--- title: Response Functions, Stability, and the Third Law module: Thermodynamics moduleNumber: 1 lessonNumber: 5 order: 105 summary: > Response functions — heat capacities, compressibilities, thermal expansion — are the second derivatives of the potentials and the quantities an experiment actually measures. This lesson derives the general relation $C_P-C_V=TV\alpha^2/\kappa_T$, shows that convexity of the potentials forces the stability conditions $C_V>0$ and $\kappa_T>0$, and states the third law: entropy approaches a constant as $T\to0$, so heat capacities and expansion coefficients vanish there and absolute zero is unattainable. topics: [Thermodynamics] sources: - book: Callen ref: "Ch. 3 §3.9 Heat Capacities, Ch. 8 — Stability, and Ch. 10 — The Nernst Postulate" - book: Reif ref: "Ch. 5 §5.8, Ch. 8 §8.6–8.9" - book: Pathria & Beale ref: "Ch. 1 §1.4, Ch. 4 §4.7" - book: Kardar (Statistical Physics of Particles) ref: "Ch. 1 — Thermodynamics; §1.9" draft: false --- The thermodynamic potentials are known only through their derivatives. First derivatives give the equation of state and the entropy; second derivatives give the **response functions** — how much the system's volume, temperature, or energy moves when a field is changed. These are the numbers a laboratory reports: a heat capacity, a compressibility, an expansion coefficient. The Maxwell relations tie them together so that a few suffice, convexity of the potentials constrains their signs, and the third law fixes their common limit as temperature approaches absolute zero. ## The response functions Three second-derivative quantities describe a simple fluid, together with the heat capacities already defined. - **Isobaric thermal expansion coefficient** $$\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P,$$ the fractional volume change per unit temperature at fixed pressure. - **Isothermal compressibility** $$\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T,$$ the fractional volume decrease per unit pressure at fixed temperature. The minus sign makes it positive, since volume falls as pressure rises. - **Adiabatic compressibility** $$\kappa_S = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_S,$$ the same at fixed entropy. - **Heat capacities** $C_V=T(\partial S/\partial T)_V$ and $C_P=T(\partial S/\partial T)_P$, using $\d S=\delta Q_{\rm rev}/T$. For an ideal gas, $\alpha = 1/T$, $\kappa_T = 1/P$, $C_V=\tfrac{3}{2}Nk_B$ (monatomic), and $C_P-C_V=Nk_B$. These four functions, measured over the accessible range of temperature and pressure, determine the full thermodynamics when combined with the Maxwell relations. ## Relations among the response functions The four response functions are not independent. Two identities, both consequences of the Maxwell relations, connect them. The **difference of heat capacities** follows from writing $S(T,V)$ and $S(T,P)$ and using the Helmholtz Maxwell relation. Start from $C_P-C_V = T[(\partial S/\partial T)_P-(\partial S/\partial T)_V]$. Expanding $(\partial S/\partial T)_P$ with $V$ as an intermediate variable and applying $(\partial S/\partial V)_T=(\partial P/\partial T)_V$ gives $$ C_P - C_V = T\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial V}{\partial T}\right)_P = -T\frac{(\partial V/\partial T)_P^2}{(\partial V/\partial P)_T} = \frac{TV\alpha^2}{\kappa_T}, $$ where the middle step used the triple product rule $(\partial P/\partial T)_V=-(\partial V/\partial T)_P/(\partial V/\partial P)_T$. The result is general — it holds for any substance, not only an ideal gas. Because $T$, $V$, and $\alpha^2$ are non-negative and $\kappa_T>0$ (established below), $C_P\ge C_V$ always, with equality only where the expansion coefficient vanishes. For an ideal gas $TV\alpha^2/\kappa_T = TV(1/T)^2/(1/P) = PV/T = Nk_B$, recovering the Mayer relation. $$ % caption: The gap $C_P-C_V=TV\alpha^2/\kappa_T$ is assembled from three measured response functions; it is non-negative because $\kappa_T>0$, so constant-pressure heating always takes more heat than constant-volume heating. \begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0] \definecolor{acc}{HTML}{4A6FA5}\node[draw,thick,minimum width=2.4cm,minimum height=1.0cm] (a) at (0,1.6) {$\alpha^2$}; \node[draw,thick,minimum width=2.4cm,minimum height=1.0cm] (k) at (0,0) {$\kappa_T$}; \node[draw,acc,thick,fill=acc!10,minimum width=3.0cm,minimum height=1.4cm] (d) at (5.2,0.8) {$C_P{-}C_V=\dfrac{TV\alpha^2}{\kappa_T}$}; \draw[->,black,thick] (a)--(d); \draw[->,black,thick] (k)--(d); \node[black,above] at (2.6,1.75) {expansion}; \node[black,below] at (2.6,-0.15) {compressibility}; \end{tikzpicture} $$ The **ratio of compressibilities** equals the ratio of heat capacities, $$ \frac{\kappa_T}{\kappa_S} = \frac{C_P}{C_V} = \gamma, $$ so an adiabatic compression is stiffer than an isothermal one by the factor $\gamma$. This is why the speed of sound, which is an adiabatic response, $c=\sqrt{1/\rho\kappa_S}$, exceeds the isothermal estimate Newton first computed; the correction that reconciles them is Laplace's factor $\sqrt{\gamma}$. The two identities leave only three of the five response functions independent. A convenient basis is $\{C_V,\alpha,\kappa_T\}$: from them $C_P$ follows by $C_P=C_V+TV\alpha^2/\kappa_T$, and $\kappa_S$ by $\kappa_S=\kappa_T C_V/C_P$. Every second derivative of the potentials is then a combination of these three plus the equation of state. $$ % caption: The five response functions reduce to three independent ones; $C_V$, $\alpha$, and $\kappa_T$ generate $C_P$ through the heat-capacity identity and $\kappa_S$ through the compressibility ratio. \begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0] \definecolor{acc}{HTML}{4A6FA5}\node[draw,acc,thick,fill=acc!10,minimum size=0.9cm] (cv) at (0,2.4) {$C_V$}; \node[draw,acc,thick,fill=acc!10,minimum size=0.9cm] (al) at (0,1.2) {$\alpha$}; \node[draw,acc,thick,fill=acc!10,minimum size=0.9cm] (kt) at (0,0) {$\kappa_T$}; \node[draw,black,thick,fill=black!8,minimum size=0.9cm] (cp) at (5.0,1.8) {$C_P$}; \node[draw,black,thick,fill=black!8,minimum size=0.9cm] (ks) at (5.0,0.4) {$\kappa_S$}; \draw[->,black,thick] (cv)--(cp); \draw[->,black,thick] (al)--(cp); \draw[->,black,thick] (kt)--(cp); \draw[->,black,thick] (cv)--(ks); \draw[->,black,thick] (kt)--(ks); \node[acc,below] at (0,-0.75) {independent}; \node[black,below] at (5.0,-0.75) {derived}; \end{tikzpicture} $$ A throttling process shows the response functions at work in a real measurement. > **Worked example.** In the Joule–Thomson (porous-plug) experiment a gas is > forced steadily through a constriction from high to low pressure with insulated > walls, a process that holds the enthalpy constant. The temperature change per > unit pressure drop is the Joule–Thomson coefficient > $\mu_{\rm JT}=(\partial T/\partial P)_H$. Expanding $H(T,P)$ and using > $(\partial H/\partial P)_T=V-T(\partial V/\partial T)_P$ (from the Gibbs Maxwell > relation) gives > $$ > \mu_{\rm JT}=-\frac{1}{C_P}\left(\frac{\partial H}{\partial P}\right)_T > =\frac{V}{C_P}\bigl(T\alpha-1\bigr). > $$ > For an ideal gas $\alpha=1/T$, so $\mu_{\rm JT}=0$ and throttling produces no > temperature change. A real gas cools ($\mu_{\rm JT}>0$) when $T\alpha>1$ and > warms when $T\alpha<1$; the locus $T\alpha=1$ is the **inversion curve**. > Precooling a gas below its inversion temperature is what makes throttling > useful for liquefaction, and the whole effect is carried by two response > functions, $\alpha$ and $C_P$. ## Convexity and thermodynamic stability Equilibrium is not merely an extremum of the appropriate potential but a **minimum**, and the minimum condition constrains the second derivatives. An equilibrium that failed the second-order test would be unstable: a small fluctuation would lower the potential and grow, and the homogeneous phase would split. Consider a system divided into two equal halves that exchange energy at fixed total. Entropy is maximized, so moving energy $\Delta E$ from one half to the other must not raise the total entropy. Expanding $S$ to second order, the first-order terms cancel at equilibrium (equal temperatures) and the second-order term gives $(\partial^2 S/\partial E^2)_V\le 0$, equivalently $$ C_V \ge 0. $$ A negative heat capacity would mean a region that gained energy grew colder, drawing still more energy from its neighbor — a runaway. The same argument applied to volume exchange at fixed temperature, using the convexity of the Helmholtz free energy in $V$, gives $$ \kappa_T \ge 0. $$ A negative compressibility would mean a region that expanded pushed harder, expanding further. These **stability conditions** are the thermodynamic content of the convexity of the potentials: $U$ is convex in each extensive variable, and each free energy is convex in its extensive natural variables and concave in its intensive ones. $$ % caption: A stable free energy is convex, lying above every tangent line; a non-convex bulge is unstable, and the common tangent (dashed) replaces it with a two-phase mixture at lower free energy. \begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0] \definecolor{acc}{HTML}{4A6FA5}\draw[->,black] (0,0)--(6.4,0) node[right,black]{$V$}; \draw[->,black] (0,0)--(0,4.2) node[above,black]{$F$}; \draw[acc,very thick] plot[domain=0.6:5.8,samples=120] (\x,{2.6 - 0.9*(\x-3.2) + 0.16*(\x-3.2)*(\x-3.2) + 0.9*sin((\x-3.2)*60)}); \draw[black,very thick,dashed] (1.75,1.63)--(4.55,1.05); \filldraw[black] (1.75,1.63) circle (2pt); \filldraw[black] (4.55,1.05) circle (2pt); \node[black] at (3.15,2.05) {common tangent}; \node[acc] at (1.15,3.0) {convex branch}; \node[black] at (4.9,2.4) {convex branch}; \end{tikzpicture} $$ Where the potential is non-convex, the homogeneous state is unstable and the system separates into two coexisting phases whose states are the two tangent points of the **common-tangent construction**. The straight tangent lies below the bulge, so the mixture has lower free energy than any homogeneous state in that range. This is the thermodynamic origin of phase coexistence, taken up quantitatively in the phase-transition module; here it is the signal that a stability condition has been violated. ## The third law The first and second laws leave the entropy defined only up to an additive constant, since only differences $\Delta S=\int\delta Q_{\rm rev}/T$ are measured. The third law fixes the constant. > **Third law (Nernst statement).** The entropy change of any isothermal, > reversible process approaches zero as the temperature approaches absolute zero: > $\lim_{T\to 0}\Delta S = 0$. Equivalently (Planck's stronger form), the entropy > of every system approaches the same constant, taken to be zero, as $T\to 0$: > $$\lim_{T\to 0} S = 0.$$ Statistically the law is transparent: $S=k_B\ln\Omega$, and as $T\to 0$ a system settles into its ground state, whose multiplicity $\Omega$ is one (or a small degeneracy contributing a negligible $\ln\Omega$ per particle). The entropy of the ground state is therefore zero, or vanishes per particle in the thermodynamic limit. Nernst reached the conclusion thermodynamically, from measurements of low-temperature reactions, before the statistical picture confirmed it. Two consequences follow immediately. **Heat capacities vanish at absolute zero.** Since $S(T)=S(0)+\int_0^T (C/T')\,\d T'$ must stay finite as $T\to 0$, the integrand cannot diverge, so $C\to 0$. Both $C_V$ and $C_P$ approach zero as $T\to 0$. This contradicts the classical equipartition prediction of a constant heat capacity, and its resolution — the freezing-out of degrees of freedom once $k_BT$ drops below the level spacing — is the recurring theme of the quantum-statistics modules. The Einstein and Debye models, the electronic $C\propto T$ of a metal, and the $C\propto T^3$ of a phonon gas all embody the third-law vanishing. $$ % caption: The classical equipartition heat capacity stays constant down to $T=0$ (dashed), violating the third law; the real heat capacity falls to zero as $T\to0$ because degrees of freedom freeze out below their level spacing. \begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0] \definecolor{acc}{HTML}{4A6FA5}\draw[->,black] (0,0)--(6.2,0) node[right,black]{$T$}; \draw[->,black] (0,0)--(0,4.0) node[above,black]{$C$}; \draw[black,thick,dashed] (0,3.0)--(6.0,3.0); \node[black,above right] at (3.4,3.0) {classical constant}; \draw[acc,very thick] plot[domain=0.05:6.0,samples=120] (\x,{3.0*(\x*\x)/(\x*\x+1.1)}); \node[acc] at (4.2,1.7) {real $C(T)$}; \filldraw[black!70] (0,0) circle (2pt) node[below right,black]{$C\to0$}; \end{tikzpicture} $$ **Thermal expansion vanishes at absolute zero.** By a Maxwell relation $\alpha=\frac{1}{V}(\partial V/\partial T)_P=-\frac{1}{V}(\partial S/\partial P)_T$, and since $S\to$ const independent of $P$ as $T\to 0$, the derivative $(\partial S/\partial P)_T\to 0$, so $\alpha\to 0$. **Absolute zero is unattainable.** Reaching $T=0$ would require removing the last increment of entropy, but with $S(T\to0)$ the same constant along every isotherm, no finite sequence of reversible isothermal and adiabatic steps can cross to $T=0$: each adiabatic step lowers $T$ by less as the isotherms crowd together near $T=0$. The unattainability of absolute zero is equivalent to the Nernst statement and is the practical face of the third law in adiabatic-demagnetization cooling, analyzed in the paramagnetism lesson. ## Summary - Response functions are second derivatives of the potentials: thermal expansion $\alpha$, isothermal and adiabatic compressibilities $\kappa_T,\kappa_S$, and the heat capacities $C_V,C_P$. They are what experiments measure. - Maxwell relations connect them: $C_P-C_V=TV\alpha^2/\kappa_T\ge 0$ (general, reducing to $Nk_B$ for an ideal gas) and $\kappa_T/\kappa_S=C_P/C_V=\gamma$, the latter fixing the adiabatic speed of sound. - Convexity of the potentials is thermodynamic stability: $C_V\ge 0$ and $\kappa_T\ge 0$. A non-convex free energy is unstable and phase-separates by the common-tangent construction. - The third law sets $S\to 0$ as $T\to 0$, forcing heat capacities and thermal expansion to vanish there and making absolute zero unattainable.