Response Functions, Stability, and Convexity
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the definitions of the thermodynamic potentials (U, H, F, G) and have reviewed the derivation of the Maxwell relations in the previous lecture.
Why this matters
Thermodynamics is not merely a bookkeeping of energy; it is a theory of stability.
While the first law tells us how energy is conserved, it does not tell us which states are physically realizable. A system with negative heat capacity would spontaneously explode into hotter and colder regions, violating the very concept of equilibrium. In this lecture, we establish the mathematical constraints that distinguish a stable physical system from a mathematical curiosity. We will see that the requirement for stability imposes strict inequalities on measurable quantities like compressibility and heat capacity, and that these constraints are deeply linked to the convexity of the thermodynamic potentials.
The Intuition of Stability
Imagine a ball resting at the bottom of a valley. If you nudge it slightly, it rolls back towards the centre. This is a stable equilibrium. Now imagine a ball balanced on top of a hill. The slightest perturbation sends it rolling away. This is an unstable equilibrium. Thermodynamic systems behave similarly. A stable system must resist spontaneous fluctuations. If a small part of the system tries to change its temperature or volume, the system must respond in a way that restores the original state. This "restoring force" is what we quantify through response functions. If the response is negative, the system amplifies the fluctuation rather than damping it, leading to instability. Therefore, the sign of our response functions is not arbitrary; it is a fundamental requirement of nature.
Stability and the Second Law
The rigorous foundation for stability comes from the second law of thermodynamics. For an isolated system, the total entropy \( S \) must be maximized at equilibrium. Consider a system divided into two parts. If we allow a small internal fluctuation where energy \( \delta U \) is transferred from part 1 to part 2, the change in total entropy is \( \delta S_{\text{tot}} = \delta S_1 + \delta S_2 \). For the equilibrium state to be stable, this change must be negative for any spontaneous fluctuation, meaning the entropy must be a local maximum.
Expanding the entropy in a Taylor series around the equilibrium point, the first-order term vanishes due to the equilibrium conditions (equal temperatures and pressures). The stability condition is therefore determined by the second-order term. This leads to the requirement that the curvature of the entropy surface must be negative. Mathematically, this translates to the positivity of the heat capacity at constant volume, \( C_V \). If \( C_V \) were negative, a small increase in temperature would lead to a decrease in energy, causing the system to absorb more energy and heat up further, a runaway process. Thus, stability demands \( C_V > 0 \).
Compressibility and Mechanical Stability
Similarly, we can analyse mechanical stability by considering volume fluctuations at constant temperature and particle number. The relevant potential is the Helmholtz free energy \( F(T,V) \). For stability, \( F \) must be a minimum with respect to volume fluctuations. This requires the second derivative of \( F \) with respect to \( V \) to be positive:
Using the thermodynamic identity \( dF = -S\,dT - P\,dV \), we can relate this curvature to the pressure. The first derivative is \( (\partial F / \partial V)_T = -P \). The second derivative is \( (\partial^2 F / \partial V^2)_T = -(\partial P / \partial V)_T \). Therefore, the stability condition becomes \( -(\partial P / \partial V)_T > 0 \), or equivalently \( (\partial P / \partial V)_T < 0 \). This means that if you increase the volume, the pressure must decrease. This is the intuitive definition of a stable fluid: it resists expansion. We define the isothermal compressibility \( \kappa_T \) as:
Since \( (\partial P / \partial V)_T < 0 \), the inverse \( (\partial V / \partial P)_T \) is also negative, making \( \kappa_T \) strictly positive. A negative compressibility would imply that increasing pressure causes the volume to increase, leading to mechanical instability.
Convexity of Thermodynamic Potentials
The stability criteria can be generalised using the concept of convexity. A function is convex if the line segment between any two points on the graph lies above the graph. For thermodynamic potentials expressed in terms of extensive variables (like \( U(S,V) \)), stability requires the function to be convex. This ensures that mixing two states of the same system results in a lower energy than the weighted average, preventing phase separation in the bulk.
Conversely, when potentials are expressed in terms of intensive variables (like \( F(T,V) \) or \( G(T,P) \)), the functions must be concave with respect to the intensive variables. For example, the Helmholtz free energy \( F(T,V) \) is concave in \( T \) because \( C_V = -T (\partial^2 F / \partial T^2)_V > 0 \) implies \( (\partial^2 F / \partial T^2)_V < 0 \). This convexity/concavity structure is not just a mathematical nicety; it guarantees that the equilibrium state is unique and stable against all possible fluctuations.
The derivation
Stability, Convexity, and Positive Response Functions
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.
It is a common misconception that all response functions must be positive. While \( C_V \) and \( \kappa_T \) are strictly positive for stable systems, other quantities like the thermal expansion coefficient \( \alpha \) or the Joule-Thomson coefficient can be negative. The stability constraints only apply to the "curvature" terms (second derivatives) that define the restoring forces, not to all first-derivative response functions. C
Worked example
Problem. Consider a simple fluid with an equation of state given by \( P = \frac{nRT}{V - nb} - \frac{a n^2}{V^2} \), where \( a \) and \( b \) are positive constants. Determine the condition on temperature \( T \) for the system to be mechanically stable at a given volume \( V \).
- Recall the stability condition for mechanical stability: \( \left( \frac{\partial P}{\partial V} \right)_T < 0 \).
- Calculate the derivative of the given equation of state with respect to \( V \) at constant \( T \): \[ \left( \frac{\partial P}{\partial V} \right)_T = -\frac{nRT}{(V - nb)^2} + \frac{2 a n^2}{V^3} \]
- Apply the inequality \( \left( \frac{\partial P}{\partial V} \right)_T < 0 \): \[ -\frac{nRT}{(V - nb)^2} + \frac{2 a n^2}{V^3} < 0 \]
- Rearrange to solve for \( T \): \[ \frac{nRT}{(V - nb)^2} > \frac{2 a n^2}{V^3} \] \[ T > \frac{2 a n (V - nb)^2}{R V^3} \]
Units check. \(a n^2/V^2\) is a pressure, so \(a\) has units Pa m6 mol-2; then \(a n (V-nb)^2/(R V^3)\) has units (Pa m6 mol-2)(mol)(m6)/((J mol-1 K-1)(m9)) = Pa m3 K/J = K, as a temperature must.
Worked number: for carbon dioxide (\(a = 0.364\ \mathrm{Pa\,m^6\,mol^{-2}}\), \(b = 4.27 \times 10^{-5}\ \mathrm{m^3\,mol^{-1}}\), \(n = 1\)) at \(V = 3b = 1.28 \times 10^{-4}\ \mathrm{m^3}\), the bound is \(T > 2a(V-b)^2/(RV^3) = 304\ \mathrm{K}\), which is exactly the van der Waals critical temperature \(8a/27bR\): below it, the isotherm has an unstable region at this volume.
Three quick tests
- Why must the isothermal compressibility \( \kappa_T \) be positive for a stable system?
Answer
Positive compressibility ensures that an increase in pressure leads to a decrease in volume, providing a restoring force against mechanical fluctuations. - If \( C_V < 0 \), what happens to a small fluctuation in temperature?
Answer
The fluctuation grows, leading to a runaway process where the system separates into hotter and colder regions, indicating instability. - Is the internal energy \( U(S,V) \) convex or concave for a stable system?
Answer
Convex. Stability requires the entropy \( S(U,V) \) to be concave, which implies \( U(S,V) \) is convex.
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.