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A two-bath step test of a temperature sensor estimates the dynamic response of a temperature sensor by transferring it between two approximately uniform, stable baths at different temperatures. The apparatus consists of the two baths, independent reference thermometers, a reproducible immersion arrangement, and a recorder fast enough to resolve the temperature sensor response time. Rapid insertion into a liquid at a different temperature is a practical approximation to an instantaneous change in the sensor's surroundings.^[ref]

Let the initial equilibrium be $T_1$ and the second bath be $T_2$. Convert the recorded sensor output to temperature, then form the dimensionless residualFor a single lumped thermal model, $r=e^{-t/\tau }$, so a plot of $\ln |r|$ against elapsed time is a straight line with slope $-1/\tau$. This provides a direct check on the assumed time constant; a curved plot suggests additional response stages or an imperfect input step. The conversion to temperature matters when sensor output is nonlinear.^[ref]

For a constructed example, transfer a sensor from $20{}^{\circ }\mathrm{C}$ into a $60{}^{\circ }\mathrm{C}$ bath. Readings of $40$, $50$, and $55{}^{\circ }\mathrm{C}$ after $2$, $4$, and $6\mathrm{s}$ give residuals $1/2$, $1/4$, and $1/8$. Each successive interval halves the residual, giving $\tau =2/\ln 2\approx 2.89\mathrm{s}$ and $t_{90}\approx 6.64\mathrm{s}$.

Record the transfer delay, immersion depth, stirring conditions, probe sheath and readout filtering with the result. These affect heat transfer or the measured transient. A temperature sensor response time measured in one medium need not apply in another.^[ref]

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.

A two-level system is a model with two accessible states or two groups of states separated by an energy gap. Choosing the lower energy as zero gives $E_0=0$ and $E_1=\Delta >0$. The approximation is useful when all other levels are inaccessible on the temperature and driving scales being studied. A spin in an applied magnetic field can provide a two-state example through Zeeman splitting.

For two nondegenerate levels in a canonical ensemble, $Z=1+e^{-\beta \Delta }$ and the upper-state probability is $p_1=1/(1+e^{\beta \Delta })$. At low positive temperature the upper level is rarely occupied; at high temperature the two probabilities approach one half. The mean energy is $\Delta p_1$, while the heat capacity is largest at an intermediate scale because occupation is then most sensitive to warming. This produces a Schottky anomaly. A coherent superposition is also possible in a quantum model, but the equilibrium occupation calculation specifies probabilities in the energy basis rather than a chosen coherent drive.

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.

An extra relativistic neutrino that decouples before ordinary neutrinos misses two stages of plasma heating. Until ordinary decoupling, entropy conservation gives a temperature ratio $(43/(4g_d){)}^{1/3}$. Electron-positron annihilation then heats photons but neither neutrino population, multiplying the ratio by $(4/11{)}^{1/3}$. Here $g_d$ counts the coupled bath at the extra species' decoupling, with no subsequent exotic entropy injection.

Here $s(m)=\frac{1+m}{2}\ln \frac{1+m}{2}+\frac{1-m}{2}\ln \frac{1-m}{2}$. This free-energy density is the energy minus temperature times entropy of independent sublattice Ising spins. Since $s^{\prime }(m)=\mathrm{atanh}m$, stationarity yields the two-sublattice Ising mean-field approximation. Writing $M_{\pm }=(M_B\pm M_W)/2$, the quadratic coefficients are $(k_{B}T+qJ)/2$ for $M_{+}^2$ and $(k_{B}T-qJ)/2$ for $M_{-}^2$. The positive quartic coefficient $k_{B}T/12$ in the staggered direction stabilizes the continuous transition. Unlike a log-cosh auxiliary potential, this variational expression can be minimized away from self-consistency.

For a constant-parameter Arrhenius equation $R(T)=A{e}^{-\beta /T}$ at positive absolute temperatures, subtracting logarithms eliminates the Arrhenius prefactor. The ratio of the two resulting equations gives the displayed prediction. It applies to a reaction rate only when composition and other factors relating rate to the rate constant are held fixed. Equal increments of temperature do not give equal multiplicative rate changes, since the linear variable is $1/T$.

For a species $s$ in two well-mixed vessels of equal volume $V$ joined by a thin opening of area $A_h$, the effusion law gives ${\dot{n}}_{s,L}=-(A_h/4V)({\overline{v}}_{s,L}{n}_{s,L}-{\overline{v}}_{s,R}{n}_{s,R})$. Each species' total inventory is conserved. With fixed temperatures and total density sum $n_{s,L}+n_{s,R}=n_{s,\Sigma }$, the left density relaxes exponentially with time constant ${\tau }_s=4V/[A_h({\overline{v}}_{s,L}+{\overline{v}}_{s,R})]$ to $n_{s,L}^{\infty }=n_{s,\Sigma }\sqrt{T_R}/(\sqrt{T_L}+\sqrt{T_R})$.

At equal temperatures, ${\tau }_s=2V/(A_h{\overline{v}}_s)\propto \sqrt{m_s}$, and initially separated species mix at different rates. At unequal temperatures, the final density ratio is $n_{s,L}/n_{s,R}=\sqrt{T_R/T_L}$ for every species, while the partial-pressure ratio is $p_{s,L}/p_{s,R}=\sqrt{T_L/T_R}$. The mass-independent final ratios do not imply mass-independent transient rates. Maintaining different temperatures requires external heat exchange even after net particle transfer vanishes.

A type-II superconductor can admit magnetic flux through quantized vortices while retaining superconductivity between their cores. Its superconducting mixed state lies between the lower critical magnetic field and upper critical magnetic field. These are equilibrium bulk distinctions; sample shape, flux pinning and preparation affect a measured magnetic response.

The motivating thin-film experiments changed an apparently minor preparation detail. Nikolay Zavaritskii evaporated metal onto glass; Alexander Shalnikov proposed cooling the substrate to helium temperatures to suppress agglomeration. The resulting critical-field behavior no longer matched the small-parameter approximation. Alexei Alexeyevich Abrikosov found a different regime within the same theory, rather than discarding its equations.^[ref]

An ultracold gas is a dilute gas cooled until atomic motion becomes slow enough that quantum mechanics is essential to its preparation, collisions or collective state. There is no single temperature boundary independent of particle mass and number density: the overlap of thermal matter waves depends on the thermal de Broglie wavelength relative to interparticle spacing.

Laser cooling can prepare a cold trapped sample, and evaporative cooling of atoms can further increase its phase-space density of an atomic gas. Expansion after releasing a trap gives information about the velocity distribution. A small measured cloud alone is insufficient evidence for low temperature, because spatial confinement also depends on the trapping potential energy.

For spin-one-half particles in three dimensions with dispersion $\epsilon =pc$, the density of states isAt zero temperature,

Ultraviolet is electromagnetic radiation with wavelengths shorter than the violet end of ordinarily visible light. The CIE convention uses $100$–$400\mathrm{nm}$, subdivided into UV-A at $315$–$400\mathrm{nm}$, UV-B at $280$–$315\mathrm{nm}$ and UV-C at $100$–$280\mathrm{nm}$. These divisions name intervals; they do not mark universal changes in every material's response. The boundary with visible light is not sharp, and other fields use near-, far-, vacuum- and extreme-ultraviolet divisions. ^[ref]. For example, NASA's Solar Dynamics Observatory observes extreme-ultraviolet bands below $100\mathrm{nm}$; the CIE interval is therefore not the full extent of astronomical ultraviolet usage. ^[ref].

The CIE ultraviolet interval corresponds to frequencies about $7.49\times 10^{14}$–$3.00\times 10^{15}\mathrm{Hz}$ and photon energies $4.97\times 10^{-19}$–$1.99\times 10^{-18}\mathrm{J}$, using vacuum wavelength and $E=hc/\lambda$ with Planck constant $h$ and speed of light $c$. Shorter wavelength means larger energy per photon; larger irradiance means more incident power per area. These are different comparisons. At the same incident power, a $254\mathrm{nm}$ ultraviolet beam has fewer photons per second than a $940\mathrm{nm}$ infrared beam, even though each ultraviolet photon has more energy.

A common ultraviolet mechanism is photon absorption promoting an electron from a ground state to an excited state of a chemical atom or molecule. By contrast, many infrared absorption bands involve stretching or bending chemical bonds. For either mechanism, both the energy difference and the coupling to the incident electromagnetic field matter. One concrete example is the $217\mathrm{nm}$ absorption discussed for butadiene: an electron moves between molecular orbitals, and the position of the absorption reveals their energy separation. Ultraviolet absorption can therefore probe molecular orbitals, while infrared spectroscopy probes many molecular motions. ^[ref].

Subsequent chemical reactions depend on the absorbing chemical substance and its environment; the label ultraviolet alone does not predict a reaction rate. At sufficiently short wavelength, a photon can cause photoionization if its energy exceeds the relevant ionization energy. There is no single wavelength that separates all ionizing from all nonionizing behavior for every chemical substance. Likewise, eventual heating after photon absorption is possible in both ultraviolet and infrared. The distinction is useful because different optical spectra expose different processes, not because the two ranges obey different laws of electromagnetism.

Ritter's 1801 observation used a chemical readout rather than a thermometer. In Caltech/IPAC's account, sunlight passed through a glass optical prism and silver chloride was placed in different parts of the optical spectrum. Darkening was weak toward red, stronger toward violet, and pronounced beyond the visible violet edge. The measured evidence was a change in the chemical substance's appearance at a spatial location with no visible color, establishing radiation beyond the eye's ordinary range. It was not a calibrated measurement of a universal ultraviolet optical spectrum. ^[ref].

A controlled modern interpretation of such a chemical test would compare equal exposure durations and matched fresh samples, keep an unexposed sample in darkness, monitor temperature, and distinguish scattered visible light from the intended short-wavelength beam with characterized filtering. These are proposed controls, not an assertion that Ritter performed them. Unequal sample preparation or heating can change a reaction rate. The optical prism, terrestrial atmosphere and sample response restrict which wavelengths the test detects. Darkening is a detector-weighted response, so its spatial maximum is not automatically the maximum of incoming radiant flux or of photon energy.

Thomas Larason and Jeanne Houston's NIST report documents a quantitative ultraviolet detector comparison. An argon source feeds a double-diffraction grating spectrometer; a shutter and mirrors deliver the selected band to a working-standard or test photodiode, successively at the same position. A beam splitter supplies a monitor photodiode. Amplifiers and voltmeters record detector and monitor simultaneously; shutter-closed readings provide dark baselines. The $200$–$500\mathrm{nm}$ scan, typically with $4\mathrm{nm}$ bandpass, measures spectral responsivity. Monitor normalization suppresses source drift; known spectral lines calibrate wavelength. The report assesses temperature effects and unwanted light outside the band, weighted by detector response. The latter need not cancel between different spectral responsivities. ^[ref].

This determines electrical response without visual perception. Bandpass and spatial nonuniformity limit the calibration; aging and changed temperature can affect later use. The scan includes visible light and ultraviolet; no response below $200\mathrm{nm}$ follows from it.

For a narrow beam, the idealized substitution calculation isHere $R$ is spectral responsivity; $I$ and $M$ are dark-subtracted detector and monitor electric currents. This follows by writing $I=RP$ for each detector and using the monitor to track changes in radiant flux. It assumes linear response, unchanged beam splitting and collection geometry, and a sufficiently narrow band. For an illustrative standard with $R_{\mathrm{std}}=0.15\mathrm{A}{\mathrm{W}}^{-1}$, ratios $I_{\mathrm{std}}/M_{\mathrm{std}}=2.0$ and $I_{\mathrm{test}}/M_{\mathrm{test}}=1.6$ give $R_{\mathrm{test}}=0.12\mathrm{A}{\mathrm{W}}^{-1}$. These invented arithmetic inputs illustrate the method; they are not NIST measurements.

An independent photon-counting model connects spectral responsivity to quantum efficiency. If $\eta (\lambda )$ is the number of collected electrons per incident photon, there is no internal multiplication, and $q$ is the elementary charge, thenAt illustrative constant $\eta =0.60$ and $P=1\mu \mathrm{W}$, this gives about $0.123\mu \mathrm{A}$ at $254\mathrm{nm}$ and $0.455\mu \mathrm{A}$ at $940\mathrm{nm}$. The ultraviolet readout is smaller in this model because there are fewer incident photons, not because their individual energies are smaller. Real quantum efficiency depends on wavelength, windows, absorption and carrier collection; a detector's band gap and construction determine its usable range. Constant $\eta$ across these two wavelengths is an analytical comparison, not a claimed specification of a real photodiode.

The curve is the calculated relation $E=hc/\lambda$, and the shaded ranges use the stated CIE conventions. The lower diagram abstracts the detector comparison rather than reproducing an instrument drawing. Together with the infrared heat-balance illustration, it shows why a calibrated detector response, rather than what looks bright to an observer, is the bridge between invisible radiation and a quantitative measurement.

An Umklapp process conserves reduced crystal momentum modulo a nonzero reciprocal lattice vector. It can occur in electron-phonon scattering as well as phonon-phonon scattering; the lattice takes up the reciprocal-lattice momentum.

In a three-phonon Umklapp scattering process, the first-zone wavevectors obey the displayed equation for a nonzero reciprocal lattice vector $\mathbf{G}$, together with conservation of energy. The crystal momentum transferred to the lattice allows these processes to relax a phonon heat current. They are suppressed at sufficiently low temperature and often determine the approximate $1/T$ high-temperature thermal conductivity of a crystalline electrical insulator.