.5\times10^{-10}\,\mathrm{Pa^{-1}}$ ; compute the fractional volume change when pressure rises by $10\,\mathrm{MPa}$ , and confirm the sign is stabilizing.

Step 1. Heat capacity: $C_V=\tfrac32 nR = \tfrac32(1)(8.314) = \mathbf{12.47\ J/K}$ per mole — positive, as required. If it were negative, adding heat would lower the temperature, and any warm patch would keep absorbing heat and heating relative to its surroundings: a thermal runaway. Stability forbids it.

Step 2. Compressibility sign. By definition $\kappa_T = -\frac1V(\partial V/\partial p)_T > 0$ means $(\partial V/\partial p)_T<0$ : raising pressure shrinks the volume. This is the stabilizing response — Le Chatelier in action.

Step 3. Fractional volume change: $\frac{\Delta V}{V} \approx -\kappa_T\,\Delta p = -(4.5\times10^{-10})(10\times10^{6}) = -4.5\times10^{-3}$ , i.e. $\mathbf{-0.45\%}$ . The minus sign is the whole point: compression reduces volume, exactly as a stable material must.

Step 4. Boundary check. As $\kappa_T\to\infty$ (approaching a critical point) the same $\Delta p$ would produce an enormous $\Delta V$ — the material becomes infinitely "soft," which is why critical fluids scatter light strongly (critical opalescence). Positive-but-finite $\kappa_T$ is the everyday stable regime.

Origins · 1 minJohannes Diderik van der Waals and how this was discovered

Johannes Diderik van der Waals

1837–1923 · Amsterdam

Van der Waals, a Dutch physicist who taught himself physics while working as a schoolteacher, gave the world the first equation of state to describe both the gas and liquid phases and the transition between them. His 1873 doctoral thesis introduced the two corrections — a finite molecular size ( $b$ ) and an attractive intermolecular force ( $a$ ) — that produce the famous S-shaped sub-critical isotherm whose middle branch is mechanically unstable, $(\partial p/\partial V)_T>0$ . It was precisely this instability that made the liquid–vapour transition thermodynamically inevitable, turning the abstract convexity conditions of Gibbs into a concrete, calculable phase diagram. Van der Waals also discovered the law of corresponding states — that all fluids, rescaled by their critical constants, obey one universal equation — an early glimpse of universality that anticipated the modern theory of critical phenomena. He received the 1910 Nobel Prize for this work. Maxwell called his thesis a landmark, remarking that van der Waals's name would "soon be among the foremost in molecular science."

History

1875

— Gibbs begins On the Equilibrium of Heterogeneous Substances, founding the theory of thermodynamic stability and phase coexistence.

1873

— van der Waals's dissertation gives the equation of state whose unstable loop illustrates mechanical instability.

1884

— Le Chatelier states the principle that a system opposes imposed changes, the physical face of the stability inequalities.

1906

— Duhem systematizes the convexity conditions and coins much of the modern stability terminology.

1960

— Callen's textbook codifies the stability and convexity postulates as the standard pedagogical framework.

Going further

References. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., Chs. 8–9 (the definitive treatment of stability and convexity) · Pathria & Beale, Statistical Mechanics, 3rd ed., §3.10 · Kardar, Statistical Physics of Particles, §1.9 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 11 · Landau & Lifshitz, Statistical Physics Part 1, §§21–22 · Gibbs, On the Equilibrium of Heterogeneous Substances (1875–78).

Provenance. The concavity of $S$ and convexity/concavity of the potentials are Callen's stability postulates (Chs. 8–9). A verifier should reproduce: $S_{UU}=-1/(T^2C_V)$ hence $C_V>0$ ; $(\partial^2 F/\partial V^2)_T = 1/(V\kappa_T)$ hence $\kappa_T>0$ ; the Sackur–Tetrode Hessian eigenvalues (both negative, negative-definite); and the van der Waals instability with $T_c=8a/27Rb$ (unstable branch below, stable above). Numerical instances (vdW $T_c=2.963$ in reduced units; ideal-gas $C_V=12.47$ J/K per mole; water $\Delta V/V=-0.45\%$ at $\Delta p=10$ MPa) are recomputed in verify_aq-sm-thermodynamic-stability-convexity.py. $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ (CODATA).

Builds on: Second Law of Thermodynamics and Entropy · Thermodynamic Potentials and the Fundamental Relation · Legendre Transforms · Maxwell Relations · Helmholtz Free Energy, Enthalpy, and Gibbs Free Energy. Leads to: Euler Relation and Gibbs–Duhem Equation · Response Functions · Phase Transitions and Coexistence · Fluctuation–Response Theorems.

Why is concavity of entropy the same thing as a stable equilibrium?

The Second Law says an isolated system in equilibrium maximizes its total entropy. A maximum of a smooth function is a point where the first derivatives vanish (the equilibrium conditions: equal $T$ , $p$ , $\mu$ ) and the second derivatives make it curve downward in every direction — that downward curvature is concavity. If the entropy surface curved upward anywhere, a spontaneous fluctuation in that direction would increase the total entropy, so the Second Law would drive the fluctuation to grow rather than decay, and the "equilibrium" would not be one. Concavity is not an extra assumption bolted onto the Second Law; it is the second-order part of "equilibrium is an entropy maximum."

How can self-gravitating systems have negative heat capacity if stability forbids it?

The stability theorem assumes the system is extensive — energy and entropy scale linearly with size, which requires short-range interactions. Gravity is long-range and unscreened, so a self-gravitating system (a star cluster, a collapsing gas cloud) is fundamentally non-extensive, and the microcanonical and canonical ensembles stop agreeing. In the microcanonical (isolated) setting such systems really do have $C_V<0$ : extract energy from a star and it contracts and heats up (the negative-specific-heat that drives stellar evolution). This is genuine physics, not a paradox — it is a flag that the assumptions behind the stability inequalities (extensivity, ensemble equivalence) have broken down. In the canonical ensemble at fixed $T$ , by contrast, $C_V=\langle\Delta E^2\rangle/k_BT^2\ge 0$ is unavoidable, which is why a self-gravitating system has no proper canonical equilibrium.

What is the difference between the spinodal and the coexistence (binodal) curve?

Both appear on a phase diagram, but they mark different things. The coexistence (binodal) curve is where two phases have equal chemical potential and pressure — the true equilibrium boundary; between it and the spinodal the homogeneous phase is metastable (locally stable but globally not the lowest free energy, so it needs a nucleation event to decay). The spinodal is where the stability Hessian's determinant first hits zero — inside it the homogeneous phase is absolutely unstable and separates spontaneously (spinodal decomposition), no nucleation barrier needed. The spinodal lies inside the binodal; the two meet only at the critical point. Stability theory locates the spinodal directly; locating the binodal additionally requires the equal-chemical-potential (Maxwell) construction.

Why is

\kappa_S

(adiabatic) smaller than

\kappa_T

(isothermal)?

Both are positive by stability, but they differ because a fast (adiabatic) compression heats the gas, and the extra thermal pressure resists further compression more than an isothermal squeeze does. Quantitatively $\kappa_T/\kappa_S = C_p/C_V = \gamma \ge 1$ , so $\kappa_S = \kappa_T/\gamma \le \kappa_T$ . Physically: hold temperature fixed and the material is "softer" (more compressible) because you let the heat of compression escape; do it adiabatically and the trapped heat stiffens it. This is exactly why the speed of sound (which depends on $\kappa_S$ ) is higher than a naive isothermal estimate — the mistake Newton made and Laplace corrected.

Does stability require the entropy to be concave, or the free energy to be convex? They sound opposite.

They are the same condition seen through a Legendre transform, and the apparent sign flip is exactly what the transform does. Entropy $S(U,V,N)$ must be concave in its extensive arguments. When you Legendre-transform from $S$ (or $U$ ) to a free energy by trading an extensive variable for its intensive conjugate slope — e.g. $U\to T$ via $F=U-TS$ — concavity in the traded variable becomes concavity in the new intensive variable, while convexity is preserved in the untraded extensive ones. The clean rule: thermodynamic potentials are convex in their extensive variables and concave in their intensive variables. So $S$ concave in $U$ and $F$ concave in $T$ and convex in $V$ are all the same physics.

Is stability a statement about a single phase or about the whole system?

About a single homogeneous phase. The convexity conditions test whether a uniform state is a local maximum of entropy against internal fluctuations. When they fail, the global maximum is no longer the homogeneous state — it is a mixture of two (or more) coexisting phases, and the true free energy is the common-tangent (convex-hull) construction lying below the unstable bump. So stability analysis tells you when a single phase is locally viable; the global phase diagram is found by then taking the convex envelope of the free energy. The unstable and metastable regions of a single-phase equation of state are precisely the parts the system avoids by phase-separating.