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Direction-dependent thermal expansion. Different crystal grain orientations can demand incompatible strains even under uniform macroscopic heating.

Differentiate Bragg's law at fixed correctly assigned diffraction order:Divide the second equality by the first. For small signed changes,The Bragg angle error $\delta \theta$ must be in radians. An error quoted for detector deflection $2\theta$ must first be halved. The expression is a first-order Taylor expansion; an unknown diffraction order is a discrete assignment error, not an infinitesimal perturbation.

For error bounds, add magnitudes. For independent zero-mean random errors with standard deviations ${\sigma }_d,{\sigma }_{\theta }$, variance addition instead givesCorrelated errors require a covariance term. A shared calibration bias does not become independent merely because many line positions are measured.

With a chosen angular error $10$ arcseconds, $\delta \theta =4.84814\times 10^{-5}\mathrm{rad}$, the angular contribution is $8.3972\times 10^{-5}$ at $30^{\circ }$ and $1.7646\times 10^{-5}$ at $70^{\circ }$. Larger Bragg angles reduce this contribution because $\cot \theta$ decreases. For a nearby spectral doublet, the inverse relation is $\Delta \theta \simeq \tan \theta \Delta \lambda /\lambda$: a fixed fractional separation spreads farther apart angularly at large Bragg angle.

This does not permit unlimited improvement. The geometry requires $n\lambda \le 2d$; a desired wavelength might not have an accessible large-angle reflection with the chosen crystal. Higher diffraction orders can reach larger Bragg angles but their strengths depend on the crystal lattice structure factor. Finite crystal response, slit widths and source size remain relevant. An error in $d$, including uncorrected thermal expansion, remains even when the angular term is small. Very near $90^{\circ }$, the leading angular derivative vanishes and the second-order Taylor expansion matters.

The volumetric coefficient of thermal expansion is the fractional volume increase per temperature increase at fixed pressure and composition:The mass density form requires fixed mass $m$. Its units are ${\mathrm{K}}^{-1}$; ${\alpha }_T$ is another notation for this coefficient. For small temperature changes $\theta$ at fixed pressure,For a finite change in a single thermodynamic phase, integrate the possibly varying coefficient: $\ln (V_2/V_1)={\int }_{T_1}^{T_2}\alpha (T,P,N)dT$. A constant-coefficient approximation gives an exponential, while the displayed linear formula additionally requires $|\alpha \theta |\ll 1$.

For an ideal gas at fixed particle number,Integrating $1/T$ recovers the exact fixed-pressure thermal expansion ratio $V_2/V_1=T_2/T_1$. Combining expansion with isothermal compressibility gives the local response of any smooth simple compressible system of fixed composition,The ratio has units $\mathrm{Pa}{\mathrm{K}}^{-1}$. For the one-mole ideal gas state in isothermal compressibility, it is $P_0/T_0=333.33\mathrm{Pa}{\mathrm{K}}^{-1}$: a $1\mathrm{K}$ increase at fixed volume raises the pressure by about $333.33\mathrm{Pa}$. Maxwell relations identify this same ratio as the entropy slope with respect to volume and give $C_P-C_V=TV{\alpha }^2/{\kappa }_T$.

Under the Boussinesq approximation, the small fractional mass density change produces buoyancy while mass density is otherwise treated as constant. A positive temperature excess $\theta$ gives reduced gravity $g^{\prime }=g{\alpha }_T\theta$ in the convention used in natural ventilation. The coefficient therefore converts a heat budget into a buoyancy budget. The same measurable expansion response supplies entropy changes and heat capacity differences through Maxwell relations, connecting fluid motion to equilibrium thermodynamics.

A crystal X-ray spectrometer selects a narrow wavelength interval by diffraction at known interplanar spacing $d$ and Bragg angle $\theta$. For fixed diffraction order $n$, a known $d$ lets an angular measurement determine $\lambda =2d\sin \theta /n$. This reverses the crystallography problem of finding $d$ from a known wavelength.

Siegbahn's laboratory developed X-ray tubes and evacuated spectrographs for longer wavelengths. Rotating the crystal selected reflections; photographic plates recorded positions. Opposite-side exposures measured angles on both sides of the direct beam. Later tube instruments mounted a beam-defining slit opposite the plate holder. A separate double instrument compared two reflected beams with two ionization chambers. These were different instruments, not one universal assembly.^[ref]

The Nobel biography describes constant-temperature spectrographs and a precision angular scale read by diametrically opposed microscopes.^[ref] Temperature control matters because thermal expansion changes $d$; an accurately read angle cannot compensate for an unknown interplanar spacing.

The schematic separates preparation, selection and readout. An X-ray tube target supplies Characteristic X-rays with a possible bremsstrahlung background. A collimator limits incident directions; evacuation reduces air X-ray attenuation. Rotating the crystal brings different wavelengths into reflection. A photographic plate stores line positions over an exposure, whereas an ionization chamber can measure collected electric charge at successive angles. Neither readout alone establishes absolute wavelength: an angular zero, angular-scale calibration, diffraction order, $d$ and any required refraction correction enter the inference. Narrower slits reduce accepted angular spread but also reduce signal; a weak line can become harder to locate despite a narrower instrumental response.

A control reported in Siegbahn's lecture photographed a line system with copper-sulphate and calcium-sulphate crystals. Irregular relative shifts around copper's characteristic frequency supported anomalous dispersion; the arrangement excluded differing absorption across that frequency as the explanation.^[ref] A finer angular scale alone cannot remove this physical refraction correction.

A gas thermometer infers temperature from the equation of state of a confined gas. The constant-volume version measures pressure; the constant-pressure version measures volume. For an ideal gas,Here $N$ is particle number, $n$ is amount of substance, $k_B$ is the Boltzmann constant, and $R$ is the molar gas constant. The reference and test states must be internally equilibrated, and $T_0$ must be known on the absolute thermodynamic temperature scale. A pressure ratio cancels the unknown gas inventory and bulb size in the ideal constant-volume limit. This cancellation is the method's appeal: a mechanical measurement transfers an absolute temperature scale between different materials. The Maxwell speed distribution and ideal gas pressure from wall collisions explain the microscopic origin of the response. ^[ref]

For an original ideal-model calculation, choose $T_0=300\mathrm{K}$, $P_0=120\mathrm{kPa}$ and $P=160\mathrm{kPa}$. Then $T=400\mathrm{K}$. These are stipulated values, not observations. Both pressures must be absolute, referred to vacuum. If the surrounding atmospheric pressure were $100\mathrm{kPa}$, the corresponding gauge pressures would be $20$ and $60\mathrm{kPa}$; their ratio is three, which would incorrectly predict $900\mathrm{K}$. A pressure difference from the atmosphere cannot be substituted into the ideal gas law.

A motivating calibration experiment. Schooley's account of the NBS/NIST work describes a helium-filled metal bulb, a capillary, a capacitance diaphragm gauge, and a mercury pressure standard. Circulating bath liquid passed the bulb assembly and platinum resistance thermometers. The resistance probes checked the surrounding temperature; gas pressure supplied the independent thermodynamic comparison. In an early series, workers filled the cleaned bulb with purified helium at the highest test temperature, checked pressure over three hours, then measured at lower temperatures and near a reference at $0{}^{\circ }\mathrm{C}$. Six filling pressures allowed extrapolation toward zero pressure. ^[ref]

To turn this principle into a readable measurement protocol, use the following controls. These steps are a proposed teaching interpretation of the cited apparatus, rather than additional reported NBS observations:

The corresponding NBS apparatus used a protective case with an outer helium layer held close to the bulb pressure, reducing pressure deformation while aiding thermal contact. Thermal expansion was measured on bulb material. Gas in the capillary remained partly outside the hot region, requiring a measured tube temperature profile and geometry. Pure working gas also mattered: contaminants that sorb on surfaces can change the freely circulating inventory. These are physical effects to measure or bound, rather than scatter that disappears by averaging. ^[ref]

For illustration, neglect connecting-tube gas and describe a uniform bulb with a dilute virial expansion in particle number density ${\rho }_N=N/V$:Here $Z=PV/(N{k}_{B}T)$ and $B_2$ is the second virial coefficient in the particle-density convention; a molar-density convention rescales the coefficient. The second relation follows exactly from the definition of $Z$ for the same $N$, and shows the signs of the corrections. If the bulb expands by $0.1\%$ and $Z=1$ at both states, the toy pressure ratio above gives $400.4\mathrm{K}$, rather than $400\mathrm{K}$. Improved expansion and helium nonideality models really changed the temperatures inferred from the 1989 NIST data: Harvey's 2023 reanalysis found expansion updates of roughly $1$–$3\mathrm{mK}$ and nonideality updates of roughly $-0.1$ to $-0.5\mathrm{mK}$. Those are reported corrections to historical results, not the much larger illustrative correction just calculated. ^[ref]

With equal corrected bulb volumes, negligible or corrected connecting-tube gas, fixed reference and test temperatures, and a dilute classical single gas phase,Repeat the reference/test comparison for several sealed fills, reducing $N$ between comparisons but keeping it fixed within each pair. The extrapolation varies density at fixed endpoint temperatures; it does not cool one fill toward zero temperature. Real gases have different finite-density corrections but share the classical ideal gas intercept under these assumptions. The limit does not remove leaks, bulb expansion, gauge offsets, or the gas thermometer dead-space correction. At very low pressures, thermal transpiration can make remote sensor and bulb pressures differ through a nonisothermal capillary; this must be corrected too. The NBS work measured this pressure correction using a separate comparison of narrow and wider tubes. ^[ref]

At $x=1$, the toy models give $T_{\mathrm{app}}=407.92\mathrm{K}$ and $388.12\mathrm{K}$, respectively, although their imposed true test temperature is the same. Their zero-density intercept is the reusable quantity. Measuring several low-density points permits a fit and a check for curvature; asserting a straight line from one finite-density point supplies neither. A practical fit must balance nonideality against increasing relative pressure error and the capillary corrections, and keep its extrapolation uncertainty.

Gas outside the uniform-temperature bulb of a gas thermometer contributes to its conserved particle inventory. An idealized calculation divides the connecting tube into small regions with volumes $V_j$ and local temperatures $T_j$. Neglect hydrostatic pressure offsets and thermal transpiration, assume a common pressure $P$ and a local ideal gas law in each region, and let the bulb have volume $V_b$ and temperature $T_b$. Adding the particle inventories givesThus even an ideal gas in a perfectly rigid sealed apparatus need not obey $P\propto T_b$: some of its particles remain in regions with other temperatures. For a stipulated bulb of $100\mathrm{c}{\mathrm{m}}^3$, a connecting volume of $1\mathrm{c}{\mathrm{m}}^3$ maintained at $300\mathrm{K}$, and bulb reference and test temperatures of $300$ and $400\mathrm{K}$, conservation of $N$ givesThese are toy dimensions and a calculated bias, not measured data. Reducing $N$ reduces both pressures proportionally and leaves this ratio unchanged: extrapolating away gas nonideality does not eliminate the connecting-volume error. Actual apparatus also requires the tube's thermal expansion, a measured temperature profile, and the pressure corrections omitted in this model.

A mechanical engineer whose work connected gas lighting to automatic operation. Chalmers records his mechanical-engineering qualification in 1896.^[ref] After an accident left him blind in 1912, he returned to AGA in 1913 and continued leading the company.^[ref]

The sun valve links radiometric absorptance, thermal expansion and mechanical motion to a useful decision: whether a beacon should receive its main gas supply. The Dalén light combines this daylight response with repeated flashes. The Nobel Prize in Physics in 1912 recognized his automatic regulators for lighthouse and buoy lighting.

For identical touching atoms represented by hard spheres of radius $r$, a face-centered cubic lattice has conventional side $a_F=2\sqrt{2}r$ and four atoms per conventional cell. A body-centered cubic lattice has side $a_B=4r/\sqrt{3}$ and two atoms per conventional cell. Their primitive unit cell volumes are respectivelyEach primitive cell contains one atom in the monatomic model, so its mass density is atomic mass divided by $\Omega$. These are contact-geometry estimates; a fixed hard-sphere radius does not encode thermal expansion or phase-dependent electronic bonding.

For a fixed-composition simple compressible system, a Maxwell relation gives $C_p-C_V=T(\partial p/\partial T{)}_V(\partial V/\partial T{)}_p$. At fixed volume, the differential equation of state gives $(\partial p/\partial T{)}_V={\alpha }_V/{\kappa }_T$, proving the displayed result. Positive absolute temperature and positive isothermal compressibility imply a nonnegative difference. It vanishes when thermal expansion does, so strict inequality requires a nonzero expansion coefficient. This stability argument includes nonideal internal-energy changes, unlike a purely ideal-gas expansion-work explanation.

A dilute particle suspension can supply both excess mass density and heat. To first order its equation of state is $\rho /{\rho }_0=1+\gamma \varphi -\theta$, where $\varphi$ is the particle volume fraction and $\theta$ is fractional thermal expansion. Particle deposition flux reduces $\varphi$, while particle heating increases $\theta$. The competition can reverse the reduced gravity and create unstable density stratification.

A nickel–iron alloy with unusually small thermal expansion over a limited temperature range. Composition and treatment matter; its name does not mean every specimen has zero expansion. The Guillaume thermal expansion experiment explains how the effect was measured.^[ref]

A cold linear plate profile has excess column mass ${\rho }_m\alpha \Delta Ta/2$ relative to hot mantle. Equal hydrostatic pressure below the plates is restored by replacing part of the rock column with water. The displayed elevation follows to first order in thermal expansion; its denominator is the rock-minus-water density difference, not the rock density alone.

The Joule-Thomson coefficient is the local temperature response to changing pressure at fixed enthalpy and fixed amount and composition:It describes the equilibrium endpoints of a Joule-Thomson effect experiment meeting the steady-flow energy balance assumptions. With $\Delta p=p_2-p_1<0$, a sufficiently small drop gives $\Delta T\simeq {\mu }_{\mathrm{JT}}\Delta p$. Thus ${\mu }_{\mathrm{JT}}>0$ means cooling, ${\mu }_{\mathrm{JT}}<0$ warming, and ${\mu }_{\mathrm{JT}}=0$ no first-order change. At zero coefficient, a finite drop can still change temperature. For a chosen illustrative local value ${\mu }_{\mathrm{JT}}=0.20\mathrm{K}\mathrm{ba}{\mathrm{r}}^{-1}$ and $\Delta p=-0.50\mathrm{bar}$, the linear prediction is $\Delta T=-0.10\mathrm{K}$; these are stipulated numbers, not measured gas properties.

For a fixed-composition simple compressible system, the measurable form isHere $C_p$ is the whole sample's heat capacity at constant pressure and $\alpha$ its volumetric coefficient of thermal expansion; alternatively use molar volume and molar heat capacity consistently. An ideal gas has $\alpha =1/T$, so ${\mu }_{\mathrm{JT}}=0$. Its enthalpy depends only on temperature; positive heat capacity at constant pressure then makes any such finite throttle preserve temperature. A real gas can have either sign. For $C_p>0$, compare $T\alpha$ with one: the thermal expansion response decides the sign, not insulation alone.^[ref]

A short derivation uses $dH=TdS+Vdp$ and the Maxwell relation $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$:Setting $dH=0$ gives the coefficient. These are equilibrium state function identities, not a claim of a reversible thermodynamic process inside the throttle. For a finite drop, solve the endpoint enthalpy equality or integratealong the constant-enthalpy equilibrium curve. The coefficient can change sign during that integration, as the Van der Waals throttling model illustrates.

A reference realizing a specified length, under specified conditions. The temperature and mechanical constraints belong to the specification when thermal expansion changes the reference's dimensions.

Awarded to Charles-Édouard Guillaume for precision measurement through anomalous properties of nickel–steel alloys. The Guillaume thermal expansion experiment explains how a reference comparison reveals low thermal expansion.^[ref]

Write the Morse potential relative to its minimum as $V(a+x)=D[e^{-2x/\xi }-2e^{-x/\xi }]$. Expanding one order beyond the harmonic term givesThe negative cubic coefficient means that stretching is softer than compression. Thermal motion therefore has positive mean displacement even though the most probable minimum is at $x=0$.

For thermal expansion of a weakly anharmonic bond, classical equipartition gives $⟨x^2{⟩}_0=k_{B}T/\kappa$. Vanishing mean restoring force, to first anharmonic order, requiresFor a bond free to expand under negligible applied tension, $L=⟨r⟩\simeq a+⟨x⟩$. Its linear thermal expansion coefficient is thereforeKeeping the leading perturbative numerical factor gives $\alpha \simeq 3k_B/(2a\kappa \xi )=5.9\times 10^{-5}{\mathrm{K}}^{-1}$. The question only requires the scale, not that factor. The thermal expansion is positive; a purely harmonic approximation would give zero mean displacement and zero thermal expansion. The local Taylor expansion describes a bound bond and is not a globally stable cubic potential.

A parcel at fixed $z$ spends time $a/u(z)$ in the light. With no thermal conduction or thermal expansion, its energy balance is $\rho cdT/dt=q_R$. Equivalently the steady advection equation is $\rho cu(z)T_x=q_R(z)$ with $T(0,z)=T_0$. ThereforeThe no-slip boundary condition makes residence time diverge as either wall is approached. The resulting unbounded temperature is a failure of the approximations, rather than a physically steady wall prediction. Even small thermal conductivity transfers heat away from these layers; conduction into real perspex adds another heat sink. Thermal expansion can alter mass density, create buoyancy-driven motion and change the background profile, while temperature-dependent dynamic viscosity changes the assumed residence time. With exactly insulating walls, finite fluid thermal conductivity still regularizes the transverse profile; an entrance region and, sufficiently near a stagnant wall, longitudinal conduction also matter.

Use the same molar amount for both thermodynamic phases. A hexagonal cell built from two basal lattice vectors at120degrees has volume $(\sqrt{3}/2)a^{2}c$, not the volume of a three-times-larger full hexagonal prism. With four atoms per stated graphite cell,Diamond has eight atoms per conventional cubic cell, soFor graphite transforming to diamond,ThereforeThe positive slope reflects diamond's smaller entropy and volume. Higher temperature favors the higher-entropy graphite thermodynamic phase, requiring greater pressure to compensate in favor of the smaller-volume diamond thermodynamic phase.

This estimate treats the supplied ambient molar entropies and volumes as applicable near coexistence and neglects their changes through thermal expansion, compressibility and heat capacity. It gives an approximate gradient, not the transition pressure or a complete high-pressure thermodynamic phase boundary; an intercept would require an additional free-energy datum.

At height $z$ above the free lower end, a uniform vertical filament carries the weight of length $z$. With cross-sectional area $A$,The maximum stress is at the top and is independent of diameter. The diameter cancels because both mass and load-bearing area scale with $A$. Neglecting self-heating gradients, evaporation and initial fabrication stress, this top point sets the first creep-strain limit.

For a clearly labelled external reference estimate, take $\rho =1.93\times 10^4\mathrm{kg}{\mathrm{m}}^{-3}$. Then ${\sigma }_{\max }=1.89\times 10^4\mathrm{Pa}=0.0189\mathrm{MPa}$; the area is $3.14\times 10^{-8}{\mathrm{m}}^2$ and the total suspended mass is about $6.06\times 10^{-5}\mathrm{kg}$. The density is independently tabulated, not invented as a supplied 2016 value.^[ref]

Use the original map's top temperature axis at $2500^{\circ }\mathrm{C}$ and its right-hand tensile-stress axis, so no uncertain shear-stress conversion or additional shear modulus is required. The point lies in C, between the $10^{-9}$ and $10^{-8}{\mathrm{s}}^{-1}$ contours, nearer $10^{-8}$. A logarithmic interpolation of the scanned contours gives about $6\times 10^{-9}{\mathrm{s}}^{-1}$, with only graphical precision. Thusor of order three months in this steady-rate, reference-density map estimate. Reading only the bracketing contours gives the appropriately broad interval $5\times 10^6$ to $5\times 10^7\mathrm{s}$; the graph does not support an exact service-life number.

For the exact prescribed input, the answer remainsThis estimate uses the cited reference density and the grain-size assumptions of the supplied map. Thermal expansion lowers density, and an evolved area or tertiary material creep changes the rate; a steady initial-rate estimate is not a guaranteed lifetime. In particular, a different filament grain size could require a different map. Other failure mechanisms could shorten useful life.

The actual ionic radii need not equal the limiting ratio, and electronic charge distributions are not rigid sphere boundaries. In the same unlike-contact hard-sphere model, put $q=r_a/r_c$ and use $a=4(r_a+r_c)/\sqrt{3}$ to getA cation somewhat larger relative to the anion than the limiting size opens gaps between the anions. Starting from $q=1.366$ and reducing $q$ modestly reduces this packing fraction; thus a lower value is expected for that departure from the ideal contact construction. The formula is not monotonic for all imaginable ratios, so no universal inequality should be inferred without specifying radii. Coordination-dependent ionic radii, polarizability, nonionic contributions to bonding, thermal expansion and point defects also limit the model. No numerical real-crystal packing fraction is determined by the supplied data alone.