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Derivation

Stability, Convexity, and Positive Response Functions

Statement

For a thermodynamic system in stable equilibrium, the requirement that the total entropy be a genuine maximum against spontaneous internal fluctuations forces the heat capacity at constant volume to be strictly positive, CV > 0, and the isothermal compressibility to be strictly positive, κT = −(1/V)(∂V/∂P)T > 0. Equivalently, the internal energy U(S,V) is a convex function of its extensive arguments and the Helmholtz free energy F(T,V) is concave in T and convex in V.

Why it matters

Equilibrium is not merely a stationary point of entropy; it must be a stable one. The first-order (Clausius/Legendre) conditions fix which state a system sits in, but they say nothing about whether a small internal rearrangement grows or decays. Stability is the second-order statement, and it is what makes matter behave sensibly: push on it and it pushes back, heat it and its temperature rises.

These inequalities are the microscope-independent guarantee of thermodynamic sensibility. When they are violated the homogeneous phase cannot exist — the system spontaneously separates into coexisting phases. The entire theory of phase transitions, spinodals, and critical points is the study of where and how these convexity conditions fail.

Assumptions
The composite system is isolated (fixed total U, V, N).Without a conserved total, the extremum principle for entropy has no fixed constraint surface and "maximum entropy" is undefined; one must switch to the appropriate potential for the actual reservoir. The system is thermodynamically large, so fluctuations are small compared with mean values.If fluctuations are not small the quadratic (second-order) expansion is insufficient and one must retain higher-order terms; near a critical point exactly this happens and mean-field stability analysis breaks down. Entropy is a smooth, twice-differentiable, extensive function of U and V.At a first-order transition S(U,V) develops a flat facet (a straight segment), the second derivative vanishes, and the strict inequalities soften to ≥; differentiability fails outright at the transition line. The system is homogeneous, so a single subsystem/reservoir split captures the relevant fluctuation.If long-range fields (gravity, unscreened Coulomb) couple distant parts, entropy is no longer additive over subvolumes and the convexity argument, which rests on additivity, must be reformulated.
Derivation
1
Stot = S(U1,V1) + S(U−U1, V−V1) = maximum
Isolate the system, imagine a partition creating subsystem 1 and reservoir 2; total entropy is additive and, in equilibrium, stationary and maximal over the internal division. A
2
Rmin ≡ ΔU − T₀ ΔS + P₀ ΔV = −T₀ ΔStot ≥ 0
The minimum work to create a fluctuation of a small part against a reservoir at (T₀, P₀) equals −T₀ times the total entropy change; a maximum of Stot means ΔStot ≤ 0, hence Rmin ≥ 0 for every fluctuation. B
3
Rmin = ½ (ΔT ΔS − ΔP ΔV) + …
Expand ΔU(S,V) to second order about equilibrium. The first-order terms (T₀−T₀)ΔS and −(P₀−P₀)ΔV cancel; the quadratic form of the Hessian of U collapses neatly into the symmetric combination of the conjugate-variable increments of the fluctuating part. C
4
ΔS = (∂S/∂T)VΔT + (∂S/∂V)TΔV = (CV/T) ΔT + (∂P/∂T)V ΔV
Choose (T,V) as independent fluctuating variables and expand ΔS; use (∂S/∂T)V = CV/T and the Maxwell relation (∂S/∂V)T = (∂P/∂T)V. B
5
ΔP = (∂P/∂T)V ΔT + (∂P/∂V)T ΔV
Expand ΔP in the same independent variables (T,V). A
6
ΔT ΔS − ΔP ΔV = (CV/T) (ΔT)² − (∂P/∂V)T (ΔV)²
Substitute steps 4–5 into step 3; the two cross terms in (∂P/∂T)V ΔT ΔV are equal and cancel, diagonalising the quadratic form. B
7
Rmin = ½ [ (CV/T) (ΔT)² − (∂P/∂V)T (ΔV)² ] > 0  ∀ (ΔT,ΔV)≠0
A diagonal quadratic form is positive-definite iff each coefficient is strictly positive. Because ΔT and ΔV are independent fluctuations, each square can be excited alone, so both coefficients are separately constrained. C
8
CV/T > 0  ⇒  CV > 0   and   −(∂P/∂V)T > 0  ⇒  κT > 0
Set each coefficient positive. With T > 0 the first gives CV > 0; the second is exactly the thermal (mechanical) stability condition, i.e. positive isothermal compressibility. A
9
(∂²U/∂S²)V = T/CV ≥ 0,   (∂²F/∂V²)T = −(∂P/∂V)T = 1/(VκT) ≥ 0,   (∂²F/∂T²)V = −CV/T ≤ 0
Re-express the same content geometrically: positive coefficients are precisely the statements that U(S,V) is convex in each extensive variable, while its Legendre transform F(T,V) is concave in the intensive variable T and convex in the extensive variable V. Convexity/concavity flips under each Legendre transform. C
Result
CV > 0   and   κT = −(1/V)(∂V/∂P)T > 0

Reading. Stability against fluctuations is two independent statements. Thermal stability (CV > 0): if a subregion spontaneously absorbs a little heat its temperature rises above its surroundings, and heat then flows back out — the fluctuation is self-correcting. Mechanical stability (κT > 0): if a subregion spontaneously compresses, its pressure rises above its surroundings, and it re-expands. A negative CV or κT would make the fluctuation grow without bound — the homogeneous state is then not equilibrium at all. Geometrically these are the convexity conditions on the potentials.

Units check. (CV/T)(ΔT)²: [J K⁻¹][K⁻¹][K²] = J.  −(∂P/∂V)T(ΔV)²: [Pa m⁻³][m⁶] = Pa·m³ = J. Both terms of Rmin are energies, as an energy must be. κT has units [Pa⁻¹]; CV has [J K⁻¹]; both inequalities are dimensionally consistent sign statements.

Limiting cases
  • Ideal gas. CV = (f/2)nR > 0 and κT = 1/P > 0 for all T,P; the ideal gas is unconditionally stable and never phase-separates.
  • Incompressible limit. κT → 0+ (a rigid solid/liquid) is the boundary of mechanical stability from the safe side — still positive, so still stable.
  • Critical point. κT → ∞ (and (∂P/∂V)T → 0): mechanical stability is marginal; the response diverges, giving critical opalescence.
  • Spinodal curve. κT = ∞ then changes sign; the locus (∂P/∂V)T = 0 marks the absolute limit of metastability of a single phase.
  • Third-law limit. As T → 0, CV → 0+; the inequality is preserved but non-strict at exactly T = 0.
Breaks when
  • Inside the spinodal region of a van der Waals–type equation of state, where (∂P/∂V)T > 0: the homogeneous phase has κT < 0 and is absolutely unstable, decaying by spinodal decomposition into coexisting liquid and vapour. The stability inequality is violated and the single-phase description is physically meaningless.
  • Self-gravitating systems (star clusters, gas clouds): gravity makes energy non-additive and produces states with CV < 0 (contract, heat up, radiate, contract further — the gravothermal catastrophe). Negative heat capacity is real there because the additivity assumption underpinning the derivation fails.
  • Small/mesoscopic systems where fluctuations are not small: the second-order expansion is inadequate, ensembles are inequivalent, and microcanonical CV can be negative even while the canonical one is positive.
  • At a first-order phase boundary the potentials lose strict convexity (a flat tie-line appears), the inequalities become non-strict, and derivatives like CP or κT can jump or diverge.
Failure modes
  • Confusing CV with CP: the stability condition that emerges directly from the (T,V) fluctuation is on CV. CP > 0 and κS > 0 follow, but are weaker; CP ≥ CV always.
  • Dropping the cross-term cancellation: forgetting the Maxwell relation in step 4, so the ΔT ΔV terms do not cancel, leaving a non-diagonal form and spurious "coupled" stability conditions.
  • Treating ΔT and ΔV as dependent: concluding only that the determinant of the quadratic form is positive, missing that each diagonal coefficient must independently be positive because the fluctuations are independent.
  • Sign slip in κT: writing κT = (1/V)(∂V/∂P)T without the minus sign, then "deriving" κT < 0 for stability.
  • Assuming stability guarantees a global minimum of free energy: convexity gives only local stability; a metastable phase satisfies the inequalities yet is not the true equilibrium (superheated liquid).
  • Using T < 0 reasoning: claiming CV > 0 "because" the coefficient is CV/T without noting the physical T > 0 needed to convert positivity of the coefficient into positivity of CV.
Discussion

The deep content is that equilibrium thermodynamics is a theory of convex functions. The fundamental relation S(U,V,N) is concave in its extensive variables; equivalently U(S,V,N) is convex. Every stability inequality — CV>0, κT>0, CP≥CV, κT≥κS — is a statement that a particular second derivative (a curvature) has the sign convexity demands. Legendre transformation, which trades an extensive variable for its intensive conjugate, flips convexity to concavity in that slot; that is why F=U−TS is concave in T but stays convex in V, and why G is concave in both T and P.

Stability is also the macroscopic face of the fluctuation–dissipation link. From the Einstein fluctuation formula w ∝ exp(ΔStot/k), the same quadratic form that must be positive-definite for stability directly sets the mean-square fluctuations: ⟨(ΔT)²⟩ = kT²/CV and ⟨(ΔV)²⟩ = kTVκT. A response function is positive because it measures the size of a fluctuation, and a variance cannot be negative. Stability and observability are the same statement seen from two sides.

A complete treatment recognises the conditions as a hierarchy of principal-minor tests on the Hessian of U(S,V,N) (or on −S). Requiring the full Hessian to be positive-semidefinite yields, minor by minor, first CV>0 (thermal), then the 2×2 determinant condition equivalent to κT>0 (mechanical), and, with particle exchange, (∂μ/∂N)T,V>0 (chemical/diffusive, or (∂μ/∂P)… giving material stability). The strongest single-variable condition CV>0 implies the others through the identities CP−CV=TVαP²/κT and CP/CV=κT/κS≥1: mechanical stability plus these relations force CP≥CV>0 and κT≥κS>0. This ordering — adiabatic responses smaller than isothermal ones — is a direct corollary of nested convexity.

Common misconceptions. (i) "Negative heat capacity is impossible." False in general — it is impossible only for additive systems in the canonical ensemble; gravitating systems and small microcanonical systems routinely show it. What the derivation forbids is a stable homogeneous phase with CV<0. (ii) "Convexity is an extra postulate." It is not independent — it is the second-order content of the entropy-maximum postulate already used to locate equilibrium. (iii) "Stability picks out the true phase." It only excludes unstable states; distinguishing stable from metastable requires comparing free energies (the common-tangent / Maxwell construction), not curvature alone.

Worked examples

Example 1 — Confirming stability of one mole of monatomic ideal gas.

1
CV = (3/2)nR
Monatomic ideal gas, f=3 translational degrees of freedom. A
2
CV = 1.5 × (1 mol) × (8.314 J mol⁻¹ K⁻¹) = 12.47 J K⁻¹
Insert numbers; positive. A
3
κT = −(1/V)(∂V/∂P)T = 1/P  (from PV = nRT)
For V = nRT/P, (∂V/∂P)T = −nRT/P² = −V/P. B
4
κT = 1/(1.013×10⁵ Pa) = 9.87×10⁻⁶ Pa⁻¹  at T = 300 K, P = 1 atm
Positive; V = nRT/P = 8.314×300/1.013×10⁵ = 0.0246 m³. A
CV = 12.47 J K⁻¹ > 0,   κT = 9.87×10⁻⁶ Pa⁻¹ > 0

Reading. Both stability conditions hold for every state of the ideal gas, consistent with its never phase-separating.

Units check. [J K⁻¹] and [Pa⁻¹] as required.

Example 2 — Locating instability in van der Waals CO₂ below Tc.

1
P = RT/(V−b) − a/V²  (per mole),  (∂P/∂V)T = −RT/(V−b)² + 2a/V³
Van der Waals equation of state and its isothermal slope. A
2
Tc = 8a/(27Rb),   a = 0.364 Pa m⁶ mol⁻²,   b = 4.27×10⁻⁵ m³ mol⁻¹
CO₂ parameters; compute the critical temperature to choose a subcritical isotherm. A
3
Tc = 8(0.364)/[27(8.314)(4.27×10⁻⁵)] = 304 K
So pick T = 280 K < Tc, where an unstable loop exists. A
4
at V = 3b = 1.281×10⁻⁴ m³:  RT/(V−b)² = 2327.9/(8.54×10⁻⁵)² = 3.19×10¹¹
Evaluate the first term; RT = 8.314×280 = 2327.9, V−b = 2b. B
5
2a/V³ = 0.728/(1.281×10⁻⁴)³ = 3.46×10¹¹
Evaluate the second term. A
6
(∂P/∂V)T = −3.19×10¹¹ + 3.46×10¹¹ = +2.7×10¹⁰ Pa m⁻³ > 0
Positive slope ⇒ κT = −1/[V(∂P/∂V)T] < 0. B
κT < 0  at  T = 280 K, V = 3b  ⇒  homogeneous phase unstable

Reading. On the 280 K isotherm the van der Waals gas has a segment with (∂P/∂V)T>0 — the stability inequality is violated, so CO₂ there cannot remain a single homogeneous phase; it splits into coexisting liquid and vapour (resolved by the Maxwell construction).

Units check. (∂P/∂V)T in [Pa m⁻³]; κT = −1/(V·∂P/∂V) in [1/(m³·Pa m⁻³)] = [Pa⁻¹].

Problems
  1. Show from (∂²U/∂S²)V = (∂T/∂S)V that convexity of U in S is equivalent to CV>0.
    Solution By definition CV = T(∂S/∂T)V, so (∂T/∂S)V = T/CV. Also (∂U/∂S)V = T, hence (∂²U/∂S²)V = (∂T/∂S)V = T/CV. Convexity of U in S means this second derivative ≥ 0. Since T>0, this holds iff CV>0.
  2. For a photon gas U = aVT⁴, P = aT⁴/3. Compute CV and verify thermal stability. Comment on κT.
    Solution At fixed V, CV = (∂U/∂T)V = 4aVT³ > 0 for all T>0 — thermally stable. For the compressibility, P = aT⁴/3 is independent of V, so (∂P/∂V)T = 0 and κT = ∞: the photon gas sits at the boundary of mechanical stability (isothermal compression at fixed T costs no pressure change), consistent with radiation exerting a fixed pressure per temperature.
  3. Using CP−CV = TVαP²/κT with αP = (1/V)(∂V/∂T)P, prove CP ≥ CV whenever the system is mechanically stable.
    Solution Mechanical stability gives κT>0. In CP−CV = TVαP²/κT, the factors are: T>0, V>0, αP²≥0 (a square), and κT>0. The whole right side is ≥0, hence CP≥CV, with equality only when αP=0 (e.g. water at 4 °C).
  4. One mole of van der Waals gas (a,b) at temperature T. Find the spinodal condition and evaluate the molar volumes at which mechanical stability is marginal for CO₂ (a=0.364, b=4.27×10⁻⁵) at T=280 K.
    Solution Spinodal: (∂P/∂V)T=0 ⇒ RT/(V−b)² = 2a/V³, i.e. RT V³ = 2a(V−b)². With RT=2327.9, 2a=0.728: solve 2327.9 V³ = 0.728(V−4.27×10⁻⁵)². Numerically this cubic has two physical roots bracketing V=3b found in Example 2 (where the slope was positive): approximately V1≈6.5×10⁻⁵ m³ (liquid-side spinodal) and V2≈3.9×10⁻⁴ m³ (vapour-side spinodal). Between them κT<0 and the phase is absolutely unstable; the marginal points are where κT→±∞. (Accept roots obtained by iteration; the key result is two real roots straddling the unstable segment.)
  5. A small system has entropy S(U) = k ln[ (U/U₀)⁵ e−U/U₀ ] for U>0. Find the range of U for which it is thermally stable, and interpret.
    Solution S(U) = k[5 ln(U/U₀) − U/U₀]. Then 1/T = (∂S/∂U) = k(5/U − 1/U₀) and (∂²S/∂U²) = −5k/U². Concavity of S(U) (thermal stability) requires (∂²S/∂U²)<0, which holds for all U>0 here — so this toy system is thermally stable everywhere it has T>0. Note T>0 only for U<5U₀; beyond that 1/T<0 (formal negative temperature). Interpretation: the concavity/stability condition and the positive-temperature condition are logically separate — a bounded-spectrum system can be "stable" in curvature yet reach negative temperature, illustrating why the physical T>0 assumption is needed to translate curvature signs into the usual CV>0 statement.