Differential thermal-expansion uncertainty

Differential thermal-expansion uncertainty is the uncertainty in the unequal thermal length changes of two compared objects. In a gauge-block thermal mismatch correction, shared temperature and coefficient information determine which contributions cancel and which remain.

For approximately equal known nominal length $L$, coefficient difference $a={\alpha }_X-{\alpha }_R$ and common temperature offset $\theta =T-T_0$, the differential correction is $d=La\theta$. Its first-order sensitivity coefficients are $\partial d/\partial a=L\theta$ and $\partial d/\partial \theta =La$. HenceLength uncertainty and other inputs require further terms when they are material to the result. General propagation must retain input covariance.^[ref]

The coefficient difference itself hasAssuming two coefficient estimates independent merely because two blocks are present can therefore misstate the uncertainty. Similarly, a common thermometer contribution is not two unrelated temperature errors. NIST's handbook explicitly treats differences in block temperatures and expansion coefficients in comparisons.^[ref]

An original example takes $L=100\mathrm{mm}$, $a=6.5\times 10^{-6}{\mathrm{K}}^{-1}$, $\theta =2\mathrm{K}$, independent $u(a)=0.5\times 10^{-6}{\mathrm{K}}^{-1}$ and $u(\theta )=0.1\mathrm{K}$. The correction is $1.3\mu \mathrm{m}$. The two first-order standard-uncertainty contributions are 0.10 and $0.065\mu \mathrm{m}$, giving combined $u(d)\approx 0.119\mu \mathrm{m}$. These are stipulated inputs, not a certificate.

A useful boundary case exposes second-order thermal mismatch uncertainty. If both central values $a$ and $\theta$ are zero, the linear sensitivities vanish even though both inputs remain uncertain. For independent inputs with those means, their product has exact variance $L^2{u}^2(a)u^2(\theta )$. With $L=100\mathrm{mm}$, $u(a)=10^{-6}{\mathrm{K}}^{-1}$ and $u(\theta )=0.1\mathrm{K}$, this gives 10 nm standard uncertainty, while first-order propagation gives zero. The GUM explicitly requires attention to higher-order terms when nonlinearity matters.^[ref]

Specify the thermal model, shared inputs and approximation used. A small nominal correction is not automatically evidence that its uncertainty is negligible.

Second-order thermal mismatch uncertainty retains product terms that a linear uncertainty propagation can miss when calculating differential thermal-expansion uncertainty. For known nominal length $L$, expansion-coefficient difference $a$ and common temperature offset $\theta$, the correction is $d=La\theta$. Both first derivatives vanish at $a=\theta =0$, but the product of uncertain inputs need not vanish.

An original exact calculation assumes independent random variables with means $a_0,{\theta }_0$ and variances ${\sigma }_a^2,{\sigma }_{\theta }^2$. Independence givesTo derive the variance, use $\mathbb{E}[a^2{\theta }^2]=\mathbb{E}[a^2]\mathbb{E}[{\theta }^2]$ and subtract $a_0^2{\theta }_0^2$. No normal-distribution assumption is required for this independent-input result. The last term is absent from first-order propagation.

For stipulated $L=100\mathrm{mm}$, $a_0={\theta }_0=0$, ${\sigma }_a=10^{-6}{\mathrm{K}}^{-1}$ and ${\sigma }_{\theta }=0.1\mathrm{K}$, the exact standard deviation is 10 nm. A zero central correction therefore does not establish zero uncertainty. The GUM specifically warns that significant nonlinearity requires higher-order terms.^[ref]

Correlated inputs introduce a further distinction between the central product and the output mean. If $(a,\theta )$ has a multivariate normal distribution with covariance $c$, expansion about its means and the Gaussian fourth-moment identity giveThe covariance term in the mean must not be silently discarded. With the same zero means and standard deviations as above, but correlation coefficient $1/2$, the mean correction is 5 nm and its standard deviation about 11.18 nm. These are probability-model calculations, not measured block specifications.

The formulas assume known $L$ and a justified joint input model. A thermal-product input-distribution check must address shared temperature calibration and coefficient estimation before assigning a correlation. NIST's physical comparison procedure treats thermal corrections through measured temperatures and material expansion properties.^[ref] Keeping the nonlinear contribution is useful only when the inputs describe that actual measurement.

A thermal-product input-distribution check tests whether the probability model used for a nonlinear thermal correction describes the available temperature and material information. In $d=La\theta$, with coefficient difference $a$ and temperature offset $\theta$, the uncertainty depends on their joint behaviour, not just on two separately quoted standard uncertainties.

Start with the physical inputs: which thermometer calibration is shared, how coefficient estimates were obtained, and whether either quantity was inferred using the other. Translate those relationships into a joint distribution or an explicit dependence model. JCGM 101 distinguishes assigning input distributions from propagating them through a measurement model; numerical sampling cannot supply missing physical information.^[ref]

An original counterexample shows why zero covariance is insufficient for the independent-input formula. Let $Z$ be standard normal and defineBoth inputs have zero mean, with variances ${\sigma }_a^2$ and ${\sigma }_{\theta }^2$, and their covariance is zero because the required Gaussian odd moments vanish. They are nevertheless dependent: $\theta$ is determined by $a$.

Their product has zero mean and varianceusing Gaussian moments $\mathbb{E}[Z^2]=1$, $\mathbb{E}[Z^4]=3$, $\mathbb{E}[Z^6]=15$. Independence would instead give ${\sigma }_a^2{\sigma }_{\theta }^2$. With $L=100\mathrm{mm}$, ${\sigma }_a=10^{-6}{\mathrm{K}}^{-1}$ and ${\sigma }_{\theta }=0.1\mathrm{K}$, the resulting standard deviation is about 22.36 nm rather than 10 nm. This is a deliberately constructed probability example, not a claim about typical thermometers.

Thermal-product dependence beyond covariance matters because higher moments enter the product variance. A multivariate normal distribution is a particular joint model whose moments are determined by its means and covariance; arbitrary inputs sharing those same summaries need not behave identically.

Check units, support, marginal shapes and dependence before simulation. Compare exact or limiting cases with a Monte Carlo method, and retain a sensitivity analysis for plausible alternative input models. The GUM's first-order rule requires higher-order treatment when nonlinearity matters.^[ref] A small simulation error measures numerical precision conditional on the model, not proof that the model is physically correct.

Thermal-product dependence beyond covariance is the influence of higher joint moments on uncertainty in a nonlinear thermal expansion correction. Two input models can have identical means, variances and covariance yet predict different uncertainty in a product. A thermal-product input-distribution check must therefore examine how the temperature offset and expansion-coefficient error occur together.

Let $a={\mu }_a+A$ and $\theta ={\mu }_{\theta }+T$, where the centered variables have expected values zero, variances $s_a^2,s_{\theta }^2$ and covariance $c=\mathbb{E}[AT]$. Direct expansion givesThe third joint moments and the mixed fourth moment of thermal inputs are extra information. For centered inputs, only the last two terms remain. A first-order uncertainty propagation rule at zero input means would miss the product entirely; the GUM explicitly requires higher-order treatment when nonlinearity matters.^[ref]

An original counterexample keeps even the marginal shapes fixed. Let $Z$ have a standard normal distribution and let $S$ independently take values $+1$ and $-1$ with equal probabilities. Define $A=s_{a}Z$ and $T=s_{\theta }SZ$. Both have centered normal distributions, the quoted variances and zero covariance. They are not independent: their absolute standardized values are identical.

Their product is $s_a{s}_{\theta }S{Z}^2$. Its expected value is zero, while its variance is $3s_a^2{s}_{\theta }^2$, because $\mathbb{E}[Z^4]=3$. Independent centered normal inputs with the same individual variances instead give product variance $s_a^2{s}_{\theta }^2$. The joint model, rather than a marginal histogram or correlation coefficient alone, determines the difference.

For an illustrative length correction $d=La\theta$ with zero means, $L=100\mathrm{mm}$, $s_a=10^{-6}{\mathrm{K}}^{-1}$ and $s_{\theta }=0.1\mathrm{K}$, these models give standard deviations $\sqrt{3}\times 10\approx 17.32$ nm and 10 nm. This constructed dependence is a mathematical test case, not a proposed universal thermometer model.

In a physical gauge block study, track paired temperature and coefficient information, shared thermometer corrections and any procedure that estimates one input using the other. NIST's thermal correction explicitly combines block length, expansion coefficient and temperature difference.^[ref] Preserve that dependence when generating joint draws. JCGM 101 distinguishes model formulation from propagation and gives a multivariate Gaussian input model as a specific choice, not a consequence of normal marginals alone.^[ref] More simulation trials reduce numerical error conditional on the chosen model; they cannot identify unmeasured joint moments.

The mixed fourth moment of thermal inputs measures how squared temperature and expansion-coefficient deviations occur together. For centered inputs $A=a-\mathbb{E}a$ and $T=\theta -\mathbb{E}\theta$, defineIt has total input degree four. It is not the fourth moment of the product $AT$, which would involve $\mathbb{E}[A^4{T}^4]$. For a centered thermal product, the exact variance isConsequently individual input variances and their covariance do not in general determine the product variance.

For jointly normal centered inputs, write $A=s_{a}U$ and $T=s_T(\rho U+\sqrt{1-{\rho }^2}V)$ using independent standard normal distributions. Expanding and using $\mathbb{E}{U}^4=3$, $\mathbb{E}{U}^2{V}^2=1$ and vanishing odd terms givesThis is a specific joint-Gaussian calculation, consistent with the Isserlis theorem; normal-looking marginals alone do not justify it.

An original bounded counterexample illustrates the extra information. Let $X$ equal $-\sqrt{2},0,+\sqrt{2}$ with probabilities $1/4,1/2,1/4$, and let an independent sign $B$ take $\pm 1$ equally. Put $A=s_{a}X$ and $T=s_{T}BX$. Both standardized marginals have mean zero and variance one, and $\mathbb{E}(AT)=0$, but $M_{22}=2s_a^2{s}_T^2$. Replacing the second standardized input by an independent copy of $X$ preserves both marginals and zero covariance while changing $M_{22}$ to $s_a^2{s}_T^2$.

For an illustrative length correction $d=LAT$, take $L=100\mathrm{mm}$, $s_a=10^{-6}{\mathrm{K}}^{-1}$ and $s_T=0.1\mathrm{K}$. The two bounded models give standard deviations approximately14.14nm and10nm. Their factor-of-two variance difference comes entirely from joint occurrence, rather than changed marginal uncertainties.

NIST's thermal correction combines block length, expansion coefficient and temperature difference, providing the physical setting for this joint-input issue.^[ref] Joint observations and the way coefficient or thermometer corrections are assigned must support the dependence model. Increasing the number of simulated draws cannot recover an unmeasured mixed moment. The GUM explicitly calls for higher-order treatment when a nonlinear response makes first-order propagation inadequate.^[ref]