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A periodic array of Abrikosov vortices. In the isotropic local theory near the upper critical magnetic field, the triangular array has a lower Helmholtz free energy than the square array.^[ref]

The useful distinction is between vortex existence, array geometry, and vortex mobility. Quantized winding does not force every array to be triangular or defect-free. A stationary array and vortices moving under an applied electric current also have different transport consequences; flux pinning matters for carrying useful electric currents.

The translational Helmholtz free energy cost of restricting a dilute ideal gas from a volume $V$ to an area $A$ at the same temperature. For identical particle number and mass,Here $\Lambda$ is the thermal de Broglie wavelength. This contribution is positive for $V/A\gg \Lambda$; a favourable binding energy must be added separately to predict net adsorption.

At chemical equilibrium, a reversible chemical reaction has equal forward and reverse reaction rates, so its macroscopic composition stays constant. Individual molecules can keep changing identity: the reaction has not stopped. This dynamic balance is the definition described by the ^[ref]. Equal reaction rates also need not mean equal concentrations. A small population converting rapidly can balance a large population converting slowly.

This distinction gives two useful questions. Which composition is favored at the specified temperature and other constraints? Thermodynamics answers through chemical potentials and an equilibrium constant. How quickly does a prepared mixture approach that composition? Chemical kinetics answers through reaction rates. Knowing the destination does not determine the journey's duration. A composition that barely changes during observation may reflect extremely slow chemical kinetics, so constancy alone does not establish chemical equilibrium.

Consider the idealized elementary chemical reactionAssume a well-mixed closed system at fixed temperature and volume, with first-order law of mass action rates in both directions and no competing chemical reactions. Write $a=[\mathrm{A}]$ and $b=[\mathrm{B}]$ for concentrations in $\mathrm{M}=\mathrm{mol}/\mathrm{L}$. The total $a+b=C=1\mathrm{M}$ is conserved because each conversion replaces one molecule by one molecule. This conservation is special to the chosen stoichiometry.

The forward and reverse reaction rates, measured as conversions per unit volume per unit time, are $J_f=k_{f}a$ and $J_r=k_{r}b$. HenceSetting this derivative to zero givesWith $a(0)=1\mathrm{M}$ and $b(0)=0$, solving the ordinary differential equation yieldsAt $t=5\mathrm{s}$, these are approximately $0.3515\mathrm{M}$ and $0.6485\mathrm{M}$. Initially $J_f=0.3M/s$ and $J_r=0$. At chemical equilibrium, both are $0.075M/s$, while the net reaction rate $J_f-J_r$ is zero. The persistent exchange is visible in the lower panel even after the upper panel has nearly flattened.

Why does chemical equilibrium also involve a free energy minimum? For a closed system at fixed temperature and pressure, without imposed non-expansion work, stable equilibrium minimizes Gibbs free energy over compositions allowed by conserved elemental inventories. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19. This is constrained Gibbs minimization for chemical equilibrium: freely minimizing over unrelated species amounts would change the inventory. At fixed temperature and volume, the appropriate potential is Helmholtz free energy. In either setting, varying an allowed extent of reaction gives the corresponding free energy slope ${\sum }_i{\nu }_i{\mu }_i$, with molar chemical potentials and signed stoichiometric vector entries. An interior equilibrium has zero slope; stability requires a minimum. For A converting to B, this says ${\mu }_{\mathrm{B}}={\mu }_{\mathrm{A}}$.

The beauty of this connection is that molecular exchange, a macroscopic composition, and a geometric minimum describe the same balance. In the ideal concentration model, let $x=b/C$ and choose the standard molar chemical potential difference consistent with $K=3$. Up to a composition-independent constant, the fixed-volume Helmholtz free energy then obeyswhere $n$ is the conserved total amount of substance and $R$ the molar gas constant. Its derivative is $\ln [x/(1-x)]-\ln 3$, vanishing at $x=3/4$. Its second derivative is $1/x+1/(1-x)>0$ inside $0<x<1$, so this balance is the unique minimum. The mixing entropy terms explain why the favored composition still contains A.

Rigid compartments impose a different constraint: fixed volume rather than fixed pressure. With Helmholtz free energy $F=U-TS$,This follows immediately by differentiating $F=U-TS$ in the chemical potential differential above. It expresses the same chemical potential through the thermodynamic potential appropriate to fixed temperature and volume.

Consider a closed system consisting of two rigid compartments of equal volume $V$, maintained at a common fixed temperature $T$. A valve permits exchange of the same ideal gas; the partition stays in place. Initially $n_A=9\mathrm{mol}$ and $n_B=3\mathrm{mol}$, with conserved total amount of substance $n_{\mathrm{tot}}=12\mathrm{mol}$. These are macroscopic moles, modeled continuously, not twelve individual particles. By the ideal gas law, $p_B/p_A=n_B/n_A=1/3$, so initiallyAt $300\mathrm{K}$ with the stipulated $R$, this is approximately $-2740.311\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, favoring transfer from A to B.

At fixed $T$ and equal $V$, integrating the ideal-gas chemical potential gives the total Helmholtz free energy, up to a term independent of the allocation,Here $n_{\mathrm{ref}}$ is a positive reference amount of substance; any omitted common term linear in $n_A+n_B$ is constant along this exchange. Let $x=n_A/n_{\mathrm{tot}}$ be the dimensionless fraction in A. The relative Helmholtz free energy then takes the particularly simple formIts derivative along the allowed transfer and its second derivative areThus the Helmholtz free energy is a strictly convex function of the allocation for positive amounts. Its unique minimum has $x=1/2$, or $n_A=n_B=6\mathrm{mol}$, and equal chemical potentials. Initially its excess above this minimum is $n_{\mathrm{tot}}RT[\frac{3}{4}\ln (3/2)+\frac{1}{4}\ln (1/2)]\simeq 3915.474\mathrm{J}$ at $300\mathrm{K}$. Relaxation decreases the total Helmholtz free energy by this amount in the model; the final pressures are also equal because both compartments have the same volume.

The beautiful connection is that the direction of matter exchange and the slope of a free energy curve are the same calculation. The entropy gained by redistributing the ideal gas makes equal allocation favorable here. As with chemical equilibrium, a free energy minimum identifies the favored state, while transport properties determine how quickly it is reached. The fixed-temperature, fixed-pressure Gibbs free energy criterion is summarized in 16.4 Free Energy; the rigid experiment instead uses Helmholtz free energy.

For a pure ideal gas at temperature $T$ and pressure $p$,where $R$ is the molar gas constant and $p^{\circ }$ specifies the reference pressure of the standard state. The ratio inside the natural logarithm is dimensionless. The chemical potential difference therefore depends on a ratio of pressures, not an absolute difference. See MIT 5.60, Lecture 14, page 4.

For a chemical species in an ideal gas mixture, replace $p$ by its partial pressure. More generally write ${\mu }_{\mathrm{m},i}={\mu }_{\mathrm{m},i}^{\circ }+RT\ln a_i$ using its thermodynamic activity $a_i$ and a consistent standard state. For an ideal dilute chemical solution, $a_i=c_i/c^{\circ }$ with reference concentration $c^{\circ }$; the thermodynamic activity section describes departures from this idealization.

As a computed comparison, take identical chemical species at $T=300\mathrm{K}$, with $p_A=2p^{\circ }$ and $p_B=p^{\circ }$. Stipulate $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. ThenHalving the pressure gives the same decrement whether starting from $2p^{\circ }$ or $20p^{\circ }$, provided the ideal gas approximation remains valid. This comparison uses different pressures; it is not the common-pressure reservoir setup of chemical potential equality for particle exchange. The fixed-volume example in chemical potential and Helmholtz free energy supplies an exchange experiment with unequal initial pressures.

The chemical relaxation time estimates how rapidly a perturbed abundance returns toward chemical equilibrium. Comparing it with the eddy mixing time identifies a chemical quench level.

For the A-to-B model in chemical equilibrium, deviations decay by exponential decay with ${\tau }_{\mathrm{chem}}=1/(k_f+k_r)=2.5\mathrm{s}$. After one such interval, $1/e$ of the initial deviation remains; reaching one percent takes $\ln (100){\tau }_{\mathrm{chem}}\simeq 11.51\mathrm{s}$. Multiplying both rate constants by two preserves their ratio and the equilibrium composition, but halves ${\tau }_{\mathrm{chem}}$ to $1.25\mathrm{s}$. The equilibrium exchange rates double to $0.15M/s$. Faster exchange therefore coexists with the same final composition and the same free energy minimum.

A coarse-grained variable retains selected macroscopic information while averaging over unresolved microscopic states. At thermal equilibrium, the statistical multiplicity of those states contributes entropy to the effective Helmholtz free energy of the retained variable.

For a system coupled to a thermal reservoir at fixed temperature $T$, prepare the forward process in the initial canonical ensemble and the reverse process in the final control parameter's canonical ensemble. Reverse the protocol, assume microscopic reversibility and matching path probability support, and count $W$ as work done on the system. With Helmholtz free energy difference $\Delta F=F_{\mathrm{final}}-F_{\mathrm{initial}}$ and $\beta =1/(k_{B}T)$,The final forward distribution need not already be in thermal equilibrium. Its reverse preparation is prescribed equilibrium preparation, rather than the actual forward final distribution used in finite trajectory entropy production. Thus the paired path logarithm $\beta (W-\Delta F)$ is dissipated work in thermal units, not automatically the instantaneous total stochastic entropy production at the end of an arbitrary driven protocol. ^{[ref] [ref]}

An entropic force is the part of a generalized force arising from the dependence of entropy on a coordinate. At fixed temperature, the system's force is $-\partial A/\partial x=-\partial U/\partial x+T\partial S/\partial x$, where $A=U-TS$ is the Helmholtz free energy. If the internal energy is independent of the coordinate, the force is entirely entropic. For an ideal chain whose configurations become less numerous on extension, it acts toward shorter extension. The external tension holding the chain extended has the opposite sign. Thus the force depends on the statistical number of available configurations, even when the energy of each configuration is unchanged.

For an ideal chain of $N$ independent statistical segments of length $a$, the end-to-end probability density is proportional to $\exp [-3R^2/(2N{a}^2)]$. Its constrained Helmholtz free energy has extension-dependent part $3k_{B}T{R}^2/(2N{a}^2)$, so the restoring force is $-3k_{B}T\mathbf{R}/(N{a}^2)$. This is an entropic spring because extension removes available chain configurations, and its stiffness grows with temperature. The Gaussian law applies at extensions small compared with the contour length $Na$; near full extension, finite extensibility matters.

For one mole with temperature-independent parameters $a,b$, the equation of state is $P=RT/(V-b)-a/V^2$. A Maxwell relation gives $(\partial S/\partial V{)}_T=R/(V-b)$, so the volume dependence is $R\ln (V-b)$. Matching the ideal gas entropy at infinite volume fixes the excess at the same $V,T$ to $R\ln (1-b/V)$. The attraction contributes $-a/V$ to the Helmholtz free energy but no excess entropy when $a$ is independent of temperature. A temperature-dependent attraction parameter would change this conclusion.

For a closed system with one specified chemical reaction, the extent of reaction $\xi$ records progress from $\xi =0$ through $n_i=n_{i,0}+{\nu }_i\xi$. Here $n_i$ is the amount of substance of species $i$, and the signed stoichiometric vector has negative entries for reactants and positive entries for reaction products. For the water chemical equation in the order ${\mathrm{H}}_2,{\mathrm{O}}_2,{\mathrm{H}}_2\mathrm{O}$, it is $(-2,-1,2)$. The normalization of $\xi$ depends on the chosen coefficients. See IUPAC Gold Book, extent of reaction.

With particle numbers instead of molar amounts, a molecular extent ${\xi }_N$ obeys $d{N}_i={\nu }_{i}d{\xi }_N$, and the chemical potentials are measured per particle. At fixed temperature and volume, the Helmholtz free energy then satisfies $dF=({\sum }_i{\nu }_i{\mu }_i)d{\xi }_N$. An interior chemical equilibrium requires ${\sum }_i{\nu }_i{\mu }_i=0$; the molar convention gives the same relation with chemical potentials per mole.

Finite-size rounding replaces a limiting discontinuity or divergence by a finite crossover or peak. For a spatial system near a continuous phase transition, the bulk description ceases to apply unchanged when the correlation length becomes comparable with the sample size. How the crossover width and location depend on size also depends on the model, geometry and boundary conditions; there is no universal $1/N$ rule for every phase transition.

An exact finite example is the Curie–Weiss model with $s_i=\pm 1$, $J>0$, and a conjugate field $h$ in energy units. Set $K=J/(k_{B}T)$ and $u=h/(k_{B}T)$. Its statistical Hamiltonian and finite partition function areThe binomial coefficient counts the microstates with $k$ positive Ising spins. The diagonal terms in the square add a configuration-independent constant to the energy; retaining them fixes the normalization used here. Divide each summand by $Z_N$ to obtain the probability mass function of $m_k=(2k-N)/N$. In particular,The variance expression makes the finite response nonnegative, while spin inversion symmetry gives $⟨m⟩=0$ exactly at $u=0$ for every plotted $N$.

Every curve is computed from the finite partition function, without selecting one minimum of the Bragg-Williams free energy of the Ising model. The field panel holds $K=1.5$ fixed and shows increasingly rapid switching between opposite spin magnetizations. The temperature panel uses $u=0$: its nonzero $⟨|m|⟩$ measures the typical magnitude, while the mean spin magnetization stays zero. The reference $k_{B}T/J=1$ marks the limiting critical temperature, not a singular point of any finite curve.

A short limiting argument explains the field panel. Combining the binomial coefficient with its leading entropy gives a Helmholtz free energy per spin whose stationarity equation is $m=\tanh (Km+u)$. For $K>1$ and $u=0$ the two ordered minima have $m=\pm m_{\ast }$, where $m_{\ast }=\tanh (K{m}_{\ast })$. For large $N$ well below the critical temperature, their relative Boltzmann weight is approximately $e^{2Nu{m}_{\ast }}$, giving $⟨m⟩\approx m_{\ast }\tanh (Nu{m}_{\ast })$. Thus this two-peak approximation has a field width of order $1/(N{m}_{\ast })$ in $u$. It is not valid at the critical temperature, where the peaks merge; the figure uses the exact finite sum throughout.

For finitely many microstates with finite energies $E_{\alpha }(h)$ that are real analytic functions of a real control $h$, the canonical partition function at $T>0$ is a finite positive sum:It and the resulting Helmholtz free energy are locally real analytic functions of $T$ and $h$. Hence such a finite Ising model has no literal nonanalytic equilibrium phase transition. A limit of these real analytic functions can nevertheless fail to be a real analytic function; the narrow but smooth crossover can become a sharp phase transition.^[ref]

Finite particle number alone is not the hypothesis of a finite sum. Classical phase space is continuous, and a system of quantum states can have infinitely many energy eigenvalues: its partition function requires convergence and local bounds that permit differentiation or an analytic extension of the integral or thermal trace. A divergent partition function, a singularly changed statistical Hamiltonian, or the zero-temperature limit falls outside the finite-positive-sum argument. Observing a steep change in a finite sample therefore motivates a limiting model; it does not measure an actual infinity.

The Gibbs free energy iswhere $H$ is enthalpy. For a closed system of fixed composition, with only pressure-volume work, spontaneous relaxation at fixed temperature and pressure cannot increase $G$; stable thermodynamic equilibrium minimizes it over the allowed states. A loaded piston can move while maintaining the imposed pressure, so the enthalpy term includes the displacement cost. Along a smooth equilibrium branch of fixed composition,Thus its temperature slope at fixed pressure is $-S$, while its pressure slope at fixed temperature is $V$. See MIT 5.60, Lecture 13, Gibbs free energy and its derivatives.

Consider two illustrative thermodynamic phases, A and B, of the same substance at one fixed pressure. Write B minus A, using quantities per mole:Treat these differences as approximately constant over a limited temperature window. ThenAt $280\mathrm{K}$, $\Delta G=+0.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, so A is preferred. At $320\mathrm{K}$, $\Delta G=-0.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, so B is preferred. At $300\mathrm{K}$, both have equal molar Gibbs free energy, permitting phase coexistence in this bulk model. B's greater entropy outweighs its greater enthalpy above the crossing.

Each plotted point compares alternatives at the same temperature; moving along the curve changes the imposed temperature. The model predicts preference, not instantaneous conversion: nucleation barriers can sustain metastability. Real enthalpy and entropy differences generally vary with temperature through heat capacities, bending this line outside the approximation.

At fixed pressure, the first law of thermodynamics gives $Q=\Delta H$ under the stated work restriction, so the same thermal reservoir accounting yields $\Delta S_{\mathrm{total}}=-\Delta G/T\ge 0$. Variable composition restores ${\sum }_i{\mu }_{i}d{n}_i$ to $dG$; constrained Gibbs minimization for chemical equilibrium handles those permitted changes.

The historical foundation is Gibbs's ^[ref]. Its equations (87) and (91) introduce the combinations now called Helmholtz free energy and Gibbs free energy, connecting entropy criteria, chemical potentials and phase coexistence within one framework.

Vitaly Lazarevich Ginzburg and Lev Landau applied an order parameter description to superconductivity in 1950. A spatially varying complex amplitude extends the fixed-amplitude description of the London equations. Their initial charge assumption preceded the later identification of the universal pair charge.^[ref]

Consider the following isotropic, local SI Helmholtz free energy model near a continuous superconducting transition temperature:Here $\beta >0$, $m_{\ast }>0$ is a phenomenological mass parameter, and $q=-2e$ fixes the electron-pair convention. The uniform superconducting minimum for $\alpha <0$ has $|{\psi }_0{|}^2=-\alpha /\beta$. The gradient term penalizes changes of the amplitude and the gauge-invariant quantum phase; the magnetic term penalizes magnetic field penetration. The gauge-coupling sign in Ginzburg-Landau superconductivity keeps the electric current and field equation consistent. This model omits material anisotropy, microscopic nonlocality and critical fluctuations.

For a simple spin-triplet p-wave pairing description, let the $3\times 3$ matrix $A$, with complex number entries, map an orbital direction $\hat{\mathbf{k}}$ on the Fermi surface into a triplet gap vector $\mathbf{d}=A\hat{\mathbf{k}}$. The indices refer to spin and orbital components. A scalar amplitude cannot retain all these relative orientations. Anthony James Leggett's account connects that extra structure to the magnetic responses of the different phases.^[ref]

Here are two explicit idealizations, with a common positive scale $\Delta$ and an overall quantum phase omitted. They show how odd orbital pairing can have either nodes or an isotropic gap. They do not calculate which state has the lower Helmholtz free energy.

The Helmholtz free energy iswhere $U$ is internal energy, also denoted $E$ elsewhere here, $T$ is absolute temperature, and $S$ is entropy. For the closed system of fixed composition and only pressure-volume work described above, spontaneous relaxation at fixed temperature and volume cannot increase $F$. Stable thermodynamic equilibrium is a global minimum over the permitted alternatives, including different thermodynamic phases. A rigid vessel fixes the total volume; it does not require each coexisting thermodynamic phase to retain its own volume.

For a constructed comparison at $300\mathrm{K}$, suppose one candidate has $1.5\mathrm{kJ}$ more internal energy but $6\mathrm{J}{\mathrm{K}}^{-1}$ more entropy than another at the same temperature and volume. Its difference isThus the candidate with higher internal energy has lower Helmholtz free energy. This is the practical reason for retaining the entropy term when selecting a state.

For equilibrium states within a smooth branch of fixed composition, the first law of thermodynamics gives $dU=TdS-pdV$, henceThese derivatives extract entropy and pressure; the differential is not an evolution equation for relaxation. If chemical species amounts $n_i$ vary, add ${\sum }_i{\mu }_{i}d{n}_i$, with molar chemical potentials ${\mu }_i$. A closed system can still change chemical species amounts through chemical reactions; the fixed-composition simplification then needs the conserved-inventory constraints of chemical equilibrium. See MIT 5.60, Lecture 14, fundamental equations and chemical potentials.

The minimum principle has a short explanation. At fixed volume, no boundary work occurs, so $Q=\Delta U$. A thermal reservoir at positive temperature $T$ loses entropy $Q/T$. Consequentlyby the Second law of thermodynamics. Here both endpoint states share the imposed temperature. See MIT 5.60, Lecture 12, criteria for spontaneous change.

The microscopic counterpart is $F=-k_{B}T\ln Z$ in the canonical ensemble, with Boltzmann constant $k_B$ and partition function $Z$. The partition function examples compute the balance between internal energy and entropy; the chemical equilibrium example already displays an ideal-mixture Helmholtz free energy minimum.

For fixed amount of substance, thermodynamic consistency givesIndeed, the first law of thermodynamics gives $U_V{|}_T=T{S}_V{|}_T-P$, and the Helmholtz free energy Maxwell relation gives $S_V{|}_T=P_T{|}_V$. For the fixed-amount convention $P=RT/(V-b)-a/V^2$ with temperature-independent $a,b$, substitution gives $U_V{|}_T=a/V^2$. Integration yields $U=h(T)-a/V$. If the dilute-limit reference is $U\to cT$ as $V\to \infty$ for every $T$, then $h(T)=cT$.

The same formula may be interpreted molarly if $V,U,a,b$ all use that convention. Mixing molar and total quantities changes the amount factors. The attraction controls the volume correction to internal energy, while excluded volume alone does not contribute to this derivative when $a,b$ are independent of temperature.

Three distinct effects spoil indefinite $T\propto B$ cooling: extra entropy outside the spin subsystem, additional magnetic energy scales, and irreversible heat or work input. They can be distinguished with a thermodynamic derivative. For a smooth equilibrium Helmholtz free energy obeying $dF=-SdT-mdB$, a Maxwell relation givesHere $C_B$ is the total heat capacity participating in the stage's thermal equilibrium. The ideal model has $(\partial m/\partial T{)}_B=-C_{B,\mathrm{mag}}/B$. If its spins share temperature with a field-independent lattice and load of combined heat capacity $C_{\mathrm{other}}>0$, thenThe positive slope still means cooling when $dB<0$, but its smaller magnitude gives less cooling per fractional field reduction. For example, a low-temperature Debye model gives a lattice entropy proportional to $T^3$. As that lattice entropy falls, the spins take up the released entropy; $x$ can no longer remain constant. A weak enough coupling to neglect its equilibrium interaction energy can still be strong enough to exchange heat on the sweep timescale.

Interactions between magnetic moments, including exchange interaction and the fields of neighboring magnetic dipole moments, introduce energy scales that do not vanish with the applied $B$. Once they compete with $k_{B}T$ or $\mu B$, the independent-site partition function fails. Correlations and possible magnetic ordering alter the available magnetic entropy and the cooling curve. Simply setting the external magnetic field to zero does not eliminate those interactions. Different refrigerants therefore need their own measured equilibrium properties; there is no universal final temperature from the applied-field ratio alone.

Finally, a real open heat switch has residual thermal conductance; supports, wires and radiation carry incoming heat. Field changes may also dissipate energy. Locally near thermal equilibrium, with incoming heat rate ${\dot{Q}}_{\mathrm{leak}}$ and internal entropy production rate ${\dot{S}}_{\mathrm{prod}}$, the stage balance becomesThis uses a common stage temperature; a non-equilibrated load requires separate balances. The sweep must allow spin-lattice relaxation while remaining fast enough to limit accumulated parasitic heat. At fixed $B$, the same balance explains warming and finite hold time. Useful refrigeration is the controlled absorption of a load's heat during that finite interval, followed by a renewed isothermal process of magnetization to export its entropy.