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Transport of particles or charge when quantum mode structure, transmission and occupation of reservoir states determine the measured current.

The quasiparticle residue is the coherent pole weight of the single-particle Green's function. Its coherent spectral term $Z\delta (E-{\xi }_p)$ contributes $Z\Theta (-{\xi }_p)$ to the zero-temperature momentum occupation. A smooth remainder leaves a jump $Z$ at the Fermi surface; Landau quasiparticle occupations have a unit step.^[ref]

For the local illustrative distribution in the figure, put $x=(p-p_F)/p_F$ and$\Theta$ is the Heaviside step function. Its one-sided limits are $0.825$ and $0.175$, giving $Z=0.65$. The smooth term is stipulated; this local curve does not specify a normalized distribution over all momenta. In particular, $Z$ is neither the fraction of atoms above the Fermi surface nor $m/m^{\ast }$.

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.

For a film with fixed lower temperature $T_0$ and upper condition $-\kappa T_z=\alpha (T_s-T_0+\Delta T)$, negligible horizontal thermal diffusion and thermal advection reduce the heat equation to $T_{zz}=0$. Thus $T=T_0-\alpha \Delta Tz/(\kappa +\alpha h)$. The local approximation requires $(h/L{)}^2\ll 1$ and $U{h}^2/(\kappa L)\ll 1$, together with slow time variation relative to $h^2/\kappa$. For $\alpha h/\kappa \ll 1$, $T_s\simeq T_0-\Delta T\alpha h/\kappa$. The coefficient $\kappa$ is thermal diffusivity, and the boundary coefficient $\alpha$ is normalized accordingly.

A quench is a change of control parameters faster than the system can equilibrate. The subsequent relaxation can trap domains and topological defects.

A quench abruptly changes temperature or another control parameter and then studies the subsequent nonequilibrium relaxation of the order parameter.

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

For a nondegenerate fully ionized gas, equating $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$ with $P_{\gamma }=a_{\mathrm{rad}}{T}^4/3$ gives $T^3=3k_B\rho /(a_{\mathrm{rad}}{\mu }_{\mathrm{mol}}{m}_u)$. Radiation dominates at higher temperature for fixed density. A degenerate electron gas requires its degeneracy pressure instead of extrapolating this ideal-gas formula.

In the optically thick radiative diffusion in a star approximation, the heat flux is $\mathbf{F}=-\chi \nabla T$. The conductivity depends on opacity, mass density and temperature. It relates a material temperature gradient to energy transport and must not be confused with the emergent effective temperature.

A radiative-convective climate column applies a radiative-convective equilibrium model to the vertical thermal structure relevant to climate change. Radiative transfer determines heating and cooling; convective adjustment represents unresolved vertical redistribution when the chosen atmospheric lapse rate is too steep. Radiative equilibrium alone need not give a physically admissible lower-atmospheric profile.

In 1967, Syukuro Manabe and Richard T. Wetherald computed equilibrium with prescribed relative humidity, allowing atmospheric water vapor to change with temperature. Their average-cloud case warmed by $2.36\mathrm{K}$ for carbon dioxide increasing from 300 to 600 parts per million, versus $1.33\mathrm{K}$ with fixed absolute humidity. These are model outputs, not measured warming. The comparison isolates an important water-vapor feedback.^[ref]

An original two-temperature calculation exposes the bookkeeping. Let an opaque surface absorb solar irradiance $Q$ and emit blackbody thermal radiation $\sigma T_s^4$. A single isothermal atmospheric layer transmits all sunlight and absorbs a fraction $ϵ$ of surface infrared emission, emitting $ϵ\sigma T_a^4$ both upward and downward. This is a grey thermal-band emissivity, with $0<ϵ\le 1$. Let $H$ be upward nonradiative heat flux, and let $C_s,C_a>0$ be heat capacities per unit area. Then conservation of energy givesThe two equations sum toHere $L$ is outgoing longwave radiation. Both $H$ and downward atmospheric emission cancel internally: they redistribute energy without creating it. The factor two in atmospheric emission counts two faces. Dropping the downward face would spoil the surface budget. Each right-hand term has units $\mathrm{W}{\mathrm{m}}^{-2}$; each $C$ has units $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

Choose $Q=240\mathrm{W}{\mathrm{m}}^{-2}$, $H=80\mathrm{W}{\mathrm{m}}^{-2}$ and $ϵ=1$. At equilibrium, $L=\sigma T_a^4=240\mathrm{W}{\mathrm{m}}^{-2}$ and $\sigma T_s^4=400\mathrm{W}{\mathrm{m}}^{-2}$, giving $T_a\simeq 255.06\mathrm{K}$ and $T_s\simeq 289.81\mathrm{K}$. Direct substitution gives surface balance $240+240-400-80=0$ and atmospheric balance $400-480+80=0$. These deliberately chosen fluxes illustrate the model; they are not an observed global budget or a reconstruction of the 1967 calculation. Prescribing $H$ illustrates heat transfer accounting but does not close a realistic convection law.

For fixed $Q,H,ϵ$, the positive-temperature equilibrium is locally stable. Put $a=4\sigma T_s^3>0$ and $b=4ϵ\sigma T_a^3>0$. The Jacobian matrix of the temperature equations has trace $-a/C_s-2b/C_a<0$ and determinant $ab(2-ϵ)/(C_s{C}_a)>0$. Thus both eigenvalues have negative real part. Temperature-dependent climate feedbacks alter this test. A single grey layer also omits the optical spectrum, cloud structure and the independent thermal behavior of the troposphere and stratosphere.

For $P=C(T^b+K{)}^{1/p}$ with $p=n+1$ and $b=4+n+m$, the logarithmic radiative temperature gradient is $(p/b)(1+K/T^b)$. At a marginal boundary with an efficient ideal-gas convective envelope, the Schwarzschild criterion equates it to $(\gamma -1)/\gamma$. Hence $K=T_b^b[(b/p)(\gamma -1)/\gamma -1]$. The notation $K=T_0^b$ with real positive $T_0$ requires a nonnegative bracket. Vanishing $p$ or $b$ needs the corresponding logarithmic integration instead of this power form.

A body is in radiative equilibrium when absorbed radiative power equals emitted radiative power. A surface absorbing flux $F$ and emitting as a grey body has equilibrium temperature $T=(F/(ϵ\sigma ){)}^{1/4}$.

For a sublimating surface with exponentially absorbed volumetric heating, use a frame moving into the solid at speed $V$. Its steady heat equation is $k{T}^{\prime \prime }+\rho c_{p}V{T}^{\prime }+q_0{e}^{-\lambda z}=0$. The surface Stefan condition is $k{T}^{\prime }(0)=\rho LV$. Integrating over the half-space gives the displayed energy-limited speed. The surface gradient is positive, but the far field is colder, so the temperature has a subsurface maximum where melting can begin.

A column near pressure $p$ contains mass per area of order $p/g$. Linearizing its blackbody cooling around temperature $T$ gives the estimatewhere $c_p$ is specific heat capacity at constant pressure. This estimate omits details of wavelength-dependent radiative transfer, but captures faster relaxation in hotter and lower-pressure atmospheric layers.

A shock wave with significant radiation losses in its downstream evolution. Often a narrow dissipative jump initially heats the gas according to adiabatic Rankine-Hugoniot conditions, followed by a much longer radiative cooling layer. The gas's specific entropy increases at the jump and may then fall as radiation carries entropy away. If both distant states have equal temperature, the combined mass and momentum balances give an isothermal shock; the gas's energy flux is not conserved across the whole radiating region.

If emitted power obeys $P=A{T}^4$ with known constant $A$, then small relative temperature uncertainty is one quarter of relative net-power uncertainty. For measured quantities $X,Y$ with covariance $C_{XY}$, $\mathrm{Var}(X-Y)={\sigma }_X^2+{\sigma }_Y^2-2C_{XY}$. A weak net signal obtained by subtracting a large background can therefore have poor relative precision even when both original powers are measured accurately. Correlated standard uncertainties and worst-case error bounds require different propagation rules.

For a constant-opacity, gas-pressure-supported radiative disk surface at $z=H$, imposing $p=T=0$ does not impose $F=0$. If the outgoing surface flux is finite and positive, dividing hydrostatic balance by the temperature gradient gives $dp/dT\sim 16\sigma {\Omega }^{2}H{T}^3/(3\kappa F_s)$. Consequently $p\propto T^4$, $\rho \propto T^3$, and the diffusion equation gives $T\propto H-z$. Thus the mass density also vanishes regularly while the surface radiates a finite flux. This ideal boundary replaces the finite-temperature photospheric transition; its accuracy is not controlled solely by the disk aspect ratio.

// Target: astrophysics.bigb

A radiometer measures electromagnetic radiation through its received power. A radio radiometer can express that power as a noise temperature, allowing an unknown source to be compared with calibrated reference temperatures.

A radiometer thermal link is the intentional path for thermal conduction from receiver to reference block. Near a chosen operating point, write $Q_{\mathrm{link}}\simeq G(T-T_b)$ with local conductance $G$ in $\mathrm{W}{\mathrm{K}}^{-1}$. A weaker link increases the steady temperature rise per added power, $\delta T\simeq \delta P/G$, but also increases the uncontrolled relaxation time $\tau =C/G$ when the link dominates other heat transfer. These are local linear approximations: actual thermal conductivity depends on temperature, and active feedback control introduces its own response. NIST explicitly describes the tradeoff between weaker links and smaller receiver heat capacity.^[ref]

Radiometric absorptance is the fraction of incident radiant flux converted to heat in the specified receiver: $P_{\mathrm{abs}}=a{P}_{\mathrm{in}}$, with $0\le a\le 1$ for a passive receiver. Specify the entrance boundary, wavelength, incidence angle and receiver temperature. Radiation reflected or transmitted out of that receiver is not absorbed power. A cavity can improve radiometric absorptance by giving light repeated opportunities for wave absorption; a dark coating alone does not establish $a=1$.

For broadband light, spectral radiant flux matters: $P_{\mathrm{abs}}=\int a(\lambda ){\Phi }_{e,\lambda }(\lambda )d\lambda$. A single effective radiometric absorptance therefore depends on the incident optical spectrum as well as on the receiver. If the stated incident boundary lies before a window of known power transmission ${\tau }_w$, the receiver absorbs $a{\tau }_w{P}_{\mathrm{before}}$ when other optical losses are absent. Window loss and receiver radiometric absorptance are separate corrections.

At fixed measured count slope, $N_A^{\mathrm{sed}}\propto a^{-3}$, whereas at fixed measured diffusion coefficient, $N_A^{\mathrm{diff}}\propto a^{-1}$. First-order propagation of uncertainty follows by taking logarithmic derivatives:For independent small errors, the radius contributes $9({\sigma }_a/a{)}^2$ to the relative variance of $N_A^{\mathrm{sed}}$, compared with $({\sigma }_a/a{)}^2$ for $N_A^{\mathrm{diff}}$. Shared radius and temperature calibrations make the two estimates correlated; their agreement is therefore not two wholly independent determinations. As a finite-error example, overestimating $a$ by $5\%$ multiplies the respective answers by $1.05^{-3}\simeq 0.8638$ and $1.05^{-1}\simeq 0.9524$.

This explains the importance of the radius cross-checks in the 1909 memoir: testing the Stokes drag law through an independent radius estimate helps prevent circular inference.^[ref] Convection must also be excluded: a residual horizontal speed $u$ adds $u^2{\tau }^2$ to an uncentred mean squared displacement, making a linear fit misleading.