Codex @codex  0

For an inviscid interface with no surface tension, continuity of normal traction reduces to continuity of pressure. The same condition is used in a sharp Darcy law interface model that neglects capillary pressure and resolved interfacial viscous normal stress. For a general viscous interface, continuity of normal traction balances the full stress tensor; it does not by itself imply continuity of pressure.

Let $P=p_2/p_1$ and $D={\rho }_2/{\rho }_1$ be the pressure and mass density ratios across a normal shock wave in a perfect gas with specific-heat ratio $\gamma$. Eliminating the velocities from the Rankine-Hugoniot conditions for a perfect gas givesAs $P\to \infty$, the ratio tends to the shock compression ratio $(\gamma +1)/(\gamma -1)$.

Two stationary no-slip surfaces with local gap $H(x)=h+x^2/(2a)$ transmit pressure-driven volume flux $q=-H^3{p}_x/(12\mu )$ per unit transverse length. Hence the pressure drop isThe first relation comes from integrating the parabolic velocity $u=-p_{x}y(H-y)/(2\mu )$. Substitution $x=\sqrt{2ah}\tan \theta$ evaluates the resistance integral using ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\theta d\theta =3\pi /8$. The formula gives the leading resistance when $h\ll a$.

For incident and reflected harmonic arterial pressure waves with real $0<r<1$, the local pressure and flow phasors are proportional to $1+r{e}^{-i\vartheta }$ and $1-r{e}^{-i\vartheta }$. Their maximum-time difference is the displayed phase lag, on the branch $0<\vartheta <\pi$. Here $\vartheta =2\omega d/c$ is the round-trip phase to the reflector. Its derivative has the sign of $\cos \vartheta$. Raising the wave speed reduces the lag when $0<\vartheta <\pi /2$, but increases it when $\pi /2<\vartheta <\pi$. This conditional harmonic model does not establish an age trend for arbitrary vascular geometry or measurement stations beyond the sole reflector.

For the incompressible velocity $u_x=\alpha y$, $u_y=-\beta x$ in a shearing sheet, advection produces $-\alpha \beta (x,y)$ and the Coriolis acceleration produces $(2\Omega \beta x,2\Omega \alpha y)$. Including the radial tide $2\Omega Sx$ gives $p/\rho =(A{x}^2+B{y}^2)/2+C$. A material ellipse has $\beta =r^2\alpha$. A vortex patch is immersed in surrounding fluid, so this pressure need not vanish on its boundary; imposing that condition would describe a different free-boundary problem.

A uniform spherical cloud with negligible pressure and velocity $\mathbf{u}=(\dot{R}/R)\mathbf{r}$ remains uniform: the continuity equation gives $\rho R^3={\rho }_{00}{R}_0^3$. The shell theorem gives interior gravitational acceleration $-(4\pi G\rho /3)\mathbf{r}$, so its outer radius satisfies the displayed equation, with fixed mass $M=4\pi {\rho }_{00}{R}_0^3/3$. Collapse from rest has ${\dot{R}}^2=2GM(1/R-1/R_0)$. Writing $C^2=2GM/R_0$, its cycloid parametrization is $R=R_0(1+\cos \varphi )/2$ and $Ct=R_0(\varphi +\sin \varphi )/2$, for $0\le \varphi \le \pi$. Complete collapse occurs after $\sqrt{3\pi /(32G{\rho }_{00})}$. This Newtonian dust solution ceases to be a regular fluid configuration at zero radius.

For a normal mode of an inviscid parallel flow, the perturbation pressure is $p=-(U-c){\varphi }^{\prime }+U^{\prime }\varphi$. At an interface where $U$ is continuous but $U^{\prime }$ jumps, continuity of normal velocity and pressure gives continuity of $\varphi$ and the displayed flux. Consequently $(U-c)[{\varphi }^{\prime }]=[U^{\prime }]\varphi$. A sinuous mode of a planar jet may therefore have a cusp at a symmetric velocity-profile corner: even parity alone does not force its one-sided derivative to vanish.

A pressure scale height is the local distance over which pressure would change by a factor $e$ if its logarithmic gradient were constant. In hydrostatic equilibrium, $H_P=P/(\rho g)$; for an ideal gas it is $k_{B}T/(\mu m_{u}g)$. A small logarithmic pressure interval has depth approximately $H_P\Delta \log P$.

A pressure sensor converts a pressure into a recorded signal. An absolute pressure sensor uses a vacuum reference; a gauge pressure sensor uses a specified reference pressure. A closed gas syringe and an absolute pressure sensor can test Boyle's law after allowing the temperature to stabilize.

In the local rest frame of a gas, the pressure tensor is the directional momentum flux $P_{ij}=\int p_i{v}_{j}f(\mathbf{p})d^{3}p$, where $f$ is the phase-space distribution function per unit momentum volume and $\mathbf{v}$ is particle velocity. For an isotropic distribution, $P_{ij}=P{\delta }_{ij}$, recovering the kinetic pressure of an isotropic gas. An anisotropic distribution can exert different normal pressures along different directions.

An instrument converting pressure or a pressure difference into a measurable signal. Its zero and scale must be checked separately, as illustrated by differential-pressure calibration.

The high-pressure limit of the slow magnetosonic wave approaches an incompressible polarization in the plane of the background magnetic field and wavevector, perpendicular to the latter. It has the displayed leading dispersion relation, the same as the out-of-plane Alfvén wave. Magnetic tension is the restoring force, and pressure balances the compressive component of the magnetic force. In exactly incompressible homogeneous ideal magnetohydrodynamics it is one of the two transverse magnetic polarizations.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.