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Consider an ideal pressure-driven closure plug with a large chamber-facing projected area $A_p$. Its opposite side seats against a smaller annular packing surface area $A_{\mathrm{seal}}$. The remaining projected part $A_u=A_p-A_{\mathrm{seal}}>0$ is the unsupported area: it has no opposing packing contact in that plane. Here $p$ is the chamber pressure relative to the exterior. Let $q$ denote the mean axial compressive contact stress on the packing, not the chamber's hydrostatic stress.

Neglect friction, the plug's weight, and other axial reactions. The plug's Newton's second law in static balance isThe load is the same on both sides; its smaller supporting surface area raises the packing's mean contact stress. For a stipulated ratio $A_p/A_{\mathrm{seal}}=1.25$ and $p=1\mathrm{GPa}$, the ideal seating stress is $q=1.25\mathrm{GPa}$. Its excess over chamber pressure is $0.25\mathrm{GPa}$. Reducing $A_u$ to zero removes this amplification. This is the elementary force balance behind self-energizing seating, consistent with the lecture's Figure 1 caption.^[ref]

The relation is a model, not a universal sealing law. The mean axial contact stress need not equal the local contact stress along every possible leakage path. Confinement and plasticity in continuum mechanics determine whether the packing closes a gap or instead extrudes into it. Friction diverts load into other surfaces; material plasticity in continuum mechanics changes contact geometry; the vessel must carry its own stress. A finite initial seating load may also be needed before chamber pressure can develop. An arbitrarily small $A_{\mathrm{seal}}$ does not yield arbitrarily useful sealing: the assumptions fail when local deformation, extrusion or structural failure takes over.

With tension taken positive, hydrostatic stress is the stress tensor $\mathbf{\sigma }=-p\mathbf{I}$, where $\mathbf{I}$ is the identity matrix. Every plane has the same compressive normal stress, and no tangential traction. More generally,where $\mathbf{s}$ is deviatoric stress. Two samples can share the same mean pressure while experiencing different deviatoric stresses. A stationary liquid cannot sustain equilibrium shear stress; a solid transmitting medium can support it. Consequently, a load divided by surface area does not by itself establish hydrostatic sample conditions. A transmitting medium's state at the actual pressure and temperature, and contact between sample and walls, matter.

Bridgman's Figure 4 compares mild steel fractured in tension at atmospheric pressure with steel stretched further in a liquid at $25000\mathrm{kgf}\mathrm{c}{\mathrm{m}}^{-2}$.^[ref] The imposed tension remains a directional load: the experiment combines confinement with deformation, rather than measuring a hydrostatic stress alone. Its comparison motivates recording the loading path and the surrounding medium as well as pressure; it does not establish a universal fracture threshold for all steels.

A single optical shift represents pressure only to the extent that the calibration's mechanical and thermal conditions apply. Under hydrostatic stress, all directions have the same compressive normal stress. Under deviatoric stress, differently oriented ruby grains can respond differently; spatially varying pressure can also superpose shifted lines. These mechanisms can produce different optical signals at equal mean pressure.^[ref]

The 2001 account describes scanning local line positions and measuring line width over an extended region. The former resolves spatial differences; the latter supplies a rapid warning.^[ref]

Neither a broadened line nor a narrow one uniquely determines the stress tensor. Instrument resolution, temperature variation and unresolved overlapping signals can also affect width. A small observed region can have a narrow line while differing from another region. Compare locations, reference traces and instrument settings before treating width as evidence for a particular mechanical explanation. These are proposed diagnostic checks, not additional historical controls attributed to the 1975 experiment.

Temperature is another independent input. A useful local expansion is $\Delta \lambda \simeq s_{p}p+s_T(T-T_0)$: one measured shift cannot determine both unknowns. Measure or control temperature, or supply a second independently calibrated observable. The coefficients need not remain constant across a large range. Even when a pressure sensitivity transfers over some temperatures, the reference line, line width and practical readout quality can change.

The historical high-temperature account reports degrading pressure measurement quality above about $300{}^{\circ }\mathrm{C}$.^[ref]

Take the trace of the constitutive relation at zero stress. It gives $0=(2\mu +3\lambda )e_{kk}-3c\alpha (T-T_0)$. With the observed stress-free volume strain this requiresThis is the bulk modulus. A stable solid resists uniform compression, so $\boxed{K>0}$; equivalently the elastic energy of a nonzero pure dilation is positive. The thermal strain is isotropic, $e_{ij}^{\mathrm{th}}=\alpha (T-T_0){\delta }_{ij}/3$, because $\alpha$ here is the volumetric, not linear, thermal expansion coefficient.

Let $\beta =K\alpha$. Using $e_{ij}=(u_{i,j}+u_{j,i})/2$, stress divergence isZero body force and equilibrium therefore giveFor an applied body force per unit volume $b$, equilibrium is $\nabla \cdot \tau +b=0$, so its right-hand side becomes $\beta \nabla T-b$. Consequentlycancels the effective thermal forcing in the displacement equation. Per unit mass the force is $\beta \nabla T/\rho$. This sign follows from the divergence of the thermal stress $-\beta (T-T_0)I$, which is $-\beta \nabla T$. Cancelling that divergence is distinct from removing the stress tensor itself: boundary tractions and thermal compatibility still matter.

For spatially uniform heating with the entire boundary fixed, $u=0$ is the unique equilibrium displacement in a stable elastic solid, andThe stress is compressive hydrostatic stress for a temperature rise. The fixed boundary supplies the tractions that prevent thermal expansion. If $T$ varies spatially and no compensating body force is imposed, fixing the boundary alone does not generally force the interior displacement to vanish: one must solve the displayed forced Navier-Cauchy equation with $u=0$ on the boundary, then recover the stress. The simple uniform hydrostatic expression should not be asserted for arbitrary temperature fields. This is uniform thermal stress in a fully constrained solid.