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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 a sphere of radius $a$ in an otherwise stationary, unbounded Newtonian fluid with dynamic viscosity $\eta$, impose the no-slip boundary condition and the slow-flow assumptions of Stokes flow. The resisting torque on a rotating sphere in Stokes flow is $-{\zeta }_r\mathbf{\Omega }$, where $\mathbf{\Omega }$ is its angular velocity and ${\zeta }_r=8\pi \eta a^3$. Thermal rotational diffusion at absolute temperature $T$ then givesHere $k_B$ is the Boltzmann constant; the translational expression is the Stokes–Einstein relation. The ratio has units of inverse length squared, so comparing the numerical values of $D_r$ and $D_t$ without their units is meaningless. At fixed temperature and dynamic viscosity, decreasing radius tenfold increases $D_r$ a thousandfold but $D_t$ only tenfold. This strong size dependence explains why small particles can lose directional memory extremely quickly.

Choose $T=298\mathrm{K}$ and $\eta =0.001\mathrm{Pa}\mathrm{s}$ as illustrative model inputs. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ gives $k_{B}T=4.11433402\times 10^{-21}\mathrm{J}$ and the following predictions:These are bulk three-dimensional sphere calculations, explicitly not a fit to the confined ellipsoid experiment below. Walls, nonspherical shape and departure from the no-slip boundary condition can change rotational resistance. A visible surface mark is still necessary to read the sphere's turning in an image, even when its rotational diffusion coefficient is well defined physically.

Write the linear rigid rotor energy eigenvalues as $E_J=k_{\mathrm{B}}{\theta }_{r}J(J+1)$, where ${\theta }_r={\hslash }^2/(2I{k}_{\mathrm{B}})=hB/k_{\mathrm{B}}$ is the rotational temperature scale, $I$ the moment of inertia, $B$ the frequency-valued rotational constant, and $k_{\mathrm{B}}$ the Boltzmann constant. Put $t=T/{\theta }_r$ and ${ϵ}_J=J(J+1)$. Define sums that include rotational multiplicity but exclude nuclear-spin multiplicity:The canonical partition function for one equilibrating internal rotor is $q_{\mathrm{eq}}=q_{\mathrm{p}}+3q_{\mathrm{o}}$, so its equilibrium orthohydrogen fraction isFor $t\ll 1$, the lowest allowed levels give $q_{\mathrm{p}}\simeq 1$ and $q_{\mathrm{o}}\simeq 3e^{-2/t}$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\simeq 9e^{-2/t}/(1+9e^{-2/t})\to 0$. For $t\gg 1$, the densely populated even and odd ladders have equal leading sums, $q_{\mathrm{p}}\sim q_{\mathrm{o}}\sim t/2$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\to 3/4$. The factor nine belongs to the low-temperature two-level approximation, and the factor three to the high-temperature spin multiplicities.

A frozen mixture requires two separately normalized probability distributions,Multiply these by the prescribed fractions $1-x$ and $x$, respectively. The nuclear-spin factor three cancels from the conditional orthohydrogen distribution. For fixed counts $N_{\mathrm{o}}$ and $N_{\mathrm{p}}$, the internal part of the many-molecule partition function is proportional to $q_{\mathrm{p}}^{N_{\mathrm{p}}}(3q_{\mathrm{o}}{)}^{N_{\mathrm{o}}}$, with combinatorial factors independent of temperature. Its logarithm yields the two conditional means weighted by the fixed counts. Using $q_{\mathrm{p}}+3q_{\mathrm{o}}$ instead would silently permit conversion and change the physical constraint.

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.

A Schottky anomaly is a peak in heat capacity caused by thermal occupation of a small set of separated energy levels. For $N$ independent two-level units with equal degeneracies and gap $\Delta >0$, write $x=\Delta /(k_{B}T)$; their heat capacity is $C=N{k}_B{x}^2{e}^{-x}/(1+e^{-x}{)}^2$, with Boltzmann constant $k_B$. It vanishes at both low and high temperature and can spoil a lattice $T^3$ fit in between. Localized electronic or nuclear levels can contribute, including in calorimeter addenda; their presence must be checked rather than attributed automatically to phonons.

Take identical rigid spheres of radius $a$, particle density ${\rho }_p$ and displaced liquid density ${\rho }_f$, at uniform absolute temperature $T$. Let $z$ increase upwards from a reference layer and write $\Delta \rho ={\rho }_p-{\rho }_f>0$. The buoyancy correction gives the effective gravitational massThe actual grain mass is ${\rho }_{p}V$; the displaced liquid subtracts ${\rho }_{f}V$ from the effective gravitational mass. In a dilute, noninteracting particle suspension, the Boltzmann distribution therefore givesHere $n$ is number density and $k_B$ is the Boltzmann constant. Equivalently, the dilute osmotic pressure $\Pi =n{k}_{B}T$ supports the grains above it: $d\Pi /dz=-n{m}_{b}g$. Dividing by $n$ and integrating yields the same exponential function. No dynamic viscosity appears in this equilibrium argument. If ${\rho }_p<{\rho }_f$, the population instead increases upwards; at equal densities this gravitational model predicts uniform $n$.

A microscope counts a finite optical slab, not an infinitesimal plane. For equal viewed area $A$ and slab thickness $\delta$, an ideal uniform detection window centred at $z$ givesThus equal optical volumes preserve the exponential function count ratio exactly, even for finite $\delta$, provided the windows lie wholly inside the accessible chamber and have the same detection response. Unequal depth response or overlapping a boundary can corrupt this inference. Repeated observations estimate the mean count; closely spaced visits need not be independent because grains remain in view.

For an original chosen model, set $a=0.250\mu \mathrm{m}$, ${\rho }_p=1200\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $T=298\mathrm{K}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. These are illustrative inputs, not Jean Baptiste Perrin's measurements. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesIf equal observation effort gives expected count $400$ in the reference layer, the expected counts $20$, $40$ and $60\mu \mathrm{m}$ higher are approximately $214.3$, $114.8$ and $61.49$. Fractions describe expected values over observations, not fractional grains. Using the actual grain mass instead of its buoyancy-corrected value would make the inferred Avogadro constant six times too small for these chosen densities.

For a p-n junction at fixed temperature, the Shockley diode equation models its electric current asHere $I_S>0$ is the diode saturation current, $n>0$ the diode ideality factor, $V_T$ the thermal voltage, $k_B$ the Boltzmann constant, and $q>0$ the elementary charge.^[ref]

For an original model example, choose $I_S=1\mathrm{nA}$, $n=2$, and $T=300\mathrm{K}$. Using $k_B=1.380649\times 10^{-23}J/K$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ gives $V_T=25.8520\mathrm{mV}$. Then $V_D=0.600\mathrm{V}$ gives $I_D=0.109591\mathrm{mA}$, whereas $V_D=0.800\mathrm{V}$ gives $I_D=5.24450\mathrm{mA}$. These are predictions of the declared model. There is no universal turn-on electric potential difference: solving the Shockley diode equation for the value needed at a chosen electric current givesAt $I_D\gg I_S$, increasing $V_D$ by $0.100\mathrm{V}$ multiplies the electric current by approximately $e^{0.100/(n{V}_T)}=6.91772$. At $V_D=-0.300\mathrm{V}$ the same model gives $I_D=-0.996979\mathrm{nA}$. Its reverse branch approaches $-I_S$; it cannot describe diode reverse breakdown.

A constructed extension adds internal series electrical resistance $R_s$ to a junction whose electric potential difference is $V_j$. Applying the Kirchhoff cycle law givesFor $R_s=20\Omega$ and the same $I_S,n,T$, sustaining $I_D=5.24450\mathrm{mA}$ needs $V_D=0.904890\mathrm{V}$: $0.800\mathrm{V}$ across the junction plus $0.104890\mathrm{V}$ across its internal electrical resistance. The derivativeshows why the current-voltage characteristic becomes nearly a straight line when the internal series electrical resistance dominates. This is still a fixed-temperature model; self-heating can change $I_S$, $n$ and $V_T$. Real diodes also have leakage and charge-storage effects omitted here.^[ref]

Simulated annealing accepts worse feasible solutions to escape a local minimum. Minimize an objective function $E(x)$, analogous to energy. Effective temperature $T>0$ has the same units as $E$, absorbing the Boltzmann constant; it controls exploration, rather than a material's thermodynamic temperature.^[ref]

Given current feasible point $x$, draw $y$ from a proposal distribution $q(x,y)$. For a symmetric proposal distribution, $q(x,y)=q(y,x)$, use the Metropolis–Hastings acceptance probabilityAccept with this conditional probability; on rejection retain $x$. Every downhill or equal-cost proposal is accepted. Draw a fresh uniform random variable $U$ on $(0,1)$ and accept if $U<a_T(x,y)$. The off-diagonal transition probability $P_T(x,y)=q(x,y)a_T(x,y)$ includes proposing $y$.^[ref]

For a maximization problem with objective function $f$, set $E=-f$: every improving proposed neighbor is accepted, and a worsening move of loss $\Delta =f(x)-f(y)>0$ has conditional probability $\exp (-\Delta /T)$ of acceptance, under the same symmetry assumption. Likelihood-power annealing uses precisely this sign convention for a log-likelihood. Retain the best feasible solution seen, since the current state can worsen even after a global minimum has been visited.

If the proposal distribution is asymmetric, use the full Metropolis–Hastings acceptance probability, for $q(x,y)>0$,A missing reverse proposal gives zero acceptance. Even a downhill proposal need not be accepted with probability one in this case. Choosing uniformly among a variable number of neighbors is generally asymmetric: $q(x,y)=1/d(x)$ and $q(y,x)=1/d(y)$ give the correction $d(x)/d(y)$, where $d$ is vertex degree.^[ref]

The useful idea is to separate exploring from insisting on improvement: an uphill move can open a route to a better feasible solution. Its usefulness depends on the proposal distribution, the scale of the objective function, and the cooling schedule for simulated annealing. Finite practical schedules carry no general guarantee of finding a global minimum.

Consider an original counting example: a square two-dimensional XY model of side $L$, a microscopic core radius $a$, and a fixed phase stiffness ${\rho }_s$. This deliberately neglects screening by other vortices. Away from a unit phase vortex, $\theta =\arg (x+iy)$ and $|\nabla \theta |=1/r$. Integrating the Goldstone-mode effective free energy over an annulus givesBoundary shape changes finite terms. There are order $(L/a{)}^2$ core positions, so the positional entropy is $S_{\mathrm{v}}=2k_B\log (L/a)+O(k_B)$, where $k_B$ is the Boltzmann constant. With $K={\rho }_s/(k_{B}T)$, subtracting $T{S}_{\mathrm{v}}$ gives the displayed free energy.

For $L/a=1000$ and $E_c/(k_{B}T)=2$, the dimensionless free energy cost is about $5.546$ at $K=0.8$ and $-0.965$ at $K=0.5$. The same logarithm appears in the cost and the positional reward: increasing the system size strengthens whichever has the larger coefficient. The corresponding one-defect Boltzmann factors, about $0.00390$ and $2.62$, are relative statistical weights, not normalized probabilities. The latter signals failure of a dilute single-defect estimate.

A single net phase vortex requires compatible boundaries: on a periodic sample use neutral pairs, as required by vortex charge neutrality. This argument identifies a possible threshold, but predicting the universal stiffness jump requires the renormalized phase stiffness, after smaller pairs have responded. It does not justify setting the transition temperature equal to a bare lattice coupling divided by a universal constant.

The Stefan-Boltzmann constant converts the fourth power of absolute temperature into blackbody radiant exitance:Here $k_B$ is the Boltzmann constant, $h$ the Planck constant and $c$ the speed of light. In the current International System of Units, these three constants have exact defining values, so $\sigma$ is exact as an expression in them and $\pi$. The displayed finite decimal is an approximation, not an exact decimal or a measured uncertainty. The 2022 recommended-constants table lists an ellipsis and an exact uncertainty for $\sigma$.^[ref]

A dimensional check uses $[k_B]=\mathrm{J}{\mathrm{K}}^{-1}$, $[h]=\mathrm{J}\mathrm{s}$ and $[c]=\mathrm{m}{\mathrm{s}}^{-1}$. Consequently $[k_B^4/(h^3{c}^2)]=\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}=\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, exactly the units needed in the Stefan–Boltzmann law.

For an equilibrium photon gas, the energy density is $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$. The radiation pressure is $p_{\mathrm{rad}}=u/3=4\sigma T^4/(3c)$. These describe the radiation filling an equilibrium cavity; $u$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas the outward radiant exitance through a small opening is $cu/4$. The factor $1/4$ comes from averaging propagation directions and their projected area over a radiation field with isotropy, not from reducing the opening's emissivity.

The Stefan–Boltzmann law gives the total hemispherical radiant exitance of an ideal blackbody at absolute temperature $T$:Here $\sigma$ is the Stefan-Boltzmann constant. Total means integrated over all wavelengths; hemispherical means integrated over all outward directions with the projected area weighting. Thus $M_b$ is emitted power per actual emitting area, rather than directional radiance or spectral radiance. Absolute temperature must be in kelvin; raising a Celsius reading to the fourth power gives the wrong result. A uniform surface of area $A$ emits radiant flux $P_{\mathrm{emit}}=A{M}_b$. A real surface instead has $M=ϵ(T)\sigma T^4$, where $ϵ(T)$ is its total hemispherical emissivity. The older symbol $F$ in the formula above denotes this emitted radiant exitance.^[ref]

For an ideal blackbody, doubling absolute temperature multiplies radiant exitance by $16$. With the approximate Stefan-Boltzmann constant $5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, the radiant exitances at $300\mathrm{K}$ and $600\mathrm{K}$ are approximately $459.300$ and $7348.805\mathrm{W}{\mathrm{m}}^{-2}$. This steep response explains why thermal radiation becomes so consequential at high temperature. It also enables heat transfer through a vacuum, where thermal conduction through a gas and convection are absent.

A short derivation integrates Planck's law in its frequency form. For a blackbody, direction-independent spectral radiance is $B_{\nu }(T)=2h{\nu }^3/[c^2(e^{h\nu /(k_{B}T)}-1)]$, where $h$ is the Planck constant, $k_B$ the Boltzmann constant and $c$ the speed of light. The outward hemisphere contributes $\int \cos \theta d\Omega =\pi$, soExpanding $(e^x-1{)}^{-1}={\sum }_{n=1}^{\infty }{e}^{-nx}$, nonnegative terms allow termwise integration. Repeated integration by parts gives ${\int }_0^{\infty }{x}^3{e}^{-nx}dx=6/n^4$; the Riemann zeta function value $\zeta (4)={\pi }^4/90$ gives the remaining integral as ${\pi }^4/15$. Hence $M_b=\sigma T^4$ with the coefficient in Stefan-Boltzmann constant. The angular integral is essential: omitting it confuses radiance with radiant exitance.

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]

For a Planck function expressed as spectral radiance per frequency interval,$h$ is the Planck constant, $c$ the speed of light and $k_B$ the Boltzmann constant. Differentiating at fixed frequency givesThe proof uses ${\partial }_{T}x=-x/T$ and the chain rule. At low frequency, the multiplier tends to one, recovering the Rayleigh-Jeans law. At high frequency it grows approximately as $x$: the same fractional temperature change has a larger fractional brightness effect on the declining spectral tail.

For an original example choose $T=2.725\mathrm{K}$ and $\delta T=1\mathrm{mK}$. At $100\mathrm{GHz}$, $x=1.76119$, so the first-order fractional residual is $7.80415\times 10^{-4}$. At $300\mathrm{GHz}$, $x=5.28357$, it is $1.94881\times 10^{-3}$. Both spectra are exact blackbodies: this residual is a temperature mismatch, not evidence of a nonthermal spectrum. Fitting temperature before assessing remaining spectral residuals therefore removes a physically distinct degree of freedom. The figure divides by a fixed peak brightness rather than by the rapidly vanishing high-frequency tail.

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.

For a positive quadratic bending operator $K$ with a real orthonormal basis of eigenfunctions, the equipartition theorem gives $C(x,y)=k_{B}T{\sum }_n{W}_n(x)W_n(y)/{\mu }_n$, where $K{W}_n={\mu }_n{W}_n$. This is the inverse-operator Green function multiplied by Boltzmann constant and temperature. Unconstrained zero-energy filament modes prevent a normalizable canonical ensemble, and negative modes signal an unstable quadratic model.

For both ends clamped and $K=A{\partial }_x^4$, the diagonal is $C(x,x)=k_{B}T{x}^3(L-x{)}^3/(3A{L}^3)$. This differs from a clamped-free tip variance. The center variance is $k_{B}T{L}^3/(192A)$.

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]

A thermal neutron belongs to a population approximately in thermal equilibrium with its surroundings. Such a population has a distribution of kinetic energies, not one fixed energy. The scale $k_{B}T$ uses the Boltzmann constant and thermodynamic temperature; it is about $0.0259\mathrm{eV}$ at $300\mathrm{K}$. The mean translational kinetic energy for a classical three-dimensional equilibrium population is $3k_{B}T/2$. Not every slow neutron has reached this equilibrium.

MIT's laboratory guide mounts resistors on an aluminum box beneath a metal beaker shield, with short twisted leads. Its SR560 amplifier uses nominal gain $1000$ before a Krohn-Hite $1$–$50\mathrm{kHz}$ band-pass filter; an oscilloscope reads root mean square voltage. Calibration sweeps an attenuated sinusoid and measures the actual gain $g(f)$. Students alternate the resistor and a short circuit, subtracting squared readings $V_R^2-V_S^2$, and disconnect the ohmmeter during noise acquisition. Lead and input capacitance enter the gain-squared integral. With settings unchanged, varying electrical resistance or temperature tests the predicted mean square. The guide prescribes repeated readings and warns against averaging noise waveforms to zero. This is a documented teaching protocol, not a supplied dataset of measured outcomes.^[ref]

The subtraction requires care: shorting the input changes the source electrical impedance. An amplifier's input current noise can therefore produce different voltage noise in the two configurations; a shorted baseline alone does not remove that contribution. Gain drift, interference and insufficient observation time also limit the inference. These are limitations of interpreting the measurement, not corrections that can be inferred from the nominal signal bandwidth alone. In an independent analysis one would compare the measured difference withwhere $C$ represents the measured shunt capacitance and $g$ excludes that separately modeled shunt. If calibration already includes the same shunt response, applying the denominator again counts the attenuation twice. The thermal noise worked example and both original figures illustrate ideal models; they claim no measured agreement or measured value of the Boltzmann constant.

Model an ideal resistor of constant electrical resistance $R>0$ at uniform absolute temperature $T$ by its Thévenin equivalent: a zero-mean random voltage source in series with a noiseless $R$. In the classical range its one-sided, open-circuit voltage power spectral density, per hertz, iswhere $k_{\mathrm{B}}$ is the Boltzmann constant. Thus a positive-frequency interval of width $df$ contributes $4k_{\mathrm{B}}TRdf$ to the mean square voltage. This is approximately white noise only where the electrical resistance is effectively constant and $hf\ll k_{\mathrm{B}}T$, with $h$ the Planck constant. The ideal constant spectrum cannot describe arbitrarily high frequencies.^[ref]

For a real voltage, an equivalent two-sided power spectral density is $S_v^{(2)}(f)=2k_{\mathrm{B}}TR$ over signed frequencies. With a real linear time-invariant system of frequency response $H(f)$, the same output mean square isAngle brackets denote an expected value; for this zero-mean noise the mean square is its variance. Do not integrate $4k_{\mathrm{B}}TR$ over both signs of frequency: that counts the same fluctuations twice. All integrals here use frequency $f$ in hertz, not angular frequency $\omega =2\pi f$. The amplitude spectral density is $\sqrt{4k_{\mathrm{B}}TR}$ in $\mathrm{V}/\sqrt{\mathrm{Hz}}$, whereas the root mean square voltage requires a specified measurement signal bandwidth.

Assume one harmonic normal mode at thermal equilibrium, weak spontaneous excitation far from electronic resonances, the same collection solid angle and wave polarization, and matched response after instrumental correction. The Boltzmann distribution givesHere $k_B$ is the Boltzmann constant, $T>0$ is temperature, and $P_v$ is the probability of vibrational quantum number $v$. Thus $P_1/P_0=r$. Linear coupling to $Q$ has squared oscillator matrix elements proportional to $v+1$ for Stokes Raman scattering and $v$ for Anti-Stokes Raman scattering. Summing over initial levels gives strengths proportional to $\overline{v}+1$ and $\overline{v}$ respectively; their ratio is $r$, even when levels above $v=1$ are occupied.

Let $I_{\mathrm{S}}$ and $I_{\mathrm{AS}}$ denote line-integrated radiant intensities, meaning power per collection solid angle. The off-resonant electric dipole radiation law adds the fourth power of emitted frequency:The photon rate divides each line's power by its own $h{\nu }_s$. Consequently, the ideal response-corrected anti-Stokes/Stokes count-rate ratio has the third power of the same frequency ratio. Peak heights alone do not give line-integrated radiant intensities if linewidths differ. Uncorrected detector counts can also include unequal detector response and unequal optical filter transmission. Resonant enhancement, overlapping spectral lines, nonthermal populations and local optical heating can invalidate a temperature inferred from this simple formula.