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For a local illustrative ice–water melting calculation, stipulate $T_m=273.15\mathrm{K}$, latent heat of fusion $L=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, solid mass density ${\rho }_s=917\mathrm{kg}{\mathrm{m}}^{-3}$, and liquid mass density ${\rho }_l=1000\mathrm{kg}{\mathrm{m}}^{-3}$. These are explicitly stipulated rounded inputs, not measurements made here. The specific volume change on melting isThe common-pressure melting-point slope therefore givesA small $1\mathrm{MPa}$ increase in the common pressure thus lowers the melting temperature by about $0.074\mathrm{K}$ in this local model. Positive latent heat does not force a positive phase coexistence curve slope: ice contracts on conversion to liquid water. Increasing pressure favors the smaller specific volume. This calculation compares equilibrium states of the same two thermodynamic phases; extending its tangent to large pressures would require updated mass densities, latent heat of fusion, and the relevant ice phase diagram.

Neutral buoyancy means a fully immersed body's upward buoyancy equals its downward force from gravity. For a body of fixed volume in a constant-mass density fluid under a uniform gravitational field, this occurs when its mean mass density equals the surrounding mass density. A $0.10\mathrm{m}$ cube of mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ in the stipulated water has mass $1.000\mathrm{kg}$, so both forces are $9.81\mathrm{N}$.

Ideally it can remain at rest at any depth for which it stays fully immersed and clear of boundaries. Raising or lowering it changes the absolute pressure, but neither its displaced volume nor the difference between the vertical pressure forces. There is therefore no vertical restoring force toward a preferred depth in this model. Neutral buoyancy is a vertical force balance, not a claim about resistance to tilting or about a compressible body whose volume changes with pressure.

For isotropic incompressible flow of a Newtonian fluid, the stress tensor is the displayed expression, with pressure $p$, dynamic viscosity $\mu$, velocity $\mathbf{u}$, and identity matrix $I$. In Couette flow, its off-diagonal component is the shear stress.

Write the non-Newtonian extra stress as $S=2({\psi }_1+2{\psi }_2)E^2-{\psi }_1\stackrel{\circ }{E}$, with the Jaumann derivative. For a steady Stokes flow, $\nabla \times \nabla \cdot \stackrel{\circ }{E}=0$. If ${\psi }_1+2{\psi }_2=0$, the extra force is a gradient and only alters pressure. In a planar incompressible flow, $E^2=q\mathrm{diag}(1,1,0)$, so its divergence is also a gradient; no coefficient restriction is needed. A Newtonian velocity with prescribed boundary velocities consequently solves the second-order model with an altered pressure. On the regular weak-flow branch, Uniqueness of Stokes flow makes each velocity correction vanish. This does not establish uniqueness of arbitrary large nonlinear branches or the absence of higher-order effects in a real material.

A finite-amplitude ideal-MHD Alfvén wave can rotate the transverse magnetic field at constant magnitude, with constant mass density, pressure and longitudinal components. Its transverse velocity obeys ${\mathbf{u}}_{\perp }=\mathbf{C}-s{\mathbf{B}}_{\perp }/\sqrt{{\mu }_0\rho }$ and its speed is constant along the wave. The constant magnetic pressure prevents compressive acceleration.

For a perfect-gas normal shock wave reflected from a stationary rigid wall, apply the Rankine-Hugoniot conditions for a perfect gas using signed velocity in each shock frame. The once-shocked state is the common left state for both applications. Its two relative speeds are roots of $v^2-(\gamma +1)u_{s}v/2-\gamma p_s/{\rho }_s=0$. Their product and the pressure jump eliminate the velocity, giving the displayed relation. A strong incident shock wave gives $p_1/p_s\to (3\gamma -1)/(\gamma -1)$.

For indentation $F=b{x}^{1/3}$, the common layer displacement is $H=b{x}^{1/3}/2$. The boundary layer stream function $\psi =x^{2/3}G(z{x}^{-1/3})$ gives velocity $U=x^{1/3}{G}^{\prime }$ and pressure $P=\bar{p}{x}^{2/3}$. Substitution into the Prandtl boundary-layer equation yields the displayed equation, with $G(0)=G^{\prime }(0)=0$ and $G^{\prime }\sim 6\eta +3b$. This is a conditional similarity construction: the constant $\bar{p}$ must permit a solution of the boundary value problem.

An optical path length measures how much propagation wave phase a path accumulates, expressed as the distance that would accumulate the same wave phase in vacuum. A centimetre in a uniform medium with refractive index $3/2$ has optical path length $1.5\mathrm{cm}$, although its geometric length remains $1\mathrm{cm}$. This makes optical path length useful whenever two contributions must arrive with matching wave phases, from an optical lens to a Michelson interferometer. ^[ref]

For a specified path $\gamma$ at fixed angular frequency $\omega$, the optical path length is the line integralHere $s$ is geometric arc length, and $n$ is the phase refractive index, evaluated at the same frequency everywhere. The propagation interpretation assumes a stationary linear medium with isotropy, a real positive scalar refractive index and negligible loss, on a valid geometrical optics branch. The path can curve; at sharp interfaces apply the appropriate matching conditions to each segment. A prescribed curve has a mathematical line integral, but it predicts a propagation wave phase only when it represents the physical branch. ^[ref] Optical fibers can require a mode wavenumber rather than the bulk refractive index along a guessed light ray. ^[ref]

Fix the vacuum wavelength ${\lambda }_0=c/f=2\pi c/\omega$, with frequency $f=\omega /(2\pi )$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$. In a locally uniform medium the wavelength is $\lambda ={\lambda }_0/n$, while the frequency stays unchanged at a stationary interface. With the convention $\Phi =k_{0}S-\omega t+{\varphi }_0$, propagation from a specified starting wave phase contributesThe relative reference ${\delta }_{\mathrm{ref}}$ includes launch wave phases and any extra wave phase from mirrors, beam splitters or coatings. Thus $k_0\mathrm{OPL}$ is a propagation contribution, not an absolute detector wave phase independent of the apparatus. If the reference contributions stay fixed, a change in optical path difference $h$ changes the phase difference by $2\pi h/{\lambda }_0$. Using the material wavelength again in this formula would count the refractive index twice. ^[ref]

A motivating measurement turns invisible optical oscillations into a slowly changing electrical signal. MIT's teaching experiment uses a laser, a beam splitter, two returning mirrors, an amplified photodiode, an oscilloscope and a calibrated mirror actuator driven by a function generator. The returning beams overlap on the photodiode; one channel records detector voltage and another the actuator drive. The measured observable is the changing light intensity, including its maxima and minima, rather than direct timing of an optical cycle. A calibrated displacement per drive voltage and the spacing of the extrema relate geometric motion to the phase difference. The guide infers the laser wavelength from known actuator calibration; conversely, with independently known vacuum wavelength and ambient refractive index, the same relation calibrates mirror motion. These two unknowns cannot both be inferred from fringe spacing alone. ^[ref]

For a proposed change-of-medium measurement, keep the mirrors fixed and put a transparent cell of internal length $L$ in one arm. Compare its contents at refractive indices $n_a$ and $n_b$, leaving its windows in place. At normal incidence the two traversals giveHere $N$ is the signed number of whole interference fringe cycles plus any resolved fractional cycle, counted continuously. A mirror displacement $x$ instead gives $\Delta \mathrm{OPL}=2n_{\mathrm{amb}}x$; distinguish this double-pass factor from a single pass in a Mach-Zehnder interferometer. OpenStax's gas-cell worked example provides the same double-pass reasoning; it is a teaching calculation rather than a published measurement dataset.^[ref]

An exact toy calculation makes the sensitivity visible. Stipulate ${\lambda }_0=600\mathrm{nm}$, $L=10\mathrm{mm}$ and $n_b-n_a=3/100000=0.00003$. Then $\Delta \mathrm{OPL}=600\mathrm{nm}$ and $\Delta \delta =2\pi$: a full interference fringe cycle, despite unchanged geometric lengths. For a continuous change $n_b-n_a=q(3/100000)$, let $0\le q\le 2$ and choose the initial detector phase difference to be zero. Equal coherent detected arm contributions of light intensity $I_0$ predictThe plot is this stipulated model: $q=1/2$ gives a minimum, $q=1$ restores a maximum, and the reference level is chosen by adjustment of the apparatus. It contains no measured values or assigned properties of a real gas. The equivalent mirror travel in vacuum is $300\mathrm{nm}$ per cycle. A single endpoint light intensity cannot distinguish $q=0$ from $q=1$; the continuous fringe count carries the missing information.

For an interpretable cell measurement, establish the initial phase difference with a small calibrated mirror scan, then track slow changes through successive interference fringes. Measure the dark signal and each arm separately, keep photodiode gain within its linear range, align both beams on the same region and match their linear polarizations. Monitor interferometric visibility and keep arm delays within the useful optical coherence time. Record cell length, contents, pressure and temperature; return to the starting condition to check drift. Changes in window deformation, cell length, external air or mirror position also change optical path length, so a reference cell or an independently monitored mechanical path is needed to isolate the contents' refractive index. These controls are deductions from the model, not a claim that the illustrated apparatus was operated. Loss of fringe count leaves a whole-cycle ambiguity; at an intensity extremum the first derivative with respect to phase difference vanishes, so a scan around phase quadrature is more sensitive to small changes. Broadband illumination, unstable optical coherence, vibration and strong loss can invalidate a simple one-frequency inference.

Osmosis is net movement of water across a boundary whose permeability differs for water and relevant solutes. The direction depends on the chemical potential of water, including both dissolved material and pressure. A cell wall can let water reach the cell membrane, yet oppose expansion through turgor pressure. Zero net movement does not mean individual water molecules have stopped crossing.

Imagine pure water on side A and a chemical solution on side B of a semipermeable membrane. At equal temperature and pressure, dissolved particles lower the chemical potential of water in B, and net water transfer initially favors A to B. Retaining the solutes lets water redistribute without simply mixing everything. Increasing B's pressure opposes entry; enough pressure gives thermodynamic equilibrium with unequal solute concentrations. This is why “water moves toward more solute” requires a pressure condition.^[ref]

At thermodynamic equilibrium, the chemical potentials of exchangeable water are equal. Individual water molecules still move and cross in both directions, with equal mean opposing transfer rates and zero net flux. A stationary liquid level alone does not prove this: a blocked route can also give no visible transfer. Selective permeability provides a route; water potential supplies the direction, and neither alone determines the speed. The cell wall, turgor pressure and plasmolysis articles show why a living boundary's mechanical support and its selective permeability must be distinguished.

Take two reservoirs separated by a semipermeable membrane that passes water and retains one neutral solute. Stipulate $T=300\mathrm{K}$, $c_A=10\mathrm{mol}{\mathrm{m}}^{-3}$ and $c_B=30\mathrm{mol}{\mathrm{m}}^{-3}$, equivalent to $10$ and $30\mathrm{mmol}{\mathrm{L}}^{-1}$. Treat both chemical solutions as dilute and ideal. These are invented model conditions, not measurements or recommended biological conditions.

For pressures measured at the same height beside the membrane, define $\Delta P=P_B-P_A$. The model predictsAt $\Delta P=0$, net water transfer favors A to B. At $\Delta P\simeq 49.89\mathrm{kPa}$, the opposing pressure balances the osmotic pressure difference, giving zero net transfer with unequal concentrations. At $\Delta P=75\mathrm{kPa}$, the sign reverses and net water transfer favors B to A, toward the more dilute reservoir. Raising pressure changes the total water potential, rather than changing the membrane into a one-way gate.

Fixed concentrations require large maintained reservoirs or an instantaneous comparison. In a closed membrane osmometer, entering water causes dilution of the receiving chemical solution and changes its volume; the final balance must use final concentrations. A supported cell wall provides another way to develop the opposing pressure, as explained under turgor pressure.