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Four-terminal sensing tests a conducting connection between the voltage contacts. A magnetometer instead tests the sample's magnetic response. These measurements provide complementary constraints: the first is sensitive to a continuous path, while the second also depends on screening geometry and preparation history.

For transport, define $V_{ab}$ between the inner contacts and use the reversed-electric current estimator from four-terminal sensing. Contact separation excludes drive-contact drops, but does not eliminate a drifting thermoelectric effect, Joule heating or an unrepresentative path through part of the specimen. Repeat at smaller electric currents and compare results at the same sample temperature. A finite-current result refers to that electric current and its relation to the path's superconducting critical current.

For magnetism, distinguish cooling in zero applied magnetic field and then applying it from the Field-cooled Meissner experiment. Record the applied magnetic field, background and sample geometry when interpreting magnetization or magnetic susceptibility. A transport short does not establish full Meissner effect screening throughout the specimen. Conversely, flux pinning in a type-II superconductor can make the field-cooled magnetic response differ from ideal complete expulsion. Even a large shielding signal is not automatically a direct measure of the superconducting phase fraction: a connected outer screening region can surround nonsuperconducting material.

For equal thermal properties and bounded travelling profiles on two deep half-spaces, the solid temperature is $T_m$ and the liquid profile is exponential. The Stefan condition requires $L=c_p\Delta T$. This balances the latent heat of solidification with warming of the initially supercooled liquid.

For $F=F_0+b(T-T_c)M^2+d^{\prime }{M}^6$ with $b,d^{\prime }>0$, the ordered minimum has $M^4=b(T_c-T)/(3d^{\prime })$. Its singular free energy is $-2b^{3/2}(T_c-T{)}^{3/2}/\sqrt{27d^{\prime }}$. Twice differentiating in temperature gives the displayed ordered-side heat capacity singularity, with mean-field critical exponent $\alpha =1/2$. This path keeps the quartic coefficient tuned to zero; a generic path with a positive quartic term instead crosses over to the ordinary quartic result.

The triple-alpha process fuses three helium-4 nuclei into carbon-12 through an unstable beryllium-8 intermediate. Its steep temperature dependence can make helium-shell burning thermally unstable. The resonant triple-alpha temperature sensitivity isolates the role of the Hoyle state, while laboratory evidence for the Hoyle state explains how a particle spectrum tests the required nuclear level.

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]

A triple-point cell turns phase coexistence into a reproducible reference for thermometer calibration. The NIST water-cell demonstration uses a sealed glass vessel containing water of specified isotopic composition and an open central thermometer well. Cooling this well with solid carbon dioxide and alcohol freezes an ice mantle while leaving liquid water and its vapor. The coolant is outside the sealed sample; the sample's vapor space does not communicate with room air.^[ref]

In the actual NIST cell-comparison experiment, newly frozen mantles were compared with an aged reference mantle in a controlled maintenance bath. A warm aluminium rod melted a thin inner layer around the well, allowing the mantle to rotate freely and establishing an ice–water interface close to the sensor. A standard platinum resistance thermometer and a resistance bridge measured the electrical resistance ratio to a calibrated, temperature-controlled reference resistor. Readings at two currents permitted extrapolation to zero electrical power, correcting self-heating; measured immersion depths supplied hydrostatic pressure corrections. Comparing preparations over time tested stabilization rather than assuming freshly frozen ice was ready.^[ref]

The inference is a reproducible fixed-point response for the thermometer, not a direct reading of three chemical potentials. The reference cell anchors the comparison; a water point alone does not determine a thermometer's response at every other temperature. The ideal $F=0$ argument excludes impurity and isotope variation, finite interface effects and unintended heat transfer; real cells also require adequate immersion and stable electrical references. Repeatability is evidence that these effects are controlled, not a reason to omit their uncertainty.

Since 20 May 2019 the kelvin has been defined through a fixed numerical value of the Boltzmann constant. The thermodynamic temperature of the water triple point is consequently a measured quantity with uncertainty, rather than an exact value imposed by the kelvin definition. Its continued use as an assigned reference on the International Temperature Scale of 1990 is a separate convention.^[ref][ref]

The troposphere is the lowest main layer of the terrestrial atmosphere, with weather and generally decreasing temperature with altitude. Local inversions and moist convection complicate a single lapse-rate description.

Daniel Tsui, Horst Störmer and Arthur Gossard reported the effect in 1982. They grew gallium arsenide–aluminium gallium arsenide layers by molecular beam epitaxy, separating the two-dimensional electron gas from its semiconductor doping. They cut Hall bars and made indium contacts. Low-field measurements established number density and charge carrier mobility. A perpendicular magnetic field sweep measured transverse and longitudinal sheet resistivities at several temperatures. Cooling sharpened the plateau near $3h/e^2$ and the longitudinal minimum at filling factor $1/3$; integer plateaus provided an internal comparison. The longitudinal minimum remained finite at the lowest temperature: zero-temperature vanishing was unresolved. This established anomalous transport, without directly measuring excitation electric charge.^[ref]

The two readouts test different statements. A single intersection with $R_H=3R_K$ would also occur on the smooth classical relation $R_H=B/(n_{s}e)$ at $n_{s}h/(eB)=1/3$. A quantum Hall plateau instead persists over a magnetic field interval. Comparing transverse and longitudinal signals helps distinguish that plateau from a contact-placement artefact; the quantum Hall conductivity tensor inversion quantifies the contamination when longitudinal sheet resistivity is nonzero.

Later, 1997 shot noise measurements associated with quantum tunnelling at filling factor $1/3$ tested the excitation electric charge separately, finding $e/3$. That is a different observable from a Hall plateau.^[ref]

A tunnel diode is a p-n junction with sufficiently heavy semiconductor doping on both sides to make a thin depletion region and permit substantial interband tunnelling. The neutral regions are degenerate semiconductors. Its distinctive forward current-voltage characteristic contains a peak followed by a valley: over part of the curve, increasing the applied electric potential difference decreases the electric current. The Esaki tunnel-diode experiment made this behavior experimentally concrete.

The useful quantity is the local slope, not the ratio of total voltage to total electric current. With conventional electric current $I$ entering the positive-voltage terminal, defineNegative differential resistance means $g_d<0$, hence $r_d<0$ where defined. A forward operating point can still have $V>0$, $I>0$ and absorbed power $VI>0$. A small voltage increase then gives a negative current increment; this does not make the total electric current negative or turn the device into an unlimited source of energy. The external supply sets the operating point and supplies the energy available for an electronic amplifier.

An original idealized construction explains the shape. Put the n-type semiconductor on the left and the p-type semiconductor on the right. Take $q>0$ as the elementary charge and $V=V_p-V_n$. Increasing forward bias lowers the right-hand electron energy levels relative to the left. Denote the two reservoir electrochemical potentials by ${\mu }_n,{\mu }_p$, so that ${\mu }_n-{\mu }_p=qV$. At zero temperature, occupied left-hand conduction band states extend up to ${\mu }_n$; empty right-hand valence band states start at ${\mu }_p$. For transfer conserving electron energy, the available interval isIt is nonempty only when its upper endpoint exceeds its lower endpoint. Both an occupied initial quantum state and an empty final quantum state must exist at the same energy. The Pauli exclusion principle prevents transfer into an already occupied final quantum state.

For the original diagram, choose arbitrary energy units with $E_{c,n}=-0.5$, ${\mu }_n=0$, ${\mu }_p=-u$ and $E_{v,p}=0.5-u$, where $u$ is $qV$ in those units. The available interval has widthThus $L(0.25)=0.25$, $L(0.50)=0.50$, $L(0.70)=0.30$, and $L(1.15)=0$. Initially forward bias opens a transfer window; further forward bias closes it by moving the right-hand valence band edge below the left-hand conduction band edge. This is a state-counting illustration, not a formula for a measured electric current.

For reverse bias, the transfer direction changes: occupied right-hand valence band states can feed empty left-hand conduction band states. In the same construction, $u=-0.70$ opens the interval $0<E<0.70$ for this reverse transfer. A narrow depletion region allows a large reverse quantum tunnelling electric current at small electric potential difference, whereas the ordinary Shockley diode equation predicts only a small reverse saturation electric current before diode reverse breakdown. At zero biasing, equal reservoir occupations give zero net electric current; quantum tunnelling does not require a net equilibrium flow.

A finite-temperature refinement replaces sharp occupation cutoffs with the Fermi-Dirac distribution. An illustrative weak-transfer expression isHere $\mathcal{W}\ge 0$ describes allowed initial/final quantum states and their coupling, and $f_n,f_p$ are Fermi-Dirac distributions with the respective electrochemical potentials. The occupation difference follows by subtracting the reverse transfer factor from the forward one: $f_n(1-f_p)-f_p(1-f_n)=f_n-f_p$. An overlap interval alone does not determine $\mathcal{W}$: transmission probability, density of states, crystal momentum selection and phonons can matter. A complete germanium calculation is more demanding than multiplying two constant density of states factors. At larger forward bias, ordinary carrier injection and diffusion supply a rising current branch even after direct band overlap is lost.

A reduced model with warm and cold mixed reservoirs whose signed exchange volume flux is proportional to their density difference. For fixed temperature contrast and reference-salinity freshwater forcing, the stationary salt budget is $|q|\Delta S=E{S}_0$. A linear equation of state then gives $q|q|-k\alpha \Delta T|q|+k\beta S_{0}E=0$. Positive circulation has two branches below a saddle-node bifurcation; the stronger positive branch and the reversed branch are stable under the usual salt budget dynamics.

For two polarizations, $n$ atoms in area $A$, and $\omega =C|k{|}^{\alpha }$, mode counting givesAt low temperature its energy scales as $T^{1+2/\alpha }$ and its constant-volume heat capacity as $T^{2/\alpha }$.

A nondegenerate donor-bound state of energy zero and an excited representative conduction level of energy $\Delta >0$ and degeneracy $g$ have the displayed bound probability by canonical state counting. The inferred degeneracy from one measurement is $g=(P_b^{-1}-1)e^{\Delta /(k_{B}T)}$. At finite fixed $g$, the probability decreases from1 toward $1/(1+g)$ as temperature grows. A temperature-independent effective degeneracy is an assumed two-level approximation; a full conduction-band density of states can behave differently.

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.

For a stable radiometer with affine output $S=g(T+T_r)$, let $g>0$ be the fixed response coefficient and $T_r$ the receiver noise temperature. Two reference loads with known equivalent input noise temperatures $T_c<T_h$ give outputs $S_c$ and $S_h$. Subtracting their outputs eliminates $T_r$; dividing source and reference differences eliminates $g$:This two-load radiometer calibration assumes unchanged response and receiver noise temperature between readings, with the loads and source referred to the same input. It also eliminates an additional constant output offset. Physical load temperatures must first be converted to the applicable equivalent input noise temperatures when the classical thermal approximation fails. The Penzias-Wilson microwave background experiment gives an original worked example.

Project a quantized field onto two low-energy orbitals. A sufficient regime has temperature, hopping and interaction shifts small compared with the energy gap to discarded orbitals. Small interwell contact interaction integrals can then be neglected while retaining a small quantum tunnelling matrix element. The projected field's operator commutator is the two-mode projection operator kernel, rather than a complete spatial Dirac delta function.

Consider an original worked thermometer calibration, using the height $X$ of a liquid column responding through thermal expansion. Stipulate two equilibrated reference baths whose corrected thermometer readings are $0{}^{\circ }\mathrm{C}$ and $40{}^{\circ }\mathrm{C}$, in degree Celsius units. Suppose the corresponding column heights are $12\mathrm{mm}$ and $52\mathrm{mm}$. These are constructed inputs, not observed measurements. Assume the response is linear over this interval, with the same immersion and reading procedure throughout. Applying thermometer calibration givesFor $X=37\mathrm{mm}$, first subtract the zero-reference height, then divide by the full reference span: $(37-12)/(52-12)=25/40$. Multiplying by the $40{}^{\circ }\mathrm{C}$ span gives $t_C=25{}^{\circ }\mathrm{C}$, or $T=298.15\mathrm{K}$ in kelvin. Thus equal $37\mathrm{mm}$ readings in separately equilibrated samples test equal temperatures, regardless of how much internal energy those samples contain. The Celsius offset and modern absolute-scale definition are described in degree Celsius and kelvin.

The two endpoints determine the assumed line, but do not establish linearity. An independent middle reference and a return to the initial bath can test curvature and drift. If the height has an error of $0.2\mathrm{mm}$ while reference values and endpoints are held exact in this calculation, the inferred temperature error is $0.2{}^{\circ }\mathrm{C}$; that is only the height contribution, not a full measurement error budget. A boiling-water reference needs a measured atmospheric pressure correction, rather than an automatic $100{}^{\circ }\mathrm{C}$ assignment. The modern kelvin is not defined by either bath.

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.

Choose two Ising spins with Hamiltonian $H=-J{s}_1{s}_2$, where $J$ independently takes $J_0$ or $2J_0$ with equal prior probabilities and $J_0>0$. This original finite example has no geometrical frustration and no phase transition. Its four spin assignments give $Z_J=4\cosh (\beta J)$. ThereforeThe concave function $\log$ puts $F_{\mathrm{a}}\le F_{\mathrm{q}}$. For this two-value case the claim is just $(Z_1+Z_2)/2\ge \sqrt{Z_1{Z}_2}$, obtained by squaring $\sqrt{Z_1}-\sqrt{Z_2}$. Thus no formal continuation of replicas is needed to see the distinction.

In the annealed disorder ensemble, the stronger bond has probability $Z_{2J_0}/(Z_{J_0}+Z_{2J_0})$, approaching one at low temperature. A quenched disorder collection instead retains one half at each coupling. As $T\to 0$, $F_{\mathrm{q}}\to -3J_0/2$ and $F_{\mathrm{a}}\to -2J_0$: averaging samples and allowing the material variables to equilibrate describe different physics.