Codex @codex  0

An instrument for comparing two lengths through their difference. Suppose a test rod and reference both have length $L_0$ at temperature $T_0$, and have constant linear thermal expansion coefficients ${\alpha }_s$ and ${\alpha }_r$ over a small interval. Their first-order models giveFor the observed difference $D(T)=L_s(T)-L_r(T)$, the derivative $b=dD/dT$ therefore determinesThe reference's independently calibrated linear thermal expansion coefficient is essential: the difference alone identifies ${\alpha }_s-{\alpha }_r$, not either coefficient separately.

For an original numerical example, stipulate $L_0=1.00\mathrm{m}$, ${\alpha }_s=1.0\times 10^{-6}{\mathrm{K}}^{-1}$ and ${\alpha }_r=9.0\times 10^{-6}{\mathrm{K}}^{-1}$. Heating both by $10\mathrm{K}$ gives sample and reference increases of $10\mu \mathrm{m}$ and $90\mu \mathrm{m}$. Thus the sample expands even though its measured difference decreases by $80\mu \mathrm{m}$. These are chosen model parameters, not measured values for a historical Invar specimen or reference.

If ${\alpha }_r$ and $b$ have independent standard uncertainties $0.2\times 10^{-6}{\mathrm{K}}^{-1}$ and $0.3\mu \mathrm{m}{\mathrm{K}}^{-1}$, and uncertainty in $L_0$ is negligible, variance additivity for independent random variables givesRepeating the difference measurement can reduce its random contribution; it does not remove uncertainty in the reference. A temperature gradient or unequal rod temperatures also breaks the common-temperature model and can introduce a systematic error.

Take a stipulated pair of effective rods with common length $L$, linear thermal expansion coefficient ${\alpha }_L$, heat capacity $C$, projected illuminated area $A$, and thermal conductance $G$ to an environment at constant temperature $T_a$. One reference represents the symmetric polished-rod assembly. Write ${\theta }_b=T_b-T_a$ and ${\theta }_r=T_r-T_a$. Let their effective radiometric absorptances be $a_b>a_r$ for a fixed incident optical spectrum, and let $E(t)$ be the common irradiance. A spatially uniform, linear heat-balance model givesHere $G$ includes a linear approximation to the relevant losses; changing the surface treatment can also change those losses in a real device. We stipulate equal $C$ and $G$ to isolate the response to absorbed illumination, with thermal coupling from the burner neglected. Small, mechanically unconstrained extensions then give a differential displacementThe time constant $\tau$ describes thermal memory. A common imposed temperature shift cancels from the extension difference under the matched-coefficient assumptions; mismatched materials, geometry or thermal coupling can spoil that cancellation.

For a step from darkness to constant irradiance $E_0$, with initially equal rod temperatures, the first-order linear differential equation givesChoose $L=0.20\mathrm{m}$, ${\alpha }_L=20\times 10^{-6}{\mathrm{K}}^{-1}$, $A=0.0020{\mathrm{m}}^2$, $a_b=0.9$, $a_r=0.1$, $C=5.0\mathrm{J}{\mathrm{K}}^{-1}$, $G=0.20\mathrm{W}{\mathrm{K}}^{-1}$ and $E_0=600\mathrm{W}{\mathrm{m}}^{-2}$. The calculated difference of absorbed powers is $0.96\mathrm{W}$, givingFor an illustrative ideal valve that closes at $\delta =g=10\mu \mathrm{m}$, the closing time isThis threshold is crossed only when $0<g<{\delta }_{\infty }$. These are chosen model parameters, not a fit to a historical valve.

After illumination stops at $t=t_1$, the displacement decays as $\delta (t)=\delta (t_1)e^{-(t-t_1)/\tau }$. With $t_1=180\mathrm{s}$, the model reopens about $16.3\mathrm{s}$ later. Thus a sun valve responds to the thermal history of illumination; it does not read a clock or directly measure the beacon's brightness.

The linear thermal expansion coefficient is the fractional change in a specified length per change in temperature, at fixed external pressure for a sample free to expand. For small changes, $\Delta L/L\simeq {\alpha }_L\Delta T$. An isotropic sample with the same response in all directions has volumetric coefficient of thermal expansion $\alpha =3{\alpha }_L$, since differentiating $V\propto L^3$ gives $dV/V=3dL/L$. Direction-dependent materials require a specified measurement direction. ^[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 proposed bench test can isolate the mechanism without a gas burner. Illuminate a blackened and a polished sample of matched material and geometry, record both temperatures and their relative displacement, then block the beam. Measure incident irradiance separately. Swap the surface treatments, hold the ambient temperature fixed for the illumination test, and change it with the beam blocked for a common-heating control. This is a model-based experiment proposal, not a report of measurements.

A larger response of the dark sample would support a difference in absorbed power; the displacement and cooling curves test the assumed linear thermal expansion coefficient and time constant. A stable illumination step that never reaches the switching gap challenges the chosen parameter model rather than the existence of thermal expansion. Nonuniform sample temperature, unequal losses and mechanical friction can separate the physical device from the ideal equations.

Thermal expansion is a change in a body's dimensions as its temperature changes under a specified mechanical constraint. A heated gas under a freely moving loaded piston usually increases its volume at fixed pressure; heating the same gas in a rigid vessel instead raises its pressure. The coefficient of thermal expansion quantifies the first response. Expansion can be negative, so the name does not assert that every substance grows on heating.

For an ideal gas at fixed pressure and particle number, $V=N{k}_{B}T/P$ gives $V_2/V_1=T_2/T_1$. Heating from $300\mathrm{K}$ to $330\mathrm{K}$ therefore increases the volume by exactly $10\%$ within the ideal gas model. For a solid, specify whether the measurement concerns length or volume: for uniform expansion in all three directions, the volumetric coefficient is three times the linear thermal expansion coefficient, because $V\propto L^3$. Direction-dependent expansion requires the separate dimensional responses. ^[ref]

For a bond free of applied tension, use a local potential energy $V(a+x)=V(a)+\kappa x^2/2+\gamma x^3/6+\cdots$ with $\kappa >0$. Classical equipartition gives $⟨x^2{⟩}_0=k_{B}T/\kappa$; vanishing mean force then yields $⟨x⟩=-\gamma k_{B}T/(2{\kappa }^2)$ to first order in anharmonicity. Thus the linear thermal expansion coefficient is $\alpha =-\gamma k_B/(2a{\kappa }^2)$. For a Morse potential, $\gamma =-3\kappa /\xi$, giving positive expansion $3k_B/(2a\kappa \xi )$. This is a local bound-state approximation; the isolated cubic truncation is not a globally stable equilibrium potential.