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The Richardson-Dushman law models the magnitude of emitted electric current density asHere $T$ is absolute thermodynamic temperature, $W$ is the work function, $k_B$ is the Boltzmann constant, and $A$ is the Thermionic Richardson constant. The signed electric current density points opposite to the outgoing electron motion; $J$ here is positive. For uniform emission over area $S$, the emitted electric current magnitude is $I=SJ$.

The Richardson-Dushman law assumes an emitter close to thermal equilibrium, a surface with a well-defined and approximately constant work function over the fitted temperature range, and predominantly thermal escape over the barrier. Interpreting a measured collector electric current with this law additionally requires an emission-limited thermionic current, known emitting area, and negligible return emission from the collector. Strong electric fields can lower the surface barrier or enable quantum tunnelling; then this zero-field expression is insufficient.

For a deliberately chosen teaching model, take $W=2.0\mathrm{eV}$ and $A=120\mathrm{A}\mathrm{c}{\mathrm{m}}^{-2}{\mathrm{K}}^{-2}$, expressing the barrier in electronvolts. These are illustrative parameters, not measured properties of a named material. Using $k_B=8.617333262\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$, at $1000\mathrm{K}$ the Boltzmann factor is about $8.33\times 10^{-11}$, yetA 20 percent increase in temperature raises the predicted electric current density by about 69 times. For a chosen emitting area $S=0.010\mathrm{c}{\mathrm{m}}^2$, these correspond to approximately $0.100\mathrm{mA}$ and $6.89\mathrm{mA}$ if collection is complete. The small escaping fraction is replenished by the external electrical circuit.

The sensitivity is especially clear from the logarithmic derivative at fixed $A,W$:In the chosen model at $1000\mathrm{K}$, a $1\mathrm{K}$ rise changes $J$ by about 2.5 percent to first order. Accurate temperature measurement matters even when electric current measurement is easy.

For the spin-free linear rigid rotor in thermal equilibrium at temperature $T>0$, assume all integer $J\ge 0$ are allowed and ignore nuclear-spin restrictions. Each $J$ level has $2J+1$ quantum states sharing one energy eigenvalue, an instance of degenerate energy levels, so the Boltzmann distribution givesHere $Q_{\mathrm{rot}}$ is a rotational partition function, $h$ the Planck constant, $k_{\mathrm{B}}$ the Boltzmann constant, and $B$ a frequency in $\mathrm{Hz}$. Thus $hB/(k_{\mathrm{B}}T)$ is dimensionless. The $2J+1$ factor counts magnetic quantum numbers; it does not mean that a whole level is one quantum state. For the stipulated $B=5\mathrm{GHz}$, the most populated levels are $J=2$ at $3\mathrm{K}$ and $J=7$ at $30\mathrm{K}$.

Population alone is insufficient to predict intensity. In weak, unsaturated absorption, the integrated absorption coefficient includes the transition frequency, the squared electric dipole moment matrix element, and the difference of lower and upper substate populations because stimulated emission returns energy to the field. A pulsed FTMW spectroscopy signal additionally depends on excitation, quantum coherence, cavity tuning, and receiver response.^[ref]

For an isotropic sample with polarization of an electromagnetic wave along $z$, write $d_z=d\cos \theta$. Normalized spherical harmonics giveThe last equality follows from ${\sum }_{M=-J}^J{M}^2=J(J+1)(2J+1)/3$. Combining this sum with the per-substate Boltzmann distribution and subtracting stimulated emission gives relative integrated absorption areasThe common factor includes $d^2/3$ and cancels upon normalization. This formula describes integrated weak absorption in a thermal, spin-free linear rigid rotor, not arbitrary peak heights. The example's strongest integrated lines originate at $J=3$ and $J=13$ for $3$ and $30\mathrm{K}$, respectively: they differ from the most populated levels. A missing line can reflect population, coupling, coverage, or sensitivity; it does not prove that the corresponding quantum state is absent.

For charge-neutral pure hydrogen in thermal equilibrium, write the Saha equation as $n_e{n}_p/n_{\mathrm{HI}}=S(T)$. A measured neutral number density fixes $n_e=n_p=\sqrt{S(T)n_{\mathrm{HI}}}$, so the neutral fraction is $f_{\mathrm{HI}}=n_{\mathrm{HI}}/(n_p+n_{\mathrm{HI}})=(1+\sqrt{S/n_{\mathrm{HI}}}{)}^{-1}$. At high ionization it scales as $\sqrt{n_{\mathrm{HI}}/S}$, rather than $n_{\mathrm{HI}}/S$. The latter mistake confuses measured neutral number density with total number density. Photoionization equilibrium with an external nonthermal field generally requires a rate equation instead of the Saha equation.

The standard conditions for baryogenesis from a symmetric initial ensemble are baryon number violation, both C violation and CP violation, and departure from thermal equilibrium. Conserved baryon number cannot develop a net value from zero; an exact conjugation symmetry pairs production with equal antiproduction. In thermal equilibrium, CPT symmetry makes the equilibrium expectation of net baryon number vanish for zero imposed chemical potentials for charges odd under CPT symmetry. These conditions assume the usual local quantum-field framework and do not alone guarantee a large asymmetry.

A semiconductor has an electronic band gap and a conductivity strongly affected by temperature, impurities and carrier injection. Optical absorption can create mobile electrons and holes, enabling a photodiode.

The useful feature of a semiconductor is that its charge carrier population can be controlled. Semiconductor doping changes the supply of mobile electrons or electron holes; carrier injection changes their population away from thermal equilibrium. A transistor turns that control into an electronic amplifier: a small input signal controls a larger flow of supplied energy.

For a uniform semiconductor in a weak electric field, a simple drift model writes its electrical conductivity asHere $q>0$ is the elementary charge, $n,p$ are the number densities of mobile electrons and electron holes, and ${\mu }_n,{\mu }_p$ their positive charge carrier mobilities. The two contributions add because opposite electric charges drift in opposite directions and therefore carry electric current in the same direction. This model assumes fixed temperature and mobilities and omits diffusion from density gradients.

An original dimensionless example isolates the effect of carrier injection. Suppose initially $p\ll n=n_0$ and ${\mu }_n=2{\mu }_p$. Adding equal excess populations $\Delta n=\Delta p=0.1n_0$ gives $\Delta \sigma /{\sigma }_0=0.15$, even though the added electron holes remain a minority carrier population. A semiconductor can respond strongly to a small injected population without changing its underlying band gap.

Shot noise is fluctuation associated with discrete transfer events. For an electric current, individual charge carriers contribute finite amounts of electric charge, even when a meter displays a smooth mean. The simplest model is a stream of equal, independently arriving charges with constant rate: its Poisson process predicts both the mean electric current and the variance around it. Discreteness alone does not imply a Poisson distribution; the timing and correlations matter.

The useful surprise is that noise can reveal something a mean hides. Two devices can carry the same mean electric current yet have different fluctuations because one transfers charges independently and the other groups or regulates them. A calibrated shot noise photocurrent measurement can connect a macroscopic voltage variance to the elementary charge. Thermal electrical noise instead exists in an unbiased resistor at thermal equilibrium; a measured background can contain both mechanisms.

Walter Schottky's 1918 paper Über spontane Stromschwankungen in verschiedenen Elektrizitätsleitern treated amplifier sensitivity limits. Part II considered independent discrete transfers, including thermionic emission through a vacuum. On p. 563 he proposed amplifying fluctuations from a separate discharge tube to determine the elementary charge; those experiments were prospective, not reported results.^[ref]

The Fano factor $F_T$ compares count variance with its mean over a specified gate. For a stationary unidirectional stream with fixed charge $q$, finite count second moments, and a generalized autocovariance with finite integral so that the long-gate limit exists,The zero-frequency relation concerns the continuous fluctuation spectrum, excluding the mean spectral line. It follows by writing $\mathrm{Var}(Q_T)$ as the double integral of the current autocovariance; for long $T$, its leading term is $T{S}_I^{(2)}(0)$. A finite-gate $F_T$, a frequency-dependent ratio $S_I^{(1)}(f)/(2qI)$, and a factor inferred from one filtered variance are not interchangeable. The last is weighted by the instrument's $|K(f){|}^2$.

Two original counting counterexamples show why electric charge discreteness is insufficient. In $n$ independent time slots, transfer one charge with probability $p$ in each slot: the binomial distribution gives mean $np$, variance $np(1-p)$ and Fano factor $1-p$. A random common origin for the slots removes a preferred time origin; the long-gate factor remains $1-p$. Alternatively, if independent Poisson process bursts each transfer exactly two charges, then $N_T=2M_T$ gives $\mathrm{Var}(N_T)=2\mathbb{E}[N_T]$ and factor $2$. Both count individual charges; only the first model allows at most one per slot, and the second groups them in pairs.

Consequently $2qI$ is a reference for independent unidirectional arrivals, not a universal noise law for all electrical conductors. Correlations, charge multiplication, and bidirectional transfer require a more specific model. For example, independent opposite streams with positive current magnitudes $I_{+}$ and $I_{-}$ have net mean $I=I_{+}-I_{-}$ but one-sided fluctuation spectrum $2q(I_{+}+I_{-})$. Using $2q|I|$ would miss their fluctuations when the net current is small. At thermal equilibrium, the full transport model must also respect thermal electrical noise; adding unrelated phenomenological formulas can double count the same underlying transfers.

Exchange of energy between magnetic spin degrees of freedom and the surrounding lattice, allowing their temperatures to approach thermal equilibrium. It permits adiabatic demagnetization to cool a solid and its attached load.

An approximation setting the time derivative of a transient population or intermediate concentration to zero when production and loss balance on the timescale considered. It differs from thermal equilibrium.

In stellar structure, stellar thermal equilibrium means that internal heating balances the surface luminosity and any energy loss through neutrinos, with negligible secular storage of thermal energy on the timescale considered. A star can satisfy this balance while its interior is hotter than its surface and heat flows outward. It is therefore different from equal-temperature thermal equilibrium between systems. Neglecting neutrino losses, the balance is $L_{\mathrm{surface}}=L_{\mathrm{heat}}$. In a slowly mass-losing donor star, a stellar thermal equilibrium radius sequence applies only when thermal adjustment is fast compared with the mass-loss timescale; rapid loss instead requires the adiabatic stellar radius response exponent.

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]

Thermal boundary resistance describes a temperature drop between the two sides of a material interface carrying heat flux. Efficient thermal conduction within each material does not guarantee efficient heat transfer between them. The solid–helium case is Kapitza resistance; the broader concept also includes solid–solid interfaces. ^[ref]

Let $T_s$ and $T_l$ be the temperatures extrapolated to the interface from its solid and liquid sides, each treated as internally in thermal equilibrium. For uniform normal heat flux $q>0$ through area $A$, define the finite-drop, area-specific thermal resistance and the resistance of the whole interface byThe macroscopic jump represents an unresolved interfacial region; it is not a claim that one point has two equilibrium temperatures. At nonzero heat flux, the two sides are outside mutual thermal equilibrium. A measured difference between distant thermometers also contains bulk temperature gradients, which must be measured or bounded before assigning all of it to the interface.

An energy balance explains why a temperature jump can coexist with continuous heat flux. For a stationary interfacial region with no local work or heat source,The first law of thermodynamics therefore requires $q_{\mathrm{in}}=q_{\mathrm{out}}=q$ in a steady state. It does not require $T_s=T_l$. For a heated sample with electrical power $P=VI$, an independently determined outward parasitic loss $P_{\mathrm{leak}}$ and no other inputs,Here $V$ is the heater's electric potential difference and $I$ its electric current. Subtracting a lead loss means subtracting power before dividing by area; $q$ is not the total heating power. Unknown losses and stored internal energy prevent that simple steady inference.

For a one-dimensional comparison, put an interface between two slabs with the same area $A$, positive constant thermal conductivities $k_s,k_l$ and lengths $L_s,L_l$. Neglect side losses. Integrating Fourier's law across each slab givesThus, for a linear interface with constant $r_B$, the series result isThese slab formulas are a chosen thermal conduction model, not a bulk transport law for moving liquid helium. If the interface law is nonlinear, substitute its own relation between $q$ and $T_s-T_l$ into the preceding balance; a single constant resistance cannot describe every load. Increasing bulk thermal conductivity has diminishing benefit when the interfacial drop dominates. Increasing usable contact area reduces the whole-interface thermal resistance if the local interface law and the other conditions remain fixed.

A bead in a liquid keeps undergoing Brownian motion even when an optical trap holds its average position fixed. For a stationary harmonic optical trap in thermal equilibrium at absolute temperature $T$, the equipartition theorem assigns mean potential energy $k_{B}T/2$ to this one quadratic coordinate. With $k_B$ the Boltzmann constant and $x=0$ at equilibrium,At the same temperature, quadrupling the spring constant divides the position variance by four and the standard deviation by two. The Boltzmann factor gives the complete equilibrium probability density function, a Gaussian distribution:The prefactor makes ${\int }_{-\infty }^{\infty }p(x)dx=1$; its second moment is ${\int }_{-\infty }^{\infty }{x}^{2}p(x)dx=k_{B}T/k$. The probability density function has units of inverse length, and its area gives probability. This density is exact for the ideal quadratic model. In a finite optical trap it approximates the occupied well when thermal excursions remain within the harmonic region and escape is negligible during observation. Driven motion, appreciable nonconservative forces or laser heating need not give this equilibrium density.

Here $\ell =\sqrt{k_{B}T/k_0}$, $q=x/\ell$ and $\kappa =k/k_0$. The plotted density per unit $q$ is $p_q(q)=\ell p(\ell q)=\sqrt{\kappa /(2\pi )}{e}^{-\kappa q^2/2}$. Each curve has total area one; their dimensionless standard deviations are $1$, $1/\sqrt{2}$ and $1/2$. The curves are exact model densities, not noisy experimental data.

Brownian motion becomes a measuring tool: a calibrated position measurement supplies ${\sigma }^2=⟨(x-⟨x⟩{)}^2⟩$, and the known temperature supplies $k=k_{B}T/{\sigma }^2$. Once $k$ is known, a slowly applied constant external force $f$ shifts the mean by $⟨x⟩=f/k$, so $f=k⟨x⟩$ in the same harmonic range. Fluctuations therefore calibrate the conversion from displacement to force. Svoboda and Block used this equipartition theorem method when calibrating their bead traps. ^[ref].

For the example stiffness $0.01pN/nm$ and $T=300\mathrm{K}$, $k_{B}T\approx 4.142\mathrm{pN}\mathrm{nm}$ and $\sigma \approx 20.35\mathrm{nm}$. Using a variance measured in uncalibrated detector units would not determine that physical stiffness. Measurement noise, finite exposure time, drift and insufficient sampling of the correlated Brownian motion can bias the estimate; record positions over many optical trap relaxation times and account for the detector response. The deterministic relaxation curve above describes relaxation without thermal forcing; an individual thermal trajectory continues to fluctuate.

Two thermodynamic systems are in thermal contact when their boundary permits heat transfer. They need not already be in thermal equilibrium: contact between unequal temperatures is precisely how equilibration can begin. Contact specifies an allowed exchange; equilibrium specifies its balance.

The thermal diffusion time $t_D=L^2/\alpha$, with $\alpha =\kappa$ the thermal diffusivity, is a characteristic scale for smoothing a temperature variation over length $L$. It is the thermal instance of the diffusive time scale. Rescaling $x=L\hat{x}$ and $t=(L^2/\alpha )\hat{t}$ removes $L$ and $\alpha$ from the constant-coefficient heat equation. Doubling length therefore multiplies comparable internal equilibration times by four; increasing thermal diffusivity by four divides them by four.

This scale is not a universal time of complete thermal equilibrium. A nonzero Fourier mode decays exponentially and never becomes exactly zero at finite time. The geometry, boundary conditions, initial pattern and accepted residual determine the numerical factor. For example, an insulated bar's slowest nonuniform Fourier mode has decay time $L^2/({\pi }^2\alpha )$, and reducing its amplitude to a fraction $\epsilon$ requiresExternal heat transfer can impose a longer cooling time even when internal temperature differences smooth quickly. Thermal diffusivity alone does not determine cooling to the room's temperature.

Thermal electrical noise is the fluctuating electric potential difference or electric current produced by thermal fluctuations in a dissipative electrical circuit, even without an applied voltage. A resistor at thermal equilibrium is both a source of noise and a sink for electrical energy. Its electrical resistance describes the dissipation, and its absolute temperature sets the scale of the fluctuations. This connects a practical limit on signal-to-noise ratio to statistical mechanics: a warm, unbiased resistor is not electrically silent.

John Bertrand Johnson measured the effect using a vacuum-tube amplifier and thermocouple; his 1928 paper reported agreement with the Boltzmann constant. Harry Nyquist's companion paper calculated the effect from thermodynamics and statistical mechanics.^[ref][ref]

Consider two rigid bodies forming an isolated system, with fixed composition and constant whole-body heat capacity at constant volume values $C_1=1\mathrm{kJ}{\mathrm{K}}^{-1}$ and $C_2=3\mathrm{kJ}{\mathrm{K}}^{-1}$. Their initial temperatures are $T_{1,0}=300\mathrm{K}$ and $T_{2,0}=400\mathrm{K}$. Assume each body remains internally uniform, their contact stores negligible energy, no phase transition occurs and no work is done or extracted. These are stipulated model values, not material measurements.

Because $d{U}_i=C_{i}d{T}_i$, write $U_i=a_i+C_i{T}_i$ with arbitrary fixed reference constants $a_i$. The conserved relative internal energy isThis is an energy reference convention, not a claim about actual heat capacities down to zero temperature. At thermal equilibrium the two temperatures equal $T_f$, so the first law of thermodynamics givesThe larger heat capacity pulls the common temperature closer to its initial value: body 1 warms by $75\mathrm{K}$, while body 2 cools by $25\mathrm{K}$.

To predict a timescale, stipulate a constant contact conductance $G=1/R_{\mathrm{th}}=0.1\mathrm{kJ}{\mathrm{s}}^{-1}{\mathrm{K}}^{-1}$, the reciprocal of the contact's thermal resistance. The rate of heat transfer into body 1 is $G(T_2-T_1)$, givingAdding proves conservation of $E$. Subtracting gives exponential decay of the temperature difference, with time constantand thereforeExact equality is approached as $t\to \infty$ in this ideal model. Changing $G$ changes the rate but not $T_f$: equilibrium and the speed of reaching it are separate predictions. Real bodies can have internal temperature gradients, changing heat capacities and contact losses, so the plotted $7.5\mathrm{s}$ is an assumed model timescale.

Two internally equilibrated systems are in thermal equilibrium if bringing them into contact that permits only heat exchange causes no net heat transfer. In ordinary nonrelativistic thermodynamics without appreciable gravitational temperature shifts, their thermodynamic temperatures are equal. Merely preventing heat transfer with insulation does not establish thermal equilibrium: two insulated bodies can have different temperatures. ^[ref]

A thermometer illustrates the definition. Its own temperature settles to the object's temperature after thermal contact, provided its effect on the object is negligible. The existing zeroth law of thermodynamics makes this comparison transitive: two objects separately in thermal equilibrium with the same thermometer are in thermal equilibrium with each other. This lets one calibrated scale compare different materials.

Equal temperatures do not imply equal internal energies or equal heat capacities. Nor does thermal equilibrium by itself imply mechanical equilibrium or chemical equilibrium; these require additional allowed exchanges. There is also a distinction between no net exchange in thermal equilibrium and a constant temperature maintained by equal incoming and outgoing heat. The latter can be a steady state with ongoing entropy production. Microscopic exchanges can continue at thermal equilibrium, but their opposing contributions balance on average.

Historically, OpenStax credits the name “zeroth law” to Ralph Fowler in the 1930s: the already numbered laws needed a logically earlier statement about thermal equilibrium. ^[ref]

A rate proportional to $T$ competes with radiation-era expansion proportional to $T^2$. The interaction-to-expansion ratio therefore grows as the universe cools, crossing unity near $T_{\mathrm{eq}}={\alpha }_Y^2{m}_{\mathrm{pl}}/(1.66\sqrt{g_{\ast }})$. If a phase transition occurs at $T_c\gg T_{\mathrm{eq}}$, this high-temperature interaction never establishes thermal equilibrium before the transition. Subsequent faster number-changing processes may still establish a thermal distribution; the initial state is not automatically Fermi-Dirac just because the particles are relativistic.

For a single mode in thermal equilibrium, its quantum harmonic oscillator energy ladder gives a geometric distributionIts probability generating function is $F(z)=(1-q)/(1-qz)$. Differentiating twice gives $F^{\prime \prime }(1)=2q^2/(1-q{)}^2=2{\mu }^2$, hence $g^{(2)}(0)=2$. At the chosen mean $\mu =2$, $q=2/3$, $p_0=1/3$, $p_1=2/9$, $p_2=4/27$, and $\mathrm{Var}N=6$. Rare high occupations increase the pair count even though the mean matches the other preparations. The geometric weights also arise from a Gaussian mixture of coherent states.^[ref]

Single-mode selection is essential to this factor of two. If $M$ independent equally populated thermal modes are collected together, let $N={\sum }_{j=1}^M{N}_j$ with $⟨N_j⟩=\mu /M$. Adding their variances gives $\mathrm{Var}N=\mu +{\mu }^2/M$, hence $g^{(2)}(0)=1+1/M$. Averaging over many unresolved modes can make thermal pair counts approach the independent-count reference.