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For a periodic thin mucus gap $h(x)$ translating as a backward wave relative to a force-free foot, put $I_j={\int }_0^L{h}^{-j}dx$. In the wave frame, the plane and foot have horizontal velocities $V-U$ and $V$. A Couette-Poiseuille flow with periodic pressure has flux $q=(V-U/2)I_2/I_3$. Zero net plane force gives the displayed crawling speed. The leading viscous dissipation per wavelength is $\Phi =2\mu UV{I}_1$.

For smooth no-slip spheres of radius scale $a$ with minimum gap $\delta \ll a$, the lateral gap scale is $(a\delta {)}^{1/2}$. Squeeze-film lubrication theory gives pressure $p\sim \mu a(-\dot{\delta })/{\delta }^2$ and the displayed normal resistance. Bounded forcing therefore gives $-\dot{\delta }=O(\delta )$, so the gap cannot vanish in finite time. The conclusion uses smooth surfaces, no slip and continuum creeping flow; it is not a contact model for rough particles.

Lunar laser ranging gives a more discriminating orbit readout. An observatory directs short laser pulses at a retroreflector array on the Moon, and a detector records returning photons. Round-trip time, statistically combined over several returns, constrains distance; pressure and temperature accompany the readout. The fitted model includes Newtonian gravity, terrestrial and lunar rotation, geophysical effects, and corrections from general relativity. Comparing orbit determination residuals across stations and model choices checks consistency. The inspected study reports $1.8\mathrm{cm}$ weighted root mean square residuals for its 2005–2011 ranges, rather than for a bare isolated inverse-square potential. It tests the equivalence principle using the Earth and Moon's response to the Sun; it is not a direct escape velocity measurement. Uneven coverage near new and full Moon, radiation pressure, and thermal expansion of reflector mounts limit that inference.^[ref]

A finite sinusoidal transverse field $B_x=a{B}_0\cos [k(z-v_{a}t)]$, $B_y=0$, has varying magnetic pressure. In a purely transverse traveling Alfvén wave, longitudinal momentum requires ${\partial }_z[p+B_x^2/(2{\mu }_0)]=0$, but the adiabatic pressure equation with $u_z=0$ requires ${\partial }_{t}p=0$. A traveling compensating gas pressure cannot satisfy both conditions. The uncompensated force is quadratic in amplitude and drives a compressive response. A regular periodic one-dimensional traveling solution cannot evade this by varying mass density: its mass, transverse momentum and induction integrals force constant mass density at the zeros of the transverse field and hence throughout the nontrivial wave.

In a thin magnetic skin layer next to a rigid no-slip boundary condition, vertical pressure balance gives $p=P-C^2{e}^{-2y/\delta }/(4{\mu }_0)$. With negligible inertia in the layer, the tangential Stokes flow balance is $\rho \nu u_{yy}=p_x$. Matching zero leading shear outside the layer gives $u=U(1-e^{-2y/\delta })$ and the displayed effective tangential slip. The physical wall still has zero velocity; the slip is a boundary condition for the outer flow extrapolated through the unresolved layer. It points toward larger squared boundary-field amplitude.

For an isolated self-gravitating fluid obeying ideal magnetohydrodynamics, let $I_{ij}=\int \rho x_i{x}_{j}dV$ and $K_{ij}=\frac{1}{2}\int \rho u_i{u}_{j}dV$. Use the force-stress convention $\rho D_t{u}_i={\partial }_j{T}_{ij}$, with pressure, magnetic and negative Newtonian gravitational stress tensor contributions. Two time derivatives of $I$ give $4K_{ij}$ plus $\int (x_i{\partial }_k{T}_{jk}+x_j{\partial }_k{T}_{ik})dV$. Integration by parts and symmetry yield the theorem when surface terms vanish. This fluid version complements the collisionless formulation of the tensor virial theorem.

The magnetohydrodynamic total pressure is $\Pi =p+B^2/(2{\mu }_0)$, the sum of gas pressure and magnetic pressure. The remaining Lorentz force is magnetic tension, so this scalar does not include the entire magnetic stress.

The inner radius at which a stellar magnetic dipole field disrupts a surrounding accretion disk. In a free fall pressure-balance estimate with fixed aspect ratio $ϵ=H/r$, $r_m\sim ({ϵ}^2{\mu }^4/(GM{\dot{M}}^2){)}^{1/7}$. Geometry and inflow structure change its numerical coefficient.

The leading force balance in a rapidly rotating conducting fluid when inertial and viscous forces are small, while Coriolis force, pressure, buoyancy and Lorentz force density remain significant. The Rossby number and Ekman number measure the omitted inertia and bulk viscosity. Rotation does not balance an arbitrary magnetic torque: impenetrable axisymmetric boundaries impose the Taylor constraint. Magnetic diffusion can remain in the induction equation even when viscosity is neglected in momentum balance.

An instrument for measuring pressure relative to another pressure. A liquid-column manometer uses hydrostatic pressure; a digital manometer uses a pressure transducer.

Continuity of temperature, pressure and mass density matches the ideal gas envelope pressure $\mathcal{R}{\rho }_b{T}_c$ to the nonrelativistic electron degeneracy pressure $K_{\mathrm{NR}}{\rho }_b^{5/3}$. Substituting this base mass density into the radiative-zero white-dwarf envelope law gives the luminosity-temperature scaling behind Mestel's cooling law. The matched point is a transition approximation; the cold degeneracy law does not represent finite-temperature corrections there exactly.

Porous-matrix drag depends on fluid velocity relative to the matrix, so it is not invariant under adding a uniform fluid velocity while leaving the matrix fixed. In a swimmer frame with stationary laboratory matrix, the drag is proportional to $\mathbf{u}-U{\mathbf{e}}_x$. A constant shift can alternatively be absorbed into a linear pressure term, whose physical interpretation must then be retained.

At fixed temperature, choose a coexistence pressure $p_{\ast }$ such that the gas and liquid have equal chemical potentials. Because $d\mu =vdp$, this gives ${\int }_{v_l}^{v_g}vdp=0$, or equivalently ${\int }_{v_l}^{v_g}[p(v,T)-p_{\ast }]dv=0$. The equal areas replace the unstable part of an isotherm by a horizontal coexistence segment.

Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives $(\partial S/\partial P{)}_{T,N}=-\alpha V$. ^[ref]

For a simple compressible system, hold particle number $N$ and composition fixed, and allow only pressure-volume work. Write $U$ for internal energy, $H=U+PV$ for enthalpy, $F=U-TS$ for Helmholtz free energy, and $G=H-TS$ for Gibbs free energy. Their natural-variable differentials areThus the first partial derivatives, with the subscript recording the variables kept fixed, areThe symmetry of second derivatives produces all four standard Maxwell relations, including their signs:For a brief derivation of the $F$ relation, differentiate $F_T=-S$ with respect to $V$ and $F_V=-P$ with respect to $T$: $-S_V=F_{TV}=F_{VT}=-P_T$. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-$N$ constraints must remain explicit. ^[ref]

For an ideal gas with fixed $N$, the ideal gas law gives $P=N{k}_{B}T/V$. At fixed volume, heating changes the pressure at rate $N{k}_B/V$. Consequently an isothermal expansion at positive temperature hasFor one mole with $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $V_2=2V_1$, this is $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. At $300\mathrm{K}$, a reversible isothermal expansion absorbs $Q=T\Delta S\simeq 1728.85\mathrm{J}$ and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same $\Delta S$ in an insulated free expansion, although there $Q=0$ and the process is irreversible.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure, $dV=\alpha VdT$, soThus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas, $\alpha =1/T$ and ${\kappa }_T=1/P$ reduce the result to $C_P-C_V=PV/T=N{k}_B$, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also givesIntegrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently $C_V$ and $C_P$ depend only on temperature at fixed $N$, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern $G$ relation gives $\delta Q_{\mathrm{rev}}=TdS=-T\alpha VdP$. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^{[ref] [ref]}