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The temperature of a planetary surface at a specified location or averaging scale. It can differ from overlying air temperature, effective radiating temperature and spectral brightness temperature.

For thermal diffusivity $\kappa >0$ and thermal conductivity $k>0$, a semi-infinite conductor satisfies $u_t=\kappa u_{xx}$ for $x>0$. Take zero initial temperature and constant incoming heat flux $-k{u}_x(0,t)=Q$. Select the physical solution by boundedness at spatial infinity together with sufficient growth control for the time Laplace transform.

Writing the time transform as $U$ gives $U_{xx}=(p/\kappa )U$. The decaying branch is $U=A(p)e^{-x\sqrt{p/\kappa }}$, where $\mathrm{Re}\sqrt{p}>0$. The transformed flux condition gives $A=Q\sqrt{\kappa }/(k{p}^{3/2})$. Since $\mathcal{L}[2\sqrt{t}]=\sqrt{\pi }{p}^{-3/2}$, inversion gives the displayed surface temperature. More generally the full profile isDifferentiation verifies the heat equation and the applied flux; for every $x>0$ it tends to zero at initial time and decays at spatial infinity. Its surface temperature grows as $\sqrt{t}$, because the depth significantly heated by diffusion has the same scale.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.

A switching-current distribution of a Josephson junction records the bias electric current at the first detected transition to a sustained nonzero electric potential difference. Interpret it using a stipulated monotone ramp $dI/dt=r>0$, a reproducible initial trapped state, and an instantaneous escape hazard function $\Gamma (I)$. Neglect retrapping and readout delay. The survival function $S$ obeys $dS/dt=-\Gamma S$, so changing variable from time to electric current givesHere $p$ is a probability density function per unit electric current, not a rate per unit time. Division by the surviving population is essential: only trials still trapped can escape. A typical switching current is consequently not identical to the noiseless superconducting critical current.

For an independent illustrative check, choose constant ${\Gamma }_0=1000{\mathrm{s}}^{-1}$ and $r=1\mu \mathrm{A}{\mathrm{s}}^{-1}$. ThenHalf the trials have switched after $\ln 2/{\Gamma }_0=0.693147\mathrm{ms}$, corresponding to $0.693147\mathrm{nA}$ of ramp. Doubling $r$ doubles the mean increment without changing the underlying escape rate. Real junction rates vary sharply with bias; this example isolates the statistical conversion.

Repeating the preparation, measuring the ramp, resolving the voltage onset, and allowing cooling after each dissipative event are prerequisites for this interpretation. Missed events, finite detection delay, changing temperature and retrapping alter the measured distribution. Even a correctly reconstructed rate needs the independent thermal and spectral checks before identifying its escape mechanism.

A broad component plus a narrow central component can have one local maximum while still displaying two distinct width scales. This is what “bimodal” means in this imaging context; it does not require two separated peaks. To isolate the population effect, construct an original normalized line profileBoth terms are normal distributions; their integrals are one, so the component areas are $1-f_0$ and $f_0$. With $f_0=0.488$, the narrow component has central height $f_0/(\sqrt{2\pi }0.20)\simeq 0.9734$, while the broad component contributes $(1-f_0)/\sqrt{2\pi }\simeq 0.2043$. A tall central feature therefore need not contain most of the particles. Holding these widths fixed deliberately isolates changing weights; it does not model how temperature, interactions or camera resolution determine actual widths.

For absorption imaging in the weak-probe regime with negligible saturation, the optical depth is $\mathrm{OD}(x,z)=-\ln [I(x,z)/I_0(x,z)]={\sigma }_{\mathrm{abs}}{n}_{\mathrm{col}}(x,z)$, where $I_0$ is the reference light intensity without atoms, $I$ the transmitted light intensity, ${\sigma }_{\mathrm{abs}}$ an effective absorption cross section, and $n_{\mathrm{col}}$ the column density. Integrating $n_{\mathrm{col}}$ over both image coordinates gives the particle number; a central pixel height alone does not. Finite resolution can obscure a narrow component, and saturation invalidates this simple logarithmic calibration.

The system-equivalent flux density is the unpolarized spectral flux density that would produce an antenna temperature equal to $T_{\mathrm{sys}}$. With effective collecting area $A_e$, $\mathrm{SEFD}=2k_B{T}_{\mathrm{sys}}/A_e$. The factor two follows from one polarization of waves measurement of an unpolarized source. State effective area and polarization of waves conventions when comparing sensitivities.^[ref]

The equivalent input noise temperature of the antenna-plus-receiver system is defined by available noise power $P_n=k_B{T}_{\mathrm{sys}}\Delta \nu$ in one matched polarization of waves mode. It includes receiver noise and the appropriate sky and instrumental contributions. It is not generally the physical temperature of a single component.

The change in overpotential per decade of current density in the activation-controlled regime of a Tafel equation. In a single-step Butler-Volmer equation with constant transfer coefficient, the anodic slope is $b_a=RT\ln 10/({\alpha }_{a}nF)$, where $T$ is temperature, $R$ is the molar gas constant, $n$ is the number of transferred electrons and $F$ is the Faraday constant. The magnitude of the cathodic slope is $b_c=RT\ln 10/({\alpha }_{c}nF)$. Inferring one from the other requires the model's relation between the two coefficients; the equality ${\alpha }_a+{\alpha }_c=1$ applies to this effective single-step model and need not describe a multistep reaction.

Temperature labels thermal equilibrium: two internally equilibrated bodies have the same temperature when contact permitting only heat exchange produces no net heat transfer. The zeroth law of thermodynamics makes the comparison transitive. If bodies A and B separately reach thermal equilibrium with the same thermometer without appreciably disturbing either body, they agree with each other. This is what lets a gas thermometer compare a metal block with water, despite their different internal energies and heat capacities. Insulation alone makes no such comparison: it can prevent heat transfer between bodies at different temperatures. ^[ref]

The zeroth law of thermodynamics establishes equality comparisons, but does not choose a numerical scale. Thermodynamic temperature supplies an absolute scale consistent with Carnot's theorem and entropy. For a simple compressible system at thermodynamic equilibrium with fixed particle number $N$ and composition,Here $U$ is internal energy, $S$ is entropy, and $V$ is volume; the second equality assumes the derivative can be inverted. Temperature is a thermodynamic conjugate variable to entropy and belongs to the intensive and extensive thermodynamic quantities as an intensive quantity. Its partial derivative tells us how much internal energy must change per unit change of entropy under the stated constraints. More general systems require their other mechanical and material variables to be held fixed too. ^[ref]

For an infinitesimal quantity $\delta Q>0$ of heat transferred from A to B, with no work or matter exchange, the equilibrium entropy changes giveFor ordinary positive temperatures, spontaneous heat transfer from $T_A>T_B$ increases total entropy; thermal equilibrium has $T_A=T_B$. This comparison uses changes in entropy, rather than the total internal energy contained in either body. A small hot object can have less internal energy than a large cool one. See entropy maximization under thermal contact for the equilibrium criterion and negative temperature for the special case of bounded energy spectra.

Express absolute thermodynamic temperature $T$ in kelvin, and a Celsius reading $t_C$ in degree Celsius:The offset is exact: $300\mathrm{K}$ corresponds to $26.85{}^{\circ }\mathrm{C}$, and $400\mathrm{K}$ to $126.85{}^{\circ }\mathrm{C}$. Their absolute temperature ratio is $4/3$; dividing their Celsius readings gives a different number. Thus gas thermometer pressure ratios and Carnot's theorem require thermodynamic temperature. The modern SI definition of the kelvin fixes the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$; a reference bath realizes or transfers the scale rather than defining that constant.^[ref]

Temperature is not universally the mean kinetic energy of a particle. In a classical, nonrelativistic ideal gas at thermal equilibrium, the equipartition theorem gives the mean translational kinetic energy relative to the gas's bulk motion:Here $m$ is particle mass, $\mathbf{v}$ its velocity, and $\mathbf{u}$ the mean flow velocity. In this regime, different particle masses at the same temperature have equal mean translational kinetic energies but different characteristic speeds. A moving gas can also carry bulk kinetic energy without being hotter in its rest frame. Internal molecular modes and quantum statistics require further care; the equipartition theorem has classical assumptions, whereas the equilibrium entropy definition applies more widely. ^[ref]

The change of reactor reactivity with a specified material temperature, with the other variables held as stated. Several component coefficients need not have the same sign.

The fractional change in electrical resistance per unit temperature change, with geometry and other conditions specified. For a small interval around $T_0$, $R(T)\simeq R(T_0)[1+{\alpha }_R(T-T_0)]$. Low $|{\alpha }_R|$ is useful in stable resistors.

Diffuse gas in photoionization equilibrium is often approximately described by the displayed temperature-density contrast relation. Photoheating and expansion cooling produce its thermal history. It is an empirical model for low-density gas, not the adiabatic equation of state of every intergalactic phase. Lyman-alpha forest line widths constrain both the temperature at a reference density and the slope through their lower envelope.

For $\varphi \ge 0$ and positive $c,B,\lambda ,T_0$, define $\gamma =2B^2/(\lambda c)$. Completing the square givesFor $0<\gamma <1$, phase coexistence occurs at $T^{\ast }=T_0/\sqrt{1-\gamma }$ with a discontinuous minimum at ${\varphi }^{\ast }=2B{T}^{\ast }/\lambda$. For $\gamma \ge 1$, a nonzero global minimum has lower free energy at every finite temperature. Following the global equilibrium branch then does not force a restoration transition.

In the absence of heat advection, an insulating sphere of radius $a$ in a uniform remote temperature gradient $\mathbf{H}$ has the displayed exterior harmonic function as its temperature field. The reflected dipole term makes ${\partial }_{r}T=0$ at $r=a$. The surface temperature is $T_0+(3/2)a\mathbf{H}\cdot \mathbf{n}$, so the tangential gradient is amplified by $3/2$. If the sphere has thermal conductivity ratio $K$ relative to its surroundings, temperature and heat-flux continuity instead give surface gradient factor $3/(2+K)$.

Equilibrium fails where the helium-3 relaxation time becomes comparable to stellar age. In a fixed-density pp-I scaling with reaction exponents four and sixteen, the equilibrium abundance scales as $T^{-6}$ and the relaxation time as $T^{-10}$. Central normalization then places the transition near six million kelvin; using the full supplied exponential factors gives a somewhat higher estimate. A precise radius or temperature requires a stellar mass density/composition profile and the second branch contribution.

The temperature gradient is the gradient $\nabla T$ of the temperature field. It points in the direction of fastest temperature increase.

A rapid temperature perturbation that shifts equilibrium, allowing relaxation toward the new state to reveal kinetic processes.

For a smooth simple compressible system, $C_V=T(\partial S/\partial T{)}_V$ and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$ give $(\partial C_V/\partial V{)}_T=T({\partial }^{2}p/\partial T^2{)}_V$. Thus an equation of state $p=a(V)T+b(V)$ implies that $C_V$ depends on temperature alone within a smooth phase. It does not require $C_V$ to be temperature independent.