Maxwell relations and thermodynamic stability

A pressure slope predicts an entropy slope

A sealed sample contains 0.100 mol of an assigned simple fluid at 350 K and 0.200 L. Hold its volume fixed and raise the temperature by 10 K. Its pressure rises by about 42.33433 kPa. Can this pressure response tell us how the sample’s entropy changes with volume at fixed temperature, without directly measuring entropy?

The connection is a Maxwell relation. It equates (∂S/∂V) at fixed T and amount to (∂p/∂T) at fixed V and amount. For this model the pressure slope is 4.233433 kPa/K, so the entropy slope at the opening state is 4.233433 J/(K L). The numerical equality uses 1 kPa L = 1 J; the derivatives still describe different controlled comparisons.

On a first reading, follow the local notation, derive one Maxwell relation, solve the pressure question, compare response functions and distinguish local from global stability. The sections marked “Further” return to integrability, relaxed coordinates, extensivity and fluctuations. Useful published preparation is “Physical systems, states, boundaries, and environments”, “Derivatives, Taylor series, and local approximation”, and “Integrals and multivariable calculus”. All thermodynamic potential definitions needed here are supplied locally, so an unpublished neighboring article is not required.

Name the coordinates and the branch

Write n for amount in moles, V for total volume, v = V/n for molar volume, T for positive absolute temperature, p for pressure and S for total entropy. Internal energy U, enthalpy H, Helmholtz energy F and Gibbs energy G are total energies in joules. A partial derivative such as (∂p/∂T)_(V,n) changes T while holding V and n fixed. Its subscript is part of the physical statement, not decoration.

The central model is a homogeneous, fixed-composition, simple compressible material: pressure–volume work is its only work coordinate. At fixed n its equilibrium identity is dU = T dS − p dV. Heat into a closed system is positive and outward work W_by is positive, giving ΔU = Q − W_by when endpoint bulk stores do not change. The equilibrium identity relates neighboring states; it does not replace actual irreversible heat by T dS or actual work by p dV.

We work within one smooth equilibrium-property branch. Continuous second derivatives allow mixed second derivatives to commute. Identities that differentiate them again need the corresponding third derivatives. At a first-order phase boundary the globally selected potential can have a kink; at a critical point a response can diverge. A formula derived for ordinary finite derivatives must not be used across those singularities by treating infinity as a number.

Metastable homogeneous branches can have smooth, internally consistent property functions too. Their derivatives may satisfy Maxwell relations while their free energy is higher than another accessible state. Integrability, local stability and global equilibrium are three separate tests.

Derive the four relations from natural variables

The natural variables of a potential are the independent coordinates appearing in its fundamental differential. Define H = U + pV, F = U − TS and G = U + pV − TS. At fixed amount and composition their differentials are listed below. For example, dF = −S dT − p dV means S = −(∂F/∂T)_(V,n) and p = −(∂F/∂V)_(T,n).

Differentiate the first equality with respect to V at fixed T,n, and the second with respect to T at fixed V,n. The same mixed derivative F_TV appears, giving −(∂S/∂V)_(T,n) = −(∂p/∂T)_(V,n). The minus signs cancel. Repeating this reasoning for U, H and G gives all four relations, with the held coordinates carried through each differentiation.

The table describes a fixed amount, fixed composition and no extra work coordinates in every row. If additional coordinates can change, retain their conjugate work terms and hold the appropriate other natural variables fixed. A memorized two-variable formula is incomplete for an unconstrained magnetic, elastic, reacting or multicomponent system.

Solve the pressure question with one consistent model

The assigned material has p = nRT/(V − nb) − a n²/V², with n = 0.100 mol, a = 0.280 Pa m⁶/mol² and b = 3.60 × 10⁻⁵ m³/mol. Its total C_V is (3/2)nR. These are hypothetical thermodynamic model parameters, not a material fit or measured-fluid accuracy claim. The domain is T > 0 and V > nb. Its model critical temperature is about 277.1709 K, so the 350 K comparisons lie above that model critical temperature.

A compatible potential is F = n{−c_v,m T ln(T/T₀) + c_v,m T − RT ln[(v − b)/(v₀ − b)] − a/v}, where c_v,m = (3/2)R, T₀ = 350 K and v₀ = 0.00200 m³/mol are fixed reference constants. Differentiating this one function gives S = n c_v,m ln(T/T₀) + nR ln[(v − b)/(v₀ − b)] and U = n c_v,m T − a n²/V. All logarithms have dimensionless positive arguments. The chosen S reference is relative and supplies no third-law or low-temperature extrapolation.

At fixed V,n, the pressure temperature derivative is nR/(V − nb). At fixed T,n, differentiating the entropy expression with respect to V gives exactly the same nR/(V − nb). At the opening state this is 4233.433 Pa/K, or 4.233433 J/(K L). A 10 K increase at fixed volume raises p by 42.33433 kPa because the assigned p is linear in T there.

For the finite isothermal comparison from 0.200 to 0.250 L, integrate the volume-dependent entropy slope: ΔS = nR ln[(V_final − nb)/(V_initial − nb)] ≈ 0.1885744 J/K. Multiplying the initial local slope by the full 0.050 L interval would give 0.2116717 J/K instead. A tangent gives a local approximation; it is not the full curved state relation. No expansion time or actual transfer history is inferred.

Relate measurable response functions

Define total C_V = (∂U/∂T)_(V,n), total C_p = (∂H/∂T)_(p,n), volumetric expansion α = (1/V)(∂V/∂T)_(p,n), and isothermal compressibility κ_T = −(1/V)(∂V/∂p)_(T,n). Their units are J/K, J/K, K⁻¹ and Pa⁻¹. These definitions have named mechanical and amount constraints; a molar or mass-specific capacity requires a corresponding change of counting basis.

Expand entropy in either convenient coordinate pair. Maxwell relations give dS = (C_V/T)dT + (∂p/∂T)_(V,n)dV, and dS = (C_p/T)dT − Vα dp. The volume differential is dV = Vα dT − Vκ_T dp. Setting dV = 0 in the latter yields (∂p/∂T)_(V,n) = α/κ_T whenever the coordinate transformation is nonsingular.

At constant p, compare the two entropy differentials: C_p − C_V = TVα²/κ_T = −T[(∂p/∂T)_(V,n)]²/(∂p/∂V)_(T,n). For a regular stable material with T,V > 0 and finite positive κ_T, this difference is nonnegative. It can be zero when α = 0. It is not universally nR; that simplification belongs to the ideal-gas model.

At constant S,n, eliminating dT gives κ_S = −(1/V)(∂V/∂p)_(S,n) = κ_T − TVα²/C_p = κ_T C_V/C_p. Thus C_p/C_V = κ_T/κ_S in this finite positive-response regime. This comparison concerns isentropic and isothermal response derivatives; it does not declare every actual adiabatic process isentropic.

Compare responses without changing the model

At the opening state, α ≈ 0.003092668 K⁻¹ and κ_T ≈ 7.305343e-07 Pa⁻¹. Substituting them into TVα²/κ_T gives C_p − C_V ≈ 0.9164823 J/K. With C_V ≈ 1.247169 J/K, this gives C_p ≈ 2.163652 J/K. The independent pressure-derivative expression gives the same result.

The response plot compares separate states at the common T = 350 K and n = 0.100 mol. At each point C_p is still the derivative defined with p held fixed; plotting its value versus the state’s V does not turn the comparison curve into a constant-pressure heating process. The assigned model’s excluded volume nb = 0.0036 L lies outside the displayed 0.006–0.050 L window.

The capacity difference reaches about 3.995738 J/K at V = 0.0108 L for this model and temperature. The ideal-gas value nR = 0.8314463 J/K is a useful reference, not a replacement for the nonideal derivative. Every displayed state has positive finite κ_T and C_V under the assigned model.

A consistency test can reject positive responses

Now deliberately propose a different fixed-amount specification: p(T,V) = γT²/2 + K₀/V, with γ = 20 Pa/K² and K₀ = 10 J, while simultaneously assigning C_V = C₀ = 2 J/K everywhere. The proposed pressures are positive and their isothermal volume slope is negative. That does not establish a consistent thermodynamic state model.

If these data belonged to a smooth entropy, Maxwell would require the candidate differential dS_star = (C₀/T)dT + γT dV. Its mixed derivatives disagree: ∂(C₀/T)/∂V = 0, whereas ∂(γT)/∂T = γ. The symbol dS_star here is a proposed one-form, not an already valid state differential.

Compare two routes from (300 K, 0.000100 m³) to (330 K, 0.000200 m³). Route A heats first at the smaller volume, then increases V at 330 K. Route B increases V at 300 K, then heats at the larger volume. Their candidate integrals are about 0.8506204 and 0.7906204 J/K. The nonzero difference 0.06 J/K rejects exactness.

These are not two possible entropy changes for the same endpoints, and their difference is not entropy production. The incompatible property specification must be repaired before it can describe a thermodynamic system. A finite rectangular loop is a practical consistency diagnostic, while a local mixed-derivative test identifies the failing coefficient.

Stability means stability under stated constraints

A stationary state has vanishing first variation with respect to the internal changes being allowed. Stable equilibrium must also resist those allowed small changes. For an isolated additive system, the total entropy is maximized at fixed total energy, volume and amount. In an equivalent energy description, the equilibrium energy surface is convex in its extensive coordinates. The reservoir and constraint choices decide which variation is physically relevant.

For example, divide a homogeneous equilibrium sample into two parts and allow a small energy or volume transfer while keeping the combined totals fixed. The first-order changes cancel between the parts. A negative entropy second variation, or the corresponding positive energy quadratic form, restores local stability in those transfer directions. Holding volume rigidly can remove a mechanical direction that would otherwise be available.

Local stability is a statement about nearby allowed states. Global stability compares the appropriate potential over all accessible alternatives under the same controls. A higher local minimum can be metastable, separated from a lower state by a barrier. Thermodynamics alone does not give its lifetime or an escape rate.

Test the constrained Hessian

For a regular simple compressible branch at fixed n, consider the 2×2 Hessian of U(S,V). A small perturbation has quadratic energy term ½[U_SS(δS)² + 2U_SV δSδV + U_VV(δV)²]. Positive definiteness requires U_SS > 0 and det H_U > 0; positive diagonal entries alone are insufficient when the cross term is present.

Write p_T = (∂p/∂T)_(V,n) and p_V = (∂p/∂V)_(T,n). From dS = (C_V/T)dT + p_T dV, one obtains U_SS = T/C_V, U_SV = −T p_T/C_V, and U_VV = −p_V + T p_T²/C_V. Their determinant simplifies to det H_U = (T/C_V)(−p_V). Thus C_V > 0 and p_V < 0, equivalently κ_T > 0, provide strict local thermal/mechanical stability on this regular fixed-amount surface.

At the opening state the Hessian entries have different units. Comparing raw eigenvalues as if all coordinates had the same unit would be meaningless. Instead set δS = (0.0100 J/K)x and δV = (1.00 × 10⁻⁶ m³)y. The quadratic form becomes ½[x,y] K [x,y]ᵀ, with every entry of K in joules. Its eigenvalues are about 0.005592559 and 0.03434483 J, both positive. This tests the assigned normalized perturbations, not a universal fluctuation amplitude.

Potential choice changes curvature signs. F(T,V) is concave in T, with F_TT = −C_V/T, and convex in V, with F_VV = −p_V; it is generally a saddle as a function of those two controls. G(T,p) has G_TT = −C_p/T, G_pp = −Vκ_T and G_Tp = Vα. Its determinant is C_V Vκ_T/T ≥ 0 in the regular stable regime. A stable state therefore does not make every potential positive definite in all its natural variables.

Separate local minima from global selection

A small original model makes the distinction visible. At fixed external conditions and amount, let an internal dimensionless coordinate x have reduced potential φ(x;h) = (x² − 1)²/4 − hx. The physical potential is a positive energy scale E₀ times φ; h is a dimensionless conjugate field. This reduced mean-field model is separate from the fluid volume coordinate and is not a measured phase diagram.

Stationary states satisfy x³ − x − h = 0. Their local curvature is 3x² − 1. At h = 0, x = −1 and +1 are equally deep global minima with curvature 2; x = 0 is a maximum with curvature −1. At h = 0.2, there are still two positive-curvature minima, but only the positive-x one is lowest. The negative-x minimum is metastable, and the middle stationary point remains unstable.

On a smooth stationary branch, differentiating x³ − x − h = 0 gives χ = dx/dh = 1/(3x² − 1). Both minima have positive χ, so local susceptibility alone cannot choose the globally preferred minimum. The central branch’s negative χ accompanies its negative curvature and does not describe stable unconstrained equilibrium.

At zero curvature, x = ±1/√3, the associated limiting fields are h = x³ − x = ∓2/(3√3). These are local loss-of-stability limits, not the equal-depth condition h = 0. At such a limit the finite-susceptibility formula cannot simply divide by zero. A zero quadratic term requires higher-order and domain analysis; it is not an automatic declaration of strict stability.

The plotted curves compare reduced potential values at h = 0 and 0.2 on the same x and φ scales. They do not trace a transition in time. The full quartic grows without bound at large |x|, so comparing all its stationary minima suffices for its global minimum. No nucleation time, hysteresis trajectory or physical barrier energy follows without further information.

Further: integrability also constrains caloric data

The entropy differential in T,V coordinates is dS = (C_V/T)dT + p_T dV. On a domain where the necessary third derivatives exist, equality of its mixed derivatives gives (∂C_V/∂V)_(T,n) = T(∂²p/∂T²)_(V,n). An equation of state and a caloric heat-capacity model therefore cannot be fitted independently without this consistency check.

For the smooth-fluid model p is linear in T at fixed V,n, so p_TT = 0. The assigned constant C_V is compatible. For the deliberately incompatible proposal, the left side is zero but the right side is γT, which is 6000 Pa/K at 300 K. It is exactly the same failure detected by the finite rectangular candidate integral.

There is a parallel relation in T,p coordinates: (∂C_p/∂p)_(T,n) = −T(∂²V/∂T²)_(p,n). Combining the fundamental identity with the entropy differential also gives dU = C_V dT + [T p_T − p]dV, and dH = C_p dT + [V − T Vα]dp. These are equilibrium property differentials, not arbitrary actual heat and work ledgers.

Mixed-derivative equality is a local exactness test. On a simply connected smooth domain it supplies the familiar sufficient condition for a one-form to be globally exact; on a domain with holes, global loop information can still matter. Thermodynamic state functions are defined on their actual state domain, so one should not extend a local branch through missing or singular states just to complete a path.

Further: relaxing an internal coordinate changes response

Suppose a smooth constrained potential Ψ(z₁,z₂,x) contains an internal coordinate x. At a nondegenerate local minimum, Ψ_x = 0 and Ψ_xx > 0. The relaxed equilibrium coordinate x*(z₁,z₂) changes when a control changes. Differentiating stationarity gives ∂x*/∂z_i = −Ψ_xi/Ψ_xx.

For the relaxed branch Ψ_eq(z) = Ψ(z,x*(z)), its first derivative is Ψ_i, because Ψ_x = 0. Its second derivative is Ψ_eq,ij = Ψ_ij − Ψ_ix Ψ_xj/Ψ_xx, evaluated at the stationary point. This is the one-coordinate Schur complement. It remains symmetric where the branch is smooth, but is generally different from the curvature with x artificially frozen.

For Φ = E₀φ(x;h), the fixed-x second derivative Φ_hh is zero, while Φ_hx = −E₀ and Φ_xx = E₀(3x² − 1). Relaxation gives d²Φ_eq/dh² = −E₀/(3x² − 1) = −E₀χ. The positive susceptibility of a stable local minimum corresponds to concavity of the minimized potential in its applied field.

If Ψ_xx vanishes, this division and the ordinary implicit-function argument fail. If the globally selected branch switches between equal minima, the minimum envelope can have a kink even though each constrained branch is smooth. Use branch derivatives and appropriate limiting descriptions; neither ordinary second derivatives across a kink nor a frozen-coordinate response replaces the relaxed equilibrium response.

Further: extensivity leaves a zero-curvature direction

In additive bulk thermodynamics, scaling the entire amount and extensive coordinates scales U: U(λS,λV,λn) = λU(S,V,n). Euler’s theorem gives U = TS − pV + μn for a pure simple material, where μ is chemical potential per mole. This amount term is absent from the earlier fixed-n differentials because dn = 0 there.

Differentiating the degree-one homogeneity relation shows that the full Hessian of U(S,V,n) has the radial vector (S,V,n) as a zero direction. The extensive surface is therefore convex with a scaling degeneracy rather than strictly positive definite in every extensive direction. This does not contradict the strictly positive 2×2 fixed-n Hessian: a nonzero radial rescaling changes n and is excluded from that constrained variation.

For the assigned potential, v = V/n and the reference molar v₀ stay fixed when n and V are doubled at fixed T. Consequently F, S and U double while p, α and κ_T stay unchanged. Redefining a total-volume reference each time the amount changes would obscure this test; the reference molar volume is held as one fixed constant.

Interface energy, finite-size corrections, long-range nonadditive systems and extra order parameters can change this simple homogeneity structure or require additional coordinates. The bulk fixed-composition result is not a universal claim about gravitating systems or every finite sample.

Further: fluctuations require the right ensemble

A response also has a fluctuation interpretation under specified statistical constraints. In the canonical ensemble, a fixed V,n system exchanges energy weakly with a reservoir at temperature T. With β = 1/(k_B T), differentiating its partition sum gives Var(U) = −∂⟨U⟩/∂β = k_B T² C_V, where C_V = (∂⟨U⟩/∂T)_(V,n). Nonnegative variance implies nonnegative canonical C_V when these derivatives exist.

For the stipulated 350 K fixed-volume case, C_V ≈ 1.247169 J/K gives Var(U) ≈ 2.109331e-18 J² and standard deviation σ_U ≈ 1.452354e-09 J. This is an assigned equilibrium-model prediction, not an uncertainty estimate for a calorimeter reading. An additive energy reference changes the mean but not the variance.

In an isothermal-isobaric ensemble, volume can fluctuate at fixed T,p,n. The factor exp[−β(U + pV)] instead gives Var(V) = −k_B T(∂⟨V⟩/∂p)_(T,n) = k_B T⟨V⟩κ_T, using the response of the ensemble mean. The fixed-volume canonical ensemble has no volume fluctuation at all. A microcanonical isolated system has fixed total energy; its total-energy variance is not the canonical value.

Curvature, covariance and response thus depend on which quantities are fixed, exchanged or allowed to relax. The full equilibrium distribution must be distinguished from a distribution restricted to a metastable basin. Finite systems can smooth bulk singularities, and ensemble equivalence has ordinary additive thermodynamic-limit conditions. These short connections do not replace the neighboring statistical-physics and fluctuation chapters.

Use consistency and stability in order

For a new property specification, name its material, amount, composition, work coordinates and smooth branch. Choose independent variables and carry every held coordinate through differentiation. Check mixed derivatives and units before combining separately supplied pressure, entropy and capacity data. Then test the allowed local quadratic form and finally compare global potentials under the same external controls.

A Maxwell identity does not prove stability; a positive local response does not prove integrability or select a global minimum; a zero second derivative does not settle higher-order behavior. A capacity difference, compressibility ratio or fluctuation formula also brings its own finite-response, branch and ensemble conditions.

The neighboring “Enthalpy, Helmholtz energy, and Gibbs energy” develops potential choice and thermodynamic uses in greater depth. “Chemical potential and phase equilibrium” develops material exchange and phase coexistence. “Maxwell relations and thermodynamic stability” supplies the derivative consistency and constrained-curvature tests used in those applications. Statistical entropy and phase-transition mechanisms remain linked follow-on subjects rather than hidden prerequisites.

Worked examples

Try it yourself

1. Which Maxwell relation follows from G(T,p,n), and why is its sign negative?

   Show the answer and reasoning

   (∂S/∂p)_(T,n) = −(∂V/∂T)_(p,n).

   G_T = −S and G_p = V. Equality of G_Tp and G_pT gives −S_p = V_T, with n and composition fixed.

2. Why does multiplying the opening entropy slope by the whole 0.050 L interval overestimate the assigned finite ΔS?

   Show the answer and reasoning

   The slope nR/(V − nb) decreases as V increases.

   The initial tangent stays above the concave entropy-versus-volume curve for larger V. The integral uses the changing slope at every volume.

3. Can C_p = C_V for a nonideal material with finite positive κ_T at a state where α = 0?

   Show the answer and reasoning

   Yes.

   TVα²/κ_T is zero there. Nonideal behavior alone does not require a strictly positive difference at every state.

4. Does a symmetric 2×2 curvature matrix with two positive diagonal entries necessarily define a stable quadratic form?

   Show the answer and reasoning

   No; its determinant must also be positive for strict positive definiteness.

   A sufficiently large cross term produces a direction with negative quadratic cost. Check AD − B² as well as a positive leading principal entry.

5. A smooth metastable branch satisfies all four Maxwell relations. Is that a contradiction?

   Show the answer and reasoning

   No.

   Maxwell relations test consistent mixed derivatives. A metastable minimum can have a consistent smooth state function while another accessible state has lower potential.

6. Why is constant C_V compatible with the assigned smooth-fluid pressure law but not with the deliberate quadratic-in-T proposal?

   Show the answer and reasoning

   The first has p_TT = 0; the second has p_TT = γ.

   Consistency requires (∂C_V/∂V)_(T,n) = T p_TT. A volume-independent constant C_V supplies zero on the left and cannot match γT in the second case.

7. May the 0.0600 J/K rectangle mismatch be called entropy generation?

   Show the answer and reasoning

   No.

   The proposed one-form is incompatible with an entropy state function. It was not established as the heat-entropy transfer of a physical process with a valid entropy balance.

8. Why does an extensive U(S,V,n) have a zero Hessian direction even when its fixed-n U(S,V) surface is strictly stable?

   Show the answer and reasoning

   Uniform scaling changes S,V,n together and is a degree-one homogeneity direction.

   The full radial scaling vector is annihilated by the Hessian. A fixed-n perturbation excludes that nonzero radial change, so its reduced Hessian can be positive definite.

9. In the reduced quartic model, are h = 0 and the zero-curvature limiting fields the same condition?

   Show the answer and reasoning

   No.

   h = 0 gives equal-depth global minima. The limiting fields ±2/(3√3) occur when a stationary local minimum loses nonzero curvature; susceptibility formulas then become singular.

10. Why can a fixed-x response disagree with the response after an internal coordinate relaxes?

    Show the answer and reasoning

    The equilibrium coordinate x* changes when the control changes.

    The relaxed second derivative includes the Schur-complement term −Ψ_ix Ψ_xj/Ψ_xx. Omitting it holds an internal coordinate artificially fixed.

11. Should Var(V) = k_B T⟨V⟩κ_T be assigned to a strictly fixed-volume canonical sample?

    Show the answer and reasoning

    No.

    That formula belongs to the isothermal-isobaric ensemble in which V can fluctuate. The exact fixed-volume constraint gives no volume fluctuation.

12. If n and V are doubled at fixed T with the same molar reference v₀, which of F, S, p and κ_T change?

    Show the answer and reasoning

    F and S double; p and κ_T stay unchanged in the additive model.

    The molar volume and intensive controls are unchanged. The reference v₀ is a fixed model constant, so the entropy reference does not acquire an artificial amount dependence.

What to carry forward

• Maxwell relations express equality of mixed derivatives on an appropriate smooth property branch with the other natural coordinates held fixed.
• Integrability, local stability and global equilibrium are distinct tests; none substitutes for the others.
• Response identities retain amount, counting basis, mechanical constraints and finite-derivative conditions.
• A constrained Hessian tests allowed perturbations; different potential variables have different curvature signs.
• A metastable local minimum can have positive susceptibility while another state has lower potential.
• Extensive scaling produces a zero direction in the full Hessian, and fluctuation formulas require their actual ensemble.