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For an energy-conserving strong spherical shock wave expanding into mass density $C{r}^{-\beta }$, $0\le \beta <3$, dimensional analysis gives the displayed radius law. Neglect ambient pressure, radiation losses and gravity in this blast-dominated approximation. Write $u=\dot{R}U(\xi )$, $\rho =C{R}^{-\beta }D(\xi )$ and $p=C{R}^{-\beta }{\dot{R}}^{2}P(\xi )$, with $\xi =r/R$. The radial continuity equation, momentum equation and adiabatic pressure equation reduce toThe Strong-shock Rankine-Hugoniot conditions give $U(1)=P(1)=2/(\gamma +1)$ and $D(1)=(\gamma +1)/(\gamma -1)$. The explosion energy fixes the radius coefficient through $E=4\pi C{R}^{3-\beta }{\dot{R}}^2{\int }_0^1[D{U}^2/2+P/(\gamma -1)]{\xi }^{2}d\xi$.

A buoyantly rising inviscid bubble in otherwise quiescent unbounded liquid has an exterior Stokeslet with $F=4\pi \mu aU=4\pi \rho g{a}^3/3$. Its tangential traction vanishes on the sphere. The dynamic normal traction is $-3F\cos \theta /(4\pi a^2)$, while the hydrostatic normal-traction variation is $\rho ga\cos \theta$. Their dipoles cancel at the terminal velocity, leaving a constant total normal stress compatible with uniform gas pressure even when surface tension is zero. This is an exact spherical solution in the stated Stokes model, not a proof of shape stability under arbitrary perturbations.

For $P_0>0$, $\alpha \ne 0$, $T\ge T_0$ and physical $T>0$, the atmospheric pressure-temperature profile $P=P_0\exp (\alpha \sqrt{T-T_0})$ is equivalent toThe branch restriction is essential: squaring does not create a second physical branch. Its temperature gradient is $dT/d\log P=2\sqrt{T-T_0}/\alpha$. Thus, in hydrostatic equilibrium, $\alpha <0$ gives an atmospheric thermal inversion, while $\alpha >0$ gives outward cooling. At $T=T_0$ the inverse branch has zero derivative; $\alpha =0$ gives constant pressure and cannot describe a nontrivial hydrostatic vertical interval.

Let the inner-centre displacement from a fixed outer centre have magnitude $r=\Delta e$, with $0\le e<1$. The leading full-film journal bearing resistance to translation in the displacement direction and perpendicular to it isFor $h=\Delta (1-e\cos \theta )$, the Reynolds lubrication equation gives radial squeezing flux $a{v}_r\sin \theta$ and tangential squeezing flux $a{v}_t[-\cos \theta +3e/(2+e^2)]$. The additive constant enforces periodic pressure. Integrating the pressure traction gives these coefficients; at $e=0$ they coincide, as required by rotational symmetry.

The pressure obtained by bringing a flow to rest without losses; in compressible flow the usual definition specifies an isentropic process for stopping. For the constant-mass density, same-height Bernoulli equation discussed under dynamic pressure, $p_0=p+q$.

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy $U$, entropy $S$ and enthalpy $H=U+pV$ are state functions; here $p$ is pressure and $V$ is volume. Transferred heat $Q$ and work $W$ belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics readsThe notation $dU$ denotes an exact differential of a state function; $\delta Q$ and $\delta W_{\mathrm{by}}$ denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,where $T$ is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function $F$ and two equilibrium states $A,B$, an integral of its exact differential givesThe proof is telescoping: add the changes $F(x_{i+1})-F(x_i)$ along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define $F(B)$ by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has $\Delta U=0$ but can have nonzero net transfers, with $\oint \delta Q=\oint \delta W_{\mathrm{by}}$. Nor does the fact that entropy is a state function make $Q/T$ its general change: the formula $dS=\delta Q_{\mathrm{rev}}/T$ uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.

A state of matter is a broad category of material behavior. A solid resists a sustained change of shape, a liquid flows while remaining difficult to compress, and a gas expands to fill its container. These distinctions are useful first descriptions, but they do not enumerate all thermodynamic phases. Plasma adds collective behavior of charged particles, while liquid crystals show that flow and structural order need not disappear together.

Distinguish three questions: what broad state of matter does the material resemble, which thermodynamic phase is present, and what are its temperature, pressure and other state variables? Two samples at different temperatures can be in the same liquid thermodynamic phase. Conversely, crystal polymorphism gives different solid thermodynamic phases of the same substance, distinguished by their crystal structures. A phase transition between them need not change the broad state-of-matter category.

For an imagined substance with crystalline forms A and B, melting A produces a solid–liquid phase transition, while converting A to B produces a solid–solid phase transition. A sample containing A and B has two thermodynamic phases even though all its material is solid. Cutting one uniform crystal into two pieces does not create a second thermodynamic phase: count distinct equilibrium forms, rather than pieces or grains. This distinction matters when applying the phase rule.

A phase diagram locates stable thermodynamic phases using temperature and pressure. Its liquid–gas phase coexistence curve ends at a liquid-gas critical point, around which a path connects liquid and gas without a first-order phase transition. Broad categories can therefore remain useful without a sharp boundary.^[ref]

For steady flow through a fixed control volume with one inlet and one outlet, the first law of thermodynamics and conservation of mass giveHere $\dot{m}$ is the mass flow rate, $\dot{Q}$ incoming heat per unit time, ${\dot{W}}_s$ outgoing non-flow work per unit time, $w$ the flow speed, $z$ elevation and $g$ gravitational acceleration. The specific enthalpy $h=u+p/\rho$ includes both the transported internal energy per unit mass $u$ and the pressure work needed to push volume through the boundary; $\rho$ is mass density. The inlet supplies that pressure work, while the outlet consumes it to displace the downstream fluid.^[ref]

For an insulated stationary throttle, require negligible heat transfer, no shaft or other non-flow work, and negligible changes in bulk kinetic energy and gravitational potential energy. Then $h_2=h_1$. With unchanged composition this is also equality of molar enthalpy. The upstream and downstream states must admit equilibrium pressure, temperature and property values, although the constriction itself is irreversible and can be far from thermodynamic equilibrium. Start-up storage of energy in the walls and fluid, heat leakage, or a large outlet speed invalidates the reduced balance. A nozzle can convert specific enthalpy into bulk kinetic energy even when it is insulated.

The same accounting for a fixed transported amount is $U_2-U_1=p_1{V}_1-p_2{V}_2$, hence $U_1+p_1{V}_1=U_2+p_2{V}_2$. This explains why dropping the two pressure work terms would give the wrong conserved quantity.^[ref]

For constant simple shear flow rate $s$, the nonpressure stress in the component-first gradient convention $L_{12}=s$ isThus the first normal-stress difference is $N_1=2(\mu -{\mu }_r)\tau s^2$, while the second is $N_2=-N_1$. Adding the incompressibility pressure changes neither difference.

A temperature versus density diagram separates classical, degenerate, nonrelativistic and relativistic stellar-matter limits. The Electron degeneracy boundary is $T\sim T_F(\rho )$, with slope $2/3$ at low density and $1/3$ at high density. Thermal relativity is controlled by $k_{B}T\sim m_e{c}^2$, while relativity of degenerate Electrons is controlled by $p_F\sim m_{e}c$. Photon pressure and thermal electron-positron pairs add further crossovers; all depend on composition and are gradual, not thermodynamic phase boundaries.

The radiative temperature gradient needed to carry a spherical stellar luminosity follows by dividing radiative diffusion in a star by the hydrostatic pressure support equation. It is a logarithmic pressure gradient, not $dT/dr$ itself. Comparing it with the adiabatic temperature gradient by the Schwarzschild criterion determines whether purely radiative transport is locally unstable at uniform composition.

A Stirling engine repeatedly heats and cools a confined gas, whose expansion and compression drive a moving mechanical load. This makes it a particularly accessible example of a heat engine: both the repeating motion and the two regions exchanging heat can be observed.

In the published Demonstration experiments with a Stirling engine by Christopher G Deacon and others, an electrical heater warms one end of a cylinder and flowing water cools the other at $12{}^{\circ }\mathrm{C}$. A moving displacer shuttles gas through copper wool that temporarily stores and returns internal energy. A laser spot traces a closed pressure-volume path on a screen: mechanical motion controls one coordinate, pressure the other. Calibrating both scales converts the enclosed area to work per thermodynamic cycle; multiplying by rotation frequency gives gas-cycle power. At $97\mathrm{W}$ heater input, the paper reports $4.4\mathrm{W}$ gas-cycle power, while a separate Prony brake measures $3.4\mathrm{W}$ shaft output. The cold-side temperature is controlled and two higher heater settings are compared. The gas has a nonuniform temperature, and heater energy also escapes to the room: neither output/input ratio measures $W/Q_h$ for an ideal two-thermal reservoir engine. These observations demonstrate work production and losses, not a direct measurement of the bound in Carnot's theorem. The inspected apparatus, readouts and results are on pp. 180–182, Figures 1–3 and Table 1.

Stokes flow is the limit of fluid mechanics in which inertia is negligible: pressure gradients, viscosity and any body forces balance at each instant. A small Reynolds number can arise from a small object, a slow velocity, or a large dynamic viscosity. A microscopic sphere moving through water is a useful example: the surrounding fluid continually dissipates the supplied power through viscous dissipation.

The picture is a meridional slice of the three-dimensional axisymmetric flow around a sphere, with the sphere held fixed and the distant fluid moving rightward at speed $U$. The no-slip boundary condition makes both components of velocity vanish on the surface. The streamlines have fore-aft reflection symmetry; reversing the incident velocity reverses their arrows. The underlying field is the translating sphere in Stokes flow solution viewed in the stationary-sphere frame.

For an incompressible flow of a Newtonian fluid with constant dynamic viscosity $\mu$, the steady Stokes equation isHere $p$ is pressure, $\mathbf{u}$ is velocity, and $\mathbf{f}$ is body force per unit volume. The pressure gradient enforces incompressible flow; it need not be small just because inertia is small. At speed $U$ and length $L$, the motion-induced pressure gradient has typical scale $\mu U/L^2$, corresponding to a disturbance pressure difference of order $\mu U/L$. See Shreyas Mandre's fluid dynamics notes, sections 9.1.1–9.1.2.

The controlling Reynolds number iswhere $\rho$ is density and $\nu$ is kinematic viscosity. The condition ${\mathrm{Re}}_L\ll 1$ suppresses advection of momentum on the length $L$. Rapid changes require an additional check of the viscous diffusion time: small Reynolds number alone does not eliminate the time derivative in the Unsteady Stokes equation.

Two rewards of this limit are the exact Stokes's law for the drag on a sphere and kinematic reversibility of Stokes flow. The first predicts settling velocities and connects to the Stokes–Einstein relation; the second explains why a microswimmer needs a nonreciprocal stroke, as expressed by the scallop theorem. Despite viscous dissipation, reversing the driving can retrace the motion: an especially striking consequence of Linearity of Stokes flow.

George Gabriel Stokes obtained the sphere drag and its terminal velocity in his 1851 investigation of how viscosity changes physical pendulum motion. He also used the slow fall of small water droplets to discuss clouds. The paper was read on 9 December 1850; 1851 is its publication date. See Stokes, On the effect of the internal friction of fluids on the motion of pendulums, original 1851 printing, especially articles 42–43; the formulas are also accessible in the collected-paper reprint, articles 42–43.