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A finite heat bath has a finite whole-body heat capacity $C_b$ and therefore changes temperature when it absorbs heat $Q_b$. For a rigid closed system with no work, fixed composition and no phase transition, the first law of thermodynamics givesIf the heat capacity at constant volume $C_b$ is constant throughout this temperature interval, the integral becomes $Q_b=C_b(T_{b,f}-T_{b,0})$. This statement concerns an internal energy difference; $C_b{T}_b$ need not be the body's absolute internal energy.

For an original example, put a rigid body 1 and a finite bath 2 into thermal contact inside an isolated system. Stipulate constant whole-body heat capacities $C_1=100\mathrm{J}{\mathrm{K}}^{-1}$ and $C_2=900\mathrm{J}{\mathrm{K}}^{-1}$, with initial temperatures $T_{1,0}=350\mathrm{K}$ and $T_{2,0}=290\mathrm{K}$. Each body remains internally uniform; the connection stores negligible energy; no external work, matter exchange or phase transition occurs. These are constructed model values. The argument in thermal equilibration of two finite bodies applies directly: equal final temperatures and conserved total internal energy giveThus the heat $Q$ transferred from body 1 to bath 2 isBody 1 loses $5400\mathrm{J}$ of internal energy and bath 2 gains exactly that amount. Body 1 cools by $54\mathrm{K}$, while bath 2 warms by $6\mathrm{K}$. Equal heat transfers produce unequal temperature changes because the heat capacities differ.

The heat bath approximation can now be assessed against a chosen allowed temperature change $\delta T>0$. Constant $C_b$ requires $|Q_b|/C_b\le \delta T$, rather than merely calling the bath large. For this example, an allowed change of $0.1\mathrm{K}$ would fail because the computed change is $6\mathrm{K}$. The exact bath shift and its ideal-reservoir limit areHold $C_1$ and both initial temperatures fixed in this limit. The transferred heat then approaches $C_1(T_{1,0}-T_{2,0})=6000\mathrm{J}$. The finite bath receives less heat because its warming reduces the available temperature difference. For heat capacities that vary, use the two integrals in the energy balance instead of this weighted average.

Conservation of energy also permits imagining the reverse transfer, so it cannot explain the direction of equilibration by itself. Entropy from heat capacity gives the additional test, using reversible comparison paths between equilibrium endpoints:All ratios inside the natural logarithms use absolute temperatures. The actual exchange across unequal temperatures is irreversible, even though reversible paths calculate the endpoint entropy differences.

There is a short general proof of positivity. Put $w_i=C_i/(C_1+C_2)$ for positive constant heat capacities and positive initial temperatures. The strict concavity of the natural logarithm giveswith equality only for equal initial temperatures. In this proof all temperatures inside the logarithms mean their numerical values in one common absolute unit. This is Jensen's inequality applied to the same weighted average that conservation of energy selected. The direction of heat transfer and the stability of its endpoint also follow from entropy maximization under thermal contact.

Why must a conducting piston move slowly? A simple model couples the first law of thermodynamics to Newton's law of cooling. Assume a fixed amount of ideal gas with uniform temperature $T(t)$ and constant whole-sample heat capacity at constant volume $C_V$, a heat bath maintained at $T_b$, negligible friction, and mechanically slow motion. Let $G>0$ be the effective thermal conductance, the reciprocal of the thermal resistance, in $\mathrm{W}{\mathrm{K}}^{-1}$. ThenPositive heat transfer must offset expansion work to keep $T$ near $T_b$. A finite $G$ and nonzero piston speed therefore require a temperature difference. This uniform-temperature model neglects spatial temperature gradients, sensor response, cylinder heat capacity and any change of $G$ with piston position; it is an illustrative calculation, not a fit to the isothermal gas syringe experiment.

For a prescribed exponential volume change $V=V_i{e}^{\lambda t}$, define $\tau =C_V/G$, $s=t/\tau$ and $ϵ=nR\lambda /G$. The dimensionless temperature $\theta =T/T_b$ obeys a linear ordinary differential equation with initial value $\theta (0)=1$:The plotted expansion uses stipulated $C_V=(5/2)nR$ and stops at $V_f=2V_i$, at $s_f=\ln 2/[(5/2)ϵ]$. During expansion, the larger $ϵ$ produces the larger temperature deficit; once the piston is held fixed, the deficit undergoes exponential decay as $e^{-(s-s_f)}$. Reducing $|ϵ|$ makes the path more nearly an isothermal process, while increasing its duration. Reversing the piston with $\lambda <0$ produces a temperature excess and outward heat transfer; the small-$|ϵ|$ regime is the intended near-isothermal approximation. Mechanical slowness alone does not ensure sufficient heat transfer.

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.

Neglecting interactions and thermal pairs, the Fermi-Dirac distribution $f(p)=[\exp ((\epsilon (p)-{\mu }_{\mathrm{kin}})/(k_{B}T))+1{]}^{-1}$ with kinetic energy $\epsilon (p)=\sqrt{m_e^2{c}^4+p^2{c}^2}-m_e{c}^2$ determines the electron contribution to a stellar equation of state. The kinetic chemical potential is fixed by the electron number density, not an independently chosen pressure fraction:The electron kinetic internal energy per volume is $({\pi }^2{\hslash }^3{)}^{-1}{\int }_0^{\infty }{p}^2\epsilon (p)f(p)dp$. In the nondegenerate limit $P_e=n_e{k}_{B}T$; at zero temperature the integrals become the equation of state of a cold electron gas. These are limits of one component and must not be added together as independent electron pressures. When electron-positron thermal pair abundance matters, both charge species must be included, with charge neutrality fixing their net density.

Finite-temperature quantum field theory represents thermal expectation values by evolution through an imaginary-time interval of length $\beta =1/T$.

Consider a finite-duration Markov process with discrete, time-even states coupled to one thermal reservoir at positive temperature $T$. Let $P_F[\gamma ]$ include the initial probability ${\rho }_0(x_0)$ of the sample trajectory $\gamma$. Define $P_R$ by reversing the control protocol and initializing it with the actual forward final probability distribution ${\rho }_{\tau }$. Write $\Theta \gamma$ for the same states visited in reverse time order. Require normalized path laws with matching probability support: a path has positive forward weight exactly when its reverse has positive reverse weight. This excludes irreversible forbidden transitions and missing initial states.

Under local detailed balance, with positive heat $Q_{\mathrm{bath}}$ delivered to the thermal reservoir, the full path-weight logarithm isThe endpoint term is the change in stochastic system entropy; the conditional path ratio alone accounts for the thermal reservoir. Here $k_B$ is the Boltzmann constant. Both terms matter even for a short observation. ^[ref]

Define the conjugate reverse functional by ${\Sigma }_R[\eta ]=\ln (P_R[\eta ]/P_F[\Theta \eta ])$, so ${\Sigma }_R[\Theta \gamma ]=-{\Sigma }_F[\gamma ]$. Summing the path identity over paths with a given value gives the exact finite fluctuation theoremfor nonzero denominator, using probability masses or corresponding probability density functions. This ${\Sigma }_R$ is defined by the paired laws: it need not be the total entropy production obtained by independently restarting the thermodynamic construction for the reverse experiment. A generic driven experiment therefore does not imply $p_F(s)/p_F(-s)=e^s$.

Freeze the magnitude of an order parameter and retain just one common orientation $\theta$ for $N$ rigidly aligned cells. Give this collective coordinate energy $E(\theta )=-Nh\cos \theta$ at positive temperature $T$. The Boltzmann distribution is the displayed normalized probability density function, where $k_B$ is the Boltzmann constant. Here $h$ is energy per cell, a different normalization from the source density in continuous vacuum selection by an external source. This is a deliberately restricted model of selection among already ordered orientations: it does not derive alignment or an ordering critical temperature from a spatial interaction.

Its mean orientation is an elementary ratio of integrals:At $h=0$ every angle has equal probability, and the numerator vanishes exactly. For small $x$, expansion of the exponential gives $m_N(h)=x/2+O(x^3)$ because the angular mean of ${\cos }^2\theta$ is $1/2$. For $x=1$ and $5$ the means are approximately $0.446390$ and $0.893383$. At fixed $h>0$, increasing $N$ concentrates the probability density function near $\theta =0$; expanding $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ gives angular variance approximately $1/x$. ConsequentlyThe uniform angular average remains a symmetric mixture if one sets $h=0$ first. The other order selects a definite direction. This is the continuous-orientation version of the symmetry-breaking order of limits; it illustrates selection without replacing the separate problem of stability against spatial fluctuations.

The first law applies conservation of energy to a thermodynamic system. For a closed system with unchanged bulk kinetic energy and potential energy, writeHere $U$ is internal energy, $Q>0$ is heat entering the system and $W>0$ is work done by the system on its surroundings. Compression therefore gives negative $W$. Some references define work positive when done on the system; their plus sign represents the same balance with a different convention. If bulk kinetic energy or potential energy changes, include it in the total energy balance. ^[ref]

Supplying $1000\mathrm{J}$ of heat while a gas does $300\mathrm{J}$ of work increases its internal energy by $700\mathrm{J}$. Compressing it with $200\mathrm{J}$ of input work in an adiabatic process instead gives $Q=0$, $W=-200\mathrm{J}$ and $\Delta U=200\mathrm{J}$.

The distinction between $d$ and $\delta$ records state versus path: internal energy is a state function, whereas heat and work depend on the process. For an ideal gas expanding reversibly at constant temperature, $W=N{k}_{B}T\log (V_2/V_1)$ and $Q=W$, since internal energy of an ideal gas depends only on temperature. Choose $N{k}_{B}T=1000\mathrm{J}$ and $V_2=2V_1$: then $Q=W=693.15\mathrm{J}$. An insulated free expansion into vacuum between the same equilibrium endpoints has $Q=W=0$. Both have $\Delta U=0$, yet their transfers differ. Their common entropy increase is $N{k}_B\log 2$. ^[ref]

For a simple compressible equilibrium system of fixed composition, combining the laws givesHere $p$ is pressure and $V$ is volume. This identity relates neighboring equilibrium states. Identifying $TdS$ with actual heat and $pdV$ with actual work requires a suitable reversible thermodynamic process; an irreversible piston calculation uses the external resisting pressure for boundary work. This distinction prevents applying a reversible ideal-gas adiabat to an irreversible adiabatic piston compression.

A lumped sample of heat capacity $C$ coupled by thermal conductance $G$ to a bath has the displayed linearized energy balance. For constant internal power, the bath-to-sample temperature response is $1/(1+i\omega C/G)$. Insulation and heat capacity suppress rapid bath fluctuations but have unity static bath-to-sample gain. A steady power change shifts the sample by $\delta P/G$, so improved isolation increases self-heating sensitivity and equilibration time. The lumped approximation requires internal equilibration faster than the variations being modeled.

A Galerkin method retaining one stream function mode, two temperature modes and two transverse-velocity modes produces five coupled amplitudes. A useful normalization gives $a^{\prime }=-\sigma a+\sigma rb-{\sigma }^{2}qd$, $b^{\prime }=a-b-ac$, $c^{\prime }=h(-c+ab)$, $d^{\prime }=a-\sigma d-ae$, $e^{\prime }=-(4-h)\sigma e+had$. Here $\sigma >0$ is the Prandtl number, $0<h<4$, and $q$ measures squared rotation. The trivial equilibrium point has characteristic polynomial $(s+\sigma {)}^2(s+1)-\sigma r(s+\sigma )+{\sigma }^{2}q(s+1)$ in its three coupled linear modes; the remaining two modes decay. Stationary bifurcations and oscillatory onset meet at $q=(1+\sigma )/(1-\sigma )$, $r=2/(1-\sigma )$ when $\sigma <1$.

A Galerkin method approximation can retain one roll streamfunction mode, two temperature modes and two magnetic-flux perturbation modes in vertical-field magnetoconvection. Projection onto these five modes, rather than pointwise equality of the truncated fields, gives a closed system of five ordinary differential equations. Products generate higher spatial harmonics that are discarded by orthogonal projection. The system has symmetry $(a,b,c,d,e)\mapsto (-a,-b,c,-d,e)$ and hence admits paired steady roll branches.

Set $y=r_{+}/|Q|\ge 1$. Then $M/|Q|=(y+y^{-1})/2$ and $|Q|T=(y^{-1}-y^{-3})/(4\pi )$. Differentiation gives a unique maximum at $y=\sqrt{3}$, proving the displayed mass and temperature. The curve is zero at extremality and asymptotic to $1/(8\pi M)$ at large mass.

For a three-dimensional parabolic band and fixed positive $\mu -E_c$, the Sommerfeld expansion gives $n(T)=n(0)[1+{\pi }^2(k_{B}T/(\mu -E_c){)}^2/8+O(T^4)]$. This increase differs from the usual fixed-density calculation, in which the chemical potential decreases with temperature. The exact density is proportional to $T^{3/2}{F}_{1/2}[(\mu -E_c)/(k_{B}T)]$.

At imposed temperature and pressure, with fixed species amounts and only pressure-volume work, increasing total system-plus-reservoir entropy is equivalent to decreasing $G=U-TS+pV$. Unconstrained particle exchange requires the reservoir chemical potentials to be included; minimizing $G$ over arbitrary particle number is a different problem.

For positive thermal diffusivity $\nu$, thermal conductivity $k$ and length $L$, subtract the stationary solution $w(x)=Hx/k$. The remainder satisfies the heat equation with a homogeneous Dirichlet boundary condition at zero and a homogeneous Neumann boundary condition at $L$. The mixed Dirichlet-Neumann modes on an interval are $\sin (k_{n}x)$, where $k_n=(n+1/2)\pi /L$. Thus for an initial temperature profile $u_0$,For zero initial temperature, $b_n=-2H(-1{)}^n/(kL{k}_n^2)$. The series gives the initial profile in the usual Fourier series sense and a smooth solution for $t>0$. If $H\ne 0$, the zero initial profile does not satisfy the prescribed endpoint derivative at $t=0$: the temperature can be continuous there without its spatial derivative extending continuously to that corner. The sign $k{u}_x(L,t)=H>0$ describes incoming heat flux, since the outward conductive heat flux is $-k{u}_x$.

Stipulate a model receiver held at $T=5.4\mathrm{K}$ with a reference at $T_b=5.0\mathrm{K}$. Choose $G=5\mathrm{mW}{\mathrm{K}}^{-1}$, so outward thermal conduction is $G(T-T_b)=2\mathrm{mW}$. Ignore other heat transfer, take $P_{\mathrm{par}}=0$ in both states, and stipulate perfect optical/electrical thermal equivalence. These values specify an original example; they are not measurements or claimed performance of a NIST instrument.

With the beam blocked, supply $P_{h,d}=2\mathrm{mW}$. With light admitted, the required heater power is $P_{h,l}=1.005\mathrm{mW}$. For stipulated radiometric absorptance $a=0.995$ at the chosen single wavelength,The illuminated receiver still gets $1.005+0.995=2\mathrm{mW}$ of heating. The remaining $0.005\mathrm{mW}$ of incident radiant flux escapes without heating it. Calling the measured heater reduction “incident power” would underestimate the stipulated incident radiant flux by $0.5\%$.

Two variations expose different pitfalls. First, if admitting the beam also adds $0.010\mathrm{mW}$ of stray heating while the same $0.995\mathrm{mW}$ is absorbed from the intended beam, the heater falls to $0.995\mathrm{mW}$. The apparent reduction is now $1.005\mathrm{mW}$; subtracting $\Delta P_{\mathrm{par}}=0.010\mathrm{mW}$ recovers the beam's absorbed power. Second, keep the original heater readings but stipulate a light-state receiver offset $\delta T=+1\mathrm{mK}$ at fixed reference temperature. The link carries an extra $G\delta T=0.005\mathrm{mW}$; the inferred absorbed power becomes $1.000\mathrm{mW}$, and incident radiant flux becomes $1.000/0.995\simeq 1.005025\mathrm{mW}$. This second variation deliberately violates exact fixed-temperature substitution.

The heater can supply only nonnegative power in this model. Consequently an absorbed beam above the available dark heater power cannot be compensated at the chosen setpoint merely by reducing the heater. The operating point, link, or admitted radiant flux must change; zero heater demand is a range boundary, not evidence of greater radiant flux sensitivity.

In 1851 Hippolyte Fizeau detected a change in light propagation through moving water using interference fringes. Sunlight entered a telescope through a partially reflecting mirror; two slits selected beams passing through parallel glass tubes, each $1.487\mathrm{m}$ long with $5.3\mathrm{mm}$ internal diameter. A second telescope and focal mirror exchanged the tubes on return, and the beams recombined at a graduated eyepiece. Compressed air drove water between reservoirs, producing opposite flows; reversing both flows reversed the interference fringe displacement. The double transit suppressed static differences in temperature and pressure; filling only one tube with water tested this compensation. Separate supports reduced mechanical disturbance. Hippolyte Fizeau estimated flow speed from volumetric flow rate divided by tube cross section.^[ref]

The Fizeau experiment turns a tiny change in phase velocity into a visible fraction of an interference fringe. The Fizeau experiment path-length convention below explains why both tube passages contribute with the same sign for one beam and the opposite sign for the other. This makes the apparatus sensitive to the part of the phase difference that reverses with the flow, while rejecting disturbances shared by both beams.