Thermodynamic Potentials and Stability

The potential minimized at equilibrium is fixed by what the surroundings hold constant. Legendre transforms replace an extensive control by its conjugate intensive variable, and a reservoir argument then turns the second law into the corresponding minimum principle. The same derivatives also expose measurable response coefficients and the curvature conditions for local stability.

Required background. Calculus and Linear Algebra supplies total differentials, mixed partial derivatives, and quadratic forms. Helpful background. Classical and Variational Mechanics supplies extremum reasoning; Quantum Mechanics supplies Hamiltonians and energy spectra used when these macroscopic potentials are reconstructed from partition functions on the next page.

Thermodynamic potentials and their natural variables

For a simple one-component system, take internal energy as $U(S,V,N)$ and write

$$dU=T\,dS-p\,dV+\mu\,dN.$$

Pressure carries a minus sign because positive $p\,dV$ is work done by the system. We set $k_{\mathrm B}=1$, so entropy is dimensionless while $T$ and $\mu$ have energy units. The natural derivatives are therefore $T=(\partial U/\partial S)_{V,N}$, $p=-(\partial U/\partial V)_{S,N}$, and $\mu=(\partial U/\partial N)_{S,V}$.

Define the enthalpy $H$, Helmholtz free energy $F$, Gibbs free energy $G$, and grand potential $\Omega$ by

$$H=U+pV, \qquad F=U-TS, \qquad G=U-TS+pV, \qquad \Omega=U-TS-\mu N.$$

Their differentials expose their natural variables:

$$\begin{aligned} dH&=T\,dS+V\,dp+\mu\,dN,\\ dF&=-S\,dT-p\,dV+\mu\,dN,\\ dG&=-S\,dT+V\,dp+\mu\,dN,\\ d\Omega&=-S\,dT-p\,dV-N\,d\mu. \end{aligned}$$

Each subtraction or addition removes one old differential and replaces it by the differential of the conjugate variable. For example,

$$dF=dU-T\,dS-S\,dT=-S\,dT-p\,dV+\mu\,dN.$$

This calculation, rather than memorization, fixes every sign. Tong derives $H$, $F$, $G$, and their natural variables in Tong, current notes, §4.4, printed pp. 128–132; the grand-potential transform appears in Tong, current notes, §1.4.2, printed pp. 29–30.

Which potential is extremized?

For an isolated system at fixed $(U,V,N)$, equilibrium maximizes $S$. In the energy representation the equivalent constrained statement is that equilibrium minimizes $U$ at fixed $(S,V,N)$. Reservoirs change which constrained combination appears.

Consider a system at fixed volume and particle number in contact with a much larger heat bath at temperature $T_0$. The combined system is isolated, so $\Delta U_{\mathrm{bath}}=-\Delta U$. Because the bath remains at $T_0$,

$$\Delta S_{\mathrm{tot}} =\Delta S-\frac{\Delta U}{T_0} =-\frac{\Delta(U-T_0S)}{T_0} =-\frac{\Delta F}{T_0}.$$

The second law, $\Delta S_{\mathrm{tot}}\geq0$, thus gives $\Delta F\leq0$ for a spontaneous change at fixed $(T_0,V,N)$. Equality characterizes equilibrium with respect to the allowed variations. Allowing the bath to exchange volume at fixed pressure $p_0$ replaces $F$ by $G=U-T_0S+p_0V$; allowing particle exchange at fixed chemical potential $\mu_0$ and volume replaces it by $\Omega=U-T_0S-\mu_0N$. The resulting criteria are

Fixed by the surroundings	Potential minimized at stable equilibrium
$S,V,N$	$U$
$S,p,N$	$H$
$T,V,N$	$F$
$T,p,N$	$G$
$T,V,\mu$	$\Omega$

“Minimize” means compare states compatible with the listed controls. It does not imply that a potential decreases during arbitrary external driving, nor does it determine an escape time from a metastable local minimum. A first-order transition is selected by comparing competing global minima, not by inspecting only one local curvature.

The controls must include every allowed exchange. If several particle species can react, the variation contains $\sum_a\mu_a\,dN_a$, and equilibrium constrains the stoichiometric combination of chemical potentials. If surface area, charge, magnetization, or another work coordinate changes, its conjugate pair must be added before selecting a potential. Omitting an allowed exchange produces the wrong stationary condition even when the algebraic Legendre transform is flawless.

Maxwell relations and response

Equality of mixed partial derivatives turns unmeasurable derivatives into measurable ones. From $F(T,V,N)$,

$$S=-\left(\frac{\partial F}{\partial T}\right)_{V,N}, \qquad p=-\left(\frac{\partial F}{\partial V}\right)_{T,N},$$

so

$$\left(\frac{\partial S}{\partial V}\right)_{T,N} = \left(\frac{\partial p}{\partial T}\right)_{V,N}.$$

The other potentials generate the remaining Maxwell relations. For example, $dG=-S\,dT+V\,dp+\mu\,dN$ gives

$$\left(\frac{\partial S}{\partial p}\right)_{T,N} =-\left(\frac{\partial V}{\partial T}\right)_{p,N}.$$

All four relations, with their natural variables and signs, appear in Tong, current notes, §4.4.2, printed pp. 131–132.

Curvature and local stability

For an ordinary additive phase on a positive-temperature, positive-volume stable branch with $C_V\gt 0$ and $\kappa_T\gt 0$, entropy is concave in its extensive variables and $U(S,V,N)$ is locally convex. In the mixed natural variables of $F(T,V,N)$, different directions have different signs:

$$\frac{\partial^2 F}{\partial T^2}=-\frac{C_V}{T}\leq0, \qquad \frac{\partial^2 F}{\partial V^2} =-\left(\frac{\partial p}{\partial V}\right)_T =\frac{1}{V\kappa_T}\geq0,$$

where $C_V=T(\partial S/\partial T)_{V,N}$ and $\kappa_T=-(1/V)(\partial V/\partial p)_{T,N}$. Thus $F$ is concave in the intensive variable $T$ and convex in the extensive variable $V$. Saying only that “free energy is convex” is incomplete.

At fixed $N$, a pure extensive representation makes the full matrix test explicit:

$$\det\nabla^2_{S,V}U =\frac{T}{VC_V\kappa_T}\gt 0.$$

Together with $U_{SS}=T/C_V\gt 0$, this is the positive-definite Hessian criterion for a locally stable phase. In the mixed variables $(T,V)$, the corresponding determinant is deliberately negative,

$$\det\nabla^2_{T,V}F =-\frac{C_p}{TV\kappa_T}\lt 0,$$

so the $F$ Hessian is indefinite even in a stable phase. With more coupled variables, the appropriate principal minors or constrained quadratic form must be checked; diagonal entries alone can miss an unstable mixed direction.

The curvature condition has a direct exchange interpretation. Let two weakly coupled subsystems trade a small energy $\delta U$ while their total energy remains fixed. Their total entropy is

$$S_{\mathrm{tot}}(\delta U) =S_A(U_A+\delta U)+S_B(U_B-\delta U)$$

The first derivative vanishes only when $1/T_A=1/T_B$. At that common temperature, the stationary point is a stable maximum when

$$\frac{d^2S_{\mathrm{tot}}}{d(\delta U)^2} =-\frac{1}{T^2}\left(\frac{1}{C_A}+\frac{1}{C_B}\right)\lt 0.$$

Positive heat capacities guarantee this condition for ordinary subsystems, but they are sufficient rather than logically necessary for a finite pair: one negative heat capacity can be stabilized if the other subsystem is small enough that the displayed sum remains positive. An effectively infinite canonical bath has $C_B\to\infty$, so it cannot stabilize $C_A\lt 0$. This is one reason a self-gravitating system cannot be diagnosed by quoting a heat capacity without naming the ensemble.

Exercise: compare the heat capacities

Define

$$\alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{p,N}, \qquad \kappa_T=-\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{T,N}.$$

Show that $C_p-C_V=TV\alpha^2/\kappa_T$, and check the result for a dilute classical ideal gas.

Solution

Write entropy as $S(T,V)$ at fixed $N$. The chain rule gives

$$C_p-C_V =T\left(\frac{\partial S}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_p.$$

The Maxwell relation replaces the first derivative by $(\partial p/\partial T)_V$. Holding $p$ fixed gives

$$0=dp=\left(\frac{\partial p}{\partial T}\right)_VdT +\left(\frac{\partial p}{\partial V}\right)_T dV,$$

hence $(\partial p/\partial T)_V=\alpha/\kappa_T$. Therefore

$$\boxed{C_p-C_V=\frac{TV\alpha^2}{\kappa_T}\geq0}$$

for a locally stable phase with $T\gt 0$ and $\kappa_T\gt 0$. The difference can vanish when $\alpha=0$.

For a dilute classical ideal gas, $\alpha=1/T$ and $\kappa_T=1/p=V/(NT)$. The result reduces to $C_p-C_V=N$, which checks both the sign and extensivity. A negative result would contradict the assumed local canonical stability rather than reveal a new dilute-gas phase.

Limits and onward use

At a critical point a susceptibility can diverge; in an extensive representation this corresponds to a vanishing eigenvalue of $\nabla^2_{S,V}U$, while a second derivative with respect to a conjugate intensive source can diverge. Finite and nonadditive systems can violate familiar concavity assumptions, while constraints can remove otherwise unstable directions. Statistical Ensembles and Partition Functions derives the same potentials from microscopic probability laws and makes the conditions for ensemble equivalence explicit.

References

• Tong, David. Statistical Physics. University of Cambridge Part II Mathematical Tripos lecture notes, current official undated-cover PDF, §1.4.2, printed pp. 29–30, equations (1.44)–(1.45), and §4.4, printed pp. 128–132, especially equations (4.19)–(4.26). Open PDF.