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A steady state does not by itself give every fluctuating observable an entropy production symmetry. If the dimensionless total entropy production ${\Sigma }_t$ has the same-law finite relation and a large deviation principle with rate function $I$ for ${\Sigma }_t/t$, then, on its common domain,The first expression denotes leading exponential scaling; the second follows by taking the scaled logarithm of the probability ratio. A finite-state, autonomous Markov process with time-even states, positive bidirectional rates and a positive stationary distribution supplies a standard setting for a finite total-entropy production relation. The stochastic system entropy is then bounded, so its endpoint contribution does not change the long-time rate function of bath entropy production. ^[ref]

Unbounded state spaces can have endpoint fluctuations on the scale relevant to a large deviation principle. Dropping system entropy, changing the measured observable to heat, or changing the reverse preparation therefore requires a fresh argument. The exact finite integral fluctuation theorem, a same-law finite relation, and an asymptotic rate function symmetry are distinct claims.

Taking the curl of axisymmetric hydrostatic equilibrium with centrifugal acceleration gives $\varpi {\partial }_z{\Omega }^2=(\nabla p\times \nabla \rho {)}_{\varphi }/{\rho }^2$. For an approximately spherical ideal gas with constant composition, use $\nabla \rho =\nabla p/c_s^2-\rho \nabla S/c_p$ to obtain the displayed relation. Latitudinal entropy differences can support noncylindrical differential rotation. A barotropic fluid gives cylindrical rotation in this leading balance; the relation does not determine the transport stresses or circulation rate.

Superfluidity is a quantum-fluid response with phase stiffness and flow without viscous dissipation in an appropriate low-speed regime. A superfluid supports a phase-gradient current; defects and excitations can limit that flow. Bose-Einstein condensation concerns macroscopic one-particle occupation and is a distinct criterion: the Gaussian phase stiffness theory can have divergent absolute-phase fluctuations while retaining a cost for gradients.

Why does cooling a liquid make it flow more freely through a tiny gap? Helium-4 supplies an experimental entrance to this question: the helium slit-flow experiment compares the motion of liquid levels on opposite sides of the transition into superfluidity. A flow measurement probes resistance to motion; a measurement of Bose-Einstein condensation probes occupation of a quantum state. The two-fluid model of superfluidity and the Landau criterion connect these questions to entropy transport and to the energy required to create an excitation.

Temperature labels thermal equilibrium: two internally equilibrated bodies have the same temperature when contact permitting only heat exchange produces no net heat transfer. The zeroth law of thermodynamics makes the comparison transitive. If bodies A and B separately reach thermal equilibrium with the same thermometer without appreciably disturbing either body, they agree with each other. This is what lets a gas thermometer compare a metal block with water, despite their different internal energies and heat capacities. Insulation alone makes no such comparison: it can prevent heat transfer between bodies at different temperatures. ^[ref]

The zeroth law of thermodynamics establishes equality comparisons, but does not choose a numerical scale. Thermodynamic temperature supplies an absolute scale consistent with Carnot's theorem and entropy. For a simple compressible system at thermodynamic equilibrium with fixed particle number $N$ and composition,Here $U$ is internal energy, $S$ is entropy, and $V$ is volume; the second equality assumes the derivative can be inverted. Temperature is a thermodynamic conjugate variable to entropy and belongs to the intensive and extensive thermodynamic quantities as an intensive quantity. Its partial derivative tells us how much internal energy must change per unit change of entropy under the stated constraints. More general systems require their other mechanical and material variables to be held fixed too. ^[ref]

For an infinitesimal quantity $\delta Q>0$ of heat transferred from A to B, with no work or matter exchange, the equilibrium entropy changes giveFor ordinary positive temperatures, spontaneous heat transfer from $T_A>T_B$ increases total entropy; thermal equilibrium has $T_A=T_B$. This comparison uses changes in entropy, rather than the total internal energy contained in either body. A small hot object can have less internal energy than a large cool one. See entropy maximization under thermal contact for the equilibrium criterion and negative temperature for the special case of bounded energy spectra.

Express absolute thermodynamic temperature $T$ in kelvin, and a Celsius reading $t_C$ in degree Celsius:The offset is exact: $300\mathrm{K}$ corresponds to $26.85{}^{\circ }\mathrm{C}$, and $400\mathrm{K}$ to $126.85{}^{\circ }\mathrm{C}$. Their absolute temperature ratio is $4/3$; dividing their Celsius readings gives a different number. Thus gas thermometer pressure ratios and Carnot's theorem require thermodynamic temperature. The modern SI definition of the kelvin fixes the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$; a reference bath realizes or transfers the scale rather than defining that constant.^[ref]

Temperature is not universally the mean kinetic energy of a particle. In a classical, nonrelativistic ideal gas at thermal equilibrium, the equipartition theorem gives the mean translational kinetic energy relative to the gas's bulk motion:Here $m$ is particle mass, $\mathbf{v}$ its velocity, and $\mathbf{u}$ the mean flow velocity. In this regime, different particle masses at the same temperature have equal mean translational kinetic energies but different characteristic speeds. A moving gas can also carry bulk kinetic energy without being hotter in its rest frame. Internal molecular modes and quantum statistics require further care; the equipartition theorem has classical assumptions, whereas the equilibrium entropy definition applies more widely. ^[ref]

A thermal relic is a surviving particle population whose earlier abundance was set by chemical thermal equilibrium with the cosmic plasma. After number-changing reactions cease and cosmic entropy is conserved, its comoving abundance is described by the yield $Y=n/s$. A relativistically decoupled species can retain an appreciable yield, whereas a species remaining in thermal equilibrium after becoming nonrelativistic is subject to Boltzmann suppression. Its later mass density, lifetime and momentum distribution determine its cosmological effects.

The temporal orientation associated with the typical increase of macroscopic entropy away from suitable low-entropy conditions. It is a statistical claim about specified initial or boundary conditions and a macroscopic description, rather than a universal monotonicity theorem for every microscopic trajectory. Time reversal in classical mechanics pairs suitable increasing trajectories with decreasing ones, while fine-grained entropy is conserved under regular volume-preserving Hamiltonian flow.

Local equilibrium associates the outgoing heat flux $q$ with outgoing entropy flux $q/\theta$ at positive temperature $\theta$. Its divergence appears in the Lagrangian entropy inequality, with the appropriate nominal flux in reference coordinates.

A thermodynamic system is in thermodynamic equilibrium when its macroscopic state has no spontaneous tendency to change through any exchange allowed by its constraints. For an isolated system, stable thermodynamic equilibrium maximizes entropy over the accessible states at fixed total energy and conserved material inventories. A wall can forbid an exchange, so always specify what can cross or move. Thermal, mechanical and chemical conditions answer different questions about that same constrained state. ^[ref]

For example, two compartments separated by a rigid, impermeable wall that transmits heat can reach thermal equilibrium while retaining different pressures and compositions. Allowing an unloaded piston to move adds a mechanical equilibrium condition. Allowing one chemical species to cross adds chemical potential equality for particle exchange; allowing a chemical reaction adds its chemical equilibrium condition. A fixed barrier can therefore support a constrained thermodynamic equilibrium that ceases to be an equilibrium when the barrier is removed.

Uniformity is not the definition of thermodynamic equilibrium. An isothermal atmosphere in hydrostatic equilibrium can have a common temperature while its pressure and mass density vary with height: the pressure gradient balances gravity. In an external field, the exchange condition must include the relevant potential energy; for charged matter this leads to the electrochemical potential. Relativistic gravitational corrections to thermal equilibrium are described separately by the Tolman temperature law.

A steady state only says that the macroscopic description is constant in time. A bar maintained between hot and cold thermal reservoirs has steady heat flux and produces entropy, so it is not in thermodynamic equilibrium. Even matching temperatures at one perfect-contact interface can coexist with heat transfer through the surrounding material; matching those two local values does not establish equilibrium of the whole apparatus. See thermal conduction for that spatial transport problem.

The useful simplification is that equilibrium conditions predict an endpoint without computing every collision or flow during relaxation. The first law of thermodynamics supplies the conserved quantities, and the Second law of thermodynamics supplies the direction of change. For a system held at fixed temperature and volume, the corresponding criterion is minimization of Helmholtz free energy; at fixed temperature and pressure, it is minimization of Gibbs free energy, with the material and exchange constraints still imposed.

The derivative of internal energy at fixed deformation gradient and entropy, or of specific Helmholtz free energy at fixed temperature, defines this force. Its product with the internal-variable rate contributes mechanical dissipation to the Clausius-Duhem inequality.

A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting $TS$ packages that accounting into a property of the chosen system. Adding $pV$ additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is $Q=\Delta U$ at fixed volume or $Q=\Delta H$ at fixed pressure, whereas neither $\Delta F$ nor $\Delta G$ generally equals $Q$.

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.

Thermodynamics studies energy transfers and the constraints on macroscopic changes. Its vocabulary connects internal energy, entropy, temperature, pressure and volume to heat and work. The first law of thermodynamics balances energy; the Second law of thermodynamics decides which changes can occur spontaneously and limits how much work a heat engine can deliver. These principles let us bound performance before choosing a material or a detailed machine design. Applications include power generation, propulsion and refrigerators. ^[ref]

Start by choosing a system boundary. A closed system can exchange energy but no matter; an isolated system exchanges neither. A state function, such as internal energy or entropy, depends on the equilibrium state. A process specifies how that state changes. Thus a body's temperature is a property it has, whereas heat records energy crossing its boundary during a process. A body contains internal energy, rather than a stored quantity of heat.

Two insulated bodies initially at different positive temperatures approach thermal equilibrium when brought into thermal contact. Their combined energy is conserved, but their combined entropy increases. They do not spontaneously separate again into a hotter body and a colder one. This familiar asymmetry makes thermodynamics interesting: conserving energy leaves many imaginable changes available, and entropy supplies a further constraint. The same distinction underlies both entropy maximization under thermal contact and the thermal efficiency limit in Carnot's theorem.

Journal of Chemical Physics 10, 51–61. doi.org/10.1063/1.1723621

Primary-paper scan: www.appliedpolymertechnology.org/uploads/1/0/8/8/108867241/jchemphys_10_51.pdf

The lattice model treats connected polymer segments and derives the nonideal combinatorial entropy before adding the energetic mixing contribution.

Let $fdx$ be reversible work supplied to stretch a band. Its Helmholtz free energy obeys $dF=-SdT+fdx$. Equality of mixed derivatives gives the displayed relation. An entropy decrease on stretching therefore corresponds to increasing tensile force on heating at fixed extension. The sign follows the force/work convention; a restoring force exerted by the band has the opposite sign.

The chosen steady-state model isHere $U$ is the terminal voltage drop along positive electric current through the module. Its Seebeck effect term opposes the cooling supply. The cold Peltier effect term removes heat; Joule heating and returning heat conduction reduce the available cooling. At zero electric current, $Q_c=Q_h=-K\Delta T$: the passive leak warms c and removes heat from h, consistent with the stated signs. Subtracting the two face equations cancels heat conduction and gives the electrical power, an independent conservation of energy check.

The factor one half is an ideal-model result, not a universal rule for every contact. To see its origin, let $x$ run from cold to hot along a uniform equivalent leg of length $L$, cross-sectional area $A$, electrical resistivity $\rho$ and thermal conductivity $\kappa$. With constant coefficients, insulated sides and no Thomson effect, steady conservation of energy givesThe Fourier's law heat current in the positive $x$ direction is $-\kappa A{T}^{\prime }$. At $x=0$ it equals $-K\Delta T-I^{2}R/2$, and at $x=L$ it equals $-K\Delta T+I^{2}R/2$, using $K=\kappa A/L$ and $R=\rho L/A$. Adding the local transported heat current $\alpha TI$ gives the two face equations. Summing equivalent legs preserves this accounting. Nonuniform material properties, side losses or asymmetric contact dissipation require a revised model.

For an original toy module, stipulate $\alpha =0.040V/K$, $R=2.0\Omega$, $K=0.40W/K$, $T_c=280\mathrm{K}$ and $T_h=300\mathrm{K}$. At $I=3.0\mathrm{A}$,The coefficient of performance is defined here only for refrigeration with $Q_c>0$ and $P_{\mathrm{in}}>0$. The hot-side assembly must remove $37.0\mathrm{W}$, not just the extracted $16.6\mathrm{W}$. These are stipulated calculations, not measurements or a commercial module specification.

At the same fixed face temperatures but $I=-3.0\mathrm{A}$, the same boundary definitions give $Q_c=-50.6\mathrm{W}$, $Q_h=-35.0\mathrm{W}$, $U=-5.2\mathrm{V}$ and $P_{\mathrm{in}}=15.6\mathrm{W}$. The formerly cold face is now heated; the formerly hot reservoir supplies $35.0\mathrm{W}$. Still $Q_h-Q_c=15.6\mathrm{W}$. Reversal changes the linear terms, while $I^{2}R$ and the leak at fixed temperatures remain unchanged. Holding the old labels fixed makes this check meaningful.

The corresponding steady entropy production of the module is the outgoing reservoir entropy minus the incoming reservoir entropy:The Peltier effect terms cancel because they carry the same entropy $\alpha I$ at each end. This cancellation explains the reversible coupling; it does not make finite-resistance operation reversible. In the toy example the two positive contributions are approximately $0.06214$ and $0.001905W/K$, giving ${\dot{S}}_{\mathrm{gen}}=0.06405W/K$ for either current direction. In cooling operation, combining this inequality with the energy balance gives the Carnot refrigerator limit $\mathrm{COP}\le T_c/(T_h-T_c)$, equal to $14$ for these reservoir temperatures. A coefficient of performance above one would describe transported heat per electrical work, not the creation of energy.

entropy associated with positions of independently translating objects. Connecting segments into one polymer leaves fewer independently translating objects than treating the same segments as separate molecules; this reduces mixing entropy per segment.

For a free molecule of mass $m$ in a large three-dimensional volume $V$, replacing the translational state sum by an integral gives the displayed partition function. For identical classical molecules the many-molecule canonical partition function is $q_{\mathrm{trans}}^N/N!$, not simply $q_{\mathrm{trans}}^N$. Differentiating its Helmholtz free energy at fixed volume gives the ideal-gas translational entropy.

A triple point is a temperature–pressure pair at which three distinct bulk thermodynamic phases of a one-component substance coexist in thermodynamic equilibrium. In the familiar solid–liquid–gas case, their three phase coexistence curves meet. Other combinations, such as two crystal polymorphs and a liquid, also qualify: “triple” counts thermodynamic phases, rather than states-of-matter categories.

For the same substance, the chemical potential in each thermodynamic phase must agree:There are two independent equality conditions for two controls, temperature and pressure. Generically they select an isolated pair $(T_t,p_t)$; the third pairwise equality follows from the first two. Equivalently, the phase rule gives $F=1-3+2=0$. This counts independent intensive controls, not phase fractions: changing the amounts of the three thermodynamic phases need not change the ideal bulk temperature and pressure, provided all three remain present. Different sets of crystal polymorphs can give different triple points for the same substance.^[ref]

At a liquid-gas critical point, the two fluid thermodynamic phases instead merge, and their mass density difference and latent heat vanish. It is the endpoint of the liquid–gas phase coexistence curve, rather than a meeting of three distinct thermodynamic phases. The regular independent-equality count used for a triple point cannot explain this merger. A triple point can retain finite mass density and entropy differences between all three thermodynamic phases.^[ref]

The two-fluid model of superfluidity gives the normal component of a superfluid and the superfluid independent velocities.

An elementary uniform-flow example makes the distinction operational. Write the mass flux per unit area and the ideal entropy flux asHere $s$ is specific entropy, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and ${\rho }_s,{\rho }_n$ are response mass densities. The mass density ${\rho }_s$ used here gives the three-dimensional phase stiffness coefficient ${\rho }_s(\hslash /m{)}^2$ when ${\mathbf{v}}_s=(\hslash /m)\nabla \theta$ is substituted into the kinetic energy density ${\rho }_s{v}_s^2/2$. For a stipulated ${\rho }_s=3\rho /4$, ${\rho }_n=\rho /4$ and $v_n=3u$ along a channel, zero net mass flux requires $v_s=-u$. Yet $j_S=3\rho su$ is positive. Opposing velocities can therefore transport entropy while transporting no net mass.

This example explains why excellent apparent heat transport need not be ordinary diffusion of heat through a stationary liquid. In the ideal zero-net-mass-flux limit, the heat flux associated with this entropy transport is $T{j}_S$. Viscous losses, quantum vortices and external heat exchange require additional terms. The model also supports second sound.

Use the independent spin-one-half two-level system with fixed $N$ distinguishable sites and Hamiltonian levels $E_{\pm }=\pm \mu B$, where $\mu >0$ is the magnitude of the field-axis magnetic moment and $B>0$ is a uniform externally controlled magnetic field. Assume positive temperature, no interactions between sites, and no appreciable lattice or load entropy. The two levels differ by $2\mu B$. With inverse temperature $\beta$, the canonical ensemble gives the one-site and total partition functionsHere the minus subscript denotes the lower energy $-\mu B$. Factorization of the partition function requires additive energy and independent sites; there is no particle-exchange factor $N!$ for these fixed distinguishable sites.

Differentiate the partition function, holding $B$ fixed when differentiating with respect to $\beta$. The internal energy, Helmholtz free energy, total field-axis magnetic moment $m$, and magnetic entropy areThe extensive moment $m$ is distinct from magnetization, the moment per unit volume. Direct substitution of $p_{\pm }$ into the Gibbs entropy gives the same $s(x)$ independently of the free-energy derivative.

At fixed $B$, the magnetic heat capacity isStrict decrease makes $s$ invertible on $x>0$. Constant magnetic entropy therefore fixes $x$, not $B$ or $T$ separately:The partition function may be used to label these equilibrium states even though the isolated cooling leg has no external heat bath. Actual thermalization of the lattice and load additionally requires a sufficiently fast spin-lattice relaxation; perfectly uncoupled spins by themselves cannot refrigerate an attached object.

The limiting cases are instructive. For $x\ll 1$, $s(x)=\ln 2-x^2/2+O(x^4)$, $m\simeq N{\mu }^{2}B/(k_{B}T)$, and $C_{B,\mathrm{mag}}\simeq N{k}_B{x}^2$. The moment response reproduces Curie's law. For $x\gg 1$, $s(x)\simeq (1+2x)e^{-2x}$ and the upper-state occupation is exponentially small. Taking $B\to 0$ at fixed $T>0$ gives $s\to \ln 2$; taking $T\to 0$ at fixed $B>0$ gives $s\to 0$. The origin has a path-dependent limit. Extrapolating a line of constant $x$ to $B=T=0$ does not construct a usable zero-temperature state or prove that a real material retains independent degenerate spins there.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.