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A diamond anvil cell compresses a sample between opposed diamond tips. The flat diamond anvil culets load a metal gasket whose hole holds the specimen and a pressure-transmitting medium. Transparent diamond admits excitation and returning fluorescence, enabling an internal ruby pressure sensor.^[ref]

The functions in the drawing answer different questions. The drive applies force, the gasket confines the sample, the medium transmits stress, and the ruby pressure scale assigns a local pressure from an optical signal. A large drive reading cannot establish either the sample's pressure or the absence of deviatoric stress. Optical access also permits repeated readings without unloading the sample: a change during a long measurement can be detected rather than silently folded into a material-property curve.

The ruby pressure scale converts a shift of a ruby fluorescence spectral line into pressure. Excitation light produces a fluorescence emission spectrum; fitting its $R_1$ line gives a wavelength ${\lambda }_{R_1}$. The reference ${\lambda }_{R_1,0}$ must be measured or assigned at specified reference pressure and temperature. The difference $\Delta \lambda$ is an observable; interpreting it as pressure requires a pressure calibration and suitable stress conditions.

The original 1972 report found sharp $R_1,R_2$ emission shifting toward lower photon energy as pressure increased; it tentatively associated broadening with nonhydrostatic conditions.^[ref] In use, excitation rejection, a calibrated wavelength axis and a fitted background matter because leakage or background can move an apparent peak. A brighter trace improves counting statistics but does not alone establish a correct pressure scale.

Construct a synthetic local calibration to show the calculation independently of the historical data. Choose reference pairs $(\Delta {\lambda }_i,p_i)=(1.00,2.69),(2.00,5.42),(3.00,8.09)$, with shifts in $\mathrm{nm}$ and pressures in $\mathrm{GPa}$. The reference zero is already subtracted. Treat the shifts as exact for this fit and give the three reference pressures equal weights. Minimizing the ordinary least squares objectiveand setting its derivative to zero givesThe regression residuals are $(-0.01,0.02,-0.01)\mathrm{GPa}$. Their weighted moment ${\sum }_i\Delta {\lambda }_i(p_i-a\Delta {\lambda }_i)$ vanishes, as required by the least-squares normal equations. An unknown reference offset requires fitting an intercept instead; forcing it to zero can bias the slope. If shift errors matter, this elementary fit is insufficient.

Use the local measurement modelHere $p$ denotes pressure relative to the reference, $a$ is calibrated sensitivity, and $k_T$ is a local wavelength sensitivity to temperature at fixed pressure. Choose $a=2.70\mathrm{GPa}\mathrm{n}{\mathrm{m}}^{-1}$, $k_T=0.010\mathrm{nm}{\mathrm{K}}^{-1}$ and a reading $\lambda -{\lambda }_0=2.00\mathrm{nm}$ at $T=T_0$. The inferred pressure is $5.40\mathrm{GPa}$. These coefficients and readings are stipulated, not measured ruby constants. With the same pressure, a $10\mathrm{K}$ increase in temperature would add $0.100\mathrm{nm}$ to the shift in this model; neglecting it would overestimate pressure by $0.270\mathrm{GPa}$.

Assign independent standard uncertainties $u_a=0.020\mathrm{GPa}\mathrm{n}{\mathrm{m}}^{-1}$, $u_{\lambda }=u_0=0.010\mathrm{nm}$ and $u_T=0.50\mathrm{K}$ for the temperature difference. Here $u_a$ is an independently assigned calibration uncertainty, not an estimate from the regression residuals above; treat $k_T$ as exact in this example. First-order propagation of uncertainty follows by differentiating the measurement model:The four standard contributions are $0.0400,0.0270,0.0270,0.0135\mathrm{GPa}$, giving $u_p=0.0569\mathrm{GPa}$. This is a standard uncertainty, not automatically a $95\%$ interval. Correlated reference and loaded wavelength readings require the corresponding covariance term; common wavelength-axis drift may cancel in their difference. Uncertain $k_T$, nonzero $T-T_0$, model inadequacy and unresolved deviatoric stress require a broader budget. Taking more optical readings reduces some random errors while leaving a common reference-scale error.