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A thermodynamic process changes a thermodynamic system between states. A quasistatic process supplies equilibrium intermediate states and therefore a path on equilibrium coordinates. An irreversible thermodynamic process can have equilibrium endpoints while lacking a uniform intermediate temperature or pressure. Endpoint values still determine changes in a state function; computing heat and work requires the actual interaction history.

Thermodynamics studies energy transfers and the constraints on macroscopic changes. Its vocabulary connects internal energy, entropy, temperature, pressure and volume to heat and work. The first law of thermodynamics balances energy; the Second law of thermodynamics decides which changes can occur spontaneously and limits how much work a heat engine can deliver. These principles let us bound performance before choosing a material or a detailed machine design. Applications include power generation, propulsion and refrigerators. ^[ref]

Start by choosing a system boundary. A closed system can exchange energy but no matter; an isolated system exchanges neither. A state function, such as internal energy or entropy, depends on the equilibrium state. A process specifies how that state changes. Thus a body's temperature is a property it has, whereas heat records energy crossing its boundary during a process. A body contains internal energy, rather than a stored quantity of heat.

Two insulated bodies initially at different positive temperatures approach thermal equilibrium when brought into thermal contact. Their combined energy is conserved, but their combined entropy increases. They do not spontaneously separate again into a hotter body and a colder one. This familiar asymmetry makes thermodynamics interesting: conserving energy leaves many imaginable changes available, and entropy supplies a further constraint. The same distinction underlies both entropy maximization under thermal contact and the thermal efficiency limit in Carnot's theorem.

An equilibrium thermodynamic state specifies sufficient independent macroscopic properties, together with material identity and constraints, to determine the other equilibrium properties. For a uniform fixed-composition ideal gas, its amount of substance, temperature and volume determine its pressure through the ideal gas law. These properties specify a state, without specifying a thermodynamic process that prepared it.^[ref]

The thermomechanical effect couples a temperature difference to a pressure difference across a superleak. For the closed-chamber model in the apparatus figure, consider neighboring equilibrium states with small velocities, negligible height difference and pure helium-4. Let ${\mu }_m$ be chemical potential per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. The Gibbs free energy differential per unit mass givesThe superleak allows the superfluid to redistribute until the two sides have equal ${\mu }_m$. Setting $d{\mu }_m=0$ therefore givesThe sign matters: heating one side at initially equal pressure lowers its ${\mu }_m$; inflow raises its pressure until chemical balance is restored. The hotter side has the higher pressure. The integral follows the constant-${\mu }_m$ path; it need not be evaluated at fixed pressure when material properties vary appreciably.

For an original local example, stipulate constant $\rho =150\mathrm{kg}{\mathrm{m}}^{-3}$, $s=100\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $T_h-T_c=0.020\mathrm{K}$. The predicted pressure difference is $300\mathrm{Pa}$. These are teaching assumptions, not helium measurements. This static thermodynamics relation predicts neither the required heater power nor the filling time: those require heat leakage, channel conductance and chamber heat capacity. An outlet would introduce kinetic energy and a flowing balance, so the static $300\mathrm{Pa}$ is not a predicted fountain height.

A thermometer measures temperature through a physical response whose relation to temperature is known. Examples include the length change from thermal expansion, a change in electrical resistance, and the pressure of a gas thermometer. If the response function $X=f(T)$ is an injective function over the working range, its measured value determines $T=f^{-1}(X)$ through the inverse function. Thermometer calibration establishes that relation for the particular instrument and measurement conditions. Two matching readings are meaningful only after calibration to a common scale and with the probes sampling comparable equilibrium states. ^[ref]

For a contact thermometer, the sensing part must reach thermal equilibrium with the region being measured. That region may differ from the rest of the object: an immersed probe near a heater can read a local temperature while the surrounding water remains cooler. A stable reading can also be a steady state set by competing heat transfer paths to the sample and the room. Good contact, adequate immersion, mixing where appropriate, and checks for spatial differences help establish which temperature the reading represents. These are measurement controls inferred from the thermal equilibrium requirement, not a guarantee that every flat trace is equilibrium.

A thermometer also changes the sample it touches. Consider an original toy calculation with an initially uniform sample at $T_o=350\mathrm{K}$, a probe at $T_p=300\mathrm{K}$, and constant whole-body heat capacities $C_o=100\mathrm{J}{\mathrm{K}}^{-1}$ and $C_p=1\mathrm{J}{\mathrm{K}}^{-1}$. Suppose together they form an isolated system, with no work or phase transition and negligible contact energy. The first law of thermodynamics and thermal equilibrium giveThe final thermometer reading underestimates the sample's original temperature by about $0.495\mathrm{K}$, although it correctly measures their new common temperature. The disturbance becomes small when the participating probe heat capacity is small compared with the sample's. A large controlled heat bath is a different limit: it supplies or removes heat while approximately maintaining its own temperature.

Waiting time is a separate issue from disturbance. Under a toy Newton's law of cooling model for a uniform probe in a maintained heat bath, a constant contact conductance $G$ givesThe time constant $\tau$ controls this exponential decay. For stipulated values $T_b=350\mathrm{K}$, $T_p(0)=300\mathrm{K}$ and $\tau =5\mathrm{s}$, the probe reads about $331.61\mathrm{K}$ after one time constant and $347.51\mathrm{K}$ after three. These are calculated values, not an instrument specification or measured data. Real probes may have multiple relaxation times and additional heat transfer paths; choose an equilibration criterion from the required accuracy, rather than assuming that a fixed waiting time works universally.

For smooth radial variation on scale $r$, $c_s\sim \Omega H$, $\nu \sim \alpha \Omega H^2$, $u_r\sim \nu /r$ and $u_z\sim H{u}_r/r$. The exact point-mass vertical gravity differs from ${\Omega }_K^{2}z$ by relative order $(H/r{)}^2$. Radial pressure support changes the rotation and dominant shear heating by the same order. Meridional viscous forces, radial radiative-flux divergence, thermal advection and compressional work are at most order $(H/r{)}^2$ relative to their retained vertical terms for $\alpha \lesssim 1$. These estimates require secular disk evolution and exclude sharp radial boundary layers or independently rapid vertical motion.

// Target: gravitational-instability-of-an-astrophysical-disk.bigb

Let a strong shock wave of radius $R$ and speed $V$ sweep a uniform mass density ${\rho }_0$ into a shell of mass $4\pi {\rho }_0{R}^3/3$ and speed $2V/(\gamma +1)$. If the cavity pressure is $\alpha$ times the immediate post-shock pressure, shell momentum balance gives $R\dot{V}=3(\alpha -1)V^2$. Constant total energy gives $R^3{V}^2=\mathrm{constant}$ and hence $\alpha =1/2$. Including shell internal and kinetic energy, and treating the cavity volume as $4\pi R^3/3$ to thin-shell accuracy, yields $R=\xi (E/{\rho }_0{)}^{1/5}{t}^{2/5}$ with ${\xi }^5=75(\gamma -1)(\gamma +1{)}^2/[16\pi (5\gamma -3)]$. This coefficient is approximate; resolving the shell's finite volume changes it.

For the uniform noninteracting Electron model with two spin states and kinetic energy of nonrelativistic particles $E=p^2/(2m_e)$, count all available one-particle states up to an energy $E\ge 0$:This extensive density of states has units of inverse energy; $g(E)/V$ is the density of states per unit volume. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero temperature those states are empty. The occupied energy distribution is $g(E)f(E)$, with the Fermi-Dirac distribution reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum integral.

With the dimensionless quantity $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt{x}/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the internal energy and its mean per particle:Thus the highest occupied energy and the mean energy differ. For the Fermi energy of a model electron gas, $U/N\simeq 1.0155\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the derivative $P=-(\partial U/\partial V{)}_N$ gives the electron degeneracy pressure $P=2n{E}_F/5$. The same state-counting geometry therefore predicts a nonzero pressure at zero temperature.

A point injection of volume at rate $Q$ produces the displayed radial potential flow in three dimensions. Its velocity potential is $-Q/(4\pi r)$, and every enclosing sphere has outward volume flux $Q$. The divergence vanishes away from the source and is $Q$ times the three-dimensional Dirac delta distribution at it. The field is also a Stokes flow with constant pressure away from its singularity.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.