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For a planar potential flow outside a bounded body, suppose its single-valued complex velocity $q=u-iv$ is holomorphic near infinity and tends to zero. Its exterior Laurent series begins $q=b_1/z+O(z^{-2})$. The outward volume flux $Q$ and counterclockwise circulation $\Gamma$ satisfy $\Gamma +iQ=2\pi i{b}_1$. Thus both vanish precisely when the leading term is absent. A nonzero $b_1$ gives $|q{|}^2\sim |b_1{|}^2/r^2$ and logarithmically divergent exterior kinetic energy. When $b_1=0$, $q=O(r^{-2})$ and the far-field energy integral converges. Zero circulation alone does not exclude the radial source field $\nabla \log r$.

A finite-energy field configuration has integrable energy density after the vacuum energy has been subtracted. The boundary conditions used for a localized classical field-theory soliton require approaching a vacuum at spatial infinity. In one dimension the two ends may approach distinct scalar-field vacua; in a gauge theory the asymptotic field can have nontrivial winding even when its local energy density tends to zero.

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.

The edges of an itinerant spin-flip particle-hole continuum follow from both excitation energy and occupation constraints. Consider a zero-temperature three-dimensional parabolic-band model with majority states occupied for $|\mathbf{k}|\le k_F$ and an entirely empty minority band. For exchange splitting $2M$ and ${ϵ}_k={\hslash }^2{k}^2/(2m)$, a majority-to-minority flip carrying wavevector $\mathbf{q}$ hasFor $q>0$, the scalar product is bounded by $\pm k_{F}q$, giving continuum edgesThe extremes occur at opposite points of the occupied Fermi surface along $\mathbf{q}$. At $q=0$ the interval collapses to the single exchange energy $2M$.

The empty-minority assumption requires sufficient splitting; otherwise Pauli exclusion principle blocks some final states and the edges must be recomputed over the allowed region. These are independent particle-hole thresholds, not a calculation of the collective ferromagnetic magnon dispersion. A collective mode below the continuum can therefore have a different low-momentum energy scale.

Take a nonrelativistic particle of constant mass $m$ with energy $E>0$, and choose the same zero of potential energy on both sides of a rectangular potential barrier:The equality of the exterior potential energies matters: it gives the same exterior wavenumber $k=\sqrt{2mE}/\hslash$ and makes $T=|t{|}^2$. Here $\hslash =h/(2\pi )$ is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.

For $0<E<V_0$, the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. We set the incident probability amplitude to one and impose no incoming wave from the right; this scattering wavefunction is not a normalizable wavefunction. The tunnelling decay constant $\kappa$ describes exponential decay, rather than an oscillatory wavenumber in the barrier. Both exponential functions are needed on the finite interval: the growing term is not excluded by a condition at infinity.

The wavefunction and its derivative are continuous at each face, giving four linear matching conditions:Their solution supplies the transmission amplitudeA short derivation is to use the two right-face equations to express $A{e}^{\kappa a}$ and $B{e}^{-\kappa a}$ in terms of $t$, then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.

For arbitrary maps $\varphi :{\mathbb{R}}^2\to S^2$, integrability of the Dirichlet energy does not imply continuous extension to the one-point compactification. For large $r$, takeand interpolate smoothly to a constant on a bounded disk. Its exterior energy isThe field has no limit because $a(r)$ keeps winding. A continuous extension at the point at infinity instead requires $\varphi (x)\to {\varphi }_{\infty }$ uniformly outside sufficiently large disks. This fixed-limit boundary condition is standard for a sigma-model lump, but is extra information beyond finite energy for arbitrary fields.

For a noninteracting ground state of energy zero, $N_0=(e^{-\beta \mu }-1{)}^{-1}$ gives $\mu =-k_{B}T\ln (1+1/N_0)$. Thus a finite condensate has a slightly negative chemical potential; exact zero pinning is a macroscopic limiting statement.

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.

A finite square well has a constant lower potential energy over a finite interval and a higher constant potential outside. For a symmetric well $V=-V_0$ on $|x|<a$ and $V=0$ outside, a bound state has $-V_0<E<0$: the wavefunction oscillates inside and decays exponentially outside.

Continuity of the wavefunction and its derivative at $x=\pm a$ gives $k\tan (ka)=\kappa$ for even parity and $-k\cot (ka)=\kappa$ for odd parity, where $k=\sqrt{2m(E+V_0)}/\hslash$ and $\kappa =\sqrt{-2mE}/\hslash$. These conditions select finitely many bound state energies. Unlike an infinite square well, the finite well allows penetration beyond the edges; a shallow one-dimensional attractive well still has an even ground state. Positive-energy scattering states lie above the outside potential.

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.

Finite-temperature quantum field theory computes field observables in thermal equilibrium, typically $\rho =e^{-\beta H}/Z$ with $\beta =1/(k_{B}T)$. The thermal trace has an imaginary time representation of length $\beta \hslash$ in physical time units, or $\beta$ after setting $\hslash =1$.

Bosonic fields are periodic around this thermal interval, while fermionic fields are antiperiodic. Their Fourier expansions produce different Matsubara frequencies, encoding the quantum statistics. Imaginary time correlations are useful for equilibrium calculations, while a real-time response or relaxation problem needs an appropriate continuation or real-time formalism. Temperature therefore changes the state and boundary prescription, rather than simply replacing every zero-temperature energy by a classical thermal value.

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.

The experiment identifies tau lepton pairs with mixed electron-muon decays and missing energy, and subtracts a nonresonant continuum contribution. This provides a concrete implementation of a leptonic tau lepton selection near the charmonium threshold.

Let $X$ be a unitary operator with $X^2=I$, and let $P$ be the orthogonal projection operator onto an unperturbed energy eigenspace. Suppose $X$ acts as the same sign $\epsilon =\pm 1$ throughout that eigenspace and $XVX=-V$. For any $u,v$ in the eigenspace, preservation of the inner product gives $⟨u,Vv⟩=⟨Xu,XVv⟩=-⟨\epsilon u,V\epsilon v⟩=-⟨u,Vv⟩$. Hence every projected matrix element vanishes, so $PVP=0$ and degenerate perturbation theory gives zero first-order energy corrections. A zero diagonal alone would be insufficient. For equal spins, the exchange parity of two coupled identical spins is constant on each total-spin sector, while their spin-projection difference changes sign under exchange.

For a piston advancing slowly into a gas, the arrival rate of molecules differs slightly from that at a stationary wall. In calculating compression work from molecular impacts to first order in piston speed $u$, however, the stationary molecular arrival flux is sufficient: the energy change per reflected molecule is already proportional to $u$.

A molecule with incoming normal velocity $v_x$ acquires leading energy $2mu{v}_x$. Multiplying by the stationary arrival weighting proportional to $v_x$ and averaging over an isotropic distribution gives the pressure work rate $pSu$, where $S$ is piston area. The first-order correction to the arrival flux multiplies an order-$u$ energy change and contributes only at order $u^2$. This ordering requires piston speed small compared with the typical molecular speed; it does not justify the same approximation for rapid compression or for the force correction itself.

In first-order Fermi acceleration, converging scattering media produce a mean fractional energy gain linear in their relative speed divided by the particle speed. At a nonrelativistic shock wave, an almost isotropic ultrarelativistic population gains $2(u_1-u_2)/(3c)$ per crossing and $4(u_1-u_2)/(3c)$ per complete return cycle. The distinction between a crossing and a cycle must also be respected when assigning an escape probability.

At a first-order phase transition, the Gibbs free energy is continuous while at least one first derivative, such as entropy or volume, jumps. A nonzero entropy jump produces latent heat $L=T\Delta S$.

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.

Angular-relaxation collision time $\tau$ distinguishes first sound ($\omega \tau \ll 1$) from zero sound ($\omega \tau \gg 1$). Collisions establish local thermodynamic equilibrium in the former.^[ref]

For the figure's stipulated law $\omega \tau =(T/T_{\times }{)}^{-2}$, cooling from $T=5T_{\times }$ to $T=T_{\times }/5$ changes $\omega \tau$ from $0.04$ to $25$, without changing the driving angular frequency. The crossover temperature is defined by $\omega \tau (T_{\times })=1$. Calling zero sound a high-frequency mode therefore means high relative to the collision rate $1/\tau$, while its excitation energy must still remain within the low-energy domain of the Fermi liquid.